curvatures of lie groups - lunds universitet · 1.2 lie groups and lie algebras deﬁnition 1.2. a...

CURVATURES OF LIE GROUPS

VALENTINE SVENSSON

Bachelor’s thesis2009:K1

Faculty of ScienceCentre for Mathematical SciencesMathematics

CE

NT

RU

MSC

IEN

TIA

RU

MM

AT

HE

MA

TIC

AR

UM

Abstract

The main purpose of this work is to write a computer program to compute thecurvatures of Riemannian Lie groups. In chapter 1 we introduce the necessarynotions and state the basis results on the curvatures of Lie groups. In chapter 2and 3 we calculate the sectional and Ricci curvatures of the 3- and 4-dimensionalLie groups with standard metrics.

Throughout this work it has been my firm intention to give reference to the stated

results and credit to the work of others. All theorems, propositions, lemmas and

examples left unmarked are assumed to be too well known for a reference to be given.

Acknowledgments

I am thankful for all the aid I have recieved from my advisor Sigmundur Gud-mundsson with writing this document and understanding the theory.

Valentine Svensson

Contents

1 Introduction 1

1.1 Riemannian manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Lie groups and Lie algebras . . . . . . . . . . . . . . . . . . . . . . . 41.3 Curvatures of Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 3-dimensional Lie Groups 19

3 4-dimensional Lie Groups 25

A The Maple program 35

Bibliography 45

Chapter 1

Introduction

1.1 Riemannian manifolds

Let M be a differentiable manifold and the tangent bundle

TM = {(p, v) | p ∈ M, v ∈ TpM}

of M be equipped with its differentiable structure induced by that of M . We definea vector field as a smooth section X : M → TM of TM and denote the set of allthese by C∞ (TM).

Definition 1.1. A real vector space V together with the bilinear map [·, ·] : V ×V →V called the Lie bracket is said to be a Lie algebra when for all X,Y, Z ∈ V

[X,X] = 0,

[X, [Y, Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0.

The latter is commonly called the Jacobi identity.

When [X,Y ] = 0 for all X,Y ∈ g the Lie algebra is said to be abelian.The set C∞ (TM) of smooth vector fields on M is a vector space and forms a Lie

algebra i.e. we have a bilinear skew-symmetric form

[·, ·] : C∞ (TM) × C∞ (TM) → C∞ (TM)

which satisfies the Jacobi identity [X, [Y, Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0 for allX,Y, Z ∈ C∞ (TM).

From now on we will assume that our differentiable manifold M is equipped witha Riemannian metric

g : C∞ (TM) ⊗ C∞ (TM) → C∞ (M) .

A connection on the tangent bundle is a map

∇ : C∞ (TM) × C∞ (TM) → C∞ (TM) ,

1

∇ : (X,Y ) 7→ ∇XY ,

such that

∇fX+gY Z = f∇XZ + g∇Y Z,

∇X(Y + Z) = ∇XY + ∇XZ,

∇XfY = f∇XY + X (f) Y,

for all X,Y, Z ∈ C∞ (TM) and f, g ∈ C∞ (M). A Riemannian manifold (M, g)has a unique connection which is torsion free and metric, that is

[X,Y ] = ∇XY −∇Y X

andg (∇XY , Z) + g (Y,∇XZ) = X (g (Y, Z)) ,

for all X,Y, Z ∈ C∞ (TM). This unique connection is called the Levi-Civita

connection and is of great importance in Riemannian geometry.The Riemann curvature tensor

R : C∞(TM) ⊗ C∞(TM) ⊗ C∞(TM) → C∞(TM)

is defined byR (X,Y ) Z = ∇X∇Y Z −∇Y ∇XZ −∇[X,Y ]Z.

There are several symmetries concerning the Riemann curvature tensor, this listwith five fundamental ones can be found in [3]:

R (X,Y ) Z = −R (Y,X) Z,

g (R (X,Y ) Z,W ) = −g (R (X,Y ) W,Z) ,

g (R (X,Y ) Z,W ) + g (R (Z,X) Y,W ) + g (R (Y, Z) X,W ) = 0,

g (R (X,Y ) Z,W ) = g (R (Z,W ) X,Y ) ,

6 · R (X,Y ) Z = R (X,Y + Z) (Y + Z) − R (X,Y − Z) (Y − Z)

+R (X + Z, Y ) (X + Z) − R (X − Z, Y − Z) (X − Z) .

We can define the curvature function

κ : C∞(TM) ⊗ C∞(TM) → C∞(M)

byκ (X,Y ) = g (R (X,Y ) Y,X)

When the curvature function is constantly zero we say that the manifold is flat.The real number

Kp (X,Y ) =κ (X,Y )

|X|2|Y |2 − g (X,Y )2

is called the sectional curvature of the 2-plane generated by X,Y in (M, g) at thepoint p.

2

The Ricci operator

r : C∞(TM) → C∞(TM)

is defined by

r (X) =n∑

k=1

R (X,Ek) Ek,

where {E1, . . . , Em} is an orthonormal basis.The Ricci curvature

Ric : C∞(TM) ⊗ C∞(TM) → C∞(M)

is defined byRic (X,Y ) = g (r (X) , Y ) .

We represent the Ricci quadratic form by a matrix Ric where

Ricij = g (r (Ei) , Ej) .

Using the symmetries for the Riemann curvature tensor we see that

g (R (X,Y ) Y, Z) = g (R (Y, Z) X,Y )

= −g (R (Y, Z) Y,X)

= g (R (Z, Y ) Y,X) .

This tells us that g (r (Ei) , Ej) = g (r (Ej) , Ei), so Ric is symmetric. Note that

Ric (X,Y ) = g (r (X) , Y ) =n∑

k=1

g (R (X,Ek) Ek, Y ) .

In the three dimensional case, with the orthonormal basis {E1, E2, E3}, we have

Ric (E1, E1) =3∑

k=1

g (R (E1, Ek) Ek, E1)

=3∑

k=1

κ (E1, Ek)

= κ (E1, E1)︸︷︷︸

0

+ κ (E1, E2) + κ (E1, E3) .

Repeating this process for all the basis elements we get this system:

1 0 11 1 00 1 1

κ (E1, E2)κ (E2, E3)κ (E1, E3)

=

Ric (E1, E1)Ric (E2, E2)Ric (E3, E3)

,

where the matrix has non-zero determinant, and we can thus solve the system forthe curvature function values. So in the 3-dimensional case the Ricci curvature givesus as much information about the manifold as the sectional cruvature.

3

1.2 Lie groups and Lie algebras

Definition 1.2. A Lie group is a smooth manifold G with a group structure · suchthat the map ρ : G × G → G with

ρ : (p, q) 7→ p · q−1

is smooth.

We say that a vector field X ∈ C∞ (TG) is left invariant if dLxX = X for allx ∈ G, where Lx is the left translation satisfying Lx (y) = xy. When all the lefttranslations Lx are isometries, we call g a left invariant metric. Geometrically aLie algebra g of a Lie group G is the set of all left invariant vector fields on the Liegroup,

g = {X ∈ C∞ (TG) | dLp (X) = X, p ∈ G} .

When X,Y ∈ g,

dLp ([X,Y ]) = [dLp (X) , dLp (Y )] = [X,Y ] ,

so [X,Y ] ∈ g. Thus g is a Lie subalgebra of C∞ (TG). The Lie algebra g ofa Lie group G is isomorphic to the tangent space TeG at the identity element ofG, thus g is finite dimensional since TeG is finite dimensional. So when we wantto perform calculations in g we can use the elements of TeG, or the other wayaround. If a Lie subalgebra h of g satisfies [h, g] ⊆ h, then we call h an ideal in g.

G G

Lp

e Lp(e) = p · e = p

TeG ∼= g TpG = TLp(e)G

dLpXe

Yy

Xp = dLp(Xe)

Yp = dLp(Ye)

Figure 1.1: Lie group and Lie algebra

When g has no non-zero proper ideals we say it is simple. We recursively defineg0 = g, g1 = [g, g] , gj+1 = [gj, gj]. The sequence

g = g0 ⊇ g1 ⊇ g2 ⊇ · · ·

is called the derived series for g. If this series eventually reaches 0 we say that g

is solvable. If g has no non-zero solvable ideals it is called semisimple. We also

4

recursively define g0 = g, g1 = [g, g] , gj+1 = [g, gj], and we call the sequence

g = g0 ⊇ g1 ⊇ g2 ⊇ · · ·

the lower central series for g. If this series eventually reaches 0 we say that g isnilpotent.

We give a few examples of Lie Groups:

Example 1.3. The real general linear group is defined by

GLn (R) ={A ∈ R

n×n | det A 6= 0}

.

The real special linear group is given by

SLn (R) ={A ∈ R

n×n | det A = 1}

.

The special orthogonal group is defined by

SO (n) ={A ∈ R

n×n | AT · A = e, det A = 1}

.

Let {E1, . . . , En} be an orthonormal basis for the vector space TeG, then bythe bilinearity of the Lie bracket it will be completely determined by the elements[Ei, Ej]. We can then look at the scalars Ck

ij such that

[Ei, Ej] =n∑

k=1

CkijEk.

These are called the structure constants of the Lie algebra. Since we have aRiemannian metric g (·, ·) we can also write these as

Ckij = g ([Ei, Ej] , Ek) .

For a linear transformation L of a Euclidean space its adjoint L⋆ is defined bythe equation g (Lx, y) = g (x, L⋆y). Whenever L⋆ = −L we call L skew-adjoint.When we are working with left invariant vector fields the function g (Y, Z) will beconstant, so X (g (Y, Z)) = 0. Remarkably, one can by using only these identitiesand combining a few permutations of variables obtain the formula

2 · g (∇XY , Z) = g ([X,Y ] , Z) + g ([Z,X] , Y ) + g ([Z, Y ] , X) (1.1)

for the Levi-Civita connection. Another identity relevant to this is

∇XY =n∑

k=1

g (∇XY ,Ek) Ek. (1.2)

This and the linearity of the metric means that we can express the connection as alinear combination of metrics of basis vectors,

∇XY =n∑

k=1

(g ([X,Y ] , Ek) + g ([Ek, X] , Y ) + g ([Ek, Y ] , X)) Ek. (1.3)

5

Example 1.4. The Lie algebra of GLn (R) is denoted gln (R) and is isomorphic to

TeGLn (R) = Rn×n

The Lie algebra of SLn (R) is denoted sln (R) and is isomorphic to

TeSLn (R) ={X ∈ R

n×n | trace (X) = 0}

.

The Lie algebra of SO (n) is denoted so (n) and is isomorphic to

TeSO (n) = TeO (n) ={X ∈ R

n×n | XT + X = 0}

.

1.3 Curvatures of Lie groups

The adjoint homomorphism ad (Z) is defined by ad (Z) (Y ) = [Z, Y ]. Let ip :G → G be defined by ip(q) = pqp−1. So ip(e) = e and we define the adjoint

representation Ad (p) of G on g by Ad (p) = d(ip)e : g → g .

Theorem 1.5 ([8]). If the linear transform ad (Z) is skew-adjoint, then

κ (Z,X) ≥ 0,

with equality if and only if Z is orthogonal to [X, g].

Proof. [8] Let us assume that we have the orthonormal basis {E1, . . . , En}. SinceCk

ij = g ([Ei, Ej] , Ek) we can insert the basis elements in (1.1) and write

2 · g(∇Ei

Ej, Ek

)= g ([Ei, Ej] , Ek) + g ([Ek, Ei] , Ej) + g ([Ek, Ej] , Ei)

= Ckij + Cj

ki + Cikj.

Applying this to (1.2) gives us

∇EiEj =

1

2

n∑

k=1

(Ck

ij + Cjki + Ci

kj

)Ek.

This we just insert into κ (E1, E2), and perform the following calculation:

κ (E1, E2) = g (R (E1, E2) E2, E1) .

∇E2E2 =

1

2

n∑

k=1

(Ck

22 + C2k2 + C2

k2

)Ek =

n∑

k=1

C2k2Ek.

∇E1E2 =

1

2

n∑

k=1

(Ck

12 + C2k1 + C1

k2

)Ek.

∇[E1,E2]E2 = ∇Pnk=1

Ck12

EkE2 =

n∑

k=1

Ck12∇Ek

E2.

6

∇EkE2 =

1

2

n∑

l=1

(C l

k2 + C2lk + Ck

l2

)El.

∇E1Ek =

1

2

n∑

l=1

(C l

1k + Ckl1 + C1

lk

)El.

∇E2Ek =

1

2

n∑

l=1

(C l

2k + Ckl2 + C2

lk

)El.

g(∇E1

∇E2E2, E1

)=

1

2

n∑

k=1

(

C2k2

n∑

l=1

(C l

1k + Ckl1 + C1

lk

)· g(El, E1)

)

=1

2

n∑

k=1

C2k2

(C1

1k + Ck11 + C1

1k

)

=n∑

k=1

C2k2C

11k.

g(∇E2

∇E1E2, E1

)=

1

4

n∑

k=1

(

(Ck

12 + C2k1 + C1

k2

)n∑

l=1

(C l

2k + Ckl2 + C2

lk

)g (El, E1)

)

=1

4

n∑

k=1

(Ck

12 + C2k1 + C1

k2

) (C1

2k + Ck12 + C2

1k

).

g(∇[E1,E2]E2, E1

)=

1

2

n∑

k=1

(

Ck12

n∑

l=1

(C l

k2 + C2lk + Ck

l2

)g (El, E1)

)

=1

2

n∑

k=1

Ck12

(C1

k2 + C21k + Ck

12

).

κ (E1, E2) =n∑

k=1

(

C2k2C

11k −

1

4

(Ck

12 + C2k1 + C1

k2

) (C1

2k + Ck12 + C2

1k

)

− 1

2Ck

12

(C1

k2 + C21k + Ck

12

))

=n∑

k=1

(

−C2k2C

1k1 −

1

4

(Ck

12 + C2k1 − C1

2k

) (C1

2k + Ck12 + C2

1k

)

+1

2Ck

12

(−C1

k2 − C21k − Ck

12

))

7

=n∑

k=1

(1

2Ck

12

(−Ck

12 + C12k + C2

k1

)

− 1

4

(Ck

12 − C12k + C2

k1

) (Ck

12 + C12k − C2

k1

)− C1

k1C2k2

)

.

We end up with the formula

κ (E1, E2) =n∑

k=1

(1

2Ck

12

(−Ck

12 + C12k + C2

k1

)

−1

4

(Ck

12 − C12k + C2

k1

) (Ck

12 + C12k − C2

k1

)− C1

k1C2k2

)

.

Now we assume Z and X are orthonormal, and that our basis is such that E1 =Z, E2 = X. Saying ad (E1) is skew-adjoint means Ck

1j = −Cj1k in the language of

structure constants. And applying this rule to the above formula we can do thesecalculations:

κ (E1, E2) =n∑

k=1

(1

2Ck

12

(−Ck

12 + C12k + C2

k1

)

−1

4

(Ck

12 − C12k + C2

k1

) (Ck

12 + C12k − C2

k1

)− C1

k1C2k2

)

=n∑

k=1

(1

2Ck

12

(−Ck

12 + C12k − C2

1k

)

−1

4

(Ck

12 − C12k − C2

1k

) (Ck

12 + C12k + C2

1k

)+ C1

1kC2k2

)

=n∑

k=1

(1

2Ck

12

(−Ck

12 + C12k + Ck

12

)

−1

4

(Ck

12 − C12k + Ck

12

) (Ck

12 + C12k − Ck

12

)− Ck

11C2k2

)

=n∑

k=1

(1

2Ck

12C12k −

1

4

(2Ck

12 − C12k

)C1

2k

)

=n∑

k=1

(1

2Ck

12C12k −

1

2Ck

12C12k +

1

4C1

2kC12k

)

=1

4

n∑

k=1

(C1

2k

)2.

And we obviously get the inequality. The sum equals 0 only when every C12k = 0.

As C12k = g ([E2, Ek] , E1) = g ([X,Ek] , Z), this equaling 0 means [X,Ek] and Z are

orthogonal. If this is true for every Ek then Z is orthogonal to [X, g].

8

Corollary 1.6 ([8]). If the linear transformation ad(Z) is skew-adjoint, then

Ric(Z,Z) ≥ 0,

with equality if and only if Z is orthogonal to [g, g].

Theorem 1.7 ([8]). If Z is orthogonal to [g, g], then

Ric(Z,Z) ≤ 0,

with equality if and only if ad(Z) is skew-adjoint.

Example 1.8. We are going to calculate the Riemann curvature and the Riccicurvature for a particular Lie group. We will define the metric by a matrix g suchthat for two basis elements Ei, Ej in the Lie algebra of the Lie group we haveg (Ei, Ej) = gij. Here we are going to use

g =

1 0 00 1 00 0 1

.

For the groups algebra we will use the basis {X,Y, Z} and the defining equations

[X,Y ] = c · Z, [Z,X] = b · Y, [Y, Z] = a · X.

We translate this to structure constants and use the technique in the proof above.We apply the identity Ck

ij = g ([Ei, Ej] , Ek) and get

CZXY = c

CYZX = b

CXY Z = a

This notation is not standard, but is an efficient shorthand for this calculation.

∇XY =1

2

∑

k∈{X,Y,Z}

(Ck

XY + CYkX + CX

kY

)k

=1

2(0 · X + 0 · Y + (c + b − a) · Z) =

c + b − a

2· Z.

When we do this for all the other combinations of basis vectors we also get

∇XZ =a − b − c

2· Y

∇Y X =b − c − a

2· Z

∇Y Z =a + c − b

2· X

9

∇ZX =b + a − c

2· Y

∇ZY =c − a − b

2· X.

With this information we can calculate the Riemann curvature tensors:

R (X,Y ) Y = ∇X∇Y Y −∇Y ∇XY −∇[X,Y ]Y

= ∇X0 − c + b − a

2∇Y Z − c∇ZY

= −c + b − a

2· a + c − b

2· X − c

c − a − b

2· X

=

(

−c + b − a

2· a + c − b

2− c · c − a − b

2

)

· X,

R (X,Z) Z =

(

−a − b − c

2· c − a − b

2+ b · a + c − b

2

)

· X,

R (Y,X) X =

(

−b − c − a

2· a − b − c

2+ c · b + a − c

2

)

· Y,

R (Y, Z) Z =

(

−a + c − b

2· b + a − c

2− a · a − b − c

2

)

· Y,

R (Z,X) X =

(

−b + a − c

2· c + b − a

2− b · b − c − a

2

)

· Z,

R (Z, Y ) Y =

(

−c − a − b

2· b − c − a

2+ a · c + b − a

2

)

· Z.

We will save some space by making a matrix K where Kij = κ (Ei, Ej). As themetric is orthonormal, we get the Riemannian curvature directly from these. Weknow that K is symmetric and that its diagonal consists of zeros, so we choosethe three ones in the upper triangle of the K-matrix and simplify by multiplyingtogether everything and completing the squares:

κ (X,Y ) = −c + b − a

2· a + c − b

2− c · c − a − b

2

=a2 + ac − ab − ac − c2 + bc − ab − bc + b2 − 2c2 + 2ac + 2bc

4

=a2 − 2ab − 3c2 + b2 + 2ac + 2bc

4

10

κ (X,Z) =−3b2 + c2 − 2ac + a2 + 2ab + 2bc

4

κ (Y, Z) =b2 − 2bc − 3a2 + c2 + 2ab + 2ac

4.

And thus

K =

0 a2−2ab−3c2+b2+2ac+2bc4

−3b2+c2−2ac+a2+2ab+2bc4

a2−2ab−3c2+b2+2ac+2bc4

0 b2−2bc−3a2+c2+2ab+2ac4

−3b2+c2−2ac+a2+2ab+2bc4

b2−2bc−3a2+c2+2ab+2ac4

0

.

To calculate the Ricci curvatures, and the Ricci quadratic form, we can use therelation between curvatures at the end of the first section.

Ric (X,X) = κ (X,Y ) + κ (X,Z)

=a2 − 2ab − 3c2 + b2 + 2ac + 2bc

4

+−3b2 + c2 − 2ac + a2 + 2ab + 2bc

4

=2a2 − 2c2 − 2b2 + 4bc

4

=a2 − (b2 − 2bc + c2)

2

=a2 − (b − c)2

2

Ric (Y, Y ) =b2 − (a − c)2

2

Ric (Z,Z) =c2 − (a − b)2

2.

So we get

Ric =

a2−(b−c)2

20 0

0 b2−(a−c)2

20

0 0 c2−(a−b)2

2

.

Lemma 1.9 ([8]). Let g be a left invariant metric on a connected Lie group G.

This metric will also be right invariant if and only if ad (X) is skew-adjoint for

every X ∈ g.

11

Proof. [8] Let us mention that a left invariant metric also is right invariant if andonly if for each p ∈ G, Ad (p) is an isometry with respect to the induced metric ong:

g (Ad(p) (X) , Ad(p) (Y )) = g (X,Y ) .

First assume that g is both left and right invariant so that Ad(p) is an isometry.For a p ∈ G close to e we have p = exp (X) for some unique X close to 0 in g. Wewill use this identity (found for example in [10], p54):

Ad (exp (X)) = ead(X).

We have

g (Ad(p) (X) , Ad(p) (Y )) = g (X,Ad(p)⋆ (Ad(p) (Y ))) = g (X,Y ) ,

and thus Ad(p)⋆ = Ad(p)−1.We get

Ad (p)−1 = Ad (p)⋆

Ad (exp (X))−1 = Ad (exp (X))⋆

e−ad(X) = ead(X)⋆

−ad (X) = ad (X)⋆ ,

and so ad (X) is skew-adjoint.Next we assume ad(X) is skew-adjoint for all X. Now Ad(exp(X)) = ead(X) acts

as isometries on g, so for any p ∈ exp(g), Ad(p) is an isometry on g. There is an openneighbourhood U in exp(g) containing the identity element e. Since G is connected,any element in G can be written as a product of elements in U . Any product ofisometries is an isometry, so we see that Ad(p) is an isometry for any p in G. So gis both left and right invariant.

Lemma 1.10 ([8]). A connected Lie group G admits a bi-invariant metric if and

only if G is isomorphic to the cartesian product of a compact group and an abelian

group.

Theorem 1.11 ([8]). Every compact connected Lie group admits a left invariant

metric such that Kp ≥ 0 for all p.

Milnor gives in [8] a couple of interesting theorems which we will have a lookat. We will use the classical Bonnet-Myers theorem in one of the proofs, so let usmention it briefly.

Theorem 1.12 (Bonnet-Myers). Let M be a complete Riemannian manifold. If

there exist an r > 0 so that Ric(Z,Z) ≥ r · g(Z,Z) at all Z, then M is compact.

Theorem 1.13 (Iwasawa, [8]). If G is a connected Lie group and H a maximal

compact subgroup of G, then G is homeomorphic to H × Rm. Furthermore, any

compact subgroup of G is contained in a maximal compact subgroup, which is a

connected Lie group.

12

These two following theorems inform us that there are non-abelian Lie groupswhich are flat.

Theorem 1.14 ([8]). If a Lie group with left invariant metric is flat, then its Lie

algebra splits into an orthogonal sum h ⊕ i. With h an abelian Lie subalgebra of g

and i an abelian ideal, and ad (Z) is skew-adjoint for every Z ∈ h.

Proof. [8] Let G be a simply connected Lie group with a flat left-invariant metric.A flat, simply connected, complete Riemannian manifold is homeomorphic to Eu-clidean space, R

n, where n is the dimension of G. Let K be any compact subgroupof G. According to Iwasawa above K ⊂ H where H is a maximal compact subgroupof G. So

Rn ∼= G ∼= H × R

m,

where n − m is the dimension of H. If H is non-trivial we can take a volume formω for H. Let ι : H → H × R

m be the inclusion and π : H × Rm → R

m be theprojection. Then π∗ω is a closed form on H × R

m ∼= Rn, and hence it is also exact:

π∗ω = dα. Thenω = ι∗π∗ω = ι∗dα = dι∗α,

meaning ω is exact on H. But that is impossible since H is compact. Thus we havea contradiction, and H must be trivial. Since K ⊂ H, K is also trivial. And so anycompact subgroup of G must be trivial.

Let g have a metric, and let us look at X 7→ ∇X which defines a mappingg → o (n).

We are assuming the curvature tensor is identical to zero, so that

∇[X,Y ] = ∇X∇Y −∇Y ∇X .

We have ∇[X,Y ] = ∇X∇Y − ∇Y ∇X = [∇X ,∇Y ] so X 7→ ∇X is a Lie algebrahomomorphism. The kernel of a homomorphism is an ideal. In this case we call thekernel i. Since

[X,Y ] = ∇XY −∇Y X

we must for any X,Y ∈ i have [X,Y ] = ∇X︸︷︷︸

0

Y − ∇Y︸︷︷︸

0

X = 0 and we understand

that i is abelian.Let us define h as the orthogonal complement of i. For a Z ∈ h and X ∈ i we

have[Z,X] = ∇ZX − ∇X

︸︷︷︸

0

Z = ∇ZX.

Due to the metric property of the Levi-Civita connection and the metric being leftinvariant we have

Z(g(X,Y )) = g(∇ZX,Y ) + g(X,∇ZY ) = 0,

or g(∇ZX,Y ) = −g(X,∇ZY ) and ∇Z is skew-adjoint. ∇ZX = [Z,X] ∈ i, so ∇Z

maps i into itself. When W ∈ h, we get g(∇ZW,X) = −g(W,∇ZX︸︷︷︸

∈i

) = 0, meaning

13

∇ZW is orthogonal to X and ∇X must therefore also map h into itself. And thush must be a Lie subalgebra of g. h maps isomorphically to a Lie subalgebra of o (n).o (n) is the Lie algebra of the Lie group O (n), which is compact, so it possesses somebi-invariant metric. And thus so do h. This means that ad (Z) is skew-adjoint forany Z ∈ h. Let us look at h with this bi-invariant metric, which we call h: Say wehave the ideal h1 ⊆ h, and some W ∈ h which is orthogonal to h1. Since ad (Z) isskew-adjoint for all Z ∈ h1 we have

h([X,W ] , Z) = −h(W, [X,Z]) = 0.

Inductively, when a Lie algebra is equipped with a bi-invariant metric, the orthogonalcomplement of an ideal in this Lie algebra is also an ideal, and we can express theLie algebra as an orthogonal direct sum of simple ideals. We write h as h1⊕· · ·⊕hk,where the hi’s are simple ideals. If a hi would be non-abelian, then the center of hi

would be trivial. We have the identity

κ(X,Y ) =1

4· g([X,Y ] , [X,Y ])

and thus

Ric(Z,Z) =1

4·

n∑

k=1

g([Z,Ek] , [Z,Ek]),

where Z ∈ hi, the Ek’s are basis elements for hi and n is the dimension of hi. Sincead(Ek) is skew-adjoint we also get

Ric(Z,Z) = −1

4·

n∑

k=1

g(Z, [Ek, [Ek, Z]]).

Let C be the Casimir-operator

C : Z 7→n∑

k=1

[Ek, [Ek, Z]] .

Say λ is an eigenvalue to C. Then ker(C − λI) is a non-trivial ideal to hi, but sincehi is simple

C − λI = 0

C = λI,

so we have

Ric(Z,Z) = −λ

4· g(Z,Z),

where λ < 0 According to Bonnet-Myers’ theorem the Lie group Bi correspondingto hi would be compact. The inclusion hi ⊂ h ⊂ g would then induce a non-trivialhomomorphism Bi → G. So G would contain a non-trivial compact subgroup. Thatis impossible. Therefore every hi must be abelian, and thus h is abelian.

So g = i⊕ h, where i is a abelian ideal, h is an abelian Lie subalgebra and ad (Z)is skew-adjoint.

14

Theorem 1.15 ([8]). Let G be a Lie group with Lie algebra g which splits into an

orthogonal sum h ⊕ i, with h an abelian Lie subalgebra of g and i an abelian ideal,

and ad (Z) is skew-adjoint for every Z ∈ h. Then the Lie group is flat.

Proof. [8] We look at equation (1.1) to make some conclusions about the Levi-Civitaconnection. Let A ∈ g, X,Y ∈ i and Z,W ∈ h. We will look at five cases, beingthe different possibilities of combinations of these in the formula for the Levi-Civitaconnection. First:

2 · g(∇ZW,A) = g([Z,W ]︸︷︷︸

=0

, A) + g([A,Z] ,W ) + g([A,W ] , Z)

= g([A,Z] ,W ) − g([A,Z] ,W )

= 0.

Second:

2 · g(∇ZX,W ) = g([Z,X]︸︷︷︸

∈i

,W ) + g([W,Z]︸︷︷︸

=0

, X) + g([W,X]︸︷︷︸

∈i

, Z) = 0.

Third:

2 · g(∇ZX,Y ) = g([Z,X] , Y ) + g([Y, Z] , X) + g([Y,X]︸︷︷︸

=0

, Z)

= g([Z,X] , Y ) − g([Z, Y ] , X)

= g([Z,X] , Y ) + g([Z,X] , Y )

= 2 · g([Z,X] , Y ).

So ∇Z = ad(Z). Fourth:

2 · g(∇XY ,A) = g([X,Y ]︸︷︷︸

=0

, A) + g([A,X]︸︷︷︸

=0

, Y ) + g([A, Y ]︸︷︷︸

=0

, X) = 0.

Fifth:

2 · g(∇XZ,A) = g([X,Z]︸︷︷︸

=0

, A) + g([A,X]︸︷︷︸

=0

, Z) + g([A,Z] , X)

= −g([Z,A] , X) = g([Z,X]︸︷︷︸

=0

, A) = 0.

So ∇X = 0. We apply these identities for three different cases of the Riemanncurvature tensor. First:

R(X,Y ) = ∇X︸︷︷︸

=0

∇Y − ∇Y︸︷︷︸

=0

∇X −∇[X,Y ]︸︷︷︸

=0

= 0.

Second:

R(X,Z) = ∇X︸︷︷︸

=0

∇Z −∇Z ∇X︸︷︷︸

=0

−∇[X,Z]︸︷︷︸

=0

= 0.

15

Third:

R(Z,W )A = (∇Z∇W −∇W∇Z −∇[Z,W ])A

= ∇Zad(W )(A) −∇W ad(Z)(A) − ad([Z,W ])(A)

= ad(Z)ad(W )(A) − ad(W )ad(Z)(A) − ad([Z,W ])(A)

= ad(Z)([W,A]) − ad(W )([Z,A]) − [[Z,W ] , A]

= [Z, [W,A]] − [W, [Z,A]] + [A, [Z,W ]]

= [Z, [W,A]] + [W, [A,Z]] + [A, [Z,W ]] = 0,

since this is the Jacobi identity. And so, the Riemann curvature tensor will alwaysbe zero.

We can look at a good example.

Example 1.16. The Euclidean motion group on two dimensional space, E(2), whoseLie algebra is defined by the structure equations

[E1, E3] = −E2, [E2, E3] = E1

in the basis {E1, E2, E3} is not abelian. But when equipped with a metric definedby a matrix of the form

s 0 00 s 00 0 t

it will be flat.

Heintze has shown in [11] that for a Lie group with left invariant metric to havestrictly negative sectional curvature there was the condition for g = g1 + RX forsome X where the eigenvalues of the restriction ad (X) |g1 : g1 → g1 have positivereal part.

Theorem 1.17 ([12]). A connected Lie group with left invariant metric with all

sectional curvatures Kp ≤ 0 is solvable.

The proof of this theorem requires many advanced topics not discussed in thispaper. It is proved in [12] by Azencott and Wilson, where it is called Corollary 7.8.

Theorem 1.18 ([14]). Let N be a transitive connected non-abelian nilpotent Lie

group. Then for p ∈ N there exist two-dimensional subspaces R, S, T ⊆ TpN such

that the sectional curvatures satisfy

Kp (S) < Kp (R) = 0 < Kp (T ) .

This theorem is difficult to prove, but we can prove this slightly weaker versionpresented by Milnor:

Theorem 1.19 ([8]). Let N be a non-abelian nilpotent Lie group. Then there exists

a direction of strictly negative and a direction of strictly positive Ricci curvature.

16

Proof. [8] let k be the smallest number so that gk+1 = 0. Choose a unit vectorX ∈ gk, which then must be central, and contained in g1. So by Corollary 1.6 weget that Ric(X,X) > 0. Say z is the center of g. If g = g1 + z then

g1 = [g, g] = [g, [g, g] + z] = [g, [g, g]] = [g, g1] = g2,

and we could not have nilpotency. So there must be a unit vector Z 6∈ z orthogonalto g1. ad(Z) is non-zero and nilpotent, so it can not be skew-adjoint. Hence we getthat Ric(Z,Z) < 0 by Theorem 1.7.

Milnor goes on to give a generalization

Theorem 1.20 ([8]). Let G be a Lie group. If there are three linearly independant

vectors X,Y, Z ∈ g so that [X,Y ] = Z, then there exists a left invariant metric so

that Ric(X,X) < 0 and Ric(Z,Z) > 0.

17

Chapter 2

3-dimensional Lie Groups

Now we will look at the Lie groups of dimension 3 equipped with the metric g (·, ·)where g (Ei, Ej) = gij and

g =

r2 0 00 s2 00 0 t2

.

For these we are using the basis {E1, E2, E3}. A classification of the Lie algebrasof dimension three and four (which are not products of lower dimensional algebras)is found in [9], where Patera et. al. lists the nine classes of three dimensional andtwelve classes of four dimensional Lie algebras. Here is the list of 3-dimensionalalgebras along with their defining Lie bracket equations.

Name Structure equations

A3,1 [E2, E3] = E1

A3,2 [E1, E3] = E1, [E2, E3] = E1 + E2

A3,3 [E1, E3] = E1, [E2, E3] = E2

A3,4 [E1, E3] = E1, [E2, E3] = −E2

Aa3,5 [E1, E3] = E1, [E2, E3] = aE2, (0 < |a| < 1)

A3,6 [E1, E3] = −E2, [E2, E3] = E1

Aa3,7 [E1, E3] = aE1 − E2, [E2, E3] = E1 + aE2, (a > 0)

A3,8 [E1, E3] = −2E2, [E1, E2] = E1, [E2, E3] = E3

A3,9 [E1, E2] = E3, [E2, E3] = E1, [E3, E1] = E2

We are going to use the Maple program to calculate the values of the curvaturefunctions and the Ricci curvatures of the groups corresponding to these algebras.The last algebra, A3,9, corresponds to that of SU (2) whose curvatures we calculatedin example 1.8 by hand. This will act as confirmation that the program works.

Example 2.1 (R3). But let us start by commenting on R3. The Lie algebra of R

3

is defined by all brackets equaling zero. And so it is simply flat in all curvatures forany left invariant metric.

The 3-dimensional Lie groups are listed in the same order as in [9] and the tableabove.

19

Example 2.2 (A3,1 = Nil). This is the Lie group consisting of matrices of the form

1 y x0 1 z0 0 1

,

so we can define the Lie algebra of the group using the basis

E1 =

0 0 10 0 00 0 0

, E1 =

0 1 00 0 00 0 0

, E1 =

0 0 00 0 10 0 0

.

The non-zero Lie brackets for this Lie algebra is

[E1, E2] = E3.

For this algebra we get

K =

0 −3/4 t2

r2s2 1/4 t2

r2s2

−3/4 t2

r2s2 0 1/4 t2

r2s2

1/4 t2

r2s2 1/4 t2

r2s2 0

and

Ric =

−1/2 t2

r2s2 0 0

0 −1/2 t2

r2s2 0

0 0 1/2 t2

r2s2

.

Eigenvalues to Ric:{

1/2t2

r2s2,−1/2

t2

r2s2,−1/2

t2

r2s2

}

Eigenvectors to Ric:{(0, 0, 1) , (0, 1, 0) , (1, 0, 0)}

Example 2.3 (A3,2). Structure equations:

[E1, E3] = E1, [E2, E3] = E1 + E2

K =

0 1/4 r2−4 s2

s2t21/4 r2−4 s2

s2t2

1/4 r2−4 s2

s2t20 −1/4 4 s2+3 r2

s2t2

1/4 r2−4 s2

s2t2−1/4 4 s2+3 r2

s2t20

Ric =

1/2 r2−4 s2

s2t2− r

st20

− rst2

−1/2 4 s2+r2

s2t20

0 0 −1/2 4 s2+r2

s2t2

.

20


1/2−4 s2 +

√r4 + 4 r2s2

s2t2, 1/2

−4 s2 −√

r4 + 4 r2s2

s2t2,−1/2

4 s2 + r2

s2t2

}

Eigenvectors to Ric:{(

−2sr√

r4 + 4 r2s2 − r2, 1, 0

)

,

(

−2sr

−√

r4 + 4 r2s2 − r2, 1, 0

)

, (0, 0, 1)

}

Example 2.4 (A3,3 = H3). The Lie algebra of H

3 is defined by the Lie brackets

[E1, E2] = E2, [E1, E3] = E3.

For this Lie group we have

K =

0 −r−2 −r−2

−r−2 0 −r−2

−r−2 −r−2 0

and

Ric =

−2 r−2 0 0

0 −2 r−2 0

0 0 −2 r−2

.

Eigenvalues to Ric:{−2 r−2,−2 r−2,−2 r−2

}



[E1, E3] = E1, [E2, E3] = −E2

K =

0 t−2 −t−2

t−2 0 −t−2

−t−2 −t−2 0

Ric =

0 0 0

0 0 0

0 0 −2 t−2

.

Eigenvalues to Ric:{0, 0,−2 t−2

}


21

Example 2.6 (Aα3,5 = Solα). This Lie group consists of matrices of the form

eαx 0 y0 e−x z0 0 1

,

so the Lie algebra of this group is defined by

E1 =

α 0 00 −1 00 0 0

, E1 =

0 0 10 0 00 0 0

, E1 =

0 0 00 0 10 0 0

.

Here the non-zero Lie brackets are

[E1, E2] = αE2, [E1, E3] = −E3,

and we have

K =

0 −α2

r2 −r−2

−α2

r2 0 αr2

−r−2 αr2 0

and

Ric =

−1+α2

r2 0 0

0 −α (−1+α)r2 0

0 0 −1+αr2

.


−α (−1 + α)

r2,−1 + α

r2,−α2 + 1

r2

}



[E1, E3] = −E2, [E2, E3] = E1

K =

0 1/4(−s2+r2)

2

s2t2r2 1/4 −3 s4+2 r2s2+r4

s2t2r2

1/4(−s2+r2)

2

s2t2r2 0 −1/4 −2 r2s2+3 r4−s4

s2t2r2

1/4 −3 s4+2 r2s2+r4

s2t2r2 −1/4 −2 r2s2+3 r4−s4

s2t2r2 0

Ric =

1/2 −s4+r4

s2t2r2 0 0

0 −1/2 −s4+r4

s2t2r2 0

0 0 −1/2 −2 r2s2+r4+s4

s2t2r2

.

22

Eigenvalues to Ric:

{

1/2−s4 + r4

s2t2r2,−1/2

−2 r2s2 + r4 + s4

s2t2r2,−1/2

−s4 + r4

s2t2r2

}

Eigenvectors to Ric:

{(1, 0, 0) , (0, 0, 1) , (0, 1, 0)}

Example 2.8 (Aα3,7). Structure equations:

[E1, E3] = αE1 − E2, [E2, E3] = E1 + αE2, (α > 0) .

K =

0 1/4 −2 r2s2+r4+s4−4 α2s2r2

s2t2r2 1/4 2 r2s2−4 α2s2r2−3 s4+r4

s2t2r2

1/4 −2 r2s2+r4+s4−4 α2s2r2

s2t2r2 0 −1/4 4 α2s2r2−2 r2s2+3 r4−s4

s2t2r2

1/4 2 r2s2−4 α2s2r2−3 s4+r4

s2t2r2 −1/4 4 α2s2r2−2 r2s2+3 r4−s4

s2t2r2 0

Ric =

1/2 −4 α2s2r2−s4+r4

s2t2r2 −(−s2+r2)α

rt2s0

−(−s2+r2)α

rt2s−1/2 4 α2s2r2+r4−s4

s2t2r2 0

0 0 −1/2 4 α2s2r2−2 r2s2+r4+s4

s2t2r2

.

Eigenvalues to Ric:

{

1/2−4 α2s2r2 +

√r8 + 4 r6s2α2 + 4 s6α2r2 − 8 α2s4r4 − 2 r4s4 + s8

s2t2r2,

1/2−4 α2s2r2 −

√r8 + 4 r6s2α2 + 4 s6α2r2 − 8 α2s4r4 − 2 r4s4 + s8

s2t2r2,

−1/24 α2s2r2 − 2 r2s2 + r4 + s4

s2t2r2

}


{(

−2(−s2 + r2) rsα√

r8 + 4 r6s2α2 + 4 s6α2r2 − 8 α2s4r4 − 2 r4s4 + s8 + s4 − r4, 1, 0

)

,

(

−2(−s2 + r2) rsα

−√

r8 + 4 r6s2α2 + 4 s6α2r2 − 8 α2s4r4 − 2 r4s4 + s8 + s4 − r4, 1, 0

)

,

(0, 0, 1)}


[E1, E3] = −2E2, [E1, E2] = E1, [E2, E3] = E3

23

K =

0 − r2t2−s4

s2r2t2−3 s4+r2t2

s2r2t2

− r2t2−s4

s2r2t20 − r2t2−s4

s2r2t2

−3 s4+r2t2

s2r2t2− r2t2−s4

s2r2t20

Ric =

−2 s2

r2t20 −2 1

rt

0 −2 r2t2−s4

r2s2t20

−2 1rt

0 −2 s2

r2t2


−2r2t2 − s4

s2r2t2, 2

−s2 + rt

r2t2,−2

s2 + rt

r2t2

}

Eigenvectors to Ric:{(0, 1, 0) , (−1, 0, 1) , (1, 0, 1)}

Example 2.10 (A3,9 = SU (2)). The Lie algebra of SU (2) we define by the Liebrackets

[E1, E2] = 2E3, [E3, E1] = 2E2, [E2, E3] = 2E1.

Here we have

K =

0 2 r2t2+2 s2t2−3 t4−2 r2s2+s4+r4

s2t2r2

2 r2s2+2 s2t2−3 s4−2 r2t2+t4+r4

s2t2r2

2 r2t2+2 s2t2−3 t4−2 r2s2+s4+r4

s2t2r2 0 −−2 r2s2−2 r2t2+3 r4+2 s2t2−t4−s4

s2t2r2

2 r2s2+2 s2t2−3 s4−2 r2t2+t4+r4

s2t2r2 −−2 r2s2−2 r2t2+3 r4+2 s2t2−t4−s4

s2t2r2 0

and

Ric =

2 2 s2t2−s4−t4+r4

r2s2t20 0

0 −2 −2 r2t2+r4+t4−s4

r2s2t20

0 0 −2 −2 r2s2+r4−t4+s4

r2s2t2

.


−2−2 r2t2 + r4 + t4 − s4

s2t2r2, 2

2 s2t2 − s4 − t4 + r4

s2t2r2,−2

−2 r2s2 + r4 − t4 + s4

s2t2r2

}


If one takes example 1.8 and set 2rst

= a, 2srt

= b and 2trs

= c we will get thesematrices. So we know that at least in this case the program calculates correctly.

24

Chapter 3

4-dimensional Lie Groups

Here we will look at the curvatures of the four dimensional Lie groups with themetric defined by

g =

q2 0 0 00 r2 0 00 0 s2 00 0 0 t2

.

As in previous chapter, the list is taken from [9].

Name Structure equations

A4,1 [E2, E4] = E1, [E3, E4] = E2

Aa4,2 [E1, E4] = aE1, [E2, E4] = E2, [E3, E4] = E2 + E3, (a 6= 0)

A4,3 [E1, E4] = E1, [E3, E4] = E2

A4,4 [E1, E4] = E1, [E2, E4] = E1 + E2, [E3, E4] = E2 + E3

Aab4,5 [E1, E4] = E1, [E2, E4] = aE2, [E3, E4] = bE3, (ab 6= 0, −1 ≤ a ≤ b ≤ 1)

Aab4,6 [E1, E4] = aE1, [E2, E4] = bE2 − E3, [E3, E4] = E2 + bE3, (a 6= 0, b ≥ 0)

A4,7 [E2, E3] = E1, [E1, E4] = 2E1, [E2, E4] = E2, [E3, E4] = E2 + E3

A4,8 [E2, E3] = E1, [E2, E4] = E2, [E3, E4] = −E3

Ab4,9 [E2, E3] = E1, [E1, E4] = (1 + b) E1, [E2, E4] = E2, [E3, E4] = bE3, (−1 < b ≤ 1)

A4,10 [E2, E3] = E1, [E2, E4] = −E3, [E3, E4] = E2

Aa4,11 [E2, E3] = E1, [E1, E4] = −E3, [E2, E4] = aE2 − E3, [E3, E4] = E2 + aE3, (a > 0)

A4,12 [E1, E3] = E1, [E2, E3] = E2, [E1, E4] = −E2, [E2, E4] = E1


[E2, E4] = E1, [E3, E4] = E2

25

K =

0 1/4 q2

r2t20 1/4 q2

r2t2

1/4 q2

r2t20 1/4 r2

s2t2−1/4 3 q2s2−r4

t2r2s2

0 1/4 r2

s2t20 −3/4 r2

s2t2

1/4 q2

r2t2−1/4 3 q2s2−r4

t2r2s2 −3/4 r2

s2t20

Ric =

1/2 q2

r2t20 0 0

0 −1/2 −r4+q2s2

t2r2s2 0 0

0 0 −1/2 r2

s2t20

0 0 0 −1/2 r4+q2s2

t2r2s2

.


−1/2r2

s2t2,−1/2

r4 + q2s2

t2r2s2,−1/2

−r4 + q2s2

t2r2s2, 1/2

q2

r2t2

}


{(0, 0, 1, 0) , (0, 0, 0, 1) , (0, 1, 0, 0) , (1, 0, 0, 0)}

Example 3.2 (Aa4,2). Structure equations:

[E1, E4] = aE1, [E2, E4] = E2, [E3, E4] = E2 + E3

K =

0 − at2

− at2

−a2

t2

− at2

0 1/4 r2−4 s2

s2t21/4 r2−4 s2

s2t2

− at2

1/4 r2−4 s2

s2t20 −1/4 4 s2+3 r2

s2t2

−a2

t21/4 r2−4 s2

s2t2−1/4 4 s2+3 r2

s2t20

Ric =

−a(a+2)t2

0 0 0

0 −1/2 −r2+4 s2+2 as2

s2t2−1/2 r(a+2)

st20

0 −1/2 r(a+2)st2

−1/2 4 s2+r2+2 as2

s2t20

0 0 0 −1/2 4 s2+r2+2 a2s2

s2t2

.


−a (a + 2)

t2,−2 s2 − as2 + 1/2

√r2a2s2 + 4 r2s2a + 4 r2s2 + r4

s2t2,

−2 s2 − as2 − 1/2√

r2a2s2 + 4 r2s2a + 4 r2s2 + r4

s2t2, −1/2

4 s2 + r2 + 2 a2s2

s2t2

}

26

Eigenvectors to Ric:{

(1, 0, 0, 0) ,

(

0,− r (a + 2) s√r2a2s2 + 4 r2s2a + 4 r2s2 + r4 − r2

, 1, 0

)

,

(

0,− r (a + 2) s

−√

r2a2s2 + 4 r2s2a + 4 r2s2 + r4 − r2, 1, 0

)

, (0, 0, 0, 1)

}


[E1, E4] = E1, [E3, E4] = E2

K =

0 0 0 −t−2

0 0 1/4 r2

s2t21/4 r2

s2t2

0 1/4 r2

s2t20 −3/4 r2

s2t2

−t−2 1/4 r2

s2t2−3/4 r2

s2t20

Ric =

−t−2 0 0 0

0 1/2 r2

s2t2−1/2 r

st20

0 −1/2 rst2

−1/2 r2

s2t20

0 0 0 −1/2 r2+2 s2

s2t2

.


−t−2, 1/2

√s2 + r2r

s2t2,−1/2

√s2 + r2r

s2t2,−1/2

r2 + 2 s2

s2t2

}


(1, 0, 0, 0) ,

(

0,− sr√s2 + r2r − r2

, 1, 0

)

,

(

0,− sr

−√

s2 + r2r − r2, 1, 0

)

, (0, 0, 0, 1)

}


[E1, E4] = E1, [E2, E4] = E1 + E2, [E3, E4] = E2 + E3

K =

0 1/4 q2−4 r2

r2t2−t−2 1/4 q2−4 r2

r2t2

1/4 q2−4 r2

r2t20 1/4 r2−4 s2

s2t2−1/4 4 r2s2+3 q2s2−r4

t2r2s2

−t−2 1/4 r2−4 s2

s2t20 −1/4 4 s2+3 r2

s2t2

1/4 q2−4 r2

r2t2−1/4 4 r2s2+3 q2s2−r4

t2r2s2 −1/4 4 s2+3 r2

s2t20

27

Ric =

1/2 q2−6 r2

r2t2−3/2 q

t2r0 0

−3/2 q

t2r−1/2 6 r2s2+q2s2−r4

t2r2s2 −3/2 rst2

0

0 −3/2 rst2

−1/2 6 s2+r2

s2t20

0 0 0 −1/2 6 r2s2+r4+q2s2

t2r2s2

.

Here the eigenvalues and eigenvectors will not fit on the page, and does not give usany useful information. So we only look at the case when r = s = t = q = 1.


−4,−3,−3 + 1/2√

19,−3 − 1/2√

19}


(0, 0, 0, 1) ,

(

−1,−1

3, 1, 0

)

,

(

18(

−1 +√

19)−2

,−6(

−1 +√

19)−1

, 1, 0

)

,

(

18(

−1 −√

19)−2

,−6(

−1 −√

19)−1

, 1, 0

)}

Example 3.5 (Aab4,5). Structure equations:

[E1, E4] = E1, [E2, E4] = aE2, [E3, E4] = bE3

K =

0 − at2

− bt2

−t−2

− at2

0 −abt2

−a2

t2

− bt2

−abt2

0 − b2

t2

−t−2 −a2

t2− b2

t20

Ric =

−1+b+at2

0 0 0

0 −a(1+b+a)t2

0 0

0 0 − b(1+b+a)t2

0

0 0 0 − b2+a2+1t2

.


−b (1 + b + a)

t2,−1 + b + a

t2,−b2 + a2 + 1

t2,−a (1 + b + a)

t2

}


{(0, 0, 1, 0) , (1, 0, 0, 0) , (0, 0, 0, 1) , (0, 1, 0, 0)}When a = b = 1 this equals the Lie algebra of H

4.

28

Example 3.6 (Aab4,6). Structure equations:

[E1, E4] = aE1, [E2, E4] = bE2 − E3, [E3, E4] = E2 + bE3

K =

0 −abt2

−abt2

−a2

t2

−abt2

0 −1/4 2 r2s2−r4−s4+4 b2s2r2

s2t2r2 −1/4 −2 r2s2+4 b2s2r2+3 s4−r4

s2t2r2

−abt2

−1/4 2 r2s2−r4−s4+4 b2s2r2

s2t2r2 0 −1/4 4 b2s2r2−2 r2s2+3 r4−s4

s2t2r2

−a2

t2−1/4 −2 r2s2+4 b2s2r2+3 s4−r4

s2t2r2 −1/4 4 b2s2r2−2 r2s2+3 r4−s4

s2t2r2 0

Ric =

−a(a+2 b)t2

0 0 0

0 − 4 b2s2r2+s4−r4+2 abs2r2

2s2t2r2 − (−s2+r2)(a+2 b)

2rt2s0

0 − (−s2+r2)(a+2 b)

2rt2s− 4 b2s2r2+r4−s4+2 abs2r2

2s2t2r2 0

0 0 0 − 4 b2s2r2−2 r2s2+r4+s4+2 r2a2s2

2s2t2r2

.

Again we get eigenvalues too complicated to print out. But we will at lest look atthe case where r = s = t = q = 1.

Eigenvalues to Ric:

{−2 b2 − a2,−a2 − 2 ab,−2 b2 − ab,−2 b2 − ab

}


{(0, 0, 0, 1) , (1, 0, 0, 0) , (0, 0, 1, 0) , (0, 1, 0, 0)}

The two last eigenvalues are not always equal. They can be written as λ3 =f(q, r, s, t) + g(q, r, s, t) and λ3 = f(q, r, s, t) − g(q, r, s, t), where g(1, 1, 1, 1) = 0.


[E2, E3] = E1, [E1, E4] = 2E1, [E2, E4] = E2, [E3, E4] = E2 + E3

K =

0 1/4 q2t2−8 r2s2

t2r2s2 1/4 q2t2−8 r2s2

t2r2s2 −4 t−2

1/4 q2t2−8 r2s2

t2r2s2 0 −1/4 3 q2t2−r4+4 r2s2

t2r2s2 1/4 r2−4 s2

s2t2

1/4 q2t2−8 r2s2

t2r2s2 −1/4 3 q2t2−r4+4 r2s2

t2r2s2 0 −1/4 4 s2+3 r2

s2t2

−4 t−2 1/4 r2−4 s2

s2t2−1/4 4 s2+3 r2

s2t20

29

Ric =

1/2 −16 r2s2+q2t2

t2r2s2 0 0 0

0 −1/2 −r4+8 r2s2+q2t2

t2r2s2 −2 rst2

0

0 −2 rst2

−1/2 8 r2s2+r4+q2t2

t2r2s2 0

0 0 0 −1/2 12 s2+r2

s2t2

.


−4 r2s2 − 1/2 q2t2 + 1/2√

r8 + 16 r6s2

r2s2t2,−4 r2s2 − 1/2 q2t2 − 1/2

√r8 + 16 r6s2

r2s2t2,

−1/212 s2 + r2

s2t2, 1/2

−16 r2s2 + q2t2

r2s2t2

}


0,−4sr3

√r8 + 16 r6s2 − r4

, 1, 0

)

,

(

0,−4sr3

−√

r8 + 16 r6s2 − r4, 1, 0

)

, (0, 0, 0, 1) , (1, 0, 0, 0)

}


[E2, E3] = E1, [E2, E4] = E2, [E3, E4] = −E3

K =

0 1/4 q2

r2s2 1/4 q2

r2s2 0

1/4 q2

r2s2 0 −1/4 3 q2t2−4 r2s2

t2r2s2 −t−2

1/4 q2

r2s2 −1/4 3 q2t2−4 r2s2

t2r2s2 0 −t−2

0 −t−2 −t−2 0

Ric =

1/2 q2

r2s2 0 0 0

0 −1/2 q2

r2s2 0 0

0 0 −1/2 q2

r2s2 0

0 0 0 −2 t−2

.


−2 t−2, 1/2q2

r2s2,−1/2

q2

r2s2,−1/2

q2

r2s2

}


{(0, 0, 0, 1) , (1, 0, 0, 0) , (0, 1, 0, 0) , (0, 0, 1, 0)}

30

Example 3.9 (Ab4,9). Structure equations:

[E2, E3] = E1, [E1, E4] = (1 + b) E1, [E2, E4] = E2, [E3, E4] = bE3

K =

0 −1/4 −q2t2+4 r2s2+4 r2s2b

r2s2t2−1/4 −q2t2+4 r2s2b+4 b2s2r2

r2s2t2− (1+b)2

t2

−1/4 −q2t2+4 r2s2+4 r2s2b

r2s2t20 −1/4 3 q2t2+4 r2s2b

r2s2t2−t−2

−1/4 −q2t2+4 r2s2b+4 b2s2r2

r2s2t2−1/4 3 q2t2+4 r2s2b

r2s2t20 − b2

t2

− (1+b)2

t2−t−2 − b2

t20

Ric =

−1/2 4 r2s2+8 r2s2b+4 b2s2r2−q2t2

r2s2t20 0 0

0 −1/2 4 r2s2+q2t2+4 r2s2b

r2s2t20 0

0 0 −1/2 4 b2s2r2+q2t2+4 r2s2b

r2s2t20

0 0 0 −2 b2+1+bt2

.


−2b2 + 1 + b

t2,−1/2

4 r2s2 + q2t2 + 4 r2s2b

r2s2t2,

−1/24 b2s2r2 + q2t2 + 4 r2s2b

r2s2t2,−1/2

4 r2s2 + 8 r2s2b + 4 b2s2r2 − q2t2

r2s2t2

}


{(0, 0, 0, 1) , (0, 1, 0, 0) , (0, 0, 1, 0) , (1, 0, 0, 0)}Example 3.10 (A4,10). Structure equations:

[E2, E3] = E1, [E2, E4] = −E3, [E3, E4] = E2

K =

0 1/4 q2

r2s2 1/4 q2

r2s2 0

1/4 q2

r2s2 0 −1/4 3 q2t2+2 r2s2−r4−s4

t2r2s2 1/4 −3 s4+2 r2s2+r4

t2r2s2

1/4 q2

r2s2 −1/4 3 q2t2+2 r2s2−r4−s4

t2r2s2 0 −1/4 3 r4−2 r2s2−s4

t2r2s2

0 1/4 −3 s4+2 r2s2+r4

t2r2s2 −1/4 3 r4−2 r2s2−s4

t2r2s2 0

Ric =

1/2 q2

r2s2 0 0 0

0 −1/2 s4−r4+q2t2

t2r2s2 0 0

0 0 −1/2 r4−s4+q2t2

t2r2s2 0

0 0 0 −1/2 −2 r2s2+r4+s4

t2r2s2

.

31


1/2q2

r2s2,−1/2

−2 r2s2 + r4 + s4

r2s2t2,−1/2

r4 − s4 + q2t2

r2s2t2,−1/2

s4 − r4 + q2t2

r2s2t2

}


{(1, 0, 0, 0) , (0, 0, 0, 1) , (0, 0, 1, 0) , (0, 1, 0, 0)}Example 3.11 (Aa

4,11). Structure equations:

[E2, E3] = E1, [E1, E4] = 2aE1, [E2, E4] = aE2 − E3, [E3, E4] = E2 + aE3

− K =

0 −q2t2+8 r2a2s2

4r2s2t2−q2t2+8 r2a2s2

4r2s2t24 a2

t2

−q2t2+8 r2a2s2

4r2s2t20 3 q2t2+2 r2s2−r4−s4+4 r2a2s2

4r2s2t2−2 r2s2+4 r2a2s2+3 s4−r4

4r2s2t2

−q2t2+8 r2a2s2

4r2s2t23 q2t2+2 r2s2−r4−s4+4 r2a2s2

4r2s2t20 4 r2a2s2−2 r2s2+3 r4−s4

4r2s2t2

4 a2

t2−2 r2s2+4 r2a2s2+3 s4−r4

4r2s2t24 r2a2s2−2 r2s2+3 r4−s4

4r2s2t20

Ric =

− 16 r2a2s2−q2t2

2s2t2r2 0 0 0

0 − 8 r2a2s2+s4−r4+q2t2

2s2t2r2 −2a(−s2+r2)

t2rs0

0 −2a(−s2+r2)

t2rs− 8 r2a2s2+r4−s4+q2t2

2s2t2r2 0

0 0 0 − 12 r2a2s2−2 r2s2+r4+s4

2s2t2r2

.

Eigenvalues to Ric:{−4 r2a2s2 − 1/2 q2t2 + 1/2

√16 r6a2s2 − 32 r4s4a2 + 16 r2s6a2 + r8 − 2 r4s4 + s8

r2s2t2,

−4 r2a2s2 − 1/2 q2t2 − 1/2√

16 r6a2s2 − 32 r4s4a2 + 16 r2s6a2 + r8 − 2 r4s4 + s8

r2s2t2,

−1/212 r2a2s2 − 2 r2s2 + r4 + s4

r2s2t2,−1/2

16 r2a2s2 − q2t2

r2s2t2

}


0,−4a (−s2 + r2) rs√

16 r6a2s2 − 32 r4s4a2 + 16 r2s6a2 + r8 − 2 r4s4 + s8 + s4 − r4, 1, 0

)

,

(

0,−4a (−s2 + r2) rs

−√

16 r6a2s2 − 32 r4s4a2 + 16 r2s6a2 + r8 − 2 r4s4 + s8 + s4 − r4, 1, 0

)

(0, 0, 0, 1) , (1, 0, 0, 0)}

32


[E1, E3] = E1, [E2, E3] = E2, [E1, E4] = −E2, [E2, E4] = E1

K =

0 1/4 s2r4−2 s2r2q2+s2q4−4 r2t2q2

r2t2q2s2 −s−2 1/4 −3 r4+2 r2q2+q4

r2t2q2

1/4 s2r4−2 s2r2q2+s2q4−4 r2t2q2

r2t2q2s2 0 −s−2 −1/4 3 q4−2 r2q2−r4

r2t2q2

−s−2 −s−2 0 0

1/4 −3 r4+2 r2q2+q4

r2t2q2 −1/4 3 q4−2 r2q2−r4

r2t2q2 0 0

Ric =

1/2 −s2r4+s2q4−4 r2t2q2

r2t2q2s2 0 0 0

0 −1/2 s2q4−s2r4+4 r2t2q2

r2t2q2s2 0 0

0 0 −2 s−2 0

0 0 0 −1/2 q4−2 r2q2+r4

r2t2q2

.


−1/2q4 − 2 r2q2 + r4

r2t2q2,−1/2

s2q4 − s2r4 + 4 r2t2q2

r2q2t2s2, 1/2

−s2r4 + s2q4 − 4 r2t2q2

r2q2t2s2,−2 s−2

}


{(0, 0, 0, 1) , (0, 1, 0, 0) , (1, 0, 0, 0) , (0, 0, 1, 0)}

33

Appendix A

The Maple program

The program relies on the Maple package DifferentialGeometry and the subpackageLieAlgebras. For information about setting up a Lie algebra to work with, oneshould look at the documentation for LieAlgebraData in the LieAlgebras package.

The metric we want to work with is represented by an n × n matrix called g,where n is the dimension of the Lie algebra. The matrix must be called g and mustbe within the scope of the procedures.

The functionality of the program is based on the fact that we can write the LeviCivita connection of two vector fields as a linear combination of metric values for thebasis vectors. Thus we must first be able to get a list of basis vectors. Maple doesnot have a predefined function for appending objects to lists, so this short procedureis used.

A procedure for adding objects to lists.

append := proc (L1 , ob ) opt ions operator , arrow ;[ seq (L1 [ i ] , i = 1 . . nops (L1) ) , ob ]

end proc

The simplest way of getting the basis is to pick it out from the multiplication tableof the internal representation of the Lie algebra. MultiplicationTable(”LieDerivative”)returns a table which has the actual vectors in it.

A procedure for getting a list of basis vectors for our Lie algebra.

ge tBas i s := proc ( )l o c a l MT, B, dim , i ;MT := Mul t ip l i c a t i onTab l e (” L i eDe r i va t i v e ”) ;B := [ ] ;dim := op (MT) [1 ] −2 ;f o r i to dim do

B := append (B, MT[ i +2, 1 ] )end do ;B

end proc

35

To understand how to use the vectors, it is important to understand how they arerepresented internally. The first basis vector of a Lie algebra is called e1. Thoughit is actually a nested list with various values. It really looks like this: [[”vec-tor”,A1,[]],[[[1],1]]]. The first sublist, [”vector”,A1,[]], tells us information about thisobject, that it is a vector in the algebra A1. The second sublist is the vector rep-resented as a linear combination of basis vectors. [[[1],1]] is a list representing asum. For a better example, we look at a = e1 &plus 5 &mult e2. (Note that vectoraddition is called &plus and multiplication by scalar is &mult). a is internally rep-resented as [[”vector”, A1, []], [[[1], 1], [[2], 5]]]. Here we can see that in the list [[[1],1], [[2], 5]] the first sublist is what basis vector we have, and the coefficient of it.And the same for the second list. So there we have the information a = 1 ·e1+5 ·e2.Both the Lie bracket and the metric is bilinear, and the vectors are represented aslinear combinations of basis vectors, so using this we only need to loop over the basisvectors used to represent the vectors and add up the results in the end. Thas is howthis procedure calculates the metric value of two vectors. It extracts the indices ofthe basis vectors in the linear combination of the given vector, and uses that as theindice in the matrix g defining the metric values. Since the metric is bilinear it addsall these up for all the basis vectors.

A procedure for calculating the metric value for two vectors

metr := proc (X, Y)l o c a l e i , e j , M;M := 0 ;f o r e i in op (X) [ 2 ] do

f o r e j in op (Y) [ 2 ] doM := M+e i [ 2 ] ∗ e j [ 2 ] ∗ g [ e i [ 1 ] [ 1 ] , e j [ 1 ] [ 1 ] ]

end doend do ;Mend proc

The most fundamental thing we need to be able to calculate to get curvatures is theLevi-Civita connection. As we saw in equations (1.1) and (1.2) it can be expressedwith only Lie brackets and metric values for basis vectors in the algebra. That iswhat this procedure does. We give it two vectors, and it finds the basis we areworking in and the metric we are using, and essentially insert equation (1.1) into(1.2) so we can get a the connection of the two given vectors as a linear combinationsof basis vectors. So here we loop first over each basis vector in the basis, and in thatover linear combinationas of vectors that express various Lie brackets. And finallywe add it all up.

The Levi Civita connection.

Con := proc (X, Y)l o c a l B, ek , N, LB, C, K, m, i , n , j ;B := getBas i s ( ) ;

36

C := LieBracket (B[ 1 ] , B [ 1 ] ) ;f o r ek in B do

N := 0 ;LB := LieBracket (X, Y) ;m := nops ( op (LB) [ 2 ] ) ;f o r i to m do

N := N+op (LB) [ 2 ] [ i ] [ 2 ] ∗ g [ op ( ek )[ 2 ] [ 1 ] [ 1 ] [ 1 ] , op (LB) [ 2 ] [ i ] [ 1 ] [ 1 ] ]

end do ;LB := LieBracket ( ek , X) ;m := nops ( op (LB) [ 2 ] ) ;f o r i to m do

n := nops ( op (Y) [ 2 ] ) ;f o r j to n do

N := N+op (Y) [ 2 ] [ j ] [ 2 ] ∗ op (LB)[ 2 ] [ i ] [ 2 ] ∗ g [ op (LB) [ 2 ] [ i ] [ 1 ] [ 1 ] , op (Y) [ 2 ] [ j ] [ 1 ] [ 1 ] ]

end doend do ;LB := LieBracket ( ek , Y) ;m := nops ( op (LB) [ 2 ] ) ;f o r i to m do

n := nops ( op (X) [ 2 ] ) ;f o r j to n do

N := N+op (X) [ 2 ] [ j ] [ 2 ] ∗ op (LB)[ 2 ] [ i ] [ 2 ] ∗ g [ op (LB) [ 2 ] [ i ] [ 1 ] [ 1 ] , op (X) [ 2 ] [ j ] [ 1 ] [ 1 ] ]

end doend do ;N := (1/2) ∗N;K := ‘&mult ‘ (N, ek ) ;C := ‘&plus ‘ (C, K)

end do ;C

end proc

Now we can calculate the Riemann curvature tensor, since it does nothing more thantake three vectors and apply them in various ways to the Levi-Civita connection.

The Riemann curvature tensor

RieCu := proc (X, Y, Z)l o c a l C, YZ, XZ, LB;YZ := Con(Y, Z) ;C := Con(X, YZ) ;XZ := Con(X, Z) ;C := ‘&minus ‘ (C, Con(Y, XZ) ) ;LB := LieBracket (X, Y) ;

37

C := ‘&minus ‘ (C, Con(LB, Z) ) ;C

end proc

Now we have all we need to calculate curvatures. First the Rici operator:

The Ricci operator

Ric := proc (X)l o c a l B, ek , R;B := getBas i s ( ) ;R := LieBracket (B[ 1 ] , B [ 1 ] ) ;f o r ek in B do

R := ‘&plus ‘ (R, RieCu (X, ek , ek ) )end do ;R

end proc

Then the curvature function:

The curvature function

CurFun := proc (X, Y)l o c a l R, K;R := RieCu (X, Y, Y) ;K := metr (R, X) ;K

end proc

To be able to show alot of information at once we will generally calculate a matrixK where Kij = κ (Ei, Ej) for basis vectors Ei, Ej. This is how we do that:

A procedure for calculating a matrix where the elements are the curvature function of basis elements

CurFunMat := proc ( )l o c a l B, M, e i , e j , K;B := getBas i s ( ) ;M := Matrix ( nops (B) ) ;f o r e j in B do

f o r e i in B doK := CurFun( ej , e i ) ;M[ op ( e j ) [ 2 ] [ 1 ] [ 1 ] [ 1 ] , op ( e i )

[ 2 ] [ 1 ] [ 1 ] [ 1 ] ] := Kend do

end do ;M

end proc

And a matrix representing the Ricci quadratic form, which we call Ric and whereRicij = Ric (Ei, Ej).

38

A procedure that calculates a matrix representing the Ricci quadratic form

RicQF := proc ( )l o c a l A, B, e i , e j , m, i , R, G;B := getBas i s ( ) ;A := Matrix ( nops (B) ) ;f o r e j in B do

f o r e i in B doG := 0 ;R := Ric ( e i ) ;m := nops ( op (R) [ 2 ] ) ;f o r i to m do

G := G+op (R) [ 2 ] [ i ] [ 2 ] ∗ g [ op (R) [ 2 ] [ i ] [ 1 ] [ 1 ] , op ( e j ) [ 2 ] [ 1 ] [ 1 ] [ 1 ] ]

end do ;A[ op ( e i ) [ 2 ] [ 1 ] [ 1 ] [ 1 ] , op ( e j )

[ 2 ] [ 1 ] [ 1 ] [ 1 ] ] := Gend do

end do ;A

end proc

These procedures for the curvatures however, assume we are working with anorthonormal basis, that is that

g =

1 0 0

0 1 0

0 0 1

.

If this was the only case we could calculate the program would not be very inter-esting, so we need to perform some process which makes the basis we are workingwith orthonormal with respect to any given metric. We firstly need this little helpfulprocedure:

Projects a given vector to a given basis vector

pro j := proc (v , u)l o c a l M, e , f , N, P;M := metr (u , v ) ;‘&mult ‘ (M, u)

end proc

Then we can do Gram-Schmidt for the vectors in the algebra we are using. Thisprocedure returns a list of orthonormalized vectors.

A Gram-Schmidt procedure for our vectors.

GrSc := proc ( )

39

l o c a l U, V, B, i , n , j , e , v , N;V := getBas i s ( ) ; U := V;n := nops (V) ;f o r i to n do

U[ i ] := V[ i ] ;f o r j to i−1 do

U[ i ] := ‘&minus ‘ (U[ i ] , p ro j (V[ i ] , U[j ] ) )

end do ;U[ i ] := ‘&mult ‘ ( 1 / sq r t ( metr (U[ i ] , U[ i ] ) ,

symbol ic ) , U[ i ] )end do ;B := U;Bend proc

These vectors we can then use as an ONB. This procedure takes that list, gets thestructure equations which define the algebra we are working in, and essentially putin the vectors of the list instead of the corresponding present vectors. It returns alist of Lie bracket products and a list of the new basis vectors. These will use thesymbol ”E”, and therefore E should always be unassigned when using this program.To use the algebra with the ONB one only need to run the result from this procedurein LieAlgebraData, and then set up with DGsetup for that result. One also need toset g to the identity matrix.

Uses the result from the Gram-Schmidt and makes new structure equations for a new orthonormalbasis.

makeONB := proc ( onb )l o c a l C, e , n , i , newBasis , tab , newSEq , eq , lb ,

prod , bb1 , bb2 , bb3 , f f f , j , m, P, k , l , t , prodpart ;C := onb ;n := nops ( onb ) ;f o r i to n do

C[ i ] := op ( onb [ i ] ) [ 2 ] [ 1 ] [ 2 ]end do ;newBasis := [ ] ;f o r i to n do

newBasis := append ( newBasis , cat (E, i ) )end do ;tab := Mul t ip l i c a t i onTab l e (” LieBracket ”) ;newSEq := [ ] ;n := nops ( tab ) ;f o r i to n do

lb := [ cat (E, convert ( Explode ( convert ( op ( tab[ i ] ) [ 1 ] [ 1 ] , s t r i n g ) ) [ −1] , symbol ) ) , cat (E, convert (

40

Explode ( convert ( op ( tab [ i ] ) [ 1 ] [ 2 ] , s t r i n g ) ) [ −1] , symbol ) )] ;

f f f := 1 ;bb1 := parse ( convert ( Explode ( convert ( op ( tab [

i ] ) [ 1 ] [ 1 ] , s t r i n g ) ) [ −1] , symbol ) ) ;bb2 := parse ( convert ( Explode ( convert ( op ( tab [

i ] ) [ 1 ] [ 2 ] , s t r i n g ) ) [ −1] , symbol ) ) ;m := nops ( op ( tab [ i ] ) [ 2 ] ) ;i f op (0 , op ( tab [ i ] ) [ 2 ] ) = ’ symbol ’ then

bb3 := parse ( convert ( Explode ( convert( op ( op ( tab [ i ] ) [ 2 ] ) , s t r i n g ) ) [ −1] , symbol ) ) ;

P := 1 ;f o r j to m−1 do

P := P∗op ( op ( tab [ i ] ) [ 2 ] ) [ j ]end do ;prod := ‘&mult ‘ (P∗ f f f ∗C[ bb1 ]∗C[ bb2 ] /

C[ bb3 ] , cat (E, convert ( Explode ( convert ( op ( op ( tab [ i ] ) [ 2 ] ) ,s t r i n g ) ) [ −1] , symbol ) ) )

e l i f op (0 , op ( tab [ i ] ) [ 2 ] ) = ‘∗ ‘ thenbb3 := parse ( convert ( Explode ( convert

( op ( op ( tab [ i ] ) [ 2 ] ) [ −1] , s t r i n g ) ) [ −1] , symbol ) ) ;P := 1 ;k := nops ( op ( tab [ i ] ) [ 2 ] ) ;f o r j to k−1 do

P := P∗op ( op ( tab [ i ] ) [ 2 ] ) [ j ]end do ;prod := ‘&mult ‘ (P∗ f f f ∗C[ bb1 ]∗C[ bb2 ] /

C[ bb3 ] , cat (E, convert ( Explode ( convert ( op ( op ( tab [ i ] ) [ 2 ] )[ −1] , s t r i n g ) ) [ −1] , symbol ) ) )

e l i f op (0 , op ( tab [ i ] ) [ 2 ] ) = ‘+ ‘ thenk := nops ( op ( tab [ i ] ) [ 2 ] ) ;f o r j to k do

bb3 := parse ( convert ( Explode( convert ( op ( op ( tab [ i ] ) [ 2 ] ) [ j ] , s t r i n g ) ) [ −1] , symbol ) ) ;

i f op (0 , op ( op ( tab [ i ] ) [ 2 ] ) [ j] ) = ‘∗ ‘ then

P := 1 ;t := nops ( op ( op ( tab [

i ] ) [ 2 ] ) [ j ] ) ;f o r l to t−1 do

P := P∗op ( op( op ( tab [ i ] ) [ 2 ] ) [ j ] ) [ l ]

end do ;prodpart := ‘&mult ‘ (

41

P∗ f f f ∗C[ bb1 ]∗C[ bb2 ] /C[ bb3 ] , cat (E, convert ( Explode (convert ( op ( op ( tab [ i ] ) [ 2 ] ) [ j ] , s t r i n g ) ) [ −1] , symbol ) ) )

e l s eprodpart := ‘&mult ‘ (

f f f ∗C[ bb1 ]∗C[ bb2 ] /C[ bb3 ] , cat (E, convert ( Explode ( convert (op ( op ( tab [ i ] ) [ 2 ] ) [ j ] , s t r i n g ) ) [ −1] , symbol ) ) )

end i f ;i f j = 1 then

prod := prodparte l s e

prod := prod+prodpart

end i fend do

end i f ;newSEq := append (newSEq , lb = prod )

end do ;newSEq , newBasis

end proc

How to use the program

As an educational example, I will show the workflow when calculating one of thecurvatures in chapter 2.

Base := [X1 ,X2 ,X3 ]

[X1 ,X2 ,X3 ]

a37aseq := [[X1 ,X3 ] = aX1 − X2 , [X2 ,X3 ] = X1 + aX2 ]

[[X1 ,X3 ] = aX1 − X2 , [X2 ,X3 ] = X1 + aX2 ]

a37a := LieAlgebraData (a37aseq ,Base)

[[e1 , e3 ] = ae1 − e2 , [e2 , e3 ] = e1 + ae2 ]

DGsetup (a37a)

‘Lie algebra: L1‘

Query (“Jacobi′′)

true

g :=

r2 0 0

0 s2 0

0 0 t2

42

r2 0 0

0 s2 0

0 0 t2

onba37a := GrSc ()

[e1

r,e2

s,e3

t]

a37aseqonb := makeONB (onba37a)

[[E1 ,E3 ] =aE1

t− sE2

rt, [E2 ,E3 ] =

rE1

st+

aE2

t], [E1 ,E2 ,E3 ]

a37aonb := LieAlgebraData (a37aseqonb)

[[e1 , e3 ] =ae1

t− se2

rt, [e2 , e3 ] =

re1

st+

ae2

t]

DGsetup (a37aonb)

‘Lie algebra: L2‘

Query (“Jacobi′′)

true

g :=

1 0 0

0 1 0

0 0 1

1 0 0

0 1 0

0 0 1

CFM := CurFunMat ()

0 −1/4 2 r2s2−r4−s4+4 r2a2s2

s2t2r2 −1/4 −2 r2s2+4 r2a2s2+3 s4−r4

s2t2r2

−1/4 2 r2s2−r4−s4+4 r2a2s2

s2t2r2 0 −1/4 −2 r2s2+4 r2a2s2+3 r4−s4

s2t2r2

−1/4 −2 r2s2+4 r2a2s2+3 s4−r4

s2t2r2 −1/4 −2 r2s2+4 r2a2s2+3 r4−s4

s2t2r2 0

RQF := RicQF ()

−1/2 4 r2a2s2+s4−r4

r2t2s2 −a(−s2+r2)t2rs

0

−a(−s2+r2)t2rs

−1/2 4 r2a2s2+r4−s4

r2t2s2 0

0 0 −1/2 4 r2a2s2−2 r2s2+r4+s4

r2t2s2

V, W := Eigenvectors (RQF )

V

43

1/2 −4 r2a2s2+√−8 r4s4a2+4 r6a2s2+4 r2s6a2−2 r4s4+r8+s8

t2s2r2

1/2 −4 r2a2s2−√−8 r4s4a2+4 r6a2s2+4 r2s6a2−2 r4s4+r8+s8

t2s2r2

−1/2 4 r2a2s2−2 r2s2+r4+s4

t2s2r2

W

−2a(−s2+r2)rs

√−8 r4s4a2+4 r6a2s2+4 r2s6a2−2 r4s4+r8+s8+s4−r4

−2a(−s2+r2)rs

−√−8 r4s4a2+4 r6a2s2+4 r2s6a2−2 r4s4+r8+s8+s4−r4

0

1 1 0

0 0 1

44

Bibliography

[1] K. Erdmann, M. J. Wildon, Introduction to Lie Algebras, Springer Undergraduate Mathe-matics Series 1, Springer-Verlag (2006).

[2] M.P. do Carmo, Riemannian Geometry, Mathematics: Theory & Applications 1, Birkhauser(1992).

[3] S. Gudmundsson, An Introduction to Riemannian Geometry, Lund University (2008).

[4] T. Christodoulakis, G.O. Papadopulos, A. Dimakis, Automorphisms of real four-dimensional

Lie algebras and the invariant characterization of homogeneous 4-spaces, J. Phys. A 36

(2003), 427-441.

[5] M.A.H. MacCallum, On the classification of the Real Four-Dimensional Lie Algebras, OnEinstein’s Path (1996), 299-317, Springer (1999).

[6] A.G. Kremlyov, Yu.G. Nikonorov, The Signature of the Ricci Curvature of Left Invariant

Riemannian Metrics on 4-Dimensional Lie Groups, to appear.

[7] G. Jensen, The scalar curvature of left invariant Riemannian metrics, Indiana Univ. Math.J. 20 (1971), 1125-1144.

[8] J. Milnor, Curvatures of Left Invariant Metrics on Lie Groups, Advances in Math. 21 (1976),293-329.

[9] J. Patera, R.T. Sharp, P. Winternitz, H. Zassenhaus, Invariants of real low dimension Lie

algebras, J. Mathematical Phys. 17, (1976), 986-994.

[10] A. W. Knapp, Lie Groups Beyond an Introduction, Progress in Mathematics 140, Birkhauser(1996).

[11] E. Heintze, On Homogeneous Manifolds of Negative Curvature, Math. Ann. 211 (1974), 23-34.

[12] R. Azencott, E. N. Wilson, Homogeneous manifolds with negative curvature, Part 1, Trans.Amer. Math. Soc. 215 (1976), 323-362.

[13] P. Petersen, Riemannian Geometry, GTM 171, Springer-Verlag New York (1997).

[14] J. A. Wolf, Curvature in Nilpotent Lie Groups, Proc. Amer. Math. Soc. 15 (1964), 271-274.

45

Bachelor’s Theses in Mathematical Sciences 2009:K1ISSN 1654-6229

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