3904.spinors and calibrations (perspectives in mathematics) by f. reese harvey

Spinors and CalibrationsF Reese HarveyDepartment of MathematicsRice UniversityHouston, Texas

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Harvey, F Reese.Spinors and calibrations/F. Reese Harvey.

p. cm. - (Perspectives In mathematics: vol. 9)BibUography: p.Includes IndexISBN 0-12-329650-11 Spinor analysis. 2 Matrix groups. I. Title. II. Series.

QA433.H3271990515'.63-dcl9 89-74

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Printed in the United States of America909192 987654321

This book is dedicated to my wife, Linda

TABLE OF CONTENTS

PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . xi

PART I: CLASSICAL GROUPS AND NORMED ALGEBRAS

1. CLASSICAL GROUPS I . . . . . . . . . . . . . . . . . . 3

The General Linear Groups . . . . . . . . . . . . . . . . 4Groups Defined by Bilinear Forms . . . . . . . . . . . . . 6

Other Miscellaneous Groups . . . . . . . . . . . . . . . . 8

Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . 13

Summary . . . . . . . . . . . . . . . . . . . . . . . . 15

2. THE EIGHT TYPES OF INNER PRODUCT SPACES . . . . 19

The Standard Models . . . . . . . . . . . . . . . . . . . 22Orthogonality . . . . . . . . . . . . . . . . . . . . . . 23A Canonical Form (The Basis Theorem) . . . . . . . . . . . 29The Parts of an Inner Product . . . . . . . . . . . . . . . 31

3. CLASSICAL GROUPS II . . . . . . . . . . . . . . . . . 41

Group Representations and Orbits . . . . . . . . . . . . . 41Generalized Spheres . . . . . . . . . . . . . . . . . . . . 42The Basis Theorem Revisited . . . . . . . . . . . . . . . . 46Adjoints . . . . . . . . . . . . . . . . . . . . . . . . . 47Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . 48

4. EUCLIDEAN/LORENTZIAN VECTORS SPACES . . . . . . 57

The Cauchy-Schwarz Equality . . . . . . . . . . . . . . . 58Special Relativity . . . . . . . . . . . . . . . . . . . . . 62The Cartan-Dieudonne Theorem . . . . . . . . . . . . . . 67Grassmannians and SOT(p, q), the Reduced Special

Orthogonal Groups . . . . . . . . . . . . . . . . . . . 75

5. DIFFERENTIAL GEOMETRY . . . . . . . . . . . . . . . 81

Real n-Manifolds: The Group GL(n, R) . . . . . . . . . 81Oriented Real n-Manifolds: The Group GL+(n, R) . . . . . . 82Complex (and Almost Complex) n-Manifolds:

The Group GL(n, C) . . . . . . . . . . . . . . . . . . 83Quaternionic (and Almost Quaternionic) n-Manifolds:

The Group GL(n, H) H* . . . . . . . . . . . . . . . . 85Manifolds with Volume: The Groups SL(n, R) and SL(n, C) . . 86Riemannian Manifolds (of Signature p, q): The Group O(p, q) . 87Conformal Manifolds (of Signature p, q): The Group CO(p, q) . 87

vii

viii Table of Contents

5. DIFFERENTIAL GEOMETRY (continued)Real Symplectic Manifolds: The Group Sp(n, R) . . . . . . . 87Complex Riemannian Manifolds: The Group O(n, C) . . . . . 88Complex Symplectic Manifolds: The Group Sp(n, C) . . . . . 88Kahler Manifolds (of Signature p, q): The Group U(p, q) . . . . 89Special Kahler Manifolds (of Signature p, q): The Group SU(p, q) 90HyperKiihler Manifolds (of Signature p, q): The Group HU(p, q) 91Quaternionic Kahler Manifolds (of Signature p, q):

The Group HU(p,q) HU(1) . . . . . . . . . . . . . . . 92Quaternionic Skew Hermitian Manifolds:

The Group SK(n, H) . . . . . . . . . . . . . . . . . . 93Coincidences of Geometries in Low Dimensions . . . . . . . . 94

6. NORMED ALGEBRAS . . . . . . . . . . . . . . . . . 101

The Cayley-Dickson Process . . . . . . . . . . . . . . . 104The Hurwitz Theorem . . . . . . . . . . . . . . . . . . 107Cross Products . . . . . . . . . . . . . . . . . . . . . 110The Exceptional Lie Group G2 . . . . . . . . . . . . . . 113

7. CALIBRATIONS . . . . . . . . . . . . . . . . . . . . 125

The Fundamental Theorem . . . . . . . . . . . . . . . . 125The Kahler Case 127The Special Lagrangian Calibration . . . . . . . . . . . . 130The Special Lagrangian Differential Equation . . . . . . . . 134Examples of Special Lagrangian Submanifolds . . . . . . . 138Associative Geometry . . . . . . . . . . . . . . . . . . 144The Angle Theorem . . . . . . . . . . . . . . . . . . . 146Generalized Nance Calibrations and Complex Structures . . 156

8. MATRIX ALGEBRAS . . . . . . . . . . . . . . . . . . 163

Representations . . . . . . . . . . . . . . . . . . . . . 163Uniqueness of Intertwining Operators . . . . . . . . . . . 167Automorphisms . . . . . . . . . . . . . . . . . . . . . 169Inner Products . . . . . . . . . . . . . . . . . . . . . 170

PART II: SPINORS

9. THE CLIFFORD ALGEBRAS . . . . . . . . . . . . . . 177

The Clifford Automorphisms . . . . . . . . . . . . . . . 181The Clifford Involutions . . . . . . . . . . . . . . . . . 182The Clifford Inner Product . . . . . . . . . . . . . . . . 184

Spinors and Calibrations ix

9. THE CLIFFORD ALGEBRAS (continued)The Main Symmetry . . . . . . . . . . . . . . . . . . 185The Clifford Center . . . . . . . . . . . . . . . . . . . 187Self Duality . . . . . . . . . . . . . . . . . . . . . . 188Trace . . . . . . . . . . . . . . . . . . . . . . . . . 189The Complex Clifford Algebras . . . . . . . . . . . . . . 190

10. THE GROUPS SPIN AND PIN . . . . . . . . . . . . . . 195

The Grassmannians and Reflections . . . . . . . . . . . . 197Additional Groups for Nondefinite Signature . . . . . . . . 199The Conformal Pin (or Clifford) Group . . . . . . . . . . 201Determinants . . . . . . . . . . . . . . . . . . . . . . 203

11. THE CLIFFORD ALGEBRAS Cl(r, s) AS ALGEBRAS . . . . 207

The Pinor Representations . . . . . . . . . . . . . . . . 210The Spinor Representations . . . . . . . . . . . . . . . 213The First Proof . . . . . . . . . . . . . . . . . . . . . 215The Spinor Structure Map on P(r, s) . . . . . . . . . . . . 217Even Dimensions . . . . . . . . . . . . . . . . . . . . 217Odd Dimensions . . . . . . . . . . . . . . . . . . . . 220

12. THE SPLIT CASE Cl(p, p) . . . . . . . . . . . . . . . . 227A Model for C1(p, p) . . . . . . . . . . . . . . . . . . . 227Pinor Inner Products for Cl(p, p) = EndR(P(p, p)) . . . . . . 230The Complex Clifford Algebras

(Continued from Chapter 9) . . . . . . . . . . . . . . 233Cl(r, s)(r + s = 2p) as a Subalgebra of

Clc (2p) = Cl(p, P) OR C . . . . . . . . . . . . . . . . 235The Pinor Reality Map . . . . . . . . . . . . . . . . . 237A Second Proof of the Classification Theorem . . . . . . . 240Pure Spinors . . . . . . . . . . . . . . . . . . . . . . 241

13. INNER PRODUCTS ON THE SPACES OF SPINORSAND PINORS . . . . . . . . . . . . . . . . . . . . . 247

The Spinor Inner Product . . . . . . . . . . . . . . . . 247The Spin Representation and the (Reduced) Classical

Companion Group Cp°(r, s) . . . . . . . . . . . . . . . 250The Pinor Inner Products t and e . . . . . . . . . . . . . 252Pinor Multiplication . . . . . . . . . . . . . . . . . . . 266Signature . . . . . . . . . . . . . . . . . . . . . . . 268

x Table of Contents

14. LOW DIMENSIONS . . . . . . . . . . . . . . . . . . . 271

Cartan's Isomorphisms . . . . . . . . . . . . . . . . . 271Triality . . . . . . . . . . . . . . . . . . . . . . . . 275Transitive Actions on Spheres . . . . . . . . . . . . . . 283The Cayley Plane and the Exceptional Group F4 . . . . . . 289Clifford Algebras in Low Dimensions . . . . . . . . . . . 298Squares of Spinors and Calibrations . . . . . . . . . . . . 308

REFERENCES . . . . . . . . . . . . . . . . . . . . . . 315

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . 317

PREFACE

This book is intended to be a collection of examples. The (simple) Liegroups, the spin groups for general signature, normed algebras for generalsignature, the exceptional groups G2 and F4, the orbit structure of thesimpler representations of these groups, and the special Lagrangian andassociative calibrations are all discussed in some detail. The underlyingand not always mentioned motivation for these examples is differential ge-ometry: Riemannian, symplectic, Kahler, hyperKahler, as well as complexand quaternionic.

The book is divided into two parts. Readers who are primarily inter-ested in the spin groups are encouraged to start with Part II-Spinors.

Part I begins with an introduction to certain specific matrix groups,with the entries real, complex, or quaternionic. Some of these groups aredefined by requiring that the matrix fix one of several types of (generalized)inner products. This leads to a discussion in Chapter 2 of eight types ofinner products.

The groups introduced in Chapter 1 are also the subjects of Chapter3. A brief discussion of group representations and orbits is followed by thecomputation of orbit structures for some of the classical representations ofChapter 1. Lie Algebras are also introduced in this chapter.

Chapter 4 is devoted to a study of inner product spaces in the usualsense, i.e., a real vector space equipped with a real nondegenerate symmet-ric bilinear form. This is just one of the eight types discussed in Chapter2. The topics vary in intensity, beginning with a very elementary but fairlycomplete discussion of the Cauchy-Schwarz "equality."

The reader is assumed to have Euclidean intuition. A discussion ofLorenzian intuition is presented based on both special relativity and theanalogue between complex numbers C and Lorentz numbers L. Some read-ers will prefer to skip the long section on the Cartan-Dieudonne Theoremuntil after reading Chapter 9, where additional motivation is provided.

The next chapter, titled "Differential Geometry," is quite differentfrom the first four chapters. Chapter 5 is intended to provide two things:first, motivation (but not prerequisites) for other material in this book;second, a skeleton or bare outline of some of the various types of geometry.

xii Preface

Hopefully the reader will be inclined to build on this outline by furtherstudy outside the book.

In contrast, Chapter 6, on normed algebras, provides a fairly completediscussion. In fact, some of the information presented here is not availableelsewhere in the literature. The chapter naturally includes a discussion ofthe exceptional Lie group G2 as the automorphism group of the octonians.

Various examples of calibrations are examined in Chapter 7. Thegroup that fixes a particular calibration provides an important ingredientin discussing the geometry of submanifolds determined by the calibration.The beautiful Angle Theorem of Lawlor and Nance is discussed. Bothhalves of the proof involve introducing an interesting class of examples: a)the Nance calibrations; and b) the Lawlor special Lagrangian submanifolds.

Chapter 8 is a brief, elementary, and somewhat tedious discussion ofmatrix algebras, but the material is essential for understanding Cliffordalgebras. Algebraists already familiar with this material are encouraged toskip to Part II.

Part II begins with a presentation of Clifford algebras that is indepen-dent of Part I. The reader may wish to start the book here. The exterioralgebra is assumed to be a familiar object, although a brief definition is pro-vided. Since the exterior algebra and the Clifford algebra are canonicallyisomorphic as vector spaces (but not as algebras), a considerable amountof intuition about the exterior algebra carries over to the Clifford algebra.Our presentation emphasizes this relationship between the Clifford algebraand the exterior algebra. For example, the natural inner product inducedon the exterior algebra immediately provides a natural inner product onthe Clifford algebra. This norm on the Clifford algebra usually appearssomewhat myteriously as a norm only defined on those Clifford elementsin the Clifford group.

Chapter 10 describes the Spin and Pin groups. Again the discussionemphasizes the connection with the exterior algebra, via the grassmanniansas subsets of the exterior algebra. Each plane through the origin, once itis oriented in one of the two possible ways, can be considered a Cliffordelement. By utilizing Clifford multiplication, both of these two orientedplanes are associated with the orthogonal transformation "reflection alongthe plane," providing the core of the double cover of the orthogonal groupby the Pin group. For the reader willing to accept the Cartan-DieudonneTheorem, this Chapter is also independent of Part I.

As algebras-forgetting the more subtle extra structure-the Cliffordalgebras are just matrix algebras. The matrices act on the vector space ofpinors. Although not canonical, this pinor space is unique up to a scalarmultiple of the identity. Chapter 11 describes the spaces of pinors andspinors.

Spinors and Calibrations xiii

Each space of pinors or of spinors can, in a natural way, be given theextra structure of one of the types of (generalized) inner products discussedin Part I. This inner product is unique up to a change of scale, and it playsan important role in understanding the spin groups and the space of spinors.Chapter 12 analyzes the case of split signature and the complex case in evendimensions. This provides a basis for determining the spinor inner productfor arbitrary signature in Chapter 13. Moreover, this spinor inner productdetermines a classical (companion) group containing the spin group thatis, probably, the smallest such group.

Chapter 14 consists of just a few of the many interesting applicationsof, as well as some very explicit models for, the various Clifford algebras inlow dimensions.

The importance of spinors in geometry (including, of course, generalrelativity) is not universally accepted. Two recent texts should play animportant role in correcting any previous oversight. The two-volume bookby Penrose and Rindler on general relativity contains many interestingtopics on general relativity and Twistor Theory. Spin Geometry, by Lawsonand Michelson, is a very exciting presentation of some of the most beautifultopics in geometry.

I wish to thank Robert Bryant, Blaine Lawson, and Roger Penrosefor very valuable discussions and frequent encouragement concerning thisbook. In addition, Jack Mealy deserves special appreciation for his carefulreading of the manuscript. Finally, I wish to thank typists Janie McBane,Anita Poley, and Janice Want.

F. Reese Harvey

Houston, Texas

PART I. CLASSICAL GROUPSANDNORMED ALGEBRAS

1. Classical Groups I

The elements of the groups defined in this chapter are matrices withentries in one of the three fields:

R the field of real numbers,C the field of complex numbers,H the field of quaternions.

Note that H, the field of quaternions (or hamiltonians), is not commutative.The quaternions will be examined in great detail, along with the octonionsO (or Cayley numbers), in Chapter 6 on normed algebras. For the purposesof this chapter, a rudimentary knowledge of H is all that is presupposed.Consult Problem 6 for the multiplication rules for quaternions.

Let M, (R), M (C), and M,, (H) denote the algebras of n x n matriceswith entries in It, C, and H respectively. Represent elements x of R", C',and H" as column n-tuples. Then each matrix A determines a lineartransformation or endomorphism x - Ax by letting the matrix A act onthe left of the column vector x, at least in the real and complex case.Special consideration is necessary for the quaternionic case since H is notcommutative. In order for the map A : H" --> H' (defined by A actingon x on left) to be H-linear, we are forced to let the scalars H act on theH-vector space H' on the right!

Although it will be convenient to consider both right H-vector spaces(where the scalars H act on the right of the vectors) and left H-vectorspaces (where the scalars H act on the left of the vectors), the space H" of

3

4 The General Linear Groups

column n-tuples will always be considered as a right H-vector space. Thenwe have

(1.1) M"(R) = EndR(R"),

(1.2) M"(C) = Endc(C"),

(1.3) M"(H) = EndH(H").

Here EndF(V) denotes the F-linear maps from a vector space V, withscalar field F, into itself. If V is a real vector space, then EndR V isnaturally a real algebra (associative and with unit). If V is a complexvector space, then EndcV is naturally a complex algebra (associative andwith unit) but may also be considered as a real algebra, which is convenientfor some immediate purposes. Finally, if V is a right quaternionic vectorspace, then EndH(V) is naturally a real algebra-in fact, a real subalgebraof the algebra EndF(V). There is no canonical way to make EndH(V) intoeven a quaternionic vector space (right or left), much less a "quaternionic"algebra (see Problem 7).

THE GENERAL LINEAR GROUPSThe group of units, or invertible elements, in the matrix algebra M"(F) iscalled the F-general linear group for F =_ R, C, or H and is denoted byGL(n, R), GL(n, C), or GL(n, H), respectively. If the group of units, orinvertible elements, in EndF(V) is denoted by GLF(V), then

(1.1')

(1.2')

(1.3')

GL(n, R) = GLR(R),

GL(n, C) GLc(R"),

GL(n, H) 25 GLH(H').

In the quaternion case, there is another important group, larger thanthe H-general linear group GL(n, H), which we will call the enhanced H-general linear group. First note that the H-general linear group GL(n, H)(which acts on the left) consists entirely of H-linear maps. However, rightmultiplication by a scalar A E H, denoted RA,, is not necessarily H-linear.In fact, RA is H-linear if and only if A commutes with all scalars µ E H

Classical Groups I 5

because RA(xp) = xµ.1, while Ra(x)p = xA . The reader should confirmthat A commutes with all µ E H if and only if A E R C H. Thus, Rais H-linear if and only if A E R C H. Let H* denote the group of rightmultiplications by nonzero scalars. Then H* is not a subgroup of GL(n, H).However, both are contained in the algebra EndR,(H") of R-linear maps.As noted above, the intersection GL(n, H) fl H* equals R* the group ofreal nonzero multiples of the identity.

The enhanced H-general linear group, denoted GL(n, H) - H*, is de-fined to be the image of GL(n, H) x H* in EndR(H") via the map sendingthe pair (A, A) to LA . RA, where - denotes multiplication in the algebraEndR(H"), i.e., composition. Thus, the following sequence of groups isexact:

(1.4) 1 -> R* -} GL(n, H) x H* -+ GL(n, H) - H* -* 1

with GL(n, C EndR,(H'). Note that the larger group GL(n,as well as the smaller group GL(n, H), maps quaternion lines to quaternionlines.

Given A E M,, (R), the real determinant of A will be denoted detK,A.Similarly, detc A denotes the complex determinant of A E M,, (C). Thelack of commutativity for H eliminates the possibility of any useful notionof "quaternionic determinant." Of course,

GL(n,R)_JA EM"(R):detR,A36 0},(1.5) and

GL(n,C)=JA EM"(C):detcA#0}.

The group

(1.6) GL+(n, R) = {A E M,, (R) : detR, A > 0}

is called the orientation-preserving general linear group.In both the real and the complex case, we have a special linear group,

defined by

(1.7) SL(n, R) _ {A E M" (R) : detR, A = 1),

(1.8) SL(n, C) _ {A E M,, (C) : detc A = 1).

Since there is no quaternion determinant, if we proceed in exact anal-ogy with the real or the complex case, the special quaternion linear group

6 Groups Defined by Bilinear Forms

does not exist. However, it is useful to retain the notation SL(n, H) byemploying the real determinant. Let

(1.9) SL(n, H) _ {A E GL(n, H) : detR. A = 1}

denote the special quaternion linear group.

GROUPS DEFINED BY BILINEAR FORMS

Some very interesting groups are best defined as subgroups of the groupsdefined above that fix a certain bilinear form.

R-symmetric

The orthogonal group O(p, q) with signature p, q is defined to be thesubgroup of GL(n, R) (n = p + q) that fixes the standard R-symmetricform

(1.10) E(x,y)=xlyl+...+xpyp-xp+lyp+l-...-xnyn.

That is,

O(p, q) {A E GL(n, R) : E(Ax, Ay) = e(x, y) for all x, y E R"}.

R-skew (or symplectic)

The real symplectic group Sp(n, R) is defined to be the subgroup ofGL(2n, R) that fixes the standard R-symplectic (or R-skew) form

(1.11) e dxl A dx2 + + dx2n_l A dx2n,

or

(1.11') e(x, y) x1y2 - x2y1 + + X2, _1y2n - x2ny2n-1.

That is,

Sp(n, R) _ {A E GL(2n, R) : e(Ax, Ay) = e(x, y) for all z, y E R2n} .

Classical Groups 1 7

C-symmetric

The complex orthogonal group 0(n, C) is defined to be the subgroupof GL(n, C) that fixes the standard C-symmetric form

(1.12) -(Z, W) - z1 w1 + + znwn.

C-skew (or symplectic)

The complex symplectic group Sp(n, C) is defined to be the subgroupof GL(2n, C) that fixes the standard C-symplectic (or C-skew) form

(1.13) E - dzl A dz2 + + dz2n-1 A dz2n,or

(1.13') e(z, w) - Z1W2 - Z2W1 + + z2n-lW2n - z2nW2n-1

C-hermitian (symmetric)

The complex unitary group U(p, q) with signature p, q is defined to bethe subgroup of GL(n, C) (n - p + q) that fixes the standard C-hermitiansymmetric form

(1.14) E(z,w)_z12i1+...+Zpwp-zp+1TWp+1-...-zn1iJ .

Remark 1.15. ie(z, w) is called the standard C-hermitian skew form.Note that the group that fixes ie is just the same group U(p,q) that fixesE. This contrasts sharply with the quaternion case.

H-hermitian symmetric

The hyper-unitary group HU(p, q) with signature p, q is defined to bethe subgroup of GL(n, H) (n =- p + q) that fixes the standard H-hermitiansymmetric form

(1.16) e(x, y) = xiyi + ... + xpyp - xp+lyp+1 - ... - xnyn-

Note: E(x, y) is H-hermitian. This means that e is additive in both vari-ables x and y, and E(xA, y) = )E(x, y), e(x, yA) = E(x, y)) for all scalarsA E H. Also note that xy is not H-linear in x. In fact, there is no standardH-symmetric or H-skew form (see Problem 8).

Remark. The group HU(p, q) is usually denoted "Sp(p, q)" and called the"symplectic group."

H-hermitian skew

The skew H-unitary group SK(n,H), or SK(n), is defined to be thesubgroup of GL(n, H) that fixes the standard H-hermitian skew form

(1.17) e(x, y) = 71iy1 -l.....+ aniyn.

Remark. This i is the quaternion i (see Problem 6). In Chapter 2, weshall see that if the i occurring in (1.17) is replaced by any unit imaginaryquaternion u E Sz C Im H, then the new form e' differs from the old forme by a coordinate change, i.e., an element of GL(n, H).

Table 1.18. The groups defined by bilinear forms

symmetric a skew ahermitian

symmetric ahermitian

skew e

O(p, q) Sp(n, R)

C O(n, C) Sp(n, C) U(p, q) U(p, q)

H HU(p, q) SK(n, H)

OTHER MISCELLANEOUS GROUPS

The subgroups defined by requiring either deter or detc to be equal to onecan also be defined by requiring that an n-form be fixed. The skew n-form

(1.19) dx=dx1A---Adxn

on Rn is called the standard volume form on R1, while the skew n-form

(1.19') dz=dz1A---Adze

on Cn is called the standard complex volume form on Cn. The volumeform transforms, under a coordinate change, by multiplication by the de-terminant:

A*dx = (deter A) dx for all A E EndRV


andB* dz = (detc B) dz for all B E Endc V.

Here A* denotes the dual (or pull back) map associated with A, whichis defined by (A*a)(u) = a(Au) if a is a form of degree one and by

ifa = al is the sim-ple product of degree one forms. This provides the most elegant definitionof the determinant. Frequently, this is also the most useful. For example,see Problem 4. This definition gives

(1.20) SL(n, R) = {A E GL(n, R) : A*dx = dx},

(1.20') SL(n, C) = {A E GL(n, C) : A*dz = dz}.

The special orthogonal group with signature p, q is defined by

(1.21) SO(p, q) _ {A E O(p, q) : deter A = 1}.

The special complex orthogonal group is defined by

(1.22) SO(n, C) _ {A E O(n, C) : detc A = 1).

The special unitary group is defined by

(1.23) SU(p, q) - {A E U(p, q) : detc A = 1}.

The various other possibilities do not lead to new groups. This is aconsequence of the facts presented below-see (1.24), (1.25), (1.26), Lemma1.28, (1.29), and (1.30).

Consult Problem 5 for proofs of the following:

(1.24) if A E Sp(n, R), then detEt A = 1;and(1.25) if A E Sp(n, C), then detc A = 1.

Forgetting the complex structure on C", the complex vector spaceC" becomes a real vector space of dimension 2n. This embeds the alge-bra Endc(C") of complex linear maps into the algebra Endit(C") of allreal linear maps. Thus, for a E M" (C), the real determinant detRA hasmeaning as well as detc A. See Problem 4 for a proof of the result:

(1.26) if A E M"(C), then detR,A = detcAl2.

10 Other Miscellaneous Groups

The quaternion vector space H" can be considered as a complex vectorspace in a variety of natural ways (more precisely, a 2-sphere S2 of naturalways). Let ImH denote the real hyperplane in H with normal 1 E H. LetS2 denote the unit sphere in Im H. Then, for each u E S2, u2 = -u-JUJ2 = -1. Therefore, right multiplication by u, defined by

Rux - xu forallxEH",

is a complex structure on Hn; that is, R,2, _ -1. This property enablesone to define a complex scalar multiplication on H" by (a + bi)x = (a +bRu)(x) for all a, b E R and all x E Hn, where i2 = -1. Note thatEndH(Hn) C Endc(Hn) for each of the complex structures Ru on H",where u E S2 C Im H. Choosing a complex basis for H" provides acomplex linear isomorphism H" = Ctn. Sometimes it is convenient toselect this complex basis as follows. Let C(u) denote the complex linecontaining 1 in each of the axis subspaces H C Hn. Thus, C(u) is the realspan of 1 and u. Let C(u)1 denote the complex line orthogonal to C(u)in H C H'. Then a

(1.27) H" - [C(u) ® C(u)1]" = C2n

Assume the complex structure on H" has been fixed, say Ri, then asnoted above EndH(H") C Endc(C2n). Moreover, given A E Endc(C2n),one can show that

A E Endf(H') if and only if ARj = R?A.

This is a useful characterization of the subspace EndH(H') of Endc(C2n).Lemma 1.28. For each complex structure R, on Hn (determined by rightmultiplication by a unit imaginary quaternion u E S2 C Im H) and for eachA E Mn(H) the complex determinant detc A is the positive square root ofdetR, A, independent of the complex structure R,,.

Proof: First, we show that the complex determinant of A E MM(H) isreal for all A E Mn(H). We will give the proof for the particular complexstructure R,. Consider the case n = 1. Let e° = 1, el = i, e2 - j, and e3k denote the standard real basis for the quaternions H. Let w°, w1, w2, w3denote the standard dual basis. Then

dzl =w°+i 1, dz2 =w2-iw3is a basis for the complex forms of type 1, 0 on H = C2 (with complexstructure Ri). Note that R (dz') = -dz2 and RR (dz2) = d 71, so thatRj(dz' Adze) =d71 Adx2.

Classical Groups I

Now

Rt A* (dzl A dz2) = Rj*(detc A dz' A dz2) = detc A d T' A dz2,

while

A*Rj* (dzl A dz2) = A* (d zl A dz2) = detc A d zl A dz2.

11

Therefore, detc A E R is real since ARC = RJA. The proof, for n > 1, thatdetc A E R for all A E M"(H) is similar and omitted. Because of (1.26),it remains to show that

detc A > 0 if A E GL(n, H).

Since detc I = 1 and GL(n, H) is connected (Problem 3), the set {detc A :A E GL(n, H)} is a connected subset of R - {0} containing 1, and henceit is contained in R+.

For elements of the subgroups HU(p, q) and SK(n, H) of GL(n, H),the real determinant is already equal to one (and hence by Lemma 1.28 allthe various complex determinants are also equal to one). That is,

(1.29) detR A = 1 if A E HU(p, q);

(1.30) detR A = 1 if A E SK(n, H).

Both of these facts follow from (1.24), since both HU(p, q) and SK(n, H) arecontained in Sp(2n, R) for a suitable choice of coordinates. For example,if A fixes the e defined by (1.16), i.e., A E HU(p, q), then A fixes thereal valued skew form Re ie(x, y), which under a coordinate change is thesymplectic form given by (1.11'). The details are provided in the nextchapter-see Lemma 2.80 and Equation (2.91).

In the quaternion case, there is always the option of enlarging thegroup by utilizing right scalar multiplications. Recall (1.4) how the groupGL(n, H) H* is an enhancement of the quaternionic general linear groupGL(n, H). For another example, consider the enhanced hyper-unitary group(perhaps a better name is the quaternionic unitary group). This group isdenoted HU(p, q) HU(1) and defined to be the subgroup of EndR(H")generated by letting HU(p, q) act on Hn on the left and the unit scalarsHU(1) - S3 act on H" on the right. Since

(1.31) 1 --> Z2 -p HU(p, q) x HU(1) -'-+ EndR(Hn)

12 Other Miscellaneous Groups

is exact, where Z2 = {1, -1} and where HU(1) = {Ry : y E S3 C H}, itfollows that

(1.32) HU(p, q) . HU(1) = (HU(p, q) x HU(1))/Z2.

See Problem 3.15 for more information about the quaternionic unitarygroup HU(p, q) HU(1). For example, this group fixes a 4-form 0 E

A4(H")*.

Remark 1.33. In the special case of n = p = 1, and q = 0, HU(1) actingon the left equals {La : Jal = 1}, while HU(1) acting on the right equals{Rb : IN = 1}. In fact, the quaternionic unitary group is just the specialorthogonal group. That is,

(1.34) HU(1) HU(1) = SO(4),

or equivalently,

(1.34') X : HU(1) x HU(1) --+ SO(4),

is a surjective group homomorphism with kernel Z2 = {-1, 1}, where themap X is defined by

Xa,b(x)=axb forallxEH.

To prove (1.34'), first note that by (1.29), or more directly, by Problem6(b), detR La = 1 if lal = 1 (similarly detf Rb = 1 if IbI = 1). Second,one can show that Jax) _ Jal JxJ under quaternion multiplication. Thus,La E 0(4) if Jal = 1 (similarly Rb E 0(4) if IbI = 1). This proves thatHU(1) HU(1) = x(HU(1) x HU(1)) C SO(4). The surjectivity of X canbe demonstrated with a topological argument based on dimension, once itis known that SO(4) is connected (see Corollary 3.31). A nontopologicalproof that x is surjective is provided by Problem 4.9.

If HU(1) denotes the diagonal copy of S3 = {a E H : Jal = 1} embed-ded in HU(1) x HU(1) and x is restricted to HU(1), then

(1.35)HU(1)

25SO(3).Z2

To prove (1.35), it suffices to note that the subgroup of SO(H) that fixes1 E H is just SO(ImH) and that the subgroup of HU(1) x HU(1) thatmaps into SO(Im H) equals {(a, b) E HU(1) x HU(1) : a b = 1} = HU(1).

The quaternionic enhancements are summarized as follows.


Group Enhanced Groupgeneral linear GL(n, H) GL(n, H) H*special linear SL(n, H) SL(n, H) HU(1)hyper-unitary HU(p, q) HU(p, q) HU(1)skew-unitary SK(n, H) SK(n, H) HU(1)

enhanced general linearenhanced special linearenhanced hyperunitaryenhanced skew unitary

Of course, one can always enhance a group G with R+, or R*R - {0}, if the nonzero multiples of the identity do not already belong toG. The groups G R+ are usually referred to as conformal groups. Forexample,

(1.36) CO(p, q) = O(p, q) - R+ = O(p, q) x R+

is called the conformal (orthogonal) group of signature p, q. This group,perhaps the most important conformal group, is usually defined by requir-ing that the inner product s (see (1.10)) be fixed up to a positive scalarmultiple (or conformal factor):

(1.36')CO(p, q) = {A E GL(n, R) : for some A E R+, e(Ax, Ay)

=.e(x,y)forallx,yER"}.

Similarly,

(1.37)CSO(p, q) SO(p, q) R+

{AEGL+(n,R):A*e=as forsome AER+}

is called the special conformal group of signature p, q.If both p, q > 1, then (see Chapter 3) SO(p, q) has two connected com-

ponents. The connected component of the identity, denoted by SOT(p, q),is, of course, a subgroup of SO(p, q). This subgroup SOT (p, q) of SO (p, q) iscalled the reduced special orthogonal group. Later, in Chapter 4, additionalsubgroups of O(p, q), denoted 0+ (p, q), and 0- (p, q) will be discussed insome detail. Briefly, if p, q > 1, then O(p, q) has four connected compo-nents. Adding any one of the remaining three components to SOT (p, q)yields three additional subgroups of O(p, q), denoted SO(p, q), O+(p, q),and 0-(p, q) . Thus, the intersection of any two of these three is alwaysSOT(p, q). See Chapter 4 for the details.

ISOMORPHISMS

The unit circle

(1.38) Sl-{zEC:Izi=1}

14 Isomorphisms

in the complex plane C is a group under complex multiplication. By def-inition, the groups U(1) and S' are the same. Of course, Sl - {en° : 0 ER} = R/2zrZ. The group of nonzero complex numbers under complexmultiplication is denoted by C*, and by definition, GL(1, C) - C*.

The set of unit quaternions

(1.39) S3-Ix EH:IxI =1}

also forms a group under quaternionic multiplication. Again, by definition,the groups HU(1) and S3 are the same.

Also, by definitions (1.20) and (1.20'), we have SL(2, R) = Sp(1, R)and SL(2, C) = Sp(1, C). The more difficult equality HU(1) - HU(1) =SO(4) has already been discussed. These and other coincidences are listedin the next proposition.

Proposition 1.40. The following isomorphisms hold

(1.41) SO(2) - U(1) = SK(1) = S1,

(1.42) CSO(2) = GL(1, C) - C* = SO(2, C),

(1.43) SU(2) HU(1) - SL(1, H) t--- S3,

(1.44) Sp(1, R) = SL(2, R) - SU(1, 1),

(1.45) Sp(1, C) = SL(2, C),

(1.46) HU(1) HU(1) - SO(4) and GL(1,H) H* - CSO(4),

(1.47) SOT (3, 1) c--- SO(3, C).

The last isomorphism (1.47) will be verified in the section on specialrelativity in Chapter 3.

The proofs of all of the other isomorphisms in Proposition 1.40 are leftas an exercise (see Problems 9, 10, and 11). One of these isomorphisms,SU(2) - HU(1), warrants the following discussion.

Let H have the complex structure R;x =_ xi (right multiplication byi). Thus, H - C2, where each p E H can be expressed as p = z + jw with


z,w E C C H. Now each A E M1(H) = EndH(H) can be considered asacting on H on the left, hence A E Endc(C2) = M2(C) is a complex lineartransformation of H = C2. Using the coordinates p = z + jw = (z, w) forp E H = C2, the complex linear map A expressed as a complex matrix isgiven by

/6A= aIb a ,

where A = a + jb, a, b E C C H. This is because

Aj = (A - 1)j = (a-1- jb)j = -b+ja.

This proves

(1.48) M1(H)- l\b a) EM2(C):a,bEC}.

The isomorphism HU(1) = SU(2) is derived from (1.48) (see Problem10).

Remark. In the standard reference (Helgason [10]), SL(n, H) is denotedby SU*(2n), SK(n, H) is denoted by SO*(2n), and HU(p, q) is denoted bySp(p, q) -

SUMMARY

The three general linear groups GL(n, R), GL(n, C), and GL(n, H) andthe seven groups described in Table 1.18 can be changed by imposing re-strictions on determinants and/or by enhancing with scalar multiplication.The connected component of the identity in SO(p, q) with p, q > 1 is alsoa group. All the groups introduced in this chapter can be obtained in thisway.

In low dimension, some of these groups coincide. One of the most inter-esting isomorphisms is SU(2) = HU(1). The topic of special isomorphismsin low dimensions will be discussed again in Chapter 14.

PROBLEMS

1. Establish EndF(F") and GL(n, F) - GLF(F'") for FR, C, H.

2. If A E EndH(H") is injective, then A-1 is H-linear.

16 Problems

3. (a) Let e denote the column vector (1, 0, ..., 0) E F". Show that foreach x E F" - {0} there exist A E GL(n, F) with Ae = x. Let Kdenote the subgroup of GL(n, F) that fixes e. Show that K has thesame number of connected components as GL(n - 1, F).(b) Use part (a) and induction to show that GL(n, C) and GL(n, H)are connected, while GL(n, R) has exactly two components.

(c) Show that SL(n, R), SL(n, C), and SL(n, H) are all connected.

(d) Show that O(n), U(n), and HU(n) are compact.(e) Show that each of the groups defined in Table 1.18 is the level setof a vector-valued polynomial, and hence is a closed set.

4. Suppose A E Endc(C") C Enda,(R2n ). Let dt denote the standardvolume form on R2' . Use the facts that:

(a) A` dt = (detR, A)dt,

(b) A*dz = (detc A)dz,(c) dt = k dz A d x for some constant k,to show that detg A = I detc Ale.

5. Let e denote the standard symplectic form on R2s, and dx the stan-dard volume form on R2n. Show that they are related by

1 E A ... AE = dx.

6. The quaternions H can be defined as the vector space R4 with

1 - ( 1 , 0, 0, 0), i = (0, 1, 0, 0), j (0, 0, 1, 0), k = (0, 0, 0,1),

and multiplication defined by

i2=.92=k2=-1ij = k and all cyclic permutations of this equation are valid,and i, j, k skew commute.

Given x = xo + xl i + x2 j + x3k, define conjugation by x - xo - x1i -Z2j - x3k.(a) Show that ry = y x, x x = 1x2, 2(x, y) = xy + yx.(b) Given a E H, left multiplication by a, denoted La, belongs toEndH(H) C Endit(H). In terms of the standard basis for H = R4,compute the 4 x 4 matrix for La, and then show that detR, La = Jal4.


7. Suppose V is a right H-vector space. Let HomH(V, H) denote thereal vector space of right H-linear maps from V to H. Given f EHomH (V, H) and A E H, define f A E HomH (V, H) by

(1.49) (fA)(v) = af(v) for all v E V.

(a) Show that (1.49) exhibits a right H-structure on HomH(V, H).The H-dual of V, denoted V*, is defined to be the vector spaceHomH (V, H) equipped with this right H-structure (1.49).(b) Exhibit a (canonical) right H-linear isomorphism (V*)* = V.(c) Given f E HomH(V, W), let f* (the dual map) be defined by

(1.50) (f * (w*))(v) = w* (f (v)) for all w* E W* and v E V.

Show that f* is right H-linear, i.e., f E HomH(W*, V*).(The real vector space HomH(V, W) cannot be (canonically) given theextra structure of either a right or a left H-vector space.)

8. Suppose : V x V -i H is H symmetric, i.e., e(x, y) = -(y, x) ands(x, y.A) = E(x, y)A. Prove that e = 0.

9. Show that

(a) U(1) - Sl,(b) SO(2) - S1,(c) SK(1) = S1,(d) Sp(1, R) = SL(2, R),(e) Sp(1, C) = SL(2, C),(f) SO (2, C) = C*,

(g) SL(1, H) = S3.

10. Show that

(a) HU(1) S3,

a -e'ab l l(b) U(2) = b ei9 I : a, b E C, (a12+ Ibj2 = 1, and 9 E R },

(c) SU(2) - (I b a f : a, b E C and 1a12+ 1b12 = 1

(d) SU(2) - HU(1).

18 Problems

11. Prove the following.

(a) U(1, 1) = { I a e'Ba) : a, c E C, B E R, and ja12 - Id12 = 1c I-

a c ):acEC and Jal2-Ic12=1}.

a J

(c) SU(1, 1) maps N into itself and M into itself, where N = {(z, z)z E C} and M = {(z, -x) E C2 : z E C} are both two-dimensionalreal subspaces of C2.

(d) C-'SU(1,1)C - SL(2, R), where C = I 1 _E) is the Cayley

Transform. \ /

2. The Eight Types of InnerProduct Spaces

The "inner products" (or "bilinear forms"), which we denoted by c inthe previous chapter, will be examined in more detail in this chapter.

Even if one is only interested in the orthogonal groups O(p, q), all typesappear (in a natural way), as will be seen in a later chapter on spinor innerproducts.

Suppose V is a vector space over F = R, C, or H and E is a biadditivemap

(2.1) e:VxV--+F.

That is, c(x + y, z) = e(x, z) + e(y, z) and e(z, x -{- y) = E(z, x) -I- e(z, y).The biadditive map is said to be (pure) bilinear if

e(Ax, y) = A,-(x, y) and -(x, Ay) = A,-(x, y)

for all scalars A = F.

The biadditive map e is said to be (hermitian) bilinear if

e(xA, y) = aE(x, y) and e(x, yA) = E(x, y)A

for all scalars A E F.

19

20 Standard Models

Frequently, when it is clear from the context, the adjective "pure" or theadjective "hermitian" will be dropped. For example, if F = R there isonly one kind of bilinear; the two notions, pure bilinear and hermitianbilinear, agree. If F = H and e is pure bilinear, then e = 0 becauseE(x, y)Ap = E(xA, yµ) = e(x, y) jA, while µ, A E H may be chosen so that[p, A] = uA - Ap 0 0. Thus, for F = H there is only one kind of bilinear,namely, hermitian bilinear.

Let BlLhei.,n(V)-or just BIL(V) when the correct adjec-tive pure or hermitian is clear from the context-denote the space of bi-linear forms on V. Note that these spaces are always real vector spaces.Suppose E E BlLpure(V). The form E is said to be F-symmetric, or F-puresymmetric if

(2.4) s(x, y) = E(y, x) for all x and y E V.

The form E is said to be F-skew or F-pure skew if

(2.5) e(x, y) = -e(y, x) for all x, y E V.

Suppose E E BlLherm(V). The form E is said to be F-(hermitian) symmetricif

(2.6) E(x, y) = s(y, x) for all x, y E V.

Finally, the form e is said to be F-(hermitian) skew if

(2.7) s(x, y) = -e(y, x) for all x, y E V.

Note that for F = C (but not for F = H) a hermitian bilinear forme E BIL(V) is skew if and only if is is symmetric.

The notations SYMpure(V), SYMherm(V), SKpure(V), and SKhe,.,,,(V)should be self-explanatory. Each E E BIL(V) has a unique decompositionE = El + E2 into the sum of a symmetric part el and a skew part E2. If E ispure, then

Ei (x, y) =2

E(x, y) + s(y, x) and E2(x, y) = 2 E(x, y) -2

E(y, x)

while ifs is hermitian then

Cl (x, y) E(x, y) +1

2e(y, x) and 62(2 , y) _ 2 E(x, y) - 2

E(y, X).

A bilinear form e E BIL(V) is said to be nondegenerate if

(2.8) E(x, y) = 0 for all y E V implies x = 0,

Inner Product Spaces

and

(2.8') e(x, y) = 0 for all x E V implies y = 0.

21

Definition 2.9 (Inner Product Space). Suppose V is a finite dimen-sional vector space of F, with F one of the three fields F = R, C, or H. Aninner product e on V is a nondegenerate bilinear form on V that is eithersymmetric or skew. If F - C, then there are two types of symmetric andtwo types of skew: pure and hermitian.

The eight types of inner products are

1. R--symmetric

e is R-bilinear and e(x, y) = E(y, x),

2. R-skew or R-symplectic

is R-bilinear and e(x, y) = -E(y, x),

3. C-symmetric

E is C-bilinear and e(x, y) = e(y, x),

4. C-skew or C-symplectic

E is C-bilinear and e(x, y) _ -e(y, x),

5. C-hermitian symmetric

e is C-hermitian bilinear and e(x, y) = e(y, x),

6. C-hermitian skew

E is C-hermitian bilinear and e(x, y) _ -E(y, x),

7. H-hermitian symmetric

e is H-hermitian symmetric and E(x, y) = e(y, x),

8. H-hermitian skew

e is H-hermitian and e(z, y) = -c(y, x).

22 Standard Models

In each of the eight cases the pair V, e is then called an inner productspace (of one of the eight types).

THE STANDARD MODELS

The standard models of inner product spaces are listed below. The readermay recall the names of the groups that fix these bilinear forms from Chap-ter 1. Please note that the conjugation in the C-hermitian symmetric casehas been changed from the second variable to (as in (1.14)) to the firstvariable z in order to be consistent with the quaternion case.

1. R-symmetric: The vector space is R, denoted by R(p, q)(n = p + q),with

(2.10) e(x, y) = xlyl + ... + xpyp - ... - xnyn

2. R-skew or R-symplectic: The vector space is R2n with

(2.11) E(z,y) = zly2 - z2y1 + + z2n_1y2n - x2ny2n-1.

3. C-symmetric: The vector space is Cn with

(2.12) E(z, w) = ziwl + ... + znwn.

4. C-skew or C-symplectic: The vectorspace is C2n with

(2.13) E(z, w) = z1w2 - z2w1 + ...+. z2n-lw2n - z2nw2n-1

5. C-hermitian (symmetric): The vector space is Cn, denoted by C(p, q)(n = p + q) with

(2.14) e(z, w) - 71w1 + ... + zpwp - ... - znwn.

6. C-hermitian (skew): The vector space is Cn(n = p+ q) with

(2.14') E(z, w) = iz1W1 + ..+ izpwp - .. iznwn.

7. H-hermitian symmetric: The vector space is Hn, denoted by H(p, q)(n = p + q) with

(2.15) e(x, y) _ Tlyi + ... + x zn yn

Inner Product Spaces 23

8. H-hermitian skew: The vector space is H'" with

(2.16) E(x, y) = xiiy1 + .....F. xniyn.

Some of the R-symmetric special cases are very important. The posi-tive definite case R(n, 0) is called euclidean space, while the case of a singleminus, R(n - 1, 1), is called Lorentzian space. If n = 4, then Lorentzianspace is also called Minkowski space or Minokowski space-time. The specialcase R(p,p) will be referred to as the split case.

ORTHOGONALITY

Definition 2.17. Suppose f : V --> V is a F-linear map from an innerproduct space V, e to another inner product space V, E of the same type.The map f is said to preserve inner products if

(2.18) E(f (x), f (y)) = e(x, y) for all x, y E V.

Since e is non degenerate such a map must be one to one. If, in addition,such a map is onto then f is called an isometry. The inner product spacesV, E and V, E are said to be isometric if there exists an isometry betweenthem.

Remark 2.19. Suppose f is an isometry from V, e to V, E. Let G denotethe e-isometry subgroup of GL(V, F), and let G denote the E-isometrysubgroup of GL(V, F). Then the groups G and G are isomorphic:

(2.20) G= f o G o f-1

For example, the model C2, E with E(z, w) - zlw2 + z2w1 is isometric tothe standard model C2, e with -(C, rl) 67]1 +6192. Using the model C2, Eit is almost immediate that SO(2, C) = C* (cf. Problem 1.9(f)).

Given a vector subspace W of the inner product space V,,-, let e1wdenote the restriction of e to W. The restriction eJw is positive definite ife(x, x) > 0 for all nonzero x E W. Similarly, one defines negative definite.A subspace W of V is said to be(2.21) positive (or spacelike) if c 1w is positive definite,

(2.22) null (or degenerate or lightlike) if e w is degenerate,(2.23) negative if -Iw is negative definite.

These three possibilities are by no means exhaustive.Positive (negative) subspaces can only occur for three types of inner

product spaces:

24 Orthogonality

R-symmetric, C-hermitian (symmetric), and H-hermitian symmetric.In these three cases, another numerical invariant (in addition to the

dimension), called the signature, is required in order to determine whentwo inner product spaces are isometric.

Definition 2.24. Suppose V, e is an inner product space of one of thethree speciAl types mentioned above. A positive subspace W of V is saidto be maximal positive if

dimZ < dim W for all positive Z C V.

A negative subspace W of V is said to be maximal negative if

dim Z < dim W for all negati ve Z C V.

The signature p, q of V is defined by

p equals the dimension of a maximal positive subspace,q equals the dimension of a maximal negative subspace.

If V, e is positive definite (on V), then the signature is n, 0, and if V, eis negative definite (on V), then the signature is 0, n. If V, e has signaturep = q, the inner product space is said to be split, or of split signature.

The case of signature p = 2, q = 1 is pictured in Figure 2.27.Obviously, if V, e and V, a are isometric, then they have the same

signature. In fact, as we will prove below, if V, c and V, E have the samedimension and signature, then they are isometric. First we must examinethe important concept of orthogonality.

Two vectors u, v E V are said to be orthogonal, written u 1 v, ife(u, v) = 0. Note that

(2.25) u .1. v if and only if v .1. u

for all eight types of inner products. Suppose W is a subset of V. Then

uIW means uIvfor all vE W.

The perp or orthogonal to W is defined by

(2.26) Wl-fu EV:uIW}.

The perp of the line [v] spanned by v E V will be denoted vl.

Inner Product Spaces

Figure 2.27

25

Since "u .1. v" just involves a pair of vectors, the key to understand-ing orthogonality lies in understanding two dimensions-more specifically,Elw where W = span{u,v}. In order to sharpen our intuition, we focusattention on inner product spaces of the R-symmetric type.

In the euclidean plane R(2, 0), two orthogonal vectors are pictured asbeing at right angles to each other.

In the Lorentzian plane R(1, 1), first picture the null or light cone,consisting of all a = (x, t) E R(1, 1) which satisfy E(a, a) = x2 - t2 = 0.Such a vector a E R(1, 1) is said to be null or lightlike. Note that a 1 afor each lightlike vector a. Given a vector b = (x, t) E R(1, 1), the vectorb' = (t, x) E R(1,1) is orthogonal to b. The vectors ±b' are obtainedpictorially by (euclidean) reflecting b through either one of the two linesmaking up the null cone (see Figure 2.28). The perp bl is just the span of±b'.

26 Orthogonality

timeliket t

spacelike

x

null

Figure 2.28

This example shows that a subspace W and its perp W1 need not becomplementary subspaces. Recall that a subspace W is null (or degenerateor lightlike) if eIw is degenerate. If eIW - 0, then W is said to be totallynull or isotropic. Note that

(2.29) W is totally null if and only if W C W.

Totally null subspaces are particularly important in the split case R(p, p)and the symplectic cases (R-skew and C-skew).

Despite these strong counterexamples to our euclidean intuition, whichsays W1 should be complementary to W, it is still true that the dimensionof W1 is complementary to the dimension of W.

The next lemma is basic for all eight types of inner products.

The Orthogonality Lemma 2.30. Suppose V, e is a n-dimensional innerproduct space, and W is a linear subspace of V. Then

(2.31) dim W+dim Wl=n,

(2.32) (W1)1 = W,


and the following are equivalent:

(2.33)

(a)

(b)

W is nondegenerate,

W n W1 = {0},

(c)

(d)

W + W-L = V,

Wl is nondegenerate.

Proof: Let F denote the field of scalars (either R, C, or H). The innerproduct e : V x V --> F determines a map b : V -* V* from V to the dualspace V* as follows. Consult Problem 1.7(b) for the definition of V* if Vis a right H-space. For each z E V fixed, let b(x) E V* be defined by

(2.34) (b(x))(y) - E(x,y) for all y E V.

Note that for all eight types of inner products b(x) E V* since e(x, y) isF-linear in y. The nondegeneracy hypothesis,

(2.35) E(x, y) = 0 for all y E V implies x = 0,

is equivalent to

(2.35') b : V -+ V* is one-to-one.

Since V and V* have the same real dimension and b is always R-linear,must be a real linear isomorphism.

Given a subspace W of V, the annihilator of W, denoted W°, is thesubspace of V* defined by

(2.36) W°- {0EV*:0(y)=0forallyEW}.

Suppose X E V and 0 E V* correspond under the isomorphism b. Then

E W° if and only if x E W.

That is,

(2.37) Wl -- W° is an R-linear isomorphism.

In particular,

(2.38) dimF Wl = dimF W°.

28 Orthogonality

Choose a basis e1,.. ., e,n, ..., en for V with e1,.. ., e,,, a basis for W. Lete*,,. . ., en denote the dual basis for V*. Then e;, is a basis forW°. Therefore, dimF W° = n dimRW, which, by (2.38), proves the firstpart of the Lemma, (2.31).

Obviously, W C (W')', and since the dimensions are the same, theymust be equal. The proof that (a)-(d) are equivalent is left as an exercise.

Remark 2.39. (a) If V is the tangent space T to a (semi)Riemannianmanifold at a point, the flat isomorphism b, defined above, between vec-tors and convectors is referred to as metric equivalence and corresponds to"lowering indices." The inverse map = b-1, called sharp, is referred to as"raising indices."

(b) If V is the tangent space to a symplectic manifold (e.g., phasespace - the cotangent bundle T*M to a manifold M), then b allows one tostart with a function (Hamiltonian function) f and associate a vector field(the associated Hamiltonian vector field) by first taking df, the exteriorderivative of f, and then applying sharp 0 = b-1.

The reader unfamiliar with these concepts may wish to glance at Chap-ter 5, "Differential Geometry."

Corollary 2.40. If W is a non degenerate subspace, then each vector u EV has a unique orthogonal decomposition:

(2.41) u=w+z with wE W andzEWJ'.

The linear map a that is zero on W1 and the identity on W is calledorthogonal projection onto W.

In the three cases R-symmetric, C-hermitian, and H-hermitian sym-metric, where positive/negative subspaces exist, the inner product space issaid to have a signature.

The Signature Lemma 2.42. Suppose V, a is an inner product spacethat has a signature, say p, q. Let W denote a subspace.

(a) W is maximal positive if and only if Wl is maximal negative.(b) If W is positive and W' is negative, then both are maximal.

Corollary 2.43. If V has signature p, q, then p + q = n the dimension ofV.

The obvious guess for the signatures of the standard models can berigorously verified using the Signature Lemma (see Problem 2).


Corollary 2.44. The standard models R(p, q), C(q, p), and H(p, q) havesignature p, q.

Proof of 2.42(a): Suppose W is maximal positive. It suffices to show thatW1 is negative, because the maximality of W1 then follows from part (b).First, we establish that W1 is negative semidefinite; that is, E(x, x) < 0 forall x E W1. Otherwise, there exists a nonzero, spacelike vector u E W1,i.e., E(u, u) > 0. Since E restricted to either of the orthogonal subspacesW or [u] - span u is positive definite, an easy calculation shows that Erestricted to W + [u] is also positive definite, contradicting the maximalityof the dimension of W. Finally, we must rule out E(x, x) = 0 for x E W1unless x = 0. If W1- contains a nonzero null vector u, then there exists avector y E W1 with E(u, y) $ 0 because W1 is nondegenerate. We mayassume that E(u, y) is real and greater than zero. Now 0 > e(u + ty, u -E-ty)/t = 2E(u, y) +tE(y, y), for all t > 0, which is impossible.

The hypothesis that V, e has a signature was used in this last equationto conclude that E(y, u) = E(u, y) from the assumption that e(u, y) is real.Can you find the other use of the signature hypothesis in this proof?Proof of 2.42(b): We shall prove that W is maximal (positive). SupposeP is another positive subspace. Since W1 is negative, {0} = P fl W1 =P fl 7r-1(0), where Tr is orthogonal projection onto W. Thus, a : P --> Wis one-to-one, so that dim P < dim W.

In a similar manner one can show:

Proposition 2.45. If V has a signature (say p, q), then each totally nullsubspace N must have dimension < min{p, q}.

A CANONICAL FORM (THE BASIS THEOREM)

Using the Orthogonality Lemma and the Signature Lemma any inner prod-uct space may be put in canonical form.

The Basis Theorem 2.46. Suppose V, e is an inner product space of oneof the eight types. Then V is isometric to the standard model of the sametype that has the same dimension and signature.

Corollary 2.47. Suppose V, e and V, 9 are two inner product spaces ofthe same type. The dimension and signature are the same if and only if Vand V are isometric.

For each of the eight types of inner product spaces, certain bases areof particular importance and will be used to prove Theorem 2.46.

30 A Canonical Form (The Basis Theorem)

Definition 2.48. Suppose V, a is an inner product space of dimension Nand signature p, q (if V has a signature). A basis lei,. . -, eN} for V is saidto be orthonormal if the linear map f from V to the standard model ofthe same type, dimension, and signature, defined by sending ej to the jchstandard basis vector for the standard model, is an isometry.

Thus, a basis {e! , ... , err} for V,,- is orthonormal ifR-symmetric (dimR. V = n):

e(ej , ej) = 1 for j = 1, . . . , p,

(2.49) E(ej,ej) -1 for p+ 1,.. ,n,e(ei, ei) = 0 fori # j;

R-skew (dimR, V = 2n):

(2.50)e(e2i-1, e2j) = 1 for j = 1, ... , n,

e(ei, e1) = 0 for all other pairs ei, ej,

(orthonormal basis = symplectic basis);

C-symmetric (dimc V = n):

(2.51)e(e,,ej)=1 for

e(ei,e1)=0 for i j,

(orthonormal basis = C-orthonormal basis);

C-skew (dimc V = 2n):

(2.52)e(e2j-l, e2j) = 1 for j = 1, ..., n,

e(ei, eJ) = 0 for all other pairs ei, e

(orthonormal basis = C-symplectic basis);

C-hermitian (symmetric) (dimc V = n):

e(ei,ej)=1 forj=1,...,p,(2.53) e(ej,ej)=-1 forj=p+1,...,n,

e(ei, ei) = 0 fori # j,

(orthonormal basis = unitary basis);


H-hermitian symmetric (dimes V = n):

1 for j(2.54) e(ej,ej)=-1 forj=p+1,...,n,

e(ei,e?)=0 for i#j,

(orthonormal basis = H-unitary or hyperunitary basis);

H-hermitian skew (dimH V = n):

(2.55)e(ej,ej) = i for j = 1,...,ne(es,ej)=0fori#j,

(orthonormal basis = H-skew basis).

Proof of Theorem 2.46: It suffices to show that V,,- has an orthonormalbasis. The proof is by induction on the dimension.

R-symplectic, C-symplectic: Choose el any nonzero vector. Since e is non-degenerate, there exists u with e(ei, u) $ 0. Let e2 denote u rescaled so thate(ei,e2) = 1. Let W = span {ei,e2}. Since W is nondegenerate, Lemma2.30 implies that V = W +W1, W n W1 = {0}, and that W1, e is a lowerdimensional (symplectic) inner product space. By the induction hypothesisW1 has a symplectic basis e3i ..., e2n. Since W 1 W1, el, e2i e3, ..., e2,ais a symplectic basis for V = W + W.Signature: Assume V, e has a signature. Choose P maximal positive andN maximal negative with N = P1 by Lemma 2.42. Since P and N areorthogonal, it suffices to show that P and N have an orthonormal basis.Thus, we may assume that V is positive if V has a signature.

The proof is completed, in the positive definite cases as well as in allremaining cases without signature, as follows. Choose a nonzero vectoru E V. Let ei denote u properly normalized so that e(ei, ei) = 1, unlesse is H-hermitian skew, in which case require e(ei, ei) = i (see Problem6). Let W = span el. Then W, and hence W1, is nondegenerate. Thus,by the induction hypothesis W1 has an orthonormal basis. Since W andW1 are orthogonal, el combined with the orthonormal basis for W1 is anorthonormal basis for V = W + W1. I

THE PARTS OF AN INNER PRODUCT

Suppose e is one of the eight types of inner products. If e is either C-valuedor H-valued, then 6 has various parts.

32 The Parts of an Inner Product

The simplest case is when a is a complex-valued inner product. Thene has a real part a and an imaginary part /3 defined by the equation

(2.56) e = a + i/3

(while requiring a and )3 to be real-valued).Suppose e is a quaternion valued inner product. There are several

options for analyzing the parts of E.First, e has a real part a and a pure imaginary part Q defined by

(2.57) e=a-}-/3,

where a = Re c is real-valued and /3 = Imc takes on values in ImHspan{i, j, k}. The imaginary part 11 has three components defined by

(2.57') 0 - i/3, + j/33 + k/3k,

where the parts /3;, /3? , /3k are real-valued.Second, using the complex structure R; (right multiplication by i) on

H, each quaternion x E H has a unique decomposition x = z + jw wherez, w E C C H. Therefore,

(2.58) e = y + jb with -y and b complex-valued

uniquely defines y (the first complex part of e) and 6 (the second complexpart of e).

Of course,

(2.58') 7=a+i,8 and b=/3j-i8k.

This section is devoted to computing the types of these parts of E.As an application, alternate definitions of the isometry group G for e arededuced. However, this material is better motivated by geometric consid-erations and should be read in conjunction with Chapter 5 on geometry.In fact, readers may wish to examine Chapters 3-5 before returning to thissection. Because of its importance in geometry, each type of inner producte is decomposed into parts using notation from geometry.

C-Hermitian Symmetric

Consider the standard C-hermitian (symmetric) form

(2.59) e(z, w) = zlwl + - . + zpwp - zp+lWp+l - ... - znwn


with signature p, q on C11. Since a is complex-valued, it has a real andimaginary part given bye = g - iw. For z = x + iy and to = l; + iq E CnR21, the real and imaginary parts g = Ree and w = -Ime are given by

(2.60) 9(z, w) = x1Si + y17)1 + ... + yprip - ... - xn. n - ynrln

and

(2.61) w(z, w) = x11 - xnrln -1-

Thus, g is the standard R-symmetric form on R2n with signature 2p, 2q.Modulo some sign changes, w is the standard symplectic form on R21. Inthis context, when q = 0, w is exactly the standard Kahler form on Cnand is usually written as

(2.62) w = 2 dzl Adz, + dzn A dzn.

Lemma 2.63. Suppose e is C-hermitian symmetric (signature p, q) on acomplex vector space V with complex structure i. Then g - Ree is R-symmetric (signature 2p, 2q) and w = -Ime is R-skew. Moreover, eachdetermines the other by

(2.64) g(z, w) = w(iz, w) and w(z, w) = g(iz, w).

Also, i is an isometry for both g and w:

(2.65) g(iz, iw) = g(z, w), and w(iz, iw) = w(z, w).

Conversely, given g R-symmetric with i an isometry, if w is defined by(2.64), then e - g - iw is C-hermitian. Also, given w R-skew with i anisometry, if g is defined by (2.64), then e = g - iw is C-hermitian.

Proof: It suffices to prove that Ree and Ime have the desired propertieswhen Cn, a is the standard model; this has already been carried out (see(2.60) and (2.61)). Alternatively, the properties of Ree and Ime can beeasily derived from the fact that a is C-hermitian symmetric with signaturep, q, providing a second proof, this one without coordinates.

Reconstructing a from g (or from w), with i an isometry is Problem8a.

Remark 2.66. If the signature is positive definite, then Lemma 2.63 maybe cryptically summarized by saying that "the confluence of any two of


(a) complex geometry,(b) symplectic geometry,(c) Riemannian geometry

is Kahler geometry." See Lemma 5.17 for more details.

This decomposition e = g - iw also provides several alternative defin-tions of the unitary group U(p, q).

Suppose V, e is a C-hermitian symmetric inner product space of sig-nature p, q. Let GL(n, C) denote Endc(V), let O(2p, 2q) denote the sub-group of GL(V, R) fixing g - Re c, and let Sp(n, R) denote the subgroupof GL(V, R) fixing w.

Corollary 2.67. The intersection of any two of the three groups

GL(n, C), Sp(n, C), and 0(2p, 2q)

is the group U(p, q).

Similar results are valid in the C-symmetric and C-skew cases, butperhaps not so interesting since in either of these cases the group that fixesa = Re e and the group that fixes /3 = Ims are just different versions of thesame group under a coordinate change. These cases are left to the reader.

H-Hermitian Symmetric

Consider the standard H-hermitian symmetric form

(2.68) E(x, y) _=71Y1 -{- ... + xp yp - ... - xnYn

on Hn with signature p, q. Note that a is H-valued. As noted earlier, it isnatural to consider H as two copies of C, H - C ® jC, or x = z + jw. Inparticular, e = h + jo- with h and o complex-valued.

Note that if x =- z -}- jw and y 1; -I- jq with z, w, 77 E C, then

(2.69)xy = (z - jw)( + j'7) = z - jwjr7 + zjr7 - jwC

= x + 017 +?(zi - w. ).

Therefore, the first complex part of h of e is given by

(2.70) h(x, y) = 91b1 + w1771 + ... + wp?7p - ... - xnSn - wn77n.

Thus h is the standard C-hermitian symmetric form on C2n = H. Be-cause of (2.69) the second complex part o- is given by

(2.71) v(x, y) = zi771 - w16 ± ... ± znr7n f

Thus, modulo some possible sign changes, a is the standard C-skew formon Ctn.


Lemma 2.72. Suppose e = h + jo- is H-hermitian symmetric on a rightH-space V. Then h, the first complex part of e, is C-hermitian symmetricand v, the second complex part of c, is C-skew. Moreover, each determinesthe other by

(2.73) h(x, y) = o(x, yj) and v(x, y) = -h(x, yj).

Also,

(2.74) h(xj, yj) = h(x, y) and u(xj, yj) = o(x, y).

Conversely, given h C-hermitian symmetric and h(xj, yj) = h(x, y), if o isdefined by (2.73), then e = h + jo- is H-hermitian symmetric. Also, givenv C-skew and o(xj, yj) = o-(x, y), if h is defined by (2.73), then e = h+ jaH-hermitian symmetric.

Proof: The equation e(x, yj) = e(x, y)j can be used to prove (2.73), whilee(xj, yj) = -je(x, y)j can be used to prove (2.74). The remainder of theproof is left to the reader (see Problem 8(b)). J

Suppose V, a is an H-hermitian symmetric inner product space of sig-nature p, q. Let GL(n, H) denote EndH(V), let U(2p, 2q) denote the sub-group of GL(n, H) that fixes the first complex part h of e, and let Sp(n, C)denote the subgroup of GL(n, H) that fixes the second complex part a ofe.

Corollary 2.75. The intersection of any two of the three groups

GL(n, H), U(2p, 2q), and Sp(n, C)

is the group HU(p, q).

It is useful to try to construct the quaternionic structure from h anda.

Suppose V is a complex 2n-dimensional vector space, h is a C-hermi-tian symmetric inner product on V, and o- is a complex symplectic innerproduct on V. Then h and o define a complex antilinear map J by

(2.76) h(xJ, y) = o(x, y).

Now o-(xJ2, y) = -v(y, xJ2) = -h(yJ, xJ2) = -h(xJ2, yJ) = -o(xJ, yJ).Therefore, j2 = -1 if and only if

(2.77) o(xJ, yJ) = v(x, y).

In this case, h and o are said to be compatible.


Lemma 2.78. Suppose h is a C-hermitian symmetric inner product anda is a complex symplectic inner product on a complex 2n-vector space V.If h and a are compatible, then they determine a right H-structure on Vand

E=h+jais an H-hermitian symmetric inner product on V.

Proof: It remains to verify that h + ja is H-hermitian symmetric. J

There is yet another description of HU(p, q). First, recall (1.27) thatfor each unit vector u E Im H, right multiplication by u (denoted R,,) actingon H" determines a complex structure on Hn and hence an isomorphismHn = C2n

Second, recall that in the C-hermitian symmetric case the Kahler formw is minus the imaginary part of E and that w is determined by the formula(2.64) in terms of g = Re E. Each of the complex structures R. on H'C2n determines a Kahler form in exactly the same manner. Let

(2.79) w,, (x, y) = Re E(xu, y), and g(x, y) = Re e(x, y).

Lemma 2.80. For all x, y E Hn,

E=g+iw;-{-jwj +kwk.

Proof: For u E ImH with Jul = u u = -u2 = 1,

(u, e(x, y)) = (1, U6(x, y)) = Re e(xu, y) = y).

For each complex structure u E Im H, Jul = 1, the complex C(u)valued form

(2.81) hu =g+uwu

is C(u)-hermitian symmetric. The group that fixes hu is a unitary groupwith signature 2p, 2q determined by the complex structure Ru.

The next corollary justifies the name "hyper-unitary" for HU(p, q).

Corollary 2.82. The hyper-unitary group HU(p, q) is the intersection ofthe three unitary groups determined by the three complex structures R;,R, , Rk on H".

Remark 2.83. Of course, HU(p, q) is also the intersection, over u E S2 CImH, of the unitary groups determined by all the complex structures Ru.


The simplest case states that HU(1) is the intersection of all the unitarygroups determined by the complex structures Ru on H = C2. Recall (1.43)the isomorphism SU(2) = HUU(1), where SU(2) denotes the special unitarygroup determined by the complex structure Ri on H = C2. Similarly, foreach complex structure R, on H the corresponding special unitary groupis again the same HU(1) and thus is independent of R.

H-Hermitian Skew

Consider the standard H-hermitian skew form on H":

(2.84) e(x, y) = xliyl + ... + x"iy".

Since i is distinguished among all the unit imaginary quaternions u, it isnatural to set H - C ®j C where each copy of C has the complex structureRi induced from Ri acting on H.

Let

(2.85) e = ih + jo,

where ih and o are the first and second complex parts of a respectively.Set x = z + jw and y =1; + jrt with z, w, , 71 E C. Then

(2.86) h(x, y) = w1r11 +....f U7"77"

is C-hermitian symmetric, and

(2.87) 0'(X, Y) = x1711+w1S1 +...+xnrln+wnS,

is C-symmetric.In summary, if e = ih + jo is H-hermitian skew then

(2.88) h is C-hermitian symmetric and o is C-symmetric.

Since the form h is equivalent to the standard°C-hermitian symmet-ric form on C2n with signature n, n via a complex coordinate change, letU(n, n) denote the group fixing h. Since o is equivalent to the standardC-symmetric form on C2n via a complex coordinate change, let O(2n, C)denote the group that fixes v.

38 Problems

Proposition 2.89. The intersection of any two of the three groups

GL(n, H), U(n, n), and O(2n, C),

is the group SK(n, H).

Note that the H-skew sphere of radius i is the intersection of the unitsphere S(2p,2p), given by Iz12 - Iw12 = 1, with the null cone >z;w; = 0,where (z, w) E C2n = H' .

For each u E S2 C ImH,

(2.90) g. (X, y) = 4X U, y) = (u, e(x, y))

defines a real symmetric inner product with split signature, while

(2.91) w=ReE

is a real symplectic inner product.Because of (2.90),

(2.92) E = w + ig= + jgi + kgk.

Finally, note that SK(n, H) is also "hyper-unitary" (but in a differentsense than HU(p, q)). The reader may wish to compute (in coordinates)w, g+, gi, and gk.

PROBLEMS

1. Give the proof of (2.33), that:(a) W is nondegenerate,(b) WnWl={0),(c)W+Wl=V,(d) Wl is nondegnerate are all equivalent.

2. (a) Prove that R(p, q), C(p, q), and H(p, q) have signature p, q.(b) If V, a has a signature, show that each positive subspace is con-tained in a maximal positive subspace.

3. Suppose V, E is an inner product space with a signature. A map f EEndF(V) is called an anti-isometry if

e(f (x), f (y)) = --c (x, y) for all x, y E V.


A map f E EndF(V) is said to be anti-conformal if, for some negativeconstant A < 0,

E(f(x),f(n)) = AE(w,y), forallx,yEV.

Show that if there exists an anti-conformal map f E EndF(V), thenthe signature must be split.

4. Suppose V, E is an inner product space of R-symmetric type with sig-nature p, q and p < q. Prove the following.(a) There exists a totally null subspace of dimension p.(b) Each totally null subspace is of dimension < p (i.e., prove Propo-sition 2.45).(c) V is split (p = q) if and only if V = Ni ® N2 with N1, N2 totallynull.

5. Suppose V is a real vector space and E : V x V -p R and b : V -> V*are related by the equation

E(x,y) = (b(x))(y)

(Thus E(x, ) is linear since #(x) E V*.) Prove the following.(a) e(., y) is linear if and only if is linear.(b) e is symmetric if and only if (Here b* is the dual map notthe adjoint.)

6. Suppose V, E is an H-hermitian skew inner product space. Show thatthere exists a vector el E V with e(e1, e1) = i.

7. A finite dimensional vector space V over F equipped with a bilinearform E satisfying all of the properties of an inner product except (pos-sibly) the nondegeneracy condition is called a degenerate inner productspace. The standard models for the degenerate inner product spaces areall of the form Fk e F'", e where F'", e is one of the standard modelsfor an inner product space. Show that each degenerate inner prod-uct space V, e is isometric to exactly one of the degenerate standardmodels.

Hint: Write V = N ® W with N = V.8. (a) Suppose g is an R-symmetric inner product on R2n and i is a

complex structure on R2n that is orthogonal with respect to g. Showthat

E(z, w) = g(z, w) - ig(iz, w)

is C-hermitian symmetric on C" = RZ", i.

40 Problems

(b) Suppose o is a C-symmetric (or C-skew) inner product on CZ"H", R; that satisfies

a(x.1, yj) = a(x, y) for all x,yEH".

Show thatE(x, y) = cr(x, y.9) +?cr(x, y)

is H-hermitian symmetric (or H-hermitian skew).9. The Basis Theorem 2.46 may also be interpreted purely algebraically.

Only the first case (R-symmetric) is discussed. Show that the twostatements (a) and (b) are equivalent.(a) (See Problem 7.) Each degenerate R-symmetric inner productspace is isometric to one of the standard models R(p, q) ® Rk.(b) (Sylvester's Theorem.) Given a symmetric matrix A E M (R),there exists an invertible matrix B E so that BAB-1 is diag-onal with each nonzero diagonal entry either +1 or -1.

10. State and prove the H-hermitian skew analogue of Lemma 2.78.11. Show that the intersection of any two of the unitary groups U; (2p, 2q),

Uj(2p, 2q), Uk(2p, 2q) (based on the three complex structures R;,R Rk respectively) is equal to the hyper-unitary group HU(p, q).

3. Classical Groups II

The classical groups introduced in Chapter 1 are examined more care-fully in this chapter. In particular, the "Lie algebra" of each of these groupsis computed explicitly.

GROUP REPRESENTATIONS AND ORBITS

Suppose G is a group and V is a real vector space.

Definition 3.1.

(a) A representation p of G on V is a group homomorphism p : GGL(V). Sometimes, the notation p is suppressed, and we say that Gacts on V, when the action p is understood from the context.

(b) Two representations p : G -+ GL(V) and o : G --+ GL(W) are said tobe equivalent if there exists a R-linear isomorphism L : V --> W witho(g) = L o p(g) o L-1 for all g E G. The linear map L is called anintertwining operator.

Since the groups defined so far were defined as subgroups of GL(n, F),each such group comes equipped with a natural action on F".

Suppose p : G -+ GL(V) is a action of G on V. A set of the form

(3.2) {p(g)v: G E G}

is called an orbit of G, or the orbit of G through v E V. The subgroupK - {g E G : p(g)v = v} is called the isotropy subgroup of G at v. The

41

42 Generalized Spheres

group G is said to act transitively on a subset X C V if X is the orbit ofG through each point v E X.

Two isotropy subgroups, K. and If,, at two points u, v in the sameorbit X are isomorphic:

(3.3) Ku=LK,, L'1,

where L E G is chosen so that Lv = u. Thus, if G acts transitively on X,we write

(3.4) G/K - X

for the quotient of the group G by the equivalence relation gl - g2 definedby g2 = glk for some k E K, where K represents the isotropy subgroup ata point u E X.

Recall from Problem 1.3 that the connectivity of GL(n, F) andSL(n, F) (for F - R, C, or H) was derived from

(3.5) GL(n, F)/K = F" - {0}(3.6) SL(n, F) f K F" - {0} (n > 2).

Thus, GL(n, F) and SL(n, F) (n > 2) acting on F" have two orbits, namely{O} and F" - {0}.

GENERALIZED SPHERES

For each of the seven types of inner products, the proof of the Basis Theo-rem can be used to compute the orbits of the isometry group G (that fixesthe inner product e) acting on F". Of course, each orbit of G is containedin one of the level sets

(3.7) {x E F" : e(x, x) = c} for some constant c E F.

The most important cases occur when e is an R-symmetric inner prod-uct.

Definition 3.8. The positive sphere of radius r > 0 in R(p, q) (p > 1) is:

(3.9) Sr -{xER(p,q):(x,x)=r2}.

The negative sphere of radius r < 0 in R(p, q) (q > 1) is:

(3.10) Sr -{xER(p,q):(x,x)_-r2}.The unit spheres S: in R(p, q) are obtained by setting the radius r = E1.

Classical Groups II 43

The standard notation for the euclidean sphere Sr in R(n, 0) is ST -1since this sphere has dimension n - I. We will use the same notation,Sr'1, for the sphere Sr- in R(0, n). S° consist of two points. Note thatthe positive unit sphere S+ in R(p, q) is the same as the negative unitsphere S- in R(q, p).

S- future timelike

S- past timelike

Figure 3.11

Proposition 3.12.

(a) (p > 1) the unit positive sphere S+ in R(p, q) is diffeomorphic to,SP-1 x R.

(b) (q > 1) the unit negative sphere S- in R(p, q) is diffeomorphic toRP x S9-1.

(c) The null cone N {z E R(p,q) : z # 0 but jjzjj = 0} in R(p,q) isdiffeomorphic to SP-1 x SQ'1 x R+.

44 Generalized Spheres

Proof: Let

(3.13) llxl 2 = 1x12 - lyl2, for all z = (x; y) E R(p, q) = RP x R9

denote the square norm of z = (x, y) where IxI and Iyl denote the euclideannorms ofx R" and yE R.

Proof of (a): If z =- (x, y) E S+ then llzll = IxI2 - lyl2 = 1. Inparticular lxl # 0 so that (x/Ixl,y) E SP-1 x R4 is well defined. Thismap from S+ to SP-1 x R9 has inverse sending (v, y) E SP-1 x RQ toz E( 1+ lyl2v, y) E S+.

The proof of (b) is similar and omitted.Proof of (c): Since

(3.14) IxI2 - IyI2 = 0, for allz = (x, y) E N,

if z = (x, y) is null but non-zero then both x and y are non-zero. Thus themap from N to SP-1 X SQ-1 x R+ sending z - (x, y) to ( , YT, r), wherer = IxI = lyl, is a well defined smooth map. The inverse of this map sends(x, y, r) E SP-1 x 59-1 x R+ to z = (rx, ry) E N.

The next result follows since S° consists of two points, S"(n > 1) isconnected, and S" (n > 2) is simply connected.

Corollary 3.15.

(a) The unit positive sphere S+ in n-dimensional Lorentzian spaceR(1, n - 1) has two connected components diffeomorphic to Ri-1.(These components are called the future and past spheres.)

(b) The unit positive sphere S+ in R(p, q) for p > 2 is connected.(c) The unit positive sphere S+ in R(p, q) for p > 3 is simply connected.

If C(p, q), E is the standard model for the C-hermitian symmetric caseof signature p, q, then R(2p, 2q) S' C(p, q) with (, ) - Re a is the standardmodel for the R-symmetric case of signature 2p, 2q. Also note that E(x, x)only takes on real values. Therefore,

(3.16) {x E C(p, q) : E(x, x) = ±r2} = S, , the sphere in R(2p, 2q).

Similarly,

(3.17) {x E H(p, q) : E(x, x) = ±r2} = S1, the sphere in R(4p, 4q).

If C', E is the standard model for the C-symmetric inner productspace, then ST'1(C) _ {z E C' : e(z, z) = r} is called the complex (orhyperquadric) sphere of radius r E C.


If H'2, E is the standard model for the H-hermitian skew inner productspace, note that -(x, X) only takes values in ImH. Given r E ImH, thelevel set

(3.18) Sr (H"-skew) _ {x E H' : E(x, x) = r} ,

fixed by the group SK(n), will be called the H-skew sphere of radius r. Notethat if u E Im H is a given unit vector then the real linear isomorphism R,' :H" - H" sends the H-skew sphere of radius r E Im H onto the H-skewsphere of radius uru E Im H. This fact can be used to yield a descriptionof the general H-skew sphere from calculating the special case r = i, whereS;(H'-skew) = {a+jb : a, b E C",1aI2-1612 = 1, and akbk+bkak = 0}.

Given an inner product, the spheres are the orbits of the associatedisometry group.

Theorem 3.19 (Orbit Structure). Let n = p + q.

(3.20)O(p, q) SO(p, q) r S:, the sphere in R(p, q).

O(p - 1, q) SO(p - 1, q)

(3.21)O(p,

O(pq

, q

-)

1) SO(p,SO(q p, q

-)

1)

,, S* the sphere in R(p, q).

(3.22)U(p, q) = SU(p, q)

- S,+, the sphere in R(2p, 2q).U(p - 1, q) SU(p, q - 1)

(3.23)U(p, q) _ SU(p, q) ,

U(p, q - 1) SU(p, q - 1)S,-, the sphere in R(2p 2q).

(3.24)HU(p,

q) , S , the sphere in R(4p, 4q).HU(p - 1, q)

_

(3.25)HU(p, q)

,,, S,-, the sphere in R(4p, 4q).HU(p, q - 1) -

O(n, C) _ SO(n, C)-1(C) for r E C' and n > 2.(3.26) 0(n - 1, C) - SO(n - 1, C)

46 The Basis Theorem Revisited

(3.27) Sp(n,R) = R2" - {0},

Kwhere K is topologically the same as R2n-1 x Sp(n - 1, R).

Sp(n, C) = C2n - {0},(3.28) Kwhere It is topologically the same as C2i-1 X Sp(n - 1, C).

SK(n, H) _ "(3.29) SK(n - 1, H) -

S,.(H skew) for each r E ImH.

Proof: Let G/If - X correspond to one of the cases in Theorem 3.19.First, the group G acts transitively on X. In each case, this is verifiedby examining the proof of the Basis Theorem 2.46. For example, O(p, q)acts transitively on Sr since, given a vector u E S* , the vector el = u/vfr-may be chosen as the first vector in an orthonormal basis for R(p, q). Theisometry f E 0(p, q) that maps the standard orthonormal basis e1, ... , efor R(p, q) to this orthonormal basis sends el to e"1 and hence \el to u.

Second, the isotropy subgroup K of G at el must be computed. In allcases, the fact that

(3.30) if f(el) = el then f : (span el)1 -i (span el)1

is of key importance. The individual cases are left to the reader. -1

Corollary 3.31. Each of the groups Sp(n, R), U(p, q), SU(p, q), SO(n, C),Sp(n, C), HU(p, q), and SK(n, H) is connected. The groups SO(p, q),p, q > 1 and O(n, C) have two connected components, while the groupsO(p, q), p, q > 1, have four connected components.

This corollary can be deduced from the theorem by induction on thedimension.

THE BASIS THEOREM REVISITED

The Basis Theorem may be interpreted as calculating an orbit of one ofthe groups GL(n, F). Let e denote the standard inner product on F".The group GL(n, F) acts on the space BIL(R") containing e as follows.Suppose rl E BIL(F'). For each A E GL(n, F), let A*ij denote the actionof A on q where A* is defined by

(3.32) (A"rl)(x, y) = q(Ax, Ay) for all x, y E F".

Let I(V) denote the subset of BIL(V) consisting of all inner products of thesame type (and signature if applicable) as e. Let G(e) denote the classicalgroup that fixes a (i.e., the isometry group of the given type).


Theorem 3.33. GL(n, F) acts transitively on I(V) with isotropy G(e) atE. That is,

(3.34) GL(n, F)/G(e) - I(V).

Note that {77 E SYM(V) : 71 is nondegenerate} is an open subset ofSYM(V) if e is symmetric; while ifs is skew, {77 E SKW(V) : r7 is nonde-generate} is an open subset of SKW(V).

Since this theorem is a reinterpretation of the Basis Theorem, the proofis an exercise (see Problem 10).

ADJOINTS

Recall from the proof of the Orthogonality Lemma 2.30 that the flat mapb : V --+ V* is a real linear isomorphism for all seven types of inner productspaces V, E. In particular, if A E EndF(V) is given and y E V is fixed, thene(y, Ax) E V* is considered as a function of x. Thus, E(y, Ax) = (b(z))(x)for a unique z E V. By definition of the flat map (b(z))(x) = e(z, x). Nowconsider z E V to be a function of the given point y E V and denote zby A* y. For each of the eight types of inner product spaces, it is easy tocheck that e(y, Ax) = e(z, x) if and only if e(Ax, y) = e(x, z). Thus themap A* : V -* V is characterized by

(3.35) either e(y, Ax) = e(A* y, x) or e(Ax, y) = e(x, A* y) for all x, y E V.

This map A* is called the adjoint (not the dual) map of A, and definitelydepends on the given inner product e. The proof of the next lemma isstraightforward and hence omitted (i.e., should be checked once carefullyenough to see that it is routine).

Lemma 3.36. The adjoint A --r A* is an anti-automorphism of the al-gebra EndFV that squares to the identity (an involution). That is, A* EEndF(V) for each A E EndF(V) and

(3.37) (AB)* = B*A* and (A*)* = A.

Note that, given A E EndF(V),

(3.38) e(Ax, Ay) = e(x, y) for all x, y E V (i.e., A fixes e),

and

(3.39) A*A = Id (i.e., A* = A-1),

48 Lie Algebras

are equivalent. Thus, each of the groups, defined by requiring that e befixed, can also be described as {A E EndF(V) : A* = A-'}.

Depending on the type of the inner product space, the adjoint is re-ferred to as the R-orthogonal adjoint, R-symplectic adjoint, C-orthogonaladjoint, C-symplectic adjoint, C-hermitian adjoint, H-hermitian symmet-ric adjoint, and H-hermitian skew adjoint. Given a matrix A = (aid) EMn(F) the matrix transpose is denoted by At = (aj2), while the matrix con-jugate is denoted by A = (a;j). Using the matrix transpose and the matrixconjugate, each of the various types of adjoints can be explicitly computed(Problem 3). For example, the C-hermitian (positive definite) adjoint ofA E is given by A* = ;P the conjugate transpose. Combining thefact that A fixes e if and only if AA* = 1 with the explicit computationsof the adjoints for A E Mn(F), it is easy to give alternative definitions ofthe groups: O(p, q), Sp(n, R), O(n, C), Sp(n, C), U(p, q), HU(p, q) andSK(n, H). For example, U(n) = {A E Mn(C) : AAt = 1}.

LIE ALGEBRAS

Each of the groups G considered so far is a subset of M, (F), with F = R, C,or H, defined implicitly by a certain number of polynomial equations. Theobjective of this section is to show that G is a submanifold of Mn(F) andto compute the tangent space to G at the identity 1 E G in Mn(F). Thistangent space is denoted g and called the Lie algebra of G.

Given two matrices A, B E Mn (F) the commutator

(3.40) [A, B] AB - BA

is also called the Lie bracket of A and B. This product provides the naturalalgebraic structure on each of the Lie algebras g.

The basic identity satisfied by a Lie algebra g, [ , ] is called the Jacobiidentity:

[A, [B, C]] = [[A, B], C] + [A, [B, C]],

for all A, B, C E g.The reader is referred to the plethora of texts on Lie algebras for more

information.The Lie algebra of a particular group is usually denoted by using the

same label used for the group-except in German lower case. For example,so(r, s) denotes the Lie algebra of SO(r, s).

First note that

(3.41) gI(n, R) = MM(R)


(3.42) g((n, C) = Mn(C)

(3.43) g[(n, H) = M,,(H),

since GL(n, F) is just an open subset of Mn(F).Now suppose that G is one of the seven classical groups that fixes one

of the standard inner products e. Then G is given implicitly as a subset ofMn(F) by

(3.44) G = {X E M,,(F) : XX* = 1},

where F is R, C, or H and X* denotes the E-adjoint.Let SymE(n, F) denote the real vector subspace of Mn(F) consisting

of the --symmetric matrices, i.e.,

(3.45) SymE (n, F) _ {A E M (F) : A = A*}.

Let SkewE (n, F) denote the real vector subspace of Mn(F) consisting ofthe --skew matrices, i.e.,

(3.46) SkewE (n, F) _ {A E Mn(F) : A* = -Al.

Proposition 3.47. Suppose e is one of the standard inner products. Thee-orthogonal group

G={AEMn(F):AA*=1}

is a real submanifold of the real vector space Mn(F) with

(3.48) dime. G = dims Skew£ (n, F).

The tangent space g at 1 E G is given by

(3.49) g = Skew, (n, F).

Proof. Consider X X * as a function from M (F) to SymE (n, F) (i.e., fromRP to R9 for some p and g). The group G is defined to be the level set ofthis function at level 1 E SymE (n, F). To prove that G is a real submanifoldof M (F) we invoke the implicit function theorem. Thus we must computethe rank of the derivative of XX* at each point B E G and show that itequals dimrt SymE(n, F).

50 Lie Algebras

The derivative (i.e., the linearization) of the function XX* at B E Gis given by

(3.50) d (B +tX)(B +tX)* It_o= XB* + BX*.

The linearized map XB* + BX* from Mn(F) to SymE (n, F) is surjective;given A E SymE (n, F), take X =

zAB and note that X B* + BX *

z ABB* +z

BB* A* = A. Thus the rank of the linear map sending X EMn(F) to XB* + BX* E SymE(n,F) is equal to com-pleting the proof that G is an implicitly defined submanifold of M,(F).

The tangent space to G at B E G is the kernel of the linear mapXB* + BX * . Thus, the tangent space g to G at 1 E G is the kernel ofX + X*, which equals F) C MM(F). Li

The submanifold G of Mn (F) defined by (3.44) can also be describedparametrically. Since, given B E G, right multiplication by B provides adiffeomorphism of Mn(F) that maps G to G and sends 1 E G to B E G, itsuffices to give a parametric description of a neighborhood of 1 E G.

First, we need a lemma about power series. The sup norm on M (F)is defined by

(3.51) sup norm A = sup I ICI I for A E Mn(F),

where I I denotes the standard positive definite norm on Fn(F = R, C,or H). The proof is an exercise.

Lemma 3.52. Suppose f (z) n=0 a,Zn is a power series with radiusof convergence r. The series

0"f (A) = >2 anAnn=0

converges in the sup norm on Mn(F) if the sup norm A < r.

Definition 3.53. Given A E M, (F), define

exp(A) = eA = 1+A+ 1 A2 +

Proposition 3.54. Let G denote the group that fixes the inner producte. The exponential map

(3.55) exp : SkewE (n, F) --+ G


provides a parameterization of a neighborhood of the identity 1 for thegroup G.

Proof: If A E Skew, (n, F), then CA E G, since (eA)(eA)* = eAeA* _eA . e-A = eA-A = eo = 1. The derivative of exp at A = 0 is given by

Wt

etx = X.t.0

This proves that exp is injective near A = 0. Now the proposition fol-lows since G has dimension equal to the dimension of the parameter spaceSkew, (n,F). Li

The special linear groups are defined implicitly by setting the polyno-mial detF X equal to one, where F =_ R or C. The derivative, or lineariza-tion, of this function at a point A is by definition

(3.56) dt detF (A + tX) .

LoAt the point A = 1 (the identity matrix), this formula provides a convenientcoordinate free definition of the trace of a linear map X E EndF(V). Thatis, for F = R or C, we define

(3.57) tracer X = dt detF(1 + tX)Lo

Of course, it is important to compute that if X - (xii) E Mn(F) forF =- R or C, then

(3.58) traceF X = xii.i.1

To check this consider detF(I+ tX)ej A . . A en = (el +tXel) A A(en + tXen), where e1, ... , e is an F-basis for V.

If X = (xii) E MM(H), then it follows thatn

(3.59) tracer- X = 4 Re xiii.1

Now suppose that A E SL(n, F). In particular, A is invertible. Thederivative of detF X at the point A is equal to

detF(A+tX) = dt detF(1 +tXA-1)A)(3.60) dt Lo t=o

52Problems

Therefore, the linearization has rank 1 for detR, and rank 2 for detc. Be-cause of the implicit function theorem, this provides a proof that the speciallinear groups are submanifolds. The tangent space at the identity 1 is thekernel of the linear map traceFX, obtained by setting A = 1 in (3.60).This proves the next proposition.

Proposition 3.61.

(3.62) st(n, R) = {A E A, (R) : traceR,A = 0},

(3.63) st(n, C) = {A E Mn (C) : tracecA = 0} ,

(3.64) sl(n, H) = {A E M (H) : traceRA = 0} .

PROBLEMS

1. (a) Prove that (AB)* = B*A* and (A*)' = A for each of the seventypes of adjoints.(b) Show that e(Ax, Ay) = e(x, y) if and only if A* = A-1.

2. (a) Show that det A* = det A if A* is the adjoint with respect to e,where e is either of R-symmetric type or C-symmetric type.(b) Show that each A E O(p, q) has deter A = ±1.(c) Show that each A E O(n, C) has detc A = ±1.(d) Suppose A E CO(p, q) with conformal factor A. Show (detR A)2 =A", where n = p + q is the dimension.

3. (Adjoints) Compute the adjoint of a matrix A E M,,, (F) in the follow-ing cases. Note that a decomposition of the vector space F'° into thedirect sum of two subspaces induces a 2 x 2 blocking of the matrices:

Case (a) R-positive definite or C-pure symmetric. Show A* = At.Case (b) Positive (or negative) definite, F - R, C, or H. ShowA* = q' (conjugate transpose). This case is a subcase of the nextcase, F-hermitian symmetric.Case (c) F(p, q) = F" x F4, with F - R, C, or H. Show

C c

d\ *

-(

_V T /


Case (d) H-hermitian skew: A* = iAt z', where i denotes i times theidentity matrix.

Case (e) R- or C-symplectic: R2i = RI (DR' or C21 = C"(BCn, writ-ten as the direct sum of the even coordinates and the odd coordinates.Show

(cd)*=

(-dtCt

at)

Hint: Reduce all cases to either Case (a) or Case (b) by finding alinear map L so that e(x, Ly) = (x, y), with (,) an inner product ofthe type in either case (a) or case (b).

4. Suppose F", e is one of the standard models for an inner product space.Let G denote the isometry group of F", e, and g = SkewE (n, F) theLie algebra of G.

(a) Suppose e is (either pure or hermitian) symmetric. Exhibit anisomorphism g = SKW(F") with the space SKW(F") of all skewbilinear forms on F" of the same type as e.(b) Formulate the analogue of part (a) for b skew.

(c) The common notation for the space of all skew bilinear forms on RIis A2(R")*, while S2(Rn)* denotes the space of all symmetric bilinearforms on R". Thus, by (a) so(p, q) = A2(R")* (n = p+q), and by (b)zp(n, R) = S2(R2n)*. Compute that dimR SO (p, q) = 2n(n + 1) andthat dimR Sp(n, R) = 2n2 + n.

5. Using the definition (3.57) for traceF X show that the formula (3.58)is valid for X E M"(R) or X E M"(C). If X E M"(H), show that(3.59) can be used to compute traceR X.

6. (a) Show that lyu(1) = ImH and that exp : ImH -* S3 = HU(1) C His surjective.

(b) Show that exp : s((2, R) -; SL(2, R) is not surjective.

7. Explicitly compute o(n), o(r, s); u(n), u(r, s); I u(n), lju(r, s); sp(n, R),.sp(n, C); o(n, C); and st(n, H).

8. The real manifold dimension of a Lie group G is the same number asthe real vector space dimension of the Lie algebra S. Show that(a)

dimR GL(n, R) = n2, dimR SL(n, R) = n2 - 1,dimR GL(n, C) = 2n2, dimR SL(n, C) = 2n2 - 2,

dimR GL(n, H) = 4n2, dimR SL(n, H) = 4n2 - 1.

54 Problems

(b) Using Problem 12, compute that (n = r + s)

dimR SO(r, s) = 1 n(n - 1),

dimR U(r, s) = n2,dimR SU(r, s) = n2 - 1,dimR HU(r, s) = 2n2 + n,dimR O(n, C) = n(n - 1),

dimR SK(n, H) = 2n2 - n,dimR Sp(n, R) = 2n2 + n,

dimR Sp(n, C) = 4n2 + 2n.

9. Give the proof of Lemma (3.52) on power series.10. Give the proof of the seven results contained in Theorem 3.33. For

example, show that

GL(n, R)/O(r, s) - I(r, s; R),

the set of R-symmetric inner products on R' with signature r, s.11. Prove that SK(n, H) is the intersection of three distinct split unitary

groups on H", corresponding to the three complex structures R;, RR,and Rk on H.

12. Show that(a) su(p, q) OR C = 5f(n, C),

(b) hu(p, 4) OR C = sp(n, C)(c) st(n) OR C - so(2n, C),(d) sf(n, H) 0 C - sf(2n, C).

13. Suppose G is one of the classical groups considered in this chapter.Given an element h E G, the mapping from G to G defined by sendingg E G to hgh'1 E G is a diffeomorphism of G that maps the identityelement 1 E G into itself. The linearization of this map, denoted Adh,is a linear map from g = T1 G into itself.(a) Show that Adh A = hAh-1, and that h Adh is a representationof the group G on the vector space g. This representation is called theadjoint representation of G.(b) Show that for B E g fixed, the map sending h E G to Adh B has thederivative at h = 1 E G given by the linear map sending A E g - T1Gto [A, B] __ AB - BA.


14. Suppose G is a Lie group, that is, a group equipped with a compatibledifferentiable manifold structure. Let Xinv(G) denote the vector spaceof right invariant vector fields, that is, vector fields on G which arefixed by the diffeomorphisms R. (right multiplication by g) for eachg E G.(a) Show that Xi,,,,(G) is closed under the binary operator [X, Y]XY -YX.(b) Under the natural map sending X E Xi,,v(G) to X1 E T1(G) .show that X;,,,(G), [ , ] and g, [ , ] are isomorphic.Hint: Consult a text on Lie groups.

15. (The quaternionic unitary group) The space of imaginary quaterionsImH has several realizations relating to H", E. First, ImH can beidentified with the coefficient space {Ru : u E ImH} of R-linear mapsof H. Second, using E, Im H can be identified with the space {wu : u EImH} of Kahler forms on H", where each 2-form wu(x, y) is definedto be (u, E(x, y)) (cf. (2.79)).(a) Show that the quaternionic unitary group HU(p, q) HU(1) can becharacterized as the subgroup of O(4p, 4q) that fixes R; A R? A Rk,where each element g E SO(4p, 4q) acts on the coefficient space {R,,u E Im H} by sending Ru to g o Ru o g-1.Hint: Recall (1.35) that HU(1)/Z2 SO(ImH) to reduce to the casewhere g E O(4p,4q) leaves the coefficient space {Ru : u E ImH}pointwise fixed.(b) Prove that the two representations of HU(1) given by

Xa(x)=axa for allxEImH

andpa (wu) = Rawu for all u E Im H

are equivalent.

(c) Prove that the following two representations of the group

HU(1) = SO(ImH)Z2

(see 1.35) are equivalent. First, pa(wu A R°a(wu A defines arepresentation on

span {wu A w : u, v E Im H} C A4 (H" )*

Second, oa(A) = X. o A o Xa defines a representation of Sym(ImH),the space of symmetric 3 x 3 matrices on R3 = Im H.

56 Problems

(d) Prove that the quaternionic unitary group HU(p, q) HU(1) fixesthe quaternionic Kahler form

1 (w; A wi + wj Awl + wk A Wk) E A4 (Hn)* .

(e)* (n > 2) Prove that if g E GLR.(H") and g*4(b _ <D, then g EHU(p, q) HU(1). Thus, the quaternionic unitary group HU(p, q) .HU(1) is characterized as the subgroup of GLR,(Hn) that fixes thequaternionic Kahler form with no reference to an inner product onHn.

16. A representation p = pl ® p2 that can be expressed as the direct sumof two representations pl and p2 is said to be reducible. Otherwise, arepresentation is irreducible.(a) Suppose p = pl ®p2 : SO(n) -- O(Vi) ® O(V2) is a reduciblerepresentation of SO(n) on a positive definite inner product space V =Vl ® V2, with inner product (, ). Show that there exists an innerproduct on V different from c(, ), where c is a constant, which is fixedby SO(n).(b) Using part (a), show that the representation of SO(n) on AkRn isirreducible unless k = n/2 and n = 0 mod 4. (In this case, the repre-sentations of SO(n) on A' R11* and Ak R2k are both irreducible, whereA}R2k . {u E AkR2k *u = ±u} denotes the space of self/anti-selfdual k-vectors, and * is the Hodge star operator.)

4. Euclidean/Lorentzian VectorSpaces

In this chapter, we will study R-symmetric inner product spaces. Ifthe inner product is positive definite, then the inner product space is betterknown as a euclidean vector space. If the signature is n - 1, 1, then theinner product space is better known as a Lorentzian vector space. Ratherthan refer to a Lorentzian vector space as "pseudo" or "semi" euclideanor to a euclidean vector space as "pseudo" or "semi" Lorentzian, we shallrefer to an R-symmetric inner product space with signature p, q simply asa euclidean vector space with signature p, q.

Remark. For the reader familiar with manifolds, a Riemannian manifoldis a smooth manifold whose tangent space at each point is equipped witha positive definite euclidean inner product, and the inner product variessmoothly. A Lorentzian manifold is a smooth manifold whose tangentspace at each point is equipped with a Lorentzian inner product, vary-ing smoothly. To be consistent with our inner product space terminology,we shall refer to a Lorentzian manifold as a Riemannian manifold with sig-nature n - 1, 1 rather than as a "pseudo" or "semi" Riemannian manifold(cf. Chapter 5).

The first topic in this chapter is very elementary: the Cauchy-Schwarzequality. The second topic is a brief introduction to special relativity, and

57

58 The Cauchy-Schwarz Equality

the third topic is the Cartan-Dieudonne theorem. This last result is crucialto our discussion of the spin groups.

THE CAUCHY-SCHWARZ EQUALITY

Suppose V, e = (,) is a euclidean vector space with signature p, q anddimension n = p + q.

Definition 4.1.(a) The norm.of a vector v E V is defined to be

IV l I(TV)I.(b) The square norm or quadratic form associated with the bilinear form

is defined by;Hull (v, v).

The inner product e = (,) is completely determined by the quadraticform 1111 by the process of polarization-namely, replace v by x + y inhull = (v, v) and use bilinearity to obtain

(4.2) (x, y) =2

(llx + yll - Ilxll - IIyIU

Recall the musical isomorphism (metric equivalence or lowering in-dices) b : V -* V*. This linear map naturally induces a linear map, alsodenoted b, on tensors:

b :O kv ®kV* = ((&kV)* (e.g., b(u ®v) = b(u) ®b(v)),

b : AkV --> AkV* S' (AkV)* (e.g., b(u A v) = b(u) A b(v)).

Similarly, 0 = b!1 : V* -+ V naturally induces a linear map # on tensors and#b = b# = Id. Thus, ®kV and AkV naturally inherit a euclidean structureV with the inner product e defined by e(x, y) _ (b(x))(y). This innerproduct on tensors will also be denoted by e - (, ). Suppose e1,. .. , e isan orthogonal basis for V with the sign o-j =_ (ej, ej) either ±1. Then, forexample,

(4.3){ei Aej : i < j} is an orthonormal basis for A2V with

(ei Aej,ei Aej) = aio-j,and more generally,

lei, :il < ... <ip}is an orthonormal basis for APV.

Note: If dim V = 2, then the one-dimensional space A2V is positive inboth the case V positive and the case V negative, while A2V is negative ifV has signature 1, 1.

Euclidean/Lorentzian Vector Spaces 59

Theorem 4.4 (Cauchy-Schwarz Equality). Suppose V is a euclideanvector space of signature p, q. For all x, y E V,

(x, y)2 + (x A y, x A y) = IIxII IIyII-

First Proof: The idea of this proof is that since the theorem only involvestwo vectors it is really a two-dimensional result. Let W denote the spanof x and y. If x and y are colinear, the proof is immediate. Thus, wemay assume that W is a plane. Because of the Basis Theorem and itsextension to the degenerate cases (Problem 2.7), the (possibly degenerate)inner product space W is isometric to one of the standard models:

(4.6) R(2, 0), R(1, 1), R(0, 2) if W is nondegenerate,

or

(4.7) R(1, 0) x R, R(0,1) x R, R x R if W is degenerate.

Finally, it is a simple matter to verify directly (4.5) in each of thesesix cases.

For example, if x, y E R(1, 1), then

(x, y)2 = (xiy1 - x2y2)2 = xiyi + x2 y2 - 2xlyix2y2,

(x A y, x A y) _ -(xiy2 - x2yi)2 = -xiy22 - x22 y1 + 2xiyix2y2,

and

IIxII IIyll = (xi - x2)(Y2 - 14) = x2y2 +x2y2 -x2y2 -x2y2

Li

Second Proof: (x A y, x A y) = (b(x A y))(x A y) = (b(x) A b(y))(x A y) =(b(x)(x))(b(y)(y)) - (b(x)(y))(b(y)(x)) = (z,x)(y,y) - (x, y)2, where eachof these equalities is an application of a definition. J

Third Proof: First, we verify the algebraic identity (Lagrange):

/2

(4.8) z=wi {

E (ziwj

- zjwi)2 =(2

z2 E WJ`:_i i<j i=1 I j_i

60

for all z, w E Cn, by noting that

The Cauchy-Schwarz Equality

n(ziwj - Zjwi)2 =

2(ziwj - wizj)2

i<j 9=1

n=2

E (ziwj2 -2ziwjzjwi+z?w;)i,j=1

2n n n

= z; wj2 - ziwii-1 i=1 i_1

Now, substituting

xj forzj and yi for wj ifj=1,...,p,

and

ixj for zj and iyj for wj ifj = p+ 1, ..., n,

with xi, yi real, yields the C.-S. equality for R(p, q). Since V, (,) is iso-metric to R(p, q), this completes the proof.

Proposition 4.9. Suppose x, y E V and W = span{x, y} is a two-dimen-sional plane. The plane W is degenerate if and only if (z A y, x A y) = 0but x A y # 0. If W is nondegenerate, then W is either positive, negative,or Lorentzian.

Positive/Negative: The following are equivalent:

(4.10) (i)

(ii)

(iii)

(iv)

W is a positive or a negative plane,A2 W is positive,

(xAy,xAy) > 0,(x,y)2 < IIzII IIyII

Lorentzian: The following are equivalent:

(4.11) (i)

(ii)

(iii)

(iv)

W is a Lorentzian plane,A2 W is negative,

(x A y, x A y) < 0,

(x, y)2 > IIxII Ilyll


Proof: The fact that (i)-(iv) are all equivalent, in either (4.10) (posi-tive/negative) or (4.11) (Lorentzian), follows from the note before Theorem4.4 and Theorem 4.4 (C.-S. Equality).

If W is a degenerate plane, then there exists a nonzero vector u E Wwith (u, z) = 0 for all z E W. Pick v E W with u A v = x A y. Then, bythe C.-S. Equality, (x A y, x A y) = (u A v, u A v) = (u, v)2 - IluJI Mvii = 0.Conversely, if x A y # 0 but (x A y, x A y) = 0, then W must be a degenerateplane, since W cannot have dimension < 2 (x A y i4 0) and W cannot be apositive, negative, or Lorentzian plane. _1

Corollary 4.12 (Cauchy-Schwarz Inequality). If V is positive eu-clidean, then

(u,v)<JulJvi forallu,vEV,

with equality if and only if one of the vectors is a positive multiple of theother.

Remark. Since -Jul lvJ < (u, v) < Jul lvJ,

(4.13) cos 0 - (u, v)Jul Ivi

uniquely defines an angle 0 between any two non-zero vectors u and v with0 < 0 < ir. Note that sin 0 = lu A vi/luJ JvJ.

A curve M in a euclidean vector space V is, by definition, a one-dimensional oriented submanifold with boundary. Since (by definition) acurve is oriented, each point on a curve possesses a well-defined tangentray (or half-line) and, if this tangent line is non-null, a well-defined unittangent vector. The length of a curve M is defined to be

(4.14) 1(M) Jx'(t)J dt,

where x : [a,#] -* M C V parameterizes M. The initial point a = x(a)and the terminal point b - x(3) form the boundary of M. If the tangentline to M is never null, then the parameterization can be chosen so thatix'(s)I = 1 and is called parameterization by arclength. The length of theline segment from a to b is I b - aI.

Proposition 4.15. Suppose V is positive definite.

Ib - al < 1(N)

62 Special Relativity

for any curve N with initial point a and terminal point b. Equality occursif and only if N = the line segment from a to b.

The proof of this result is fundamental in the theory of "calibrations"(see Chapter 7).

Proof: Let M denote the line segment from a to b. Let M denote the unit

tangent vector to M (i.e., M= b - a/lb - al). Let 0 = b(M) denote thecorresponding 1-form tangent to M. Then lb - aI = 1(M) = fm 0, whichequals JN 0 by the fundamental theorem of calculus for paths since do = 0.Choose the parameterization x(s) of N by arclength. Then

j(Mz'(s))f =N

By the C.-S. Inequality, this is < fo ds = 1(N). Li

SPECIAL RELATIVITY

The terminology and intuition provided by special relativity is very helpfulin understanding Lorentzian vector spaces, and, in particular, in under-standing dimension four (Minkowski space).

A Lorentzian vector space V (signature n - 1, 1) is said to be timeoriented if one of the two components of the timecone {v E V : (v, v) < 01is designated as the future timecone. Vectors in the future timecone aresaid to be future timelike vectors.

Corollary 4.16 (Backwards Cauchy-Schwarz Inequality). SupposeV is a time-oriented Lorentzian vector space. If u and v are future timelikevectors, then

-(u, v) > lul M,

with equality if and only if u and v are positive multiples of each other.

Remark. For u, v future timelike

(4.17) cosh 0 =-(u'v)

lul IvI

uniquely determines an angle 0 < 0 < +oo between u and v. Note thatsinh 0 = juAvj/jul Ivj.

Suppose V is a time-oriented Lorentzian vector space. A curve M inV is called a worldline of a particle if its tangent is future timelike at each

.Euclidean/Lorentzian Vector Spaces 63

point. The arclength parameter is called proper time and usually denoted-r. The length of M is called the proper time of the worldline or particle. Ifthe worldline M is a line segment, then the particle is said to be in free fall.Of course, the proper time of a particle in free fall from a to b is lb - al.

Figure 4.18

Proposition 4.19 (Twin Paradox). The proper time of a particle ismaximized by free fall.

Proof: See Problem 3. LJ


Corollary 4.20 (Backwards Triangle Inequality). If u and v are fu-ture timelike vectors, then

Iu+vI >_ Iul+ Ivh

with equality if and only if u and v are positive multiples of one another.

The null lines in a Lorentzian vector space V are called light rays (orthe worldline of a photon). Thus, the union of all the light rays is thelightcone (or nullcone)

{x E V : IxII = 0}.

In Minkowski space R(3, 1), the lightcone is given by

{(x, t) E R(3, 1) : IIxII = t2}.

The speed of light has been normalized to be one. One of Einstein's axiomsof special relativity, referred to as the Constancy of Light, states that

If a linear coordinate change A E GL(R(3, 1)) on Minkowski spaceis physically admissible (or provides a new inertial coordinate sys-tem), then A must preserve the set of light rays.Any invertible linear map A on V, II ' II determines a new square norm

IIby

IIxII=IIAzil for all z E V.

If A preserves the nullcone for II- II, then II - II' has the same nullcone as

II II Under this hypothesis, the next theorem states that II- II' and II - II

differ by a constant c. That is, A must preserve II . II up to a constant. Insummary, the next theorem has as a consequence:

Einstein's axiom, the Constancy of Light, implies that each iner-tial coordinate system on R(3, 1) is obtained from the standardcoordinate system by a coordinate change A that preserves II . II

up to a constant.If additionally A preserves futuretime, then it is easy to show that the

constant is positive, i.e., A must be conformally Lorentzian, A E CO(3, 1).(An additional axiom, called the Axiom of Relativity, implies that the con-stant is 1; that is, A E SOT(3, 1) the Lorentz group.)

Lemma 4.21. Suppose V, II . II has signature p, q with p, q > 1 and II II'is another quadratic form on V. If each null vector for II II is also null for

1111', then for some constant c,

IIxII' = c11x11-

Euclidean/Lorenizian Vector Spaces 65

Proof: First assume that V has dimension 2. Then V, II - II is isometricto the Lorentz plane R(1, 1). Let R ® R denote R2 with the square normIIxII = xlx2. Then R(1, 1) and RED R are isometric. Using the modelR ® R, the hypothesis that II

-II' vanishes on the null cone for II . II, says

that IIxII' = Ax2+Bxlx2+Cx2 must vanish when xi = 0 or x2 = 0. ThusIIxII'=BIIxII.

Now assume V has dimension > 2. It suffices to prove that for anypair of vectors u, v with one spacelike and one timelike the ratios IIuII'/IIuIIand IIuII'/IIuII are the same. Repeated use of this will yield IIxII' = cIIxIIfor all x. Since span{u, v} is a Lorentz plane, there exists a constant c(depending on u and v) such that

IIxII' = cJIxII for all x E span{u, v}.

A future timelike vector might also be called the instantaneous world-line of a particle. This particle can also be thought of as an "observer."It is useful to define an observer to be a curve whose unit tangent is al-ways future timelike. That is, "observer" and "worldline of a particle" aremathematically identical concepts. Finally, an instantaneous observer isanother name for a (unit) future timelike vector. To say that an instanta-neous observer u is provided or given should be thought of as the same asto say that an orthogonal decomposition of Minkowski space is provded orgiven. Namely,

R(3,1) = S ® span u,

where S =_ (span u)1 is a spacelike 3-plane referred to as rest space. Armedwith this expanded definition of an instantaneous observer, it is usuallypossible to guess the classical concepts that are "observed" or "deduced"from a relativistic concept. Two examples are discussed. For both, supposean instantaneous observer u is given at a point z in Minkowski space withassociated rest space S.

In the first example, suppose M is the worldline of a particle. Let

M denote the unit (future timelike) tangent vector to M at the point z.H H

Then M has the orthogonal decomposition M = to + to, and the observedvelocity of the particle is v = w/t E S.

As a second example, consider the relativistic concept of an electro-magnetic field on Minkowski space. By definition, this is just a two-formF on R(3, 1); that is, for each point z E R(3, 1), F(z) E A2R(3,1). Now,given an instantaneous observer u at z E R(3, 1), what can be "observed"?The vector field E that is metrically equivalent to the one-form u L F iscalled the (observed) electric field.


Recall that if a euclidean space V, (,) is oriented, then there exists aunique unit volume element A E An V defined by A = el A ... A e wheree 1, ..., e is an oriented orthonormal basis for V. Moreover, the equation

a A (*/3) = (a, /3)A

defines a linear isomorphism

*: A'kV --+ A"-kV

called the Hodge * operator.Now consider the unit volume element A = dx' A dx2 A dx3 A dt

along with the standard inner product (, ) on R(3, 1). Note that * mapsA211(3, 1) isomorphically on A2R(3, 1). Let b(u) denote the one-form met-rically equivalent to u. The vector field B that is metrically equivalent to*(F Ab(u)) is called the (observed) magnetic field. If the instantaneousobserver u = e4 is taken to be the standard unit timelike basis vector inR(3, 1), then

F = B3dx1 A dx2 + B1 dx2 A dx3 + B2dx3 A dx'

+E'dx1Adt+E2dx2Adt+E3dx3Adt.

The classical Maxwell equations are most succinctly written

dF = 0 and d(*F) = 0.

We conclude this section with a sketch of the proof of the isomor-phism between the Lorentz group SO1(3,1) and the special complex or-thogonal group SO(3, C). First note that SOT(3, 1) can be defined as thesubgroup of SO(3, 1), which preserves the future timecone. In particular,-10 SO1(3,1). Later, the alternate description of SO1(3, 1), as the con-nected component of the identity in SO(3, 1), is used.

The standard action of SO1(3,1) on R(3, 1) induces an action ofSOT (3, 1) on A2R(3,1) by the usual "pull back" map on forms. The follow-ing facts are straightforward to verify. The square, *2, of the star operator* on A2R(3,1) is minus the identity. Consequently, * provides A2R(3, 1)with a complex structure J - *. The real valued symmetric bilinear form*(aAQ) on A'R(3,1) is nondegenerate, and J is an anti-isometry withrespect to this bilinear form. Consequently,

e(a, 6) = *(a A /3) + i * ((*a) A fl)

is a C-symmetric inner product on the complex vector space A2R(3, 1), J*. Since SO1(3,1) commutes with J =_ *, we obtain a group homomorphism


from SOT(3, 1) to SO(3, C), the subgroup of GLc(AZR(3,1)) (where Jthat fixes E.

THE CARTAN-DIEUDONNE THEOREM

Given u E V nonnull, then ul is a nondegenerate hyperplane and byCorollary 2.40, each vector x E V has a unique orthogonal decompositionz w+Au with WE u1 and A ER.

Definition 4.22. The map Ru defined by w - Au is called thereflection through (the hyperplane) ul. or reflection along u. That is, R isthe identity on ul and minus the identity on the line through u. Therefore,

&x (x, u)-x- 2

Ilull

Note that R. E O(V ), det R _ -1, and R,,', = I.By definition, the identity I E O(V) is said to be the product of r = 0

reflections along lines.

Theorem 4.23 (Cartan-Dieudonne).(a) (Weak Form) Each orthogonal transformation A E O(V) can be ex-

pressed as the product of a finite number r of reflections

(4.24) A=R,,o...oR,,,.

along none ull lines, so that I Iui 11 : 0 for all j = I,-, r.(b) (Strong Form) The number of reflections required is at most n; that

is, r < n.(c) (Sherk Version) First note that the fixed set (or axis of revolution)

FA =- {x E V : Ax = x} = ker(A - I) is always orthogonal toimage(A - I). Now suppose that (4.24) is any representation of A (asthe product of r reflections) with r minimal. Then

(1) If A - I is not skew, then r = n - dim FA. In fact, ul, ... , ur is abasis for image(A - I).

(2) If A - I is skew, then r = n - dim FA + 2, but FA must be totallynull so dim FA < n/2. This case cannot occur if V is positive ornegative definite, or positive or negative Lorentzian.

Remark 4.25. The part of this theorem that will be used later to studythe Spin and Pin groups is simply the weak form: each A E O(V) can beexpressed as a (finite) product of reflections. Consequently, some readersmay wish to skip the proof of parts (b) and (c).

68 The Cartan-Dieudonne Theorem

The proof of the theorem begins with the case of dimension n = 2.As a prelude to Chapter 6 on normed algebras, this case provides us withan excuse to discuss the complex numbers C and the less familiar Lorentznumbers L. In order to emphasize the analogy between C and L, a briefdiscussion of the well-known complex numbers is included (for later com-parison with L).

The Complex Numbers C

The complex numbers C are defined to be R(2, 0) with the extra struc-ture of multiplication, given by

(a, b)(c, d) _ (ac - bd, ad + bc).

Let I =- (1, 0) and i (0,1), so that (a, b) = a + bi and i2 = -1. Conjuga-tion is defined by 7 = a - ib for z = a -{- ib. Note that zw = T W, zz = 1IzIand hence JJzwJJ = JJzJi JJwJJ. If z # 0, then z-1 = x/JIzJJ, so that C is a(commutative) field. Also, note that

(z, w) = Re zw (zw -{- 7w) .

Let eie = cos 0 + i sin 0 denote a point on the unit circle and note thatMere, multiplication by e19, is an orthogonal transformation since eseAs a 2 x 2 real matrix,

Me.e = (cosO -sing)`sin 0 cos 0 J

so that M. E SO(2). Since Meiseio = Mei(e+o), the map 0 r-, Meie inducesthe group isomorphism,

R - SO(2).27rZ -

Lemma 4.26.. Suppose A E 0(2) and define 0 by Al = e'B. If det A = 1,then

(4.27) Rie;e/2 o A = Ri,

where Ri (reflection along i) is just conjugation Cz = z. If det A = -1,then

(4.28) Rieie/2 o A = Id.


Proof: It suffices to show that Riei0/2 o A fixes 1, because Ri and Idare the only two orthogonal transformations fixing 1. The reflection ofeie = Al along ieie/2 (or through span eie/2) is 1 since both unit vectorseie and 1 have the same inner product with ei9/2, namely, cos 0/2. J

Corollary 4.29. Suppose A E SO(2). Then

(4.30) A = Rie0/2 o Ri

is the product of two reflections. Moreover, A = Mere .

Suppose A E 0(2) and det A = -1. Then

(4.31) A = Rieie/2

is a reflection. Also, A = C o M,-;e.Proof: Solving (4.27) and (4.28) for A, we obtain A as the product ofreflections proving the first equalities in (4.30) and (4.31). To prove A =Mete, note that both satisfy det = 1 and map 1 to eie so that both satisfy(4.27). The proof of A = C o Me_;e is similar. J

The Lorentz Numbers L

The Lorentz numbers L are defined to be the inner product spaceR(1,1), (, ) with the extra structure of multiplication given by

(a,b)(c,d) _ (ac+bd,ad-{-bc).

Let 1 = (1, 0) and r =- (0,1), so that (a, b) = a+br and r2 = 1. Conjugationis defined by z = a - br for z = a + br. Note that FT = T U Y and z z = IIzII,so that IlzwJJ = IIzII JIwII. Thus, if IIzII :0 (z nonnull), then z-1 = x/IlzlIexists, while for IIzII = 0 (z null) z cannot have an inverse. Also note that(z, w) = Re zw =

2(zw + zw).

Let ere = cosh B + r sinh B (calculate the formula power series for eTeto see that this definition is appropriate). Note that Me,e, multiplicationby ere, is an orthogonal transformation since IIeT°II = 1. As a 2 x 2 realmatrix,

Mere = cosh 0 sink 0 )sinh 9 cosh 0 J

so that det Mere = 1. Define a timelike vector z = a + br to be futuretimelike if b > 0. Since Me .e (r) = sinh 0 + r cosh 0, multiplication by

70 The Cart an-Dieudonn 6 Theorem

ere preservesreserves the futuretime cone. Thus, Mere E SO1(1, 1). In fact, sinceere erW = e`(8+) the map sending 0 i- Mere determines the group iso-morphism

(4.32) R S--- SOT(l, 1).

Note that the set of unit spacelike vectors {z = a + br : IJzIJ = a2 -b2 = 1} consists of two disjoint curves {ere : 0 E R} and {-ere : 0 ER). Therefore, given A E 0(1, 1), the unit spacelike vector Al uniquelydetermines an angle 0 by Al = ±ere

L ~\Figure 4.33

Lemma 4.34. Suppose A E 0(1, 1) and Al = ±ere defines 0 E R.

(a) If Al = ere and det A = 1, then

RTer@12 OA = R.

(b) If Al = -ere and det A = 1, then

RTere/2 o A = R1.

Euclidean/Lorenizian Vector Spaces

(c) If Al = ere and det A = -1, then

RTere/2 o A = Id.

(d) If Al = -ere and det A = -1, then

RTere/2 o A = -Id.

71

Proof: The reflection of ere along the line rere/2 (or through the lineere/2) is 1 since both ere and 1 have the same inner product with ere12(since ere = e-re and (ere , ere/2) = Re eTee- re/2 = (ere/2 , 1)). Thus,(RTere/2 o A)(1) = ±1, where Al = ±er9 has the same ± sign. SinceRTere/2 o A is orthogonal, it must also map the 7--axis into itself and hencebe ± Id on the r-axis. The four cases now follow.

Corollary 4.35. Suppose A E 0(1,1). Ifdet A = 1, then A is the productof two reflections. Either

A = RTere/2 c Rr or A = RTere/2 o R1.

If det A = -1, then A is a reflection. Either

A = RTere/2 or A = Rere/a.

This completes the proof of Theorem 4.23 in the special case of dimen-sion two using the complex numbers C and the Lorentz numbers L.General Proof of Theorem 4.23: The general proof proceeds by in-duction on the dimension n of V. Given A E O(V), there are cases toconsider.

Case 1: A is said to have a nonnull axis of rotation if there exists a nonnullvector x with Ax = x.Proof of Case 1: In this case, the fixed set FA can be written as FA =N ® H, where N is totally null and H is nondegenerate. Since A is theidentity on H, A maps H1 to H-L. By the induction hypothesis, thetheorem is true for AIH-L E O(H1). In particular, if (4.24) is valid forAIHI E O(H -), then the same representation is valid for A E 0(V). Notethat R,,, reflection along a line u in Hl, naturally extends to reflectionalong the same line u, but now considered a line in V.Case 2: Suppose u = Av - v and v are nonnull vectors. Then fixesv.

72 The Cart an-Die udonne Theorem

Proof of Case 2: Note that u = Av - v and to = Av + v are orthogonalsince A E O(V). Therefore, Av has the orthogonal decomposition

(4.36) Av = 2 u + 2 w

with respect to V span u ® (span u)-L. Thus, -1 u + iw = vfixes v. By the induction hypothesis, can be expressed as theproduct of < n-1 reflections. Therefore A can be expressed as the productof < n reflections.

Now the proof of the weak form (i.e., each A E O(V) is the product ofa finite number of reflections) can be completed. Pick any nonnull vectorv E V. If u = Av - v is nonnull, then and hence can beexpressed as a product of a finite number of reflections. If to = Av + v =-((-A)v - v) is nonnull, then R. (-A) Ispaz, vim, and hence Rv, (-A), can b eexpressed as a product of a finite number of reflections. Finally, note thatboth u and w cannot be null, since they are orthogonal and v = - au+ 1w .

Before preceeding with Case 3, we need to make several observations-Recall that for any linear map L E End (V),

(4.37) image L = (ker L* )1

and

(4.38) dim image L + dim ker L = n.

Since A* = A-1, the equations Ax - x and A*x = x are equivalent.Thus, ker(A - I) = ker(A* - I). Combined with (4.35) this proves that

(4.39) image(A - I) = FA for A E O(V).

Note that if A E O(V) does not belong to Case 2, then u - Av - vis null for each nonnull vector v. By continuity, this implies that eachu = Av - v is null. Therefore, if A E O(V) does not belong to Case 2, thenimage(A-I) is totally null. If Case 1 is not applicable, then FA = ker(A- I)is totally null.

Case 3: Suppose both image(A - I) and ker(A - I) are totally null.Proof of Case 3: Since by (4.38) each is the orthogonal compliment ofthe other, they must be equal, i.e.,

N =- image(A - I) = ker(A - I).

By (4.38) this implies that dim V = 2p, where p = dim N.


Since N is totally null, Proposition 2.45 implies that V has split sig-nature p, p.

Note that A is the identity on N. Pick M complimentary to N. GivenY E M, Ay = (Ay - y) + y is the direct sum decomposition for Ay E V =N ®M since Ay - y E N. Let b : M - N denote A - I restricted to M.Then V = N® M determines the 2 x 2 blocking of A as

(4.40) A= I b).(oI

Therefore det A = 1. This proves that if A belongs to Case 3, thendetA = 1.

Choose any reflection R through a nondegenerate hyperplane. Thendet RA = -1. Thus, RA must be in Case 1 or Case 2. Therefore RA =Rl . Rkk < n. Now det RA = -1 implies k is odd. Since n = 2p iseven, we must have k < n - 1. Therefore A = RR, ... Rk with k < n - 1.This completes the proof of what is usually called the Cartan-DieudonneTheorem, i.e., part (b).Remark. By choosing a null basis for V = N ® M, i.e., n1, ..., np a basisfor N and M = span{ml,..., mp} totally null with the planes span{ni, mi}nondegenerate and orthogonal, one can show that A E O(V) belongs toCase 3 if and only if A is of the form (4.40) with b a skew matrix. SeeProblem 10 for an example of an orthogonal transformation A E O(V)that belongs to Case 3 (dimension of V equals four).

Now we complete the proof of the theorem. Since A satisfies part (c)of the theorem if and only AIH1 satisfies part (c), we may assume thatthe fixed point set FA is totally null. Several statements equivalent to"L = A - I is skew" are provided in Problem 11. In particular, A - I isskew if and only if image(A - I) C ker(A - K) _- FA. Therefore, becauseof the assumption that FA is totally null, the cases where A - I is skew fallunder Case 3. This case is discussed in Problem 11.

Now assume that A -.1 is not skew, FA is totally null, and hence Case2 is applicable. Suppose j = Lv is nonnull with L = A - I and IIvil # 0.Then fixes v.

We must show that, for some choice of u, R,,A - I is not skew:

(4.41)If RA - I is skew then, for all v E V,

IILvII(L + L*)(y) = -4(Lv, y) (L*v, y) for all y E V.

Proof of (4.41): Assume u = Lv is nonnull. Note

HUM

74

Therefore,

The Cartan-Dieudonne Theorem

RUA - I =L-2(u,A( ))n

IIuIi Mull

This gives

(442) R A-I=L+2 (v,

U (lull

since L*A = (A-' - I)A = I - A = -L. Thus, RUA- I is skew if and onlyif

(4.43) (L + L*)(y) + 4jjujj-1(L*v, y) (Lv, y) = 0.

Since (4.41) is valid for all v E with jjLvjj # 0, by continuity it is valid forall v V.

Note that {v E V : Lv = v} _ {v E V : L*v = V} = FA.quently, if RUA is skew, then

(4.44) Lv null implies Lv = 0,

Conse-

or each nonzero u E image L is also nonnull. Consequently, the subspaceimage L = image(A - I) is either positive or negative definite. In this case,the theorem is valid for A. Therefore, we may choose u = Av - v with uand v nonnull so that RUA - I is not skew.

It remains to show that

(4.45) rank(RUA - I) < rank(A - I).

Now RUA still fixes FA since u E Im(A-I) = FA,.. In addition, RUA fixesv and v 0 FA. Thus

dim dim FA,

which proves (4.45). This completes the proof of Theorem 4.23. 'J

Definition 4.46. Given a non degenerate subspace W of V, reflectionalong W (or through W1), denoted Rw, is defined to be -1 on W and+1 on Wl. I f u1, ... , uk is a basis for W, then reflection along W will alsobe denoted by RU,n...AU,,.

Just as O(V) is generated by reflections along lines, SO(V) is generatedby reflections along planes.


Theorem 4.47. If n = dimR V > 3, then each f E SO(V) can be ex-pressed as the product of an even number k < n of reflections along non-degenerate planes:

f = Ru1Ao1 0... o RukAvk, k even and < n

Proof: By the Cartan-Dieudonne Theorem, f = Ru, o . o R, witheach uk E V nonnull and with k even. If n = 3, then either k = 0and f = 1, or, k = 2 and f = Rut o Ru2. Now -Ru = R,AW, wherespan(v A w) = (span u)1. Thus

f = Ru, o Rue = (-R+1,) o (-RT1z) = Rv1AW, o R.,A.,.

If n > 4, then given a pair of nonnull vectors u1, u2 there exists a nonnullvector v orthogonal to both ul and u2 (see Problem 4.10). Therefore,

f = Rut o Rut = R,,, o R o Ru o Ru2 = Ru1Aa o RuaAo

as desired.

GRASSMANIANS AND SOt(p,q), THE REDUCEDSPECIAL ORTHOGONAL GROUPS

Suppose V, (,) is an R-symmetric inner product space with signature p, qthat is not definite, i.e., p, q > 1. Let G+(p, V) denote the grassmanian ofall oriented p-planes through the origin that are maximally positive. Given1 E G+(p, V), represent ty =_ el A A ep E ApV by choosing an orientedorthonormal basis for V. Let P span 4 denote the p-plane withoutorientation.

If l E G+(p, V) is given, then each ij E G+(P, V) can be graphed overP = span 4 because 7r, the orthogonal projection onto P along Pl, is injec-tive when restricted to Q = span 77 (ker a nQ = {0}). Consequently, thereis a well-defined notion of when , q are compatibly oriented. If el, ..., epis an oriented orthonormal basis for 4 and Q = span n is the graph ofa linear map A : P - Pl, then either {el + Ae + 1, ... , ep + Aep} or{-el - Ae1, e2 + Ae2,.. . , ep + Aep } is an oriented basis for r). In the firstcase, we say that 4 and n have the same (spacelike) orientation. Note thatin terms of the natural inner product (,) on ApV, f and rt have the sameorientation if and only if (l;, 17) is positive and the opposite orientation ifand only if (4, 77) is negative. Thus, G+(p, V) is naturally divided into twosets.

76 Grassmanians and The Reduced Special Orthogonal Groups

It is also easy to see that G+(p, V) has two connected components.Shrink the graphing map A to zero. That is, if l; and rl are compatiblyoriented with Q = span r7 the graph of A over P span ., then rltgraph to defines a path in G+(p, V) connecting q to e.

Similarly, the grassmannian, G_(q, V), of oriented maximal negativesubspaces of V has two connected components. Selecting G+ (p, V) as one ofthe two components of G+(p, V) is called the choice of spacelike orientation,while selecting Gt (q, V) as one of the two components of G_(q, V) is calledthe choice of timelike orientation (or future orientation).

An orthogonal transformation A E 0(V) is said to preserve the space-like orientation if A preserves the two components of the grassmannianG+(p, V) of oriented maximal positive subspaces. An orthogonal transfor-mation A E O(V) is said to preserve the timelike orientation if A preservesthe two components of the grassmannian G_ (q, V) of oriented maximalnegative subspaces. In this context, A E O(V) with det A = 1 is said topreserve the total orientation of V.

Definition 4.48. The orthogonal group O(V) in the cases that are notdefinite (p, q > 1) has subgroups defined by

(4.49) SO(V) _ {A E O(V) : A preserves the total orientation},(4.50) O+(V) _ {A E O(V) : A preserves the spacelike orientation},

(4.51) 0-(V) _ {A E O(V) : A preserves the timelike orientation}.

(The notation Ot (V) is also used for 0-(V). The arrow indicates thatthe direction of time is preserved.)

The intersection of any two of these three subgroups is the same groupand is denoted by

SO1(V) __ {A E O(V) : A preserves the total, spacelike,(4.52)

and timelike orientation}.

O+(V) is called the spacelike reduced orthogonal group, 0-(V) is called thetimelike reduced orthogonal group, and SO1(V) is called the reduced specialorthogonal group.

Proposition 4.53.

(a) 0+(p, q)/SO(p) x O(q) - GI (p, V).

(b) O- (p, q)/0(p) x SO(q) - GL (q, V)

The straightforward proof is omitted.In summary, the group O(p, q) (with signature that is not definite) has

four connected components given by the four possibilities for A E O(V) : Apreserves (or reverses) space orientation and A preserves (or reverses) timeorientation.


Lemma 4.54. Suppose A E O(V) is expressed as the product of reflectionsA=Ru,o...oRur.

(a) The number of spacelike vectors in {ul,..., u,.} is even if and only ifA E 0+(V).

(b) The number of timelike vectors in {ul, ..., u,.} is even if and only ifA E 0-(V).

In particular, the parity of the number of reflections along spacelike lines(or along timelike lines) is independent of the representation of A as theproduct of reflections.

Definition 4.55. Given A E O(V), let IIAIIs denote the parity of thenumber of reflections along timelike vectors when A is expressed as theproduct of reflections.

Note that

(4.56) II'11s:0(V)- Z2

is a group homomorphism. The number IIAIIs is called the spinorial normof A.

The two group homomorphisms det : O(V) --+ Z2 and II-

I Is : 0(V) --3Z2 distinguish the four components of O(V ). In addition, each of the threegroups

SO(V), Ot(V), and 0+(V) 21

consist of the identity component SOT(V) plus one additional componentgiven by

(4.57) det = 1 and(4.58) det = -1 and(4.59) det = -1 and

for SO(V),for 0-(V) = OT (V),for 0+(V).

PROBLEMS

1. (a) Verify (4.3) directly from the definition of the inner product onA2V.

(b) Show that A2 V has signature (2) + (2 ), pq. In particular, A2 V hassignature 3, 3 if V has signature 3, 1 or 1, 3.

2. (A canonical form for 0(n) acting on o(n)) Given a E A2R*(n), find anorthonormal basis w', ..., w" for R*(n) and al > )2 > > h, > 0,so that

a = Alwl A w2 + ... + A,w2r-1 A w2r.

8 Problems

Hint: If the maximum value al of a(x A y) taken over all Ix A yl = 1is attained at el A e2, then show that a(el A u) = 0 for all u E R(n)with u 1 e2 by considering the function a(el A (cos 0 e2 + sin 0 u)) of0.

3. Give the proof of the twin paradox, Proposition 4.19.

4. Suppose V is a time-oriented Lorentzian vector space. The backwardstriangle inequality says that lu + vl > Jul + Ivl for all future timelikevectors, and equality holds if and only if u and v are positive multiplesof one another. Deduce this result(a) algebraically from the backwards C.-S. Inequality;(b) as a special case of the twin paradox (for piecewise smooth curves).Given a, b with b - a future timelike, show that there exist (piecewisesmooth) particles from a to b with arbitrarily small proper time.

5. Suppose V, II II is a Lorentzian vector space. If II II' is a square normon V with Ilxll' < M for all unit timelike vectors, then show thatII II' = cll . II for some constant c.

6. (a) Show thatzw=zwandthatz'z=zz=llzllforallz,wEL.(b) Show that

SO(1,1)={±MM,e :OER}.

(c) Let Cz = z denote conjugation on L. Show that C E 0(1, 1) andthat

0(1,1)_{±Me,e :0ER}U{±CMe,e :O E R}.

7. (Double Numbers) Consider the algebra R ® R of diagonal 2 x 2 ma-trices

ul 0

0 u2)'with matrix multiplication, and with the inner product determined bythe quadratic form u1 u2. Define conjugation by

C (0vv ) - (0 uu)

.

Exhibit an isomorphism between L and R e R preserving the multi-plication, the inner products, and the conjugations. Exhibit eTa as a2 x 2 matrix.

8. Suppose V is a euclidean vector space of signature p, q. Recall thatthe unit sphere is defined to be {u E V : Jul = 1}. Show that if x is apoint on the unit sphere, then x is normal to the unit sphere at x.


9. Consider R(4, 0) = H.(a) Show that reflection along a line spanned by a E H with Ial = 1 isgiven by Rax = -aia.(b) Use part (a) to give a proof that X : S3 x S3 -+ SO(4) defined byXa,b(x) = axb for all x E H, as in Remark 1.33, is a surjection.

10. (a) Exhibit a pair of nonnull vectors u1i u2 E R(2, 1) that do not havea nonnull vector v E R(2, 1) orthogonal to both of them.(b) If n - dim V > 4, show that each pair u1i u2 of nonnull vectorshas a nonnull vector v E V orthogonal to both of them.(c) Find A E SO(2, 1) with the fixed axis FA =- {x E R(2, 1) : Ax = x}a null line.

11. Let V - M2(R) with JJvll = detv. Let A E End(V) denote left

multiplication by the matrix C 1 110 1

(a) Show that A E SO(V).(b) Show that image (A - Id) is totally null of dimension 2.(c) Show that FA = {x E V : Ax = x} is totally null of dimension 2.(d) Show that A cannot be written as the product of less than fourreflections along (nonnull) lines.

12. Given A E O(V), show that the following are equivalent:(a) A - I is skew,(b) 2(A + A*) = I,

(c) A'1 is the identity on image(A - I),(d) image(A - I) C FA - ker(A - I).(e) FA C FA.

13. Suppose A, B E O(V) both leave the subspace F C V fixed and agreeon F.(a) Show that A = B if V is positive definite or Lorentzian.(b) Find a counterexample to A = B if V = R(2,2).

14. Suppose V, (, ) is positive definite euclidean space of dimension n.Given a subspace W of V, let Rw denote reflection along W throughWl and let Pw denote orthogonal projection from V onto W. Showthat Pw = 2 (I - Rw) or Rw = I - 2Pm.

15. Let G(k, V) denote the grassmanian of (not necessarily oriented) k-planes in V through the origin.(a) Show that G(k, R"), the grassmannian of non-oriented k-planes inR', can be identified with a subset (actually a submanifold) of the

80 Problems

vector space of self-adjoint linear maps Sym(R') _ {A E EndR,(R")A' = A}. More precisely, identify G(k, R") with

.{ A E Sym(R') : A2 = A and traces A = k)

(b) Similarly, identify G(k, C"), the grassmannian of complex k-planesin C", with

{A E Herm(C") : A2 = A and traces A = k} .

(c) Identify the grassmannian of quaternionic k-planes in H", denotedG(k, H") with the set of projection operators

{ A E Herm(H") : A2 = A and trace, A = 4k).

In the special case k = 1, G(k, F") is called projective space anddenoted P"(F) _ G(1, F")-see Chapter 5 and Problem 6.16 for moreinformation.

16. Suppose V, II .II has split signature p, p. Given N, totally null of di-

mension p, show that there exist M complimentary to N, also totallynull of dimension p, and an isometry from V to RP x RP, with squarenorm II(x, y) 11 = x1y1 + +xpyp, which sends N to RP x {0} and Mto {0} x RP.

17. Show that each A E O(n, C) can be expressed as the product

A = o . . . o R,,,, with r < n,

where each uj is nonnull. The complex reflection Ru along a nonnullline u is defined by exactly the same formula as in the real case.

5. Differential Geometry

This chapter provides a brief description of several types of geometries.Its purpose is to add motivation for the other chapters. As such, it may beeither browsed over quickly or studied carefully in conjunction with some ofthe standard texts in geometry. The chapter should only be considered asintroductory and motivational, despite the fact that the reader is assumedto be familiar with the notion of a manifold.

While the novice in geometry may wish to use this chapter as a ref-erence point for further study, the more experienced student may be dis-appointed in its introductory nature, especially after reading the followingtable of geometries.

Each of the geometries discussed is labeled with a particular group (seeTable 5.1). However, a particular group may be associated with one, two, oreven three different but closely related geometries. The connection betweenthe group and the geometry is meant only to be suggestive. A deeperrelationship is provided by the notions of principle G-bundles, reduction ofthe structure group, and integrability. The reader is encouraged to pursuethese topics.

REAL n-MANIFOLDS: THE GROUP GL(n, R)

Briefly, for the sake of completeness, the general notion of a (smooth) realn-dimensional manifold M is defined as follows.

81

82 Oriented Real n-Manifolds

Definition 5.2. M must be a topological space equipped with a countableatlas that is smoothly compatible. An atlas is a collection of charts, wherein turn each chart consists of two things: an open subset of M and ahomeomorphism of this open set with an open subset of R". Of course,each point of M is required to belong to at least one chart. Two overlappingcharts induce a homeomorphism from one open set in R' onto another,called a transition function. The atlas is said to be smoothly compatible ifeach transition function is a diffeomorphism.

This completes the definition of a real n-manifold. The group GL(n, R)is associated with this geometry, since the linearization, at a point, of eachtransition function belongs to GL(n, R). In later discussions in this chapter,the reader is assumed to be familiar with the notion of the tangent spaceTyMtoMatapoint xEM.

Table 5.1ManifoldReal n-manifoldsOriented real n-manifoldsComplex (and almost-complex) n-manifoldsQuaternionic (and almost-quaternionic)

n-manifoldsManifolds with volumeRiemnnian manifolds (of signature p, q)Conformal manifolds (of signature p, q)Real symplectic manifoldsComplex Riemannian n-manifoldsComplex symplectic manifoldsKahler manifolds (of signature p, q)Special Kahler manifolds (signature p, q)HyperKahler manifolds (of signature p, q)Quaternionic Kahler manifolds

(of signature p, q)Quaternionic skew hermitian manifolds

GroupGL(n, R)

GL+(n, R)GL(n, C)

GL(n, H) H*SL(n, R) and SL(n, C)

O(p, q)CO(p, q)Sp(n, R)O(n, C)

Sp(n, C)U(p, q)

SU(p, q)HU(p, q)

HU(p, q) HU(1)Sk(H)

ORIENTED REAL n-MANIFOLDS: THE GROUP GL+(n, R)Suppose M is a real n-manifold equipped with an atlas so that each transi-tion function preserves orientation; that is, the linearization of each transi-tion function, at each point, belongs to the subgroup GL+(n, R), consistingof all invertible matrices with positive determinant. Then M is called an

Differential Geometry 83

oriented real n-manifold. Note that each tangent space TxM carries a nat-ural orientation as follows. Choose any chart with open set containing x.Then the associated homeomorphism between this open set and an opensubset of R" induces an orientation on TxM, and this orientation on TxMis independent of the choice of chart since the transition functions havederivatives in GL+(n, R).

An oriented real n-manifold may also be (equivalently) defined asfollows. Assume M is a real n-manifold. Suppose there exists a never-vanishing n-form 0 on M (but Q is not necessarily distinguished), then Mis said to be orientable. In this case, any other never-vanishing n-form S2'is a multiple f of S2 where f is a never-vanishing function. If f is positive,then we say 52' and S2 are equivalent. Thus, on an orientable manifoldthere exist precisely two equivalence classes of never-vanishing n-forms. Ifa choice of one of these two equivalence classes is made, then M is calledan oriented manifold.

COMPLEX (AND ALMOST COMPLEX) n-MANIFOLDS:THE GROUP GL(n, C)

The general notion of a complex n-manifold is defined by mimicking thedefinition of a real n-manifold, except R° is replaced by C' and the tran-sition functions are required to be biholomorphisms (from an open subsetof C" to an open subset of C'). There are several useful, and of courseequivalent, definitions of a holomorphic map. The simplest, from our pointof view, requires that the linearization or derivative of the map, at eachpoint, should be complex linear. Thus, a complex n-manifold M is, first ofall, a real 2n-dimensional manifold, but with the additional property thateach transition function (from an open subset of C" - R2" to anotheropen subset of C" = R2n) has its derivative (_ differential) in the sub-group GL(n, C) of GL(2n, R). The obvious choice is to label the geometryof all complex n-manifolds with the, group GL(n, C).

On a complex vector space T, complex scalar multiplication by i is areal linear map (denoted I) that squares to minus the identity. Conversely,given any real linear map I (on a real vector space) that squares to minusthe identity, it is easy to convert the real vector space to a complex vectorspace by setting the given linear map I equal to multiplication by i. Sucha map I is called a complex structure. Now, on a complex manifold M eachtangent space TM (considering M as a real 2n-manifold) is naturallyequipped with a complex structure due to the fact that the linearization ofa transition function commutes with the complex structure on C".

84 Complex (and Almost Complex) n-Manifolds

An almost-complex n-manifold is a real 2n-manifold that is equip-ped with a complex structure I on each tangent space TrM (that variessmoothly with the point x E M).

The differential forms (with complex coefficients) on an almost com-plex manifold are particularly interesting. The space of degree one forms(at a point) is naturally the direct sum of two vector spaces: the spaceT*1,0 of forms of type 1, 0 and T*O,l the space of forms of type 0, 1 (seeProblem 3). More generally the space of degree k forms, denoted AkT*,is the direct sum EP+q-k ®AP,QT*, where AP,9T* is the space of forms ofbidegree p,q.

There are convenient nontrivial ways of testing whether a given almostcomplex n-manifold is a complex n-manifold, which are provided by the"Newlander-Nirenberg Theorem."

There is a further weakening of the concept of an almost-complex n-manifold that is not of much interest itself but does provide a basis fora useful quaternion analogue. A "weak" almost-complex n-manifold is areal 2n-manifold that is equipped with two complex structures dI, one thenegative of the other, on each tangent space TrM (and they should locallyvary smoothly with x). The point is (see Problem 2) that there may notbe any way globally to make a choice of I versus -I.Remark 5.3. For the reader comfortable with the notion of a vector bun-dle on a manifold, the distinction between almost-complex and the "weakerform" of almost-complex can be described as follows: Let End(TM) de-note the endomorphism vector bundle on M, i.e., the vector bundle withfiber End(TrM) at each point x E M. A two-dimensional subbundle C ofEnd(TM) is said to be a bundle of C-algebras on M if, for each point x,the fiber Cx is real algebra isomorphic to C. Now a "weak" almost-complexn-manifold is a real 2n-manifold equipped with a subbundle C C End(TM)of C-algebras. Since z --*Y is a real algebra automorphism of C sending ito -i and the isomorphism Cr C is not canonical, the set of operators{I, -I} C C., corresponding to {i, -i} C C is distinguished, but not theindividual operators I and -I.

If, in addition, a global choice of I is possible, then by identifyingI E Cz with i E C (and by identifying the identity in each Cr with 1 E C)we have a global trivialization of the bundle C. Thus, we have the followingequivalent definition of an almost complex manifold.

Definition 5.4. An almost-complex n-manifold is a real 2n-manifoldequipped with a trivial subbundle M x C C End(TM) of C-algebras.


QUATERNIONIC (AND ALMOST-QUATERNIONIC)n-MANIFOLDS: THE GROUP GL(n, H) H*

Now the situation is more delicate, and the concept that is analogous ina straightforward manner to the concept of a complex n-manifold is bar-ren, i.e., there are essentially no examples. The example of quaternionicprojective space Pn(H) provides some guidance.

Let [x] denote the quaternionic line [x] =_ {xA : A E H} throughthe origin and the point x = (xo, ..., xn) E Hn+1 - {0}. By definition,P"(H) E {[x] : x E Hn}1 - {0}}.

The standard atlas for P'(H) consists of n + 1 charts Uk, Ok, k =0,.. ., n, defined as follows. Let Uk = {[x] E Pn(H) : xk # 0}, and definemaps 4k : Uk --+ Hn by sending [x] E Uk to Ok([x]) = (xoxk

1, , 1, ,

xnxk 1) E H'. Now the transition functions are all of the same form.For example, the transition function : 0o(Uo) --> 01(U1) is given by

ti(yl, yn) = (yl1, 2y1

1,. . ., ynyl 1). The linearization 0'(y) (or deriv-ative) of z/i at y sends u E Hn to

(Y1 lulyl u'2y11.... ,yny11th y11 -unyl1).

In particular, 0'(y) __ Ay o Rye where Ayu = -O(y)ul + (0, u2, ..., un)

is H-linear and Rye u = uyi1 is simply right multiplication by yi 1

In summary, the linearizations ik'(y) belong to GL(n, H) H* but notGL(n, H). Thus, P'(H) is a "quaternionic n-manifold" in the sense thatthe transition functions for the standard atlas have linearizations in theenhanced quaternionic linear group GL(n, H) H*. Unfortunately, even

adopting this more general notion of "quaternionic n-manifold," it fol-lows that each such manifold is locally equivalent to P"(H) (see Besse[3], p. 411). Consequently, it is most useful to reserve the terminology"quaternionic n-manifold" for yet another notion due to Salamon. Since

the precise definition involves the concept of a torsion-free connection, we

only state the definition and refer the reader to the literature (see Salamon

[16] or Besse [3]).A quaternionic n-manifold M is an almost-quaternionic n-manifold

whose GL(n, H).H* reduced frame bundle admits a torsion-free connection.

If n = 1, this condition is automatic.The useful notion of "almost-quaternionic" is simpler to define. First

consider Pn(H) again. Suppose a point [x] E Pn(H) is fixed. Right multi-plication on the tangent space T1x1Pn(H) by a quaternion A can be defined

in several ways using different charts. Using Uk, 4, it is natural to de-fine uA by representing the tangent vector u E Tx3Pn(H) as a vector inOk(Uk) C Hn and multiplying by A on the right. Let this multiplication

86 Manifolds with Volume

on T1t1P"(H), for x E Uk, be denoted by R. Then

(5.5) Ra = Rk 3atkSJ 1

(see Problem 5). Therefore, right multiplication by a scalar A E H is notwell-defined on T[ ]P"(H). However, the vector space H C End(T[ ]M)defined by H = {Ra E End(T[,-]M) : A E H} for each k with xk $ 0is independent of k and (non-canonically) isomorphic as a real algebra tothe algebra H, since A i aAa-1, with a = xjxk1 fixed, is an algebraautomorphism of H.

This example makes it clear, since a productive definition of "almost-quaternionic n-manifold" should include P"(H), that one should elect toutilize the quaternionic analogue of the notion of "weak" almost-complex.

Definition 5.6. An almost-quaternionic n-manifold M is a real 4n-mani-fold equipped with a subbundle C of End(TM) with C5 = H (as real al-gebras) for each point x E M. The subbundle C is called the coefficientbundle or the almost-quaternionic structure bundle. Suppose this subbun-dle C is trivial and M is equipped with a particular trivilization (a globalbundle isomorphism)

C=MxHso that operators I, J, K equal to right multiplication by i, j, k E H areglobally defined. Then M will be referred to as an almost- quaternionicmanifold with trivialized coefficient bundle. This stronger motion of analmost-quaternionic manifold with trivialized coefficient bundle is the exactanalogue of almost-complex.

MANIFOLDS WITH VOLUME: THE GROUPSSL(n, R) AND SL(n, C)

A real n-manifold M equipped with a never-vanishing n-form S2 is called a(real) manifold with volume. Of course, such a manifold is automaticallyoriented. A interesting result of Moser says that if a compact n-manifoldM is a manifold with volume in two different ways (i.e., two different never-vanishing n-forms Q and ' are prescribed), then M, 0 and M, SZ' are equiv-alent (i.e., there exists a diffeomorphism f of M with f' (Q') = S2) if andonly if the l volume of M is the same as the S2' volume of M' (see Problem10).

Similarly, a complex n-manifold M equipped with a never-vanishingn, 0-form Sl is called a complex manifold with (complex) volume. If, inaddition, 0 is closed under exterior differentiation d, then M is said to bea complex manifold with trivialized canonical bundle.


Remark 5.7. If an almost-complex manifold M of complex dimension n isequipped with a never-vanishing d-closed form SZ of bidegree n, 0, then thehypothesis of the Newlander-Nirenberg Theorem can be verified so that Mmust, in fact, be a complex manifold. (And hence S2 must be a holomorphicn-form.)

RIEMANNIAN MANIFOLDS (OF SIGNATURE p, q):THE GROUP O(p, q)

A Riemannian manifold M of signature p, q is a real n-manifold (withn - p+q) equipped with a real symmetric inner product of signature p, q oneach tangent space TTM (and varying smoothly with the point x E M). Inparticular, this includes the usual concept of Riemannian manifold (q = 0)and the concept of Lorentzian manifold (q = 1) as particular cases. Referto Chapter 4 for additional information.

CONFORMAL MANIFOLDS (OF SIGNATURE p, q):THE GROUP CO(p, q)

A conformal manifold of signature p, q is a real n-manifold with n = p + qequipped with a (smoothly varying) ray [g] of R-symmetric inner productsof signature p, q, where g is a Riemannian inner product and [g] denotesthe ray {.\g : A E R+}. The group associated with an oriented conformalmanifold is CSO(p, q).

REAL SYMPLECTIC MANIFOLDS: THE GROUP Sp(n, R)

We shall give two definitions of a symplectic manifold. A theorem of Dar-boux states that the two definitions are equivalent. First, a symplecticmanifold is a real 2n-manifold -hose atlas has the property that each tran-sition function is a symplectomorphism from an open subset of RZ", w (thestandard symplectic vector space) to another such open subset of R`, w.A symplectic map is a map whose linearization at each point belongs tothe subgroup Sp(n, R) of GL(2n, R), or equivalently a map that fixes thedegree 2 form w under "pull back." Thus, the global degree 2-form it ob-tained by pulling back the standard symplectic form w on RZ",w (for eachchart) is well-defined. Moreover, the 2-form S2 on M has the propertiesthat

(5.8)At each point x E M, S2 defines a real skew inner product

on TTM, the tangent space to M.

88 Complex Symplectic Manifolds

(5.9) S2 is closed under exterior differentiation, i.e., dQ = 0 on M.

The second definition of a symplectic manifold M requires that Mbe a real 2n-manifold equipped with a (smoothly varying) real skew innerproduct 0 on T.M at each point x E M (i.e., a nondegenerate 2-form onM) with dS2 = 0 on M.

Symplectic geometry is important for two reasons. First, it providesa framework for classical mechanics. As noted in Remark 2.39 (b), a func-tion h (the Hamiltonian) can be converted to a vector field by first takingthe exterior derivative dh and then (using the symplectic form) convertingdh to a vector field Vh. Many important flows are the solutions to suchHamiltonian vector fields. The main point is this fact: Another function fis constant on the flow lines of Vh (i.e., a first integral or conserved quan-tity) if and only if the original Hamiltonian function h is constant in thedirections Vj, i.e., Vj(h) = 0. There are far-reaching global consequences.See the beautiful book by Arnold entitled Classical Mechanics [1] for moreinformation. The second reason symplectic geometry is important is hintedat in Problem 8. In particular, symplectic manifolds provide a geometricframework for solving first order nonlinear partial differential equations.As one might suspect, these two reasons for the importance of symplecticgeometry are closely related.

COMPLEX RIEMANNIAN MANIFOLDS:THE GROUP O(n, C)

A complex Riemannian n-manifold is a complex manifold equipped with acomplex symmetric inner product on each tangent space TM (and varyingsmoothly with the point x E M). These manifolds arise naturally by"complexifying" a (real analytic) Riemannian manifold.

COMPLEX SYMPLECTIC MANIFOLDS:THE GROUP Sp(n, C)

A complex symplectic manifold is a complex 2n-manifold equipped with acomplex symplectic inner product o- on each tangent space (and varyingsmoothly), where the bidegree 2, 0 form o- is required to be closed underexterior differentiation. These manifolds arise naturally by "complexifying"a (real analytic) symplectic manifold. Compact examples occur in algebraicgeometry.


Remark 5.10.(a) On a complex symplectic manifold M, o the 2, 0-form o is automati-

cally 8-closed (i.e., holomorphic) so that M, o is also called a holomor-phic symplectic manifold.

(b) An almost-complex symplectic manifold M, v is automatically a com-plex manifold because of Remark 5.7: since an is d-closed, never-vanishing and of bidegree n, 0.

KA.HLER MANIFOLDS (OF SIGNATURE p,q):THE GROUP U(p, q)

A Kahler manifold M is a complex n-manifold equipped with a com-plex hermitian inner product h on each tangent space TrM (and varyingsmoothly), with the property that the 2-form

(5.11) w = -Im h (the Kahler form) is d-closed.

Note that

(5.12) g=_Reh

is a Riemannian structure with signature 2p, 2q on M. The hermitianstructure is given by

(5.13) h = g- iw.

Given a Riemannian structure g, a real symplectic structure w, andan almost-complex structure J on a manifold M:

(5.14) w and J are compatible if J is an w isometry,

(5.15) g and J are compatible if J is a g isometry,

w and g are compatible if the linear operator J defined by(5.16) w(x, y) = g(Jx, y) is an isometry for either

(or both) w and g.

Lemma 5.17. Given any two of the following three structures-(1) a Rie-mannian structure, (2) a real symplectic structure, and (3) an (almost)

90 Special Kahler Manifolds (of Signature p, q)

complex structure, that are compatible-they determine the third struc-ture. In fact, they determine an (almost) Kahler structure.

Proof: Apply Lemma 2.63. -

Kahler manifolds have been studied extensively.

SPECIAL KAHLER MANIFOLDS (OF SIGNATURE p,q):THE GROUP SU(p, q)

Suppose M is a Kiher manifold with signature p, q and complex dimensionn. If, in addition, in a neighborhood of each point, there exists a form a ofbidegree n, 0 that is both

(5 18)(a) d-closed, and(b) of constant size dal = c,

then M is called a special Kahler manifold of signature p, q. Note thatdo = 0 and ao = 0 (i.e., a holomorphic) are equivalent since there are noforms of bidegree n + 1, 0.

If M is a simply connected Kihler manifold, then the definition of spe-cial Kahler-can be simplified. In this case, M is a special Kihler manifoldif and only if there exists a global holomorphic n, 0 form a of constant sizelo'l = 1. In particular, note that in this case the canonical bundle A"'0 isholomorphically trivialized (by a).Remark 5.19. Suppose M is a Kihler manifold of complex dimension nand that a is an n, 0 form on M. One can show that either of the twoconditions:(a) a is parallel,(b) the first chern form of A"'° vanishes (i.e., Ricci flat),

are equivalent to the condition (5.18) required in the definition of specialKahler manifolds.Remark 5.20 (Calabi-Yau). A beautiful theorem of Yau provides ameans of constructing special Kihler manifolds in the important positivedefinite case. Suppose M is a (positive definite) compact Kihler manifoldwith Kahler form w. Assume that the canonical bundle A"'0 is trivial.Let a denote the global, never-vanishing holomorphic n-form (unique upto a constant). Yau proved that there exists a new Kihler form Co thatis homologous tow (i.e., w - w = da for some global 1-form a) such thatM, w is a special Kahler manifold, i.e., a has constant size with respect tothe new Kahler metric based on (Z. (Yau's theorem can be restated in astronger form if M is not simply connected.)

Differeniial Geometry 91

HYPERKAHLER MANIFOLDS (OF SIGNATURE p, q):THE GROUP HU(p, q)

There are quite a few useful ways of defining a hyperKahler manifold.The next definition contains in a certain sense the maximal amount ofinformation.

Definition 5.21. A hyperKahler manifold M is a real 4n-manifoldequipped with the following extra structure. First, assume that the quater-nions H act (smoothly) on the tangent bundle on the right, giving eachtangent space TT the structure of a right quaternionic vector space (i.e., Mis equipped with an almost quaternionic structure with trivial coefficientbundle, cf. Problem 7). Second, supposes is an H-hermitian inner prod-uct of signature p, q on each tangent space (which varies smoothly withthe point x E M). Thus, g = Rer provides M with the structure of aRiemannian manifold of signature 4p, 4q; and each operator R,+ (right mul-tiplication by u) with u E S2 C Im H a unit imaginary quaternion, providesM with the structure of an almost-complex manifold. Third, require thateach associated (Kihler) form w,,, defined by

(5.22) z) - Rer(wu, z) for all w, z E TM,

be closed under exterior differentiation. Fourth, require that the almost-complex manifold M, R,,, with u E S2 C Im H, be, in fact, a complexmanifold. This completes the definition of a hyperKahler manifold.

Remark 5.23. The fourth and final requirement can be proved to be aconsequence of the first three requirements, by utilizing Remark 5.10b-seethe definition of au given by (5.27) below.

Recall from the section in Chapter 2 on "The Parts of an Inner Prod-uct" that an H-hermitian symmetric inner product s has first and secondcomplex parts h and v defined by

(5.24) e=h+jo.

Further, with respect to the complex structure R;, h is a C-hermitiansymmetric inner product and o is a C-skew (complex symplectic) innerproduct. In particular,

(5.25) h = g + iwl

provides a natural Kahler structure on M, and

(5.26) 0'= wJ - iwg

92 Quaternionic Kahler Manifolds (of Signature p, q)

provides a complex symplectic structure (with almost-complex structureI). More generally, for each u E S2 C ImH, h = g + iwu provides aKahler structure on M (thus the name hyperKahler), while

(5.27) E = hu + uou

defines a complex symplectic structure ou (with complex structure Ru).Also note that

(5.28) c=g+iwi+jwj +kwk.

On a hyperKahler manifold, with the complex structure I chosen,i.e., distinguished, the complex hermitian structure h and the complexsymplectic structure o defined by (5.26) are compatible in the sense that

(5.29) h(xJ, y) = o(x, y)

and

(5.30) o(xJ, yJ) = o(x, y).

Conversely, suppose a complex hermitian structure h and a complexsymplectic structure o are given. Define a linear map J by requiring (5.29)to be valid. In Chapter 2, it was shown that (5.30) is valid if and only ifJ2 = -1 (see Equation (2.77)). Now once (5.29) and (5.30) are satisfiedone can define c to be h + jo and show that c is H-hermitian symmetric.This proves that the next definition of a hyperKahler manifold is equivalentto the earlier definition.

A hyperKahler manifold M is a Kahler manifold equipped with a com-plex symplectic structure o that is compatible with the hermitian structureh=g+iw.

QUATERNIONIC KAHLER MANIFOLDS(OF SIGNATURE p, q): THE GROUP HU(p, q) HU(1)

A quaternionic Kahler manifold of signature p, q is, by definition,

(a) an almost-quaternionic manifold (with coefficient bundle C),(b) equipped with a Riemannian structure of signature 4p, 4q, with each

coefficient u E Cs of unit length an orthogonal map,(c) such that the coefficient bundle C is parallel with respect to the rie-

mannian structure.


Recall by Problem 3.15 (a) that the quaternionic unitary group,IIU(p, q) HU(1) can be defined to be the subgroup of SO(4p, 4q) thatfixes H c--- C,.

By choosing a particular algebra isomorphism Cy = H between thefiber of the coefficient bundle C and the quaternions H, we may define thequaternionic 4-form 4D by

FE 6 (W1 -{- w -- wk) ,

where the two forms w,, with u E ImH are defined by

wu(x, y) _ (xu y) for all tangent vectors x, y,

using the Riemannian metric (, ).Since any other isomorphism CC = H differs by an element a E HU(1)

acting on H by Xa(x) = axa, Problem 3.15 (c) can be used to show thatdoes not depend on the choice of isomorphism Q, c--- H.

Thus, the quaternionic 4-form is a well-defined global 4-form on anyquaternionic Kahler manifold. Also note that is a nonzero volume form.Moreover, is parallel since the coefficient bundle is parallel.

Based on Problem 3.15, one can show that (n > 2) an almost-quater-nionic n-manifold equipped with a 4-form 4i, that (at each point) can beexpressed as

s(w; +w? +w,) for some choice of metric (, ), in fact, uniquely

determines this metric. If -1 is parallel, then the manifold is quaternionicKahler. In fact, for n > 3, the weaker condition d(k = 0 implies that P isparallel (see Salamon [16]). The case n = 2 is unknown.Note: Since HU(p, q) . HU(1) is the quaternionic analogue of the unitarygroup U(p, q), quaternionic Kahler geometry is the analogue of Kahler ge-ometry. (Similarly, since HU(p, q) is the quaternionic analogue of SU(p, q),hyper-Kahler geometry is the quaternionic analogue of special Kahler ge-ometry.) However, a quaternionic Kahler manifold need not be Kahlermanifold.

QUATERNIONIC SKEW HERMITIAN MANIFOLDS:THE GROUP SK(n, H)

An H-hermitian skew manifold M is a real 4n-manifold equipped withthe following extra structure. First, assume that the quaternions H act(smoothly) on the tangent bundle on the right giving each tangent spaceT.,M the structure of a right quaternion vector space (cf. Problem 7). Sec-ond, suppose c is an H-hermitian skew inner product on each tangent space

94 Coincidences of Geometries in Low Dimensions

(that varies smoothly). Third, require that the 2-form defined by

(5.31) w=Ree

be closed under exterior differentiation. Fourth, and last, require thateach of the almost-complex structures R,,, right multiplication by a unitimaginary quaternion, be, in fact, a complex structure on M.

For each u E S2 C Im H,

(5.32) g. (z, w) = Re -(zu, w) for all z, w E TxM

defines a split Riemannian structure on M. In fact, gu-uw is a C-hermitiansymmetric inner product on the complex manifold M, Ru with Kahler formw = Re e independent of u.

Note that

(5.33) E =w + igt -f jgg +kgk.

Consult the H-hermitian skew case in the section The Parts of an InnerProduct in Chapter 2.

COINCIDENCES OF GEOMETRIES IN LOW DIMENSIONS

In this section, we discuss the coincidences of certain geometries in lowdimensions. These coincidences can be labeled with group coincidences(cf. Proposition 1.40, see Problem 9).

Sp(1, R) 25 SL(2, R) and Sp(1, C) SL(2, C)

Suppose M is a real (respectively complex) manifold of real (complex)dimension 2. Then the notion of a real (complex) symplectic manifold Mis exactly the same as the notion of a real (complex) manifold with volume.That is, M must be equipped with a particular never-vanishing 2-form.

Suppose M is a complex manifold of complex dimension 2. Then thenotion of an almost-complex symplectic manifold (or equivalently a holo-morphic symplectic manifold-see Remark 5.10b) is exactly the same asthe notion of a complex surface with trivialized canonical bundle. That is,M must be a complex surface equipped with a never-vanishing holomorphic2, 0-form.


SO(2) - U(1) and CSO(2) - GL(1, C)

Theorem 5.34.(a) The notion of an oriented conformal manifold of real dimension 2 is

the same as the notion of a complex manifold of complex dimension 1.(b) The notion of an oriented Riemannian manifold of real dimension 2 is

the same as the notion of a Kahler manifold of complex dimension 1.

Proof: First, assume that M is a one-dimensional complex manifold withcomplex structure J. Then M is oriented (apply (1.26) to the linearizationof a holomorphic transition function). Let w denote a choice of orientationpreserving volume form. All other choices must be of the form qw where¢ is a smooth positive function on M. Each skew 2-form w determines aRiemannian structure g by

(5.35) g(x, y) - w(x, Jy) for all tangent vectors x, y.

Since Ow determines fig, the orientation class [w] {4w : 0 > 0} determinesa conformal structure [g] _ log : 0 > 0} on M.

If M is Kahler, then M is equipped with a particular w that, by (5.35),determines a particular Riemannian structure g on M.

Conversely, assume that M is equipped with an orientation class [w]and a conformal class [g]. Given a choice of Riemannian structure g E [g],renormalize w so that w is a unit volume form. Define a linear map J by

(5.36) w(x, Jy) = g(x, y) for all tangent vectors x, y.

If ¢g is any other representative of the conformal class [g], then ow is theunit volume form (because the dimension is 2) in the metric ¢,g. Thus, ¢gand g determine the same map J. To prove that J is an almost-complexstructure, i.e., j2 = -1, fix g and select an oriented orthonormal basisel, e2. Then w(el, e2) = 1 so that (5.36) implies Jel = e2,Je2 = -el.(Thus, J is counterclockwise rotation by 90°.) Therefore j2 = -1. It isa standard classical result that, for complex dimension one, each almost-complex manifold is a complex manifold. This result is a special case of theNewlander-Nirenberg Theorem. The reader is referred to the literature.

If M is an oriented Riemannian 2-manifold, then the metric g yields thealmost-complex structure J by (5.36). Note that in addition to the propertyJ2 = -1, J is an isometry with respect to both g and w. Therefore,

h_g - iw

is a complex hermitian inner product. Of course, dw = 0 since dw is a3-form on a 2-manifold. This proves that M is a Kahler manifold. U

96 Coincidences of Geometries in Low Dimensions

SO(4) = HU(1) HU(1) and CSO(4) = GL(1, H) H*

Theorem 5.37.

(a) The notion of an oriented conformal manifold of real dimension 4 isthe same as the notion of an (almost) quaternionic manifold of quater-nionic dimension 1.

(b) The notion of an oriented Riemannian manifold of real dimension 4 isthe same as the notion of a quaternionic Kahler manifold of quater-nionic dimension one.

Proof: Suppose M is an almost quaternionic manifold of quaternionic di-mension one. Let C C End(TM) denote the coefficient bundle (i.e., thealmost-quaternionic structure) so that, at each point x E M, Cx = H asreal algebras. This isomorphism is not canonical, but if fl : Cx -* H andf2 : Cx --* H are two such isomorphisms, then f2 o fi 1 E Aut(H) is anautomorphism of the algebra H. Later, in Chapter 6, in the context ofnormed algebras, it is easy to compute that Aut(H) - SO(ImH). Assum-ing this fact, it follows that the standard real inner product (,) and thestandard orientation (on H) induce a real inner product and orientationon C.

Now pick any nonzero tangent vector u. Then any other nonzerotangent vector.v is of the form Rau = v for a unique choice of Ra E C.Thus, we may define

(5.38) 9(V1, v2) = (Rai, Rae) = (al, a2),

where vj = R,,, u. The inner product depends on u, but if u is replacedby u = uA, then v = ua = ua)'1a = uA-la, so that a is replaced bya - A-1a. Thus, the new inner product g equals JAI-2 times the old innerproduct g, so that we have determined a conformal structure [g] on thetangent space. Similarly, the orientation {1, i, j, k} on H determines anorientation on the tangent space.

If, in addition, M is quaternionic Kiihler, then M is automaticallyequipped with a volume form ' =

s(w; + w + wk). Now, there exists a

unique choice of Riemannian structure g in the conformal class determinedby the quaternionic structure, so that the given 4-form is of unit norm.

Next suppose M is an oriented conformal manifold of real dimension4 with conformal structure [g]. Given a choice of g E [g], let 4D denotethe unit oriented volume form. The inner product g and the volumedetermine a linear map * on forms by

(5.39) a A*,3 = g(al3),


where g also denotes the inner product induced by g on forms. If g ontangent vectors is rescaled by A so that g = Ag, then g on forms of degreek rescales by g = A-2kg, so that the new unit volume form 1i = A41b.Therefore, if a, E A2 are of degree 2, g(a, Q)41 = g(a, Q)db is independentof the rescaling A. This proves that the * operator on A2 is uniquelydetermined by the oriented conformal structure on M. Using an orientedorthonormal basis e1, . . . , e4, it is easy to prove that * : A2 -> A2 squaresto 1, and that the ±1 eigenspaces are both 3-dimensional. Thus, A2 =A+ ® A- decomposes into the space of self-dual forms and the anti-self-dual forms. Each self-dual form a E A+ determines a linear map Ja on thetangent space by

(5.40) a(u, v) = g(J, u, v) for all tangent vectors u, V.

The coefficient bundle C C End(TM) is defined by CI = span 1 ®tJaa E A+} C End TIM. Note that if g = )g is a rescaling of g then Ja =A-'J,,,. Therefore, the bundle C only depends on the conformal structure[g] and not the choice of g E [g]. In terms of an oriented orthonormal basisa1, a2, a3, a4 for the cotangent space (based on a choice g E [g]), it is easyto compute that ,31 - a1 A a2 + a3 A a4, (32 = a1 A a3 - a2 A a4, and/33 =- a1 A a4 + a2 A a3 provide a basis for A+. The operators JJ11 Jp Jp,satisfy Jp1 = J 3 = J 3 = -1 and Jp1Jp2 = Jp etc., so that CI = Has real algebras. This proves that M is naturally an almost quaternionicmanifold. In this dimension n = 1, each almost-quaternionic manifold is aquaternionic manifold.

Now, if M is an oriented Riemannian manifold with metric g, theprevious discussion applies yielding an almost-quaternionic structure onM. In addition, the 4-form

(Qi + Q2 +,6,1)

is just the unit volume element and hence is independent of the choice oforiented orthonormal basis a1, a2, a3, a4. Combining the coefficient bundleC, the form -1), and the metric g provides the quaternionic Kahler structureon M.

Remark 5.41. The right H-structure determined by C C End(TM) deter-mines EndH(TM), the space of quaternionic linear maps. This subbundleEndH(TM) of End(TM) is exactly the bundle span {1} ® {Ja : a E A-}of operators obtained from the space of anti-self-dual 2-forms, i.e., A-.

98 Problems

SU(2) t--- HU(I)

Theorem 5.42.(a) Suppose M is a hyperKahler manifold of quaternion dimension one.

Then M is naturally a special Kahler (positive definite) manifold fora 2-sphere of different complex structures on M.

(b) Suppose M is a (positive definite) special Kahler surface. Then M isnaturally a hyperKahler manifold.

Proof:(a) Suppose M, e is a hyperKahler manifold of quaternionic dimension

one. For each complex structure R,, on M (with u E S2 C ImH a unitimaginary quaternion), M, h is a Kahler manifold with complex hermitianinner product h = g - uw and M, a,, is a complex holomorphic symplecticmanifold with complex symplectic form a,, defined by e = h + uo-,,. (SeeLemma 2.72.) Moreover, to is globally constant since it equals the normof dzldz2 on the standard model when u = i.

(b) Suppose M is a special Kahler surface with complex structure I,hermitian structure h = g + iwj, and unit holomorphic volume form a-.Define a linear map J on the tangent space at each point p by

(5.43) h(xJ, y) = Y (x, y) for all tangent vectors x, y.

Choose any isomorphism TpM = C2 and note that a = e'Bdzl A dz2. If yis replaced by a - dz' A dz2 in (5.43), then

(5.44) h(xJ, y) = A(x, y)

obviously defines a standard quaternionic structure I, J, K = IJ on C2.

Since or = e"), J must equal e'BJ. Therefore, J satisfies the sameidentities, namely IJ = -JI and j2 = -1, as J. This proves that I, Jprovides a quaternionic structure on the tangent space TpM.

Now one can show that e = h + ja defines an H-hermitian symmetricinner product on each tangent space.

PROBLEMS

1. (a) Prove that the two definitions of oriented (real) manifold given inthe text are equivalent.


(b) Prove that each complex n-manifold is automatically oriented asa real 2n-manifold.

2. Prove that each non-orientable Riemannian 2-manifold (of signature2, 0) is a "weak" almost-complex manifold as described in the text.

3. Suppose a real 2n-dimensional vector space V with complex structureJ is given. Let Vc = V OR C = V ® iV denote the complexificationof V. That is, Vc E {u + iv : u,v E V}. Vc is naturally equippedwith a conjugation that fixes V C Vc, i.e., u + iv = u - iv. DefineRe z =

z(z--z) and Imz = (z - x)/2i for all z E Vc. Extend J to be

complex linear on V. Let V1'0 denote the +i eigenspace of J and Vo,l,denote the -i eigenspace of Jon Vc, so that Vc = V1,o® Vo,l Let 9rdenote the projection of Vc onto V',o along V°'1. Show that V withcomplex structure J and V1-° with complex structure i are complexisomorphic via the map 7rlv and that 2R.e provides the inverse.

4. When V, J is taken as TM, the tangent space to a complex manifold,then utilizing the coordinate functions z1,.. ., z' with z - x + iy, areal basis for V is given by: aa, , ... , a=n , ay l , ... , aa . Show that(a) / l

11

, ... , OZn , where a = 2 ( - i ay

is a complex basis for Vl,o \(b) The isomorphism V, J = V1-1, i of complex vector spaces exhibitedin Problem 3 does not preserve the bracket [ , ] of vector fields.

5. Verify (5.5), i.e., Rya =Rkt=k

1.

6. Prove that 0 =_ da, where a is a 1-form defined by a(V) E ¢(a*V) ateach point Or E T*M, provides a symplectic structure on the cotan-gent bundle T*M to a manifold M.

7. Suppose M is an almost-quaternionic manifold equipped with an H-hermitian inner product a that is either symmetric or skew. Provethat the coefficient bundle C C End(TM) is trivial, i.e., exhibit r, J, Kglobally on M.

8. (Lagrangian submanifolds) Consider the standard model RZn, w of asymplectic vector space as a symplectic manifold, where

w=_dxl

Suppose M is the graph of a smooth function y = f (x) over anopen simply connected subset U of Rn, i.e., f : U -* R' and M{(x, f (x)) E R2n : x E U). Prove that the following are equivalent.

100 Problems

(a) f = V4 for some scalar-valued function 0 defined on U.(b) The 1-form E fi(x)dx' is d-closed on U.(c) The 1-form a - E y'dx' on R21 restricts to a d-closed form on M.(d) The 2-form w vanishes when restricted to M (i.e., M is an isotropicor totally null submanifold of R 2 ' , ).

In particular, note that 0 satisfies a nonlinear partial differential equa-tion of the form P(x, VO(x)) = 0 if and only if both of the followinggeometric conditions on the graph M are satisfied:

(1) M is contained in {(x, y) E U x R" : P(x, y) = 0}, and(2) M is isotropic.

9. (a) Extract a sublist of the list of group isomorphisms provided byProposition 1.40 with the stronger property that the correspondinglowest dimensional representations are isomorphic. For example,SO(2) acting on R(2) and U(1) acting on C(1) are isomorphic rep-resentations, but the representation Sk(1) on H is different, so thatSO(2) L, U(1) should be included on the new list but SO(2) - Sk(1)should be deleted.(b) Compare this new list with the list of low dimensional coincidencesof geometries presented in this chapter.

10. Prove Moser's Theorem on compact manifolds with volume.Hint:(a) Show 1' = 11 + da for some global (n - 1) form a.(b) Define a vector field V by a = V .i 0. Make use of the correspond-ing flow Ot and the beautiful formula

Lv (a) = d(V _j a) + V _j da

(valid for any form a), where Lv denotes the Lie derivative with re-spect to the vector field V.

6. Normed Algebras

Suppose V, (, ) is a euclidean vector space of signature p, q. The normof a vector Ivi . fj(v v)I is always positive. In this section, we shall dealwith the square norm or quadratic form

(6.1) lixll = (x,x).

By a normed or euclidean algebra, we mean a (not necessarily associative)finite dimensional algebra over R with multiplicative unit 1, and equippedwith an inner product (, ) of general signature whose associated squarenorm I I I I satisfies the multiplicative property

(6.2) lixyil = Ilxll Ilyll for all x, y.

The inner product (x, y) can be determined from the square norm lullby polarization. That is, replace z by x + y in the formula ilzil = (z, z) andobtain

Ilx+yll = (x+y,x+y) = Ilxll+2(x,y)+Ilyll

or

2(x, y) = lix + yli - Ilxii - 11Y11-

Similarly, the identity Iizwjj = ilzli Ilwil can be polarized.

101

102 Normed Algebras

Lemma 6.3. The following identities are all equivalent:

(6.2) Iixyll = IIxII IIyII

(6.4) (xw) yw) = (x, y)IIwII

(6.5) (wx, wy) = IIwII(x, y)

(6.6) (xz, yw) + (yz, xw) = 2(x, y) (z, w)

Note that the final identity is linear in all of the variables x, y, z, andw.

Proof: Setting z = w in the last identity yields (6.4), and one obtains(6.5) similarly. Setting x = y in (6.4) yields the multiplicative identity(6.2). Thus, it remains to deduce all the identities from the multiplicativeproperty lixyll = IIxII Ilyll via polarization.

Replace x by x + y and y by w in (6.2). This proves that

II(x + y)wII = ((x + y)w, (x + y)tv) = (xw, xw) + 2(xw, yw) + (yw, yw)

= IIxwII + Ilywll + 2(xw, yw) = IIxII IiwII + IIyII IIwII+ 2(xw, yw)

must equalIx + yll IiwII = (IIxII + 2(x, y) + IIyII)IIwII,

yielding (6.4). The proof of (6.5) is similar.The identity (6.4) is quadratic in w and hence may be further polarized

by replacing w by z + w. Thus,

(x(z + w), y(z + w)) = (xz, yz) + (xz, yw) + (xw, yz) + (xw, yw)

=I 1Z11(x, y) + IIwlI(x, y) + (xz, yw) + (xw, yz)

must equal

1k+wII(x,y) = I1zII(x,y)+2(z,w)(x,y)+IIwII(x,y),

yielding (6.6).In summary, the defining property Ilxyll = IIxII IIyII, which is quadratic

in x and y, may be re-expressed as an identity, (6.6), linear in x, y, z, andw. L

Given an algebra, let Rw denote the linear operator right multiplica-tion by w. Similarly, let Lw denote left multiplication by w. The identity(6.4) may be rewritten

(6.4') (Rwx, Rwy) = Ilwll(x, y)

Normed Algebras 103

That is, the key axiom for a normed algebra may be stated as:

Right multiplication by w is conformal with conformal factor IIwII.

Similarly,

(6.5') (L. x, Lwy) = IIwII(x,y)In terms of the operators RZ and R,,, the fully polarized identity (6.6)becomes

(6.6') (R-- x, Rwy) + (R. x, R. y) = 2(x, y)(z, w)

Note: If IIwII = -1, then R,,, V O(A).

Interchanging x with z and y with w in (6.6) yields

(6.6") (L,2x, Lwy) + (Lwx, Lzy) = 2(x, y)(z, w)

We will adopt the following notational conventions. Let Re V denotethe span of 1 E V. Since IIxII = III . xII = 11111 IIxII, the multiplicativeidentity 1 cannot be null. Let Im V denote the orthogonal compliment ofRe V. Then, by Lemma 2.30, Im V is a nondegenerate hyperplane, andeach x E V has a unique orthogonal decomposition:

x=x1+x', with xiEReV, x'EImV.

Occasionally, we let Rex denote x1 and Imx denote x'. Conjugation isdefined by

(6.7) x = xl - x'.

Thus

xlRex=(6.8)

x'Imx=The adjoints of R,, and L. are

(6.9) Ra, = R. , L*, = Lw.

To prove (6.9), first note that these identities are linear in w, and obviouslytrue when w E Re V. Thus, we may assume w 1 Re V or w E Im V. Nowsetting z = 1 in (6.6') yields

(x, R. y) + (Rwx, y) = 0

or R*, _ -R, = Rw. Similarly, Lv, = L.The elementary facts concerning conjugation are contained in the next

lemma.

104 The Cayley-Dickson Process

Lemma 6.10.

(a) -'X= = x and (x, y) _ (x, y)(b) (x, y) = Re xy = Re xy.(c) xy = y T.(d) xx = xx = JIxJJ.

Proof: (a) is true because conjugation is reflection through a hyperplane.Part (b) is a special case of Ry = Ry. Part (c) is true because

(xy, z) = (xy, z) = (y, x x) _ (yz, x) = (z' y x)

Part (d) is now an easy direct calculation.

The associator

[x, y, z] - (xy)z - x(yz)

measures the lack of associativity in an algebra. For a normed algebra,there is always a weak form of associativity, namely

Lemma 6.11. The associator [x, y, z] vanishes if any two of the argumentsare set equal, or equivalently, the trilinear form [x, y, z] is alternating.

An algebra on which the associator is alternating is called alternative.Lemma 6.11 states that any normed algebra is alternative.

Proof: Note that the associator vanishes if one of its variables is real.Hence, it suffices to show that the associator vanishes when two of itsvariables are set equal to w E Im V pure imaginary. Note, we show that[x, w,] = 0. Since ww = JJwJJ, [x, w, w] = (xw)w-xJJwJJ. Since Rw = R,and R. is conformal,

((xw)w, z) = (xw, zw) = IIwII(x, z),

which proves [x, w, w] = 0. Similarly, [w, w, z] = 0. Since [x, w, w] = 0,polarization yields [x, y, z] _ -[x, z, y]. Therefore [w, y, w] -[w, w, y]0.

THE CAYLEY-DICKSON PROCESS

Using the facts that Lw = Lo- and Rw = Ru, (as well as Lemma 6.10a), theidentities (6.6') and (6.6") can be written as

Normed Algebras 105

Lemma 6.12.x(yw) + y(7w) = 2(x, y)w,

(wy)x + (wx)y = 2(x, y)w.

In particular, we obtain the key identities that motivate the Cayley-Dickson process:

Corollary 6.13. If x 1 y, then xy = -yx and

(6.14) x(yw) = -y(xw) and (w'y)x = -(wx)y, for all w.

These crucial equations enable us to interchange a pair of orthogonalvectors x and y.

Lemma 6.15. Suppose A is a normed subalgebra (with 1 E A) of thenormed algebra B and that E E A-' is a unit vector orthogonal to A with11611 = ±1. Then Ae is orthogonal to A and

(6.16) (a + bE)(c + dE) = (ac+ db) + (da + bc)E for all a, b, c, d E A.

Note that if IIEII = 1, then -db occurs, while if Ilel I = -1, then +db occurs.

Proof: First, we prove A J. Ae. Since 1 E A, x E A if and only if F E A.Thus, if a, b E A, then (a, be) = (ba, e) = 0 because ba E A.

Note that e2 = T1 if and only if IIehl = ±1 since e 1 1 is pure imaginary(i.e., e = -e and eE = IIEII implies e2 = -ee = -IIEII).

Finally, to prove (6.16) we use the key identities (6.14) several times,applied to the terms in

(a + be)(c + de) = ac + (be)(de)-I-a(de) + (be)c.

That is,

(be)(de) = -d((be e) = d((eb)E) = -d(& )b) = -IIeIIdb

a(de) = a(ed) = e(ad) = (ad) e = (da)E(be)c = (b'c)E.

Lemma 6.15 mandates the following doubling process.

Definition 6.17 (Cayley-Dickson). Suppose A is a nonmed algebra.Motivated by Lemma 6.15, we define two algebras A(+) and A(-). Asvector spaces both

A(±) = A ®A.

106 The Cayley-Dickson Process

Multiplication is defined by

(6.18) (a, b)(c, d) _ (ac + db, da + bc),

with the coefficient of db taken to be -1 for A(+) and +1 for A(-).

Both A(+) and (A-) are easily seen to be algebras with multiplicativeunit 1 = (1, 0). The algebra A is naturally a subalgebra of A(±). In fact,given a E A, we also let a denote (a, 0), embedding A as a subalgebra ofboth A(+) and A(-). Let e denote (0, 1). Then

(a, b) = a + be.

Note that e2 = -1 for A(+) and e2 = +1 for A(-).There is a natural square norm or quadratic form to impose on A(+)

and on A(-). If (a, b) = a + be E A(±), let

(6.19) II(a, b)II = IIail + IIbii.

Note that IIeII = 1 for A(+) and IIell = -1 for A(-).The associated inner product is given by

(6.19') (x, y) _ (a, c) f (b, c),

if x = a + be and y = c + de. Conjugation is defined by (a, b) = (d, -b)or a + be = i - be. Real and imaginary parts are defined in terms ofconjugation or equivalently in terms of the orthogonal splitting of A(±)into R and RJ-.

Next, we collect together some of the formulas valid in A(+). Theproofs are straightforward calculations using the definition (6.18) of multi-plication.

Lemma 6.20. Suppose A is a normed algebra. Let A(±) denote thealgebra defined via the Cayley-Dickson process. Suppose x = a + ae, y =b+/3e, z=c+yeEA(+).

(6.21) xy = y T.(6.22) xx = xx = II xII

(6.23) (xy + yx) = Rexy = (x, y).

(6.24)

2

[x, y] =

2

[a, b] ± Im aQ + (Q Im a - a Im b) c.

Normed Algebras 107

(6.25)[x, y, z] = ± [a, TO] ± [b, ay] ± [c, 8a]

+ a[b, C]e + Q[a, 4e + y[a, b]e ± (ofy - yQa)e,

assuming A is associative.

(6.26)[x, g, y] = ±[a, Q, a,] + [a, b, ale.

jlxii Ilyil - Iixyll = ±2(a, (Q, a, Ll) -

Corollary 6.27. Suppose A(+) is the algebra defined via the Cayley-Dickson process from a normed algebra A.

(6.28) A(±) is commutative if and only if A = R.

(6.29)

(6.30)

A(±) is associative if and only if A iscommutative and associative.

A(±) is alternative, A(±) is normed, andA is associative are all equivalent.

THE HURWITZ THEOREM

Corollary 6.27 can be used to deduce the properties of the euclidean al-gebras obtained via the Cayley-Dickson process. First, we list (some of)these algebras.

Definition 6.31.

C = R(+)

H = C(+)

0=H(+)L = R(-)

M2(R) = C(-)

0 = H(-)

the complex numbers.the quaternions or Hamiltonians.the octonians or Cayley numbers.the Lorentz numbers.real 2 x 2 matrices.

the split octonians.

Remark. The normed algebra M2(R) of real 2 x 2 matrices (with IIAIIdetp. A defined to be the square norm or quadratic form) occurs in severaldifferent ways via the Cayley-Dickson process. By Corollary 6.38(a) below,

M2(R) = C(-) = L(+) = L(-).

108 The Hurwitz Theorem

To show M2(R) = C(-), identify C with the 2 x 2 reals of the form

C b ab

and choose

= a + ib

-1

to be the matrix corresponding to conjugation on R2 = C. Check that1 C and E2 = 1. Then by Lemma 6.15, C(-) and M2(R) are algebra

and norm isomorphic.

Corollary 6.32.

(6.33)C =R(+) and L = R(-) are commutative, associative,

and normed.

(6.34)H =C(+) and M2(R) are not commutative but associative

and normed.

(6.35)O =H(+) and O = H(-) are neither commutative

nor associative but are alternative and normed.

Proof: Apply Corollary 6.27.

Remark 6.36. O(+) and O(-) also fail to be alternative or normed.

Theorem 6.37 (Hurwitz). The only normed or euclidean algebras overR are

R, C and L, H and M2(R), 0 and O.

Proof: Suppose B is a normed algebra. Let Al =- Re B = R. If Al = B,we are finished. If not, choose El E Ai a unit vector, i.e., -Ei = I IEl I I = +1Let A2 = Al + Al 1. By Lemma 6.15, A2 is a normed subalgebra of Bisomorphic to either C = R(+) or L = R(-). If A2 = B, we are finished.Suppose not. Choose E2 E AZ a unit vector, and let A3 =- A2 + A2E2.

Case R(+): Suppose A2 = C = R(+). Then by Lemma 6.15, A3 is anormed subalgebra isomorphic to H = C(+) or M2(R) = C(-).Case R(-): Suppose A2 = L = R(-). Then E2 = 1, E2 = ±1, andE3 . 1e2 satisfies E3 = f1 since 6162 + E2E1 = In particular,

Normed Algebras 109

either e2 = -1 or 2 = -1. Say e3 = -1. Then interchange el and e3 (notethis does not change A3) in the above constructions so we are back in CaseR(+). Thus, either A3 = C(+) or A3 = C(-). If A3 = B, we are finished.Suppose not. Choose e3 E A3 a unit vector, and let A4 =- A3 + A3e3.Case C(+): If A3 = C(+) = H, then by Lemma 6.15 A4 is a normedalgebra isomorphic to either 0 = H(+) or 0 = H(-).Case C(-): If A3 = C(-) = M2(R), then interchanging either e3 ore4 - e2e3 with 62 reduces us to Case C(+). If A4 = B, we are finished.If A4 # B, then by repeating the above process one more time we obtaineither 0(+) or 0(-) as a normed subalgebra A5 of B. Since 0(+) andO(-) are not normed algebras, this is impossible. J

Corollary 6.38.(a) As normed algebras M2(R) = R(-, -) = R(+, -) = R(-, +).(b) The seven normed algebras R(f, f, f), excluding R(+, +, +) = 0,

are all the same, namely O.

Remark. Because of the Hurwitz Theorem, all normed algebras are sub-algebras of either 0 or O. Consequently, it is frequently clearer to stateresults that are valid for all normed algebras as results for 0 and O. Thenext theorem is an example, as are "cross products" in the next section.

The weak form of associativity for 0 and O, corresponding to the factthat the associator vanishes if any two of the three variables are set equal,can be strengthened.

Theorem 6.39 (Artin). A subalgebra with unit, generated by any twoelements of either 0 or 0, is associative.

Proof: Suppose A is generated by 1, x, y. Let S =_ span{Imx, Im y}.If dimS = 0, then A = R and the proof is complete. If dimS = 1,choose a nonzero vector e1 E S and note that since 62 = -IIc1Il is real,A = {a + bet : a, b E R}. Now it is easy to verify directly that A isassociative. If dim S = 2, choose an orthogonal basis 61, 62 for S and notethat by repeated use of Corollary 6.13 as in the proof of Lemma 6.15, Ais spanned by (as a vector space) 1, e1 i e2, and El 62. Now using the factsthat [e1, e2, e162], [el, El, e21, etc. vanish, we see that A is associative.

Actually we can prove much more.

Proposition 6.40. If S = span {Imx, Im y) is nondegenerate, then thesubalgebra A is determined as follows:

(6.41) If dimS = 1, and S is positive, then A = C.

110 Cross Products

(6.42) If dim S = 1, and S is negative, then A = L.(6.43) If dim S = 2, and S is positive, then A = H.

If dim S = 2, and S is negative or Lorentz,(6.44) then A = M2(R).

Further, if dimS = 1 and S is null, then

l(6.45) A= { l 0 6) :a,bER}.

Proof: If dim S = 1 and S is nondegenerate, choose E1 E S to be a unitvector. Then by Lemma 6.15, A = R+Rrl is either C or L depending onII-'111.

Suppose S is degenerate of dimension one. Choose a nonzero nullvector u in S and a pure imaginary unit vector e with (u, E) 1 0. Thenspan{,-, u} is a Lorentz plane = L containing S as a null line. Therefore,we may choose an orthonormal basis El, E2 for L with El + E2 E S. Thealgebra generated by El, E2 is isomorphic to M2(R) with

E1

Therefore,

E2 =

S2{(0 ):bER}and

A{ (a b) :a,bER}.

Now assume dim S = 2. If S is nondegenerate, then S has an or-thonormal basis 61,E2, and by the above arguments and Lemma 6.15, ei-ther A = L + Lee or A = C + Cr2. It follows that either A = M2(R) orA=H.

CROSS PRODUCTS

The results of this section are formulated for the octonians O. However,all of the results are valid with 0 replaced by the split octonians O. First,we consider the cross product of two octonians.

Normed Algebras 111

Definition 6.46. Given x, y E 0, the cross product of x and y is definedby

(6.47) xxy=2(yx-xy)=Imyr.

The next lemma justifies the term "cross product".

Lemma 6.48.

(a) x x y is alternating on 0.(b) Ilxxyll=lJxAyllforallx,yE0.

Proof:(a) The cross product is alternating because x x x = 0 for all x E 0.(b) Since both sides are alternating, it suffices to prove (b) when x 1 y.

Then 0 = (x, y) = Re iy = (xy + yx) so that x x y = yx. Therefore,IIx

X

yII = Ilyxll = IIyII llxll = IIx A yll because of the Cauchy-Schwarzequality.

Remark. The proof that Ilx x yll = IIx A yll explains why we needed theconjugates in the definition of x x y; namely, we wanted the vanishing of(x, y) to imply that x x y could be expressed as one term yx.

Note that

(6.49) Re x x y= O for all x, y E 0.

Frequently, the cross product is restricted to Im 0, in which case x x y canbe expressed in a variety of interesting ways.

Lemma 6.50. If x, y E Im O, then

(6.51) xxy = 2[x,y] = xy+(y,x),

where [x, y] = xy - yx is the commutator of x and y.

Proof: (x, y) = -2

(xy + yx) for X, y E Im O. i

Remark 6.52. If x, y E R3 =_ Im H C Im O, then the cross product x x yis just the usual vector cross product on R3 based on the "right hand rule."This explains the choice of sign in the definition of x x y.

112 Cross Products

Lemma 6.53. If x, y E Im O, then

(a) x x y E Im O is orthogonal to span{x, y}.

(b) x x (x x y) = -IIxIIy + (x, y)x.

Part (b) says that (for IIxII = 1) the square of left cross product by xequals minus orthogonal projection from Im 0 to (span x)-L. See Problem13, where this propety is used to reconstruct octonian multiplication fromthe cross product on ImO.

Proof:(a) (x, x x y) =

2(x, xy) -

a(x, yx) = 2IIxII(1, y) -

aIIxII(1, y) = 0.

(b) Let Cs y = x x y. If y = x, then CC = 0. If Y I x, then xy = -yx,so that Cxy = -IIxIIy

The natural extension of the cross product to three octonians x, y, z isthe alternation of x(yz). (Note that this is not equal to x x (y x z)!) How-ever, by repeated use of Corollary 6.13, the six-term expression obtainedby alternating x(17z) can be simplified to two terms. Therefore, we adoptas our definition of x x y x z the following two-term expression.

Definition 6.54. Given x, y, z E 0, the triple cross product of x, y, and zis defined by

(6.55) xxyxz = 2 1--07z) - z(-9x)] .

Lemma 6.56.

(a) x x y x z is alternating on O.(b) IlxxyXZII=IxnyAz1Iforallx,y,zEO.

Proof: (a) Since the subalgebra generated by any two elements is associa-tive,

x x y x z= 2 [x(x'z) - zIIxII] = 0,

and

z x x x x= 1 [zIIxII-x(xz)] =0.

Obviously, xxzxx=0.(b) Since both sides are alternating, we may assume that x, y, and z

are pairwise orthogonal. Repeated use of the key identities in Corollary6.13 yield -z(yx) = x(yz). That is,

(6.57) x x y x z = x(yz) if x, y, z are orthogonal.

Normed Algebras 113

Therefore,

(6.58) jjx x y x zjj = llx(yz)ll = lixil IlyII ilzi _ lix A y A zil. _1

Now consider the real-valued trilinear form <k on Im 0 defined by

(6.59) ci(x, y, z) = (x, yz) for all x, y, z E Im O.

This form is called the associative 3-form for O. The terminology is ex-plained in the next chapter on calibrations. Since 0 vanishes when any twoof the variables x, y, z E Im 0 are set equal, 0 must be alternating. Thatis,

(6.60) 4 E A3(Im O)*.

Perhaps Gureirch [8] was the first to consider this 3-form (in coordinatesas in (6.74)).

It is natural to consider the triple cross product restricted to Im O.

Lemma 6.61. If x, y, z E Im O, then

(6.62) Rexxyxz=qS(xAyAz),

(6.63) Imxxyxz= 1[x,y,z].

Proof: It suffices to prove the lemma when x, y, z are orthogonal. Inthis case, x x y x z = x(Vz) = -x(yz) by (6.57). Thus Rex x y x z =-(1, x(yz)) = (x,yz) = 4'(xAyAz). Also, x x y x z = -z x y x x = z(yx) by(6.57). Thus F x -y x z = (xy)z. Consequently, Imx x y x z = 2[-x(yz) +(xy)z] = 2 [x, y, z].

THE EXCEPTIONAL LIE GROUP G2

Some of the automorphism groups of the normed algebras are exceptional.Let

(6.64) Aut(A) _ {g E GL(A) : g(xy) = g(x)g(y) for all x, y E A}

denote the automorphism group of a finite dimensional algebra A. Thegroup G2 is most naturally defined as the automorphism group of the oc-tonians

(6.65) G2 Aut (0),(6.66) GZ Aut(O) (the split case).

Before examining G2 and G2, we discuss the automorphism groups ofthe other normed algebras besides 0 and O. These other automorphismgroups will be seen to be orthogonal groups or special orthogonal groups.

114 The Exceptional Lie Group G2

Lemma 6.67. If A is a normed algebra, then Aut(A) C O(ImA) theothogonal group of Im A.

Proof: Let g E Aut(A). First note that g(1) = 1 since g(x) = g(1)g(x)for all x E A. Second, since x2 = (Re x)2 + (IM X)2 + 2(Re x)(Im x)and (IM X)2 = -IJIm xJJ E ReA, it follows that x2 E Re A if and onlyif x is either real or pure imaginary. Thus, if x E ImA, then g(x)2 =g(x2) = x2g(1) E Re A, so that g(x) is either real or pure imaginary. Thus,Aut(A) C GL(ImA) and g(x) = g(x). Therefore,

11g(x)11 = g(x)g(x) = g(x)g(x) = g(x9) = g(IIxIj) = 11x11g(1) _ 114-

This proves g E O(ImA).

Note that this proof also shows that Aut(O(+)) C O(O(+)).

Corollary 6.68. Aut(C) --- Z2 and Aut(L) = Z2 each Z2 consist of theidentity and conjugation.

Lemma 6.69. If g E Aut(A), then g fixes the associative 3-form 0 for A.

Proof: Here O(x, y, z) - (x, yz) for all x, y, z E ImA. Suppose g E Aut(A).Then

(g(x),g(y)g(z)) = (g(x),g(yz)) = (x,yz)

since g is orthogonal.

If A - H, then the associative form 0 is the unit volume form onImH corresponding to the orientation determined by {i, j, k}. Thus, ifg E Aut(H) C (Im H), then, by Lemma 6.69, det g = 1.

Let

C = ( 10 -0) (conjugation),

J= ` 1 I (complex structure),

and

R == (° 1) (reflection or Lorentz structure)

denote the standard orthonormal basis for ImM2(R). Then 0 (the asso-ciative form) is the unique unit volume form determined by the orientation{C, J, R}, since cb(C A J A R) = (C,JR) = -(C,C) = C2 = 1. Thus, ifg E Aut(M2(R)) C O(ImM2(R)), then detg = 1, because of Lemma 6.69.

Normed Algebras 115

Proposition 6.70.

Aut(H) = SO(ImR) = SO(3, 0).Aut(M2(R)) = SO(ImM2(R)) = SO(1,2).

Proof: See Problem 6.7. J

This completes our discussion of the automorphism groups of thenormed algebras except for G2 and G2. Sometimes, in order to avoidrepetition, results will only be stated for G2. However, these results will bestated in such a way that no modifications will be necessary for G2 otherthan replacing 0 by O.

The 4-form 0 on Im 0 defined by

1(6.71) i,b(x, y, z, w) =

2(x, y(zw) - w(zy)) for all x, y, z, w E Im 0

is called the coassociative 4-form for O. If any two of the four variablesx, y, z, w E Im O are set equal, then O(x, y, z, w) = 0. (See Problem 6.8).Thus

(6.72) 0 E A4(Im O)*

is a skew tensor of degree 4.

Lemma 6.73. If g E Aut(O), then g fixes the coassociative 4-form 0 EA4(Im 0)*.

Proof: Apply the facts that g E O(Im O), g E Aut(O), and g(x) = g(x)to the definition of b.

One can choose an orthonormal basis e1,.. . , e7 for Im O, with dualbasis W1, ... , W7 for (Im O)*, so that direct computation yields

(6.74) O = W123 - W156 - W426 - W453 - W147 - W257 - W367and

(6.75) O = W4$67 - W4237 - W1537 - W1267 - W2536 - W1436 - W1425,

where wijk = wi Awj A Wk, etc..Therefore,


Lemma 6.76. 0 A 0 = 7C'1234567

Corollary 6.77.(a) G2 C SO(Im O) = SO(7),(b) G2 C SO(Im O) = SO(3,4).

Proof: g E G2 = Aut(O) has det g = 1 since g fixes 0 and 0 and 0 A 0 isa nonzero volume form. The proof for G2 is analogous. J

Remark 6.78. The unit volume element ?4,A1' provides an orientation forIm O. Using this orientation the Hodge star operator * : A Im O -} A Im Ois well-defined. One can easily check that

(6.79) 0 _ 4Similarly, 0 = *0 in the split case O.

As noted above, if g E G2, then g*4, = 0. The converse is moredifficult.

Theorem 6.80 (Bryant).

(6.81) G2 = {gEGL(ImO):g'O=0}.

Proof: Assume for the moment that

(6.82) if g E GL(Im0) satisfies g' then g E O(Im0).

Then, for all x, y, z E Im O,

(g(x), g(y x z)) = (x, y x z) = 4,(x A y A z) _ (9*4,)(x A y A z)

= 4,(9(x) A g(y) A 9(z)) = (g(x),9(y) x g(z)).

This proves that g preserves the cross product:

(6.83) g(y x z) = g(y) x g(z) for all y, z E Im O.

Therefore, by Problem 11, g E Aut(O) - G2. It remains to prove (6.82).The identity (Problem 14 (c))

(6.84) (x _j 0) A (x _j 0) A ¢ = 6IIxIIa,

where A is the standard volume element for Im O, can be polarized to yield

(6.85) (x j 4,)A (y j 4,)A 4, = 6(x, y)A for all x, y E Im O.

Normed Algebras 117

Now suppose g E GL(Im 0) satisfies g*4, Applying g* to (6.85)yields

(g-'(X) i 0) A (9-1(y) .j 0) A 0 = 6(x, y) (det g).\,

while replacing x by g-1(x) and y by g-1(y) in (6.85) yields

(9-1(x) _j 0) A (g-1(y) i 0) A 0= (9-1(x), 9-1(y))A.

Therefore,

(6.86) (g(x), g(y)) = (det g)-1 (x, y) for all x, y E Im O.

Therefore, g is either conformal (det g > 0) or anticonformal (det g < 0).Since the signature of 0 is not split, g cannot be anticonformal (see

Problem 2.3). Thus, g is conformal with conformal factor A = (det g)-1 >0. Recall (Problem 2.9(d)) that if g is a conformal transformation withconformal factor A, then

(6.87) (detg)2 = a",

where n is the dimension. Substituting .\ = (det g)-1 and n = 7, we obtain

(detg)9 = 1 or det g = 1.

Thus, we have shown that

(6.88) if g E GL(Im O) satisfies g*qs = ¢, then g E SO(Im O),

completing the proof of Theorem 6.80.

This theorem describes G2 as a level set of the function from End(Im 0)to A3(ImO)* that maps A E End(Im O) to A*4, E A3(Im O)*. The lin-earization of this map at the identity I E End(Im O) is the map sendingA E End(Im O) to

Now

D(A) = dt (I + tA)*Ol t=o

D(A)(x A y A z) = dt 4,((x + tAx) A (y + tAy) A (z + tAz))(6.89) LO

_ qS((Ax) A y A z) + q5(x A (Ay) A z) + 4,(x A y A (Az)).

Thus, D(A) is just the action of A on 0 E A3(Im O)*, considering A as aderivation.


Lemma 6.90. The linear map (6.89),

D : End(Im O) --+ A3(Im O)*,

is surjective with 14-dimensional kernel.

Because of the implicit function theorem, we have several interetingconsequences.

Corollary 6.91. G2 is a closed 14-dimensional submanifold of End(Im 0)implicitly defined by G2 = {A E End(ImO) : A*q = 0). The tangentspace g2 to G2 at the identity is ker D.

Corollary 6.92. The orbit of 0 under GL(ImO) is an open subset ofA3(Im O)*.

Proof of Lemma 6.90: Since dime. End(Im 0) = 49 and dim A3(Im O)*= 35, the map D must have a kernel of at least 14 dimensions; and thekernel is exactly 14-dimensional if and only if D is surjective. Therefore, itsuffices to prove that

(6.93) dim ker D < 14.

First, we derive another formula for D(A) in the special case whereA = u ®a E (Im O) ®(Im O)* - End(Im O) is simple, i.e., Ax = a(x)u,where a E (Im 0)*, u E Im O.

D(a(9 u)(xAyAz)=a(x).(uAyAz)+a(y)O(x A U Az) +a(z)O(x AyAz)

= (u -j ')(a(x)y A z - a(y)x A z + a(z)x A y)

= (a A (u j qS))(x A y A z).

That is,

(6.94) D(u ®a) = a A (u q).

Consider the linear map

B : A3(Im O)* -+ S2 (Im O)*

defined by

(6.95) *B(,3) (x, y) _ (x -1 0) A (y -j 0) A,3 for x, y E Im 0 .

Normed Algebras 119

Assume, for the moment, that

(6.96) BD(A) = A + A* + trace A

has been verified. Then, if A E ker D,

(6.97) A + A* + trace A = 0.

Taking the trace, 2 trace A + 7 trace A = 0, so that

(6.98)

That is,

(6.99)

A+A* = 0 if A E kerb.

ker D C Skew(Im O).

Let L {Au : u E ImO), where

forallxEImO.

Then L is a 7-dimensional subspace of the 21-dimensional vector spaceSkew(ImO). It is straightforward to verify that

(6.100) A Y A z) 2 (u, [x, y, z])

(see Problem 6.15). The associator [x, y, z] takes on all values in Im O.Therefore, D(A,,) = 0 implies u = 0. That is,

(ker D) n L = {0}.

To complete the proof of the Lemma 6.90, the identity (6.97) must beverified. It suffices to assume A = u ® a is simple. Then

1 BD(u 0 a) = 2 B(a A (u]qf),

applied to x x E Im 0 is equal to

(6.101) 2 (x j 0) A (x J 0) A a A (u J ).

However, A + A* + trace A applied to x x E Im 0 is equal to

(6.102) a(u)jjxjj + 2a(x)(u,x).

120 Problems

Finally, (6.101) and (6.102) are equal; see Problem 6.14(b). J

PROBLEMS

1. Suppose A is an algebra with an inner product. Assume 1 is not null,and use this fact to define conjugation. Show that if A has the twoproperties

(a) IIxIJ = x x, xy = y(b) any subalgebra generated by two elements is associative,then A is normed.

2. Verify the Moufang identities:

(xyx)z = x(y(xz)) (Lxyx = LZLYLr)z(xyx) _ ((zx)y)x (Rxyx = R.RyR.)

(xy)(zx) = x(yz)x

for all x, y, z in 0 or O.Hint: First show that the difference of the two sides vanishes if anytwo of the variables are equal.

3. Suppose A is a normed algebra.(a) Show that each nonnull element of A has a unique left and rightinverse.

(b) Given x, y E A with IIxII 0, show that the equations xw = y andwx = y can be uniquely solved for w with w = Yy/JJxJJ and w = y-x/llxllrespectively.Note that (a) does not automatically imply (b), since (a) is true forO(+) but not (b).

4. A complex algebra with unit, equipped with a nondegenerate complexsymmetric bilinear form (, ) satisfying JJxyJJ = IlxII Ilyll, where JJxJJ _(x, x), is called a complex nonmed algebra.(a) Suppose A, (,) is a normed algebra. Show that A ®R C is acomplex normed algebra, where (,) is extended from A to A OR C tobe complex bilinear.(b) Consider the following four complex normed algebras:

C, C ®C, M2(C), O OF, C.

Show that each complex normed algebra is isomorphic and isometricto one of these four.

Normed Algebras 121

5. Assuming that each prime number p can be written as the sum of foursquares, p = ni + n2 + n3 + n4 (n I, n2, n3, n4 E N), show that eachnatural number n E N can be written as the sum of four squares.

6. Show that, for all x, y, z E 0,

(a) x x y = 1[x, y] - (xiy' - yix'),(b) x x y x z = (x', y'z') +

2[x, y, z] +

zxl[z, y] +

2yi [x, z] +

2zl [y, x]

7. Use the Cayley-Dickson process to complete the proof of Proposition6.70.

8. Show that the coassociative form , defined by (6.71), is skew.

9. (a) Let Ss denote the unit sphere in Im O. Show that G2 acts transi-tively on S6 with isotropy subgroup SU3:

G2/SU3 = S6 C IM O.

(b) Let V7,2 denote the Stiefel manifold of ordered pairs of orthonormalvectors in Im O. Show that G2 acts transitively on V7,2 with isotropysubgroup SU2:

G2/SU2 = V7,2.

(c) Let V7,3(¢ = 0) denote the collection of ordered orthonormal triplesel, e2, e3 E Im 0 with O(el A e2 A e3) = 0 (i.e., e3 1 ele2). Show thatG2 acts transitively on V7,3(o = 0) with no isotropy:

G2 = V7,3(0 = 0) -

10. Consider the algebra homomorphism

0:H®R.H--*M4(R)

induced on H OR H by defining ¢, on simple tensors to be

qi(p ®q)(x) = pxq for all x E R4 = H.

Let H OR H have the inner product defined by (a (& b, c ® d) _(a, c) (b, c), and let M4(R) have the inner product defined by (A, B)4 trace ABt.Prove that ¢ preserves inner products and hence is an algebra isomor-phism (cf. Remark 1.33).Hint: Show that trace O(a ® b) = 0 if b equals i, j, or k.

11. Show that

G2 = {g E GL(Im O) : g(x x y) = g(x) x g(y) for all x, y E Im O}.

122 Problems

Hint: x' x y' = x'y' + (x', y') for all x', y' E Im O.12. Suppose u E Im O and lull = 1. Show that, for all x, y, z E 0,

(a) (xu) x (yu) = u(x x y)u,(b) (xu) x (yu) x (zu) = (x x y x z)u.

13. A cross product on an inner product space V, (,) is usually defined tobe a bilinear map x x y from V x V to V that satisfies not only

(a) llx x yII _ lIx A ylIbut also(b) x x (x x y) = -11xIly+ (x, y)x.Adopt this (stronger) standard definition. Given a cross product onan inner product space V, (, ) define a product on R ® V by

(xo +x)(yo+y) = xoyo+xoy+yox+ (x,y)+x x yfor all xoyo ERandx,yE V.

Define (xo + x, yo + y) = xoyo + (x, y). Show that R ® V, (,) is anormed algebra.

14. Let A denote the standard volume element on Im O.(a) Show that (x -j ¢) A (x .1 0) A (x 4,) = 6xI I (x i A).Hint: Set x .= el = i and note that w x .i 0 is the Kahler form on[i]1 under the complex structure right multiplication by i, while x i Ais the unit volume element on [z]t _ C3.(b) Show that (u -1 ¢) A (x -1 0) A (x i ¢) = 2 (11zliu + 2(x, u)x) i A.(c) Show that (x 0) A (x -1 0) A 0 = 6lIxllA.

Hint: Use (a) and x A (x j A) = jjxjjA.15. (a) Given a 3-form 0 E A3(R7)* in 7-variables and a volume element

A on R7, let

(x 0) A (x j 0) A 4, = (x, x)ma

define a real bilinear form on R7. If 0 is nondegenerate (i.e., (, ) 0 isnondegenerate), exhibit either an octonian structure with R7 = Im 0,or a split octonian structure with R7 = Im O, so that 4,(x A y A z) _(x, yz)O is the associative 3-form.(b) (Gureirch [8]) Show that GL(7, R) acting on A3(R7)* has two openorbits with isotropy G2 and G2(at the associative three form ¢ andthe split associative three form 0, respectively).

16. Suppose N is one of the normed algebras C, H, or O. Let Herm(2, N){A E M2(N) :IT' =A}. Define P'(N) _ {A E Herm(2, N) : A2 =

Normed Algebras 123

A and trace A = 1}. Given a unit vector a = (al, a2) E N2, define

[a]=a`a= 1a1I2

a2a1 Ia212a1a2

(a) Show that [a] E PI(N), and that [b] = [a] if and only if b = as forsome A E N.(b) Let S(NP) denote the unit sphere in NP and verify that

S(N) -+ S(N2) -' P1(N)

is a fibering of spheres by spheres. That is,

Sl -* S 3 - P1(C) = S2 (Hopf fibration),

S3 -> S 7 -f P1(H) = 54,

S7---*S15- P1(O)S8.

(c) Show that the Hopf fibration

S1 -;53-}

can also be defined by sending

xES3CH to xixES2CImH.

Note: Setting x - z + j w with z, w E C yields a formula for anisomorphism from

P1(C)=C2-{O}/- to S2CImH=R3

in terms of quaternion multiplication. Namely,

[(z, w)] is mapped to(z -f jw)i(z - jw)

Iz12 + IwI2 E H C ®jC.

17. Consider the representations of G2 induced on Ak(ImO)* by the stan-dard action of G2 on Im 0. Show that these representations decomposeas

(a) A2(Im O)* = A7 ®A14,(b) A3(Im O)* = Ai ® A7 ®A27,

124

where

A7={w., uEImH},wu = u - or equivalently wu(x, y) _ (u, x x y),A2 ,..4 = 92,

A7 = {u j 0: u E ImH},A3 = span 4,

and

A3 ®A27 = {D(A) : A E Sym(Im O)},

Problems

with D the linear map defined in (6.89) or (6.94).

7. Calibrations

Throughout this chapter, we restrict attention to R" with the standardpositive definite inner product (, ). The concept of a "calibration" has beenalluded to in the previous discussion of Proposition 4.15. This result statesthat a straight line segment in R" minimizes length. The "FundamentalTheorem" of the theory of "calibrations" is a straightforward generalizationof this proposition replacing curves by higher dimensional submanifolds inR". In this chapter, it is assumed that the reader is familiar with topicsfrom advanced calculus, such as submanifolds of R" and Stokes' Theorem.

THE FUNDAMENTAL THEOREM

Again, we identify an oriented p-dimensional linear subspace of R" withthe element __ el A ... A eP E APR" where el,..., e1, is an orientedorthonormal basis for the p-dimension subspace. Thus, G(p, R") _ { EAPR" : = e1 A . A eP for some el, ..., eP orthonormal in R"} denotesthe grassmannian of oriented p-dimensional subspaces of R".

The dual space of APR" is AP(R")*, the space of p-forms (with con-stant coefficients). If ei,... , e, is an orthonormal basis for R" andall ... , an is the dual basis for (R")* (for example el =_ -9/ax', ... , e" __a/ax" the standard basis and ai dxl, ..., a" = dx" the standard dualbasis), then each p-form 0 on an open subset of R" can be expressedas 0 = E111=P 01a', where I = (ii, ... , i,) is a multi-index of length

125

126 The Fundamental Theorem

III = p, aI = a'1 A A a'a, and E' denotes summation over strictlyincreasing multi-indices, i.e., it < - < ip. The coefficients Oj are func-tions on the open subset of R', given in terms of ¢ by 0, = O(ej).

Definition 7.1. A p-form ¢ on an open subset U of R" is said to be acalibration if

(a) d¢ = 0 on U, and,(b) for each fixed point x E U, the form 0x E AP(Rn) satisfies 4)x() < 1

for all E G(p, R"), with the contact set

nonempty.

Condition (b) says that the maximum of 0 on the compact set G(p, R°)is one.

Recall that a p-form 0 can be integrated over an oriented p-dimensionalsubmanifold M of R". By using a partition of unity (so that one can assume0 is supported in a coordinate chart), this integral fm 0 can be defined bypulling back 0 (with a parameterization map) to an open subset of RP andthen performing ordinary RP integration.

Using the inner product (, ) on R" tfiis integral can be expressed interms of volume measure on M. For each point x E M, let AM E.P(R")*

denote the unit volume form for M, and let M denote the unit volumeelement for M; that is, M = el A A ep and Am - al A A as', wheree1, ... , e, is an oriented orthonormal basis for TxM and al , ... , a' is anoriented orthonormal basis foror Tz M. Then

(7.2)

i.e., fm ¢ equals the integral of the scalar valued function O(M) over Mwith respect to volume measure over M.

Definition 7.3. A closed oriented submanifold M of R" is said to be(area) volume minimizing if, for each relatively compact open subset U ofM with smooth boundary 8U,

(7.4) vol(U) < vol(V)

for all other compact oriented p-dimensional submanifolds V with the sameboundary as U, i.e., 6U = 8V.

A calibration 0 can be used to distinguish a class of oriented submani-

folds M; namely, those with ME G(4)) for each point x E M. Equivalently,0 restricted to M equals AM, the unit oriented volume form on M.

Calibrations 127

The Fundamental Theorem 7.5. Suppose 0 is a calibration of degreep on R. Each closed oriented p-dimensional submanifold M distinguishedby 0, i.e.,

O (M)

is volume minimizing.

Proof:

for all points x E M

vol(U) = JU 0 = ff()Av0 = 0V

where

f c

-f0=Ju-v c= j do f(U_V)IG=0,

because of Stokes' Theorem and the fact that:

(7.6) if do = 0, then 0 = do for some p - 1 form b on R.

THE KAHLER CASE

Consider C" equipped with the standard positive definite C-hermitian in-ner product h. Then h = (,) - iw, where (,) is the standard positivedefinite real inner product on R2,(R5 CI) and

(7.7) w = 2 d? A dz' .... + 2 dz' Adz'r

is the standard symplectic inner product on R2' (_s2 C"), which is alsocalled the standard Kii.hler form on C". Note that w may be viewed as thesum of the complex axis lines

(7.8) Ai = 2 dzi Adzi =dxi Ady', j = 1,...,n.

Since forms of degree 2 commute under wedge product,

(7.9) wP = L/, aIp! 111=P

is the sum of the complex axis p-planes AI. The unitary group U(n) fixesh, and hence w. Therefore, U(n) also fixes 1,wP. Also note that this formis d-closed since it has constant coefficients.

128 The Kdhler Case

Theorem 7.10. The 2p-form = 10 on R2n(- C") is a calibration.The contact set equals the complex grassmannian. That is,

(7.11) G(q) = Gc(p, C") C GR(2p, C").

Since do = 0, the theorem is equivalent to

Theorem 7.12 (Wirtinger's Inequality).

1 for all f E GR(2p, C"),P

with equality if and only ifl; E Gc(p, C").

Corollary 7.13. Each complex submanifold of C" is volume minimizing.

First Proof (Federer): The case p = 1 is particularly easy. Let u, vdenote an oriented orthonormal basis for e, i.e., 4 = u A v. Then

(7.14) w(u, v) = (Ju, V) < IJUI IvI = 1

by the C.-S. inequality. Moreover, equality occurs in (7.14) if and only ifv = Ju. That is, precisely when = u A Ju is a complex line (with theorientation induced from the complex structure on the line).

The general case reduces to the p = 1 case. Suppose E GR(2p, C")and let P = span l; denote the real 2p-subspace corresponding to . Re-striction of w to P yields a 2-form wp E A2P* on P. This 2-form canbe put in canonical form (see Problem 4.2) with respect to the real innerproduct on P inherited from R" 5--- Cn:

(7.15) WP = alai A a2 + A2a3 A a4 + ... + A,,a2r-1 A e2-r

where A1 > A2 > ... > A,. > 0 and a1, ... , aP is an orthonormal basis forP*. Let e1i ..., eP denote the dual basis for P. Then, by (7.14), for eachj = 1,...,r,

(7.16) Aj =w(e2i_l) e2j) < 1,

with equality if and only if e2j_1 Ae2j represents a complex line. NowlwP restricted to P is the same as p,wp, which equals Al ...APa'A Aar'.''herefore,

(7.17) Al ...Ap.

Combined with (7.16), this completes the proof.

A second proof of Wirtinger's Inequality utilizes a canonical form fora real 2p-plane in C" under the action of the unitary group U(n) .

Calibrations 129

Lemma 7.18 (Harvey-Lawson). Each oriented 2p-dimensional realsubspace 1; E GR(2p, Cn) of R2n = Cn can be expressed in terms of aunitary basis a 1, ... , e as(a) for 2p < n,

(7.19)=e1 A (cos01 Jel +sin01 e2) A...

A e2p_1 A (cos 0p Je2p_1 + sin Op e2p),

where the angles satisfy

0 < 91 < < 9p _ 1 2 and 9p _ 1 < 9p < 7r;

(b) for 2p > n,

1;-ejAJe1A...Ae,.AJe,.(7.19') A er+1 A (cos 01 Jer+l + sin 01 e,.+2)

A(cosOq Oq en

where the angles satisfy

0 < 91 < < 9q_1 < 2 and 9q_1 < eq < 7r.

Here r=2p-nand q=n-p.Second Proof of Wirtinger: The Kahler form can be expressed as

(7.20) anAJan

for any orthonormal basis a1, Jal, .,a', Ja" dual to the orthonormalbasis e1, Je1, ..., en) Jen (i.e., e', ..., en is a unitary basis for Cn). Then,by (7.19) and (7.20),

(7.21)11

WP(l;) = cos 91 . . . cos Bp, 2p:5 n,p

which is less than or equal to one and equal to one if and only if each9i = 0, j = 1, ..., p. That is, if and only if = el A Jet A A ep A Jep EGc(p, Cn).

Proof of Lemma 7.18: The proof is by induction on p. First, maximize(Ju, v) over all orthonormal pairs of vectors u, v E span 1; and set cos 91(Ju, v) equal to the maximum value with 0 < 91 < 7r/2. Since the functionf (O) _ (Ju, (cos O)v + (sin O)w) has maximum value at TP = 0 for each w E

130 The Special Lagrangian Calibration

span with w L span{u, v}, the derivative f'(0) = 0 vanishes. However,f'(0) _ (Ju, w), so that w 1 Ju. Similarly, w L Jv. Now define el = u.Then v = cos 01 Jel + sin 01 z, z a unit vector orthogonal to both el andJet. Define e2 - z. As noted above, if w E spank and w L {u,v}, thenw L {Ju, Jv}. Therefore w L {e1, Je1, e2, Je2}.

Consequently, = elA(cos&jJe1+sin 9ie2}Ar) with rl E GR.(2p-2, V)where V = lei, Jel, e2, Je2}l. In the special case 01 = 0, the proof mustbe modified: = eiAJeiArl with r) E GR.(2p-2, V), where V = lei, Jell-L.Also note that in the last step of the induction, v cannot be replaced by -vso that one must allow the possibility of cos O, negative, or Op_1 < Op < a.

THE SPECIAL LAGRANGIAN CALIBRATION

Consider C" with the standard positive definite C-hermitian inner producth = (, ) - icy, exactly as in the last section. However, also assume that C"is equipped with a complex volume form

(7.22) dz = dz1 A . A dz"

of norm IdzI2 =In this section, we shall examine the real n-form

(7.23) Redz (dz + dz).

Note that SU(n) fixes this form ¢. Of course, do = 0, since ¢ has constantcoefficients on C" 25 R2". In the coordinates zi = xi + iyi, nfor C",

(7.24) = Re (dxl + idyl) A - A (dx" + idy")

can be expanded out and is the sum of the 2"-1 real axis n-planes of theform ddxl Adyj with I U J = {1, ..., n} and I J I even.

Definition 7.25. A real oriented n-plane t E GR,(n, C") is said to bespecial Lagrangian if l; A(el A ... A e") for some A E SU(n), wheree1, ... , e" is the standard unitary basis for C", i.e., = el A . A e"represents the standard oriented subspace R" C C" = R" ® iR". LetSLAG C GR,(n, C") denote the subset of special Lagrangian n-planes.An oriented n-dimensional submanifold M of C" is said to be special La-

grangian if ME SLAG for each point x E M.

Theorem 7.26. The n-form 0 _- Re dz is a calibration on C", and thecontact set G(4) = SLAG consists of all the special Lagrangian n-planes.

Corollary 7.27. Each special Lagrangian submanifold M of C" is areaminimizing.

Calibrations 131

Remark 7.28. For each fixed angle 9, consider the n-form

(7.29) Oa = Ree-iedz.

For example, if 9 = it/2, then 0 = Im dz. Each such 09 is also a calibrationreferred to as the special Lagrangian calibration of phase 0.

The proof of Theorem 7.26 is via several lemmas that clarify the notionof special Lagrangian subspace. First, the weaker notion of Lagrangian isinvestigated. A subspace L of dimension n in C' = R2n is said to beLagrangian if L is totally null with respect to the Kahler (symplectic) formw. However, there are many equivalent ways of expressing this fact. Let 1denote orthogonality with respect to (, ) = Re h.

Lemma 7.30. Suppose L is an n-dimensional real subspace of Cn, h. Thefollowing are equivalent.

(a) w (u, v) = 0 for all u,vE L.(b) The 1-form E". =1 yjdxj restricted to L is d-closed.(c) If u E L, then Ju I L.(d) L={Ax:xERnCCn=R"ED iRn} for some A E U(n) unitary.(e) jdz(ul A ... A un)I = 1 if ul, . , u,, is an orthonormal basis for L.

Further, if L = graph f can be graphed over Rn C Rn ® iRn,f : Rn -> Rn, then the following can be added to the equivalence.

(f) L = graph V is the graph of a gradient y = (0¢)(x) of a scalarvalued function O(x).

Proof: (cf. Problem 5.8) Condition (a) has been taken as the definition ofL Lagrangian, i.e.,

w=EdxjAdyj

restricted to L should vanish. Since da = -w with

n

a = yjdxj,j=1

and restriction to L commutes with exterior differentiation, (b) is equivalentto (a).

IfL=graph f=_{(x,f(x)):xERn},then,

n

alL = E.fj(x)dxj.j=1

132 The Special Lagrangian Calibration

Condition (b) says that this 1-form is d-closed, i.e.,

afj _ Ofi

axi axi .

This symmetry is true if and only if

fi (x)axi

(x), n,

for some scalar-valued function 0 on Rn. Thus, (b) and (f) are equivalent.To prove that (a) and (c) are equivalent use the formula

(Ju, v) = w(u, v).

Since unitary transformations preserve (, ), J, and w, and since (c) isobviously valid for L = R' C Cn, the condition (d) implies condition (c).Conversely, if L satisfies (c), then an orthonormal basis u1i... , u for Lcan be used to define a unitary map A E U(n) sending the standard basisel,...,en for Rf1 into u1i...,u, so that (d) follows.

Let uj - Aej denote the image of ej under any complex linear trans-formation A. Then

(7.31) (dz)(ul A ...A un) _ (dz)(A(el A ...A en)) = detc A.

If (d) is valid, then !; = A(e1 A...Aen) with A unitary, so that I detc Al = 1,which verifies (e).

To prove that (e) implies (c) and thus complete the proof of Lemma7.30, we shall prove the following lemma.

Lemma 7.32. I(dz)(t;)I < 1 for all 1; E GR,(n, Cn), with equality if andonly if L = span is Lagrangian.

Proof: Assume Al; = ul A .. A u with A E R and u1, ... , un any orientedbasis for L = span (not necessarily orthogonal). Define A E GL(n, C) tobe the complex linear map sending the basis e 1 , . .. , en for R'i to ul, , 'fin

Then, exactly as in (7.31),

(7.33) A detc A.

However,

(7.34)1

=IA(elA...AenAJelA...AJen)I=A2I AJI

Calibrations

Therefore,

(7.35) I(dz)(e) I2 = 1 A Jt; I.

133

The classical Hadamard inequality (see Corollary 9.33 in Part II fora proof) states that awl A ... A wkl2 < Iw1I2 . Iwkl2 with equality if andonly if w1, ... , wk are orthogonal. In particular,

(7.36) If A JAI <_ lull ... IJuil ...IJu"I

with equality if and only if

u1i...,u",Jul,...,Ju"

are orthogonal (i.e., if and only if u1i ..., u is a unitary basis for C").Now assume that uii..., u" is an oriented orthonormal basis for l;.

Combining (7.35) and (7.36) yields the proof of Lemma 7.32, since L isLagrangian if and only if u1, ... , u,,, Jut,... , Ju is orthonormal (by (c)above).

Corollary 7.37. Suppose span 1; is Lagrangian, E GR(n, C"). Then

(7.38) (dz)(e) = detc Al detc Al',

where A is any complex linear map sending the standard basis e1,.. . , e"for R" into an oriented basis u1i ... , u" for e.

Proof. By (7.33), a(dz)(.) = detc A with A > 0. Since I(dz)(C)l = 1, thecorollary is immediate.

Proof of Theorem 7.26. Consider E GR,(n, C"). By Lemma 7.32,0(!;) = Re(dz)(t') < 1(dz)(£)j < 1 so that 0 is a calibration. If equal-ity 0(C) = 1 holds, then I(dz)(t:)l = 1, so that . must be Lagrangian.Therefore, by condition (d) of Lemma 7.30, l; = A(e1 A ... A e,,) forsome unitary map A. Corollary 7.37 implies that detc A = 1. There-fore, C E SLAG is a special Lagrangian. Conversely, if 1; E SLAG (bydefinition 1; = A(e1 A. Ae") some A E SU(n)), then Corollary 7.37 impliesthat (dz)(£) = 1 so that 0(f) = 1.

Corollary 7.37 provides several useful criterion for a manifold to bespecial Lagrangian.

134 The Special Lagrangian Differential Equation

Corollary 7.39. Suppose { E GR,(n, Cr) and span is Lagrangian.

(a) f£ is special Lagrangian if and only if Im (dz)(e) = 0.(b) t is special Lagrangian if and only if for any (some) complex linear

map A sending the standard basis e1, ..., e for R" into an orientedbasis for t;:

(7.40) deter A > 0 (real and positive).

THE SPECIAL LAGRANGIAN DIFFERENTIAL EQUATION

In pursuing the analogues between the Kahler (form) calibration and thespecial Lagrangian calibration, one question naturally comes to mind.What is the special Lagrangian analogue of the Cauchy-Riemann equa-tions? More precisely, suppose

(7.41) M= graph f={x+if (x):xEU0'U'P'" C R'}

is the graph of a smooth function f : U --> W. Then the question becomes:What is the differential equation(s) imposed on f by requiring that M bea special Lagrangian submanifold?

The n vectors

(7.42) u1=_ej +ia n,

form a basis for the tangent space to M at each point. Thus, one answerto this question is the equation

(7.43)

-Lf ) A((e1+i!r)A...A(en+iL))-Lf(el+i 8x1) A (en+i of !

This answer is not very useful. In fact, in the Kahler case C2, w a similaranalysis yields the equation

(7.44)

/e1 -i e2 + axl Je2i Jel + 5 e2 + y Je2/

2 2 au av av 8u 2[1+ Du + v (8x1 8y1 8x1 8y')

Calibrations 135

which is not the elegant Cauchy-Riemann equations:

(7.45)au = av au = av with f = u + iv.(9xi ayi ' ayl ax'

After this unproductive analysis, we can make a fresh start by using Corol-lary 7.39(a). Since

AMei+i ax)A...Ae"+i_xP

with A E R, this corollary states: Either M or -M (M with orientationreversed) is special Lagrangian if and only if

(7.46) M is Lagrangian

and

(7.47) (Im dz) I f ei + i ax ) A A (en i x 1 1 = 0.

The first condition, M Lagrangian, can be combined/ /with the (firstorder) partial differential equation (7.47) by using a potential (or generat-ing) function 0.

Lemma 7.48. Suppose M = graph f = {x + if(x) : x E Uopen C R"}and U is simply connected. Then M is Lagrangian if and only if

(7.49) f=vO

is the gradient of a scalar function 0 on U.

The proof of this lemma is exactly the same as the proof that parts(a) and (f) of Lemma 7.30 are equivalent.

Now, if f = V O, then of/ax' is the ith column of the Hessian matrix

(7.50) Hess(O)(9..).

That is, of/ax' = Hess(q)(ei).Therefore, (with a real)

(7.51)a M = ( e i + i ax) A ... A (en + i

ax"= (1 + i Hess(c))(ei A ... A e"),

136 The Special Lagrangian Differential Equation

so that

(7.52) (dz)(A M) = detc(I + i Hess (q )).

Theorem 7.53. Suppose M - graph f = {x+i f (x) : x E U C R' C Cn }is the graph of a smooth function over a simply connected domain U in R".Then M (with one of the two possible orientations) is special Lagrangianif and only if

(7.54) f = VO for some potential 0 on U,

and either of the following equivalent conditions hold:

(7.55) Im detc(I -I- i Hess(¢)) = 0 on U,

or

(7.56)[(n-1)/2]

E (_1)ka2k}1Hess(O) = 0 on U.k=0

Remark 7.57. The equation (7.56) will be referred to as the specialLagrangian differential equation. Here aj (A) denotes the jth elementarysymmetric function of a symmetric matrix A (such as A = Hess(c)). Interms of the eigenvalues A1i ... , An of A, recall that

(7.58) aj (A) = E Ail AijIII=j

If n = 2, then the special Lagrangian differential equation is not a newequation but is just i = trace Hess(O) = 0. The first nonclassical inter-esting case is n = 3. In this case, the equation is

(7.59) AO = MA(C),

where AO = trace Hess(O) and MA(C) - det Hess(O) is the Monge-Ampere operator on ¢.

Proof of Theorem 7.53: The theorem is derived from the characteriza-tion of M as special Lagrangian given by (7.46) and (7.47). By Lemma7.48, M is Lagrangian if and only if there exists a potential function 0 asin (7.54). The equation (7.55) has already been verified; see (7.52). Thus,it remains to show that (7.55) and (7.56) are the same equation. Let Adenote the symmetric matrix Hess(O) at a fixed point x E U. The matrix

Calibrations 137

A can be put in canonical form with respect to the orthogonal group O(n).This standard result is often referred to as the Principle Axis Theorem.Its proof is very similar to the proof of Problem 4.2 (a canonical form forO(n) acting on Skew(n, R) = AR), and hence is omitted. The result says:Given

A E Sym(n, R) C Mn(R)

a symmetric matrix, there exists g E O(n) such that

(7.60) gAgt = D

is a diagonal matrix with nonzero entries the eigenvalues Al, ..., An of A.Now detc(I + i Hess(O)) = detc(g(I + i Hess(q))gt) = detc(I + iD),

and hence (7.55) is equivalent to

n

(7.61) Im 11(1+i)3)=0,J=1

which is the same as (7.56).

Theorem 7.62. Suppose M is a connected oriented n-dimensional realsubmanifold, of an open subset U of C", which is defined implicitly bysmooth functions on U:

(7.63) M E {z E U : f1(z) = cii ..., fn(z) = cn}.

Then M (with one of the two possible orientations) is special Lagrangianif and only if

(7.64)

and

n

C

8fafg + afp afq \11 _ 0OZk OZk 8zk 8xkky-

(7.65) Im I detc (i' I I = 0

for all points zEM. //Proof: The normal n-plane to M is Lagrangian if and only if the tangentn-plane to M is Lagrangian (c.f., condition (c) of Lemma 7.30). Thisnormal n-plane is spanned by the vectors

\(7.66) up = 2 I O p+ i ayp a

138 Examples of Special Lagrangian Submanifolds

Note that

(7.67) Jug = 2 (i ag a4) = 2 aafq

z. .y

Therefore, M is Lagrangian if and only if

(7.68) (up, Jug) = 0 for each 1 < p, q < n.

Equivalently,

(7.68') Re n afp afg = 0,k=1

194 axk

which is Equation (7.62).Now assume M is Lagrangian. Then

Jup=i afpa-

p=1,...,n,

is a basis for the tangent space. The complex matrix

A = (iaafp)

\ xg

\/

maps the standard basis e1,. .. , e for Rn to the basis Jut, ... , Jun forN H HM, i.e., A M= Jul A ... A Jun with A > 0. Therefore, Im (dz) (A M)AIm detc A by (7.31). Thus, A satisfies (7.65) if and only if d M is specialLagrangian by Corollary 7.39.

EXAMPLES OF SPECIAL LAGRANGIAN SUBMANIFOLDS

Rather than discuss general classes of examples, specific examples are dis-cused in this Section. The first example is invariant under the action ofthe n - 1 torus = {diag(e191, ..., eie^) : 01 + 02 + + On = 0} containedin SU(n).

Theorem 7.69. Let MM denote the locus of the equations:

(7.70) lzj l' - Ix1l2 = cj, = 2, ..., n,

and

(7.71) Re z1 ... zn = c1 if n is even,

Calibrations

or

(7.71') Im z1 . zn = c1 if n is odd.

139

Then M, (with the correct orientation) is a special Lagrangian submanifold.

This provides perhaps the simplest example of an area minimizing ob-ject (with a singularity at the origin) that is not a real-analytic variety, thatis, not the zero set of a family of real analytic functions. The key propertydistinguishing a real analytic zero set is that if it contains a half-ray thenit must contain the full line. This is because a real-analytic function of onevariable t that vanishes for t > 0 must vanish identically.

Corollary 7.72. Suppose n is odd and let

M+={(re'B',... re'°'):01+...+0,=0and r>0}

M- _ { (re`B' ... , re'e-) : 01 + .....{- On = it and r > 0}

Then the cones M+ and M- are both special Lagrangian, and hence bothvolume minimizing. Note that M+ and M- are disjoint cones with onethe image of the other under the map - Id, so that neither M+ nor M- isa real-analytic variety.

Proof: Set cl = = cn = 0 in Theorem 7.69 and note that Mo =M+ U M- on Cn - {0}.

Proof of Theorem 7.69: Set fl(z) = Re inz1 zn and fq(z) = Izg12

I z1 1 2 for q = 2, ... , n. Then (see Problem 2a)

n(7.73) detc t Z afp/

(-1)n j_1

Now verify (see Problem 2a) the conditions (7.64) and (7.65) of Theorem7.62.

The next example provides a useful tool for the study of pairs of min-imizing planes in euclidean space; see the following section entitled AngleTheorem.

Given a1, ..., an > 0, let

(7.74) P(a, y) = y2 [(1 + aly2) ... (1 + apy2) - 1] .


Now for each k = 1, ..., n, and each y E R, define

(7.75) Zk(y) =rk(y)e:ek(y)

with

(7.76) rk(y) = ak 1 + y2

and

(7.77) Bk(y) = aky dy

o (1 + aky2) 1'(a, y)

It is convenient to use the calibration 4 Imdzl A A dz" ratherthan Re dz.

Theorem 7.78. The manifold

Ma {wEC" :uwk=tkzk(y) withyER, IER" and Etk=1ll k=1

is = Im dz) special Lagrangian (if correctly oriented).

This example (see Lawlor [11]), introduced by Lawlor and completedby Harvey, is used in the proof of the Angle Theorem 7.134.Remark 7.79. Let r denote the curve

(7.80) r = {8(y) : y E R},

defined by (7.77). Let

(7.81) Pe = eie R" = J (tie1B1, t E R" }

denote the n-plane in C" obtained by rotating R" by e'9. The key propertyof the special Lagrangian manifold M. is that each nontrivial intersection(i.e., 0 E r) of Ma with PB is the boundary of a bounded open subset ofP9. In fact, for 0 E r,

(7.82) Ma n Pg . (tle181 ... t"et9") E,2(y) 1

is an ellipsoid.

Calibrations 141

Proof of Theorem 7.78: In addition to giving a proof of Theorem 7.78,we shall sketch the proof that the submanifolds Ma are the only possiblespecial Lagrangian submanifolds with each nontrivial intersection Ma fl Pea compact hypersurface of P8, thus removing some of the mystery of thisexample.

If M is to consist of the union of hypersurfaces in a family of the n-planes P9, and M is required to be Lagrangian, then M must be of theform (set wj = Rj eaei )

(7.83) M(r, h) = w E C" : R?d8j _ Ads dsj=1

for some curve r in the 9-space and some function h(s) defined on r (here9(s) parametrizes r). This can be reformulated as a general fact about La-grangian submanifolds with degenerate (one-dimensional) projection ontoone of the Lagrangian axis planes, using the alternate symplectic coordi-nates pj = 2R and qj = Bj, j = 1, , n. (Note that dpj Adgj = dxjndyj.See Harvey-Lawson [9] for the proof of (7.83) (cf. Problem 3).

Note that for 9 E r, EB = M(r, h) fl Po is an ellipsoid with radiir1, , r defined by

(7.84)1 _ dOj

ifdO

> 0r? A A ,

and that otherwise the hypersurface M(r, h) fl Pa is not compact.The vectors

(7.85) Vk= (-tkz1,0,...,0,t1zk,0,...,0), k =2,...,n,

with tk Rk/rk and zk = rkeiek, provide tangent vectors to the ellipsoid

EBwEC':wk=tkzk, 7tk=1}

since N (tIz1i...,tnz,l) is normal to E9. In addition,

(7.86) V1 = (tizi, ...,

is also tangent to M.Let L denote the complex linear map sending ej to V j, n,

when e1, ... , e is the standrad basis for R" C C'. Since M is Lagrangian,M is 0 = Im dz special Lagrangian if and only if

(7,87) detc L = is with A > 0,


because of Corollary 7.37.

One can compute that (Problem 2(b))

2

(7.88) detc L = ti-1 I1I+ ... zlzri I2 1 (zl ... Z") .

Thus, M is special Lagrangian if and only if each zkzk is of the form

(7.89) zkzk = fki z1 z,,, with fk real and > 0.

In polar coordinates zk = rkeiek, with ik__10k, Equation (7.89)becomes

(7.90) rkrk = fkrl . . r sin

and

(7.91) rkOk = fkr1 r,i cos'.

Since rk(d9k/dh) = 1, Equation (7.91) implies that all the fk must beequal. Reparametrizing, we may assume fk = 1, yielding the system (with0=E9 )(7.92)

and

(7.93) rka ek = r1 ...rn cos

Let cj = zz (0) > 0, i.e., rj(0) = cj > 0 and 0j(0) = 0, j = 1,. .. , ndenote the initial conditions at time t = 0.

It remains to solve this system (7.92), (7.93). First, note that ifr(t), 0(t) is a solution, then r(t) = r(-t), 0(t) _ -0(-t) is also a solu-tion. Hence, we need only consider t > 0.

Since rkrk is independent of k.

(7.94) rk=ck+u and u(0)=0

for some function u. Also (7.92) and (7.93) imply that r1 r cos 0 hasderivative zero so that

(7.95) r1rcos0=c1c,a.

Calibrations 143

Remark. Alternatively, note that the solution curves to (7.92), (7.93)provide the flow lines of the Hamiltonian vector field associated with theHamiltonian function H - rl r cos = Re z1 - - - z (it is interesting tocompare with Theorem 7.69).

Since B(0) = cl . c,,/c,2E > 0, the functions Ok(t) and ii(t) must bestrictly increasing for t small. Since u = 2rkrk = 2r1 r sin 0, it is alsostrictly increasing for t > 0 small. Now, in order forr to vanish, sin V = 0or cos 7P = 1, i.e., r1 .. r = c1 - - - c by (7.95). This proves that Tk nevervanishes, and hence it is strictly increasing for all t > 0. Assume t > 0 anddefine y by it - y2. Since ii = 2y' = 2rkik = 2r1 - - - r sin 0, we have

(7.96) yy = r1 r sin 0.

Substituting cos cl r into (7.96) and (7.93) yields thesystem

(7.97)(Cl+y2)...(C2i+y2)-C2...cn

y2 1

(7.97') Bk = c1 ... c, / (c2r + y2) k = 1, ... n.

Eliminating t and replacing ck by 1/ak yields

(7.98)

as desired.

flak ak

dy (1 + aky2) P(a, V)

Remark 7.99. Equation (7.97) implies that

(7.100) t= fy

dyM'

where

(7.101) Q(y) = y-2 [ (C1 + y2) ... (Cn + y2) - C2 ... C21 .

If n = 2, the integral diverges as y oo, and the system is solved for alltime; while if n > 3, the integral converges and the system is solved fort E [-T,T], where the terminal time T is given by

(7.102) T =JOcO dy

Q(c,y)

144 Associative Geometry

Note that as cl approaches +oo, the terminal time T approaches zero.

ASSOCIATIVE GEOMETRY

In this section, consider the 3-form 0 on Im 0 = R7 defined by

(7.103) Ox, y, z) =- (x, yz) for all x, y, z E Im 0

that was examined in Chapter 6. Recall that 0 E A3(IMO)*, sincevanishes when any two of the variables x, y, z are set equal. The analogueof the Wirtinger inequality for 0 is most elegantly described by an equality.

Theorem 7.104 (Associator Equality). For all x, y, z E Im O,

(7.105) O(xAyAz)2+ 1I[x,y,z]12= IxAyAz12.

Proof: Recall that the triple cross product satisfies

(7.106) Ix x y x zj _ Ix A y A zj,

(7.107) Rex x y x x = O(x,y,z),

(7.108) Imxxyxz= I[x,y,z].

The associator equality (7.105) then follows.

Definition 7.109. If 1; E GR.(3, Im 0) is the canonically oriented imag-inary part of any quaternion subalgebra of 0, then l; is said to be anassociative 3-plane. Let ASSOC denote the set of all associative 3-planes.

A real oriented 3-manifold M C ImO with ME ASSOC at each pointx E M is said to be an associative submanifold of Im O.

Given l; = ul A U2 A u3 with u1, u2, u3 an oriented orthonormal basisfort E GR(3, Im O), Artin's Theorem 6.39 implies that the associator[ul, u2i u3] vanishes if and only if span{1, ul,'u2i u3} is a subalgebra of 0,algebra isomorphic to H.

Consequently, given t;' E GR(3, Im O),

(7.110) fit; E ASSOC if and only if [u1i u2, u3] = 0

for some (or equivalently all) basis u1, u2, u3 for span £. Therefore, as aconsequence of the associator equality we have that ¢ is a calibration andthat

(7.111) ft; E ASSOC if and only if q5(t;) = fl.

This proves the next result.

Calibrations 145

Theorem 7.112. The associative form 0 E A3(ImO)* is a calibrationwith the contact set

G(O) = ASSOC.

Corollary 7.113. Each associative submanifold of Im 0 = R7 is volumeminimizing.

Definition 7.114. An oriented 4-plane 71 E G-(4, Im 0) is said to be coas-sociataive if the 3-plane 771 is associative. The 4-form *0 E A4(Im O)*is called the coassociative calibration.

Theorem 7.115. The coassociative 4-form 0 E A4(ImO)* is a calibra-tion, and any real 4-dimensionalsubmanifold of Im 0 whose normal 3-plane(at each point) belongs to ASSOC is volume minimizing.

This theorem can be deduced from Theorem 7.112 (see Problem 4).Interesting systems of partial differential equations arise in associative

geometry (see Problem 5) and coassociative geometry, which in some waysare analogous to the Cauchy-Riemann equations. However, we turn nowto examples.

A particularly interesting example is the Lawson-Osserman coassocia-tive cone.

Theorem 7.116. The cone

(7.117) MVr5xix

+2 xl xi:xEH

on the graph of the Hopf map

x%xrl(x) = 2 xl

where i maps the unit 3-sphere S3 C H to the 2-sphere S2 of radius v'5/2in Im H, is coassociative, and hence volume minimizing.

Remark 7.118. The function rl : H -+ ImH defined by

rl(x) = 2 jxj

is Lipschitz continuous but not of class Cl (continuously differentiable).This makes the fact that its graph is volume minimizing particularly inter-esting, since a classical result of C. B. Morry says that if the graph of aCl-function is volume minimizing, then the function is real analytic.

146 The Angle Theorem

A 3-dimensional cone M in R7 = Im O with vertex at the origin is areal 3-dimensional submanifold of R7 - {0} of the form

M-{tx:tER+andsEL},

where L is a 2-dimensional submanifold of the unit 6-sphere S6 C IMO.The real surface L C Ss is called the link of the cone M. Associative conesM have an elegant description in terms of their links L given by Corollary7.121 below.

Lemma 7.119. At each point x E Ss C Im 0, left multiplication by x,denoted L, is a real linear map from the tangent space T,S6 into itselfand Lx = -1 so that Lx is an almost complex structure on T, S"6.

Proof: If u E TxS6, then ux + xu = 2(u, x) = 0. Therefore, ux E Im O.Since (xu, x) _ (u,1) = 0, this proves xu E TiS6.

Lemma 7.120. A two plane i E TxS6 is almost complex if and only ifthe three plane x A 27 is associative.

Proof: p= u A Lxu implies x A 77 = x A u A xu is associative. Conversely,if x A u A v is associative then v = xu (if x, u, v are orthonormal).

Corollary 7.121. A 3-dimensional cone in Im 0 is associative if and onlyif the link L is an almost-complex curve in S6 C Im O.

R. Bryant proved that each compact Riemann surface occurs as thelink on some associative cone. The proof, using the Newlander-NirenbergTheorem, is beyond the scope of this book.

Theorem 7.122 (Bryant). Given a compact Riemann surface S thereexists a almost-complex embedding of S in the 6-sphere S6 E Im O.

THE ANGLE THEOREM

This beautiful result uses a general class of calibrations as well as specialLagrangian geometry in its solution. But before presenting the result, themotivation will be discussed in a heuristic manner.

A standard method for understanding a singularity (of, for example,a real analytic variety M) is to study the associated "tangent cone." Ata point of M, a "tangent cone" is obtained by considering the intersectionof M with successively smaller balls (centered at the point in question)and then enlarging each ball with a rescaling from the small ball to the

Calibrations 147

unit ball. In the limit, one obtains a cone in the unit ball, which can beconsidered a cone in Rn, "the tangent cone to M at the point."

Thus, it is important to have a good understanding of tangent cones.In associative geometry, Corollary 7.121 reduces the study of (tangent)cones to the study of complex curves in S6 C ImO.

In the study of general area minimizing cones, one of the simplestquestions one can ask is:

(7.123)When is the union of a pair of oriented p-dimensional

planes = , i E GR(p, R") minimizing?

This question has a remarkably simple answer, as conjectured by FrankMorgan. This answer involves certain characterizing angles 01,. . ., Bp which

distinguish one pair of planes l;, q from another such pair 4', rl' (see (7.133')below).

Lemma 7.124. For each pair of oriented planes 4, 71 E GR(p, R2P), thereexists an orthonormal basis e 1, ... , e2p for Rep and angles

(7.125) 0<01 8p, Bp <7r-Op_1i

so that

(7.126) 1; = el A A ep,

(7.127) 77 _ (cos 01 e1 + sin 01 ep+1) A .. A (cos Bp ep -}- sin Op e2p) .

Remark. The statement and proof are easily adapted to the more generalcase where span 1 = Rm x {0} and r) e GR(p, R' x R). For example, ifm < p < n, then rl can be expressed as

27 =(cos01 el+sin01 em+sin Om fm)Afm+iA...Afp

Proof of Lemma 7.124: Choose e1 E P - span 4 and u1 E Q - span rl,both unit vectors, so that (e1, u1) _- cos 01 is maximized. (Note that 0 <(el, u1) < 1 so we may choose 0 < 01 < it/2.)

Since (el, ul) is maximized, the orthogonal projection of u1 onto Pmust be equal to cos01 e1. In particular,

(7.128) u1 = cos 01 e1 + sin 01 ep+l


defines a unit vector ep+l E Pl. The equation (7.128) says that

(7.129) if eEP and e ..L e1, then elul.

By symmetry,

(7.130) if u E Q and u 1 ul, then u 1 e1.

Next choose e2 E P, e2 1 e1 and u2 E Q, u2 1 u1 so that (e2, u2)cos 02 is maximized, with 0 < 01 < 02 < 7r/2. Because of (7.130), theprojection of u2 onto P only has an e2 component. Similarly, the projectionof u2 onto P1 must be orthogonal to ep+1. Thus,

(7.131) U2 = cos 92e2 + sin Beep+2

defines ep+2 E P1, ep+2 1 ep+1. Continue in this manner. At the last step,choose ep E P and up E Q so that e1, ..., ep is an oriented orthonormalbasis for , and ul, ... , up is an orthonormal basis for 17. Since the angle Bpdefined by cos 9p = (ep, up) must satisfy I cos Bp < cos 9p-1, the angle 9pcan be chosen so that 9p-1 < 0p < it - 9p-1.

Intuitively, the union of the pair!;, 77 is volume minimizing if and onlyif a and -r/ are not too close together. (Picture two large disks very closetogether, almost parallel, and with opposite orientation. Then a narrowstrip will have the same boundary and smaller area.)

If 91i ... , Bp are the characterizing angles for the pair £, 71, then thecharacterizing angles for the pair !;, -rl are given by

(7.132) 01=01i...,/ip-1Bp-1, and ?'p=7r-9p.

To prove this fact, replace up by -up in the proof of Lemma 7.124. Thus,-up = - cos 9pep - sin 9pe2p = cos ,ipep - sin ipe2p. Finally, replace e2p by-e2p and note that Op < 7r - Op_1.

The next theorem states that the union off and 71 is not volumeminimizing if and only if i;, -rl are "close together" in the precise sensethat

(7.133) 'Y1+.....FVjp <9r,

or, in terms of the characterizing angles 01i ..., 9p for and al,

(7.133') <9p.

This result was conjectured by Frank Morgan.

Calibrations 149

Angle Theorem 7.134 (Lawlor-Nance). The union of a pair 1, 77 EGR(p, Rep) of oriented p- dimensional su bspaces of RP is volume minimizingif and only if the characterizing angles satisfy the angle criterion:

(7.135) Op < 01 + - - + Op_1 (or equivalently Y'1 + - +'p > ir).

Proof: First, assume that , rt do not satisfy the angle criterion (7.135), orequivalently, assume that the characterizing angles 451, ..., ipp for the pair4, -r7 satisfy (7.133) (i.e, 4 and -9 are "close"). We must show that 4 U rjis not volume minimizing.

It is convenient to make 4 and -77 symmetrical about a fixed p-plane.Let 9j denote VJj /2, j = 1, - - -, p, and pick a new orthonormal basisel, ., e2p so that

(7.136) 4 = 4(8) and - 77

where 4(e) is defined to be

(7.137) 4(8) _ (cos B1 e1 + sin B1 ep+1) A .. A (cos 0 ep + sin Op e2p )

The hypothesis (7.133) now takes the form

p

(7.138) E 8j < 9j=1

Identify RIP with Cp by identifying the ej axis in Rep with the xjaxis in Cp (j = 1, ... , p), and identifying the ep+j axis in Rep with theyj axis in CP (j = 1,. .. , p). Assume, for the moment, that for somechoice of the constants a =- (al, ..., a,,), the special Lagrangian manifoldM. constructed in Theorem 7.78 has nontrivial intersection with PBspan 4(B). Then one can show that the portion M of M cut off by Pespan 4 and P_B _ span r7 satisfies

(7.139)

vo1(M) = J Im dz

= J Imdz - J Imdz < vol(UB) + vol(U_a),ve v-e

where Ue denotes the solid ellipsoid in Pe = span 4(0), with the boundarythe ellipsoid MaflPe. Thus UB -U_B, and hence 4Ur7 is not area minimizing.


Note that

(7.140) eae, ... ese'.

Therefore, both and -(Imdz)(l (-B)) equal sin> 0,,, whichis strictly less than one by the hypothesis (7.138). This proves the strictinequality in (7.139).

It remains to prove that the constants al, ... , an > 0 can be chosen sothat Ma fl PT is nonempty. Fix a value of y > 0 and consider the functionB(y) defined by (7.98) as a function 8(a) of the parameters a, i.e., set

v dy(7.141) 9(a) = ak

Jo (1 + aky2) P(a, y)

with

(7.142) P(a, y) = y-2 [(1 + aly2) ... (1 + any2)

It remains to prove that the map a F--* 0(a) is surjective onto the set{0 : E O j < it/2 with 0 < 0j, j = 1, ... , n}. The proof is by induction onn. If n = 1, then 0(a) = arctan (/y) surjects onto 0 < 0 < 7r/2. Notethat a i- 0(a) extends continuously to the closed positive quadrant sincesetting any one of the variables aj equal to zero in the formula for 9 simplyresults in the formula for 0(a) in fewer variables.

Because of (7.93), the inverse image of the hyperplane (in the positivequadrant)

(7.143) H,p =- {8 : E 0j = zb and 0 j > 0, 1, ... , n}

under the map 9(a) is contained in the surface

S,, = {a:(1+aiy2)...(1+any2)=cos-2

It suffices to show that

(7.144) 9 : S,p -> H,p is surjective for 0 < 0 < it/2.

The proof of (7.144) is obtained by standard topological methods usingthe notion of degree of a map. By the induction hypothesis, the degree ofthe boundary map (0 restricted to the boundary), 0 : 8S,p -+ BH,p, canbe shown to be one. Now a standard topological result says that the map0 : S,1, --} H,p must have the same degree one and hence is surjective.

Calibrations 151

To prove the remaining half of the theorem, assume that the charac-terizing angles 01 i ... , Bp for , q satisfy the angle criterion (7.135).

Assume that the orthonormal basis e1 ,-- , e2p of Lemma 7.124 is thestandard basis for Rep. Let z _ (x, y) with x, y E RP denote coordinates.Following D. Nance [13], consider the p-form

(7.145) c'u _ Re (dx1 + uidyl) A ... A (dxp + updyp) E AP (R2p)*

for each p-tuple ul, ..., up E S2 C ImH of unit imaginary quaternions.Note that if each Uk = i for k = 1, ...,p, then 0 is the special Lagrangiancalibration on Cp = Rep.

Proposition 7.146. Each ¢ is a calibration. Moreover,

(7.147) 1; (B) _ (cos 01 e1 + sin 01 ep+1) A A (cos Bp ep + sin Op e2p)

belongs to if and only if the quaterion product v1 ...vp equals 1,where vj = cos O2 + sin Oj uj E H, j = 1, , p.

Proof: Note that

(7.148) q ( ( 9 ) ) = Rev1 ...vp :5 Ivl ...vpj = Ivii ...Ivpl = 1,

with equality if and only if

(7.149)

It remains to show that the maximum of ¢,, over the full grassmanniancannot be larger than the maximum of 0. over the special p-planes 1;(B).This is a consequence of Frank Morgan's Torus Lemma, which is discussedbelow.

First we use this proposition to calibrate both f and £(9) with 0,,.We must find unit imaginary quaternions U]. .., up so that v1 vp = 1,where vp = cos Bj + sin Bj uj, 1, ... , p. If we choose vl,... , vp to be ofthe form

(7.150) V1 = W1 'w2, V2 = W2 W3, ... , vp = Wpw1,

with w1 i ... , wp unit quaternions then v1 vp = 1 is automatic. The onlyother requirement on v1, ..., vp is that

(7.151) Revj = cos9j, j = 1,...,p.

152

The u1's are defined by

(7.152) u - Im vjp-- JImvj I

The Angle Theorem

Note that with vi given by (7.150) the final condition (7.151) is that

(7.153) (w1iw2) =cos01, ..., (wp,wl) =cosop.

For simplicity, suppose w1i ... , wp E S2 are unit imaginary quaternions.Then the problem is reduced to the following standard problem in sphericaltrigonometry: Given real numbers 0 < 01 < < op, op < r - op_1,satisfying the condition op < 01 + . + Op-1, construct p disjoint pointsWi, ... , wp on the 2-sphere so that the spherical distance between the pointsequals the 0's. That is,

(7.154) d(wl, w2) = oi, d(w2, wa) = 03, ..., d(wp, wl) = op.

A picture of a p-sided spherical polygon is helpful (the sides are arcs ofgreat circles).

The useful torus lemma can be described as follows. Let T denote thesubset of GR,(p, Rep) consisting of all p-planes of the form

4(0) = (cos 01 el {- sin 01 ep+1) A . . A (cos op ep + sin op e2p).

Given a calibration 0, let GT(c) = {l;(o) E T : 0(1;(0)) = 1} denote thetorus contact set as opposed to the (full) contact set

G(O) = {l; E GR(p, RZp) : q5(4) = 1}

of Definition 7.1. In coordinates, ej = 8/8xj and dxj = et j = 1, , p,

while ep+j = 8/8yi and dy' = ep+,i j = 1, -,p. A form 0 E AP(RZP)* inthe span of the forms dx' A dy1, where I and J are disjoint and I U J ={1, ...,p} is called a torus form.

Torus Lemma 7.155 (Morgan). A torus form 0 E Ap(R2p)* is a cali-bration if and only if

0(t;(O)) < 1 for all t;(o) E T.

Proof: The proof is by induction on p. Renormalize 0 so that M = 1is the maximum value of 0 on G(p, R2p). Choose t; E G(p, R2p) to be a

Calibrations 153

maximum point, i.e., 4)(1;) = 1. We must show that the torus contact setGT(4') is nonempty.

Put 1; in canonical form (see the remark following Lemma 7.124) withrespect to Rep = R2 x R2p-2, with R2 x {0} the span of el and ep+l . Thatis,

(cos01 u1+sin01 fl)A(cos02 u2+sin02

where 0 < 01 < 02 < a/2; u1i u2 is an orthonormal basis for R2 x {0} andfl, , f2p-2 is an orthonormal basis for {0} x R2p'2. Since 0 is a torusform, ¢ = ei A4D+ep+i A with (D and 0 torus forms on Rep-2. Expressingel and ep+l as linear combinations of ui and u2 yields

0= uina+u2A/i,

with a and /j torus forms on R2p-2. Now

4)(1;) = acos01sin02+bsin01cosB2i

where a = a(f2A...Afp) _ 0(uinf2A .Afp) and b = -p(fl A f3A...A fp) _4)(flAu2Af3A...Afp). Therefore,

1 = !5(1;) < a2 cost 61 + b2 sin2 01 < max{lal, IbI} < 1,

so that we must have equality. Therefore 02 = a/2-01 i so that with 0 = 01,

e= (cos 0 ul + sin 0 fi)A(sin Afp

In particular, a = 1 and/or b = 1 so that by the induction hypothesis eitherthe torus form a or the torus form Q has attained its maximum value 1 onthe p - 1 torus, say a(q) = 1. Then 4)(ui A'i) = a(t7) = 1, proving that 0attains its maximum value 1 on the p-torus T.

The Torus Lemma has several interesting reinterpretations and conse-quences. Note that the space of torus forms is the dual space of span T,so the Torus Lemma can be stated as

(7.156a)ET max,,) 001

for all 0 E (span T)*.


Given a set A, let ch A denote the convex hull of A. Let P denoteorthogonal projection onto the linear subspace span T of the vector spaceAPR2P. The Torus Lemma implies

(7.156b) P(ch(G(p, RIP))) = chT

and

(7.156c) ch T = ch(G(p, R2P)) fl span T.

In fact, both (7.156b) and (7.156c) are equivalent to the statement (7.156a)of the Torus Lemma.

Assuming (7.156a) we prove (7.156b). Of course,

ch T C P(ch (G(p, R2P))).

Now suppose E G(p, R2P). If PC V ch T, then by the Hahn-BanachTheorem there exists 0 E (span T)* with ¢(r7) < 1 for all n E ch T butwith O(1;) > 1. This is impossible by (7.156a). Next note that (7.156b)implies (7.156c) and that (7.156c) implies (7.156a).

For a torus calibration the torus contact set GT(4) actually determinesthe full contact set.

Corollary 7.157. Given a torus calibration 0 and 1; E G(p, R2P),

E G(0) if and only if Pi; E ch(GT(q5)).

Proof: If P. E ch(GT(0)), then Pt; = Ail j with each j E GT(4) andEAj = 1. Therefore, 4(1;) = ¢(Pl;) = Aj 1. Conversely, given

E G(O), O(PC) = 1.

By (7.156b), P E ch T. It remains to show that

{'iEchT:¢(7)=1}Cch(GT(¢)).

Since the extreme points of a convex set generate the convex set via convexcombinations, it suffices to show that the extreme points of {q E ch T :0(77) = 1} are contained in ch(GT(O)). However, if rl is such an extremepoint, then one can easily verify that 77 is also extreme in ch T. Sincethe extreme points of ch T are just the elements of T this proves that17 E GT(O)-

An alternate method for determining G(O) from GT(4), due to FrankMorgan, is described next.

Calibrations 155

Suppose w E A2 R4 can be written as w = w l A w2 + w3 A w4, wherew1, w2, w3, w4 is an orthonormal basis. Let u1i u2, u3, u4 denote the dualbasis. Then w has contact set G(w) = {f E GR,(2, R4) : = u A Ju EGc(1, C2)} = CP1 with respect to the complex structure on R4 definedby Jet = e2, Je3 = e4

More generally if 0 E AP (R')* is of the form ' = wAw5 A AwP-2,with w = w1 A w2 + w3 A w4 as before, then G(O) = G(w) A u5 A A up-2will also be referred to as a CPI-contact set.

A subset B C G(p, R') with the following property will be calledCP1-closed.

If K is any CP1-contact set containing two points of B,

then the whole (CP1-contact) setK is a subset of B.

Proposition 7.158. The contact set G(c/) of a calibration ¢ E AP(Rn)*is CP1-closed.

This is a consequence of the "First Cousin Principle"; see Problem 8.Note that an arbitrary intersection of CP' closed sets is again CP1-closed.

Proposition 7.159. Suppose 0 E AP(R2P)* is a torus calibration. Thecontact set G(O) is the smallest CPI-closed set containing the torus contactset GT(O).

Proof: Given any set B with the two properties; B is larger than GT(cf)and B is CP'-closed, we must show that G(q5) is a subset of B. Again,the proof is by induction on p. Suppose C E G(q5), i.e., O(£) = 1. We mustshow that 1; E B. We proceed exactly as in the proof of the Torus Lemmato show that

l:= (cos9u1-I-sin0 f1)A(sin0u2-1-cos0 f2)Af3A.. A fp.

Case 1:If9=0,then= u1Af2A...Afpand0(.)=a(f2A...Afp)=1,so that AfpEG(a).

Let B' _= {rl E G(p- 1, R2P'2) : u1 Al E B}. Then B' contains GT(a)and B' is CPI-closed. Therefore, by the induction hypothesis B' containsG(a). Since Oo E G(a) C B', this proves that 1; = u1 A rlo E B.Case 2: If 0 < B < a/2, then a = b = 1 and both u1 A f2 A . A fpand f1 A u2 A f3 A A fp belong to G(O). As in the previous case, byinduction both belong to B. Both also belong to the CP1-contact setK = G((u1 A f2 + f1 A U2) A f3 A A fp). Therefore, the hypothesis on Bimplies that B contains all of K. Since 1; E K the proof is complete.

156 Generalized Nance Calibrations and Complex Structures

GENERALIZED NANCE CALIBRATIONS AND COMPLEXSTRUCTURES

The Nance calibrations 0,,, defined in (7.145), provide a large class of "toruscalibrations" using the quaternions, or more precisely, a collection of unitimaginary quaternions ul, ... , up. This construction can be generalized asfollows.

Consider the vector space M2n(R) of matrices with positive definiteinner product given by

(A, B) = 2n trace AY.

Let P : M2,,(R) -+ span 1 denote the orthogonal projection onto the spanof the identity matrix 1 E M2,,(R). The orthogonal complex structures onR2n are given by

Cpx(R2n) - {J E M2n(R) : Jt = J-1 and J2 = -1} .

Definition 7.160. The p-form 0. E AP(R2p)* defined by

(7.161) ou = P [ (dx1 + u'dy1) A ... A (dxP + updyp)] ,

where u1i ..., up E Cpx(R2n), is called a generalized Nance form.

Proposition 7.162. Each generalized Nance form ¢u is a calibration.Moreover,

(7.163) 1;(B) E G(giu) if and only if Al . . Ap = 1,

where Aj = cos Bj + sin 6j uj E M2, (R), j = 1, ... , p.

Proof: If u E Cpx(R2s) C M2n(R) and B E R, we set

A=cos0+sin6u.Then

and

At = cos 6 - sin B u,

AAt = cost B + sine B = 1,

so that A E O(2n) is also an orthgonal transformation. If u1, ..., up ECpx(R2') and 01 i ... , Bp E R are given, define Aj = cos Bj + sin Bj uj, j =1,.. . , p. Then each Aj E O(2n) and hence the product Al . Ap E O(2n).

Calibrations 157

Therefore, Al . AP is a unit vector in M2n(R). To complete the proof,note that

P[(cos Bl + sin B1 u1) .. (cos Bp + sinO up)]

= (A, ... Ap, 1) < 1,

with equality if and only if Al Ap = 1; and then apply the Torus Lemma.

In order to understand why this construction-using complex struc-tures Cpx(R2n) rather than unit imaginary quaternions S2 C ImH-generalizes the Nance construction, we examine the set Cpx(R2n) of or-thogonal complex structures in more detail.

Since each complex structure determines a canonical orientation onR2n, the set Cpx(n) naturally divides into two disjoint sets of complexstructures

(7.164) Cpx(n) = Cpx+(n) U Cpx (n),

with Cpx+(n) the subset of Cpx(n) consisting of complex structures thatinduce a given orientation on R2n, and Cpx (n) the complex structuresthat induce the opposite orientation on R2n.

The group O(2n) acts on Cpx(n) by

(7.165) J E Cpx(n) maps to gJg'1 E Cpx(n) for each g E O(2n).

The isotropy subgroup of O(2n) at a particular J E Cpx(n) is just

(7.166) U(n) = GL(n, C) fl O(2n),

where GL(n, C) = {A E GL(2n, R) : JA = AJ}.Given a complex structure J E O(2n) on R2n, there exists a (complex)

unitary basis for C' = R2n. Thus,

(7.167) el,... , en, Jel,... , Jen

is an (real) orthonormal basis for R2n. With respect to this orthonormalbasis the matrix for the complex structure J is

(7.168) J = ( _0 1)1 0

where Rn = V ® JV with V = spanf{el,..., en} induces the 2 x 2 block-ing of J. This proves that 0(2n) acts transitively on Cpx(n). SinceO(2n) has two connected components and U(n) is connected, the quo-tient O(2n)/U(n) has two connected components. In summary we havethe following proposition.

158 Generalized Nance Calibrations and Complex Structures

Proposition 7.169. The set Cpx(n) of orthogonal complex structures onR21 has two connected components Cpx+(n) and Cpx (n), with

Cpx(n) = O(2n)/U(n),(7.170)

Cpxt(n) = SO(2n)/U(n).

In low dimensions, the space of complex structures Cpx(n) can be rep-resented in terms of the quaternions and octonians. For example, Ru, rightmultiplication by a unit imaginary quaternion, is an orthogonal complexstructure on H - R4.

Proposition 7.171.

Cpx+(2)={Lu:uES2CImH}=S2and

Cpx-(2)={Ru:uES2CImH}= S2,

where H has the standard orientation.

Proof: First note that, for each u E S2 C ImH, Lu E Cpx+(2). Thisis because the complex structure Li induces the standard orientation onH. Hence, each L. also induces the standard orientation on H, since S2 isconnected.

Now suppose J E Cpx+(2) is given and identify R4 = H. Definee1 J 1 and choose a unit vector e2 1 1, e1. Then, because of the Cayley-Dickson process, 1, e1, e2 and e3 = e1e2 is an oriented orthonormal basisfor H. Now Jet must equal ±e3. However, since 1, J1, e2, Jet must inducethe standard orientation on H, Jet = e3. This proves J = Let.

Similarly, Cpx (2) {Ru : u E S2 C ImH} - S2. J

Alternatively, Proposition 7.171 can be proved using the fact (7.170)that each J E Cpx+(2) is of the form J = gLig-1 with g E SO(4). Becauseof (1.34'), g(x) = pxq' for some unit vectors p, q E H. Thus J = Lpip-.

Replacing the quaternions H by the octonians 0, each Ru with u ES6 C Im 0 determines a (6-sphere of) complex structures on Rs = 0(compatible with the standard orientation on 0). However, Cpx+(O)SO(8)/U(4) has dimension 12, so that Ss is too small.

Definition 7.172. Given l; = u A v E G(2, 0), let J£ =z

RuR-).

First, note that Jg is well-defined (independent of the choice of u, vwith = u A v) since the right hand side vanishes when u = v. Second, thebasic octonian formula

R,X + R,, R,-. = 2(u,v)

Calibrations 159

implies that, for u, v orthonormal, J£ = R Rr = -R R,-,, so that

(7.173) J R,,RV = -1.

This proves that JJ E Cpx(4).

Proposition 7.174.

(n = 3) Cpx+(3) = u E S6 C Im O} - P6(R)(n = 4) Cps+(4) = {Je : 1; E G(2, O)} = G(2, 0).

The proof will be given in Chapter 14 Low Dimensions.

Remark 7.175. Consider the case n = 4. Given 1; = u A v E G(2, 0) withu, v orthonormal.

(7.176) Jfu = v and Jfv = -u.

Thus, in the plane span{u, v}, Jf is just counterclockwise rotation by 7/2.Suppose x E span{u, v}1. Then v(ux) = -v(xu) = x(vu), so that

(7.177) Jfx = if x E span{u, v}1,

where u x v =i

(vu - Vv). In summary, the complex structure A is

(1) counterclockwise rotation by 7r/2 on the 2-plane span !;', and(2) right multiplication by u x v on (span l;)1-, where u, v is any oriented

orthonormal basis for C.

PROBLEMS

1. Use the canonical form given by Lemma 7.18 to give a second proof ofTheorem 7.26:

(Re 1 for all f E GR,(n, C"),

with equality if and only if 1;' E SLAG.

2. (a) Complete the proof of Theorem 7.69, computing

dete \\liqP /

J,

and then verifying conditions (7.64) and (7.65) so that Theorem 7.62is applicable.

160 Problems

(b) Compute the complex determinant detc L of the linear map Lused in the proof of Theorem 7.78.

3. (HL [9]-assumes knowledge of harmonic functions on surfaces in R3,minimal surfaces, and normal bundles.)Given a surface S in R3 and a smooth function F on a neighborhoodof S, let

A(S,f)-{(x,tnx+VF(x))ER6:xES, tER}.

Here n,x is the unit normal to S at x E S and f ___ Fps.(a) Show that A(S, f) depends only on f - FIs and not on the ambientextension F of f.(b) Show that if S is a minimal surface and f is an harmonic functionon S, then A(S, f) is a special Lagrangian 3-manifold in Rs = C3.

4. Deduce the Coassociative Theorem 7.115 from the Associative Theo-rem 7.112.

5. Show that M=Ix +f(x)e:xEUCImH}, the graph off:U -;Hover an open domain U in ImH, correctly oriented, is associative ifand only if

Lz

+ axl e, 9 ax2 e, k + a s eJ = 0 on U.

6. (a) When n = 2, explicitly solve (using elementary functions) thesystem of differential equations (7.92), (7.93) for zl(t),z2(t).(b) Find a real orthogonal coordinate change in R4 - C2 so thatthe special Lagrangian manifold M,, described in Theorem 7.78 equals{wEC2:alwi-a2wz=1}.

7. (a) At the point/

rI 2 i+el, rER,

compute a basis for the normal 3-plane to the cone M defined by(7.117) (the cone on the graph of the Hopf map). Verify that thisnormal 3-plane is associative.(b) Show that M is fixed by HU(1) acting on 0 = H ® He, withp E HU(1) sending a + be to pap + (bp)e.(c) Prove Theorem 7.116.

Calibrations 161

8. (a) (The First Cousin Principle) Suppose 0 E AP (R")* is a calibrationand q(el A . A ep) = 1, where el,..., e is an orthonormal basis.Prove that 0(t;) = 0 for all first cousins:

=eaA(ei](eiA...Aep)), 1 =1,...,pandi=p+1,.

ofelA...Aep.(b) Prove Proposition 7.158.

9. Consider a 7-torus M = Im O/A where A is a rank 7 lattice in R7Im O equipped with the associative calibration 0 E A3(M). Provethat the lattice may be chosen so that M has a compact associativesubmanifold.

8. Matrix Algebras

This chapter collects information about matrix algebras that will beneeded in our discussions of spinors and pinors. More knowledgeable read-ers may wish to treat this as a reference chapter rather than reading it indetail.

The matrix algebras

MN(R), MN(C), MN(H)

are examined in this chapter. The algebras MN(R) and MN(H) will alwaysbe considered real algebras, while the algebra MN(C) may be considereda real algebra or a complex algebra.

REPRESENTATIONS

At first, consider all three of the algebras MN(R), MN(C), MN(H) to bereal algebras.

Definition 8.1. Given a real algebra A and a vector space V over FR, C, or H), a real linear map

(8.2) p : A -+ Endp(V)

sending 1 to the identity on V that satisfies

p(ab) = p(a)p(b) for all a, b E A,

163

164 Representations

is called an F-representation of A. Two F-representations, pl : A --->EndF(Vl) and P2 : A --> Endp(V2) of the real algebra A are, equivalent ifthere exists an invertible F-linear map

(8.4) f : Vi -+ V2

satisfying

(8.5) P2(a) = f o pi(a) o f-' for all a E A.

Such an operator f is called an intertwining operator.

The kernel of a representation p is a two-sided ideal in A (which isproper since 1 ker p).

Lemma 8.6. Each two-sided ideal in one of the matrix algebras MN(R),MN(C), or MN(H) is either {O} or the entire matrix algebra.

The proof is a straightforward calculation, left as an exercise. Analgebra with this property is said to be simple.

Corollary 8.7. Each representation of MN(R), MN(C), or MN(H) isinjective.

Examples (The Standard Representations)

(8.8) p(a)x = ax for a E MN(F) and x E FN

defines an F-representation of MN(F) called the standard representationof MN(F). Here F is either R, C, or H.

Given a matrix a E MN(C), let a denote the conjugate matrix. TheC-representation of the real algebra MN(C), defined by

(8.9) p(a)z = az for a E MN(C) and z E CN,

is called the conjugate representation of MN(C).Note that each H-representation, or each C-representation of A, is

automatically an R-representation. As R-representations, the standardrepresentation of MN(C) and the conjugate representation of MN(C) areequivalent; the intertwining operator C : CN --> CN is conjugation on Cextended to CN by acting on each component. However, since p(i) = iand p(i) = -i, the intertwining operator cannot be chosen to be com-plex linear. Therefore, these two representations are not equivalent asC-representations of the real algebra MN (C).

Matrix Algebras 165

Definition 8.10. Suppose pl : A -+ EndF(Vl) and P2 : A --r EndF(V2)are two F-representations of A. Then p = PI ® p2 : A --+ EndF(Vl ® V2),defined by p(a) = pl(a) ® p2(a), is also an F-representation of A. AnF-representation of this form p = p1 ® p2 is said to be F-reducible. If anF-representation p of A is not reducible, then it is called F-irreducible.

Theorem 8.11. Suppose F is it, C, or H. The standard representationof MN(F) on FN is the only irreducible R-representation of MN(F), upto equivalence.

Proof: Let I denote the subspace of MN(F) consisting of those matriceswith nonzero entries confined to the first column, and identify I with FN.Note that I is a left ideal in MN(F). Also note that

(8.12)if b E I (nonzero) is given, then each a E I can

be written as a = cb for some c E MN(F).

In particular, if K is a left ideal contained in I, then either K = {0} orK = I. (Thus, I is called a minimal left ideal.)

Let E denote the matrix with all entries zero except for the entry 1in the position first row-first column. Since p is injective (Corollary 8.7),p(E) # 0. Suppose v1 E V satisfies p(E)vl # 0. Define f : I ? FN -+ Vby

(8.13) f (b) = p(b)vl for all b E I.

First, we show that f is injective. Let

J - {a E MN(F) : p(a)ve = 0}.

Note that J, and hence ker f = Ifl J, is a left ideal in MN (F). Since E E I,but E V J, the left ideal In! is proper in I. However, since I is a minimalleft ideal, this proves that I n J = {0}, i.e., f is injective.

Now define V1 = image f = {p(b)vl : b E I}. Note that p(a)Vl C V1for each matrix a, since p(a)p(b)vl = p(ab)vl. Let pi(a) = p(a)jv,. Then

(8.14) pi(a) = f o a of-'.

This is because for each u = p(b)vl E V1, p(a)u = p(a)p(b)vi = p(ab)vl =f(ab) = f (a f -1(u)). This proves that pl is equivalent to the standardR-representation of MN(F) on FN -_ I. If V1 = V the proof is complete.

Next, we prove that

(8.15) if p(E)V C V1, then V1 = V.

166 Representations

If p(E)V C V1, then for any matrices a and b,

p(aEb) V = p(aE)p(b)V C p(aE) V = p(a)p(E)V C p(a)V)i C V1.

Since the two-sided ideal in MN(F) generated by E is all of MN(F), thisproves that p(1)V C V1 or V = V1.

Now choose v2 so that p(E)v2 V1. Exactly as above, define V2{p(a)v2 : a c I}, note that p(a)V2 C V2, and prove that p2(a) - p(a)lv,defines an R- representation P2 of MN(F) equivalent to the standard rep-resentation.

Next, since p(E)v2 ¢ Vi it follows that Vi n V2 = {O}. If Vi n V2 # {O},say p(ai)vl = p(a2)v2 with ai, a2 E I-{O}, then by (8.12) E = cat for somematrix c. Therefore, p(E)v2 = p(ca2)v2 = p(c)p(a2)v2 = p(c)p(ai)vi =p(cai)vi E V1 since cal E I.

Continuing in this manner, we obtain V1 ® ... ® Vk = V, so thatPi ®' - - ®Pk = p is reducible unless V = V1. L

Corollary 8.16. The real algebra MN(C) has exactly two irreducibleC-representations-the standard C-representation and the conjugate C-representation.

Proof: Suppose p : MN(C) -+ Endc(V) is any C-representation ofMN(C). The operator p(i) E Endc(V) has square -1. Therefore, theeigenvalues are ±1 where I is the complex structure on V. Let

Wf={xEV:p(i)x=±Ix}

denote the eigenspaces, and note that if p(i)x = ±Ix, then p(i)p(a)x =p(a)p(i)x = ±Ip(a)x, so that p(a)WW C Wt for all a E MN(C). Letp (a) = P(a) l wf . Then

p = p+ ®p_ as C-representations.

Thus we may assume that either p(i) = I or p(i) = -I if p is an irreducibleC-representation. If P = P1 ® P2 is reducible as an R-representation, withV = Vi ® V2i then each subspace V is complex with respect to I, sinceI = ±p(i) and p(i) maps Vj to V j. That is, p = P1 ® P2 is also reducible asa C-representation.

Therefore, we may assume that p is R-irreducible, and hence R-equivalent to the standard representation of MN(C) on CN via an inter-twining operator f : CN -+ V, which is real linear. Now f o i o f -1 = p(i) =±1, or f o i = ±1 o f. That is, f is C-linear if p(i) = I and C-antilinear ifP(i) = -I.

Matrix Algebras 167

Corollary 8.17. The standard representation of MN(H) on HN is theonly irreducible H-representation, up to equivalence.

Proof: Suppose p : MN(H) --r EndH(V) is an H-representation of MN(H)on a right H-space V. Recall that

(8.18) MN(H) ®a. H - EndR(HN) - M4N(R),

with (a (9 A)(x) - axX for a E MN(H), X E HN, and A E H. Nowp : M4N(R) - EndH(HN) --} EndR,(V), defined by

(8.19) p(a ))(v) = p(a)J, for all v E V,

is an R-representation of M4N(R) on V. Thus by Theorem 8.11,

P=P1Ej...EPk, V =Vi®...®Vk,

where each representation pj : M4N(R) - Enda(HN) -t Enda(V') is R-equivalent to the standard representation of M4N(R) via an intertwiningoperator fj : R4N - HN -; Vj. That is,

(8.20) pj(A)=fjoAof11.

If A E H, then right multiplication by A on R4N = HN is given by

rA,=1®a,

and p(ra) = p(1 0 a) is RA, right multiplication by A on V (see (8.19)).Therefore, by (8.20), fj o ra = RA o fj. That is, each fj : HN --+ V isequivalent, as an H-representation, to the standard H-representation onMN(H). LJ

UNIQUENESS OF INTERTWINING OPERATORS

The center of an algebra A is, by definition,

(8.21) cenA={aEA:ab=bafor all bEA}.

Let R C MN(R) and C C MN(C) denote all scalar multiples of the iden-tity, and let R C MN(H) denote all real multiples of the identity. Thecenters of the matrix algebras are listed in the next lemma.

168

Lemma 8.22.

Uniqueness of Intertwining Operators

Algebra CenterMN (R) RMN(C) CMN(H) R

Proof: The proof for N = 2 is Problem 1 (note that H has center R).The general case follows immediately, because, for a - (a;2) in the center,each 2 x 2 submatrix of the form

a ash }a1i ail

must belong to the center of M2(F).

Corollary 8.23.(a) The intertwining operator for two irreducible R-representations of

MN(R) is unique up to a nonzero scale A E R*.(b) The intertwining operator for two equivalent irreducible C-representa-

tions of MN (C) is unique up to a nonzero scale A E C*.(c) The interwining operator for two irreducible H-representations of

MN(H) is unique up to a nonzero real scale A E R*.

Proof: (a) Suppose p'(a) = f1 o p(a) o fi 1 = f2 o p(a) o f2 1, where f1 andf2 are both intertwining operators. Then f2 1 0 f1 E EndF(V) commuteswith all p(a) E EndF(V). Since p : MN(F) -* EndF(V) is an isomorphism,the corollary follows.

Suppose A is a subalgebra of B. The centralizer of A in B is definedto be

(8.24) {bEB:ab=baforallaEA}.

Lemma 8.25.(a) The centralizer of Endc(CT') in EndR(CN) is C.(b) The centralizer of EndH(HN) in EndR,(HN) is H-acting on the right

as scalar multiplication.(c) The centralizer of Enda(RN) in EndFt(CN) is M2 (R).

Proof: (See Problem 2.)

Matrix Algebras 169

Corollary 8.26.(a) The intertwining operator for two irreducible R-representations of

MN(C) is unique up to a nonzero complex scale A E C.(b) The intertwining operator for two irreducible R-representations of

MN(H) is unique up to a nonzero right multiplication RA, A E H*.

Proof: Analogous to the proof of Corollary 8.23. Ll

AUTOMORPHISMS

The automorphisms of the algebras MN(R), MN(C), and MN(H) can becomputed as a special case of Theorem 8.11 and its two corollaries. If a isa real automorphism of one of these algebras, then

(8.27) p(a) = a(a)

defines an F-representation of the algebra on FN. If V is a complex vectorspace, let Endc(V) denote the space of complex antilinear maps from Vto V.

Corollary 8.28 (Automorphisms).(a) (Existence) Each automorphism a of the algebra MN(R) is inner.

That is, there exists h E MN(R) with

(8.29) a(a) = hah-1 for all a E MN(R).

(Uniqueness) The intertwining operator h in (8.29) is unique up to a

real scalar multiple.(b) Suppose a is an automorphism of the algebra MN(C) considered as a

real algebra. Then a is either complex linear or complex antilinear.(Existence) If a is complex linear, then a is inner. That is, there existsh E MN(C) with

(8.30) a(a) = hah'1 for all a E MN(C).

If a is complex antilinear, then there exists h E Endc(C") satisfying(8.30).(Uniqueness) In both cases the intertwining operator h is unique upto complex scalar multiples.

(c) (Existence) Each automorphism a of the real algebra MN(H) is inner.That is, there exists h E MN(H) with

170 Inner Products

(8.31) a(a) = hah-' for all a E MN(H).

(Uniqueness) The intertwining operator h is unique up to a real scalarmultiple.

INNER PRODUCTS

An inner product e (of one of the seven types presented in Chapter 3) on avector space V over F determines an antiautomrophism of Endp(V) calledthe adjoint (and denoted by a* or Ad(a)):

(8.32) e(ax, y) = e(x, Ad(a)y) for all x, y E V.

The adjoint in each of the seven cases is explicitly given in Problem 2.7.Up to a scalar multiple no other inner product can determine the sameadjoint.

Theorem 8.33. Suppose V is an F-vector space with F = R, C, or H.Given two inner products el and e2 on V that determine the same an tiau-tomorphism of the algebra EndF(V) by taking adjoints, el and e2 must beof the same type and differ by a constant multiple c }e 0:

(8.34) E2 (X, y) = c el (x, y),

where c E R* if the type is R-symmetric, R-skew, C-hermitian symmetric,H-hermitian symmetric, or H-hermitian skew, and c E C* if the type isC-symmetric or C-skew.

Proof: Let a, = Reei and a2 = Ree2. Note that if a is an inner producton a complex vector space V, then

(8.35) (e(x, y), i) = - Re ,-(x, y)i = - Re e(x, iy);

while if a is an inner product on a right H-vector space, then

(8.36) (e(x, y), u) = Re e(x, y)u = Re e(x, yu)

for all scalars u E H.In particular, al and a2 are real inner products on V. Let bi : V --+ V*

be defined by

(bi (x)) (Y) = ai (x, y), j = 1, 2.

Matrix Algebras 171

Given a E EndF(V), let a* E EndF(V*) denote the dual map from V*to V*. Let Adi(a) denote the adjoint of a with respect to aj. That is,aj (ax, y) = aj (x, Adi (a)y). Let b : V - (V*)* --.> V* denote the mapdual to bj. Note that aj(ar,y) = (bj(ax))(y) = (ax)(b*(y)) = x(a*bj* (y)),while aj (x, Adj (a)y) = (bj (x))(Adj (a)y) = x(bp Adj(a)(y)). Therefore,

(8.37) Adj (a) =

By hypothesis, Ad,(a) = Ad2(a) for all a E EndF(V) C EndF(V). Thus,by taking the dual of (8.37),

blabi 1 = b262 1 for all a E EndF(V).

Therefore, b2b1 E EndF(V) is in the centralizer of EndF(V), and the theo-rem follows from Lemma 8.25 as described below.

If F - R, then bl = cb2 with c E R*, and hence El = cE2. If F - C,then bl = b2c with c E C*. Therefore,

aI(x, y) = al(cx, y),

and it follows easily that el and 62 differ by a constant using (8.35).Further, if El, 62 are C-hermitian, then the constant must be real. IfF - H, then bzbl = R, right multiplication by a scalar c E H*, be-cause of part (b) of Lemma 8.25. Therefore, al(x, y) = a2(xc, y), andhence El(x, y) = E2(xc, y) = c e2(x, y) by Corollary 8.26. Since both el and62 satisfy E(xa, y) = Ae(x, y), the constant c must be in the center of H,i.e., c E R*.

An F-inner product E on a vector space V with scalar field F inducesan R-symmetric inner product on EndF(V).

Definition 8.38. Suppose V, E is an inner product space with scalar fieldF and real dimension N. Define

(8.39) (a, b) = N traceR a*b for all a, b E Endp V,

where a* denotes the E-adjoint of a.

Lemma 8.40. (, ) is an R-symmetric inner product on Endp(V), and

(8.41) (ab, c) = (b, a*c) for all a, 6, c E EndF(V).

Proof: Note that (,) is symmetric if traceR. a = traceR a*. If V, e is a realinner product space, then traceR a = traces a* by Problem 2.7. In general,

172 Inner Products

Ree is an R-inner product on V and (Re e)(ax, y) = (Re e)(x, a* y), sincee(ax, y) = E(x, a*y). Therefore, the general case reduces to the real case.To show that (, ) is nondegenerate it suffices to prove that if traceR, ab = 0for all a E MN(F), then 6 = 0, which follows by considering matrices awith all but one entry zero. Finally, (8.40) is immediate since (ab)* = b*a*.J

Definition 8.42. Suppose V, E is an F-inner product space. Given x, y EV, define the product x O Y E EndF(V) by

(8.43) (x 0 y)(z) = xE(y, z) for all z E V.

Note that xOy=0ifandonlyifx=y=0.Lemma 8.44. For all a E EndF(V),

(a) (ax) O y = a(x (D y), x O (ay) _ (x O y)a*,

(b) (x O y)(z (D w) = (xe(y,z)) 0 w,

(c) (x O y)* = y O x if e is symmetric,(d) (x O y)* = -y O x if a is skew.

Proof:

(a)

and

((ax) O y)(z) = axe(y,z) = a((x O y)(z)),

(x 0 (ay))(z) = xe(ay, z) = xe(y, a*z) = (x (D y)(a* z).

(b) (x (D y)(z (D w)(u) _ (x (D y)(Xe(w, u))= xe(y, ze(w, u)) = xe(y, z)e(w, u)

_ ((xe(y, z)) (D W) (U).

In all cases, but e C-symmetric or C-skew, (c) and (d) are proven bynoting

4U, (x (D y)*v) = e((x 0 y)(u),v) = e(xe(y,u),v) = e(y,'u)E(x,v)

= e(x,v)e(y,u) = e(ye(x,v),u) = e((y (D x)(v),u)

_ E&(u, (y (D x)(v))

The cases a C-symmetric or e C-skew are proven by deleting the conjuga-tions in the previous proof. J

Theorem 8.45. Suppose V, e is an F-inner product space. Then

(1, x O y) = (dimF V)-1 Ree(y, x) for each x, y E V.

Matrix Algebras 173

Proof: We give the proof in the most difficult case F = H.

Let (, ) denote a positive definite H-hermitian inner product on V.Choose an H-orthonormal basis with respect to (, ), say U1, ... , u)n E V,and let v1, ..., v4n denote the corresponding real orthonormal basis. Thatis vl = U1, V2 = u1i, v3 = U1, j, v4 = ul k, v5 = u2, v6 = u2i, etc. Thenthe real dual basis vi, ... v*,, is given by v,*(x) = Re (vj, x), j = 1, ... , 4n.Therefore,

4n 4n

traceR, x O y = vj* [xe(y, vj )] Re (vj, (xe(y, vj ))j=1 j=1

4n 4n

_ Re .j(vj,x)e(y,VA = Re [e(y, vj)(vi, x)]j=1 1

j=1

1

4n

=Ree (YVi(VIZ))j=1

However, since ui(ui, x) = u(u, x), etc.,

4n n

Evj(vj,x) = 4Euj(uj,x) = 4x,j=1 j=1

completing the proof, since dimR, v = 4 dimH V.

Corollary 8.46. Suppose V, e is an F-inner product space of F-dimensionn.

(8.47) (x (D y, a) = 1 Re e(ay, x) for all a E EndF(V).n

Proof:(x O y, a) _ (a*(x 0 y), 1) = ((a*x) 0 y, 1)

= 1 Ree(y,a*x) = n Ree(ay,x).

Corollary 8.48. Suppose V, c is an F-inner product space of F-dimensionn. If c is F-symmetric or F-skew, then

(8.49) (x0y,zcw) = n Re [e(x,z)e(y,w)]

174 Problems

If E is F-hermitian, then

(8.50) (xOy,z(D w) = 1 Re [e(x,z)e(y,w),n

Proof: We give the proof of (8.50). The proof of (8.49) is exactly the sameexcept for the last equality. By Corollary 8.46,

(x O y, z O w) =1

Re E((z (D w) y, x) =n1

Re E(zE(w, y), x)

= 1 Re [e-(X-,Z)'-(Y, w), . -Jn

PROBLEMS

1. Show that M2(R) has center R, M2(C) has center C, and M2(H) hascenter R.

2. Prove Lemma 8.25.3. Show that MN (F) x MN (F), F = R or H, has exactly two irreducible

F-representations, up to equivalence. Namely, given (a, b) E MN(F) xMN (F),

and

p+(a, b)(x) = ax

p_ (a, b)(x) = bx

for all x E FN,

for all x E FN.

4. Let Vl denote HN with the usual right scalar multiplication RA, A EH. Let V2 denote HN with right multiplication by A E H given byRµaµ-1, where µ E H is a fixed unit quaternion. Let p, (a = 1,2)denote the R-representation of MN(H) on Va defined by p,,,(a)v = anfor all v E Va. Find (explicitly) the intertwining operator f for thesetwo H-representations.

PART II. SPINORS

9. The Clifford Algebras

We start with V, (, ), 1111, a euclidean vector space of dimension nand signature p, q, and with the orthogonal group O(p, q) acting on V.

The Clifford algebra Cl(V) is constructed in this chapter. It has con-siderably more structure than just that of an algebra. In fact, if algC1(V)denotes the underlying associative algebra with unit, then each algebraalg Cl(V) is nothing more than one, or occasionally two, copies of a matrixalgebra, i.e., MN(R),MN(C), or MN(H). Some patience is required, asthis important fact is not proven until Chapter 11!

The reader is assumed to be familiar with the exterior algebra

n

AV AVP=O

and the tensor algebra

®VEOPVP=O

over V. Since the extent of this familiarity may vary, we include a briefreminder of these constructions.

For example, the tensor algebra is defined by OPV = FIR. Here Fis the vector space of all real-valued functions on V x x V (p times)that vanish except at a finite number of points. Let 4(vl, ... , vP) E Fdenote the function that vanishes everywhere except at the single point

177

178 The Clifford Algebras

(vi, ... , vp), where it has the value one. Thus, F consists of all expressionsof the form L..N1 aiq(vi,..., v,), with each ai E R. These expressions aresubject to certain rules. This is made precise by quotienting out by all therules or expressions one wishes to vanish. Thus, R is defined to be thevector subspace of F generated by

(vii...)Vi_l,x+y,Vi+1i...,Vn)

- 0 (vi.... ) vi-1, x1 vi+1, ... , vn) - (vi, ... , Vi-1, Y, Vi+1) ... , Vn)

and

0 (vi, ... , vi-11 cx) Vi+l, ... , vn) - CO (v1, ... , vi-1, x, Vi+l, ..., vn)

with c E R. The equivalence class ¢(v1,. .. , vn)+R is denoted vie .. -®vp E®PV.

The tensor algebra ®V is graded, associative, with unit, and has aninner product that we also denote by (, ). Let I(V) denote the two-sidedideal in ®V generated by all elements x ® x E ®2V with x E V. Thequotient algebra AV - ®V/I(V) is called the exterior algebra over V, andthe equivalence class of vi ® -. 0 vp is denoted v1 A ... A vp. In fact,AV = Zp=0 APV is a graded, associative, anticommutative algebra withunit (and with inner product also denoted (, )).

The exterior algebra AV may also be identified with a vector subspaceof ®V, the space of skew tensors. For example, x A y = a x 0 y- y ®x],x, y E V V. More generally,

(9.1) x1A Ax,=Alt(x1® ®xp)=-Esignoxo(1)0.

.®xo(p).

Elements of APV are called p-vectors and those of the form v1 A.. .Avpare said to be simple or decomposible. The set of all unit (i.e., lull = ±1)simple p-vectors is called the grassmannian of oriented, nondegenerate, p-planes in V, and denoted by G(p, V). Of course, when p = 1, this is justthe unit sphere in V.

Given w E V, let Ew : AV -+ AV, left exterior multiplication by w, bedefined by

Ewu=to Au foralluEAV.The adjoint I. : AV -> AV, defined by

(w A u, v) _ (u, Iw v) for all u, v E Av,

is called interior multiplication by w and sometimes is denoted by

I(u)=wLU.

Clifford Algebras 179

Observe thatI,,,v=(w,v) ifvEV,

and that interior multiplication I. is an antiderivation on AV; that is,

P

(9.2) Iw (Vi A ..Avp) = E(-1)k-1(w,vk) V1 A...nvk_lnvk+lA...Avpk=1

on simple p-vectors, and

II (u A v) = Iw(u) A V + (-1)deguu A I,, (v)

for all u, v E AV with u of pure degree.In particular,

(9.3) IIwIlu=wA(wLu)+wL(wAu) or IIwlI=EwoI.+I,,,oEE

for w E V and u E AV. This is called a chain homolopy for E. if IIwII 0 0.

Definition 9.4. Given a euclidean vector space V, (,) of signature p, q,the Clifford algebra Cl(V) is the quotient ®V/I(V), where I(V) is thetwo-sided ideal in ®V generated by all elements:

(9.5) x ®x + (x,x) with x E V.

If V - R(p, q), then Cl(V) will be denoted Cl(p, q). The multiplication onC1(V) will be denoted by a dot.

Polarizing x ® x + (x, x) E I(V) yields

(9.6) x ®y + y 0 x + 2(x, y) E I(V).

The Fundamental Lemma for Clifford algebras follows.

Lemma 9.7. Suppose ¢ : V -+ A is a linear map from V into A, anassociative algebra with unit A. If

(9.8) O(x)O(x) = -IIxII for all x E V,

then 0 has a unique "extension" (also denoted 0) to an algebra homomor-phism of Cl(V) into A.

V A(9.9)

1

Cl(V) 4'

180The Clifford Algebras

Proof: The fundamental lemma for tensor algebras says that 0 has aunique extension to ®V as an algebra homomorphism, which we also denote

by 0: ®V -+ A.The hypothesis O(x)O(x) = -JJxJJ is equivalent to requiring that the

kernel of 0 : ®V -+ A contains I(V). Thus ¢, descends to a well-definedmap on the quotient Cl(V) - ®V/I(V). I

Frequently, it is more convenient to use Lemma 9.7 in the followingform.

Remark 9.10. Suppose e1,.. . , e, is an orthonormal basis for V andO(ei) E A satisfy

(a) O(ei)2 = -IIei!I, and(b) q(ei)4(ej) + ¢(ej)g(ei) = 0 for all i # j.

Then r¢ has a unique linear extension 0 : V --+ A that satisfies q(x)2 =-JJxJ], and hence Lemma 9.7 is applicable.

Proposition 9.11. The subspace AV of ®V is isomorphic (as a vectorspace, not as an algebra!) to Cl(V) - ®V/I(V) under the quotient map.In particular, the natural map

V i-+C1(V)

is an inclusion.

Moreover, if x E V and U E AV Cl(V), then (Clifford product)

(9.12)

That is, Clifford multiplication is an enhancement of exterior minus interiormultiplication.Proof: The quotient map restricted to AV maps onto C1(V); that is, eachtensor can be expressed modulo 1(V) as a skew tensor. This is automaticfor tensors of degree zero and one. Each 2-tensor x (D y is the sum of asymmetric 2-tensor and a skew 2-tensor. The symmetric part belongs toI(V) modulo a tensor of degree zero. The proof is completed by induction.

To prove that AV injects into Cl(V), consider the map ¢ : V -+End(AV) defined to be exterior minus interior multiplication. Given x E V,

¢(x) _- E,x - Ix E End(AV).

Note that

[5(x) o O(x)] (u) _ (E. - Ix) (E. - Ix) (u)_ - (E., o Ix + Ix o E,x) (u) = -IlxIJu


by (9.3). That is, O(x) o O(x) _ -11x1j, so by the Fundamental Lemma, 0extends to an algebra homomorphism 0 : Cl(V) --> End(AV). Now considerthe linear map 0 : End(AV) , AV defined by evaluation at 1 E AV. Themap 0 o 0 : Cl(V) , AV is a right inverse to the natural map from AV toCl(V). To prove this, choose u E AV of the form

(9.13) u = ui A . . A up, with u 1 , . . .,up orthonormal.

Let [u] denote the corresponding element of Cl(V). Since

u=

0 ([u]) =

sign o uo(1) ®... ® uo(p),

sign Qo (uo(i)) o ... o 0 (uo(p))

sign v (Eu,(,) - I-,(,)) o ... o (Eu,(P)

Since ul, ... up is orthonormal, evaluating at 1 E AV yields

1

- sign vuo(1) A ... A ua(p) = ul A ... A up.0

Because V has a basis with each basis vector of the form (9.13), this provesthat the composite

AV - Cl(V) 4 End(AV) - AV

is the identity map on AV.Finally, we prove (9.12). Note that cS(x u) = ¢(x) o 0(u) and hence,

(0 o 0) (x u) = 0(x)(0(u)(1)) _ 0(x)(u) _ (Ex -.1,,,) (u),

since 0(u)(1) = (b o 0)(u) = u.

THE CLIFFORD AUTOMORPHISMS

An important special case of the Fundamental Lemma of Clifford algebrasis where the target algebra A is also a Clifford algebra.

182 The Clifford Involutions

Proposition 9.14. Suppose V1, (, )1 and V2, (, )2 are two inner productspaces. Each linear map f : V1 -+ V2 that preserves the inner products hasa unique extension (also denoted f) to an algebra homomorphism

(9.15) f : Cl(V1) -+ CI(V2)-

Proof. f(x) - f(x) _ -(f(x), f(x))2 = -(x,x)1 = -IIxII1 for all x E V1, sothat f satisfies (9.8). -1

Remark 9.16. Each linear map f : V1 --> V2 also has a unique extensionto an algebra homomorphism

(9.17) f : AV1 --> AV2.

The extensions (9.15) and (9.17) are the same under the canonical identi-fications Cl(V1) - AV1 and CI(V2) - AV2.

Remark 9.18. Suppose that V3i (, )3 is another inner product space andg : V2 -+ V3 is a linear map preserving the inner products then the com-posite h = g o f : V1 -+ V3 has a unique extension H to an algebra homo-morphism of C1(V1) to Cl(V3). By the uniqueness part of Proposition 9.14,h must be the composite of the extension of f with the extension of g.

In order to emphasize the fact that Cl(V) is much more than an alge-bra, let alg CI(V) denote Cl(V) considered just as an associative algebrawith unit. In particular, Aut(alg Cl(V)) denotes the algebra automor-phisms of C1(V).

Definition 9.19. A Clifford automorphism of C1(V), f E Aut(Cl(V)), isan algebra automorphism ofalg Cl(V) that also maps V to V.

Theorem 9.20. Aut(Cl(V)) = O(V).

_Proof: Suppose f E Aut(Cl(V)). Then f E O(V), since (f(x),f(x))-f(x)f(x) = -f(x2) = f((x,x)) = (x,x)f(1) = (x,x), for all x E V. Con-versely, each f E O(V) has a unique extension to an algebra automorphismof Cl(V) by Proposition 9.14 and Remark 9.18 applied to g = f'1.

THE CLIFFORD INVOLUTIONS

The Clifford algebra C1(V) comes equipped with a particularly distin-guished automorphism and two anti-automorphisms.


Definition 9.21. The isometry x r+ -x on V extends to an automorphismof the Clifford algebra Cl(V), because of Theorem 9.20. This automorphismof C1(V) will be denoted by x i-+ x and is referred to as the canonicalautomorphism of CI(V).

Note that x= x because of Remark 9.18.Define the even part of Cl(V) by

(9.22) Cleven (V) {x E C1(V) : i = x

and the odd part of C1(V) by

(9.23) Clodd(V) = {x E C1(V) : x = -x).

Under the canonical vector space isomorphism AV - C1(V),

(9.24) Cl even (V) = Aeven V

and

(9.25)Clodd(V) - AoddV

Defintion 9.26. The anti-automorphism of ®V defined by reversing theorder in a simple product, i.e., sending vi ® ®vp to vp ® ®vl, mapsI(V) to I(V), and hence determines an anti-automorphism of the Cliffordalgebra C1(V) _ ®V/I(V). This anti-automorphism obviously squares tothe identity (i.e., is an involution). It will be denoted by z and referredto as the check involution.

Note that the composition V o - o V is an automorphism of Cl(V)that equals minus the identity of V. Thus, by the uniqueness in Lemma9.7, V o - o V =-. This proves that V and -. commute. The secondanti-automorphism is defined to be the composite A = V o _ o V andis referred to as the hat involution x x.

We say that u E C1(V) is of degree p if u E APV C AV Cl(V).

Proposition 9.27. Ife E Cl(V) is of degree p, then ii = fu, u = fu, andu = ±u with the plus or minus depending only on p mod 4 as indicated inthe following table.

p mod 4(9.28)

AV

0 1 2 3

+ - + -+ ++ +

184 The Clifford Inner Product

Proof: It suffices to verify the last row of the table for check V. Also, wemay assume that u is an axis p-plane. Since ei ej = -ej - ei for i # j, andp - 1 + p - 2 -I- + 1 =

2p(p - 1), the check of the Ith axis p-plane is

(ei, ... eip) v = eip . .. . lei, = (-1)2P(P-1)ei1 .. eip.

THE CLIFFORD INNER PRODUCT

The canonical vector space isomorphism AV = Cl(V) provides the Cliffordalgebra Cl(V) with a natural inner product (, ), namely, the natural innerproduct on AV. This will be referred to as the Clifford inner product. Thus,if el, ... , e is an orthonormal basis for V, then lei : I increasing}, withei ei, ... eip, is an orthonormal basis for Cl(V). Note ei1 ... ei, =el A A eip are the same. The square norm on Cl(V) determined by theinner product will be denoted 11 11, as usual for AV. The inner product inCl(V) can be computed using the hat involution:

Proposition 9.29.

(x,y) forallx,yEC1(V).

Proof: Let (x, y)' _ (1, i y). The bilinear forms (,) and (, )' on Cl(V)are equal because lei : I increasing} is an orthonormal basis for (, )' aswell as (, ).

(el, ei)' = (1, eip ... ii, ei, ... eip)

= I1ei1II ... Ileipll = Ilei, A ... A eipll = (ei, ei),

since it u = -u2 = IIulI for all u E V. Similarly, (ei, ej)' can be seen tovanish if I J, completing the proof. J

The adjoint of Clifford multiplication by a is just Clifford multiplica-tion by a.

Proposition 9.30. Given a E CI(V),

and

for all u, v E Cl(V).

(a - u, v) = (u, a v)

(u - a, v) = (u, v a)

Proof: (a u, v) = (1, (a u)^ v) _ (l, u a v) = (u, a v).

We shall say u E Cl(V) is a simple product if u = ulu1,...,up E V.

... at,, with


Proposition 9.31. If u E C1(V) is a simple product u = ul ... up then

(9.32) Hull

Proof: up

llxll for all x E V. In particular, it u is a scalar multiple of 1, so thatProposition 9.29. J

Given it E Cl(V), let (u)k E AkV denote the orthogonal projection ofu on AkV.

Corollary 9.33 (Hadamard). F o r all u1, ... , up E V,

P-1

Iluill ... Ilupll = Ilui A ... A upll + E II(ul ... up)kll-k=0

Thus, if V, (,) is positive definite,

(9.34) Iul A ... A upl <- lull ... lupl

with equality if and only if u1i..., up are orthogonal.

Proof: Note that u1i ... , up are orthogonal if and only if u1 ... up =u1 A...Aup.

The Clifford inner product can also be expressed in terms of a trace(see Theorem 9.65).

THE MAIN SYMMETRY

Suppose e1,. .. , e is an orthonormal basis for R(r, s) (n = r + s), andE, E1,. .., E is an orthonormal basis for either R(r + 1, s) (with E2 =-IJEJl = -1) or R(s, r+1) (with E2 = -IJEll = 1). Define q5(ei) =-EE1, j =1,...,n. Then ¢(ei)5(ei) = EE;EEj = -E2EiEE = if i # j.Also, 4(ei)O(ei) = -E2EE = E2IIEill. We wish to apply Lemma 9.7 (in theform discussed in Remark 9.10) and extend 0 to an algebra homomorphism.The remaining condition to verify is

(9.35) O(ei)c(ei) = -Ileidl for i= 1,...,n.

There are two cases to consider. First, if E2 = -1 and IlE.ill = lleillfor all j, then (9.35) is valid and ¢ extends to an algebra homomorphism

(9.36) 0 : Cl(r, s) -* Cl(r + 1, s).

186 The Main Symmetry

Second, if E2 = 1 and IIE,JI _ - II ejII for all j, then (9.35) is valid and 0extends to an algebra homomorphism

(9.37) ¢ : Cl(r, s) - Cl(s, r + 1).

Theorem 9.38. The maps defined above provide algebra isomorphisms:

(9.39)

and

(9.40)

C1(r, S) - Cl(r + 1, s)even

Cl(r, s) = Cl(s, r + 1)even,

which preserve the hat involutions.

Proof: Certainly the image of ¢ is contained in the even part of thetarget Clifford algebra. Each product EEti belongs to the image, and henceE;Ej = ±EE;EEE also belongs to the image. Thus, 0 surjects onto theeven part. By counting dimensions, 0 must be an isomorphism. Note that

(9.41) ¢ preserves even degrees.

(9.42). 0 adds one to odd degrees

Therefore, the hat involutions are preserved.

Combining these two isomorphisms yields the main symmetry for bothCl' eveand Cl.

Cl(r, Seven - CI(S' r) even

Cllr -1, s) = Cl(s - 1, r).

Moreover, these isomorphisms preserve the hat involutions.

Remark 9.46. In fact, the isomorphism Cl(r, seven = Cl(s, r) even pre-serves degree, while the isomorphism Cllr - 1, s) = Cl(s - 1, r) preservesA2mRn

® A2.n-'R".


THE CLIFFORD CENTER

The center of the Clifford algebra Cl(V) is

(9.47) cen Cl(V) - {a E C1(V) : a x = x a for all x E Cl(V)}.

Since Cl(V) has a basis of simple products of vectors in V,

(9.47') cen Cl(V) = V}.

The twisted center of the Clifford algebra Cl(V) is

(9.48) twcen Cl(V) {a E Cl(V) : a x = x a for all x E Cl(V)}.

Note that

(9.48')

Of course, the real scalars R = A° V always belong to the center.

Lemma 9.49. If n = dim V is even, then

cen Cl(V) = A°V and twcen Cl(V) = A"V.

If n = dim V is odd, then

cen Cl(V) = A°V ® A'V and twcen CI(V) _ {0).

Remark 9.50. Perhaps a more convenient way to remember Lemma 9.49is in terms of a unit volume element A E A" V. First, note that the centerand the twisted center always belong to A° V ® An V. Second, note that

(9.51) (n odd) A commutes with all of the Clifford algebra Cl(V).

(9.52)(n even) A commutes with Cleven(V) and anticommutes

with Clodd(V)

All the information in Lemma 9.49 is contained in this remark.

Proof. First note that for any non-null vector e E V, we have the following

(9.53) If a is even and a e = e a, then a does not involve e.(9.54) If a is even and a e = -e a, then a does involve e.(9.55) If a is odd and a e = -e a, then a does not involve e.(9.56) If a is odd and a e = e a, then a does involve e.

188 Self Duality

For example, to prove (9.53), express a as a = x + e y, where x, y ECl(V) do not involve e. Then a e = x e + e y - e = x e - e2 - y, since yis odd; and e - a = e x + e2 y = x e + e2 y, since x is even. Therefore,y = 0, so that a = x does not involve e.

Now choose an orthonormal basis for V. An element abelongs to the center if and only if a commutes with each vector e if andonly if both the even and the odd part of a commute with each vector ej.Thus, it suffices to consider the even and odd parts of a separately.

Now if a is even and in the center, repeated application of (9.53) impliesthat a E A°V = R does not involve any of the e1, ..., e,,. The other casesare similarly handled.

SELF DUALITY

A choice of orientation for R(r, s) is equivalent to a choice A of one of thetwo unit volume elements for R(r, s). However, the conditions A2 = 1 andA2 = -1 are independent of the choice, A or -A.

Proposition 9.57. Suppose A is a unit volume element for R(r, s).

(9.58) A2 = 1 if r - s = 0, 3 mod 4,(9.59) A2=-lifr-s=1,2mod4.

Proof: First note that as = (-1)r for A - el er+,. Second, recall fromProposition 9.27, that

A=Aif n=r+s= 0,1 mod 4,-A if n-r+s=2,3mod 4,

Thus, if n = 2p is even, then A = (-1)PA, so that A2 = (-1)r-p =(-1)(r-°)/2; while if n = 2p + 1 is odd, then A = (-1)PA, so that A2 =(-l)r-P = (-1)(r-3+1)/2. 0

Definition 9.60. Suppose r - s = 0, 3 mod 4, or equivalently A2 = 1. Anelement a E Cl(r, s) is said to be self-dual if as = a and anti-self-dual ifas = -a. Let

(9.61) Cl(r, s)f = {a E Cl(r, s) : as = ±a}


denote the self-dual and anti-self dual parts of the Clifford algebra Cl(r, s).

Proposition 9.62. Suppose r - s = 3 mod 4. Then

(9.63) Cl(r, s) = Cl(r, s)+ ® Cl(r, s)-,

and both Cl(r, s)+ and C1(r, s)' are two-sided ideals in C1(r, s).

In fact, the only case in which Cl(r, s) has a nontrivial proper two-sided ideal is this case r - s = 3 mod 4 (see Problem 11.4 and Lemma8.6).

Remark 9.64. Cl(r, s)±, with the multiplicative identity defined to beI (I ± a), is an algebra (assuming r - s = 3 mod 4).

TRACE

The Clifford inner product can be expressed in terms of the trace given arepresentation of the Clifford algebra.

Theorem 9.65. Suppose p : Cl(r, s) --+ EndR(V) is an R-representationof the Clifford algebras Cl(r, s) on a vector space V of real dimension N.Then

(9.66) (a, b) =

N

trace, p(ab) for all a, b E Cl(r, s),

unless r - s = 3 mod 4 and trace, p(a) # 0.

Proof: Since (a, b) = (1, ab), it suffices to show that

(9.67) (1, a) =

N

traceR p(a) for all a E Cl(r, s).

Because traceR p(a) is a linear functional on C1(r, s), there exists an elementc E Cl(r, s) such that

(9.68) N traceR, p(a) = (c, a) for all a E Cl(r, s).

The property traceR AB = traceR, BA implies that (c, ab) = (c, ba) forall a, b E Cl(r, s). Therefore, (c&, b) = (ac, b), or c& = ac for all a, so thatc E center Cl(r, s). Consequently, if the dimension n = r + s of R(r, s) iseven, then c E R, so that

(9.69) (r + s even)1

traceR, p(a) = c(1, a).

190 The Complex Clifford Algebras

Setting a = 1, yields (9.66). While if the dimension n = r + s is odd, thenc = a + /3a with a,,3 E R. Thus,

(9.70) (r -l- s odd) N trace, p(a) = a(1, a) + 3(A, a).

Setting a = 1 yields a = 1, while setting a = A yields

_ (I) traces p(A)

Since r + s is odd, r - s is also odd and hence equals either 1 or 3 mod 4. Ifr - s = 1 mod 4, then A2 = -1 and hence p(.X)2 = -1. In this case, p(A) isa complex structure on V and therefore traces, p(A) = 0 is automatic. U

Remark 9.71. This proof shows that

(9.72) N traceR, p(a) _ (1 a) + (N)s

(tracea p(A))(A, a)

for all a E Cl(r, s), no matter what the signature r, s or the representation.In particular, for any representation p of a Clifford algebra Cl(r, s),

(9.73) traceR, p(v) = 0 for each u E R(r, s).

In fact, for any u E Cl(r, s) - AR(r, s) which does not have a degree 0 ordegree n = r + s part, (9.73) also holds.

THE COMPLEX CLIFFORD ALGEBRAS

Suppose V is a complex vector space with nondegeneate symmetric complexbilinear form (, ). Then, proceeding exactly as in the real case, the complexClifford algebra Clc(V) is defined by

(9.74) Clc(V) - OV/I(V),

where I(V) is the two-sided ideal in ®V generated by

(9.75) x ®x + (x, x) with x E V.

All of the results of this chapter carry over, with the field R replacedby the field C, except for the following two items. First, the HadamardInequality (9.34) for the positive definite case must be deleted. Second, aunit volume element A for V complex is only determined up to a multiple


e" A, so that Proposition 9.57 loses any significance. However, the conceptsof self-dual and anti-self-dual are now valid in all dimensions and indepen-dent of the choice of volume element. Some of the results that remain validfor Clc(V) have been reformulated as exercises.

Since all of the complex inner product spaces C(r, s) of the same di-mension n = r + s are isometric, and the isometry has a unique extensionto a complex Clifford algebra automorphism, there is (essentially) only onecomplex Clifford algebra for each dimension, denoted Clc(n).

Proposition 9.76. Let R denote conjugation (or the reality operator) onE (r, s) = R(r, s) OR C, the complexification of R(r, s). Since (Rx, Rx) _(x, x) for all x E C(r, s), R has a unique extension to a complex antilinearisomorphism of Clc(n), which is the conjugation for

(9.77) Clc (n) = Cl(r, s) OR C.

Proof: The fundamental lemma for complex Clifford algebras does notapply to R since R is not complex linear and does not preserve (, ). How-ever, R has a unique extension to a complex antilinear algebra automor-phism of ®V (define R(i) _ -i). This extension R maps x ® x + (x, x)to R(x ® x + (x, x)) = Rx ®Rx + (x, x) = Rx ®Rx + (Rx, Rx), andhence maps I(C(r, s)) to I(C(r, s)). Therefore, R descends to the quotientyielding a complex antilinear algebra automorphism of Clc(n). Since Rfixes R(r, s), R fixes the real subalgebra Cl(r, s) of Clc(n) generated byR(r, s). Since R is complex antilinear, R is -1 on iCl(r, s) C Clc(n). Now,counting dimensions,

and hence

Clc(n) = Cl(r, s) ® iCl(r, s),

Clc(n) = Cl(r, s) ®R C. iJ

PROBLEMS

1. Use (9.6) to show directly that the natural map of V to CI(V) isinjective.

2. Under the canonical isomorphism of vector spaces AV = Cl(V), showthat, for all vectors xl, x2i x3, x4 E V

(a) x1 x2 = x1 A x2 - (z1, x2),(b) x1 x2 ' x3 = x1 A X2 A x3 - (x2, 53)x1 + (x1, 53)x2 - (x1, 52)x3,

192

(c)

Problems

X1 X2'x3'X4=x1AX2AX3Ax4-- (x1, x2)23 A x4 + (x1, 23)x2 A x4 - (x1 i 24)x2 A x3

- (x3, x4)x1 A Z2 + (x2, 24)21 A 23 - (52i 23)x1 A x4

+ (x1, 52) (x3, 54) - (Si, 53)(52, x4) + (Si, 54)(52, X3)-

3. (a) Show that CI(1, 0) and C are isomorphic as algebras with innerproduct and that A (hat) on C1(1, 0) corresponds to bar (conjugation)on C.(b) Show that Cl(0, 1) and L are isomorphic as algebras with innerproduct and that A (hat) on C(0, 1) corresponds to bar (conjugation)on L.Note that V (check) is the identity on Cl(1, 0) and Cl(0,1) so thatA =- on both Cl(1, 0) and Cl(0,1).

4. (a) Show that as algebras with inner products

Cl(2, 0) = H,

and that the isomorphism can be chosen so that

Cleven (2, 0) - C C H, Clodd(2, 0) = jC C H,

and the hat on Cl(2, 0) corresponds to the bar (conjugation) on H.(b) Show that as algebras with inner products

Cl(1,1) = M2(R),

and that the isomorphism can be chosen so that

Cleven (1,1) '" R® R C M2(R) as diagonal 2 x 2 matrices,

and the A (hat) on Cl(1, 1) corresponds to the bar (conjugation) onthe normed aglebra M2(R).Alternatively, show that the isomorphism Cl(1,1) = M2(R) can bechosen so that

r / \ lCleven(1,1)-Lj `b b a) :a,bER}

the `space of/L-linear maps,

Clodd(1,1)- { I ab ba I :a,bER}

the space of L-antilinear maps,


and, as before, the A (hat) on C1(1, 1) corresponds to the bar (conju-gation) on the normed algebra M2(R).(c) Show that as algebras with inner product

Cl(O, 2) = M2(R)

and that the isomorphism can be chosen so thatl

CI even(0, 2) C 16 ab ) : a, b E R } (C-linear)

Clodd(1 1) ((6 a) : a, b E R y (L-antilinear),

and the A (hat) on Cl(O, 2) corresponds to bar (conjugation) on thenormed algebra M2(R).

5. Suppose u E V is nonnull. Given x E APV show that u L x = 0 if andonly if x=tLyforsome yEAP+1V.

6. Give the proof of Remark 9.16.7. Complete the proof of Lemma 9.49.8. The centralizer of A in B is defined to be

{bEB:ab=baforallaEA}.

(a) Show that a E Cl(p, q) belongs to the centralizer of Cl(p, q)even inCl(p, q) if and only if a x = x a for all elements x E G(2, R(p, q)) CAZR(p, q).(b) Show that the centralizer of Cl(p, q)even in Cl(p, q) is A°R(p, q)A"R(p, q) = span {1, Al.(c) Show that the center of Cl(p, q)even is A°R(p, q) = R if n = p + qis odd and A°R(p, q) ® A"R(p, q) if n = p + q is even.

9. Give the proof of Proposition 9.62.10. Suppose that p : C1(V) --* EndR(W) is an injective R-representation

of Cl(V) on W, where W, e is an inner product space; and consider p asthe identity so that Cl(V) C EndR(W) is a subalgebra of EndR(W).(a) Assume that, given a E Cl(V),

,-(ax, y) = e(x, ay) for all x, y E W.

If V is not positive definite, show that there exists an anti-isometry ofW, and hence that the signature of W is split (if W has a signature).

194 Problems

(b) Assume that, given a E Cl(V),

e(ax, y) = s(x, ay) for all x, y E W.

If V is not negative definite, show that there exists an anti-isometry ofW, and hence that the signature of W is split (if W has a signature).

11. Suppose R(p, q) is oriented and let A denote the (unique) unit volumeelement. The Hodge star operator* on AR(p, q) is defined by

a A (*b) = (a, b) A, for all a, b E AR(p, q).

Show that*b = bA, for all b E AR(p, q) = C1(p, q).

That is, * and (right) multiplication by A agree on elements of degree0 or 3 mod 4, while * equals (right) multiplication by -A on elementsof degree 1 or 2 mod 4.

10. The Groups Spin and Pin

In this chapter, we construct several groups contained in the Cliffordalgebra Cl(V). First, consider the multiplicative group Cl*(V) of invertibleelements in the Clifford algebra Cl(V). (Later we shall see that this groupis just one or two copies of a general linear group over either It, C, or Hdepending on the signature.) Since

u2 = -11ull for each vector u E V C C1(V),

a nonnull vector u E V belongs to Cl*(V), with inverse u-1 = -u/11ull,while a null vector u E V has no inverse in Cl(V).

Definition 10.1. The Pin group is the subgroup of Cl*(V) generated byunit vectors in V. That is,

(10.2) Pin eachuj EVandFujJ=1}.

Note that each a E Pin is either even or odd, since it is a simpleproduct of vectors.

Defintion 10.3. The spin group is defined by Spin = Pin fl Cleven(V)That is,

(10.4)Spin ={aECl*(V):a=ul u2, with eachuj E V

and Juj1 = 1}.

195

196 The Groups Spin and Pin

The alternate notation Pin (V), Spin (V) is slightly more precise. IfV - R(p, q) is the standard model, then the notation used will be Pin (p, q)and Spin (p, q).

Clifford multiplication by a vector on the left is just exterior minusinterior multiplication, therefore;

(10.5) xy = -yx if x, y E V are orthogonal,while

(10.6) xy = yx if x, y E V are colinear.

This proves the next lemma.

Lemma 10.7. If u E V is nonnull, then R,,, reflection along u, is given interms of Clifford multiplication by

(10.8) Rux = -uxu-1 for all x E V.

_ Motivated by this lemma consider the twisted adjoint representationAd of the group Cl*(V) on C1(V):

(10.9) Adax =_ axa-1 for all x E Cl(V).

Note that (Ad,, Adb)(x) = abx6'la-1 = Adab(x), so that Cl*(V) aa}

GL(Cl(V)) is a group representation.The adjoint representation Ad of Cl*(V) on C1(V) is defined by

(10.10) Adax =- axa-1 for all x E Cl(V).

Theorem 10.11.(a) The sequence

(10.12) 1 -+ Z2 -+ Pin O(V) -.. I.

is exact.(b) The sequence

(10.13) 1 --> Z2 --+ Spin naf SO(V) -+ I.

is exact.

The Groups Spin and Pin 197

Proof: If u E V is nonnull, then Adu = R.,, E O(V). Therefore, if a E Pin,then Ada E 0(V). The weak form of the Cartan-Dieudonne Theorem (Re-mark 4.25) states that each orthogonal transformation can be expressed asthe product of reflections along nondegenerate lines, so that id is surjec-tive. _

Now suppose a E Pin and Ad,, = Id. That is, assume ax = xa forall x E V. First, suppose a is odd. Then -ax = xa for all x E V. Thusai = xa for all x E Cl(V), so that a belongs to the twisted center. But(Lemma 9.49) the twisted center of Cl(V) cannot have an odd part. Second,suppose a is even. Then ax = xa for all x E V and hence for all x E C1(V);that is, a belongs to the center of C1(V). Therefore, by (Lemma 9.49),aEA°V=R.

Finally, since a = ul u, is a simple product of unit vectors, byProposition 9.29,

(10.14) IIall _ IIu1Il .. Ilurll = ±1 for all a E Pin.

This proves a E ZZ {1, -1} C Pin, completing the proof of part (a).Part (b) follows easily from part (a) and the fact that det Ru = -1 for

each reflection R . Cl

Remark 10.15. Note that the full strength of the Cartan-DieudonneTheorem 4.23 implies that

(10.16) Pin = {a E Cl* (V) :a = ul ur, r < n}.

THE GRASSMANNIANS AND REFLECTIONS

Recall that G(r, V), the grassmannian of all unit, oriented, nondegenerater-planes through the origin in V, consists of all simple vectors in A' V thatare of unit length. That is,

uEG(r,V)if u =with u1i...,ur E V and Jul = 1 (Ilull = f1).

Of course, we may choose u1,. .. , ur orthonormal without changing u. Thespan of u, denoted span u, is just the span of {u1,. .., ur}.

Given u E G(r, V), reflection along u, denoted R,,, is defined by

(10.17) RuxE -x ifxEspan u,and(10.18) if xE(span u)1.

Lemma 10.7 can be generalized.

198 The Grassmannians and Reflections

Lemma 10.19. If u E G(r, V) is a unit nondegenerate r-plane in V, then

Ruv=Adu x forallxEV.

Remark 10.20. Therefore, each reflection Ru E O(V) along a subspacespan u of V is replaced in the double cover Pin(V) of O(V) by either ofthe two elements Eu E G(r, V) C A''V C Cl(V) in the Clifford algebra.Proof of Lemma 10.19. Each u E G(r, V) can be expressed as u =ul A ... A u,., where ul, -, u,- is chosen to be an oriented orthonormalbasis for the nondegenerate subspace span u C V. Since ul, ..., U. areothogonal, R = R,,, o ... o Rug and u = ul A - - A u,. = ul u,., so that

Adu = Adu, o. oAdur.. Finally note that&, = Adu, for each j by Lemma10.7. J

By definition, the group Pin is generated by the element G(1, V) CPin. Although the definition of Spin appears to be "generative," it suffersfrom the defect that the generators u C G(1, V) are not in Spin. Thisdefect can be corrected in several ways. First, if e is any unit vector andS"-1 denotes the unit sphere in V, then e S"-1 generates Spin, since

w=±eue ES"-1.In addition, we have

Proposition 10.21. (n = dim V > 3). The group Spin is the subgroupof Cl*(V) generated by G(2, V). That is,

(10.22)Spin = with each

u; E G(2, V) C AV - Cl(V)}.

Moreover, r can be chosen even and < n.

This proposition is the "double cover" of a theorem in Chapter 4.However, the proof is so short that it is repeated here.Proof: First, suppose n = 3. Each a E Spin can be expressed as a = u1u2with u1i u2 E V unit vectors. Let A denote a unit volume form for V. Thenvi - u1A E G(2, V) and V2 - Au2 E G(2, V). Thus, a = ulu2 = fvlv2iwhere A2 = fl.

Second, suppose n > 4. Then, given a product u1u2 of unit vectors111, u2 E V, there exists a unit vector v orthogonal to both ul and u2 (seeProblem 4.10(b)). Hence u1u2 = ±ulvvu2 = (±ul A v) (v A u2). Li

If n is odd, then Spin is generated by another sphere as well as by thegrassmannian G(2, V) and e Sn-1.


Proposition 10.23. (n > 3) Suppose n - dim V is odd. The group Spinis generated by the sphere G(n - 1, V) in A"-1V

See Chapter 14 for applications (for example, Spin(7) C Cleven(8) =M8(R) is generated by {Ru : u E S6 C ImO}).Proof: Since n is odd, the unit volume element A is in the center of Cl(V).Therefore, for u, v an orthonormal pair and A2 = f 1,

u A v = uv = (±Au)(Av), with ± Au, Av E G(n - 1, V).

ADDITIONAL GROUPS FOR NONDEFINITE SIGNATURE

If the signature p, q is not definite (i.e., p > 1 and q > 1), then, in additionto the groups Spin and Pin, there are three other closely related groups (cf.Definition 4.48).

Note that the square norm (or Clifford norm) IIaII - (a, a) = (aa,1) _(aa, 1) is multiplicative on Pin because of (10.14), i.e.,

IIabII = IIaII IIbil for all a, b E Pin.

That is, the Clifford norm

(10.24) II - II : Pin -+ Z2

is a group homomorphism. The twisted Clifford norm, defined by

(a, b)' - (a, b) for a, b E Cl(p, q),

provides another group homomorphism from Pin to Z2. The kernels ofthese group homomorphisms are subgroups of Pin.

Definition 10.25.

(10.26) Pin^ {a E Pin as = 11(10.27) Pin' {a E Pin : as = 1}.

Note that the intersection of any two of the three subgroups of Pin,

(10.28) Spin, Pin^, and Pin",

is the same as the intersection of all three subgroups. This common inter-section is denoted

200 Additional Groups for Nondefinite Signature

(10.29) Spino = {a E Spin : a& = 1}

and called the reduced spin group.

Remark 10.30. If V is positive definite, then

(10.31) Pin' = Pin, Pin" = Spin, and Spin° = Spin;

while if V is negative definite, then

(10.31') Pin" = Pin, Pin" = Spin, and Spin° = Spin,

so that the only groups are Spin and Pin.

Since loll = II -all, the Clifford norm II II on Pin descends to thehomomorphism

(10.32) II IIs : O(V) --* Z2,

called the spinorial norm (of an orthogonal transformation-see Definition4.55). To prove this fact, assume that

o...oR1,,, EO(V)

is expressed as the product of reflections. By definition, the spinorial normof A is equal to the parity of number of reflections that are along timelike(11u11 = -1) directions, which of course is also equal to the Clifford normof u l u,.. Similarly, the twisted Clifford norm on Pin descends to thehomomorphism (det A)IIAIIs on O(V).

The next result is immediate from Theorem 10.11 (cf. Lemma 4.54).

Proposition 10.33. (p, q > 1) The following sequences are exact.

(10 34) Z -*S1- ° SOO *1i. 2 p n -(10.35) 1-*Z2-*Pin "-b01--*1

(10 36) vAd ).O+ *1-_.Pi. n - .2

There are certain distinguished classical groups containing Spino, Spin,Pin", or Pin, with equality occurring in low dimensions. We shall concen-trate on the case Spino, leaving the other cases to the reader.


Definition 10.37. The classical companion group, denoted Cp(V), is de-fined by

(10.38) Cp(V) - {a E Cl.... (V) : as = 1}.

The connected component of the identity element in Cp(V) is a subgroupthat is called the reduced classical companion group and is denoted byCp°(V).

Note: Spin°(V) C Cp°(V). By Problem 2(d), Spin°(r, s) is connected(unless p < 1 and q < 1). Therefore Spin°(V) C Cp°(V).

The identification of the groups Cp°(r, s) as classical groups is a majorpart of Chapter 13.

THE CONFORMAL PIN (OR CLIFFORD) GROUP

Definition 10.39. The conformal pin group (or Clifford group) consistsof all simple products of nonnull vectors. That is,

(10.40)with each uj E V nonnull}.

The proof of Theorem 10.11 applies to the conformal Pin group yield-ing

Proposition 10.41. The sequence

(10.42) 1 -+ R* --+ Cpin na CO(V) -+ Id

is exact.

The conformal Pin group has a very useful alternative description,as the subgroup of Cl*(V) that maps V to V under the twisted adjointrepresentation.

Theorem 10.43.

(10.44) Cpin = {a E Cl*(V) : Ada(V) C V}.

The proof of this result depends on the next lemma.

202 The Conformal Pin (or Clifford) Group

Lemma 10.45. The following are equivalent:

(a) as = IIaII,(b) as = Ilall,(c) as E R,(d) Qa E R.

Proof: For each a E C1(V),

hall = (1, ad) = (1, aa),

so the lemma follows easily. _J

Corollary 10.46. If a E Cl(V) satisfies the equivalent conditions ofLemma 10.45, then

IlabIl = IIaII IIbII for all b E Cl(V).

Proof: Ilabll = (1, abab) = (1, baab) = I1all(1, bb) = IIail IIbII u

Proof of Theorem 10.43: Let

I'- {aECI*(V):axa-1 EV ifxEV}

denote the group occurring on the right hand side of (10.44).We wish to show that the subgroup Cpin of F is all of I'. Because of

the exact sequence (10.42),

1--+R*-*CpinTd CO(V)-+Id,

it suffices to prove that

(10.47) 1 -* R* -+ F - CO(V)

is exact.Assume a E F and Ada Id. That is,

(10.48) ax=xa forallxEV.

Let a+ denote the even part of a, and a_ denote the odd part of a. Then(10.48) is equivalent to

(10.48') a+x = xa+ and a_x = -xa_ for all x E V.


Thus, the even part of a is in the center of Cl(V), and the odd part of ais in the twisted center of Cl(V). However, referring back to Lemma 9.49,the twisted center of C1(V) cannot have an odd part, and the even part ofthe center is always R*. Thus,

1-.R*-}F GL(V)

is exact.Next, we show that

(10.49) aaER forallaEF

Because of (10.47), as E R* if as E ker Ad. The hypothesis a E r says thataxa-1 E V if x E V. Therefore, the check anti-automorphism fixes axa-1if x E V. However, the check of axa-1 is a-1x& so that axa-1 = a-1xa or(aa)x = xaa, proving that as E ker Ad = R*.

Now by Lemma 10.45, if a E F, then a-1 = a/11all. Thus,

11axa-'11

= IIaII-2IIaxaII IIxII,

because of Corollary 10.46 and the fact that-and -are isometries. Thisproves that Ad : r --r CO(V), thus completing the proof of Theorem 10.43.U

Sometimes it is useful to adopt the following as alternate (equivalent)definitions of the groups Pin, Spin, and Spin°.

Corollary 10.50.

Pin = {a E Cl*(V) : Ada(V) C V and 1lall = ±1}Spin = {a E Cl*(V)even Ada(V) C V and h ai = ±11

Spin° = {a E Cl*(V)even : Ada(V) C V and hall = 11.

DETERMINANTS

In several different situations it is useful to compute the determinant of anelement of Spin or Pin acting on a vector space W.

The idea of the proof is the same in the two cases considered below.Suppose p : G -+ EndR,(W) is a real representation of a group G on avector space W. Further, suppose G is generated by elements u in a subsetS with

(10.51) u2=±l for each u E S.

204 Problems

Note that this is true for G - Pin with S - G(1, V); and true for G - Spinwith S - G(2, V) by Proposition 10.21 (if n > 2). Then,

(10.52) p(a)2 = ±1 for all a E S.

Consequently,

(10.53) detRp(a) = ±1 for all a E S,

and hence for all a E G.To prove (10.53) consider two cases.

Case 1:

(10.54)If p(a)2 = -1, then detR p(a) = 1,

because p(a) is a complex structure on W.

Case 2:(10.55)

If p(a)2 = 1, then p(a) is reflection along W_ through W+,where Wf - {x E W : p(a)z = ±x} are the eigenspaces of p(a).

Therefore,

(10.56) detR p(a) = (_l)dim W_

PROBLEMS

1. Verify Remark 10.30.

2. Suppose V, (,) has signature p, q, and el and e2 are orthonormal inV.

(a) Show that

cos 0 + sin 0el e2 = (- cos 0/2 el + sin 0/2 e2)(cos 0/2 el - sin 0/2 e2),

ifllelll=lle2ll=1,

and that

cos 0 + sin 0e1e2 = (cos 0/2 el - sin 0/2 e2)(cos 0/2 el + sin 0/2 e2),

if llel ll = Ile2ll = -1-

(b) If either p > 2 or q > 2, find a subalgebra of Cl(V) that is isomor-phic to C with Spin(V) fl c Sl the unit circle in C C Cl(V).

The Groups Spin and Pin

(c) If either p > 2 or q > 2, show that

Spin°(V) -+ SOT(V)

205

is a nontrivial double cover, i.e., show that -1 can be connected to +1in Spin°(V).(d) If either p > 2 or q > 2, show that Spin°(p, q) is connected.

3. Give the proof of Proposition 10.41.

4. Compute, explicitly as (classical) subgroups of Cl*(V),(a) Pin and Spin if p = 2, q = 0, or p = 0, q = 2.(b) Pin, Pin^, Pin", Spin, and Spin° if p = 1, q = 1.

5. (a) Show that1 --+ Z2 --+ Pin

Ad O(V) -> 1

is exact if n is even.(b) Show that for n odd,

O(V)-}1

is exact where K = {±1, ±A} with A a unit volume element. Notethat

K = Z2 X Z2 if p - q = 3, 7 mod 8, whileK=Z4 ifp-q=1,5mod8.

6. (a) Show that the natural inclusion of R(p, q) into R(p + 1, q) inducesan inclusion of algebras C1(p, q) C Cl(p + 1, q).(b) Prove that Pin(p, q) is included in Pin(p + 1, q) by

Pin(p, q) {a E Pin(p + 1, q) : Ada(ep+1) = ep+i

where e1, ... , ep, ep+i; ep+2, ... , is the standard basis forR(p + 1, q). Thus,

Pin(p, q) C Pin(p + 1, q)

1Ad 1AdO(p,q) C O(p+1,q).

7. Suppose p : Spin(r, s) --* Endc(W) is a group representation withdimension n = r + s > 3. Show that

deter p(a) E {±1, ±i} for all a E Spin(r, s).

206

8. Define the complex Pin group by

Pin(n, C) = {a E C1 (n) : a = ul u, with

uj E C" and Ijj11 = 1, j = 1,...,r}

Problems

and the complex Spin group by

Spin(n, C) - Pin(n, C) n Cleven(n)

Show that (see Problem 4.17)

(a) 1 -+ Z2 -+ Pin(n, C) - O(n, C) -+ 1 is exact,

(b) 1 --+ Z2 - Spin(n, C) Ad SO(n, C) --+ 1 is exact,

(c) Pin(n, C) _ {a E C1c(n) : Ada(Cn) C C" and IIall = 1}.9. Suppose A E O(V) can be written as the product of reflections A =

R,,, o ...o R,,, with ul, . . ., u, linearly independent. Let a = ul u,. E

Pin(V). Let (a)k denote the degree k component of a in AkV C AVCl(V). Note (a)k = 0 fork > r and (a),. = ul A . A u,..

Suppose A = R , o o R, is any other representation of A as theproduct of reflections.(a) Show that k < r implies ul,..., u,. are linearly dependent, so thatk > r (i.e., r is the Cartan-Dieudonne rank of A).(b) Show that if k = r, then ul A . Au, = ±vl A A v,. (Thus, thereis a well-defined vector space span{ul, ... , u,} = span{vl,... , v,.} as-sociated with A.)

10. Show that, for n > 3, Spin°(p, q) is generated by G+(2, V) U G- (2, V)(the definite 2-planes).

11. Show that the centralizer of Spin°(p, q) in Cl(p, q) is A°R(p, q)A"R(p, q), and hence the center of Spin°(p, q) is equal to(a) {±1} = Z2 if n is odd or q is odd.(b) {f1, ±A} = Z2 ®Z2 if p - q = 0 mod 4 and n, q are even.(c) {±1, ±A} = Z4 if p - q = 2 mod 4 and n, q are even.

11. The Clifford Algebras Cl(r, s)as Algebras

Recall the algebra isomorphisms

(11.1) Cl(1,0)= C, C1(0,1)=L=R®R

(11.2) Cl(2, 0) - H, Cl(1, 1) = M2(R), Cl(0, 2) - M2(R)

established in Problems 9.3 and 9.4. Each of the Clifford algebras Cl(r, s) isisomorphic, as an associative algebra with unit, to a matrix algebra. Someof the important extra structure of Cl(r, s), that is hidden in the corre-sponding matrix algebra will be exposed in terms of "pinors" in Chapter13.

Theorem 11.3. The Clifford algebras Cl(r, s) of signature r, s are iso-morphic, as associative algebras with unit, to the matrix algebras listedbelow.

r - s mod 8:0,6 Cl(r, s) MN (R)2,4 Cl(r, s) = MN (H)1,5 Cl(r, s) = MN (C)3 C1(r,s) = MN(H) ® MN(H)7 Cl(r,s) MN(R) ® MN (R)

207

208 Clifford Algebras Cl(r, s) as Algebras

Remark 11.4. The integer N is, of course, trivial to compute in terms ofthe dimension n = r + s, since the Clifford algebra Cl(r, s) has dimension211 as a real vector space.

Before giving the proof of Theorem 11.3, we examine some reformula-tions and corollaries.

Some readers may find the following tabular form of Theorem 11.3convenient for reference.

Table 11.5. Matrix Algebras Isomorphic to Cl(r, s)

M16(H) M128(R)M16(R) M16(C) M16(H) e M32(H) M64(C) M128(R) a 256(R)

M16(H) M12,3(R)

M5(H) M64(R)M8(C) M8(H) a M16(H) M32(C) M64(R) e M12,3(R) M12s(C)

Ms(H) M64(R)

M4(H) M32(R)M4(H) e Ms(H) M16(C) M32(R) a M64(R) M64(C) M64(H)

M4(H) M32(R)

M2(H) M16(R) M32(H)e M4(H) M8(C) M16(R) e M32(R) M32(C) M32(H) e

M2(H) M16(R) M32(H)

M8(R) M16(H)M2(H) M4(C) M8(R) e M16(R) M16(C) M16(H) M32(H)(R)

M Ml (H)

MMM2(C) M4(R) ED M8(R) Ms(C) Ms(H) ED M16(H) M32(C)

M4(R) M8(H)

M2(R) M4(H)M2(R) M4(R) M4(C) M4(H) M8(H) M16(C) M32(R)

M(R)

M H)M2(H) M16(R)

ReR M2(R) M2(C) M2(H) M4(H) M8(C) M16(R) eM2(H)H) (R)

Ms (R)R C H H E) H M2(H) M4(C) Ms(R) M16(R)

M(R)

r

Clifford Algebras Cl(r, s) as Algebras 209

The even part of the Clifford algebra, Cl(r, seven may be read of fromthe entry to the left of Cl(r, s) in this table since (see Theorem 9.38)

(11.6) Cl(r, )even - Cl(r - 1, s), r > 1.

For an entry

(11.7) Cl(0, seven Cl(s -1, 0), s > 1.

Note that (11.7) is equivalent to extending the table one column tothe left, in the obvious manner, so that (11.6) can be applied in the caser=0.Corollary 11.8. The even Clifford algebras Cl(r, seven are isomorphic, asassociative algebras with unit, to the matrix algebras described below.

r - 2 mod 8:0 Cl(r, s)even - MN (R) ® MN(R)

1,7 Cl(r, S)even - MN(R)3,5 Cllr, seven - MN(H)2,6 Cl(r, )even L-- MN (C)

4 Cl(r, ,)even - MN (H) ® MN (H)

Given a choice of unit volume element A for R(r, s), recall that anelement a E Cl(r, s) is said to be self-dual if Aa = a and anti-self-dual ifas = -a. Also recall that Cl+(r, s) denotes the space of self-dual Cliffordelements and Cl- (r, s) denotes the space of anti-self-dual Clifford elements.If Cl(r, s) or Cl(r, seven is isomorphic to the sum MN(F) ® MN(F) of twosimple matrix algebras, the summands can be easily identified in Cliffordterms as the self-dual and anti-self-dual parts, as follows.

Lemma 11.9.

(a) In the following cases,

r-smod8:3 Cl(r, S) MN(H) ® MN(H)7 Cl(r, s) = MN (R) 0 MN(R)4 Cllr, Steven = MN(H) ® MN(H)0 Cl(r, S)even S-:: MN ® MN (R)

a unit volume element A for R(r, s) must correspond to either (1, -1)or (-1, 1).

210 The Pinor Representations

Selecting A - (1, -1) yields

r-smod8:3 C1(r, s)+

C1(r, s)-- MN(H) ®{0}

{0} ® MN(H)7 Cl(r, s)+

C1(r,s)-= MN (R) ® {0}= {0} ® MN(R)

4 C1(r, s)even = MN(H) ® {0}Cl(r, s)even {0} ® MN (H)

0 Cllr, s)+ en - MN(R) ®{0}Cl(r, s)even - {0} ® MN (R) .

(b) In the cases where Cl(r, s) = MN(C)(i.e., r - s = 1, 5 mod 8) or thecases where Cl(r, seven - MN(C)(i.e., r - s = 2, 6 mod 8), a unitvolume element A for R(r, s) must correspond to ±i E MN(C). Inthese cases, the standard choice is A - i.

Proof: (a) Suppose A = M is one of the isomorphisms listed in this lemma.Of course, the identity 1 E A must correspond to the identity 1 E M, and-1 E A must correspond to -1 E M. The remaining elements in the centerof M that square to 1 are (1,-1) and (-1, 1). The remaining elements inthe center of A that square to 1 are ±A since A M and M has centerRE) R.

(b) The elements of the center of MN(C) that square to -1 are ±i.Hence ±i must, correspond to ±A E A.

THE PINOR REPRESENTATIONS

The irreducible representations of the Clifford algebra Cl(r, s) are calledthe pinor representations, while the irreducible representations of the evenparts Cl(r, seven are called the spinor representations.

Definition 11.10. The pinor representation of Cl(r, s) and the space ofpinors P(r, s) are defined as follows. (If the dimension n = r + s is odd,assume a choice of unit volume element for R(r, s)-an orientation-hasbeen made.)

For r - s = 0, 6 mod 8, an irreducible representation

Cl(r, s) = EndR,(P),

is called the pinor representation, and the real vector space P is called thespace of pinors.

Clifford Algebras C1(r, s) as Algebras 211

For r - s = 2,4 mod 8, an irreducible H-representation

C1(r, s) = EndH(P),

is called the pinor representation, and the right H-space P is called thespace of pinors.

For r - s = 1, 5 mod 8, an irreducible C-representation

C1(r, s) - Endc(P),

with the unit volume element in C1(r,s) corresponding to the complexstructure on P, is called the pinor representation, and the complex vectorspace P is called the space of pinors.

For r - s = 3 mod 8, an irreducible H-representation

C1(r, s)+ = EndH(P+)

is called the positive pinor representation, while an irreducible H-represen-tation C1(r, s)- - EndH(P_) is called the negative pinor representation.The right H-space P = P+ $ P_ is called the space of pinors. Note that

C1(r, s) - EndH(P+) ® EndH(P_),

and that the unit volume element is

0 E EndR(P+) ®EndH(P-).1

-10

For r - s = 7 mod 8, an irreducible representation

C1(r, s)+ = EndR(P+)

is called the positive pinor representation, while an irreducible representa-tion

C1(r, s)- - EndR,(P_)

is called the negative pinor representation. The real vector space PP+ ® P_ is called the space of pinors. Note that

C1(r, s) - End(P+) ® End(P_)

and that the unit volume element is

1 0 E EndR(P+) ®EndR(P_).0 -1

212 The Pinor Representations

Now Theorem 11.3 may be rewritten:

Theorem 11.3 (Pinor Version). The Clifford algebras Cl(r, s) are iso-morphic, as an associative algebras with unit, to the endomorphism alge-bras listed below,

r - s mod 80, 6 C1(r, s) = EndH(P)2, 4 C1(r, s) = EndH(P)1, 5 Cl(r, s) = Endc(P)

3 Cl(r, s) = EndH(P+) ® EndH(P_)7 C1(r, s) = EndR,(P+) ® EndR,(P_ )

Remark 11.11. If r-s = 1, 5 mod 8, the isomorphism Cl(r, s) Endc(P)is required to map the unit volume element in Cl(r, s) to i E Endc(P).That is, this is the standard irreducible C-representation of Cl(r, S). Theconjugate C-representation of Cl(r, s) defines the conjugate pinor space P:

(11.12) Cl(r, s) = EndH(P),

where the unit volume element in Cl(r, s) maps to -i E EndH(P). In allodd dimensions, Cl(r, s) has exactly two irreducible representations (withthe appropriate notion of equivalence of representations) that define twospaces of pinors, P, P or P+, P_ as the case may be.

The question of how unique or canonical the space P of pinors ismay be answered by the uniqueness results of Chapter 8 for intertwiningoperators, which are summarized in the next remark.Remark 11.13. Suppose that A is one of the real algebras MN(F) withF =_ R or F = H, and that pl : A EndF(P1), P2 : A --* EndF(P2) aretwo irreducible representations of A. Then there exists an F-linear mapf : P1 --* P2 (the intertwining operator) with

p2(a)=fop1(a)of-' for allaEA.

Moreover, f is unique up to a real scalar multiple c E R*. If it were not forthe flexibility of this scale c E R*, the concept of pinor would be canonical.

Next, consider the algebra A = MN(C). First recall that, for anyC-representation,

p : A -> Endc(W)

sends i E A to either i E Endc(W) or -i E Endc(W), and that (up toC-equivalence) there are exactly two irreducible C-representations:

p : A - Endc(W) where p(i) = i (standard), andp : A- Endc(W) where p(i) = -i (conjugate).


Suppose pl : A - Endc(P1) and P2 : A Endc(P2) are two equivalentC-representations of A. Then by definition there exists a C-linear mapf : P1 -> P2 (the intertwining operator) with

p2(a) = f o pl(a) o f-1, for all a E A.

Moreover, f is unique up to a complex scalar multiple c E C*. Again, if itwere not for the flexibility of this scale c E C*, the map f would be unique,making the concept of the pinor space P canonical.

In summary, the pinor space P is "projectively canonical."

THE SPINOR REPRESENTATIONS

The definition of the space of spinors is analogous to Definition 11.10 forpinors, except that now a choice of volume element is required for evendimensions.

Definition 11.14. The spinor representation ofCl(r, s)eve" and the spaceof spinors S(r, s) is defined as follows. (If the dimension n = r + s is even,assume that a choice of unit volume element for R(r, s)-an orientation-has been made.)

For r - s = 1, 7 mod 8, an irreducible representation

p : Cl(r, seven = Endrt(S)

is called the spinor representation, and the real vector space S is called thespace of spinors.

For r - s = 3, 5 mod 8, an irreducible H-representation

p : Cl(r, seven = Ends(S)

is called the spinor representation, and the right H-vector space S is calledthe space of spinors.

For r - s = 2, 6 mod 8, an irreducible C-representation

p : Cl(r, seven - Endc(S),

with the unit volume element A equal to the complex structure i on S, iscalled the spinor representation, and the complex vector space S is calledthe space of spinors. An irreducible C-representation

p : Cl(r,s)even = Endc(S),

214 The Spinor Representations

with the unit volumeelement A equal to -i where i is the complex structureon S, is called the conjugate spinor representation, and S is called the spaceof conjugate spinors.

For r - s = 0 mod 8, an irreducible representation

p+ : CI(r, s)+ en = EndH(S+)

is called the positive spinor representation, while an irreducible represen-tation

p_ : C1(r, seven - End1 .(S_)

is called the negative spinor representation. The real vector space SS+ a S_ is called the space of spinors.

For r - s = 4 mod 8, an irreducible H-representation

p+ : C1(r, s)+en EndH(S+)

is called the positive spin or representation, while an irreducible H-represen-tation

p_ : C1(r,s)even = EndH(S_)

is called the negative spinor representation. The right H-vector space SS+ a S_ is called the space of spinors.

The spinor representation(s) of C1(r, seven restrict to representationsof the group Spin(r, s) C Cl(r, Seven It is convenient to use the samesymbol for these representations of Spin(r, s) (i.e., p or p,;5 or p+, p_, asthe case may be). Now Corollary 11.8 may be rewritten:

Corollary 11.15. (The spin representations).

r-smod8:0 p+ e p_ : Spin(r, s) C Cllr, Seven

pfEndH(S+) a EndR,(S_)

C1(r, s)f en = EndH(S±)

4 p+ e p_ : Spin(r, s) C Cl(r, S) evenEndH(S+) a EndH(S-)

pf Cl(r,.)even = EndH(S±)2,6 p : Spin(r, s) C Cllr, seven Endc(S)

1,7pp

Spin(r, s) C Cl(r, s)even

: Spin(r, s) C C1(r, s)even

Endc(S)

= EndR,(S)3, 5 p : Spin(r, s) C Cl(r, s)even = EndH(S)


THE FIRST PROOF

We shall give two proofs of the important Theorem 11.3. The first proofis by induction on the dimension n - r + s and is based on the followingthree lemmas.

Lemma 11.16. Cl(1, 0) - C, C1(0,1) = R ® R, Cl(2, 0) = H, Cl(1, 1)M2(R), Cl(0, 2) = M2(R).

Proof: A set of "'y matrices" for each one of these five Clifford algebrasCl(r, s) is listed below:

For Cl(1, 0),

For Cl(0, 1),

For Cl(2, 0),

For Cl(1, 1),

For Cl(0, 2),

Lemma 11.17.

e1iEC.e1=rEL.e1 j, e2=kEH.

e1

(01

e21 0

E M2(R).

U

e2 = (10 -10) E M2(R).

(11.18) Cl(r + 1, s + 1) = Cl(r, s) ® Cl(1,1) = Cl(r, s) OR M2(R).

(11.19) Cl(s, r + 2) = C1(r, s) ® Cl(0, 2) = Cl(r, s) OR M2(R).

(11.20) Cl(s + 2, r) = Cl(r, s) ® CI(2, 0) = Cl(r, s) OR H.

Proof: Suppose is an orthonormal basis for R(r,s) C C1(r, s).Let u1i u2 denote an orthonormal basis for V C Cl(V), where V is a two-dimensional inner product space, and let '1= U1u2 denote the unit volumeelement.

Define W C Cl(r, s) OR C1(V) to be the inner product space withorthonormal basis

(11.21) e1 ®72, ..., e ®7J, 1 (2) U1, 10 U2,

and signature given by

(11.22)Ile2 07711 = i2IIeiII, j =1,...,n,II1®u.II= IlujIl, j=1,2.

Note that any two of the basis vectors in (11.21) anticommute and that

(e.i®,)2=e2®772=-77211e111

_-lief ®7711,11 2( . 3)

(1®v1)2=1 ®uj=- ilu.ilj=-ll1 ®uiil, j=1,2.

216 The First Proof

Therefore, by the Fundamental Lemma of Clifford algebras the naturalinclusion 0 : W -; Cl(r, s) ® Cl(V) extends to an algebra homomorphism

4 : Cl(W) -} Cl(r, s) 0 C1(V).

This map 0 must be surjective, since the basis for W generates the algebraCl(r, s) ® CI(V). Therefore, by a dimension count, 0 must be injective andhence provides the desired isomorphism

Cl(W) = Cl(r, s) ® C1(V).

The signature of W follows from (11.22), reversing that of R(r, s), whenV - R(2, 0) or V - R(0, 2), since in both these cases ,772 = -1. J

Lemma 11.24.

(11.25) Mk(F) OR Mm(R) = Mkm(F) for F = R, C, or H.

(11.26) H OR H M4(R).

(11.27) C OR H M2(C).

(11.28) COR CC®C.

Proof: The isomorphism (11.25) is a consequence of these two specialcases:

(11.25a) Mk(R) 0 M,, (R) = Mk,,,(R)

and

(11.25b) F®Mm(R)-M,(F) forF - CorH.

The algebra homomorphism ¢ : H OR H -> M4(R), defined by

(11.29) q(p®q)(x)-pxq forallxER4-H,

is an isomorphism since 0 is an isometry. (See Problem 1.)Let 0(i®1) (left multiplication by i) determine the complex structure

on H - R4 - C2. Then the subalgebra C OR H of H OR H correspondsto the subalgebra M2(C) of M4(R) under the map 0. This proves (11.27).

Since C2 = C ® C j = H, via the complex structure 0(i (9 1) (leftmultiplication by i), the matrix 0(l (9 i) (right multiplication by i) equals


the 2 x 2 complex matrix (i i ). (Note (z + wj)i = iz - i(wj). Thus,

C 0 C = C ® C, contained in M2(C) as the diagonal 2 x 2 matrices. U

The reader is invited to give the proof of Theorem 11.3, based onthese three lemmas, in Problem 2. A second proof of Theorem 11.3 will begiven in the course of developing additional information about the Cliffordalgebras in the next chapter.

THE SPINOR STRUCTURE MAP ON P(r, s)

Definition 11.30. An invertible real linear map s on P(r,s) is called aspinor structure map if s2 = ±1 and

(11.31) a = sas` z for all a E Cl(r, s) C EndR(P).

Thus, a spinor structure map on the space P(r, s) of pinors is sufficientextra structure to determine Cl(r, s)even as a subalgebra of EndF(P(r, s)).Recall that, if the dimension n = r + s is even, then

(11.32) a = AaA-' for all a E Cl(r, s),

while if the dimension n =- r + s is odd, then

(11.33) Aa = as for all a E Cllr, s);

where A is a unit volume element.

EVEN DIMENSIONS

If the dimension n =- r + s is even, then the spinor structure map s isdefined to be a choice A of the two unit volume elements. The next theoremdescribes how s = A determines the spinor space S(r, s) from the pinorspace P(r, s). This theorem is stated so that each of the four cases r - s =0, 2, 4, 6 mod 8 can be referred to independently of the other three cases.Consequently, there is a certain amount of repetition.

Theorem 11.34 (The Spinor and Pinor Representations in EvenDimensions). Define the pinor structure map s to be a choice of unitvolume element A.

For r - s = 0 mod 8: Consider the pinor representation

(11.35) Cl(r, s) = EndR,(P).

218 Even Dimensions

Define S+ and S_ to be the eigenspaces of A:

(11.36) St = {x E P : Ax = -

±X}-The decomposition

(11.37) P = S+ ®S_

determines a 2 x 2 blocking

(11.38) A= ( a d) for each A E EndR,(P),c

and isomorphisms:

(11.39) C1(r, seven (O b) : a E EndR,(S+), b E EndR,(S_)} ,

a(11.40) C1(r, s)""' = {(0 0) : a E EndR,(S+) } = Enda(S+),

(11.41) C1(r, seven - 1 ( 0

b): b E Enda(S_)} = Enda(S_)

that are the spinor representations. Note that

(11.42) A= 1 0

0 -1)

and that

(11.43) dimR, S+ = dimR, S_.


(11.44) C1(r, is) c--- EndH (P).

Define S+ and S_ to be the eigenspaces of A:

(11.45) Sf={xEP:Ax =±x}.

Clifford Algebras C1(r, s) as Algebras

The decomposition

(11.46) P = S+ ®S_

determines a 2 x 2 blocking

(11.47) A = (2d) for each A E EndH(P),

and isomorphisms

(11.48)

Cl(r, s)even = f (ab

) : a E EndH(S+), b E EndR(S_) }

EndH(S+) ® EndH(S_), JJJ

(11.49) C1(r, s)+en (O 0 ) : a E EndH (S+) } - EndH(S+),

(11.50) Cllr s)even f(O b) : b E EndH(S_) } = EndH(S_)

that are the spinor representations. Note that

A- C1 -10 1

219

and that

(11.52) dimH S+ = dimH S_.


(11.53) C1(r, s) = EndH(P),

with the right H-structure on P given by I, J, K. Utilizing the complexstructure I on P, define S and S to be the eigenspaces of A:

(11.54) S={xEP:Ax=xI}, S={xEP:)x=-xI}.Then C1(r,s)even C EndH(P) maps S to S and S to S, and the inducedisomorphisms

(11.55) p : C1(r, seven = Endc(S), C1(r, s)even = Endc(

220 Odd Dimensions

are the spinor representations. As R-representations, p and p are equivalentvia the intertwining operator

(11.56) J : S


(11.57) Cl(r, s) = EndR(P).

Define S and '9 to be the eigenspaces of A:

(11.58) S={sEP®RC:ax=ix},

Then Cl(r, s)even C EndR,(P) C Endc(P ®R C) maps S to S and '9 to S,and the induced isomorophisms

(11.59) p : Cllr, seven = Endc(S), p : Cl(r, s)e°eII = Endc(S)

are the spinor representations. As R-representations, p and p are equivalentvia the intertwining operator

(11.60) C:S

the natural conjugation on P OR C.

Proof: The cases r - s = 0, 4, 6 mod 8 are exercises (see Problem 6).Suppose r - s = 2 mod 8. Since A E Cl(r, seven commutes with

A, A preserves the eigenspaces S and S of A. Let a = p(A) denote A ECl(r, Seven restricted to S, and a = p(A) denote A E Cl(r, seven restrictedto S. Note that J : S --+ S is an anticomplex linear isomorphism. Since Acommutes with J,

(11.61) a(xJ) _ (ax)J for all x E S.

The theorem now follows easily. Note that p(A) = I and p(A) = -I.

ODD DIMENSIONS

Now we turn to the odd dimensional cases. Several questions arise. Doesa spinor structure map exist? How (non-) unique is it? Finally, how doess determine the spinor space S(r, s)? These questions are most easily an-swered by turning to the inverse question first: How to construct the pinorrepresentation from the spinor representation? The unit volume elementA provides the key to the answer to this question, since A commutes withCl(r, S)even


Lemma 11.61. If r - s = 1, 5 mod 8, then

(11.63) Cl(r, s) - C1(r, seven OR C, with A = i.

If r - s :=3,7 mod 8, then

(11.64) C1(r, s) = Cl(r, )even OR L, with A = r,

or equivalently,

(11.64') Cl(r, s) = Cllr, seven ® (R ®R), with A = (1, -1).

These algebra isomorphisms are the identity on Cl(r, S)even.

Proof: Since Cl(r, s) = Cl(r, s)even ® A Cllr, seven and A commutes withCllr, Seven

Cl(r, s) = Cl(r, s)even ®R A,

where A is the two-dimensional algebra spanR{1, A}.If r - s = 1, 5 mod 8, then A2 = -1 and A - C with A = i; while if

r - s= 3,7 mod 8, then A2=1 and A=L=RED Rwith A=r=(1,-1).

This lemma provides part of the second proof of Theorem 11.3 that isgiven in the next chapter.

Corollary 11.65. The even dimensional cases of Theorem 11.3 imply theodd dimensional cases of Theorem 11.3.

Proof: Of course, MN(F)®R(RED R) - MN(F)®MN(F) and MN(R)®RC - MN(C). Exactly as in Lemma 11.24, MN(H) OR C - M2N(C).

Theorem 11.66 (The Spinor and Pinor Representations in OddDimensions).For r - s = 1 mod 8: Given the spinor representation

(11.67) Cl(r, s)even EndR,(S),

define

(11.68) P = S OR C.

Since Cl(r, s) = Cl(r, s)even OR C, with J1 = i, there is an induced isomor-phism

Cl(r, s) - Endc(P)

222 Odd Dimensions

that is the pinor representation. The natural conjugations on P = S®R,Cis a spinor structure map. Note

SE{x-P:sx=x}.

All other spinor structure maps are of the form ete s, 0 E R.

For r - s = 3 mod 8: Given the spinor representation

(11.69) C1(r, s)even - EndH(S),

define P± - S so that P - S ® S. Since C1(r, s) ^' C1(r, seven ® (R ®R)with A = (1, -1), there is an induced isomorphism

C1(r, s) = EndH(S) ® EndH(S)

that is the pinor representation. The natural reflection

(0 111 0)

that maps (x, y) E P to (y, x) E P is a spinor structure map. All otherspinor structure maps are of the form

Peers=± e_er 0C

0

eer

onP=_S®S.For r - s = 5 mod 8: Given the spinor representation

(11.70) C1(r, s)even - EndH(S),

define P to be S with the complex structure I. Since

C1(r, s) = C1(r, s)even ® C,

with A = I, there is an induced isomorphism

(11.71) C1(r,s) - Endc(P)

that is the pinor representation. Given the isomorphism (11.70), the iso-morphism (11.71) induced by (11.70) is given explicitaly by

(11.72) (a 9 z)(x) = axz for all x E P,

Clifford Algebras C1(r, s) as Algebras 223

where a E Cl(r, seven - EndR(S) and z E C = spanR{1, I}. The mapJ (right multiplication) is a spinor structure map and all other spinorstructure maps are of the form eIBJ.

For r - s = 7 mod 8: Given the spinor representation

(11.73) Cl(r, s)even = EndR(S),

define P± - S so that P = S ® S. Since Cl(r, s) - Cl(r, s)even (g (R ® R)with A = (1, -1), there is an induced isomorphism

Cl(r, s) - EndR(S) ® Endri(S)

that is the pinor representation. The natural reflection

s=

that maps (x, y) E P to (y, x) E P is a spinor structure map. All otherspinor structure maps are of the form

0 eeT.eSTs = ± (e-8T 0

on P=S®S.

Proof: Except for the uniqueness results for a spinor structure map s, allother parts of the theorem are immediate consequences of Problem 1(c)and Problem 7.

Note that (n odd) a spinor structure map s for Cl(r,s) C EndR(P)must lie outside of cen Cl(r, s) = spanR{1, A} but still lie inside of thecentralizer of Cl(r, s)even in EndR(P). This leaves very little choice for s,and the precise range of possibilities for a spinor structure map s can bededuced from the next lemma (Problem 8). iJ

Lemma 11.74.r - s cen Cl(r, s) - spanR It, Al centralizer of Cllr, seven

mod 8 in EndR(P(r, s))

1 C with A i M2(R)

3 R® R with A r (1, -1) M2(R)

5 Cwith A I H7 R ®R with A - r (1, -1) M2(R).

224 Problems

Proof: In the case r - s = 1 mod 8, the lemma is just the statement that

(11.75) the centralizer of EndR(S) in EndR,(S OR C) is M2(R).

In fact, 1, i, s, and is form a vector basis for this centralizer, where s denotesthe natural conjugation on S OR C.

In the case r - s = 5 mod 8, the lemma follows from the fact that

(11.76)the centralizer of EndH(S) in EndR(S) is H, the field of

right scalar multiplications,

which is part (b) of Lemma 8.25.The cases r - s = 3 mod 8 and r - s = 7 mod 8 are similar. The

centralizer of

EndF(S)a

: a E EndF(S)(° ) }

in EndR(S ® S) is M2(R) for both F = R and F = H since H has centerR.

PROBLEMS

1. Let H OR H have the inner product defined by (a 0 b, c ® d)(a, c) (b, d). Let M4(R) have the natural positive definite inner prod-uct defined by

(A, B) = 4 trace ABt.

(a) Prove that the map H ®R H --> M4(R) defined by (11.29)preserves the inner products.

(b) Prove that 0 is an isomorphism.(c) Suppose VH is a right H-space with I, J, K a standard basis forthe scalars. Let VC denote VH with the complex structure I. Showthat (aOz)(x) = axz for a E EndH(VH), z E C - span {1, I}, x E V,defines an isomorphism

EndH(VH) OR C EndC(VC).

2. Using Lemma 11.16, Lemma 11.17, and Lemma 11.24, give the proofof Theorem 11.3.


3. Prove that Table 11.5 is symmetric about the lines q = p + k withk = 3 mod 4, and that CI(p, q) = Cl(p - 4, q + 4) as algebras.

4. Equip C" with the standard C-symmetric inner product

(a) Define the associated complex Clifford algebra, denoted Clc(n).(b) Prove that Clc(1) - C ® C and Clc(2) = M2(C) as complexassociative algebras with unit.(c) Prove that, as complex associative algebras with unit,

Clc(2p) - MN(C) with N = 2P,and

Clc(2p+ 1) = MN(C) ® MN(C)by showing

Clc(n + 2) Clc(n) ® Clc(2).

with N = 2P,

(d) Prove that Cl(p, q) OR C - Clc(n) for all signatures in a givendimension n = p + q.

5. Prove that Table 11.5 has the following periodicity isomorphisms:

Cl(p+1,q+1) - C(p,q)®M2(R),C1(p + 8, q) C(p, q) 0 Mis(R),Cl(p, q + 8) 25 C(p, q) 0 M16(R).

6. Give the proof of Theorem 11.34 for the case r - s = 0 mod 8. More-over, verify that the isomorphism Cl(r, s) = EndR(P) determines thefollowing isomorphisms:

Cil(r, Seven 251( a

Cl(r, S)odd - r ( 0b

Cl(r, s)+ -1(

ab

Cl(r, s)_ - 01(0(a

Cl(r, S)+en=

Cl(r, Seven = f ( 0l 0

a E Enda(S+), b E EndR,(S_) I ,

a E IIomR,(S_, S+), b E HomR,(S+, S_)

a E EndR(S+), b E HomR,(S+, S_) } .

a E HomR,(S+, S_), b E EndR,(S_)

a E EndR,(S+)

bEEndR,(S_)}.

226 Problems

7. (a) Given an algebra isomorphism A °-' EndR(S), show that there isan induced isomorphism A OR C '- Endc(S OR C). If s denotes thenatural conjugation on S OR C, then show that for a E A, z E C,

s(a ®z)s = a ®z.

(b) Given an algebra isomorphism A = EndR(S), show that

A®(R(D R)=A®A-{ (0 b) :a,bEA}CEndH(S®S),

and that

111 \ JJJ

8. Using Lemma 11.74, verify that the only possible spinor structure maps(for odd dimensions) are those described in Theorem 11.66.

9. (a) Suppose V is one of the 2-dimensional spaces R(2, 0), R(1, 1), orR(0, 2), and that W is the vector space defined in the proof of Lemma11.17. Recall from Lemma 11.17 that

Cl(W) = C1(r, s) 0 Cl(V) (as algebras).

Prove that the extra structure in C1(W) of the three involutions andthe inner product is given by

(1) (a ®b) = a ®b,

(2) (a ®b)^ = a ®

(3) (a ®b)v = a ®(4) (a (9 b, c ®d) = (a, c) (b, d).

(b) Show that, as associative algebras with unit,

Cl(p - 1, q) = CI(q - 1, p) (symmetry about the line q = p + 1).

Moreover, show that this isomorphism can be chosen to preserve thehat involutions and the inner products.

12. The Split Case C1(p, p)

The case Cl(r, s) with r = s is particularly easy to analyze, indepen-dent of the inductive method used in the last chapter. Furthermore, thesplit case can be used to determine information about the complex Clif-ford algebra Clc(2p) = C1(p, p) ®R. C and the other real Clifford algebrasCl(r, s) (r + s = 2p) as subalgebras of Cl(p, p) ®R, C. This chapter isindependent of Chapter 11 and may be read first.

A MODEL FOR Cl(p,p)

Let R(p, p) = R(p, 0) x R(p, 0) with z = (u, v) E R(p, p) and IIzII Iu!I - IIvII. Let Eux =- uAx denote left exterior multiplication by u E R(p, 0)for x E AR(p, 0). Let 1,,x = u L x denote left interior multiplication byu E R(p, 0) for x E AR(p, 0). Then Lu Eu - Iu denotes left Cliffordmultiplication by u in Cl(p, 0) L ARP, and Lu = Eu + Iu denotes leftClifford multiplication by u in Cl(0, p) = ARP.

Lemma 12.1. Lu and Ly anticommute for all u, v E BY = A'RR.

Proof 1:

Lu Lv + LU Lu = (E,, - Iu)(E, + Iv) + (E, + II)(Eu - Iu)= Eu IU + I Eu - (I Eu + E 1u)=(u,v)-(v,u)-0.

227

228 A Model for Cl(p, p)

Here we use the fact that

E. I,, + I, E. = (u, v),

obtained from (9.3) by polarization (i.e., replace w by u + v in

Proof 2: By Problem 1, Lv x = R+1. Therefore

(LV Lv +LvLu)x=uxv+uxv=0

for all x.

(9.3)). J

Lemma 12.1. Let e1,.. . , ep denote the standard orthonormal basis forR(p, 0). The algebra generated by E,,,.. ., Eej, lep is all ofEndn.(ARR).

Proof: The proof is by induction on p. If p = 1, then computing the four2 x 2 matrices for the operators IEEe, Iej Ee, Eele with respect to the basis1, e for AR yields the result. Now assume the lemma is true for p and letAli E Enda(ARP) denote the linear map that sends e j to e j and eK tozero for K # J (i.e., the matrix with a single 1 in the I, Jth position). Lete =- ep+l and observe that

(12.3)

IeAijEe maps e; to el,Aji1. maps aAej to el,

EeAli maps e; to e A ej,EeAijIe maps eAej to eAej,

while mapping all other eK or e A eK to zero.

Theorem 12.4. There exists an algebra isomorphism

(12.5) Cl(p, p) = Enda(AR'),

extending the map : R(p, p) --* EndR,(ARR) defined by

(12.6)

Proof: Because of Lemma 12.1,

(Lu + (Lu + (Lu )2+ (LU )2+Lu Lv +L Lu = - (Ilull - Ilvll)

The Split Case Cl(p, p) 229

for all u, v E RP. The Fundamental Lemma of Clifford Algebras impliesthat 0 extends to an algebra homomorphism in the following diagram:

R(p, p) - Enda, (ARP)(12.7) n

Cl(p, P)

Because of Lemma 12.1, the image of the map 4 : Cl(p, p) -+ Enda(ARP)must be all of Endf,(ARP). Therefore, by a dimension count, 0 must bean isomorphism. _1

Consider the map 0 : C1(p, p) -> EndR,(ARP) as an identification.

Corollary 12.8. Let el, ..., eP denote the standard orthonorrnal basis forR(p, 0).

(12.9) L...... Lep, Lei ... I Len

provides the standard basis (of y matrices) for R(p, p) C EndR,(ARP)Cl(p,p)-

Definition 12.10 (The Split Case). The vector space P(p,p) of pinorsis defined to be ARP, and the isomorphism

(12.11) Cl(p, p) = Enda(P)

is called the pinor representation of Cl(p, p).

Since the dimension n = 2p is even, the canonical automorphism ofCl(p, p) is given by

(12.12) a=Aaa-1 for all aECl(p,p),

where A is a unit volume element for R(p, p). Recall that there are twochoices for A and that making a choice of A is equivalent to choosing anorientation for R(p, p).

Definition 12.13 (The Split Case). Assume that an orientation forR(p, p) is given, and let A denote the positive unit volume for R(p, p).

(12.14) S+ E {xEP:Ax=x}

is the space of positive spinors, and

230 Pinor Inner Products for C1(p, p) - Enda(P(p, P))

(12.15) S_= {xEP:.1x=-x}is the space of negative spinors.

The map x = .1x for x E P is sometimes called the canonical pinorinvolution. This notation x -+ x =- Ax is consistent with the notationa a for a E Cl(p, p) because

(12.16) ax = Aax = )aA-lax = ax.The decomposition

(12.17) PS+®S_determines a 2 x 2 blocking

(12.18)

Note that

(12.19)

Lemma 12.20.

\b)d

for each A E EndR,(P).

(12.21) Cl even(p, p) = 1(0 6) : a E Endit(S+), b E EndR,(S_) }

(12.22)J 11

Clodd(p, p)C

°

0a E Enda(S_, S+), b E EndR,(S+, S_)

Proof: Consult (12.12) and (12.19). J111

Corollary 12.23. dimS+ = dimS_.

Proof: If s± = dim St, thens+ + s? = dim Cl even(p, p) = dim Cl odd (p p) _ 2s+s_. J

PINOR INNER PRODUCTS FOR Cl(p, p) - EndR,(P(p, P))

The pinor space has inner products i and E that determine the hat andcheck involutions as adjoints with respect to these inner products.

Let (, ) denote the inner product on P =- AR(p, 0) induced by thestandard inner product on R(p, 0). Let o = el A A ep denote the unitvolume element for R(p, 0). Utilizing the Clifford multiplication on PAR(p, 0) - Cl(p, 0), given x E P, let cx E P denote the Clifford productof a and x in CI(p, 0).


Definition 12.24. The hat bilinear form e on P = AR(p, 0) is defined by

(12.25) (p odd) t(x, y) _ (x, o y) for all x, y E P,

(12.26) (p even) e(x, y) _ (.i, oy) for all x, y E P.

The check bilinear form e on P = AR(p, 0) is defined by

(12.27) e(x y) E(x,y)

Hat/Check Theorem 12.28. Both e and a are nondegenerate bilinearforms on the space P of pinors. The Clifford hat anti-automorphism, AA, is equal to the adjoint on EndR(P) with respect to the bilinear form e.That is, given A E Cl(p, P) End(P), A is determined by

(12.29) e(Ax, y) = e(x, Ay) for all x, y E P.

Similarly,

(12.29') e(Ax, y) = e(x, Ay) for all x, y E P.

The bilinear forms 9 and e on P are either symmetric (with split signature)or skew, as indicated in the following table.

Table 12.30

pmod4 9 e

0 symmetric symmetric1 skew symmetric2 skew skew3 symmetric skew

Moreover,

(12.31) S+ and S_ are nondegenerate and orthogonal if p is even, and

(12.32) S+ and S_ are totally null if p is odd.

Proof: Since (,) is nondegenerate on P =- AR(p, 0), and both x + xand x E--* ax are invertible, the bilinear forms e and e are nondegenerate.Suppose (12.29) is true. Then (12.29') follows directly:

e(Ax, y) = e(Ax, y) = e(Ax, y) = e(x, Ay) = e(x, Ay).

232 Pinor Inner Products for Cl(p,p) = Endp,(P(p, p))

Let A - At denote the anti-automorphism of End(P) determined byt. Suppose we can show that

(12.33) if A E R(p,p) C Cl(p,p) = End(P), then At = -A.

Then (At)" is an automorphism of Cl(p,p) that extends the identity onR(p, p), so that by the uniqueness part of the fundamental lemma of Clif-ford algebras (At)^ = A for all A E Cl(p,p). Therefore, At = A^ for allA E Cl(p, p).

The next lemma implies, as a corollary that At = -A for A E R(p,p),completing the proof of (12.33).

Lemma 12.34. Suppose u E R(p, 0) and x E AR(p, 0).

(12.35) (p odd ) u A (ox) = -o(u L x) and u L (ax) = -o(u A x).

(12.36) (p even) u A (cx) = o(u L x) and u L (cx) = o(u A x).

Note that Lemma 12.34 implies

(12.37)

(12.38)

(p odd (Eu-lu)o=o-(Eu-Iu),

(Eu + Iu)o = -o(Eu + Iu),

(p even) (Eu - Iu)o = -o(Eu - Iu),(Eu + Iu)o = a(Eu + Iu)

Corollary 12.39.

and

for all u E R(p, 0).

(E.-II)t=-(Eu-Iu)

(Eu+Iu)t =-(Eu+Iu)

Proof: Suppose p is odd. Then

and

e((Eu - I3)x, y) = ((E. - II)x, oy) _ -(x, (Eu - Iu)cy)_ -(x, o(EE - .t )y) = -E(x, (Eu - Iu)y),

E((Eu + I1)x, y) ((Eu + II)x, oy) = (x, (Eu + Iu)oy)_ -(x, o(Eu + lu)y) = -E(x, (Eu + Iu)y)

The Split Case Cl(p,p) 233

The proof for p even is Problem 2(a). J

Proof of Lemma 12.34: Suppose uAx = 0. Then x is of the form x = uAyand hence u L ax = 0. Therefore, u A (ax) = uax and a(u L x) = -oux.Suppose U L x = 0. Then u A ax = 0. Therefore, U L (ax) = -uax anda(u Ax) = aux. If p is odd, then ua = au and (12.35) follows. If p is even,then ua = -au and (12.36) follows.

The proof of the Hat/Check Theorem 12.28 is completed as follows:If either f or E has a signature (i.e., is symmetric), then by Problem

9.10 the signature must be split.Table 12.30 is deduced as follows. First, suppose p is even so that

E(x, y) _ (x, ay). Then

(12.40) E(y, x) (y, ax) = (ax, y) = (ax, y) _ (x, &y).

Here & is the CI(p, 0) hat of the unit volume element a for R(p, 0). Since& = a if p = 0 mod 4, and & = -a if p = 2 mod 4, this verifies that e issymmetric if p = 0 mod 4 and that E is skew if p = 2 mod 4. The analysisof t for p odd is Problem 2 (b).

Using the facts that

(12.41) A2 . 1,

(12.42) A = A = A, for p even,

(12.43) A = A = -A, for p odd,

both (12.31) and (12.32) follow. For example, if p is even and x E S+, Y ES_, then

e(x, y) = -E(Ax, Ay) = -E(x, AAy) = -E(x, y),

proving that S+ and S_ are orthogonal.

The bilinear form e on P can also be understood in terms of (,) anda. Alternatively, t can be analyzed in terms of e using the relationship

(12.44) E(x, y) = E(Ax, y), with A = (1 -0) .

0 1

The details are left as Problem 3. [J

THE COMPLEX CLIFFORD ALGEBRAS(CONTINUED FROM CHAPTER 9)

The complex Clifford algebra Clc(2p) is obtained from the complex vectorspace Cep equipped with a square norm (see (9.74)).

234 The Complex Clifford Algebras

Theorem 12.45. As complex associative algebras with unit,

(12.46) Clc(2p) = Mv(C) with N = 2p.

Also, given an isomorphism

Clc(2p) - Endc(Pc),

there exist complex inner products ec and ec on PC, so that

(12.47) ec(ax, y) = 9C (X, ay)

and

(12.48) ec(ax, y) = 1c(x, dy)

for all a E Clc(2p) and x, y E Pc. Both ec and ec are C-symmetricor C-skew exactly when their counterparts, a and t in the split case, areR-symmetric or R-skew as indicated in Table 12.30.

Proof: It is convenient to use the square norm on C2p = C(p, p) parallelingthe real split case. Let

z - (u, v) E C(p, 0) x C(p, 0) = Cep

have square norm

llzll= -vp.

Let

(12.49) Lu E - I E Endc(AC")

(12.50) Lu = E, + I E Endc (ACp)

with interior multiplication I,, determined by the square norm llull = u2 ++u2 on Cp.

Exactly as in Lemma 12.1,

Lu Lu + Lv Lu = 0

since L,+, and LV acting on ACp are just the complexifications of the realoperators Lu and LV acting on AR". The map 0: C(p,p) --> Endc(ACP)defined by O(z) _ O(u, v) = Lu + Lv has a unique extension

0 : Clc(2p) --+ Endc(AC")


that is a complex algebra isomorphism. This proves that Clc(2p)MN(C).

Because of the results of Chapter 8, any other isomorphism Clc(2p)Endc(Pc) of complex algebras is equivalent, via a complex linear inter-twining operator, to the isomorphism

(12.51) Clc(2p) = Endc(ACP).

Thus, we need only construct ec and ec on ACE. Now simply take ec(and tc) to be the complexifi cations of the real inner products e and e onARE. H

Corollary 12.52. The complex Clifford algebras Clc(2p), in even dimen-sions, contain no proper nontrivial two-sided ideals.

Proof: Because of (12.46), it suffices to show that MN(C) has no propernontrivial two-sided ideals. This was done in Chapter 8.

Corollary 12.53. The real Clifford algebras Cl(r, s), in even dimensionsr + s = 2p, contain no proper nontrivial two-sided ideals.

Proof: Since Clc(2p) Cl(r, s) ®R C is the complexification of the realalgebra Cl(r, s), Corollary 12.53 is an immediate consequence of Corollary12.52.

Cl(r, s)(r+s = 2p) AS A SUBALGEBRA OF Clc(2p) = Cl(p,p)®RC

Each Clifford algebra Cl(r, s) of even dimension n = r + s = 2p is of theform Cl(p + k, p - k) with r - s = 2k, k E Z. These Clifford algebras arereal subalgebras of Clc(2p). By Proposition 9.76,

(12.54) Cl(p + k, p - k) OR C = Clc (2p).

The conjugation or reality operator fixing Cl(p + k, p - k) in Clc (2p)will be denoted R, when the signature r, s = p+k, p-k has been prescribed.

No matter the signature, the model taken for Clc(2p) = Cl(p+ k, p -k) OR C will be the complexification of the split case Cl(p,p). Thus,

(12.55) Clc(2p) - Cl(p, P) OR C

is realized as the complexification of the matrix algebra

(12.56) Cl(p, p) L EndR(ARE).

236 Cl(r, s)(r + s = 2p) as a Subalgebra of Clc(2p) - Cl(p, p) OR C

Namely,

(12.57) Clc(2p) - Endc(ACP),

where the pinor space for Clc(2p),

(12.58) ACP = ARP OR C,

is the complexification of the pinor space ARP for Cl(p, p).The conjugation, or reality operator for Cl(p, p), contained in Clc(2p)

Cl(p, P) OR C will be denoted C. The complexification ACP t--- ARP (DRC also comes equipped with a conjugation, the natural extension of theconjugation on

(12.59) CP - RP ® C.

This conjugation on ACP that fixes ARP will be denoted C (or x = Cx forx E ACP). In turn, the conjugation C on ACP induces a conjugation onEndc(ACP):

(12.60) a - CaC for all a E Endc(ACP),which fixes EndR(ARP) in Endc(ACP). Here EndR(ARP) is naturallyconsidered a subspace of Endc(ACP) by extending each real linear map ofARP to a complex linear map of ACP - ARP ® C. Thus,

(12.61) Ca ="d for all a E Clc(2p) - Endc(ACP).

Remark 12.62. Note that the map a -+ Ca = a is a real algebra auto-morphism of Clc(2p) = Endc(ACP), that is complex antilinear.

Also note that each element A E EndR(ACP) has a unique decompo-sition

(12.63) A = a + bC

into a complex linear map a E Endc(ACP) and an anticomplex linear mapbC, where 6 E Endc(ACP) is complex linear.

Let a+, ... , e+, e 1- , . . . , ep denote the standard basis for R(p,p) con-sidered as a subspace of EndR(ARP) - C1(p, p). That is,

et = L , left Clifford multiplication by ej(12.64)

on Cl(p,0) = ARP,

and

e- __ Lei, left Clifford multiplication by ej(12.65)

on Cl(0, p) = ARP.

The signature of R(p, p), considered as a subspace of

C(p,p) C Endc(ACP) = Clc(2p)

can easily be modified.


Definition 12.66. Let

(12.69) (k positive) ei , ... , ep , ie1 , ... iek ; ek}1, ..., ep

(12.70) (k negative) e+ , ... , eP+k , ieP+k.+1 , ... , ie4 , el , ... , ep

define a standard basis for R(p + k, p - k) as a subspace of C(p,p) CEndc(ACP) = Clc(2p).

Proposition 12.69. The real subalgebra of Endc(ACP) generated byR(p + k, p - k) is isomorphic to Cl(p + k, p - k).

Proof: The inner product (, ) on C(p, p), restricted to R(p + k, p - k) isreal-valued with signature p + k, p - k. Thus the Fundamental Lemma ofClifford Algebras applies, yielding an algebra homomorphism

(12.70) 0 : Cl(p+k,p- k) --* Endc(ACP)

that extends the inclusion

R(p + k, p - k) C Endc(ACP)

given by Definition 12.66. Since the dimension n - 2p is even, Cl(p +k, p - k) contains no proper nontrivial two-sided ideals, by Corollary 12.53.Therefore, the kernel of 0 is trivial. J

THE PINOR REALITY MAP

In order to identify Cl(p + k, p - k) as a matrix algebra, it is useful tocompute the centralizer of Cl(p + k, p - k) in EndR.(ACP), i.e., all reallinear maps of ACP that commute with Cl(p + k,p - k). Of course, thecentralizer always contains C, all complex multiples of the identity. Notethat EndR(ACP) is larger yet than ClC(2p) ; Endc(ACP).

Theorem 12.71. The centralizer of Cl(p+k, p- k) in EndFt(ACP), whichcanonically contains C, is the algebra

(12.72) M2 (R) if k = 0,3 mod 4 (r - s = 0,6 mod 8),

(12.73) H if k = 1, 2 mod 4 (r - s = 2, 4 mod 8).

Remark 12.74. Recall from Chapter 6 that both M2(R) and H arenormed algebras and that the square norm is uniquely determined bythe algebraic structure.

Theorem 12.71 will be proved after some lemmas are established.

238 The Pinor Reality Map

Lemma 12.75. A E Enda(ACP) commutes with Cl(p + k, p - k) if andonly if A is of the form A = a+bpC, with a, b E C and i defined as follows:

(12.76) (k even positive)y = ei ek

(12.77) (k odd positive) y = ei ...e+ ek+1 .. eP ,(12.78) (k even negative) µ = ep k+i...ep

(12.79) (k odd negative) p = ei epkei ep .

Proof: Given A E EndR,(ACP ), it has a unique decomposition (see (12.63))

A = a + 6C, with a, b E Endc(ACP).

Suppose A commutes with all u E Cl(p + k, p - k), or equivalently, with allu E R(p + k, p - k). Since Cu = iC, this is equivalent to

(12.80) au = ua and bu = ub for all u E R(p + k, p - k).

Since a commutes with i and

Endc(ACP) = Clc(2p) - Cl(p + k, p - k) ® i Cl(p -f k, p - k),

(12.81) a belongs to the center of Clc(2p).

Thus, a E C is a complex scalar (because of either the complex version ofLemma 9.49 or the fact that Clc(2p) L--- MN(C)).

Referring back to the choice (Definition 12.66) of basis for

R(p + k, p - k) C Cl(p + k, p - k) C Endc(ACP) - Clc(2p),the condition bu = ub becomes

(12.82)(k positive) b commutes with ei , ... , ep , ek+l, ... , eP

and anticommutes with e1 , ..., ek .

(12 83)(k negative) b commutes with ei , ..., ep k, ei , ..., ep

.

and anticommutes with epk+1 .. , eP

Since these conditions are valid for b if and only if they are valid forboth the even part of b and the odd part of b, it suffices to consider thecases of b even and b odd separately.

Suppose k is positive and b is even. Then, by (9.53), b does not involvee+ l ..., ep , ek+l, ., ep , while by (9.54), b does involve ei , ..., ek. Thus,b is a scalar multiple of ei ... ek , and hence k must be even.

Suppose k is positive and b is odd. Then by (9.56), 6 does involveei , ... , ep , ek+1, ... , eP , while by (9.55), b does not involve el , ... , ek .

Thus, b is a scalar multiple of ei . . . . eP ek+l ep , and hence 2p - kand k must be odd.

The proofs for k negative are similar and so are omitted.


Definition 12.84. Suppose p is defined as in Lemma 12.75. Then

(12.85) R= e'apC E EndR(ACP), for 9 E R

is called a choice of pinor reality map for Cl(p + k, p - k).

Lemma 12.86. Suppose R is a pinor reality map for Cl(p+k, p-k). Then

(12.87) Ri = -iR, i.e., R is complex antilinear.

(12.88) Ra = RaR-1 for all a E Clc(2p) = Cl(p + k, p - k) OR C,

(12.89)R2 = 1 for k = 0, 3 mod 4 (r - s = 0, 6 mod 8),

R2=-1 fork =1,2mod4(r-s=2,4mod 8).

Proof: Since e'Bp E Endc(ACP) Clc(2p), R = e'BpC is complexantilinear. By Lemma 12.75, RaR-1 = a if a E Cl(p + k, p - k), and sinceRi = -iR, RbR-1 = -b if b E i Cl(p+k, p- k). Therefore (12.88) is valid.Finally, to prove (12.89), note that R2 = e'BpCe1 ItC = p2, and that p isa unit volume element for

(k even positive)

(k odd positive)(k even negative)

(k odd negative)

R(0, k) C C1(0, k) c--- AR(O, k),

R(p, p - k) C Cl(p, p - k) = AR(p, p - k),R(k, 0) C Cl(-k, 0) AR(-k, 0),R(p + k, p) C Cl(p + k, p) = AR(p + k, p).

Now (12.89) follows immediately from Proposition 9.57.

Proof of Theorem 12.71. The centralizer of Cl(p + k, p - k) inEndR(ACP) is C ® RC, where R anticommutes with i and R2 = ±1 de-pending on k mod 4, so the theorem follows. O

It is convenient to choose the reality map to have an additional prop-erty. Recall the hat bilinear form e on the pinor space ARP for Cl(p, p)(the split case), as well as the complex extension, denoted Ec, which is acomplex inner product on the pinor space ACP = ARP 0 C for Clc(2p).

240 Pure Spinors

Lemma 12.90. The pinor reality map R for Cl(p+ k, p - k) can be chosenso that

Ec(Rx,Ry)=ec(x,y) forallx,yEACP.

Proof: Let R = e'BpC denote a reality map. Then

F(Rx, Ry) = E (Ce-'Opx,Ce-te,jy)

_ e (e-aepx e-4epy) = e2tee(izx,Ay)

= e2ieII/'lIe(x, y).

Since p is the product of unit simple vectors, IIpII = ±1. If IIµII = 1, takeR = ±IX, while if IIFiHI = -1 take R = ±ipC. J

A SECOND PROOF OF THE CLASSIFICATION THEOREM

A second proof of the important Theorem 11.3 can be given using thereality map R on the pinor space ACP for Clc(2p) = Cl(r, s) ®n, C.

Theorem 12.91.

If r - s = 0, 6 mod 8, then Cl(r, s) = Ends (ARCP) .

If r - s = 2,4 mod 8, then Cl(r, s) EndH (ACP).

Remark 12.92. Combined with Corollary 11.65 this theorem provides asecond proof of Theorem 11.3.Proof:

Case 1 (r - s = 0, 6 mod 8): Choose a reality map R for CI(r, s).Then R2 = 1. Therefore, R is a conjugation. In general, conjugations on acomplex vector space V are characterized by two properties:

(12.93) Ri = -iR and R2 = 1.

This is because the negative eigenspace for R is just i times the positiveeigenspace for R, which implies that these two eigenspaces are of the samedimension. Let

(12.94) ARC {x E AC :Rx=x}

denote the set fixed by the reality map R. Since R is a conjugation andRa = RaR-1 (for all a E Endc(ACP)) fixes Cl(r, s),

(12.95) Cl(r, s) = Ends(ARCP).


Case 2 (r - s = 2, 4 mod 8): The centralizer of Cl(r, s) in EndR,(ACP )C Endc(ACP) - Clc(2p) is H C ® RC. Therefore, H (the centralizer)defines a right quaternion structure on ACP, and since Cl(r, s) is exactlythe subset of Endc(ACP) that commutes with R,

(12.96) Cl(r, s) - EndH(ACP).

PURE SPINORS

In general, the orbit structure of Pin acting on the pinor space P is ex-tremely complicated. However, there is one particular orbit that warrantsspecial attention, the so-called pure spinors. We shall examine the splitsignature case and the complex case. See Lawson-Micheleson [12] for aslightly different treatment of the complex case and some nice applicationsto geometry. Also consult Proposition 13.78 and Problem 13.8 in the nextchapter for a further elaboration of the material in this section in terms ofsquares of spinors.

Let N denote the space of all p-dimensional totally null vector sub-spaces of R(p,p). Recall, by Problem 2.4, that the maximal dimensionpossible for a totally null subspace of R(p, p) is p. Consider a pinor repre-sentation:

(12.97) Cl(p, p) - EndR,(P).

Given a pinor x E P, let

(12.98) N. = {z E R(p, p) C Cl(p, p) : z(x) = 01.

Note that for x # 0, Nx is always totally null, since 0 = z2(x) = -jjzjjx forall z N.

Definition 12.99. A pinor x E P(p, p) is said to be pure if dimN, = p,i.e., NN E N. Let PURE(p) or PURE denote the set of all pure pinors.

Consider Pin(p) _- Pin(O,p) as a subgroup of Pin(p,p), where O(p)is the subgroup of 0(p, p), which acts as the identity on the first factorR(p, 0) C R(p, p). That is, A E O(p) sends (u, v) to (u, Av), and Pin(p) isgenerated by the unit sphere in R(0, p) C R(p, p).

See Proposition 13.78 for a further elaboration of the next theorem interms of the square of pure spinors.

242 Pure Spinors

Theorem 12.100. Pin(p) acts transitively on PURE/R* with trivial iso-tropic subgroup. O(p) acts transitively on N with trivial isotropic sub-group. The map sending x E PURE to Nx E N induces an equivarientisomorphism X : PURE/R* -* N:(12.101)

Xa(NN) = Nax = aNxa-1 for all a E Pin(p)and all x E PURE.

The orbit PURE consists of two connected components of dimensiona p(p - 1); one, denoted PURE+, is a subset of S+, the space of posi-tive spinors, while the other, denoted PURE-, is a subset of S_, the spaceof negative spinors.

Proof: First we show that O(p) acts transitively on N with trivial isotropysubgroup. Suppose N E N is totally null of dimension p. Then N fl ({0} xRP) = {0} implies that N = {(u, Au) : u E RP} is the graph over RP x {0}of a linear map A E End(RP). A graph N = graph A is totally null if andonly if

((u, Au), (v, Av)) = (u, v) - (Au, Av) = 0

for all u, v E R(p, 0), i.e., if and only if A E O(p). Now it is obvious thatO(p) acts transitively on N with trivial isotropy subgroup. In particular,since O(p) has exactly two components, this proves that N consists of twoconnected components, each of dimension 2p(p - 1). The p-planes in onecomponent are called a -planes, while the p-planes in the other componentare called Q -planes.

Suppose a E Pin(p, p) and x E PURE are given. Let z, w E R(p, p) berelated by z = a-1wa or w = aza-1. Then the following are equivalent:w E Nax, wax = 0, azx = 0, zx = 0, z E N, w E aNxa-1. This provesthat Pin(p) C Pin(p, p) maps the set of pure pinors PURE into itself andthat the map from PURE/R* to N is equivariant, i.e., Na., = aNxa-1

Next we shall show that there exists a positive spinor s+, which ispure, and that if Nx = N,+ for any other pure pinor, then x = A x+with A E R*, is a scalar multiple of s+. Assuming this fact, the proof ofthe theorem is completed as follows. Each totally null plane N E N isof the form N = aN,+a-1 = Na,+ for some a E Pin(p), since O(p) actstransitively on N. Therefore, the map from PURE to N is surjective. Theequivariance, plus the fact that Nx = N,+ if and only if x = As+ withA E R*, implies that PURE/R* 25 N.

Since Spin(p) C Spin(p, p) has two connected components, oneSpin°(p) = Spin(p) n Cleven(p p) consisting of even Clifford elements andthe other Spin(p)flClodd(p p) consisting of odd Clifford elements, the orbitof Spin(p) through s+ E S+ must consist of a component PURE+ in S+and a component PURE- in S_.


To prove the existence of a pure positive spinor s+, we use the modelCl(p, p) = EndR(P) of this chapter with P =_ AR(p, 0). Consider thepositive spinor s+ - 1 E AevenR(p, 0). A vector z - (u, v) E R(p, p) CCl(p,p) = End(P) acts on s+ by

z(s+) _ (Eu+0 , - Iu-v) (s+) = u + v.

Therefore, N,+ _ {(u, -u) : u E RP}, which is p-dimensional (i.e., s+ EPURE). If x E PURE and Ny = N,+, then for all z = (u, -u) EN,+, z(x) = -I2u(x) = -2u -i x vanishes, i.e., u i x = 0 for all u cR(p, 0) . This proves that the element x E AR(p, 0) P must belong toA°R(p, 0), i.e., x = )s+ for some A E R*, completing the proof of thetheorem.

Now we consider the complex case. Suppose Vc is a complex innerproduct space and assume that the dimension n - dimc VC = 2p is even.Let NC denote the space of all totally null p-planes through the originin V. (Note that Nc is a complex submanifold of Gc(p, VC) and henceis projective algebraic.) Assume that Clc(Vc) 25 Endc(Pc) is a pinorrepresentation. A pinor x E Pc is said to be pure if

(12.102) NN __ {z E VC C Cl(Vc) : z(x) = 0}

is of complex dimension p, i.e., Nx E N. Let PUREC denote the set ofpure spinors.

The complex inner product space Vc may be expressed as VcV(p, p) OR C with V(p, p) = R(p, p) split. The model

Cl(p, p) = EndR.(AR(p, 0))

can be complexified, and some parts of the proof of Theorem 12.100 carryover to the complex case. For example,

(12.103)(a) there exists a pure positive spinor s+ E S+, and(b) Nx = N,+ if and only if x = As+ for some ,1 E C*

are proved exactly as in the split case. However, just because N E Ncis totally null N fl ({0} x CP) need not be equal to {0}. That is, it is nolonger true that each N E Nc can be graphed over CP x {0} C CP x CP,as in the split case.

Consequently, the model

(12.104) Vc V(2p) OR C, with V(2p) = R(2p) positive definite,

244 Pure Spin ors

is more convenient. Now each N E NC can be graphed over V = V(2p) CV ® iV = Vc, since iV cannot contain any null vectors other than 0.Suppose N = graph A = {u + iAu : u E V} is the graph of A E Endlt(V).Then N is totally null if and only if (u+iAu, v+iAv) = (u, v) - (Au, Av)+i((Au, v) + (u, Av)) = 0. That is, A E O(2p) and At = -A, or equivalently,A E O(2p) and A2 = -Id. Thus, A E Cpx(2p) is an orthogonal complexstructure. This proves that

Nc = Cpx(2p).

Recall that Cpx(2p) - 0(2p)/U(p) with g E O(2p) sending J E Cpx(2p)to gJg-1. Therefore, NC O(2p)/U(p) with g E O(2p) sending N E NCto gNg-1. The space

(12.105) NC = Cpx(2p) = O(2p)/U(p)

is called the twistor space (at a point on a manifold), i.e., the twistor fiber.Remark 12.106. Suppose J E Cpx(V). Note that N = graph J = {u +iJu : u E V} is just V1"0(J), the +i eigenspace of J (extended to VC). Asin the split case, it follows easily that Na. = aNxa-1 for z E PUREC anda E Pin(2p, C). The group Pin(2p) is naturally a subgroup of Pin(2p, C),since V(2p) C Vc. This completes the proof of the following result.

Theorem 12.107. Pin(2p) C Pin(2p, C) acts transitively on PUREC/C*.The map x sending x E PUREC to NN E NC - Cpx(2p) is an equivariantisomorphism x:

(12.108) PUREC/C* - NC - Cpx(2p) - O(2p)/U(p).

Also, PUREC = PURE+ U PURE- with PURE a connected subset ofSC.

Remark 12.109. In particular, given s+ E PUREC (with associatedcomplex structure J) and an element a E Pin(2p):

(12.110) Xa(s+) E U(p) if and only if as+ = cs+ with c E C*.

Proposition 13.79 has a complex analogue which says that

(12.111) if 8 E PUREC then s 6 s= cdz1 A ... A dzp,

with c E C* and z1,. .. , zp a choice of complex linear coordinates forVc(2p). The square of s is defined by (s o s) (x) = se(s, x). It then followseasily that

(12.112) {a E Pin(2p) : as = s} = SU(p),


with the isomorphism given by the vector representation X. That is, theisotropy subgroup of Pin(2p) at a pure spinor is SU(p) (cf. Problem 8).

PROBLEMS

1. Given u E V and x E Cl(V), show that

-Fua x.

This formula has the following interesting interpretation: Right Clif-ford multiplication on x by a vector u is the same as first replacing xby i and then left multiplying by u using Clifford multiplcation basedon the inner product -(, ) of signature s, r rather than r, s.

2. (a) Give the proof of Corollary 12.39 if p is even.(b) Complete the proof that to is skew if p = 1 mod 4, t is symmetricif p = 3 mod 4.

3. Complete the proof of the e portion of Table 12.30.

4. (a) Prove that, for r - s = 1 or 5 mod 8, Cl(r, s) contains no propernontrivial two-sided ideals.(b) Prove that, for r - s = 3 or 7 mod 8, the only proper nontrivialtwo-sided ideals of Cl(r, s) are Cl}(r, s).(c) Suppose

R(r, s) A (nontrivial)n /

Cl(r, s)

is given exactly as in the Fundamental Lemma of Clifford Algebras.Show that 0 is injective if r - s # 3, 7 mod 8, while the kernel of 0 iseither {0}, Cl(r, s)+, or Cl(r, s) - if r - s = 3, 7 mod 8.

5. In the model P (p, p) = AR(p, 0), Cl(p, p) EndR,(P) of this chapter,show that the set of pure positive spinors PURE+ is equal to the subsetR* Spin(0, p) of AevenRP = S+

6. Prove that the span of PURE+ is S+(p, p).

7. Show that a positive spinor x E S+(4, 4) is pure if and only if x is null,i.e., r(x, x) = 0.

8. Suppose J E Cpx(2p) is an orthogonal complex structure on the pos-itive euclidean space R(2p). Suppose el, Jet, ... , eP, Jep, is an or-thonormal basis for R(2p) and let al, 01, ... , aP, PP denote the dualbasis. Then w = al A ,61 + ... + aP A /3P is the standard Kii.hler

246 Problems

form. Let X : Spin(2p) --+ SO(2p) denote the vector representa-tion. Identify AR(2p) and AR(2p)* using the inner product. LetA - a1 A Q1 A ... A ap A /3p denote the unit volume.

(a) Show that

0 = 2-p/2(1 + a1 A )1) . (1 + Cep A/3P) E Spin(2p) C Cl(2p)

is independent of the orthonormal basis.(b) Show that xm = J, 02 = A , and that 0-1 = OA = ¢.

(c) Define U(p) {a E Pin(2p) : X. E U(p)}. Show that

U(p) _ {a E Spin(2p) : a4 = Oa}

and

U(p) = {a E Spin(2p) : aw = wa}.

(d) Using (12.111), show that the isotropy subgroup of Pin(2p) at apure spinor is SU(p).(e) Show that x-'(SU(p)) has two connected components.

9. (a) Suppose N E N is a p-dimensional totally null plane in R(p,p)and a E GLR,(N). Show that there exists A E 0(p, p) with AIN = a;i.e., show that a extends to an isometry of R(p, p).Hint: Use the model N = RP x {0} C RP x RP with I I(u, v)II = uv.(b) Show that any linear isomorphism a : N, -+ N2 of totally nullp-planes N1, N2 E N can be extended to an isometry A of R(p,p), i.e.,AEO(p,p).(c) Prove Witt's Theorem: Suppose V and W are isometric euclideanvector spaces (i.e., same dimension and signature). Then each isometrya : V --> W, from a subspace V of V onto a subspace W of W extendsto an isometry A from 7 to W.

13. Inner Products on theSpaces of Spinors andPinors

The hat and the check anti-automorphism of Cl(r, s) can be describedin terms of extra structure on the space P(r, s) of pinors. These two anti-automorphisms agree on Cl(r, s)even and will be referred to as the canonicalinvolution of Cl(r, seven. This canonical involution on Cllr, seven can bedescribed in terms of extra structure on the space S(r, s) of spinors.

THE SPINOR INNER PRODUCT

The spinor inner product, denoted c, is an inner product on the spaceS(r, s) of spinors with the property that the adjoint with respect to a isthe canonical involution of Cl(r, s)eVen. The next result establishes the ex-istence of spinor inner products and classifies the types. It is interesting tonote that all of the different types of inner products introduced in Chapter2 can occur as spinor inner products.

Theorem 13.1. There exists an inner product e on the space of spinorsS(r, s) called the spinor inner product with the property that, given a ECl(r, Seven

(13.2) (ax, y) = r(x, ay) for all x, y E S.

247

248 The Spinor Inner Product

Table 13.3. Spinor Inner Products

r - s = 0 mod 8:

S-S+®S_ real

Cllr, S)even =EndR(S+) (D EndR(S_)

p mod 4

0

1

2

3

r - s = 1,7 mod 8: p mod 4

0

C1(r, S)even = EndR(S) 1

2

3

r- s= 2,6 mod 8: p mod 4

0

C1(r, Seven = Endc(S) 1

2

3

r-s=3,5 mod 8: p mod 4

0

Cl(r, S)even - EndH(S) 1

2

3

r - s = 4 mod 8: p mod 4

0

S=-S+eS_1

C1(r, s) 2

EndH(S+) e EndH(S_)3

e =- c+ E) e-, bothof R-symmetric

e reflective, S_ = S+e - e+ ® e_, both et R-skew

e reflective, S_ = S+

e

R-symmetricR-skewR-skewR-symmetric

e

C-symmetricC-hermitian (symmetric)C-skewC-hermitian (skew)

e

H-hermitian skewH-hermitian symmetricH-hermitian symmetricH-hermitian skew

e

e - e+ e_, both,-:k H-hermitian skew

e reflective, S_ S+

E =- -+ E) e-, bothE± H-hermitian symmetric

e reflective, S_ = S+

Inner Products on Spaces of Spinors and Pinors 249

The type of the spinor inner product a is described in the Table 13.3. Heren - r + s = 2p defines p if the dimension n is even, and n - r + s - 2p + 1defines p if the dimension n is odd. If the signature of R(r, s) is definite(i.e., positive, in case s = 0, or negative in case r = 0), then a is definite.(In particular, a must be of the type that has a signature.) In all othercases, where e has a signature, the signature ofe is split.Remark. Since the spinor inner product e is only determined up to achange of scale (see Theorem 8.33), it is not the signature r', s' of a that isan invariant but the absolute value Ir'-s'I. However, we continue to use theterminology "e has a signature." In the cases e - e+ ®E-, another abuse ofterminology occurs in Theorem 13.1. Here either e+ or e_ may be changedby a nonzero real constant, so that, strictly speaking, e - e+ ® e_ does nothave a signature even in the absolute value sense, Ir' - s'I, discussed above.However, if e± has a signature r±, s±, then I r± - st I is a true invariant.

The proof of Theorem 13.1 (constructing the spinor inner products e)will be given as a corollary of Theorem 13.17 (constructing the pinor innerproducts e and e).

The next remark applies to the algebra A = Cl(r, s)even for eitherr - s = 0 mod 8 (with F-= R), or r - s = 4 mod 8 (with F - H). It givesthe definition of the term e-reflective used in Table 13.3.Remark 13.4. Suppose that the algebra A is of the form

A = MN(F) x MN(F) --- EndF(V+) ® EndF(V_).

Suppose e : V+ x V_ -* F is a nondegenerate hermitian F-bilinear form (orequivalently, b(u)(v) - e(u, v) for all u E V+, v E V_ defines an F-linearisomorphism b : V+ --+ V-*). Then e is said to be reflective, and the anti-automorphism of the algebra A sending (a, 6) E A to (b*, a*) E A, definedby

(13.5)e(au, v) = e(u, a*v) and

e(u,bv)=e(b*u,v) foralluEV+,vEV_,

is called the (reflective) a involution of A.See Problem 1.7 for the definition of W*, when W is a right H-space.Suppose a reflective e : V+ x V_ --> F is given. Then there exists

an (unique up to a change of scale) F-hermitian symmetric inner producton V - V+ e V_ (also denoted e) with the property that the reflective einvolution is the adjoint with respect to this inner product on V. Namely,(13.6)

c(z1, x2) = e(x1, y2) + 6(x2, yi)for all z1 = (x j, yr), x2 = (x2, y2) E V - V+ ®V_,

250 Spin Representation and (Reduced) Classical Companion Group

defines the inner product on V. Note that we could have defined this innerproduct on V by

E(zl, Z2) = E(xl, Y2) - E(x2, yl)

so that E would be F-hermitian skew. Thus, the property of the innerproduct on V being symmetric or skew is of no significance, and we simplysay E is reflective on V - V+ ® V_.

THE SPIN REPRESENTATION AND THE (REDUCED)CLASSICAL COMPANION GROUP Cp°(r, s)

As noted after (10.38), the reduced spin group Spin°(r, s) is a subgroup ofthe classical companion group

(13.7) Cp(r, s) - {a E Cleven(r, s) : a& = 1}.

Since a is the adjoint of a with respect to the spinor inner product E, thisgroup Cp(r, s) is indeed a classical group, namely, the group that fixes theinner product e. Thus Theorem 13.1 provides a classical description of thiscompanion group Cp(r, s) (see Problem 1).

The subgroup of the classical companion group, equal to the connectedcomponent of the identity element in Cp(r, s), is called the reduced classicalcompanion to Spin°(r, s) and denoted by Cp°(r, s).

Recall from (10.53) and Problem 10.7 that for the spin representationsp in dimension n > 3, the determinant (real or complex) of p(a), witha E Spin, is in the finite set {f1, ±i}. Therefore, for a E Spin°(r, s), eachsuch determinant must be one, because Spin° (r, s) is connected.

Each of these reduced classical companion groups (is a classical groupthat) can be read off from the next theorem. For example, if r-s = 0 mod 8and n = rd-s = 2p with p, = 0 mod 4 then Cp°(r, s) = SO1(S+) X SOt(S_)is the reduced classical companion of Spin°(r, s) (cf. Problem 5). Themain result of the section Cartan's Isomorphisms in Chapter 14 is obtainedby reading off the (reduced) classical companion group Cp°(r, s) in lowdimensions.

Inner Products on Spaces of Spinors and Pin ors

Theorem 13.8 (The Spin Representations).r - s = 0 mod 8:

251

= 0 mod 4 1 -+ Z S)P + SO1(S{1 in°(rAl -+ Sp 2 , ,p +),

1 -+ Z2 {1, -Al -+ Spin°(r, s) -°_-+ SOT (S_),

p = 2 mod 4 1 Z2 {1, A) -+ Spin°(r, s) -°+ Sp(S+, R),1 -* Z2 {1, -A} -+ Spin°(r, s) -°__+ Sp(S_,R),except for Spin° (2,2)

p = 1, 3 mod 4 1 - Z2 {1, A} -* Spin°(r, s) °--> SL(S+, R),

1 --* Z2 { 1, -A) _+ Spin°(r, s) -°_+ SL(S+, R),except for Spin(1,1) = GL(1,R).

r - s = 4 mod 8:

p = 0 mod 4 1 -+ Z2 1, A} ---+ Spin°(r, s) P* SK(S+),

1 -+ Z2 {1, -A} -+ Spin°(r, s) -°-+ SK(S_),

p = 2 mod 4 1 --+ Z2 {1, Al -+ Spin°(r, s) -°-+ HU(S+),1 -+ Z {1 -A} - + S in°(r s) -° =+ HU(S2 , ,p _ ),except for Spin(4,O) = Spin(0,4),

p = 1 3 mod 4 1 A} -+ S in°(r= {1 ) °-+* SL (S H- Z, , p , s +, ),2

1 = {1 in°(r- Z -A} -+ S ) -° =* SL(S H)r - s= 1 7 mod 8:

2 , , sp +, .

,

p= 0,3mod4 Spin°(r, s) C SO1(S),p = 1,2 mod 4

r - s= 2 6 mod 8:Spin°(r, s) C Sp(S, R).

,

p = 0 mod 4 1 --* Spin°(r, s) -°- SO(S, C),

1 -+ S iri°( ) -° + SO S Cp r, s ( , ),p= 1,3mod4 1 - Spin°(r, s) -°-+ SU(S),

1 -* Spin°(r, s) -°-+ SU(S),except for Spin(2,O) = Spin(0,2) = U(1),

= 2 mod 4 1 -+ S in°( ) -f + S S Cp p r, s ,p( ),

1 -+ S i °( ) -° S S Cp n r, s + p( , ).r - s = 3,5 mod 8:

p= 0,3 mod 4p= 1,2mod4

Spin°(r, s) C SK(S),Spin°(r, s) C HU(S).

252 The Pinor Inner Products a and e

Proof: The only part of Theorem 13.8 that does not follow from the spinorinner product theorem and the extra determinant restrictions discussedabove is the fact that

(13.9) ker p+ = {1, A}, kernel p_ = {1, -A}

for r - s = 0 mod 4 and the dimension n - r + s > 6. This follows fromthe next theorem (cf. Problem 10.11).

Theorem 13.10. If the dimension n - r + s > 5 and

p : Spin°(r, s) ---; EndR,(W)

is a nontrivial representation of the (reduced) spin group Spin°(r, s), then

(13.11) ker p C cen Spin°(r,s).

This is a standard result proved in most textbooks on Lie algebras (thecomplex Lie algebra so(n, C) has no nontrivial two-sided ideals for n > 5,i.e., so(n, C) is simple for n > 5).

THE PINOR INNER PRODUCTS a AND

The extra structure on P(r, s) required to describe the hat anti-automor-phism, mapping a to a, is an inner product on P(r, s), denoted e, with theproperty that given a E Cl(r, s),

(13.12) e(ax, y) = e(x, ay) for all x, y E P.

The extra structure on P(r, s) required to describe the check anti-automorphism, mapping a to a, is an inner product on P(r, s), denoted e,with the property that given a E Cl(r, s),


Recall from Definition 11.30 and Theorems 11.34 and 11.66 that thereexists a spinor structure map s E Enda(P) that determines the canonicalautomorphism of Cl(r, s) by

(13.14) a = sas-1 for all a E Cl(r, s).

Consequently, either one of E or a can be used to determine (define) theother by the formula:

(13.15) E(x, y) - E(sx, y) for all x, y E P.


For example, if e satisfying (13.12) is given and t is defined by (13.15),then

(13.16) e(ax,y) - E(sax,y) = E(asx,y) = E(sx,ay) - E(x,ay).

Because of Theorem 8.33, 9 and z are unique up to a change of scale(as usual, the cases r - s = 3,7 mod 8 will be somewhat different). Inparticular, if both 9 and e are given, then a change of scale may be requiredin order for (13.15) to be true.

The next theorem describes the pinor inner products e and e.

Theorem 13.17. There exist inner products e and ton the space of pinorsP(r, s) with the property that, for each a E Cl(r,s)

(13.18) E(ax, y) = E(x, ay) and E(ax, y) _ ff(x, ay) for all x, y E P.

The type of the pinor inner products t and e is described in Table 13.19.Here n = r + s = 2p defines p if the dimension n is even, while n - r + s2p + 1 defines p if the dimension n is odd. If R(r, s) is positive definite(i.e., s = 0), then 9 is (positive) definite. If R(r, s) is negative definite(i.e., r = 0), then is (positive) definite. In all other cases, if t or t has asignature it must be split.

The proofs of the spinor Theorem 13.1 and the pinor Theorem 13.17are intertwined in the following five steps.

Step 1. Suppose n is even and that the E portion of the Pinor Theorem isvalid. Then the t portion of the Pinor Theorem is valid.

Step 2. Suppose n is even. The inner product t is constructed from theinner product Ec of Theorem 12.45 using the pinor reality mapR.

Step 3. Suppose n is odd. The Spinor Theorem is deduced from the PinorTheorem for even dimensions.

Step 4. Suppose n is odd. The Pinor Theorem is deduced from the SpinorTheorem.

Step 5. Suppose n is even. The Spinor Theorem is deduced from the PinorTheorem for odd dimensions.

254

y H

M >, x

C+~ C4 C4 C

V

C

The Pinor Inner Products E and E

'w

<w

V

V

U

M0

o CO

a0

o -I

oo

oU wb®W

aa

0.-,o" E

1-+

Ul

211 IIII I

Co

II u a NII

Inner Products on Spaces of Spin ors and Pinors 255

w ® hIII II

J v 'w 'w 'W 'ww &I

cad cad cad J Ce 2

a acd cd

y Q+ IIM 401

-HOII .,

c, U ti U

N 211 III aW u) a 3 W to ° -

256 The Pinor Inner Products E and e

bo A bo A+'S'.arya

v o 0 0 0 >w

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cad cad Cy

w .dM -

xi x :- -G

cd RI.. ii3 3 I. 3 I~a a w a o

rn-

<w ' > S. Np

W W W W 0e-i N M

Q.

b0 O GV MFR'i

r-100 xv' 00

0 nal (D V0

2 + 5211 II

I II I ya

UII

-c

I I


Cl ClT! N

w wII v II N,,w >w ,w w

O

cad

s~

0

O

cad 0

y N N

V

L

N M y h

,w III

w

w'w

II

w

wm n m .n <w -

w w

v

vb0

II v

4

II

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a.

b-4 t14 cl

a

258 The Pinor Inner Products t and e

Proof of Theorems 13.1 and 13.17.Step 1.

Assume n = r + s = 2p is even. Also, for the moment assume that 9has been constructed and satisfies Theorem 13.17. Let A denote the choiceof unit volume element for R(r, s).

Definition 13.20. (n even). The (check) pinor inner product t on P(r, s)is defined by

(13.21) e(x,y)=e(Ax,y) for allx,yEP.As noted above in 13.16, if a E Cl (r, s), then

e(ax, y) = e(x, ay) for all x, y E P,

because of the corresponding fact for e. Also note that

(13.22) e is F-bilinear if and only if e is F-blinearand

e is F-hermitian bilinear if and only if(13.23) t is F-hermitian bilinear.

Lemma 13.24. If n = r + s = 2p is even, then

(a) (p even)

e is symmetric if and only if a is symmetric, ande is skew if and only if a is skew,

(b) (p odd)

e is symmetric if and only if a is skew, ande is skew if and only if a is symmetric.

We give the proof for the nonhermitian cases. The proof for the her-mitian cases is essentially the same.Proof: Because of Proposition 9.27,

(13.25) A = .\ if p is even and A _ -A if p is odd.

Assume that e(y, x) = ±e(x, y) with the + if a is symmetric, and the - ife is skew. Then

e(y, x) _ e(Ay, x) = e(y, Ax) = ±e(ax, y),

which, because of (13.25), equals ±&(Ax, y) = fe(x, y) if p is even, andequals ::Fe(ax, y) = ::Fe(x, y) if p is odd.

If r - s = 0, 4 mod 8, then A2 = 1 and the eigenspaces

St= {xEP:Ax=±x}are (by definition) the spaces of positive and negative spinors.


Lemma 13.26. If r - s = 0, 4 mod 8 (and n = r + s = 2p), then

(a) (p even) S+ and S_ are orthogonal for both e and e,(b) (p odd) S+ and S_ are totally null for both e and e.

Proof: Suppose p is even. Then, for x E S+, y E S_,

s(x, y) _ -e(Ax, ay) _ -e(x, aAy) _ -e(x, A2y) = -e(x, y),

since A = A. The proof of (a) for a is identical since A = A.Suppose p is odd. Then, for either x, y E S+ both positive, or for

x, y E S_ both negative,

e(x, y) = e(Ax, Ay) = e(x, 5 Ay) = -e(x, A2y) _ -e(x, y),

since A = -A. The proof of (b) for is identical since A = -A.

If r - s = 2,6 mod 8 then A2 = -1. Note that

(13.27) A is an a isometry if and only if A is an e isometry.

(13.28) A is an a anti-isometry if and only if A is an e anti-isometry.

Both of these statements are true because; if e(Ax, Ay) = ±e(x, y),then

'(ax, Ay) = e(A2x, Ay) = ±e(Ax, y) = ±e(x, y).

Lemma 13.29. If r - s = 2,6 mod 8 (and n = r + s = 2p) then

(a) (p even) A is an anti-isometry for a and e,(b) (p odd) A is an isometry fore and 9.

Proof: e(Ax, Ay) = e(x, AAy), which equals e(x, A2y) = -e(x, y) if p iseven, and which equals e(x, A2y) = e(x, y) if p is odd, because of (13.25).

Step 2.

In the even dimensional cases, it remains to construct a satisfying thePinor Theorem 13.17.

Recall from Chapter 12 that Cl(r, s) can be realized as a subalgebra ofEndc (ACP) - Clc(n) when n =- r + s = 2p. Also recall that there existsa reality operator R that has the property that

(13.30) Ra RaR-1, for all a E Endc(ACP) - Cl(n) = Cl(r, s) ®a C,

260 The Pinor Inner Products E and e

where 7Z is the natural conjugation on Cl(r, s) OR C. Both the fact thateach of the three involutions sending a to a, a, and a on Clc(n) leaveClc(r, s) C Cl(n) fixed, and the fact that each of these involutions whenrestricted to Cl(r, s), yield the corresponding involutions on Cl(r, s), will(implicitly) be assumed in the following.

The complex inner product ec on ACP constructed in Chapter 12 willprovide the basis for the construction of a on R(r, s). The facts that weshall need about ec are listed here for convenience.

Given a E Clc(n) = Endc(ACP),

(13.31) ec(ax, y) = ec(x, ay) for all x, y E ACP.

The reality operator R E EndR,(ACP) for Cl(r, s) may be chosen so that

(13.32) ec(Rx, Ry) = ec(x, y) for all x, y E ACP.

Also,

(13.33)if p = 0, 3 mod 4, then ec is C-symmetric;if p = 1, 2 mod 4, then ec is C-skew.

Case r - s = 0, 6 mod S. Here R2 = 1 and Cl(r, s) = EndR,(P),where the space of pinors is defined to be P - {x E ACP : Rx = x} (see(12.94) and (12.95)). Note that because of condition (13.32), ec restrictedto P is real-valued and nondegenerate.

Definition 13.34. (r - s = 0, 6 mod 8) The hat inner product on P(r, s)is defined to be

e = ecIp.

The condition (13.18) fore follows immediately from the analogouscondition (13.31) on ec. Finally, because of (13.33), e is R-symmetricwhen p = 0, 3 mod 4 and a is R-skew when p = 1, 2 mod 4.

Case r - s = 2, 4 mod S. Here R2 = -1 and Cl(r, s) = EndH(P),where the space of pinors P(r, s) is defined to be ACP equipped with theright H-structure given by I =- i and J = R (see (12.96)).

Definition 13.35. (r - s = 2,4 mod 8). The hat inner product a onP(r, s) is defined to be

(13.36) E(x, y) _ -ec(xJ, y) + jec(x, y) for all x, y E P.

The condition (13.18) fore follows from the analogous condition(13.31) on ec because J commutes with all a E Cl(r, s).

Inner Products on Spaces of Spin ors and Pinors 261

Lemma 13.37. Suppose W is a right H-space and cc is a complex bilinearform on W with respect to the complex structure I (right multiplication)on W, that satisfies

(13.38) ec(xJ, yJ) = ec(x, y).

Let

(13.39) h(x, y) _ -Ec(xJ, y) + je(x, y) for all x, y E W.

Then

(13.40) h is H-hermitian symmetric if and only if cc is C-skew;

(13.41) h is H-hermitian skew if and only if cc is C-symmetric.

Proof: Since cc is complex bilinear and I, J as well as i, j anticommute,it is easy to show that

(13.42) h(xI, y) = -ih(x, y) and h(x, yI) = h(x, y)i.

Also,

(13.43) h(xJ, y) = ec(x, y) + jec(xJ, y) = -jh(x, y)

is automatic.Finally, to prove that

(13.44) h(z, yJ) = h(x, y)j,

the hypothesis (13.38) must be used:

h(x, yJ) = -Ec(xJ, yJ) + jec(x, yJ) _ -ec(x, y) - jec(xJ, y)= jec(x, y)j - cc(xJ, y).7 = h(x, y)j.

Thus, h is H-hermitian.The proofs of the remainder of (13.40) and (13.41) are similar and so

are omitted (also cf. Lemmas 2.72 and 2.78).

Because of Lemma 13.37 and (13.33), e is H-hermitian skew whenp = 0, 3 mod 4 and is H-hermitian symmetric when p = 1, 2 mod 4.

262 The Pin or Inner Products t and 9

Combining Step 1 and Step 2, this completes the proof of the PinorTheorem 13.17 when the dimension n = r + s is even.

Step 3.Next we give the proof of the Spinor Theorem 13.1 when n = r + s

2p + 1 is odd. The proof is based on two ingredients. The first ingredientis the pair of isomorphism:

(13.45) Cllr, seven CI(r - 1, s) if r > 1,and

(13.45') Cl(r, s)even = C1(s - 1, r) if s > 1,

from Theorem 9.38. (The important fact is that these isomorphisms pre-serve the hat anti-automorphisms.) The second ingredient is the d portionof the Pinor Theorem 13.17 for even dimensions.

Case r - s = 1, 7 mod 8. Then r - 1 - s = 0,6 mod 8, so thatCl(r, seven = CI(r - 1, s) 25 EndR(P(r - 1, s)), and we may take S(r, s) _-P(r - 1, s). The spinor inner product a on S(r, s) is, by definition, the hatinner product 9 on P(r - 1, s). Finally, a is R-symmetric if p = 0, 3 mod 4,and a is R-skew if p = 1, 2 mod 4 by Theorem 13.17 applied to 9 onP(r - 1, s). If r = 0, consider the isomorphism C1(r, s)even - Cl(s - 1, r)EndR(P(s - 1, r)) instead.

Case r - s = 3,5 mod S. Then r - 1 - s = 2,4 mod 8, so thatCl(r, $)even e Cl(r - 1, s) EndH(P(r - 1, s)), and we may take S(r, s) tobe the right H-space P(r-1, s). The spinor inner product e on S(r, s) is, bydefinition, the hat inner product a on P(r -1, s). Finally, a is H-hermitiansymmetric if p = 1, 2 mod 4, and a is H-hermitian skew if p = 0, 3 mod 4by Theorem 13.17 applied to P(r-1, s). If r = 0, consider the isomorphismCl(r, s)even CI(s - 1, r) EndH(P(s - 1, r)) instead.

Step 4.

Next the Pinor Theorem 13.17 for odd dimensions will be deducedfrom the Spinor Theorem 13.1 for odd dimensions.

Case r - s = 1 mod 8. Recall from Theorem 11.66 that P(r, s)S(r, s) ®R, C and that Cl(r, s) = Endc(P); and Cl(r, s)even 25 EndR,(S),where A = i is a unit volume element and s, the spinor structure map, isthe natural conjugation sending x to Y on P S OR C. Let cc denotethe complexification of the spinor inner product a on S. Consulting ther - s = 1 mod 8 entry in Table 13.3 associated with Theorem 13.1, provesthat

(13.46)cc is C-symmetric for p = 0, 3 mod 4,cc is C-skew for p = 1, 2 mod 4.


It is natural to require that e and a be related by e(x, y) = e(x, y), forall x, y E P, because of (13.15).

Definition 13.47. (r-s = 1 mod 8) Let n = r + s 2p+1 define p. Thepinor inner products a and e on P(r, s) are defined by: First, if p, is even

(13.48) e(x, y) - ec(x, y), e(x, y) = ec(x, y) for all x, y E P,

and, second if p is odd

(13.49) e(x, y) = ec(x, y), e(x, y) = ec(x, y), for all x, y E P.

Now we prove that

(13.50) e(ax, y) = E(x, ay) for all x, y E P.

If a E Cl(r, $)even, then (13.49) is automatic. Suppose is E Cl(r, s)odd, thena E Cllr, seven. Then

(13.51)(p even) e(iax, y) - ec(iax, y) _ -iec(ax, y)

_ -iec(x, ay) = -ec(x, iay) = -e(y, iay),

where -ia = is since i = -i (the dimension n - 2p + 1 = 1 mod 4).

(13.52) (p odd) e(iax, y) = ec(iax, y) = ec(x, iay) = e(x, iay),

where i& = is since i = i (the dimension n 2p + 1 = 3 mod 4).The type of a and a listed in the r - s = 1 mod 8 portion of Theorem

13.17 follows from (13.46) and the definitions of e and e.

Case r - s = 3 mod 8. Recall from Theorem 11.66 that S(r, s) isa right H-space and that P(r, s) = P+(r, s) ® P_(r, s) with P±(r, s)S(r, s). The unit volume element is A = (o _°) and the spinor structuremap is s = (° o). Also recall that Cl(r, s)'- EndH(P+) a EndH(P_) andCl(r, s)even - EndH(S) embedded diagonally in EndH(P+)® EndH(P_). Let e denote the spinor inner product on S. Because ofTheorem 13.1,

(13.53)a is H-hermitian skew if p = 0, 3 mod 4, ande is H-hermitian symmetric if p = 1, 2 mod 4.

264 The Pinor Inner Products t and e

Defintion 13.54. The pinor inner products t and a on P - S ® S aredefined by

(13.55) (p even) e(zl, z2) = t(szl i z2) and e = e ®E.

(13.56) (p odd) E = E ®e and E(z1i Z2) = E(szli z2).

It remains to prove that

(13.57)

and

(13.58)

(p even) e(Azi, z2) = e(z1, Az2)

(p odd) E(Azl, z2) = e(z1, Az2)

for A = (a, b) with a, b E EndH(S) and z1 = (x1, y1), z2 = (x2, y2) E PS®S.

Both cases are automatic if A (a, a) E Cllr, seven. Each B(a, -a) E Cl(r, s)odd is of the form B = AA with A = (a, a) E Cl(r, sevenIf p is even, the dimension 2p + 1 = 1 mod 4 and A = -A, while if p is oddthen 2p+ 1 = 3 mod 4 and ) = A. Now (13.57) and (13.58) follow easily.

Case r - s = 5 mod S. Recall from Theorem 11.66 that S(r, s) is aright H-space and that P(r, s) equals S(r, s) equipped with the complexstructure I. The spinor structure map on P is s - J and the unit volumeelement is A - I. Let a denote the spinor inner product on S. Because ofTheorem 13.1,

(13.59)e is H-hermitian skew if p = 0, 3 mod 4, ande is H-hermitian symmetric if p = 1, 2 mod 4.

Definition 13.60. The pinor inner products E and t are defined to becomplex-valued inner products by

(13.61) (p even)

ande(x, y) = E(x, y) + it (x, y),

(p odd) e(x, y) = e(x, y) - je(x, y)(13.62) (or je(x,y) E(x,y) + je(x,y))

forallx,yeP - S.Note that in both cases,

(13.63) E(x, y) = E(xJ, y) for all x, y E P.

Inner Products on Spaces of Spinors and Pin ors 265

Now we prove that


If a E Cl(x, S)even then (13.64) is automatic. Each 6 E Cl(r, s)odd is of theform b = as with a E Cl(r, $)even. Therefore,

E(bx, y) = e(aax, y) = E(.Xx, ay) = E(xI, ay) = -iE(x, ay).

If p is even, this implies that e(bx, y) _ -ie(x, ay) = -e(x, Aay), which(since the dimension n = 2p + 1 = 1 mod 4 implies a = -a) equalse(x, aay) = e(x, by). If p is odd, this implies that e(aAx, y) = ie(x, ay),which, since the dimension 2p + 1 = 3 mod 4, equals e(x, )ay) = e(x, aay).

The type of e and e follows from the type of E listed in (13.59) becauseof the next lemma.

Lemma 13.65. Suppose h - a + j/3 is an H-hermitian form and a and,8are complex valued. If h is H-hermitian symmetric, then a is C-hermitian(symmetric) and /3 is C-skew. If h is H-hermitian skew, then a is C-hermitian (skew) and /3 is C-symmetric.

Remark 13.66. The inner product a is said to be the first complex partof h while ,3 is said to be the second complex part of h (cf. Chapter 2).

Case r - s = 7 mod 8. This case is so similar to the case r - s =3 mod 8 that the proof is omitted.

This completes the proof of the Pinor Theorem 13.17. It remains togive the proof of the Spinor Theorem 13.1 in the even dimensional cases.

Step 5.Using the hat preserving isomorphisms

Cl(r, seven N Cl(r - 1, s)

Cl(r, seven Cl(S - 1, r),

the r - s odd mod 8 portion of the Spinor Theorem 13.1 can be read offfrom the even portion of the Pinor Theorem 13.17 (Problem 3).

This completes the description of the types for the spinor inner prod-ucts E and the pinor inner products a and e. The discussion of the signaturesof these inner products will be given later in the section entitled Signature,as an applicatian of pinor multiplication.

266 Pinor Multiplication

PINOR MULTIPLICATIONThe inner products t and a on P(r, s) enable one to multiply pinors, asdescribed in Definition 8.42.

An application to calibrations is given in Chapter 14, and an applica-tion to pure spinors in the split case is given at the end of this section. Theproduct x O y induced by e will be denoted x 6 y, and the product x 0 yinduced by t will be denoted x 6 y. Therefore, if x, y E P are given, thenx 6 y and x 6 y E EndF(P) are defined by

(13.67) (x a y)(z) xE(y,z)

and

(13.68) (x 6 y)(z) = xe(y,z)

for all z E P.For all signatures r, s, the pinor representation enables us to consider

C1(r, s) as a subalgebra of EndF(P(r, s)), with equality

C1(r, s) = EndF(P)

unless r-s= 3 mod4. Ifr-s=3 mod 4, then

Cl(r, s) EndF(P+) ® EndF(P_) C EndF(P).

However, in all cases, p(a) (which we also denote by a in the presentcontext) has real trace zero. Therefore Theorem 9.65 implies that

Lemma 13.69. Consider Cl(r, s) C EndF(P(r, s)). The Clifford innerproduct (,) can be expressed as a real trace:

(13.70) (a, b) = (dim,, P)-1 trace,, ab for all a, b E Cl(r, s).

Remark 13.71. Note that this implies that the twisted Clifford innerproduct is also given by a trace:

(13.72) (a, b) = (dim,, P)-1 trace,, ab.

This lemma states that the natural R-symmetric inner product onEndF(P) induced by t (see Defintion 8.38) is exactly the same as the Clif-ford inner product on Cl(r, s). Therefore, all the results of the section InnerProducts in Chapter 8 are applicable to P,E and EndF(P), (, ). These re-sults are summarized in the next two theorems. Let N = dimF P.


Theorem 13.73. (x 6 y, a) =N

Re e(ay, x), for all a E Cl(r, s) andx,y E P.

Note that e(ay, x) has the order of x and y reversed.

Theorem 13.74. For signatures where t is F-hermitian on P(r, s),

(13.75) (x 6 Y' Z 6 w) = N Re z)E(y, w))

For signatrues where a is F-symmetric or F-skew on P(r, s),

(13.76) (x 6 Y' Z 6 w) =

N

Re ((x, z)E(y, w) I .

Here x, y, z, w E P(r, s).

Remark 13.77. Because of (13.72), if the Clifford inner product (a, b) isreplaced by the twisted Clifford inner product (a, b), the hat product x 6 yby the check product x 6 y, and t by t, then both Theorem 13.73 andTheorem 13.74 remain valid.

Other properties of pinor multiplication are easily read off Lemma8.44.

Proposition 13.78. For all a E Cl(r, s)

(i) (ax) 6 y = a(x 6 y), and x 6 (ay) = (x 6 y)a,(ii) (x 6 Y) (z o w) = (xe(y, z)) 6 w,(iii) (x o yj = y 6 x if e is symmetric (hermitian or pure),(iv) (x 6 y) = -y 6 x if a is skew (hermitian or pure).

Recall the notion of a pure spinor in the split case P(p, p). The squareof a pure spinor represents the associated null plane in AR(p, p).

Proposition 13.79. Suppose s E PURE C P(p, p) is a pure spinor. Thenthe squares o s E Cl(p, P) AR(p, p) is of the form s o s = z1 A Azp withN. = span{zl,..., zp} = {z E R(p, p) : z(s) = 0} totally null. Conversely,each E C1(p, p) = AR(p, p) of the form = zi A . A xp with N =span{zl, , zp} totally null is the square of a pure spinor.

Proof: Since, for all a E Pin(p), a(s 6 s)a-1 = (as) 6 (as), we need onlyshow that for some s+ E PURE+, s+ o s+ = zl A. ..A zp with zl, , zp EN,+ (see Theorem 12.100). As in the proof of Theorem 12.100, note thats+ = 1 E AR(p, 0) = P(p, p) is a pure positive spinor. By (13.67) andDefinition 12.24,

(13.80) (s+ 6 s+)(x) = (1 6 1)(x) = f(1, x) 1= (1, ox) 1.

268 Signature

Therefore, s+ o s+ = Pa, where P is orthogonal projection onto the linethrough 1 E AR(p, 0) = P. Let zj -1(ei, -ej) so that z1,.. . , zp is abasis for N,+. Recall that zj = lei . Therefore, zl zp = 1e, o o Iermaps the volume element c el A A ep E AR(p, 0) = P to ±1 and theorthogonal complement of v to zero. This proves that s+ 6 s+ = ± . J

SIGNATURE

First note that, by Problem 9.10,

(13.81)If s > 1 (i.e., R(r, s) not positive definite), then 9 must have

split signature (if it has a signature).

(13.82)If r > 1 (i.e., R(r, s) not negative definite), then a must have

split signature (if it has a signature).

Consequently, unless r = 0 or s = 0, the spinor inner product a must havesplit signature (if it has a signature)

Case s = 0 (i.e., R(r,O) positive definite). Then the Clifford innerproduct (, ) on Cl(r, 0) ARr is positive definite. Suppose x is a nullvector in P, i.e., &(x, x) = 0. By Theorem 13.74,

(x6y,x&y)=0 for all YEP.

Since (,) is positive definite, x 6 y = 0 for all y. Thus x = 0. Thisproves that e has no nonzero null vectors, so that t must be definite (andby adjusting the scale we may assume that t is positive definite).

Case r = 0 (i.e., R(O,s) negative definite). Let (,) denote theClifford inner product on Cl(0, s) AR'. The twisted Clifford inner prod-uct (a, b) (a, b) is positive definite. Using the product x 6 y and Theorem13.74 it follows that a cannot have nonzero null vectors.

The signature part of Theorem 13.1 (cf. the Remark that follows thisTheorem), for the spinor inner product a on S(r, s), follows because of thehat preserving isomorphism

Cl(r, 0)even - CI(r - 1, 0).

Table 13.19 can be used to determine exactly when e and e have asignature.

Inner Products on Spaces of Spinors and Pin ors 269

Proposition 13.83.

(i) The pinor inner product t has a signature if and only if either s =0 mod4orr=3mod4.

(ii) The pinor inner product t has a signature if and only if either r =0 mod4ors=1mod4.

PROBLEMS

1. List the classical companions Cp(r, s), defined by (13.7), as classicalgroups.

2. Deduce the r - s = 0, 4 mod 8 cases of the Spinor Theorem 13.1 fromthe r - s = 0, 4 mod 8 cases of the Pinor Theorem 13.17.

3. (a) Deduce the r - s = 0 mod 8 case of the Spinor Theorem 13.1 fromthe r - s = 7 mod 8 cases of the Pinor Theorem 13.17.(b) Deduce the r - s = 2, 6 mod 8 case of the Spinor Theorem 13.1from the r - s = 1, 5 mod 8 cases of the Pinor Theorem 13.17.(c) Deduce the r - s = 4 mod 8 case of the Spinor Theorem 13.1 fromthe r - s = 3 mod 8 case of the Pinor Theorem 13.17.

4. Suppose r 36 0 mod 4. Show that the check inner product i on P(r, 0)never has a signature, using Table 13.19.

5. Suppose r- s = 0 mod 8 and r+ s = 2p with p = 1 or 3 mod 4. Showthat the (reduced) classical companion (p > 1) is

SL(p, R) a f\ 0

(at)-1) : a E SL(p, R) } .

6. Suppose Cl(V) EndF(P) is the pinor representation for a positivedefinite inner product space V.(a) Prove that G - {a E Cl(V) : as = 1} is a subgroup of O(Cl(V)),and hence compact.(b) Pick any positive definite inner product on P and define E to bethe average of this inner product over the compact group G. Showthat e is the pinor inner product (up to a scale).

7. A real vector space V equipped with a linear map J, with j2 = 1 anddim V.. = dim V_, where Vf = {x E V : Jx = Ex} is called a L-vectorspace. Assume r - s = 0 mod 8.(a) Show that P(r, s) is an L-vector space.

270 Problems

(b) Combine 9 and a to obtain an "L-valued inner product" on P(r, s).Define L-hermitian symmetric, etc., and describe the "type" of thisinner product (depending on p mod 4 with r + s = 2p defining p).

8. State and prove a complex analogue of Proposition 13.79 concerningsquares of pure spinors.

14. Low Dimensions

The results of the previous chapters are particularly interesting in thelow dimensional cases. These special cases are examined in some detail inthis chapter.

CARTAN'S ISOMORPHISMS

Recall from Chapter 1 some low dimensional isomorphisms of the groupsdefined in that chapter (see Proposition 1.40):

SO(2) = U(1) - SK(1) - S' and CSO(2) = GL(1, C) = SO(2, C),

SO(4) - HU(1) HU(1) and CSO(4) = GL(1, H) H',

Sp(1, R) = SL(2, R) - SU(1, 1) and SL(2, C)!-:- Sp(1, C),

SU(2) - HU(1) = SL(1, H) - S3 and SOT (3, 1) 2_-' SO(3, C).

The final isomorphism SO? (3, 1) - SO(3, C), which was proved inChapter 3, is also a consequence of two of the spin isomorphisms presentedbelow-see Spin°(3, 1) and Spin(3, C), and note that both are SL(2, C),which has center Z2 = {±1}. Also SU(2) = HU(2) followsfrom Spin(3, 0)HU(1) and Spin(0, 3) = SU(2) (see Remark 14.113).

271

272 Cartan's Isomorphisms

In low dimensions, the spin groups Spin° are not new but are isomor-phic to classical groups. Recall from Problem 10.2(d) that Spin°(r, s){a E Spin(r, s) : as = 1} is the connected component of the identity inSpin(r, s) (n > 2), except when both r = 1 and s = 1.

Theorem 14.1 (The Spin Isomorphisms).n=r+s=2:

Spin (2) = U(1)

t =Spin°(1, 1) . f 0

Spin (3) = HU(1) (? SU(2))Spin°(2,1) SL(2, R)

Spin (4) = HU(1) x HU(1)Spin°(3, 1) = SL(2, C)Spin°(2, 2) SL(2, R) x SL(2, R)

Spin (5) HU(2)Spin°(4,1) HU(1, 1)Spin°(3, 2) Sp(2, R)

n=r+s-8:

Spin (6) SU(4)Spin°(5,1) = SL(2, H)Spin°(4, 2) SU(2,2)

pf : Spin°(3, 3)/Z2 - SL(4, R)

pf : Spin°(6, 2)/Z2 = SK(4).

Proof: These results can be read off from Theorem 13.8 by comparingdimensions and connectivity. Recall the type of the spinor inner productimplies:

Spin°(6) C U(4)

Spin°(5,1) C GL(2, H),

Spin°(4, 2) C U(2,2),

Spin°(3, 3) /Z2 C GL(4, R).

The extra information required in these cases is that each element ofSpin° has determinant equal to one. This follows easily from the fact that

Low Dimensions 273

Spin° is connected and the fact that the determinant takes on a finite num-ber of values for any representation of Pin (see the section Determinantsin Chapter 10).

In summary, using the spinor inner products (and in the four caseslisted above, information about the determinants), the group Spin° is asubgroup of the particular classical group (the (reduced) classical compan-ion) described in Theorem 14.1. To complete the proof, equality is obtainedby counting dimensions and using connectivity of this (reduced) classicalcompanion associated with the group Spin°. The last case, signature 6,2in dimension 8 is a special case of Theorem 14.3 presented below. Ll

Theorem 14.2 (Dimension Seven).Spin(7) C SO(8) C Ms(R)

Spin°(6,1) C SK(4) C M4(H)

Spin°(5, 2) C SK(4) C M4(H)

Spin°(4, 3) C SOT (4, 4).

Proof: This result follows from Theorem 13.8 exactly parallel to the proofof the previous theorem.

Theorem 14.3 (Dimension Eight). First, Spin°(7,1) C SO(8, C) andSpin°(5, 3) C SO(8, C).

Second, Spin(8) C SO(8) x SO(8). Moreover, the

the positive spin representation

1 --+ Z2 = {1, Al -+ Spin(8) P±, SO(8) -+ 1,

the negative spin representation

1 --+ Z2 = {1, -A) -+ Spin(8) -_+ SO(8) -+ 1,and the vector representation

1 --+ Z2 = {1, -1} -a Spin(8) --* SO(8) -+ 1

are exact sequences of groups.Also, Spin°(4,4) C SO(8) x SO(8). Moreover,


1 --+ Z2 =_ {1, A} -+ Spin°(4, 4) °+ SOT(4, 4) _+ 1,

the negative spin representation

1 -+ Z2 =- {1, -A} -+ Spin°(4, 4) _+ SOT (4,4) 1,

and the vector representation

1 -+ Z2 = {1, -1} --f Spin°(4, 4) SOT(4, 4) -+ 1

274 Cartan's Isomorphisms

are exact sequences of groups.In addition, Spin°(6, 2) C SK(4) x SK(4). Moreover,


1 -+ Z2 = { 1, A} -+ Spin°(6, 2) °-++ SK(4) -- 1

and the negative spin representation

1 --+ Z2 - {1, -a} -+ Spin°(6, 2) p-+ SK(4) , 1

are exact sequences of groups.For each one of these Spin groups in dimension eight, all of the group

representations listed above are inequivalent.

Proof: Most of the information in this theorem follows from Theorem 13.8in a parallel manner to the proof of Theorem 14.1. It remains to prove thatthe representations listed above are not equivalent. For example, suppose

p+ : Spin(8) --+ Ms(R) S--- EndR(S+)

and

p_ : Spin(8) --+ M8(R) - EndR,(S_)

were equivalent, with intertwining operator f : S_ -+ S+. Then p+(g) =f o p_ (g) o f -1 for all g E Spin(8). In particular, this implies that ker p+ _ker p_ which is false.

Theorem 14.1 has a complex analogue.

Theorem 14.4.

Spin(2, C) = C*Spin(3, C) SL(2, C)

Spin(4, C) = SL(2, C) x SL(2, C)Spin(5, C) - Sp(2, C)Spin(6, C) = SL(4, C).

Hint of Proof: For n = 2p even, complexify the split case, C1(2p)Cl(p,p) ®R, C, and refer to the split cases Spin°(p, p), p = 2 and 3, inTheorem 14.1. For n = 2p + 1 odd, refer to the Spin°(p + 1, p) case inTheorem 14.1. The details are omitted.

There is an alternate elementary proof of this theorem. This proof forSpin(6, C) is outlined in Problem 1.

In addition to the isomorphisms presented above:

Low Dimensions

Theorem 14.5. The following groups are isomorphic.

SK(2) = SU(2) x SU(1,1)/Z2SK(3) = SU(3,1)/Z2.

Proof: See problem 2. -J

TRIALITY

275

In dimension eight, for both the positive definite case and the split case,the spin group Spin° has three inequivalent 8-dimensional representations:the positive spin representation p+, the negative spin representation p_,and the vector representation X. There are automorphisms of the groupSpin called the triality automorphisms, which interchange these threerepresentations. Of course, they cannot be inner automorphisms since thethree representations are distinct (Theorem 14.3). Thus, by definition, theautomorphisms are outer.

There is an elegant description of these triality automorphisms in termsof the octonians 0 for Spin(8) and the split octonians O for Spin°(4,4).We shall pursue the positive definite case Spin(8) using O. However, exceptfor notational changes, the development will apply to Spin°(4, 4) and O aswell.

As an added bonus, we discuss a very useful concrete model for Cl(8, 0)(or alternatively, Cl(4, 4)). As will be showm, all three of the 8-dimensionaleuclidean spaces V - R(8), S+, and S_ can be identified with O.

First, the space of vectors, V(8) C Cl(8, 0), is identified with 0 asfollows.

Definition 14.6. Let V(8) C Enda(O (D 0) denote the positive definite8-dimensional euclidean space defined by

\

The square norm on V(8) is defined by jjA(u)jj 1juJll, where

A(u) 0 R,-R-U 0 )

Note that

(14.7) A(u)A(v) . I- OR" _Ru

276 Triality

so thatA(u)A(u) = -Ijull 1 for all u E 0.

Thus, by the Fundamental Lemma of Clifford Algebras, the map A0 , EndR,(O (D 0) extends to an algebra homomorphism A : CI(O) --}EndR(O ® 0). By Lemma 8.6 and Theorem 11.3, CI(8) = CI(O) has notwo-sided nontrivial ideals. Thus the algebra homomorphism is injective. Adimension count shows that this map A is onto, so that A is a isomorphism:

(14.8) CI(O) c--- EndR(O (D 0).

Therefore, the space of pinors may be taken as

P=0®O.

The isomorphism (14.8) will frequently be composed with the natural iso-morphism CI(O) = A(O) (as vector spaces).

Lemma 14.9. A unit volume element ,\ for V(8) - 0 is given by:

A= (' -' ) E End,,(O ®O) = CI(O) = A(O).

Proof 1: Consider the standard orthonormal basis

eo = 1, el = i, e2 = j, e3 = k, e4 = e, e5 = ie, e6 - je, e7 - ke

for O. Using octonian multiplication compute,

A(eo) ... A(e7) = (1-Ol) .

Proof 2: Note A belongs to the twisted center of CI(O) by (9.48'). Lemma9.49 implies that A E A130 and Theorem 9.65 implies that A is of unit length.

Remark. Right multiplictaion by i, denoted R;, induces a complex struc-ture on 0. The orientation induced by this complex structure is the sameas the orientation induced by this standard basis.

Corollary 14.10. If 0 is identified with V(8) C EndR,(O ® 0) by

(14.11) V(8) - (I -& 0 J : u E O } ,

then

CI(O) - Enda(S+ ® S_)

Low Dimensions 277

and

Cleven(0) = EndR(S+) ® EndR(S_) = a : a, b E EndR(O)1(0b) } ,

where S+ _-- 0 is the first copy of 0 and S_ = 0 is the second copy of 0in P=0®0.Proof: Note that A = (o _°) commutes with (d n) if and only if c = d = 0.

Lemma 14.12. The standard inner product on 0,

(14.13) (x, y) = Reiy for all x, y E 0,

when applied to P = O®O so that the two factors S+ = 0 and S_ = 0 areorthogonal, can be adopted as the pinor inner product t on P = S+ ® S_ .

Proof: It suffices to show that, for z = (x, y) and z' =_ (x', y') E 0 ® 0,

(14.14) t (A(u)z, z') = -9(z, A(u)z') for all u E O,

where t is defined by

(14.15) (z, z') - (x, x') + (y, y').

Since A(u)z = (yu, -xu) and -A(u)z' = (-y'u, x'u), the result (14.14)follows from RU = R-.

Corollary 14.16. Under the isomorphism Cl(O) = EndR(O (D 0) deter-mined by Definition 14.6, the hat anti-automorphismis equal to the adjoint(transpose) with respect to the standard norm on 0 ® 0. In particular,Theorem 14.3 for Spin(8) yields

(14.17) Spin(8) C SO(O) x SO(O) SO(S+) x SO(S_).

Note that

{g E Cl(O)even : 99 = 1) 0(0) X 0(0).

We shall adopt both the notation

(14.18} 9 = (9+, g-) E Spin(8), as well as

9 = (go,9+,9-) E Spin(8),

where by definition g+ = p+(g) and g_ _- p_(g), while go = X9 denotesthe vector representation of g. The vector representation X will also bedenoted by po = X.

Octonian multiplication can be used to give an important characteri-zation of Spin(8).

278 Triality

The Triality Theorem 14.19. Suppose (go, g+, g_) is a triple of or-thogonal linear maps on O. Then (9+, g-) E Spin(8), with go the vectorrepresentation of (g+, g_) if and only if

(14.20) g+(xy) = g_(x)go(y) for all x, y E O.

Remark 14.21. If (go, g+, g_) is a triple of orthogonal transformationson 0 satisfying (14.20), then each of the three must have determinant one,because the theorem implies that g - (go, 9+, g_) E Spin(8) and Spin(8) isconnected.

Proof: Given

A = (0b

) E EndR.(O) ® Endp,(O) - Cl(O)e°e°,

then, by Corollary 10.50, A E Spin(8) if and only if

AA = 1 (i.e., a, b E 0(0)), and XA(u) E 0 for all u E 0,

where

XA(u)=A(_p 0)Ais the vector representation. Thus, for A (0 6) to belong to Spin(8), wemust have that a, b E 0(0) and that: for each u E 0, there exists a v E 0such that

(14.22)

That is,

(14.22')

(-R 0)=A(_R

0 )A.

R = a

Considering this as a map sending u to v, it must be the vector represen-tation go = po(A) of A. Applying (14.22') to 1 E 0, we have

v = a(bt(1)u).

Therefore, if g = (g+, g-) E Spin(8), then

(14.23) v = go(u) __ g+(gt (1)u) for all u E 0

Low Dimensions 279

defines the vector represenation go = Xg. Now (14.22') with a = g+ andb = g_ can be rewritten as

(14.24) wgo(y) = g+(g' (w)y) for all w, y E 0.

Setting x = gt (w) yields the desired equation (14.20).Conversely, suppose (14.20) is satisfied by (go, g_, g+). Set

/g+0A__

I .0 g-

If u E 0 is given, then (with v = go(u)) the equation (14.22) is satisfied,since the equations (14.22), (14.22'), and (14.24) are all equivalent. iJ

Since conjugation, c(x) = x, is an orthogonal transformation, the rep-resentation po = CO po o c is 0(8) equivalent to po. Similarly, p' = c o p+ o cis 0(8) equivalent to p+. Note that if h . R and h' _- c o h o c thenh' = Lu. The identity g+(xy) = g_(x)go(y), for g E Spin(8), implies that

9+(xy) = g+(xy) = g+(9x) = 9-(9)go(x) = go(x)9 (y)-

Therefore,

(14.25) a(go, g+, 9-) _ (g'_, g+, 90)

defines an automorphism a : Spin(8) --+ Spin(8). Similarly,

(14.26) Q(go, g+, 9-) _ (go, g-, 9+)

defines an automorphism 3 : Spin(8) -> Spin(8).Note that the product r = a/3 is given by

(14.27) r(go, 9+, g-) = (g+, g''-, 90),

and is of order three. r is called the triality automorphism of Spin(8)-Since a2 = j32 = r3 is the identity homomorphism of Spin(8) and

cep = r, , a = r2, it follows that a and ,3 generate a group of automorphismsof Spin(8) that is isomorphic to S3, the symmetric group on three letters.These automorphisms are all outer since they act nontrivially on the center,{±1, ±A} = {(1,1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1,1)}, of Spin(8).

This concrete realization of Spin(8) using the octonians has many ap-plications. We give one example involving complex structures on R$ (seeProposition 7.174). Let Reff denote the orthogonal reflection along theplane (in order to avoid confusion with right octonian multiplication).

280

Proposition 14.28. Spin(8) is generated by triples

(Refg, Jt , J£) with £ E GR,(2, O),

where the complex structures J and J£ are defined by

J£

and

J£ =1

2 (Rv Ru - Ru Ro)

Triality

if s=uAv.

Proof: Proposition 10.21 states that GR,(2, V) C Spin(8) generatesSpin(8). Given g = E GR,(2, V) C Spin(8), we must show that

(14.29) go=Reff, and

Given two unit vectors it, v E V = 0, identify it with

0 RuRu 0 '

and v with0 R,

a0 JC-R

as in the above discussion. Then the Clifford product is given by

(14.30) u v = R. Ru 00 -R-R

Note g = it v E Spin(8), so that (14.30) can be rewritten as

(14.31) (Ref ,, .Ref,,, -R,,&, E Spin(8) for all u, v E O.

=Finally, given l; E G-(2, 0), choose it, v orthonormal with 1; = u AVit v and apply (14.31) and (6.14), to complete the proof of (14.29).

Now we can prove Proposition 7.174 (n = 4):

(14.32) Cpx+(4) = {JE : E GR,(2, O)} = Gx(2, 0).

Low Dimensions 281

Proof of Proposition 7.174 (n = 4): First note that the complex struc-ture Ri = J+i induces the same orientation on 0 as the standard basiseo, ..., e7 defined above. Therefore, R, E Cpx+(4). Since GR(2, 0) isconnected, each complex structure J+ induces the same orientation on 0.

It remains to show that the map sending E GR(2, 0) to J+ ECpx+(4) is an isomorphism. This map is just the positive spinor rep-resentation p+ restricted to GR(2, 0). Since ker p+ = {1, A) and AGR(2, 0), p+ restricted to GR(2, 0) is injective.

Consider the action of Spin(8) on GR(2, V) given by sending togeg-' for each g = (go,g+,g_) E Spin(8). Let K denote the isotropysubgroup at a point o. Then considering the vector representation and thepositive spinor representation yields an equivarient isomorphism betweenthe quotient

(14.33) GR(2, 0) = SO(8)/(SO(2) x SO(6))

and the quotient

(14.34) Cpx+(4) = SO(8)/U(4).

In the following Remark this isotropy subgroup K of Spin(8) at .o iscomputed explicitly for fo = 1 A i (see (14.41)).Remark 14.35. The octonian description of Spin(8) presented in theTriality Theorem 14.19 naturally leads to octonian descriptions of the twoisomorphisms

(14.36) Spin(6) = SU(4) and Spin(5) = HU(2)

as follows.

Recall (Problem 10.6) that the subgroup of Spin(n) that fixes a vectorin the vector representation is Spin(n - 1). Thus

(14.37) Spin(7) = {g E Spin(8) : go(1) = 1}(14.38) Spin(6) _ {g E Spin(8) go(1) = 1 and go(i) = i}(14.39) Spin(5) = {g E Spin(8) go(1) = 1, go(i) = i, and go(j)

Also, let Spin(2) denote the following subgroup of Spin(8):

(14.40) Spin(2) = {g E Spin(8) : go(x) = x if x 1 span{1, i}}.

Then it follows that

(14.41) K = Spin(2) x Spin(6).

282 Triality

The triality identity (14.20) can be used to compute p+ (G) and p- (G)where G is one of the above subgroups of Spin(8). First, we obtain

(14.42) Spin(7) = {g E Spin(8) : g+ = g_}

from (14.20) by setting y = 1.Next, note that each element g of the group Spin(2) defined by (14.40)

must have go a rotation Rote through an angle 0 in the 1, i plane. Thus,go is the product of two reflections along lines in span{1,i}. For example,go = Ref 1 o Ref, Consequently, either g or -g is equal to the Cliffordproduct of 1 E V(8) with e'`B/2 E V(8), i.e.,

0 1 O Re-;e/a _ Re:e/2 0

( -1 0) (-Reie/2 0 0 -Re_ie/2

Therefore, with V) = z + a,

(14.43) Spin(2) = {(Rote,,, Reio, Re_i*) E R.) .

Next we show that the positive spinor representation p+ yields anisomorphism

(14.44) (Spin(2) x Spin(6))/Z2 = U(4),

when S+ = 0 is given the complex structure Ri, and Z2 = {1,a). If g ESpin(6) C Spin(7), then the triality identity (14.20) with y - i implies thatg+ = g_ commutes with the complex structure Ri, so that g+ = g_ E U(4).Now a dimension count combined with the connectivity of U(4) completesthe proof, showing p+ from Spin(2) x Spin(6) to U(4) is surjective.

Using Problem 10.7 and the fact that ker p+ = {1, A} proves that

(14.45)Spin(6) = {(ho, h, h) : h E SU(4), and ho is given by

ho(x) - h(1)h(x), for all x E V(8) - 0).

In particular, Spin(6) S--- SU(4).Finally, the triality identity implies that if g E Spin(5), then g+ = g_

commutes with both Ri and R. Thus, g+ = g_ E HU(2), where 0 = H2is provided with the (right) quaternionic structure inducted by I - Ri, JR and K RiRj. Again connectivity and a dimension count show that

(14.46) p+ = p_ : Spin(5) - HU(2).

Low Dimensions 283

TRANSITIVE ACTIONS ON SPHERES

In low dimensions, n < 6, the spin groups agree with their (larger) classicalcompanions-see Theorem 14.1. For slightly larger dimensions, vestiges ofthese isomorphisms remain. Consider, for example, Spin(7) C SO(8). Inthis case, SO(8) is the larger classical companion (determined by the spinorinner product e on S = Its). The group Spin(7) is "almost" as big as itsclassical companion SO(8). One way of making this precise is to show thatSpin(7) also acts transitively on spheres in R.

This action as well as the other transitive actions to be discussed inthis section are listed below for convenient reference.

(14.47) Spin(7)/G2 S7.

(14.48) Spin(8)/G2 = S7 X S7.

(14.49) Spin(8)/Spin(7) = S7.

(14.50) Spin(9)/Spin(7) = S15.

Also, recall

(14.51) G2/SU(3) - S6

from Problem 6.9(a).First, a model for C1(7) is constructed, using the octonians.

A Model for Cl(7)

Identify V(7) = ImO with the following subspace of EndR(O)EndR(O):

Let

(14.52) A(u) ` _ -I ) for u E ImO.

This map A : V(7) = Im 0 --+ EndR(O) ® EndR(O) extends to an algebraisomorphism

(14.53) Cl(7) = EndR(O) ® EndR(O) = EndR(P+) ® EndR(P_)

by the Fundamental Lemma of Clifford Algebras because A(u)A(u) _-I1uII 1.

284 Transitive Actions on Spheres

The natural inner product on P = 0 ® 0 can be adopted as thepinor inner product E = e ®c, since R;, = -R,, for u E Im 0 (one mustverify that E(A(u)z, z') = -t(z, A(u)z')). The unit volume element .A forV(7, 0) = Im O can be chosen to be (o °1) = A(el) A(e7), consistentwith the orientation on Im O determined by the basis e1, ... , e7 definedabove (see the proof of Lemma 14.9). Obviously, CI(7)even is contained inEnda(O), embedded diagonally in EndR(O) ® EndR(O) = C1(7). Count-ing dimensions yields

(14.54) Cl(7)even = Enda(O).

Because of Corollary 10.50, g E Cl(7)even - EndR(O) belongs toSpin(7) if and only if

(14.55a) gg = 1 (i.e., g E 0(0))

and

(14.55b) given u E Im O - V(7), there exists w E Im O = V(7)

such that

(14.56) Rw.

The only possibility for w (apply both sides of (14.56) to 1 E 0) is

(14.57) w = g(g-1(1)u).

Therefore, if g E Spin(7) C 0(0), then

(14.58) Xg(u) = g(g-1(1)u) for all u E ImO

is equal to the vector representation

(14;59) X = Ad : Spin(7) -+ SO(7).

Moreover, (apply (14.56) to g(v))

(14.60) g(vu) = g(v)X9(u) for all u E ImO and v E O.

(Note that (14.58) can be used to extend xg to 0 with xg(l) = 1, and then(14.60) is valid for all u E 0.)

Conversely, if (14.60) is satisfied by g E 0(0), with xg defined by(14.58), then (14.56) is valid with w = X5(u) (take v = g-1(z) and applyboth sides of (14.56) to z). This proves the following lemma.

Low Dimensions 285

Lemma 14.61. An element g E 0(0) belongs to Spin(7) if and only if

(14.62) g(uv) = g(u)X9(v) for all u, v E 0,

where

(14.63) X9(v)=g(g-1(l)v) forallvEO

defines the vector representation of Spin(7) on V(7) = Im 0.

Now because of the Triality Theorem 14.19

Corollary 14.64.

(14.65)Spin(7) _ {g = (g+, g-) E Spin(8) : g+ = g-}

_ {g = (go,g+,g-) E Spin(8) :go(1) = 1}.

To complete our list of characterizations of Spin(7), note that if

(R, ?) E G(1, Im O),0 -Ru

then

(0 )EGIm0).

Lemma 14.66. Spin(7) is generated by {R. : u E Ss C Im 0}.

Proof: By Proposition 10.23, Spin(7) is generated by

AG(1, Im O) = G(6, Im O) C Cl(7)even 5 EndR(O).

Theorem 14.67. Consider the spin representation of Spin(7) on S R8.Then Spin(7) acts transitively on the 7-sphere

S7=Ix ER8:IIxII=1},

and the isotropy subgroup at a point is G2:

(14.68) Spin(7)/G2 = S7.

Proof: To compute the isotropy subgroup of Spin(7), the octonian modelfor Cl(7) is useful. Let K denote this isotropy subgroup of Spin(7) at1 E S 25 0. Suppose g E K. Then, by (14.63), X.9(v) = g(g-1(1)v) = g(v).Therefore, by (14.62), g(uv) = g(u)g(v), so that g E G2. Conversely, if


g E G2, then, by Lemma 6.67, g E 0(0). Since g(1) = 1 it follows thatX9 = g, and (14.62) is valid. Therefore, conditions (14.55a and b) aresatisfied and g E Spin(7). This proves that K = G2. We give two proofsthat Spin(7) acts transitively on S7.

First, recall that Spin(6) = SU(4). The spin representation of Spin(7)on Rs, when restricted to SU(4) _-- Spin(6) C Spin(7), is the standard rep-resentation of SU(4) on C4, where the volume element A for R(6, 0) pro-vides the complex structure on Rs = C4. Note that Cl(6)e°e° = Endc(C4)and CI(6)even C Cl(6) = CI(7)even = EndR,(Rs). Since SU(4) acts tran-sitively on S7 C C4, this proves that the larger group Spin(7) also actstransitively on S7.

Second, the orbit of Spin(7) through 1 E S7 C 0 is Spin(7)/G2 whichhas dimension 7 = 21 - 14. Since S7 is connected and Spin(7) is compact,the orbit must be all of S7. J

Theorem 14.69. Consider the spin representation p = p+ ®p_ of Spin(7)on S+ ®S_ = Rs ®Rs. The orbit through a point (a, b) E S+ ® S_ is

{(x, y) E S+ ® S_ : IxI = Jai and IyI = IbI}.

(a) If a, b # 0, the isotropy subgroup is G2 and

(14.70) Spin(8)/G2 = S7 X S7.

(b) If exactly one of a, b vanishes, then the isotropy subgroup is a copy ofSpin(7) and

(14.71) Spin(8)/Spin(7) = S7.

Proof of (a): Spin(7) is diagonally embedded in Spin(8) by Corollary14.64. Thus, an element g E Spin(8) can be chosen that sends a to jaibecause Spin(7) acts transitively on the 7-sphere. Consequently, we mayassume that a E 0 - S+ is nonzero and real.

Consider the subgroup

(14.72) H = {g = (go, g+, g_) E Spin(8) : g+(1) = 1}

of Spin(8). Applying the triality automorphism r(go, g+, g-) = (g+, g'_' go)shows that

rH = {(go, 9+, g-) E Spin(8) : go(1) = 1}.

Low Dimensions 287

Now go(1) = 1 if and only if g+ = g_ (see Corollary 14.64). Therefore

(14.73) TH = Spin(7).

Thus, H = r2(Spin(7)) is an embedding of Spin(7) into Spin(8) send-ing h E Spin(7) to the triple (h, Xh, h') E Spin(8). In particular, h' ESpin(7) can be chosen to send b to jjbjj since Spin(7) acts transitively onspheres. This proves that Spin(8) acts transitively on

S7xS7={(x,y)ES+ED S_:lxj =jalandjyI=bbl}.

Let K denote the isotropy subgroup of Spin(8) at

(a, b) - (r cos 9, rsin 9) E S7 X S7 C O ®O,

with r > 0 and 0 < 0 < ir/2. Suppose g = (g+, g_) E K. Recall thatg+(uv) = g_(u)go(v) for all u, v E 0. Since g_(1) = 1, g+(v) = go(v).Therefore, go(l) = g+(1) = 1 also. Consequently, g+(u) = g_(u). Thisproves go = g+ = g_ E G2. Conversely, if go = 9+ = g_ E G2, theng - (g+, g_) fixes (a, b) since a, b are real. Therefore, K = G2. J

Proof of (b): Since p+ : Spin(8) -+ SO(8) is surjective, Spin(8) certainlyacts transitively on S7.

Suppose (a, 0) E S. x {0}, with a real and nonzero. The isotropy Hof Spin(8) at (a, 0) E S+ ® S_ is given by (14.59) so that H = r2(Spin(7))as desired.

The proof for (0, a) E {0} x S_, with a real and nonzero, is similarand so omitted. J

A Model for Cl(9)

Let S=-0®0 and PP=S®R,\ C. Choose

(14.74) V(9) = j Z 1 Ru I: r E R, u E 01 C Endc(P).

If

A(r,u)= i(Ru

r R"l ,

rthen A(r, u)A(r, u) = -(r2 + 1u12)Id. Therefore, by the FundamentalLemma of Clifford Algebras

(14.75) Cl(9) = Endc(P).


Also note that

(14.76) Cl(8)even = EndR(S),

and the unit volume element for V(9) may be chosen to be i E Endc(P),the complex structure on P = S OR C.

This model can be developed further, yielding a characterization ofSpin(9) via Corollary 10.50 (see Problem 5). However, what we shall needbelow is a characterization of Spin(9) using Proposition 10.23. This Propo-sition says Spin(9) is generated by .G(1, V(9)) = G(8, V(9)), where A - iis the unit volume element. That is,

Lemma 14.77. The group Spin(9) is generated by the 8-sphere

(14.78) f ( ) :rER,uE0, andr2+Iu12=1}

in EndR(O ® 0) - CI(9)even

This lemma can be used to prove that Spin(9) acts transitively on S's

Theorem 14.79. Consider the spin representation

(14.80) p : Spin(9) --> SO(16),

of Spin(9) on S - R's. Then Spin(9) acts transitively on the 15-sphere

S15_Ix ER16:IIxii=1},

and the isotropy subgroup at a point is Spin(7):

(14.81) Spin(9)/Spin(7) Sis

Proof: Because Spin(8) C Spin(9), Theorem 14.69 says that a unit vectorin 0 ® 0 can be mapped to the arc (cos 9, sin 9), with 0 < 0 < a/2 by anelement of Spin(8). Lemma 14.77 says, in particular, that

con B sin 0)E Spin(9) for each 0.

sin O - cos B J

This implies that each point on the arc (cos 9, sin 0), 0 < 0 < a/2, can bemapped to the point (1, 0) by an element of Spin(9). This proves Spin(9)acts transitively on the unit sphere S's on R's

Low Dimensions 289

Let K denote the isotropy subgroup of Spin(9) at (1, 0) E S = 0 ® O.Since Spin(8) C Spin(9), part (b) of Theorem 14.69 says that

(14.82) K fl Spin(8) = r2(Spin(7)).

This proves that

(14.83) r2(Spin(7)) C K.

Therefore,it : Spin(9)/r2(Spin(7)) -+ Sls

is a covering map, since dim Spin(9) - dim Spin(7) = dim S15. Since S15is simply connected, it must be one to one.

THE CAYLEY PLANE AND THE EXCEPTIONAL GROUP F4

Consider the relation - on F" - {0} defined by

(14.84) a - 6 if as = b for some scalar A E F.

If F = R, C, or H then - is an equivalence relation and Pr-1(F)is the projective space of all F-lines through the origin in F". However,since 0 is not associative the "Cayley projective spaces P"-1(O)" are notwell-defined. Alternate (equivalent) definitions of P"-1(F), for F =- R, C,or H, are available (cf. Problem 4.15).

Instead of considering a line L - [a] in F", consider the orthogonalprojection A from F" onto L. (Equip F" with the standard F-hermitiansymmetric inner product.) Then A satisfies

(14.85) At = A, A2 = A, and traceF A = 1.

Recall the notation Herm(n, F) _ {A E M"(F) : At = A). Nowprojective space can be described as a subset of the real vector space

Herm(n, F).

(14.86) P"-1(F) = {A E Herm(n, F) : A2 = A and tracer A = Q.To complete the proof of (14.86), note that if A2 = A with A E Herm(n, F),then A has eigenvalues 0 and 1, so that A is orthogonal projection fromF" onto L = {x E F" : Ax = x}. In particular, tracer A = dimp L.

If A E M,, (F) represents orthogonal projection onto L and dimF L =1, then, choosing any unit vector a E L,

(14.87) A = a a or Ax = a(a, x) for all x E F".

Here a E F" represents a column vector.Now we turn to the Cayley plane.

290 Cayley Plane and Exceptional Group F4

Definition 14.88. The Cayley plane is defined by

(14.89) P2(O) _ {A E Herm(3, 0) : A2 = A and traceo A = 1}.

Lemma 14.90.

(14.91)P2(O) _ {a at : at = (a1i a2, a3) E 03 with

all = 1 and [al, a2, aa] = 0},

which is a 16-dimensional compact submanifold of Herm(3, 0).

Proof: Suppose

Then

rl 23 22A x3 r2 x1 E Herm(3, 0).

x2 21 r3

r1+112211+11x311=A 2 (r1 + r2)23 + 21x2

(r1 + r3)x2 + 21x3

(rl + r2)x3 +'E211 (rl + r3)72 + 23x1

r2 + lIx1II + lIx311 (r2 + r3)x1 + x3x2(r2 + r3)71 + 22x3 r3 + Ilx1II + IIx21i

Now suppose A2 = A and traceo A = 1. Then r1x1 = x322i r2x2 = 71x3,and r3x3 = 21x2. Since not all of r1i r2, r3 can vanish, Artin's Theoremimplies that for A E P2(0) the entries x1i x2, x3 belong to a quaternionsubalgebra H C 0. Thus, A E Herm(3, H) C Herm(3, 0). Because of thisfact, (14.87) is applicable. Choose at E H3 with A = a Qt and hail = 1.This proves that P2(O) is contained in the right hand side of (14.91).

Finally, if at = (a1 i a2, a3), jjajj = 1, and [a1, a2, a3] = 0, then A =_ a -d'satisfies traceA = 1 and A2 = A so that A E P2(0).

To prove that P2(O) is a 16-dimensional manifold, note that it iscovered by the three charts

U1 _-aa-I

. at _ (1, a2, a3) E 03 = 02,

U2 _ Dollat (al, 1, a3) E 031 = 02,

U3= jll II LIa

Low Dimensions 291

Definition 14.92. The group F4 is defined to be the automorphism groupof the (Jordan) algebra Herm(3, 0), equipped with the symmetric (or Jor-dan) product A o B =

a(AB + BA).

A linear map g E GL(Herm(3, 0)) is an automorphism if

(14.93) g(A o B) = g(A) o g(B) for all A, B E Herm(3, 0).

Note that A o A = AA, so that the notation A2 is unambiguous.In particular, (14.93) implies that

(14.93') g(A2) = g(A)2.

This simpler condition of "preserving squares" suffices to guarantee thatg E F4, because polarization of (14.93) yields (14.93).

Givenrl 73 12

A = x3 r2 xl E Herm(3, 0),(X2 xl rs

define

(14.94)

traceA rl + r2 + r3,

IIAII = rl + r2 + r3 + 211x112 + 21x212 + 21x312,

o2(A) = (rlr2 - (x312) + (tits - (x212) + (r2r3 - 1x112),

det A = rlr2r3 + 2x1, x2x3) - rl Ix1I2 - r2Ix212 - r3Ix312,

in analogy with Herm(3, R).Note that (i.e., calculate directly)

(14.95) IIAII = trace (A2), r2(A) = 1((trace A)2 - trace A2)

and

(14.95') det A = 6 (trace A)3 - 2 (trace A2)(trace A) + 3 trace A3,

for all A E Herm(3, 0). Here and in the following A3 = A o A o A denotesthe Jordan cube, so that A3 = "(A2 A + A A2) in terms of ordinarymatrix multiplication.


Lemma 14.96. If g E F4, then

trace g(A) = trace A, for all A E Herm(3, O).

Consequently, F4 fixes 11 - 11, 0-2, and det.

Proof: It suffices to show that F4 fixes trace because of (14.95) and(14.95'). Since each g E F4 maps 1 to 1, it suffices to show that iftrace A = 0, then trace g(A) = 0. One can show by direct calculationthat each A E Herm(3, 0) satisfies

(14.97) A3 - (trace A)A2 + 0-2 (A)A - det A = 0.

Let

and

C - {A E Herm(3, O) : A3 = aA + b for some a, b E R},

Q - {A E Herm(3, 0): A2 = aA + b for some a, b E R}.

Note both C and Q are invariant under F4. Now assume that trace A = 0.Because of (14.97) this implies A E C. Therefore B - g(A) E C so that

(14.98) B3 = aB + b for some a, b E R.

Subtracting (14.97) (with A replaced by B) from (14.98) yields

(trace B)B2 - (o2(B) + a)B + (det B - b) = 0.

Therefore, either trace B = 0 or B E Q. If B E Q, then A = g-1(B) E Q.This proves that g maps {A : trace A = 0} N Q into {A : trace A = 0}.Finally, note that {A : trace A = 0} . Q is a nonempty open subsetof {A : trace A = 0} (choose A diagonal with trace A = 0 and distincteigenvalues).

Theorem 14.99. F4 acts transitively on the Cayley plane P2(O) withisotropy subgroup equal to (an isomorphic copy of) Spin(9) at the point

1 0 0

El =- 0 0 0 E P2(O),0 0 0

(14.100) F4/Spin(9) = P2(O).

Low Dimensions 293

Proof: Spin(9) is generated by

r Ru) :rER,, uEO, andr2+JU12=1}l Ru r) J

because of Lemma 14.77. Let C denote conjugation on both factors of

0 ® O. It is more convenient to consider the copy C C. Spin(9) C of Spin(9).Since

r RuC' (R).c=(;u-rthe group C Spin(9) C is generated by

Suppose

r Lul :rER., uEO,

r

Lu rJJJ

9= )(Lu-ris a generator for C Spin(9) C. Then

rxl + ux2lg(x) = I for

uxl - rx2JJJ (x2)E O®O,

determines the spinor representation of C Spin(9) C. Also,(14.101)

p Lv - (r2 + lul2)p + 2r(u, v) L2rpu-rev}uvuXg k Lv -p L2rpu-rev-uvu - (1u12 + r2) p - 2r(u, v)

determines the vector representation of C Spin(9) C on

p L" ): P E R, V E 0

Lu- -P

Extend Xg to act on

r2 L \ ll

}(Lu r3) :r2 r3ER,vE0

by defining

Xg(0 0)-(0 1 0)


The action of C C. Spin(9) . C on Herm(3, 0) is defined as follows. Given

rl zl 72)(14.102) A - x1 r2 v E Herm(3, O)

X2 'U r3

and

(14.103)

define

Ta generator for C Spin(9) C,

t

g(A) =9(r

1x) Xg ()) '

where a =(Lu-

r2 L" I and x = (X2)x1\ a) r3 /

Now we can prove that

Lemma 14.104. C C. Spin(9) C C F4.

Note that this implies that Spin(9) C IsotE, (F4).Proof: Given A E Herm(3, 0) and g a generator for C Spin(9) C as in(14.102) and (14.103), let

(14.105)

1 0 0G- 0 r u

0 u -r

Then, using (14.101) for Xg(a), a direct calculation shows that

(14.106) GAG= r1 g(x)t g(A).(9(x) Xg(a)

Suppose U - (u;j) E M,,(O), with span{uij, 11 contained in a subal-gebra isomorphic to C C O. (For example, U - G defined by (14.105).)Then for any element z E 0, span{uij,1, z} is contained in an associa-tive subalgebra of O. In particular, ulzu2u3zu4 is well-defined for allU1i U2, U3, U4 E span{u;j, 1), so that

(14.107) (ulzu2)(u3zu4) = ul[z(u2u3)z]u4

Polarization yields

(u1 Zu2)(U3wU4) + (ulwU2)(u3ZU4)(14.107')

= u1[z(u2u3)w]u4 + ul[w(u2u3)Z]u4

Low Dimensions 295

This implies that

(14.108) (UAU)(UBU) + (UBU)(UAU) = U(AU2B)U + U(BU2A)U,

for all A, B E Ilerm(n, 0). Applying this identity, with U - G defined by(14.104), yields

g(A)g(A) = (GAG)(GAG) = GA2G = g(A2),

since G2 = 1. Thus g E F4. This proves that C Spin(9) C is a subgroupof F4. J

Since F4 preserves the trace, both the conditions, A2 = A andtrace A = 1, defining P2(O) are preserved by F4. Next we prove thatF4 acts transitively on P2(O). Suppose A E P2(O):

A (x a)' x (x2)and a= (v2 V

r3

Choose g E C Spin(9) C C F4 so that x9 (a) = (o ° ). This is possiblebecause x : Spin(9) --> SO(9) acts transitively on the 8-sphere in each 9-plane {a E Herm(2, 0) : trace a = constant). The isotropy of C Spin(9) C

at (o °) is the same as the isotropy at (o °) , which contains

C End(O ®O) has orbits S7 x {0}, S7 X S7, and{0) x S7. In all cases, there exists h E Spin(8) so that h(xi, x2) = (yi, y2)with yl, y2 E R. This proves that the orbit of F4 (in fact, C Spin(9) C CF4) through any point A E P2(O) contains a point B E P2(O) with eachentry bij E R, i.e., B E Herm(3, R),

Given g E SO(3) C M3(R), define

(14.109) p9(A) = gAg` for all A E Herm(3, 0).

This defines SO(3) as a subgroup of F4. If A is real, then we may chooseg so that B =_ p9 (A) is diagonal. The conditions B2 = B and trace B = 1imply that B has eigenvalue 1 with multiplicity one and eigenvalue 0 withmultiplicity 2. Finally, by applying a permutation zr E SO(3) to B, weobtain El, proving that F4 acts transitively on P2(O).

It remains to show that each g E F4, which fixes El, belongs toC Spin(9) C.

Given A (r' 3Y) E Herm(3, 0), note thatx a

(14.110) 2AoEl -A= (r' 0 /)0 a


Thus, if g E F4 fixes El, then g maps the subspace of Herm(3, 0) definedby a = 0 into itself. Therefore,

(14.111) g(A) = ( r f(x)t

f(x) h(a)

where h E SO(9), acting on Herm(2, 0) while leaving the identity fixed.Now by utilizing an element of Spin(9), we may assume that h = 1. Itremains to show that f is the identity.

Suppose A is of the special form

0 71 72

xi 0 0

X2 0 0

Since h - 1, g fixes the basic diagonal matricies E2 and E3. Note that

0 71 0

2AoE2= xi 0 0

0 0 0

Thus, 2A o E3 = A if and only if x2 = 0. Consequently if x2 = 0, theny2 = 0, where y = f (x). Similarly, if xl = 0, then yl = 0. This proves thatyi = fl(xi) and Y2 = f2(x2). Since

Ix12 0 0

A2 = 0 0 '1x20 x271 0

the condition g(A2) = g(A)g(A) becomes

21x2 = fl(xl)f2(x2) for all x1i x2 E O.

If u = fi(1), then u-1 = f2(1), and it follows that fl(xl) = xiu andf2(x2) = x2u-1. The above condition on f becomes ('lu)(u-172) = 21x2for all 21,x2 E 0, which implies that u = 1. Therefore f(x) = x. Thiscompletes the proof of Theorem 14.99.

The complex Lie group F4 can be defined to be the automorphismgroup of the complexified Jordan algebra Herm(3, O)®R,C - Herm(3, OC).Here Oc - Oc ®R, 0 denotes the algebra of complexified octonians. Notethat Oc contains the split octonians O as a subalgebra. The noncompact

Low Dimensions 297

or split case of F4, denoted F4 is defined to be the automorphism group ofthe Jordan algebra Herm(3, O).

In addition to F4 and F4, there is a third real form of F4 c, which wedenote by F4. Let

f r1 -ix1 -i22 1

Herm'(3, 0) = 1A E M3(Oc) : A = ix1 r2 x3Zx2 x3 r3

and x1i x2, x3 E O, rl, r2, r3 E R1-

(Here i denotes the i E C in the complexification Oc = 0 (9R C of 0.)Define F4 to be the automorphisms of the Jordan algebra Herm'(3, 0).Note that all three Jordan algebras:

Herm(3, 0), Herm(3, O), and Herm'(3, 0)

have the same complexification Herm(3, Oc).The results of this section have analogues for F4 and F4. For example,

define

P2(O) -= {A E Herm(3, O) : A2 = A and traceo A = 11.

Then the orbit of F4 through

1 0 0

0 0 0

0 0 0

is P2(O) with isotropy subgroup Spin°(5,4):

(14.112) P2(0) a-- F4/Spin°(5,4).

Also,

(14.112)' F4/Spin(9) L'5 {A E Herm'(3, O) : A2 = A and traceA = 1}

The proofs of (14.112) and (14.112)' can be adapted from the proof ofP2(O) L, F4/Spin(9), with suitable modifications and additions.

298 Clifford Algbebras in Low Dimensions

CLIFFORD ALGEBRAS IN LOW DIMENSIONS

The models presented in the previous sections for Cl(7), Cl(8), Cl(3, 4),Cl(4, 4), and Cl(9) are quite useful. Similar models for the other low dimen-sional Clifford algebras are available. Some of these models are collectedin this section (dimensions 2, 3, and 4 and the various signatures). jr,these low dimensions, the algebraic structure of Cl(r, s) along with thecanonical involutions uniquely determines the choice of vectors A'R(r, S)and hence uniquely determines the full Clifford structure. This is becausethe hat involution is minus one on A'R(r,s), and plus one on A3R(r, S).Thus, in dimensions n < 4, A'R(r, s) is uniquely defined as the subspaceof Clodd(r, s) fixed by the hat involution, sending a to a.

Dimension 2 and Signature 2, 0 or 0, 2

Spinors.Cl(2, 0)even = Cl(O1 2)even = C,

where a choice A of unit volume element corresponds to i E C. A canonicalchoice for the spinor space S (a complex 1-dimensional vector space withCl(2, 0)even = Endc(S)) is possible once an orientation has been selected.Just take S = Cl(2, 0)even with complex structure A, and let Cl(2, O)evenact on S by left multiplication. Thus

Spin(2)={aEC:hail=1}=Sl

and the canonical involution (hat = check) on Cleven C is just conjuga-tion on C.

Pinors. The two larger algebras Cl(2, 0) and Cl(0, 2) containing Cl evenC are the two Cayley-Dickson doubles of C.

First,

can be rewritten as:Cl(2, 0) = Cl(2, 0)even ® Cl(2, 0)odd

H C ® Cl

where Cl = span{ j, k}. In particular, the space of vectors R(2, 0) _A'R(2, 0) C Cl(2, 0) is just C'. One can take P = Cl(2, 0) with rightH-structure H - Cl(2, 0).

Second, the pinor space P(0, 2) is a real 2-dimensional vector spacewith Cl(0, 2) - EndR(P). Now

Cl(2, 0) - Cl(2, 0)even ® Cl(2 0)oddcan be rewritten as:

Low Dimensions 299

M2(R) = C ® C'

where Cl = span { C, R}, C = (o _°) conjugation, R = (° o) reflection,and i = J = (° -o). In particular, C' = span{C, R} = A1R(0, 2), i.e., theset of complex antilinear maps in M2(R) is the space of vectors.

In both cases, the vector representation of Spin(2) S' is equivalentto the representation X : Spin(2) -. Enda(C) defined by X,,(z) = a2z.Remark. Suppose M is a 2-dimensional (positive definite) Riemannianmanifold. Then, via the canonical isomorphism Cl(2, 0) = A(R(2, 0), theexterior bundle ATM is naturally a bundle of H-algebras, with the sub-bundle AeVe"TM = A°TM ® A2TM a bundle of C-algebras_ Since in thisdimension AevenTM acts on TM by Clifford multiplication, AevenTM isnaturally a subbundle of End(TM). Note M is oriented if and only ifAevenTM = M x C is trivial. (Similarly, ATM is naturally a bundle ofM2(R)-algebras.) In this dimension (if M is oriented), AeVe"TM = M x Ccan be taken as the bundle of spinors M x S with the bundle isomorphism

Cleven(2) Endc(S).

Dimension 2 and Signature 1,1

Spinors. Cl(1, 1)even R ® R - L (double numbers of Lorentznumbers), where a choice of unit volume element corresponds to r E L.A canonical choice for the spinor space S = S+ ® S_ (two real vectorspaces S+ and S_ with Cl(1, 1)even = Enda(S+) ® EndF(S_)) is possibleonce an orientation has been selected. Just take St = Cl(1,1)fen = {a ECl(1, 1)even :Aa = ±a}, and let Cl(1,1)even act on S by left multiplication.The canonical involution on Cleven - L is conjugation on L. Now

Spin°(1, 1) = {fe''e : 9 E R} = R ® Z2

Spin(1,1) = {±er9 0 E R} U {±rere : 0 E R}

Pinors. The pinor space P is just S and Cl(1, 1) - M2(R) -EndR,(P), with

Cl(1, 1) - Cl(1,1)even ® cl(1,1)odd

corresponding to

M2(R) = L ® L1

where r =- (o _°) and Ll = span{R, J} with R = (° o) and J (° o).That is, L C M2(R) consists of the L-linear maps and Ll C M2(R)consists of the L-antilinear maps.



Signature 3, 0.

(14.114) Cl(3, 0) - M1(H) ® M1(H),

with r = (o _°) a choice of unit volume element. Also S =_ H withP = H ® H. The hat inner product t on P = H ® H is the standardH-hermitian symmetric inner product. Thus, the hat anti-automorphismapplied to (o

°)equals (o °) . Now Cl(3, 0)even 25 M1(H) considered to be

diagonally embedded in M1 (H) ® M1 (H) - Cl(3, 0). Thus,

Spin(3) {a E H : 1Iall = 1} = S3 C H = C1(3, 0)even

That is, Spin(3) = HU(1). The vectors A1R(3, 0) can be distinguished as asubspace of AR(3, 0) - Cl(3, 0) by noting that A E A1R(3, 0) if and only ifa is odd and a is orthogonal to the unit volume element A = (o _°) . Thus,

Al R(3, 0) p _0 I : u E Im H } C M1(H) ®M1(H) Cl(3, 0).

Conversely, adopting this as the definition of A'R(3, 0) C M1(H)®M1 (H),it is easy to deduce (from the Fundamental Lemma for Clifford Algebras)the isomorphism (14.114) and all of the rest of the information above aboutCl(3, 0).

The vector representation Ad : Spin(3) --* SO(3) is given by Adau =aua, for all u E ImH, where a E Spin(3) - S3 = HU(1) C H.Remark. If M is a 3-dimensional (positive definite) Riemannanian man-ifold, then the vector bundle E = Aeve.TM = Cleven(TM) is naturally abundle of H-algebras. Letting E act on itself on the left embeds E as asubbundle of EndR(E). Let C denote the centralizer of E in Enda(E).Then C is a bundle of H-algebras that acts on E on the right, giving E thestructure of a right H-vector bundle with coefficient (or scalar) bundle C.Because of dimensions, Cleven(TM) - EndH(E), so that E may be takenas the global bundle of spinors for M.

Signature 0, 3.Cl(0, 3) - M2(C),

where A = i is a choice of unit volume element. Thus P = C2. The checkinner product t is the standard C-hermitian symmetric positive definiteinner product on C2; thus A = A* = At. Therefore, consulting the Table

Low Dimensions 301

in Lemma 9.27, we see that A1R(0, 3) must consist of those A E M2(C)that satisfy

A is orthogonal to 1 (i.e., traces A = 0)and

A=A*, i.e., A is hermitian symmetric).

This provesA1R(0, 3) = Hermo(C2),

the space of trace free hermitian symmetric 2 x 2 matrices.Note that the Pauli spin matrices

_ (0 1 _ 0 i _ 1 0U1- (1 0 '

0.2i 0 ' Q3- (0 -1)

form a basis for A'R(0, 3) C M2 (C) = Cl(0, 3), with A = i = U1620'3 thevolume element. Since A1R(0, 3) = Hermo(C2),

A2R(0, 3) = i Hermo(C2)

so thatCI(0, 3)even = { ( z

I. wz I :z,wEC}.

Therefore,

Spin(3) = {(zI : z, w E C with Iz12 + 1w12 = 1 } SU(2).

w z/JJJ

The vector representation Ad : Spin(3) - SO(3) is given by

AdAh = AhA* for all h E Hermo(C2),

where A E Spin(3) = SU(2).Remark 14.113. The isomorphism Cl(3, 0)even - CI(0, 3)even induces anisomorphism of the two realizations of Spin(3), namely,

HU(1) t--- SU(2).

Thus, the Clifford algebra point of view leads naturally to an isomorphism,which was first discussed in Proposition 1.40; the isomorphism between

Cleven(0 3) =

((Wz : z, w E C} C M2(C) = C1(0, 3),

and


Cleven (3, 0) = Mi(H).

Remark. One can show that

(a) Pin(3, 0) = Z2 x S3 C H ® H with Z2 = {(1, 1), (1, -1)},(b) Pin(0, 3) = {A E U(2) : detc A = f1} C M2(C),,(c) Center Pin(3, 0) = Z2,(d) Center Pin(0, 3) = {f1,±i} = Z4.

Thus by (c) and (d), Pin(3, 0) and Pin(0, 3) are not isomorphic as abstractgroups.


Spinors.C1(2, 1)even - Cl(1, 2)even - M2(R).

The space of spinors S = R2, and the spinor inner product e is thestandard volume form (R skew) on S = R2. Thus, the canonical involutionon Cleve. = M2(R) is the cofactor transpose, and

Spin°(2,1) SL(2, R)

Spin (2,1) (A E M2(R) : detA = ±1}.

Pinors for Signature 2, 1.

Cl(2, 1) - M2(C) = C1(2, 1)even OR C.

The pinor space p is C2 and the unit volume element A equals i E M2(C).The hat inner product t on C2 is the standard complex volume elemente = dzl A dz2 on C2. Thus, the hat involution on Cl(2, 1) = M2(C) isthe cofactor transpose. Consequently, A'R(2, 1) C M2(C) Cl(2, 1) isdistinguished as

{iA:AEM2(R)andtrace A=0}={ Cix -iyl \` iz ix

l :x,y,zER}.) JJ

JJ1

Pinors for Signature 1, 2.

Cl(1, 2) - M2(R) ® M2(R) = Cl(1, 2)even ®R L

The pinor space is given by P = P+ ® P_ with P± = R2, and the hatinner product E on P equals e ®e, where e = dxl A dx2 is the standard

Low Dimensions 303

volume element on R2. Thus, the hat involution is the cofactor transposeapplied to both 2 x 2 real matrices in C1(1,2) = M2(R) ® M2(R). Thespace A1 R(1, 2) of vectors can be distinguished in Cl(1, 2) and equals

Al R(l, 2) = { I 0 -A ) : A E M2(R) and trace A = 0 } .

Remark. Thus, in both cases, signature 2, 1 and 1, 2, the vector represen-tation

Ad : Spin(2, 1) -> SO(2, 1)

is given by

AdAB =_ ABA-' for all B E M2(R) such that trace B = 0,

where A E Spin(2, 1) = SL(2, R).

Dimension 4, Signature 4, 0 and 0, 4

Spinors.

Cl(4, 0)even - Cl(0, 4)even - M1 (H) ® M, (H).

The space of spinors S =_ S+ ® S_ with S+ = H, so that Cleve.EndH(S+)®EndB(S_). The spinor inner products of are the standard H-hermitian symmetric (positive definite) inner product on S± = H. Thus,the canonical involution on Cleve. LI M1(H) ® M,(H) is conjugation ofboth factors. Therefore,

Spin(4) = S3 X S3 a

The unit volume element is A

0) :IIail=IIbhl=1} C M1(H)ED M1(H).

_ (1 0lo_11

Pinors.As algebras both

C1(4, 0) - M2(H), and Cl(0, 4) = M2(H).

Thus, P(4, 0) and P(0,4) are both H2. The standard H-hermitian sym-metric positive definite inner product on H2 corresponds to the hat innerproduct t on P(4, 0) = H2 while it corresponds to the check inner product


t on P(0, 4) - H2. Consequently, V(4, 0) = A1R(4, 0) is distinguished asthe subspace

(/ l 1

(14.115) V(4, 0) = S l -Lu _ u E H) C Cl(4, 0) = M2(H),

i.e., those odd elements̀ that are H-hermitian skew; while V(0,4)A1R(0, 4) is distinguished as the subspace

(14.116) V(0, 4) _ f0u1 ) : u E H } C Cl(0, 4) = M2(H).

Conversely, adopting (\14.115) as the definition of V(4, 0) C M2(H) and(14.116) as the definition of V(0, 4) C M2(H), it is easy to deduce (fromthe Fundamental Lemma of Clifford Algebras) the isomorphisms C1(4, 0)M2(H) and C1(0, 4) = M2(H) and all the rest of the information aboveabout Cl(4, 0) and Cl(0, 4).

Dimension 4, Signature 1, 3 and 3,1 (Special Relativity)

Spinors.C1(3, 1)even - Cl(1, 3)even = M2(C),

and the unit volume element A for R(3, 1) or R(1, 3) can be chosen to bei E M2(C). The space of spinors S is a 2-dimensional complex vector

space with isomorphisms

Cl(3,1)even - Cl(1, 3)even t--- Endc(S).

The spinor inner product c on S is C-skew. (Hence, we may take S = C2and s = dzi A dz2, the standard complex volume form on C2.) There-fore, the canonical involution on Cleven = M2(C) is the cofactor transposesending

Thus,

and

A=`a b)\c dto A*

Spin°(3,1) = Spin°(1, 3) = SL(2, C)

Spin (3,1) = Spin (1, 3) = {A E M2(C) : detc A = ±11.

The reduced spin group Spin°(3,1) = SL(2, C) acts on the space S, eof spinors preserving the C-skew inner product e. The conjugate represen-tation of SL(2, C) is on the space S of conjugate spinors. (Pc = S ®S isthe space of complexified pinors.)

Low Dimensions 305

In this important case, the spinor representations yield a classificationof all representations of SL(2, C) with important implications in generalrelativity. A brief description is included.

Start with S, J a complex vector space with complex structure J. Asusual, consider the complexification Sc = S®R,C = 510®S°,1 decomposedinto the +i eigenspace Sl,0 and the -i eigenspace S°,1 for J (extended tobe complex linear on S), along with the canonical conjugation on Sc.Now assume dimc S = 2, let S = S1'°, S = S°'1, and assume that Sis equipped with a complex volume element (a C-skew inner product e).Representations of the group SL(S) = SL(2, C) will be denoted by thevector space alone, i.e., the action is assumed to be obvious.

For example, SL(2, C) has a natural induced action on SP the pthsymmetric tensor product of S, and on S the qth symmetric tensor productof S. It is a standard fact in representation theory that

(1) SP ®Sq (the space of p, q spinors) is irreducible, p, q > 0,(2) there are no other irreducible representations of SL(2, C).

For example, a defines an isomorphism b of S with dual space S* that isthe intertwining operator for S = S*. Also, e provides an isomorphismbetween A2S and C, the trivial representation. The vector representationof Spin°(3, 1) SL(2, C) is obtained as follows. Let H = (S ®S)R denotethe subspace of S ® S A1,1Sc fixed by conjugation. Let

2 (Im h) A (Im h) _ (h, h)e A E E A2'2SC

define a bilinear form ( , ) on H. One can identify H with the space ofhermitian symmetric 2 x 2 matrices with square norm equal to the determi-nant. In particular, H = R(1, 3) is Minkowski space. Note that replacinge by eae dos not change the metric on H.

Using e, the space of p, q spinors (with the SL(2, C) action) can beidentified with the space Cp q [z, z] of polynomials which are homogeneousof degree p in z E S and homogeneous of degree q in x E S (with the naturalaction of SL(2, C) on C[z, z]).

Remark. On a Lorentzian manifold M equipped with a spinor bundleS, e, note that: (i)T*M®C-S®S, and s®S-S2®A2S-S2®C-S2'0 ®S°'°. More generally, one can show that: (ii) S ®SP = SP+1 ®Sp-1

These facts, (i) and (ii), enable one to define four different (if p > 1 andq > 1) differential operators on p, q spinor fields s based on decomposingthe covarient deviative Ds of s as follows:

Ds E TTM 0 Sp,q = Sp-1,q-1 ® SP+l,q-1 ®Sp+l,q+l ® SP-1,q+1.


q

.p

Figure 14.117

If q = 0, then there are just two differential operators. the one takinga p, 0 spinor field to a p - 1, 1 spinor field is called the Dirac operator anddenoted by ill. The other, taking a p, 0 spinor field to a p+ 1, 1 spinor fieldis called the twistor operator. The Dirac equation

P f = 0, where f is a p, 0 spinor field,

is called the equation for a massless particle of spin s = p/2. The casep = 2 (s = 1) is that of an electron.

Pinors for Signature 1, 3: The pinor space P is a real 4-dimensional vec-tor space. A choice of unit volume element A provides a complex structureI A on P. The spinor space S can be taken to be P with the complexstructure I. The spinor inner product a is complex skew and

e=E+ie, ore=Ree.It is possible to choose a conjugation c on P so that

e(cx, cy) = e(x, y) for all x, y E P.

Note thatAoddR(1, 3) = Endc(P)

(the space of all complex antilinear maps of P) can be expressed as

Endc(P) = {Ac: A E Endc(P)} .

Low Dimensions 307

Now c2 = 1 and c is complex antilinear. Therefore, given A E Endc(P),t(Acx,y) = Ree(Acx,y) = Res(cx,A*y) = E(cx,A*y) = e(x,cA*c2y) _9(x, A * cy). This proves that

(Ac)' = A*c for all A E Endc(P).

Therefore,

That is

A'R(l, 3) = {Ac : A E Endc(P) and A* = -A} .

V(1, 3) - A1R(1, 3) must be the subspace

(14.118) V(1,3)=1 z -xc:zEC,andx,yER}l\y zl J

of EndR,(P).The square norm on V(1, 3) is given by

(14.119) IlAcIl = (Ac)(Ac)^ = AcA*c = AA* = detc A,

which equals -jjzjj - xy if

A= (y -y )

Conversely, adopting (14.118) as the definition of V(1, 3) C EndR,(P)M4(R), where V(1, 3) is equipped with the square norm given by

(14.119), it is easy to deduce the isomorphism Cl(1, 3) - EndR,(P) andall the rest of the information presented above about Cl(1, 3).

Pinors for Signature 3,1: The pinor space P is a 2-dimensional(right) quaternionic vector space; so that CI(3, 1) - Endc(P). One canshow that, under an isomorphism Endc(P) - M2(H), the space of vectorsA'R(3, 1) is isomorphic to

-w/ :x,yERandwEC}

Left multiplication by i, denoted Li, corresponds to a choice of unit volumeelement.

308 Squares of Spinors and Calibrations

SQUARES OF SPINORS AND CALIBRATIONS

The pinor multiplication introduced in Chapter 13 is a useful toolfor constructing calibrations. The Clifford element 0 - xoy E C1(r, s)obtained from multiplying two pinors x and y may also be considered as adifferential form 0 E AR(r, s)*; under the natural identifications of Cl(r, s)with AR(r, s), and AR(r, s) with AR(r, s)*. In this section, we considerthe positive definite case.

The special case n = 8p is particularly interesting. Recall

CI(8k) - Endn.(P),

with P = S+ ® S_ decomposing into the real vector spaces S±, of positiveand negative spinors. The spinor inner product c on S+ and S_ is positivedefinite and real. Given a differential form 0 E AR(n)*, let cfk E AkR(n)*denote the degree k part of 0.

Theorem 14.120. Consider R(n) with n = 8p. Given a unit positivespinor x E S+, let

(14.120) 0 = 161'x OX E Cl(8p) - AR(n)*.

Then

(a) 0 E > A4mR(n)* only has components of degree 4m.(b) Each ck is a calibration, i.e., ¢k(') < 1 for all t: E G(k, R(n)).(c) Equality Ok(1;') = 1 occurs if and only x,(d) The isotropy subgroup of 0(n) that fixes 0 is isomorphic to the sub-

group Kx of Spin(n) that fixes the spinor x.

Corollary 14.122. Under the hypothesis of Theorem 14.120, the degreek-form ck is a nontrivial calibration (k = 4m) if and only if there exist an1; E G(k, R(n)) such that x = x. The contact set

G(4) - { E G(k, R(n)) : q(4) = 1}

is equal to {!; E G(k, R(n)) C Spin(n) : t;x = x}, which via the map X isisomorphic to {Refspa=,t : E G(k, R(n)) and RefBpan£(0) = 4'} C SOW-

Proof. By Theorem 13.73 (dims P = 161'),

(14.123) q(t;) = 16P(xox,t;) =s(l;x,x) for all 1; E Cl(n).

This formula is very useful.

Low Dimensions 309

(a) If E Cl(n)°dd then x E S+ implies E S_ so thatIf has degree = 2 mod 4, then = - and hence

x) = 0.

e(x, x) = x),

so that again 0({) = 0.(b) By the Cauchy-Schwarz inequality, Equation (14.123) implies that

(14.124) jxj = 1 for all E G(k,R(n)) C Cl(n)

where y12 = e(y, y) is the spinor norm on S+.(c) Also,

(14.125) 1 if and only if ex = x.

Note that Jex! = 1 in (14.124) because £ E G(k, R(n)) C Spin(n) fork = 0 mod 4, which implies that = 1.(d) Consider the vector representation x : Pin(n) --> O(n). Given a EPin(n) and v E R(n) then Xa(v) = ava-1. More generally, given a EPin(n), Xa E O(n) acts on AR(n)* = AR(n) by Xa(O) = a0a-1 =16Pa(x 6 x)a. By Lemma 8.44(a), this equals 16(ax) 6 (ax). Therefore, ifXa(0) = 0 then (ax) o (ax) = x o x. By the definition of spinor multiplica-tion, this implies that ax = cs for some scalar c E R. Since a preserves thepinor norm, c = ±1. The element a E Pin(n) must be either even or odd.If a is odd, then ax E S_ and ax = ±x is impossible. Thus ax = ±x and

{a E Pin(n) : X.,(') = ¢} _ {a E Spin(n) : ax = ±x}.

Since, under the vector representation X, both a and -a have the sameimage, this proves that

(14.126) X:Kx={aESpin(n):ax =x}--*{gEO(n):g*c6=¢}

is surjective. Since -1 ¢ Ks,, this is an isomorphism.

Using the octonian model for C1(8) = EndR,(O ®O) with S+ = 0, thesquare of a positive spinor is easily calculated. First note that Spin(8) actstransitively on the unit sphere in S+, so that any two squares are SO(8)-equivalent in AR(8)*. Therefore, we may choose the multiplicative unit inS+ = 0, denoted 1+, as a unit positive spinor without loss of generality.

310 Squares of Spinors and Calibrations

Theorem 14.127.

(a) 16 1+ o 1+ = 1 + 4D + a, where ID E A4O* is defined by

(14.127) c(u1 A U2 A U3 A u4) _ (u1, u2 X U3 X U4),

and a is the unit volume element determined by the orientation on O.(b) The 4-form is a calibration (called the Cayley calibration) on Rs

0, that is fixed by the group Spin(7) {g E Spin(8) : g+(1) = 1}.

Proof: By (14.123)/,

p 716(1+ 6 1+)(u1 A U2 A U3 A U4) = E(RuR1u,R,,3Ru,1+, 1+),

which equals

(((U4 u3)'U2)'U1, 1) = ((U4U3) u2, 'U1)

= (U1, u2 ('U3u4)) =' (ui A U2 A u3 A U4)

if u1i u2, U3, U4 are orthogonal. Therefore, the degree 4 component of16 1+ 6 1+ is C Similarly, one shows that the degree 8 component isA.

See Bryant and Harvey [6] for other examples of pinor multiplicationin dimension 8. In the next multiple of 8, dimension 16, all unit positivespinors are not the same under the action of Spin(16). The orbit structureof Spin(16) acting on S+ is very interesting. Each orbit can be used toconstruct calibrations by squaring a spinor in that orbit. See Dadok andHarvey [7] for the details.

The case dimension n = 0 mod 8 can be used to understand the casedimension m = 7 mod 8 (see the next Theorem and Problem 9).

Recall that if a unit vector eo E V(n) is chosen so that V(n - 1) = eo ,

then sending a vector u E V(n-1) to eo - u E Cl(n) induces an isomorphismof Cl(n - 1) 25 Cl(n)even. Furthermore, if Cl(n) = EndR,(P) is a pinorrepresentation with P - S+®S_, then Cl(n)e°eII = Enda(S+)®EndR,(S_).Therefore, an identification S - S+ - S_ induces an isomorphism

Cl(n - 1) = EndR,(S) ® EndR(S),

which is a pinor representation of Cl(n - 1).Now, a unit positive spinor x E S+(n) can also be considered a unit

pinor y-(z,0)EP(n-1)-S®S.

Low Dimensions 311

Theorem 14.129. Let

(14.130) 16" x o x E CI(n), n= 8p,

and

(14.131) 16" y 6 y E Cl(n - 1).

Let the orthogonal decomposition

0-eona+define a, Q in AR(n - 1)*. Then

(14.132) +)3.

Proof: By Theorem 13.73

(14.131) y) = E(ex, z) for all E Cl(n - 1).

If i; has degree 1 or 2 mod 4, then _ - so that 0. We must showthat

(14.134) if degree l; = 0 mod 4,

and

(14.135) a(eo A t;) if degree = 3 mod 4.

Let u1, ..., denote an orthonormal basis for V(n - 1) and eo, elul,..., en- 1 - an orthonormal basis for V (n). Then Cl(n-1) embedsin C1(n) by sending ui E Cl(n-1) to eo e3 E Cl(n). If E' _- ui, ui, ui, ui4 ECl(n - 1) - AR(n - 1), then the corresponding element, also denoted e, inCl(n) = AR(n) is equal to

eoui, eoui, eoui, eoui, = uil ui, ui, ui,.

Therefore, O(C) = E(cx, x) = /3(e). If C = uil ui, ui, E C1(n - 1) '=AR(n - 1), then the corresponding element of Cl(n) AR(n) is eoui,eoui, eoui, = eoui, ui, ui,. Therefore, b(£) = ¢(eo ) = a(e).

312 Problems

PROBLEMS

1. A complex volume element A on C4 determines a nondegenerate com-plex inner product (, ) on A2C4 by

Show that the induced action x of SL(4, C) on A2C4 provides a doublecover

SL(4, C) x 0(6, C).

2. Verify the isomorphisms in Theorem 14.5.3. (a) Show that the (triality automorphism) r defined by (14.27) is an

automorphism of Spin(8) of order three that permutes the representa-tions x, p+, and p-.(b) Show that (go, g+, g_) E Spin(8) if and only if one (all) of thefollowing triples belongs to Spin(8):

9o,9+,9- , 9_,9+,90

(9+,9-',9o) , (go, 9-, 9+)

(9-190,9+) (9+, 90, 9- )

,

4. (a) Show that Spin(3, 4) acts transitively on S7 = S+ U S? ,

St -{xE0:IIxII=dd}CO-S,

with isotropy subgroup at 1 E O - S equal to split G2.(b) Show that Spin°(3, 4)/G2 25 S.

5. Show that, under the spinor representation p = p+ ® p_ of Spin(4, 4)on S-S+®S_ 0®0,(a) Spin(4, 4)/G2 {(x, y) E O x o : IIxII = ±1 and IIyII = ±1}.(b) Spin (4, 4) /Spin(3, 4) 95 {x E O : IIxII = ±1}.

6. Show that

Spin(5, 4)/Spin(3, 4) {x E R(8,8) : IIxII = ±1}.

7. Use Corollary 10.50 to characterize Spin(9) in the model Cl(9)evenEndR,(O ® 0) defined by (14.74).

Low Dimensions 313

8. (a) Show that, under the action of F4 on Herm(3, 0), each A EHerm(3, 0) may be put in the canonical form

11rl 0 0

A = 0 r2 0 with rl, r2i r3 E R and rl < r2 < r3-0 0 r3

(b) If two (but not all three) of the eigenvalues rl, r2i r3 are equal,then the orbit through A is

F4/Spin(9) = P2(O).

(c) If all three of the eigenvalues ri, r2, r3 of A are distinct, then theorbit through A is

F4/Spin(8) - An Ss bundle over P2(0).

9. Prove that 16 1+ 6 1+ considered as an element of C1(7) = AR(7)* isequal to 1 + ¢ + V) + A, where 0 is the associative calibration, 0 is thecoassociative calibration, and A is the unit volume form on Im O.

REFERENCES

1. Arnold, V. L, Mathematical Methods of Classical Mechanics, Springer-Verlag, New York, 1978.

2. Atiyah, M. F., R. Bott, and A. Shapiro, Clifford modules, Topology, 3(1964), 3-38.

3. Besse, A. L., Einstein Manifolds, Springer-Verlag, Berlin, 1987.4. Bryant, It. and It. Harvey, Submanifolds in hyperkahler geometry,

Jour. Amer. Math. Soc., 2 (1989), 1-31.5. Bryant, R. and It. Harvey, Stabilizers of calibrations, Rice University

preprint, 1989.6. Bryant, R. and R. Harvey, Geometry of G2 and Spin(7) structures,

Rice University preprint, 1989.7. Dadok, J. and It. Harvey, Calibrations and spinors, Rice University

preprint, 1989.8. Gureirch, G. B., Foundations of the Theory of Algebraic Invariants,

P. Noordhoff LTD., Groningen, The Netherlands, 1964.9. Harvey, R. and Lawson, H. B., Jr., Calibrated geometries, Acta Math.

148 (1982), 47-157.10. Helgason, S., Differential Geometry, Lie Groups, and Symmetric

Spaces, Academic Press, New York, 1978.11. Lawlor, G., The Angle Criterion, Invent. Math., 95 (1989), 437-446.12. Lawson, H. B., Jr. and M. Michelsohn, Spin Geometry, Princeton Uni-

versity Press, Princeton, New Jersey, 1989.13. Nance, D., Sufficient conditions for a pair of n-planes to be area-

minimizing, Math. Ann. 279 (1987), 161-164.14. O'Niell, B., Semi-Riemannian Geometry, Academic Press, New York,

1983.15. Penrose, It. and W. Rindler, Spinors and Space-Time (2 vols.), Cam-

bridge University Press, Cambridge, 1986.16. Salamon, S., Riemannian Geometry and Holonomy Groups, Wiley,

New York, 1989.

315

Subject Index

AAdjoint map, 47, 52, 170Adjoint representation, 54Algebracenter of, 168centralizer, 168complex Clifford, 190, 225,

233exterior, 178Jordan, 291Lie, 48, 53representation of, 163simple, 164, 235tensor, 177

Almost-complex manifold, 84"weak", 84

Almost-quaternionic manifold, 86coefficient bundle, 86

Angle between vectorsLorentzian, 62positive definite, 61

Angle theorem, 140, 149Annihilator of a subspace, 27

Anti-automorphism, 47, 182Anti-self dual, 97, 188, 209Artin's theorem, 109Associative3-plane, 144calibration, 113, 144submanifolds, 144

Associator, 104Atlas, 82Automorphism groupof complex numbers, 114of Lorentz numbers, 114of octonians, 116of quaternions, 115of split octonians, 116

BBackwards Cauchy-Schwarz

inequality, 62Backwards triangle inequality, 64Basis theorem, 29, 40, 46Bilinear formcheck, 231

317

318

hat, 231hermitian, 19hermitian skew, 20hermitian symmetric, 20nondegenerate, 20pure H-symmetric, 17pure skew, 20pure symmetric, 20

Bryant's theorem, 146

CCalibration, 126, 308associative, 113, 144, 313Cayley, 310coassociative, 115, 145, 313contact set, 126fundamental theorem, 127generalized Nance, 156Nance, 151, 156special Lagrangian, 130

Canonical automorphism, 181, 247Cartan's isomorphisms, 14, 271,

274Cartan-Dieudonne theorem, 67Cauchy-Schwarz inequality, 61Lorentzian, 62positive definite, 61

Cauchy-Schwarz equality, 59Cayley calibration, 310Cayley plane, 290, 292Cayley-Dickson process, 104Centerof an algebra, 167of Clifford algebra, 187

Centralizer of subalgebra, 170Clifford algebra, 223, 237

Check involution, 183, 231Classical companion group, 201,

250Clifford algebracanonical automorphism, 183,

247

Subject Index

center of, 187centralizer of, 223, 237check involution, 183, 252classical companion group, 201

250Clifford group, 201conformal pin group, 201degree of an element, 183even dimensional, 235even part, 183, 230fundamental lemma, 179hat involution, 183, 252inner product, 184, 189irreducible representations, 212isomorphisms with

endomorphism algebras, 212isomorphisms with matrixalgebras, 207models in low dimensions, 275,

287, 298-307odd part, 183, 230, 232split, 227twisted center of, 187volume element, 187

Clifford automorphisms, 181Clifford group, 201Clifford multiplication, 180Clifford norm, 199Coassociative4-plane, 145calibration, 115, 145

Coefficient bundle, 86Commutator (octonian), 111Complex Clifford algebra, 190,

225, 233Complex manifold, 34, 83with volume, 86

Complex normed algebras, 120Complex Riemannian manifold, 88Complex structure, 83orthogonal, 156, 271

Complex symplectic manifold, 88

Subject Index

Conformal manifold, 87Conformal pin group, 201Conjugate representation, 164Constancy of light axiom, 64Cross product, 110, 122, 124triple cross product, 112

Curvelength of, 61proper time, 63

DDeterminant, 9for spin representation, 203, 205

Dirac operator, 306Double, or Lorentz, numbers, 69,

107Dual map, 9

EElectromagnetic field, 65Equivalence of representations, 41,

164Euclidean vector space

with signature p, q, 57Cauchy-Schwarz equality, 59Euclidean positive definite, 57Lorentzian, 57

Exceptional Lie groupF4, 289G2, 113

Exponential map, 50Exterior algebra, 178Exterior multiplication, 178

FFibrations of spheres, 123First cousin principle, 155, 161Free fall, 63Future timelike vector, 62

Gy matrices, 180, 215

Grassmannian, 75, 79, 125, 178,197

Groupassociated geometry, 82classical companion, 201, 250Clifford, 201complex general linear

GL(n, C), 4complex orthogonal O(n, C), 7complex symplectic Sp(n, C), 7complex unitary U(p, q), 7, 34conformal, 13conformal orthogonal CO(p, q),

13conformal pin, 201connected components, 16, 46enhanced H-general linear

GL(n, H) H*, 5enhanced hyper-unitary (or

quaternionic unitary)HU(p, q) HU(1), 11

exceptional, F4, 289exceptional G2, 113hyper-unitary HU(p,q), 7, 35Lorentz SO1(3,1), 66low-dimensional coincidences,

14, 271orientation preserving general

linear GL+(n, R), 5orthogonal O(p, q), 6Pin, 195, 197Pin, 199Pin', 199quaternionic general linear

GL(n, H), 4quaternionic unitary (or

enhanced hyper-unitary)HU(p, q) HU(1), 11

real general linear GL(n, R), 4real symplectic Sp(n, R), 6reduced classical companion,

201, 250

319

320

reduced special orthogonalSOt(p, q), 13, 76

reduced spin, Spin°(p, q), 200skew H-unitary SK(n, H), 8, 38spacelike reduced orthogonal

O+(V), 76special C-linear SL(n, C), 5, 9special complex orthogonal

SO(n, C), 9special conformal SCO(p, q), 13special H-linear SL(n, H), 6special orthogonal SO(p, q), 9,

78special R-linear SL(n, R), 5, 9special unitary SU(p, q), 9Spin, 195, 198symplectic Sp(p, q), 8timelike reduced orthogonal

0-(V), 76

HHadamard's inequality, 133, 185Hat involution, 183, 231Hodge star operator, 66, 96, 116,

194Hopf fibration, 123, 145Hurwitz theorem, 107HyperKahler manifold, 91

IInner product space, 21basis theorem, 29canonical form, 29standard models, 22

Inner productC-hermitian skew, 21C-hermitian symmetric, 7, 21,

32C-skew, 7, 21C-symmetric, 7, 21Clifford, 184degenerate, 39

Subject Index

exterior algebra, 58, 184H-hermitian skew, 8, 21, 37H-hermitian symmetric, 21,

34on EndF(V), 171pinor, 253-257polarization of, 58R-skew, 6, 21R-symmetric, 6, 21spinor, 247tensor algebra, 58

Instantaneous observer, 65Interior multiplication, 178Intertwining operator, 41, 166,

169Irreducible representation

algebra, 165

group, 56Isometry, 23anti-isometry, 38

Isotropy subgroup, 41

JJordan algebra, 291

KKahler form, 33, 89, 127Kahler manifold, 34, 89

LLagrangian submanifold, 99, 131, 135Lie algebra, 48, 55Lie bracket, 48Light cone, 43, 64Light rays, 64Lorentz, or double, numbers, 69, 109Lorentzian manifold, 57, 87, 304Lorentzian vector space, 23, 57time orientation, 62triangle inequality, 64

Low dimensional coincidencesgeometries, 96

Subject Inder

groups, 14, 272

MMagnetic field, 65Matrix algebras M,, (R), M, (C),

M,, (H), 3Maximal negative subspace, 24, 28Maximal positive subspace, 24, 28,

38Maxwell's equations, 66Metric equivalenceflat isomorphism b, 27, 47sharp isomorphism #, 28

Minkowski space, 23Monge-Ampere operator, 136Morgan's torus lemma, 152Moser's theorem, 86

NNance calibrations, 151Negative subspace, 23NormClifford, 199of vector, 58spinorial, 200twisted Clifford, 199

Normed algebras, 101alternative, 104Artin's theorem, 109associator, 104Cayley-Dickson process, 105commutator, 111complex normed algebra, 120complex numbers C, 68, 107conjugation, 103cross products, 110, 122fibrations of spheres, 123Hurwitz theorem, 108Lorentz numbers L, 69, 107octonians 0, 107polarization, 101quaternions H, 107

real and imaginary parts, 103split octonians O, 107split quaternions M2(R), 107Null basis, 73Null cone, 64Null subspace, 23

0Observer, 65Octonians, 107associative 3-plane, 144associative calibration, 113, 144associative submanifold, 144automorphism group, 116coassociative 4-plane, 145coassociative calibration, 115cross product, 111triality, 275

Orbit, 41Orientationspacelike, 76timelike, 76total, 76

Oriented real manifold, 82Orthogonal complex structures,

156, 281Orthogonal projection, 28Orthogonal transformation, 6, 67Orthogonal vectors, 24Orthonormal basis, 30

PPauli spin matrices, 301Pin group, 195, 197, 203Pin group, 199Pin - group, 199

Pinor

canonical involution, 230inner product, 254-257reality map, 239representation, 210

Polarization

321

322

for normed algebras, 101of inner product, 58

Positive subspace, 23Projective space, 80, 123, 289Cayley plane, 290, 292quaternionic, 85

Proper time, 63

QQuaternionic Kahler form, 93Quaternionic Kahler manifold, 92Quaternionic manifold, 85cofficient bundle, 86

Quaternionic n-space H", 3complex structures on, 10

Quaternionic skew hermitianmanifold, 93

RReal manifold, 82with volume, 86

Real symplectic manifold, 34, 87Reality operator, 239, 259Reducible representationalgebra, 165group, 56

Reflection along a subspace, 67,74, 197

Representation of a group, 41Representation of an algebra, 163conjugate standard, 164equivalence, 164irreducible, 165reducible, 165

Rest space, 65Riemannian manifold, 34, 87

SSelf-dual, 97, 188, 209Sherk theorem, 67Signature, 24, 28of standard models, 29

Subject Index

positive definite, 23split, 24

Simple algebra, 164, 235Simple p-vector, 178Simple product, 178, 184Skew tensor, 178Spacelike orientation, 76Special Kiihler manifold, 90Special Lagrangiancalibration, 131differential equation, 136submanifold, 130, 136, 137,

138subspaces, SLAG, 130, 134

Special relativity, 62spinors, 304twin paradox, 63

Spherecomplex (or hyperquadric), 44H-skew, 45negative, 42positive, 42unit, 42

Spin group, 195action on spheres, 283classical companion, 201, 250coincidences with classical

groups, 272-275, 281generating set, 198reduced, 200, 203, 273spin(7), 282spin(8), 275-281spin(9), 288

Spin representations, 251Spinor inner product, 248signature of, 249, 268table of, 248

Spinor multiplication, 266of pure spinors, 267

Spinorconjugate, 213inner product, 247

Subject Index

negative, 214, 230positive, 214, 229products, 172, 266pure, 241, 243representation, 213, 217, 221space, 213structure map, 217

Spinorial norm, 77, 200Split Clifford algebra, 227Split signature, 24Squares of spinors, 172, 266, 308Standard representation, 164Symplectic manifoldcomplex, 88real, 34, 87

TTangent cone, 146Tensor algebra, 177Timecone, 62Timelike orientation, 76Torus lemma, 152Total oreintation, 76Totally null subspace, 26, 29rTrace, 51

323

Transitive action, 42, 283Triality, 275, 278, 285Twin paradox, 63Twisted center 187Twisted Clifford norm, 199Twistor operator, 306Twistor space, 244

VVectornorm of, 58

Volume form, 8, 187Volume minimizing submanifold,

126

W"Weak" almost complex manifold,

84Wirtinger's inequality, 128Witt's theorem, 246Worldline of a particle, 62Worldline of a photon, 64

YYau's theorem, 90

3904.spinors and calibrations (perspectives in mathematics) by f. reese harvey

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