differential geometry (2006) - jimmie lawson

7/28/2019 Differential Geometry (2006) - Jimmie Lawson

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CHAPTER 1

BACKGROUND

The prerequisites for this course are a background in general topology, advanced calculus(analysis), and linear algebra.

The background in general topology is a knowledge of the notions of a topology as givenin a one semester or quarter course: the relative topology, the quotient topology, compactsets and topological spaces, second countable topological spaces, and Hausdorff spaces.

At several points in the notes connections are made to algebraic topology, e.g., thefundamental group, but algebraic topology is not required in the logical dependence ofthese notes.

The assumed analysis background is a course in Calculus of several variables that in-cludes the The Inverse Function Theorem, The Implicit Function Theorem, The Changeof Variables Formula from integration, and The Mean Value Theorem for Integrals. Inaddition, we will also use the theorem of Existence and Uniqueness of solutions to firstorder differential equations. We now state and discuss these Theorems as well as settingsome notation.

Suppose O Rm and f : O Rn. If p O and v Rm, then the directional

derivative off at p in the direction v is

Dvf(p) = limh0

f(p + hv) f(p)

h, if the limit exists.

Some authors require that v is a unit vector. This assumption is almost universal incalculus books, but not in analysis books or literature. We do not assume that v is a unitvector.

If f is differentiable, then the derivative at p is the linear map Df(p) : Rm Rn

defined byDf(p)(v) = Dvf(p).

The matrix for Df(p), in the standard basis, is the Jacobian matrix Jf(p) and

Jf(p) =

f1x1

f1xm

......

fnx1

fnxm

The basic theorems that we assume from a background of analysis are: The InverseFunction Theorem, The Implicit Function Theorem, The Change of Variables Formulafrom integration, and The Mean Value Theorem for Integrals. In addition, we will also usethe theorem of existence and uniqueness of solutions to first order differential equations.

We now state and discuss these Theorems.

copyright c2002

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2 CHAPTER 1 BACKGROUND

Inverse Function Theorem.

Implicit Function Theorem. Let f : Rn+m Rm be a C function. Write pointsin Rn+m = Rn Rm as (t1, , tn, x1, , xm) = (t; x) and f = (f1, , fm). Let

f(t0; x0) = p for p Rm. Suppose that det(fjxi

(t0;x0)

) = det(Df|Rm

(t0; x0)) = 0.

Then there is an-dimensional open set T0 containing t0, an m-dimensional open set X0containing x0, and a unique function g : T0 R

m such that

(1) g isC

(2) g(t0) = x0, and(3) for (t; x) T0 X0, f(t; x) = p if and only ifg(t) = x.

There is a corollary of the Implicit Function Theorem that will, for us, be the usefulform of the Implicit Function Theorem. It is sometimes called the Rank Theorem.

The Rank Theorem. SupposeO Rn+m is an open set and f : O Rm is C.Supposep Rm, q O and f(q) = p. IfDf(q) has rankm, then there is an open setU O, q U; open setsU1 R

n, andU2 Rm;and, a diffeomorphism H : U U1 U2

such that y U f1(p) if and only ifH(y) U1 {0}.

Proof. The kernel ofDf(q) is n-dimensional, so let Ker(Df(q)) = span{v1, ,vn} = Rn

and let span{vn+1, ,vn+m} = Rm, so that vn+1, ,vn+m is an ordered basis of

Rn+m = Rn Rm. Let K : Rn+m Rn+m be defined by K(y) =n+m

i=1 yivi + qfor each y = (y1, , yn+m) R

n+m. The map K is a C diffeomorphism, it is

an isomorphism and a translation. Also Df K(0)(Rm) = Df(q) DK(0)(Rm) =Df(q)(span{vn+1, ,vn+m}) = R

m, and so det(Df K(0)|Rm

) = 0. We have thatK1(O) is an open set, f K : K1(O) Rm is a C function, f K(0) = p, anddet(Df K(0)|

Rm) = 0, and hence we can apply the Implicit Function Theorem. We con-

clude that there are open sets T Rn, X Rm, T X K1(O), and a C functiong : T X such that f K(t; x) = p if and only ifg(t) = x, i.e., (f K)1(q) is the graphofg.

Take an open set U1 T with 0 U1 and an open set U2 X such that O1 ={(t, x) | x g(t) + U2} T X. It is an exercise in general topology that this canalways be arranged (one first takes U1 so that its closure is a compact subset of T). Let

h : O1 U1 U2 by h(t, x) = (t, xg(t)). The map h is a C

diffeomorphism, its inverseis (t, x) (t, x + g(t)). Under this diffeomorphism, Gg O1 maps to U1 {0}.

To complete the proof, take H= hK1 and U= hK1(O1). Then H : U U1 U2is a C diffeomorphism and x U f1(p) if and only ifH(x) U1 {0}.

Change of Variables Formula.

Mean Value Theorem for Integrals.


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CHAPTER 2

MANIFOLDS

In this chapter, we address the basic notions: What is a manifold and what is a mapbetween manifolds. Several examples are given.

An n dimensional manifold is a topological space that appears to be Rn near a point,i.e., locally like Rn. Since these topological spaces appear to be locally like Rn, we mayhope to develop tools similar to those used to study Rn in order to study manifolds. Theterm manifold comes from many fold, and it refers to the many dimensions of space thata manifold may describe.

Later in this section we will also need structure to discuss C functions on manifolds.Our notion of smooth will be C, i.e., continuous partial derivatives of all orders. Theterms C and smooth are usually synonymous, however, in this chapter and the next,we will use C to describe maps between real vector spaces (as in advanced Calculus)and smooth for maps between manifolds. We use this distinction since many confusingcompositions occur and the reader is assumed to be familiar with C maps between realvector spaces while results on smooth maps between manifolds must be proven. We beginwith the notion of a topological manifold.

Definition 2.1***. A topological manifold of dimension n is a second countable Haus-dorff space M for which each point has a neighborhood homeomorphic to an open set inRn.

This notion appears to capture the topological ideal of locally looking like Rn, but inorder to do Calculus we will need more structure. The term n-manifold is usually writtenfor n dimensional manifold, and the dimension n is often suppressed.

Definition 2.2***. Let U M be a connected open set in a topological n-manifold M,and : U Rn be a homeomorphism to its image (U), an open set in Rn.

The pair (U, ) is called a coordinate system or chart. If x0 U and (x0) = 0 Rn,

then the coordinate system is centered at x0. Call the map a coordinate map.

Definition 2.3***. Suppose M is a topological manifold. For k N, k = 0, or k = ,a Ck atlas is a set of charts {(Ui, i) | i I} such that

(1)iIUi = M, and

(2) i 1j is C

k on its domain for all i, j I.

The k denotes the degree of differentiability. If k = 0, then maps are just continuous.

Definition 2.4***. If A = {(Ui, i)|i I} is a Ck atlas for an n-manifold Mn andf : U Rn is a homeomorphism onto its image with U M open, then (U, f) is

compatible with A ifi f

1

: f(U Ui) (U Ui) is C

k

and f

1

i isC

k

for alli I.

copyright c2002 Larry Smolinsky

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2 CHAPTER 2 MANIFOLDS

Theorem 2.5***. If A = {(Ui, i)|i I} is a Ck atlas for Mn then A is contained in a

unique maximal atlas for M where atlases are ordered as sets by containment.Proof. Let

M ={(V, f)|V M is open, f : V Rn

is a homeomorphism onto its image, and (V, f) is compatible with A}.

Since M contains all compatible charts it is the unique maximal atlas if it is an atlas. Wenow show that it is an atlas. If (V, f) and (V, f) are in M then we must show thatf f

1 : f(V V) f(V V) is C

k. Suppose m V V. Take i I such thatm Ui. Then on f(V V Ui), f f

1 = (f

1i ) (i f

1 ). Since f and f are

compatible with A, f 1i and i f

1 are C

k on open sets of Rn to open sets of Rn.

Therefore f f1 is Ck on its domain.

Definition 2.6***. A maximal Ck atlas is called a Ck differential structure.

Definition 2.7***. A Ck n-manifold is a topological n-manifold M along with a Ck

differential structure S. By Theorem 2.5***, a single atlas is enough to determine thedifferential structure.

The reader should note that this definition for a C0 structure agrees with the definitionof a topological manifold. A C n-manifold is also called a smooth manifold. The wordmanifold, without other adjectives, will denote a smooth manifold as thesewill be the subject of the remainder of this manuscript.

Usually the notation for the structure S is suppressed. However, the phrase a manifoldM supposes that there is an unnamed differential structure S in the background. Inparticular this means that if A is an atlas, then A S; and, if (U, ) is a chart, then(U, ) S. The differential structure contains all compatible charts. For example, if(U, ) S and V U is open, then (V, |V) S.

Example 2.8.a***. A real n-dimensional vector space is an n-manifold.

Let V be an n-dimensional vector space. Pick an ordered basis v1, , vn and define

charts (V, fv1, ,vn), where fv1, ,vn(ni=1

aivi) = (a1, , an) is an isomorphism ofV to Rn.

The image is all ofRn

, which is an open subset of itself. These charts are compatible sincefw1, ,wnf

1v1, ,vn

is a linear automorphism on Rn.Recall that every real vector space is isomorphic to Rn, however Rn comes equiped with

a standard ordered basis or set of coordinates. A real vector space may appear in severalguises. We now mention a few instances of vector spaces that relate to matrices.

Example 2.8.b***. Let the set of n n real matrices be denoted Matnn. The setMatnn is a real vector space of dimension n

2.

Let e1, e2, , em be the standard ordered basis for Rm. If eij is the matrix with 1in the i-th row and j-th column and zero elsewhere, then {eij | 1 i, j n} is a basis.

The isomorphism of Matnn to Rn2 is determined by the linear map f(eij) = en(i1)+j

for 1 i, j n, and allows one to think of the matrix entries as coordinates in Rn2

. Incoordinates, if A = (aij), then f(A) = (a11, a12, , a1n, a21, , ann).


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CHAPTER 2 MANIFOLDS 3

Example 2.8.c***. The symmetric matrices Symnn = {A Matnn | At = A} form a

vector subspace of dimension n(n + 1)2 .

The entries on and above the diagonal can be arbitrary. Below the diagonal, the entriesare determined by symmetry, i.e., aij = aji .

Example 2.8.d***. The skew symmetric matrices Skewnn = {A Matnn | At = A}

form a vector subspace of dimensionn(n 1)

2.

The entries on the diagonal must be zero, since aij = aji. The entries above thediagonal can be arbitrary and the entries below are determined by the antisymmetry, i.e.,aij = aji .

Example 2.9a***. The sphere Sn = {x Rn+1 | |x| = 1} is an n-manifold.

We construct an atlas {(U1, 1), (U2, 2)} with the aid of a standard well-known mapcalled stereographic projection. Let U1 = S

n\{(0, , 0, 1)} and U2 = Sn\{(0, , 0, 1)}.Note that U1 U2 = Sn. Let 1(x1, x2, , xn+1) = (

x11xn+1

, , xn1xn+1 ). The map

1 : U1 Rn is called stereographic projection. The inverse map 11 : R

n U1 isdefined by

11 (y1, , yn) =

2y1

ni=1

y2i

+ 1,

2y2

ni=1

y2i

+ 1, ,

2yn

ni=1

y2i

+ 1, 1

2

ni=1

y2i

+ 1

.

Both 1 and 11 are continuous and hence 1 is a homeomorphism.

The second coordinate chart (U2, 2), stereographic projection from the south pole,is given by 2 = 1 (1) where (-1) is multiplication by 1 on the sphere. Sincemultiplication by -1 is a homeomorphism of the sphere to itself (its inverse is itself), themap 2 : U2 Rn is a homeomorphism.

Checking the compatability conditions, we have 211 (y1, , yn) =

1ni=1 y

2i

(y1, , yn)

and 2 11 = 1

12 . Hence, S

n is shown to be an n-manifold.

Example 2.9b***. Another atlas for the sphere Sn

.We use 2(n + 1) coordinate charts to construct this atlas. For each i {1, , n + 1}

let Ui,+ = {(x1, , xn+1) | xi > 0} and Ui, = {(x1, , xn+1) | xi > 0}. Define i,+ :Ui,+ R

n by i,+(x1, , xn+1) = (x1, , xi1, xi+1, , xn+1) and i,+ : Ui, Rn

by i,(x1, , xn+1) = (x1, , xi1, xi+1, , xn+1). The coordinate xi is a function ofx1, , xi1, xi+1, , xn+1 on the sets Ui,+ and Ui,.

The atlases in Examples 2.9a*** and 2.9b*** are compatible and give the same differ-ential structure. See Exercise 1***.

Example 2.10***. Suppose U1 Rn and U2 Rm are open sets. Iff : U1 U2 is a

C

function, then the graph of f,

Gf = {(x, y) Rn+m | y = f(x)}


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is a manifold.

There is only one coordinate neighborhood required. Let : U1 U2 U1 be theprojection (x, y) = x and let if : U1 U1 U2 be defined by if(x) = (x, f(x)). The onecoordinate neighborhood is (Gf, |Gf). Both and if are C

maps. The composites are|Gf if = IU1 and if |Gf = IGf. Hence, |Gf is a homeomorphism.

Proposition 2.11***. An open subset of an n-manifold is an n-manifold.

Proof. Suppose M is an n-manifold and U M is an open subset. If (, V) is a chart ofM, then (|VU, V U) is a chart for U.

While the next example uses Proposition 2.11***, it is just an open subset of Rn2

.

Example 2.12***. GL(n, R) = {n n nonsingular matrices} is an n2

-manifold.Consider the function det : Matnn R. We use the usual coordinates on Matnn, the

entries as was described in the first example. In these coordinates,

det(A) =Sn

(1)signa1(1)a2(2) an(n).

This function is an n-th degree polynomial. Hence, it is a continuous map. The setdet1(R \ 0) is an open set, the set of nonsingular matrices, GL(n, R).

If a vector space V does not have a natural choice of basis or we do not wish to focus onthe choice of basis, then we write its set of automorphisms as GL(V). By picking a basis

for V, it becomes GL(n, R).It is useful to see that the basic operations in GL(n, R) are C.

Proposition 2.13***. The following maps are C:

(1) GL(n, R) GL(n, R) GL(n, R) by (A, B) AB(2) GL(n, R) GL(n, R) by A AT

(3) GL(n, R) GL(n, R) by A A1

Proof. Recall that GL(n, R) is an open subset of Rn2

and that GL(n, R) GL(n, R) is

an open subset of Rn2

Rn2

. The notion ofC is from advanced Calculus.The first map, multiplication, is a quadratic polynomial in each coordinate as the i, j

entry of AB is

nk=1 aikbkj .

The second map, transpose, is just a reordering of coordinates. In fact, transpose is alinear map.

The third map, inverse, is a rational function of the entries of A. The numerator is thedeterminant of a minor of A and the denominator is detA, a polynomial that is nonzeroon GL(n, R).

Proposition 2.14***. The product of an n-manifold and an m-manifold is an (n + m)-manifold.

Proof. Suppose M is an m-manifold with atlas AM = {(Ui, i) | i I} and N is an

n-manifold with atlas AN = {(Vj, j) | j J}. An atlas for M N is

A = {(Ui Vj , i j) | i I, j J},


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where i j(x, y) = (i(x), j(y)) Rn Rm for (x, y) Ui Vj .

It is easy to see that A is an atlas, since the union

(i,j)IJUi Vj = M N, and it iseasy to check compatibility. If (Ui1 , i1), (Ui2 , i2) AM and (Vj1 , j1), (Vj2 , j2) AN,then

(1) (i1 j1) (i2 j2)1 = (i1 (i2)

1) (i1 (i2)1)

on the set (i2 j2)((Ui1 Vj1) (Ui2 Vj2)) = i2(Ui1 Ui2) i2(Vj1 Vj2) an openset in Rm Rn. Since AM and AN are atlases, the right side of (1) is C

, hence the leftside is also, and A is an atlas.

When can a manifold be pieced together from abstract charts? The question is of

philosophical and practical interest.

Theorem 2.15***. Let X be a set. Suppose A = {(Ui, i)|i I} satisfies

(1) Ui X for each i I.(2)iIUi = X.

(3) i : Ui i(Ui) Rn is a bijection for all i.

(4) i(Ui), i(Ui Uj) Rn are open for all i, j I.

(5) j1i : i(Ui Uj) j(Ui Uj) is C

for all i, j I.

Then there is a unique topology on X such that each Ui is open and each i is a home-omorphism. If the topology is second countable and Hausdorff then X is an n-manifold

and A is an atlas.

Remark 2.16***. If for every x, y X there are i, j I with x Ui, y Uj andUi Uj = or there is an i I with x, y Ui, then X is Hausdorff.

If the open cover {Ui|i I} has a countable subcover, then X is second countable sincethe countable union of second countable spaces is a second countable space.

Proof of Theorem 2.15***. For each x M we give a neighborhood basis. Induce atopology on Ui by taking O Ui is open if and only if O =

1i (W) for W R

n open.The map i is then a homeormorphism. We must check that this gives a well definedneighborhood basis for a topology. Suppose x Ui Uj . The neighborhood basis is then

defined by both i and j . Since j1i is a homeomorphism of open sets, j = (j

1i )i

defines a neighborhood basis ofx in a manner consistent with i. This construction definesa neighborhood basis of each point and so a topology on X. This is the only topology witheach i a homeomorphism as the basis determines the topology. We now see that A satisfiesthe conditions for an atlas: {Ui|i I} is an open cover by 2, i is a homeomorphism by3 and the construction above, and the compatibility condition is 5. If X is Hausdorff andsecond countable, then X is an n-manifold.

Now that we know the definition of a manifold, the next basic concept is a map betweenmanifolds.

Definition 2.17***. Suppose f : Mm

Nn

is a function between manifolds. If for allcharts(U, ) and(W, ) in the differential structures ofM andN respectively, f 1

is C on its domain, then f is a smooth map or function.


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The reader should note that Definition 2.17*** agrees with the notion of a C map of

functions from Calculus, i.e., if O Rm

, then f : O Rn

is C

as defined in Calculusif and only if it is smooth as a map between manifolds.In order to check if a function between manifolds is smooth, one does not have to check

every chart in a differential structure. It is enough to check one chart about each point asis proved in the following theorem.

Proposition 2.18***. Let f : Mm Nn be a function between manifolds. Furthersuppose that for each x M there are charts (Ux, x) about x for M and (Wf(x), f(x))about f(x) for N such that f(x) f

1x is C

on its domain. Then f is smooth.

Proof. Suppose that (U, ) and (W, ) are in the differential structure for M and N re-

spectively. We wish to show that f 1

is C

on its domain, (f1

(W) U).Take a point in the domain of f 1, say (x) for x f1(W) U. Then(f1(W) U f1(Wf(x)) Ux) is an open neighborhood of (x) in R

m, and on thisopen set,

f 1 = ( 1f(x)) (f(x) f

1x ) (x

1)

The compositions in parentheses are maps between real spaces. The first and third areC since (Wf(x), f(x)), (W, ) and (Ux, x), (U, ) are compatible pairs of charts. The

second composition is C by the hypothesis of the theorem. Hence, f 1 is C, soby Definition 2.17***, f is smooth.

Proposition 2.19***. The composition of smooth functions is a smooth function. Sup-pose f : Mm Nn and g : Nn Kk are smooth functions between manifolds. Theng f : Mm Kk is a smooth function.

Proof. Suppose p M. Suppose that (U, ), (V, ), and (W, ) are chart on M, N, Krespectively; and p U, f(p) V, and g(f(p)) W. The composite g f 1 isdefined on a neighborhood of p, and we need to show that this composite is C on someneighborhood of p. The composite function

g 1 f 1 = g f 1

on some neighborhood of p. Note that ( g 1) ( f 1) is the same functionand it is C since each of the functions in parentheses is C because both f and g aresmooth.

The notion of equivalence between differential manifolds is diffeomorphism.

Definition 2.20***. Suppose M and N are differential manifolds. If there is a smoothmap f : M N with a smooth inverse f1 : N M, then

(1) f is called a diffeomorphism, and

(2) M and N are diffeomorphic.

We make the following simple but useful observation.


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Proposition 2.21***. Suppose that M is an n-manifold, U M, : U Rn, and

(U) is open in Rn

. The pair(U, ) is a chart of the manifold M if and only if is adiffeomorphism.

Proposition 2.22***. If M is an n-manifold, then the set of diffeomorphisms of M is agroup under composition.

Let Diff(M) denote the group of diffeomorphisms of M.

Proof. The identity map on M, the map IM, is a smooth map, which we easily check. If(U.) is any coordinate chart for M, then IM

1 is just the identity on (U) Rn.By Proposition 2.18***, IM is smooth.

A diffeomorphism f has a smooth inverse by its definition.

The composition of two diffeomorphisms is again a diffeomorphism. If f and g arediffeomorphisms, the composition is smooth by Proposition 2.19***. The inverse of f gis g1 f1.

Finally note that the composition of functions is associative. Hence, Diff(M) i s agroup.

We now consider a method of constructing manifolds. Those students who have al-ready learned of covering spaces will recognize the construction, although no backgroundis assumed in these notes.

The construction involves a subgroup G of the diffeomorphisms of an n-manifold M.We can consider the quotient space M/G defined by the equivalence relation on M thatx y if and only if y = g(x) for some g G. Notice that is an equivalence relation so

that M/G makes sense as a topological space:(1) Ifx x since x = IM(x)(2) Ifx y then y = g(x) and x = g1(y), so y x(3) Ifx y and y z, then y = g(x) and z = h(y), so z = h g(x) and x z

We know that M/G is a topological space under the quotient topology and that thequotient map : M M/G is continuous. Can we guarantee it is a manifold? In general,the answer is no, but we do have the following theorem.

Theorem 2.23***. Suppose M is an n-manifold, andG is a finite subgroup of Diff(M).Suppose that G satisfies one of the following, either (a) or (b):

(a) If for some x M and g G, g(x) = x, then g is the identity, or(b) There is atlas A for M such that if (U, ) A,

(1) then (g(U), g1) A, and(2) h(U) g(U) = for all g, h G, g = h.

Then M/G is an n-manifold and the quotient map : M M/G is a smooth map.

The condition if for some x M and g G, g(x) = x, then g is the identity, says thatno g, other that the identity can fix a point. The group G is said to operate without fixedpoints. Before proving Theorem 2.23***, we first prove the following lemma.

Lemma 2.24***. Suppose M is an n-manifold, and G is a finite subgroup of Diff(M).Suppose that G satisfies the following property: If for some x M and g G, g(x) = x,then g is the identity. Then there is atlas A for M such that if (U, ) A,

(1) then (g(U), g1) A, and


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(2) h(U) g(U) = for all g, h G, g = h.

Proof. We first show that for every g G and x M there is an open set Og, a neighbor-hood ofg(x), such that

(1) if g, h G and g = h then Og Oh = .

Fix x M. For each pair g(x), h(x) there are open sets Ogh and neighborhood of g(x)and Ohg and neighborhood of h(x) such that Ogh Ohg = . These sets exist since M isHausdorff. Let Og =

hG Ogh. The intersection is finite, so Og is open. Given g, h G,

g = h, then Og Ogh and Oh Ohg, so

Og Oh Ogh Ohg = .

The next step is to show there is a neighborhood of x, call it O, such that

g(O) h(O) = for any g = h.

We have g(x) Og, so g1(Og) is a neighborhood of x. Let

O =gG

g1(Og).

Now, g(O) g(g1(Og)) = Og, and therefore, g(O) h(O) Og Oh = .

We now produce the atlas. Take a chart (Ux, x) about x with Ux O. Then (g(Ux), xg1) is a chart about g(x). Note that h(U) g(U) h(O) g(O) = . Hence, the atlas{(g(Ux), x g

1) | x M, g G} satisfies the required properties.

Proof of Theorem 2.23***. We first construct the coordinate charts. Take A an atlas thatsatisfies the two items in Lemma 2.24*** and take (U, ) A. The quotient map is acontinuous and open map. By the second item in Lemma 2.24***, is also one-to-oneon U. Therefore |U is a homeomorphism. Denote its inverse by iU : (U) U. Let = iU, then the pair ((U), ) is a coordinate chart for M/G.

We have constructed charts about any point and it remains to show that these chartsare compatible. Suppose z M/G and z (V) (U) where (V, ) A and ((V), )

is a chart ofM/G constructed as in the previous paragraph. We wish to show that

1 : ((U) (V)) ((U) (V))

is a diffeomorphism. Since it as an inverse (of the same form) we only have to show it isC. Now, for some y M, 1(z) = {g(y) | g G} a set of |G| points. There are twogroup elements g, h G with g(y) V and h(y) U. Let O be a neighborhood ofh(y) inhg1(V) U. A neighborhood of g(y) in V gh1(U) is gh1(O). The map 1 onthe open set (O) is

(1)

1 = ( iV)( iU)1

= (iV |U) 1

= (gh1) 1


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since gh1 : O gh1(O) is a diffeomorphism, so is the composite (1). This completes

the proof that M/G is a manifold.That is a smooth map is almost a tautology. Given x M take ((U), iU) a chart inA about x and ((U), ) the chart about (x) constructed above in the first paragraph.The map which we are required to show is C is 1 = (iU |U)

1 whichis just the identity on (U). Therefore is smooth.

We give two examples.

Example 2.25***. Real Projective Space RPn.

The n-sphere, Sn Rn+1 admits the action of the group Z2 = {1, 1}. Multiplicationby 1 is the identity which is smooth, Theorem 2.22***. We check -1 using the charts from

Example 2.9b***. Multiplication by -1 is a map

1 : Ui,+ Ui, and 1 : Ui, Ui,+.

Now, both i,+(1)1i, and i,(1)

1i,+ are the same as maps from B1(0) B1(0).

They are (y1, , yn) (y1, , yn) which is C on Rn. Hence, {1, 1} Diff(M).

Therefore, by Theorem 2.23***, Sn/Z2 is an n-manifold. It is called real projective space.Denote it RPn.

Let q : Sn RPn be the quotient map. Notice that if U Sn that is entirely in somehemisphere of Sn then q(U) can serve as a neighborhood for a chart since U U = .

Example 2.26***. Configuration Space.Imagine that n-particles are moving in Rm. These particles are ideal particles in that

they are points, i.e., they have no diameter. What are the possible arrangements? If theparticles are labeled, they we can write the arrangement as an n-tuple, (x1, x2, , xn).Since no two particles can occupy the same point in space, the manifold that describes sucharrangements is {(x1, x2, , xn) | xi = xj for i = j}, i.e., Rmn with several subspacesremoved. It is a manifold since it is an open subset of Rmn. It is ordered configurationspace. Denote it Cnm

If the n-particles are not labeled, then we can only know the location of the n pari-ciles and not which particle occupies which location, e.g., we cannot distinguish between

(x ,y ,z) and (y ,x ,z). The symmetric group Sn acts on Cnm by (x1, x2, , xn) =(x11, x12, , x1n). The space Cnm/Sn is configuration space.

We show that Cnm/Sn is an mn-manifold. We use Theorem 2.23***. The manifoldM is Cnm and the subgroup of the diffeomorphisms is Sn the permutation group. Eachpermutation is the restriction of a linear map on Rmn to Cnm, and so each permutationis C, i.e., smooth. The inverse of a permutation is again a permutation. Hence, Sn isa subgroup of Diff(Cnm). If fixes x = (x1, , xn), then xi = x(i) for i = 1, , n. If(i) = j and j = i, then xi = xj for an i = j. Therefore x is not in Cnm. If (i) = i forall i = 1, , n, then is the identity. We can now apply Theorem 2.23***. By Theorem2.23***, Cnm/Sn is an nm-manifold.

In physics, the term configuration space is used for to describe the space of physicalconfigurations of a mechanical system. Ifm = 3, then Example 2.26*** is the configurationspace of n particles in ordinary 3-dimensional space.


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ExercisesExercise 1***. Verify the calculations of Example 2.9a***. Show the two atlases givenfor Sn in Example 2.9a*** and Example 2.9b*** give the same differential structure andso may be merged.

Exercise 2***. S1 S1 is a 2-manifold, S2 S1 is a 3-manifold,and S2 S1 S1 is a4-manifold.

Of course these all follow from Proposition 2.13***. The reader should note, however,that there is an ambiguity in S2 S1 S1, is it (S2 S1) S1 or S2 (S1 S1)? Thereader should show that the atlases are compatible and so these are the same manifold.

There is also a second approach that is sometimes used to define smooth functions. Inthis approach, one first defines a smooth function for f : M R only. The statement ofthe next exercise would be a defintion in some textbooks, e.g., Warner and Helgason, butfor us, it is a proposition.

Exercise 3***. Show that a function f : Mm Nn between manifolds is smooth if andonly if for all open sets U N and all smooth functions g : U R, g f is smooth on itsdomain.

Exercise 4***. Consider R with the following three atlases:

(1) A1 = {f | f(x) = x}

(2) A2 = {f | f(x) = x3

}(3) A3 = {f | f(x) = x3 + x}

Which of these atlases determines the same differential structure. Which of the manifoldsare diffeomorphic?

Exercise 5***. Let M, N, and Q be manifolds.

(1) Show that the projections 1 : M N M and 2 : M N N are smooth.(2) Show that f : Q M N is smooth iffif is smooth for i = 1, 2.(3) Show forb N that the inclusion x (x, b) : M M N is smooth.

The following is a difficult exercise.

Exercise 6***. Prove that the set of all n n matrices of rank k (where k < n) is asmooth manifold. What is its dimension?

If this is too hard, then prove that the set of all n n matrices of rank 1 is a smoothmanifold of dimension 2n 1.


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Warmup Exercises, Chapter 2Exercise 1*. Suppose that {(Ui, ) : i I} is an atlas for M. Argue that a chart (V, )is compatible with the atlas if for each x V, there exists an open set W, x W Vand an ix I such that ix (|W)

1 and |W 1ix

are C.

Exercise 2*. Suppose that {(Ui, ) : i I} is an atlas for M, J I andiJUi = M.

Argue that {(Ui, ) : i I} is an atlas that generates the same differentiable structure onM.

Exercise 3*. Let : U Rn be a chart for a smooth manifold M and let V be anonempty open subset ofU. Argue that |V : V Rn is also a chart in the differentiable

structure of M.

Exercise 4*. Let U be a nonempty open subset of a manifold M. Show that the chartsof M with domain contained in U form a differentiable structure on U. Show that therestriction of any chart on M to U belongs to this differentiable structure.

Exercise 5*. Prove Proposition 2.21***.


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CHAPTER 3

SUBMANIFOLDS

One basic notion that was not addressed in Chapter 1 is the concept of containment:When one manifold is a subset on a second manifold. This notion is subtle in a mannerthe reader may have experienced in general topology. We give an example.

Example 3.1. Let Qd be the rational numbers with the discrete topology and R the

usual real numbers. The inclusion : Qd R is a one-to-one continuous map.Is Qd a subspace ofR? The usual answer is no, it is only a subspace if is a homeomor-

phism to its image, (Qd) = Q. This map is not a homeomorphism to its image since thetopology on Q, the subspace topology, is not discrete. the same issue arises in manifoldtheory. In fact Qd is a 0-dimensional manifold and is a smooth map. We, however, willnot refer to Qd as a submanifold ofR, i.e., submanifold is not quite the correct relationshipofQd to R. This relationship will be studied in Chapter 6*** (immersed submanifold). Forthe notion N is a submanifold ofM, we require that N inherits its differential structurefrom M. Some authors refer to this relationship as embedded submanifold.

We give the definition of submanifold in Definition 3.2***. At first glance, it may appear

to be overly restrictive, however, it turns out not to be the case. It is analogous to thenotion of subspace topology. This assertion is justified in a later chapter, Theorem 6.3***.

Definition 3.2***. Suppose m > n and write Rm = Rn Rmn. Let M be an m-manifold and N M. Suppose that for each x N there is a chart of M

(1) (U, ) centered at x such that 1(Rn {0}) = U N

Then N is an n-dimensional submanifold of M.

Charts that satisfy property (1) from Definition 3.2*** are called slice charts or slicecoordinate neighborhoods.

We next observe that if N M is an n-dimensional submanifold, then N is an n-manifold.

Proposition 3.3***. Suppose that N is an n-dimensional submanifold of them-manifoldM. Then N is an n-manifold and

A ={(Ux N, |UN) | (Ux, ) is a chart of M centered at x N such

that 1(Rn {0}) = U N}

is an atlas for N.

Proof. We first show that N is a topological manifold in the induced topology on N. The

topology on N is the subspace topology. Therefore, N is Hausdorff and second countable

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2 CHAPTER 3 SUBMANIFOLDS

since M is Hausdorff and second countable. Also since the topology on N is the subspace

topology, Ux N is an open set in N. The set (Ux N) = (Ux) (Rn

{0}), since(Ux, ) is a slice neighborhood. The map is a homeomorphism to its image, so |UxNis a homeomorphism to its image. Now, since there is a neighborhood Ux N around anyx N, the requirements of Definition 2.1*** are shown and N is a topological manifold.

To show that N is an n-manifold, we must show that the charts satisfy the compatibilitycondition of Definition 2.3*** item 2. Suppose (Uy, ) is another coordinate neighborhoodin the potential atlas A. Then

|UxN(|UyN))

1

|(UxUyN) = (1)|(UxUyN)

is C since it is the composition of the inclusion of Rn into Rm composed with 1,two C functions. This completes the proof.

Theorem 3.4***. Suppose O Rn+m is an open set andf : O Rm is aC map. Letq Rm and M = f1(q). IfDf(x) has rankm for allx M, then M is an n-dimensionalsubmanifold of Rn+m.

Proof. For each x M, there is a neighborhood of x, Ux Rn+m; an open set U1 R

n;an open set U2 R

m; and a diffeomorphism H : Ux U1 U2 such that H(M Ux) =U1 {0} as was laid out in the Rank Theorem. The pair (Ux, H) is a chart for the smoothmanifold Rn+m since H is a diffeormorphism, Proposition 2.21***. This chart satisfies therequired property given in Definition 3.2***. Hence, M is an n-dimensional submanifold

of Rn+m.

Example 3.5***. Sn is a submanifold of Rn+1.

Let l : Rn+1 R by l(x1, x2, , xn+1) =

x21 + x2n+1. Then S

n = l1(1). The

map l is C on Rn+1 \ 0. In fact, each partial derivative is a rational function ofx1, , xn+1, and l, i.e.,

lxi

= xil . Therefore the partial derivative of a rational func-tion of x1, , xn+1, l is another such function and l is C

. To check that the rank ofDvl(x) is one, it is enough to show that some directional derivative is not zero. Hence, forx Sn, we compute,

Dxl(x) =d

dtl(x + tx)|t=0

=d

dt(1 + t)

x21 + x

2n+1

t=0

=d

dt(1 + t)|t=0

= 1.

Since l has rank 1 on Sn, Theorem 3.4*** applies.

Proposition 3.3*** along with the Rank Theorem*** gives instructions for computingan atlas. The atlas essentially comes from the Implicit Function Theorem. The chartsinclude the atlas in Example 2.9b***. The reader should check this fact.


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CHAPTER 3 SUBMANIFOLDS 3

Example 3.6***. Let SL(n, R) = {A Matnn | detA = 1}. Then

(1) If A SL(n, R), then Ddet(A) has rank 1.(2) SL(n, R) is an (n2 1)-manifold.

First, det : Rn2

R is a polynomial (as shown in Example 2.12***) and so C. Toshow that Ddet(A) has rank 1 it is only necessary to show that some directional derivativeDBdet(A) is nonzero. We compute for A SL(n, R),

DAdet(A) =d

dtdet(A + tA)|t=0

=d

dt(1 + t)ndet(A)|t=0

=d

dt(1 + t)n|t=0

= n.

Hence item 1 is shown.To see item 2, we use Theorem 3.4***. The set GL(n, R) Rn

21R is an open set anddet : GL(n, R) R is a C map. The set det1(1) = SL(n, R), and Ddet(A) has rank 1for each A SL(n, R). Therefore, by Theorem 3.4***, SL(n, R) is an (n2 1)-manifold.

Example 3.7***. Let O(n, R) = {A Matnn | AAT = I} and let SO(n, R) = {A

Matnn

| AAT = I and detA = 1}. Also let f : Matnn

Matnn

by f(A) = AAT

(1) f(A) Symnn and f is C.

(2) If A O(n, R), then Df(A) has rankn(n + 1)

2.

(3) O(n, R) and SO(n, R) aren(n 1)

2-manifolds.

First note that AAT = I implies that (det(A))2 = 1 so A is invertible.The map f is a composition the identity cross the transpose and multiplication. These

maps are C maps by Proposition 2.13***. Since f is the composition of C maps, itis C. Since f(A) = AAT and (AAT)T = AAT, f(A) Symnn. This argument shows

item 1.To show item 2, it is enough to show that Df(A) is surjective since dimSymnn =

n(n + 1)

2. Now,

Df(A)(M) =d

dt(A + tM)(A + tM)T

t=0

= AMT + M AT

= AMT + (AMT)T

Hence Df(A) is a composition of two maps

Matnn Matnn

M M ATand

Matnn Symnn

X X+ XT


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The first map is onto if AT is invertible, which it is if A f1(I). The second map is

onto, since, if Y Symnn then Y = YT

and Y =1

2Y + (1

2Y)T

, i.e., if X =1

2Y, thenY = X+ XT. Therefore, Df(A) is a surjection and has rank

n(n + 1)

2.

The third item follows from Theorem 3.4***. Since f1(I) = O(n, R), O(n, R) is a

manifold of dimension n2 n(n + 1)

2=

n(n 1)

2. The determinant function is continuous

on Matnn and O(n, R) Matnn. If AAT = I, then detA = 1. Therefore SO(n, R)

is an open subset of O(n, R) and hence a manifold of dimensionn(n 1)

2, by Proposition

2.11***.The next example requires notation and a lemma.

Lemma 3.8***. Suppose that n is even and let J be the n n matrix

J =

0n/2n/2 In/2n/2In/2n/2 0n/2n/2

Then JT = J = J1.

Proof. Just compute the transpose, the negative, and the inverse.

.

Example 3.9***. Suppose that n is even and let Sp(n, R) = {A Matnn | AJAT = J}.

Also let f : Matnn Matnn by f(A) = AJ AT

(1) f(A) Skewnn and f is C.

(2) If A Sp(n, R), then Df(A) has rankn(n 1)

2.

(3) Sp(n, R) is ann(n + 1)

2-manifold.

First note that AJAT = J implies that (det(A))2 = 1 so A is invertible.The map f is a composition the identity cross the transpose and multiplication. These

maps are C maps by Proposition 2.13***. Since f is the composition of C maps, itis C. Since f(A) = AJ AT and (AJAT)T = AJTAT = AJAT by Lemma 3.8***,

f(A) Skewnn. This argument shows item 1.To show item 2, it is enough to show that Df(A) is surjective since dimSkewnn =

n(n 1)

2. Now,

Df(A)(M) =d

dt(A + tM)J(A + tM)T

t=0

= AJMT + M JAT

= AJMT (AJ MT)T

Hence Df(A) is a composition of two maps

Matnn Matnn

M M JATand

Matnn Symnn

X X XT


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The first map is onto if JAT is invertible, which it is if A f1(J). The second map is

onto, since, if Y Skewnn then Y = YT

and Y =1

2Y (1

2Y)T

, i.e., if X =1

2Y, thenY = X XT. Therefore, Df(A) is a surjection and has rank

n(n 1)

2.

The third item follows from Theorem 3.4***. Since f1(I) = Sp(n, R), Sp(n, R) is a

manifold of dimension n2 n(n 1)

2=

n(n + 1)

2.

Proposition 3.10***. Suppose that N is an n-dimensional submanifold of them-manifoldM. Suppose that U is an open neighborhood of N in M and g : U P is a smooth mapto a manifold P. Then, g|N : N P is a smooth map.

Proof. Suppose x N is an arbitrary point and (O, ) is a chart of P about g(x). By

Proposition 2.18***, it is enough to check that there is a chart of N, (V, ), with x Vand g|M

1 is C. By the definition of submanifold, Definition 3.2***, we can alwaysfind a product chart (of M) centered at x. Suppose the chart is (W, ), so we can takeV = W N and = |WN. Then

g|M 1 = g|M (

1)|Rn{0}(W),

the restriction of a C function of m variables to its first n variables by setting the lastm n variables to zero. This map is C.

Example 3.11***.

Suppose that M is either of O(n), SO(n), SL(n, R), or Sp(n, R). Then multiplication

M M M

(A, B) AB

and inverseM M

A A1

are smooth maps.

Just as M is a submanifold of GL(n, R), M M is also a submanifold of GL(n, R) GL(n, R). Both GL(n, R) and GL(n, R) GL(n, R) are open subsets of real spaces. Mul-tiplication and inverse are C functions as explained in Example 2.13***. By Proposition3.10***, these maps are smooth maps between manifolds.


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Exercises

Exercise 1***. Suppose O Rn+m is an open set, f : O Rm is a C map, q Rm,and M = f1(q). Further suppose U Rs+k is an open set, g : U Rk is a C map,p Rk, and n = g1(p). If Df(x) has rank m for all x M and Dg(x) has rank k forall x N, then show that M and N are manifolds; and, (f g)1(q, p) is the manifoldM N.

Exercise 2***. Show that the atlas for Sn

constructed in Example 3.5*** does includethe 2(n + 1) charts constructed in Example 2.9b***.

Exercise 3***. Let M be defined by

M = {(w,x ,y ,z) R4 | w3 + x2 + y3 + z2 = 0 and yez = wex + 2}.

Show that M is a two dimensional submanifold of R4.

The reader may wish to compare the following Exercise to Example 2.10***.

Exercise 4***. Suppose U1 Rn and U2 R

m are open sets. If f : U1 U2 is a C

function, then show the graph of f,

Gf = {(x, y) Rn+m | y = f(x)}

is a submanifold of Rn+m.

In the exercise below, we use that R4 = C2 under the isomorphism of real vector spaces,(w,x ,y ,z) (w + xi,y + zi).

Exercise 5***. a. Suppose p and q are relatively prime integers. Let = e2ip . Show

that : S3 S3 by (z1, z2) = (z1, qz2) is a smooth map. Let G = {,

2, , p}.Show that G is a subgroup of Diff(S3).

b. Show that S3 is {(z1, z2) C2 | z1z1 + z2z2 = 1}, and a submanifold of C2.

c. Show that S3/G is a 3-manifold. It is denoted L(p, q) and is called a lens space.


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Warmup Exercises, Chapter 3

Exercise 1*. Let F : R2 R be defined by

F(x, y) = x3 + xy + y3.

Which level sets are embedded submanifolds of R.

Exercise 2*. Consider the map F : R4 R2 defined by

F(x1, x2, x3, x4) = (x2

1

+ x2, x2

1

+ x2

2

+ x2

3

+ x2

4

+ x2).

Show that F restricted to M = F1(0, 1) has rank 2 at every point of M.

Exercise 3*. Which level sets of

f(x1, . . . , xn+1) = x1x2 xn+1 + 1

are submanifolds, according to Theorem 3.4***.


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CHAPTER 4

PARTITIONS OF UNITY

AND SMOOTH FUNCTIONS

In this section, we construct a technical device for extending some local constructionsto global constructions. It is called a partition of unity. We also use the opportunity todiscuss C functions. We begin with examples of C functions on R and Rn. Some of

these are required for the construction of partitions of unity. At the end of the chapter, wereturn to C funcitons and apply our new techniques. The first examples we construct,Examples 4.1abc***, are standard and we follow Warner.

Example 4.1a***. A function on R which is C but not analytic. Let

f(t) =

1

e1/tif t > 0

0 if t 0.

We show that for each n = 0, 1, 2, 3, there is a polynomial pn such that

f(n)(t) =

pn(1/t) 1e1/t if t > 00 if t 0.

We use mathematical induction to show the claim. The claim is true for n = 0 as f(0) = f.Here p0(x) = 1. For k > 0 we separately handle t > 0, t < 0 and t = 0. If f

(k)(t) =pk(1/t)

1e1/t

for t (0, ), then

f(k+1)(t) = pk(1/t)(1

t2)

1

e1/t+pk(1/t)(

1

t2)

1

e1/t

and pk+1(x) = x

2

p

k(x) + x

2

pk(x). For t (, 0), f is zero and so are its derivativesof all orders. At t = 0 we use the definition of the derivative.

limh0+

f(k)(h) f(k)(0)

h= lim

h0+

pk(1/h)1

e1/h

h

= limx

xpk(x)

ex

= 0.

The substitution was x = 1/h, and the last limit was a polynomial divided by an exponen-

tial: a standard LHopitals Theorem example from Calculus. The limit, limh0f(k)(h)f(k)(0)

h

is zero as the numerator is identically zero.

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2 CHAPTER 4 PARTITIONS OF UNITY AND SMOOTH FUNCTIONS

Example 4.1b***. Let

g(t) =f(t)

f(t) + f(1 t)=

0 if t 0

1

1 + e1/t

e1/(1t)

if 0 < t < 1

1 if t 1.

Then g is C on R and g is strictly increasing on [0, 1].

The function g is C since it is a quotient of C functions and the denominator is

never 0. Ift 0, then f(t) = 0 so g(t) = 0. If t 1, then f(1 t) = 0 so g(t) = f(t)f(t) = 1.

Recall that f

(t) =1t2 f(t) for t > 0, and so,

d f(t)dt = f

(1 t). Now, for t (0, 1),

g(t) =f(t)(f(t) + f(1 t)) f(t)(f(t) f(1 t))

(f(t) + f(1 t))2

=f(t)f(1 t) + f(t)f(1 t)

(f(t) + f(1 t))2

=f(t)f(1 t)

(f(t) + f(1 t))2

1

t2+

1

(1 t)2

This expression is positive on (0, 1). Hence g is strictly increasing on [0, 1]. The graph of

g is as follows.

Example 4.1c***. The bump function on (2, 2). Let

h(t) = g(t + 2)g(2 t) =

0 if t 2

g(t + 2) if 2 < t < 1

1 if 1 t 1

g(2 t) if 1 < t < 2

0 if 2 t

The function h is C on all of R, h(x) = 0 if x (, 2] [2, ), h(x) = 1 ifx [1, 1], h is strictly increasing on [2, 1], and strictly decreasing on [1, 2].

Its graph is shown below.Let C(r) denote the open cube in Rn which is {(x1, , xn)|xi (r, r) for i = 1, 2, 3, , n}

and let C(r) be its closure.

Example 4.2***. The bump function on C(2) Rn. Let

b(x1, , xn) = ni=1b(xi).

Then b(x1, , xn) = 0 on the complement of C(2), b(x1, , xn) = 1 for (x1, , xn) inthe closure of C(1), and 0 < b(x1, , xn) < 1 otherwise.


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CHAPTER 4 PARTITIONS OF UNITY AND SMOOTH FUNCTIONS 3

Definition 4.3***. If U = {U| A} is an open cover of a manifold M, then a subset

of U which is also a cover is called a subcover.Definition 4.4***. If U = {U| A} is an open cover of a manifold M, then the opencover V = {V| } is a refinement if for all there is an A such that V U.

Definition 4.5***. A collection of subsets U = {U| A} of a manifold M is calledlocally finite, if for all m M there is an neighborhood O of m with U O = for onlya finite subset of A.

Definition 4.6***. A partition of unity on a manifold M is a collection of smooth func-tions {i : M R | i I} such that

(1) { the support of i | i I} is locally finite

(2) i(p) 0 for all p M, i I, and,(3)

iIi(p) = 1 for all p M.

Note that the sum is finite for each p.

Definition 4.7***. The partition of unity on a manifold M {i | i I} is subordinateto the open cover U = {U| A} if for all i I there is an A such that the supportof i is in U.

Lemma 4.8***. Suppose M is a connected manifold. Then there is a sequence of opensets Oi such that

(1) Oi is compact(2) Oi Oi+1

(3)i=1

Oi = M

Proof. Take a countable basis for the topology ofM (as M is second countable) and for eachx M pick a compact set Kx that contains x in its interior (as M is locally compact). SinceM is Hausdorff, we obtain another basis for the topology ofM by keeping only those basiselements which are in some Kx. We now have a countable basis U = {Ui|i = 1, 2, 3, }such that if Ui U then Ui is compact.

Let O1 = U1. Each of the other open sets will have the form Ok = jki=1 Ui. Suppose wehave constructed Ok. We show how to construct Ok+1. Since Ok is compact let jk+1 bethe smallest counting number with Ok

jk+1i=1 Ui. We establish the required properties.

Since Ok =jki=1 Ui is a finite union of compact spaces, it is compact. By construction,

Ok Ok+1. If Ok Ok then Ok = M as Ok is open and closed, otherwise jk+1 > jk.

Thereforeki=1 Ui Ok so

i=1

Oi = M, and (3) follows.

Proposition 4.9***. Suppose U = {U| A} is a basis for the manifold M with Ucompact for all A. Suppose W = {W | B} is any open cover. Then there is acountable locally finite refinement of W, V = {Vi|i I}, with Vi U for all i I.

Proof. If the manifold has more than one component then we separately handle eachcomponent, hence we assume that M is connected. By the previous lemma, there is a


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collection of open sets {Oi|i = 1, 2, 3, } which satisfy the conditions of the lemma.

Now

Oi+1 \ Oi is compact and contained in the open set Oi+2 \

Oi1. Note that O ={Oi+2 \ Oi1, O4 | i = 3, 4, } is an open cover of M.We construct V by reducing U in two steps. Let U be the set of all U U such that

there is a with U W and that U Oi+2 \ Oi1 or O4. The set U is a basis sinceO and W are open covers and M is Hausdorff, i.e., for each x M, U contains a nbhdbasis. Take a finite subset of U each of which is in O4 and covers O3, a compact set. Foreach i > 2 take finite subsets ofU each of which is in Oi+2 \ Oi1 and covers Oi+1 \ Oi, acompact set. The union of these various finite collections V = {Vj |j I} is locally finitesince an open set in the ith collection can only meet open sets from the (i 2)nd collectionup through the (i + 2)nd. These are each finite collections. The set V is a countable unionof finite sets and so countable. The cover V is subordinate to W since U is subordinateto W.

Lemma 4.10***. SupposeM is a manifold. Then there is a basis {U| A} such that

(1) U is compact and(2) For each A there is a smooth function : M R such that (x) = 0 if

x / U and (x) > 0 if x U.

Notice that the function guaranteed in the lemma cannot be analytic but must be C.For example on R the support of is compact and so the function is zero on (m, ) for

some m. Hence ifx > m, then (n) (x) = 0, for n = 1, 2, 3, .

Proof. First note that R(m) = {1(C(2))|(U, )is a chart centered at m and C(3) (U)} is a neighborhood basis at m. If fact, if (U, ) is any chart centered at m thecharts (U,k) for k large suffice. Now, we produce the function. If R R(m), then thereis a C function R : M R with R(x) = 0 ifx / R and R(x) > 0 if x R. If (U, )is a chart with C(3) (U) and R = 1(C(2)) R(m), then take

R(x) =

b (x) if x U

0 if x / R

Here b is the C

function we produced onRn

with b(x) > 0 for x C(2) and b(x) = 0for x / C(2). The function R is smooth since b and 0 agree on the overlap of theirdomains, U \ R an open set.

Theorem 4.12***. If M is a manifold and W is any open cover, then M admits acountable partition of unity subordinate to the cover W with the support of each functioncompact.

Proof. Apply Proposition 4.9*** to the basis constructed in Lemma 4.10***. We obtaina locally finite collection {Ui|i = 1, 2, 3, } with i : M R as in the lemma. Let

(x) =

i=0

i(x). Recall that for each x there is an open set O with x O and i = 0 on

O for all but finitely many i. The sum is finite for each x. It is the fact that locally finiterequires an open set about each x that meets only a finite number ofUi (say i = 1, , m)


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CHAPTER 4 PARTITIONS OF UNITY AND SMOOTH FUNCTIONS 5

that gives the C differentiablity, since on the open set O the function is a finite sum

of smooth functions: the composition of

M(1, ,m)

Rm

R.

If only each x were in a finite number of Ui, then we could only guarantee well-defined

but not smooth. Let i =i

. Then {i|i = 1, 2, 3, } form a partition of unity with the

support of i being Ui.

It is interesting to note that there is no special field to study the zeros of C functions.Algebraic geometry studies the sets which are zeros of polynomials, and the nature of thezeros of analytic functions is also studied, but the sets which are zeros of smooth functions

does not give a new area of study. The reason is:

Theorem 4.13***. If X M is closed, then there is a smooth function f : M R withf(x) = 0 if and only if x X.

Proof. Let N = M \ X. N is an open subset of M and so a manifold. Let {U| A}be the basis of M produced in the lemma. Let U = {U| A and U N}. Thecollection U is a basis for N with each U compact for U U. We apply Proposition4.9*** to the manifold N with W = {N} to get a countable locally finite subcollection of

U, {Ui|i = 1, 2, 3, }. Each Ui is equipped with i : M R whose support is exactly

Ui, as in Lemma 4.10***. Let f =

i=1

i. The function f is C and is nonzero exacty on

i=1Ui = N

Theorem 4.13*** demonstrates a dramatic array of possible behavior of C functionscompared to analytic functions. To observe the restricted nature of the zeros of analyticfuctions, we consider functions on the real line.

Theorem 4.14***. Suppose f : R R and X = {x | f(x) = 0}.

(1) If f is a polynomial, then X is a finite set.(2) If f is an analytic function, then X is a discrete set.(3) If f is a C function, then X can be an arbitrary closed set.

Proof. An n-th degree polynomial has at most n zeros, which shows the first item. Thethird item follows from Theorem 4.13***.

To show the second item we suppose that the zeros off are not discrete and show thatimplies f is identically zero. Suppose that pm X for m = 1, 2, 3, and lim

mpm = p.

Now,

f(x) =n=0

1

n!f(n)(p)(x p)n.

In order to show f is the zero function, it is enough to show that f(n)(p) = 0 for all wholenumbers n.

We show that f(n)(p) = 0 for all whole numbers n by mathematical induction. Wefirst observe it is true for n = 0, f(0) = f. By continuity, lim

mf(pm) = f(p) and each


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f(pm) = 0, and therefore, f(p) = 0. We now assume that f(k)(p) = 0 for k < n and show

this implies f(n)

(p) = 0. First note that

limm

n!f(pm)

(pm p)n= lim

mn!

0

(p pm)n= 0.

Now compute the same limit using LHospitals rule,

0 = limm

n!f(pm)

(pm p)n= lim

xpn!

f(x)

((x p)n

= limxp

n!f(x)

n(x p)(

n 1)...

= limxp

n!f(n1)(x)

n!(x p).

By the induction hypothesis, these limits are all of the indeterminate form 00

. One lastapplication of L Hopitals rule yields,

limm

n!f(pm)

(pm p)n

= limxp

n!fn(x)

n!

= f(n)(p)

the last equality by continuity of the n-th dervative. Therefore f(n)(p) = 0 for all wholenumbers n, and f is identically zero.

Exercises

Exercise 1***. Show that the bump function b on Rn has the following property: ifb(x) = 0, then all of the derivatives of b at x are also zero. Show that if X Rn is aclosed set, then there is a function f : Rn R such that f(x) = 0 if and only if x X

and all partials of f vanish on X.Exercise 2***. Assume the following version of the Stone-Weirerstrass Theorem: IfK Rn is compact and f : K R is continuous, then given any > 0 there is a polynomialfunction g such that | g|K (x) f(x)| < for all x K.

Prove that is M is a smooth manifold and f : M R is continuous, then given any > 0 there is a C function g : M Rn such that |f(x) g(x)| < for all x M.


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CHAPTER 5

TANGENT VECTORS

In Rn tangent vectors can be viewed from two perspectives

(1) they capture the infinitesimal movement along a path, the direction, and(2) they operate on functions by directional derivatives.

The first viewpoint is more familiar as a conceptual viewpoint from Calculus. If a point

moves so that its position at time t is (t), then its velocity vector at (0) is

(0), a tangentvector. Because of the conceptual familiarity, we will begin with the first viewpoint,although there are technical difficulties to overcome. The second interpretation will bederived as a theorem. The second viewpoint is easier to generalize to a manifold. Forinstance, operators already form a vector space. It is the second viewpoint that ultimatelyplays the more important role.

Suppose M is an n-manifold. If m M, then we define a tangent vector at m asan equivalence class of paths through m. Equivalent paths will have the same derivativevector at m and so represent a tangent vector. The set of all tangent vectors at m forms thetangent space. The description and notation of tangent vectors in Rn from the advanced

Calculus setting and in the present setting is discussed in Remark 5.9***.Definition 5.1***. Suppose M is a manifold. A path is a smooth map : (, ) M,where > 0.

As was mentioned, if M = Rn, then (0) is the velocity vector at (0). We also recall,from advanced Calculus, the relationship between the derivative map and the directionalderivative,

(1) D(0)(1) = D1(0) = (0)

Definition 5.2***. Suppose M is a manifold and m M A tangent vector at m is an

equivalence class of paths with (0) = m. Let (U, ) be a coordinate chart centered atm, two paths and are equivalent if

d (t)

dt

t=0

=d (t)

dt

t=0

.

Denote the equivalence class of a path by []. We can picture [] as the velocity vectorat (0).

We next observe that the equivalence class doesnt depend on the specific choice ofa coordinate chart. If (W, ) is another coordinate neighborhood centered at m, then = 1 , and, we use formula (1),

d (t)

dtt=0

= D( 1)((m)) D( )(0)(1).

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2 CHAPTER 5 TANGENT VECTORS

The diffeomorphisms 1 and are maps between neighborhoods in real vector

spaces, so

d (t)

dt

t=0

=d (t)

dt

t=0

if and only ifd (t)

dt

t=0

=d (t)

dt

t=0

.

Therefore the notion of tangent vector is independent of the coordinate neighborhood. If : (, ) M is a path in M with (0) = m, then [] is a tangent vector to M at m andis represented by the path . Consistent with the notation for Rn, we can denote [] by(0).

Let T Mm denote the set of tangent vectors to M at m. Other common notations areMm and TmM.

Theorem 5.3***. Suppose M, N, and R are manifolds.

(1) If : M N is a smooth map between manifolds and m M then there is aninduced map m : T Mm T N(m).

(2) If : N R is another smooth map between manifolds then ( )m = (m) m. This formula is called the chain rule.

(3) If : M M is the identity then m : T Mm T Mm is the identity. If : M N is a diffeomorphism and m M then m is 1-1 and onto.

(4) T Mm is a vector space of dimension n, the dimension of M, and the induced maps

are linear.

The induced map m is defined by

m([]) = [ ].

Notice that if M = Rm, N = Rn, then we have a natural way to identify the tangentspace and the map . We have coordinates on the tangent space so that

[ ] =d (t)

dtt=0

and

m([]) = D(m)((0)).

The induced map m is also commonly denoted T or d. These results follow forneighborhoods in manifolds since these are manifolds too. Also note that if there is aneighborhood U of m M and |

Uis a diffeomorphism onto a neighborhood of (m) then

m is an isomorphism.

Proof.

(1) If : M N is a smooth map and m M then there is an induced mapm : T Mm T N(m) defined by m([]) = [ ]. We need to show this map is


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CHAPTER 5 TANGENT VECTORS 3

well-defined. Take charts (U, ) on N centered on (m) and (W, ) on M centered

on m. If [] = [], then

d (t)

dt

t=0

=d (t)

dt

t=0

( 1)(d (t)

dt

t=0

) = ( 1)(d (t)

dt

t=0

)

d

dt( 1 )(t)

t=0

=d

dt( 1 )(t)

t=0

d

dt

( )(t)t=0

=d

dt

( )(t)t=0

so m is well defined on equivalence classes.(2) If : M N and : N R are smooth maps, then ( )m([]) = [ ] =

(m)([ ]) = (m) m([]).

(3) IMm([]) = [IM ] = []. If 1 = IM then (

1) = IM = ITMm .Also, if 1 = IM, then (

1) = ITMm . Therefore is a bijection and()

1 = (1).(4) Let (U, ) be a coordinate neighborhood centered at m. We first show that TRn

0

is an n-dimensional vector space. Since Rn requires no coordinate neighborhood

(i.e., it is itself), [] is equivalent to [] if and only if

(0) =

(0): two pathsare equivalent if they have the same derivative vector in Rn. Every vector v isrealized by a path v, v(t) = tv. This identification gives TR

n0

the vector spacestructure. We show that the linear structure is well defined on T Mm. The linearstructure on T Mm is induced by the structure on TR

n0

(where [] + k[] = [ + k]and induced maps are linear) via the coordinate maps. If (V, ) is another chartcentered at m, then the structure defined by and agree since ( 1) is anisomorphism and ( 1) = .

We can give explicit representatives for linear combinations of paths in the tangent spaceT Mm. In the notation of the proof of Theorem 5.3*** part 4,

k[] + c[] = [1(k + c )]

Note that the coordinate chart serves to move the paths into Rn where addition andmultiplication makes sense.

Before we turn to the second interpretation of a tangent vector as a directional derivative,we pause for a philosophical comment. We first learn of functions in our grade schooleducation . We learn to speak of the function as a whole or its value at particular points.Nevertheless, the derivative at a point does not depend on the whole function nor is it

determined by the value at a single point. The derivative requires some open set about apoint but any open set will do. If M is a manifold and m M, then let Gm be the set offunctions defined on some open neighborhood of m.


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Definition 5.6***. A function : Gm R is called a derivation if for every f, g Gm

and a, b R,(1) (af + bg) = a(f) + b(g) and(2) (f g) = (f)g(m) + f(m)(g)

Denote the space of derivations by D. The product rule which occurs in the definitionis called the Leibniz rule, just as it is in Calculus.

Proposition 5.7***. Elements of T Mm operate as derivations on Gm. In fact there is alinear map : T Mm D given by v v.

The theorem is straightforward if the manifold is Rn. If v TRnx

, then the derivationv is the directional derivative in the direction v, i.e., v(f) = Df(x)(v). On a manifoldthe argument is really the same, but more technical as the directions are more difficult torepresent. We will see in Theorem 5.8*** that the derivations are exactly the directionalderivatives.

Proof. If : ((, ), {0}) (M, {m}) represents v then define v(f) =df (t)

dt

t=0

.

The fact that v is a linear functional and the Leibniz rule follow from these properties ofthe derivative.

To show that is a linear map requires calculation. Suppose (U, ) is a coordinate chart

centered at m. If [] and [] are equivalence classes that represent tangent vectors in T Mmand c, k R, then 1((k(t) + c(t))) represents k[] + c[]. Hence,

k[]+c[](f) =df(1((k(t) + c(t))))

dt

t=0

= f1

d(k(t) + c(t))

dt

t=0

= f1

k

d((t)

dt

t=0

+ cd((t))

dt

t=0

= kf1

d(t)dt

t=0

+ cf1

d(t)

dt

t=0

= kdf(1(((t))))

dt

t=0

+ cdf(1(((t))))

dt

t=0

= kdf(((t)))

dt

t=0

+ cdf(((t)))

dt

t=0

= k[](f) + c[](f)

Lines 3 and 4 respectively follow from the linearity of the derivative and the total derivative

map. Therefore is linear.

The second interpretation of tangent vectors is given in the following Theorem.


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Theorem 5.8***. The linear map : T Mm D given by v v is an isomorphism.

The elements of T Mm are the derivations on Gm.Proof. We first note two properties on derivations.

(1) If f(m) = g(m) = 0, then (f g) = 0

Since (f g) = f(m)(g) + g(m)(f) = 0 + 0.

(2) If k is a constant, then (k) = 0

Since (k) = k(1) = k((1) + (1)) = 2k(1), (k) = 2(k) and (k) = 0.We now observe that is one-to-one. Let (U, ) be a coordinate chart centered at m.

Suppose v = 0 is a tangent vector. We will show that v = 0. Let (v) = w1 Rn. Note

that w1 = 0. Then [1

(tw1)] = v where t is the real variable. Let w1, , wn be a basisfor Rn and (n

i=1 aiwi) = a1. Then

v( ) = [1(tw1)]( )

=d((1(tw1)))

dt

t=0

=dtw1

dt

t=0

= w1.

Next we argue that is onto. Let (U, ) be a coordinate chart centered at m and let

ei for i = 1, , n be the standard basis for Rn. We consider the path t 1(tei) andcompute some useful values of , i.e., the partial derivatives.

[1(tei)](f) =df 1(tei)

dt

t=0

=f 1

xi

0

Let xi(a1, , an) = ai. Suppose d is any derivation. We will need to name certain values.Let d(xi ) = ai. These are just fixed numbers. Suppose f is C

on a neighborhood of

m. Taylors Theorem says that for p in a neighborhood of0 Rn,

f 1(p) = f 1(0) +ni=1

f 1

xi

0

xi(p) +n

i,j=1

Rij(p)xi(p)xj(p)

where Rij(p) =10

(t 1) 2f1

xixj

tp

dt are C functions. So,

f = f(m) +ni=1

f 1

xi

0

xi +n

i,j=1

(Rij ) (xi ) (xj ).

We now apply d. By (2), d(f(m)) = 0. Since xj (m) = 0, the terms d((Rij ) (xi )

(xj)) = 0 by (1). Also, d( f

1

xi0 x

i) = ai[1(tei)](f). Hence, d = P

ni=1

ai[1(tei)],and is onto.


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Remark 5.9***. Tangent vectors to points in Rn.

The usual coordinates on Rn give rise to standard coordinates on TpRn. Let ei =(0, , 0, 1, 0, , 0) with the only nonzero entry in the i-th spot. The path in Rn definedby i(t) = tei + p is a path with i(0) = p. Its equivalence class [i] is a vector in TpR

n

and we denote it

xi

p

. In Advanced Calculus, the ordered basis

x1

p

, ,

xn

p

is

the usual basis in which the Jacobian matrix is usually written and sets up a naturalisomorphism TpR

n = Rn. The reader should notice that the isomorphism is only naturalbecause Rn has a natual basis and is not just an abstract n-dimensional vector space.If is a path in Rn, then (0) T(0)R

n via this isomorphism. This notation is alsoconsistant with the operator notation (the second interpretation) since,

xi

p

(f) = [f i]

=d

dtf(tei + p)

t=0

=f

xi

x=p

Rn = TpRn

In the first line, the tangent vector

xip

operates via the second interpretation on the

function f.

Example 5.10***. T Mx for M an n-dimensional submanifold of Rk.

Suppose M Rk is a submanifold and i : M Rk is the inclusion. Take (Ux, ) aslice coordinate neighborhood system for Rk centered at x as specified in the definitionof a submanifold, Definition 3.2***, : Ux U1 U2. Under the natural coordinates ofTRkx

= Rk, T Mx = (U1 {0}) Rk and ix has rank n.

To see these facts, note that i (|UxM)1 : U1 {0} U1 U2 is the inclusion.

So, rank(i) = rank(( i |UxM)) = n. Under the identification TRkx

= Rk, x(Rn

{0}) = D(x)(Rn{0}) Rk. This is the usual picture of the tangent space as a subspaceof Rk (i.e., shifted to the origin) that is taught in advanced Calculus.

Example 5.11***. T Snx for Sn Rn+1, the n-sphere.

This is a special case of Example 5.10***. Suppose (x1, , xn+1) Sn, i.e.,

n+1i=1

x2i =

1. One of the xi must be nonzero, we assume that xn+1 > 0. The other cases areanalogous. The inclusion from the Implicit Function Theorem is |Rn(x1, , xn) =

(x1, , xn,

1 n

i=1 x2i ) so

D|Rn(x1, , xn)(v1, , vn) = (v1, , vn,

ni=1 xivi

1 n

i=1x2i )

).


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Since xn+1 > 0, xn+1 = 1

ni=1 x

2i ) and the tangent space is

T(x1, ,xn+1)Sn = {(v1, , vn,

ni=1 xivi

xn+1) | vi R}

= {(w1, , wn+1) |n+1i=1

wixi = 0}

Example 5.12***. Recall that O(n) Matnn = Rn2 is a submanifold of dimension

n(n1)2 which was shown in Example 3.7***. Then, we claim,

X TAO(n) Matnn

if and only if XA1 is skew.

This computation is a continuation of Example 3.7***. Suppose A O(n). SinceO(n) = f1(I), TAO(n) Ker(Df(A)). The dimension of the kernel and the dimension

of TAO(n) are bothn(n1)

2. Therefore TAO(n) = Ker(Df(A)). It is enough to show that

Ker(Df(A)) {X | XA1 is skew} since the dimension of {X | XA1 is skew} is the

dimension of Skewnn =n(n1)

2 (from Example 2.8d***). So it is enough to show that

XA1 is skew.Again, from Example 3.7***, Df(A)(X) = AXT + XAT. If Df(A)(X) = 0, then

AXT = XAT. Since A O(n), A1 = AT. So,

(XA1)T = (XAT)T = AXT = XAT = XA1

Therefore XA1 is skew.

Example 5.13***. Recall that Sp(n, R) Matnn = Rn2 is a submanifold of dimension

n(n+1)2

which was shown in Example 3.9***. Then, we claim,

X TASp(n, R) Matnn

if and only if J XA1 is symmetric.

This computation is a continuation of Example 3.9***. Suppose A Sp(n, R).Since Sp(n, R) = f1(J), T

ASp(n, R) Ker(Df(A)). The dimension of the kernel and

the dimension of TASp(n, R) are bothn(n+1)

2. Therefore TASp(n, R) = Ker(Df(A)). It

is enough to show that Ker(Df(A)) {X | J XA1 is symmetric} since the dimension

of {X | J XA1 is symmetric } is the dimension of Symnn =n(n+1)

2(from Example

2.8c***). So it is enough to show that J XA1 is symmetric.Again, from Example 3.9***, Df(A)(X) = AJ XT + XJ AT. If Df(A)(X) = 0, then

AJ XT = XJ AT. Since A Sp(n, R), A1 = J ATJT. So,

(J XA1)T = (J X J ATJT)T = (JAJXTJT)T = J XJTATJT

= J X J ATJTas JT = J by Lemma 3.8***

= J XA1

Therefore XA1 is symmetric.


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Remark 5.14***. Notation for Tangent vectors

The space Rn comes equipped with a canonical basis e1, , en which allows us to picka canonical basis for TRnx . For an n-manifold M, T Mp doesnt have a natural basis. Wecan give coordinates on T Mp in terms of a chart. Suppose that (U, ) is a chart for aneighborhood of p U M. Write = (1, , n) in terms of the coordinates on R

n.Hence, i = xi . We can import the coordinates TR

n(p). Let

i

p

= 1

(

xi

(p)

)

As a path

ip is the equivalence class of

1

(tei + (p)). As an operator,

i

p

(f) =f 1

xi

(p)

.


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ExercisesExercise 1***. Suppose F : R4 R2 by

F((w , x , y , z)) = (wxyz, x2y2).

ComputeF and be explicit in exhibiting the bases in the notation used in Remark 5.9***.Where is F singular?

The reader may wish to review Example 2.10** and Exercise 4*** from chapter 3 forthe following exercise.

Exercise 2***. Let g((x, y)) = x2 + y2 and h((x, y)) = x3 + y2. Denote byGg and Ghthe graphs of g and h which are submanifolds of R3. Let F : Gg Gh by

F : ((x , y , z)) = (x3,xyz,x9 + x2y).

The reader may wish to review Example 2.10** and Exercise *** from chapter 3.

a. Explicitly compute the derivative F and be clear with your notation and bases.b. Find the points of Gg where F is singular. What is the rank of Fp for the various

singular points p Gg.

Exercise 3***. Let F : R3 S3 be defined by

F(( , , )) = (sin sin cos , sin sin sin , sin cos , cos ).

Use the charts from stereographic projection to computeF in terms of the bases discussed

in Remark 5.9*** and Remark 5.14***.


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CHAPTER 6

IMMERSIONS AND EMBEDDINGS

In this chapter we turn to inclusion maps of one manifold to another. If f : N M isan inclusion, then the image should also be a manifold. In chapter 3, we saw one situationwhere a subset of f(N) M inherited the structure of a manifold: when each point off(N) had a slice coordinate neighborhood of M. In this chapter, we show that is the only

way it can happen if f(N) is to inherit its structure from M.We first review the situation for functions f : Rn Rm for n m. The infinitesimalcondition of a function to be one-to-one is that the derivative is one-to-one. That thederivative is one-to-one is not required for the function to be one-to-one, but it is sufficientto guarantee the function is one-to-one in some neighborhood (by the Inverse FunctionTheorem). On the other hand, if f(y0) = f(z0), then there is a point x0 on the segmentbetween y0 and z0 where Df(x0) is not one-to-one. This last statement is a consequence ofRolles Theorem. This discussion, perhaps, serves as some motivation to study functionswhose derivative is injective. A second justification is that if f is to be a diffeomorphismto its image, then the derivative must be invertible as a linear map.

While the phrase f(N) inherits manifold structure form M is vague, it certainlyincludes that f(N) inherits its topology from M which is precise.

Definition 6.1***. Suppose f : N M is a smooth map between manifolds. The mapf is called an immersion if fx : TxN Tf(x)M is injective for all x N.

The derivative is injective at each point is not enough to guarantee that the func-tion is one-to-one, as very simple example illustrate. Take f : R R2 by f(x) =(sin(2x), cos(2x)). This function is infinite-to-one as f(x + 1) = f(x), but Df(x) isinjective for all x R. Hence it is clear that we will need some other condition to obtainan inclusion. An obvious first guess, that turns out to be inadequate, is that f is alsoone-to-one.

Example 6.2***. A one-to-one immersion f : N M in which f(N) is not a topologicalmanifold.

Let N = (

4,

3

4), M = R2, and f(x) = (cos(x) cos(2x), sin(x) cos(2x)). The image

f(

4,

3

4) is two petals of a four leafed rose. The map is one-to-one: only x =

4maps

to (0, 0). Note that if

4or

3

4were in the domain, then they would also map to (0, 0).

Df(x) is rank one, so f is a one-to-one immersion. However, f(N) is not a topologicalmanifold. Suppose U B1/2((0, 0)), then U f(N) cannot be homeomorphic to an open

interval. An interval with one point removed has two components, by U f(N) \ (0, 0) has

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2 CHAPTER 6 IMMERSIONS AND EMBEDDINGS

at least four components. Hence no neighborhood of (0, 0) f(N) is homeomorphic to an

open set in R.Definition 6.3***. Suppose f : N M is a smooth map between manifolds. The mapf is called an embedding if f is an immersion which is a homeomorphism to its image.

This extra topological condition is enough to guarantee that f(N) is a submanifold inthe strong sense of Definition 3.2***.

Theorem 6.4***. Suppose Nn andMm are manifolds and f : N M is a smooth mapof rank n. If f is a homeomorphism to its image, then f(N) is a submanifold of M and fis a diffeomorphism to its image.

Proof. To show that f(N) is a submanifold ofM, we suppose x0 N and we must producea slice neighborhood off(x0) f(N) M. We produce this neighborhood in three steps.The first step is to clean up the local picture by producing coordinate neighborhoods ofx0and f(x0) that properly align. The second step is to produce a coordinate neighborhoodof f(x0) in M in which f(N) looks like the graph of a function. The graph of a functionwas already seen to be a submanifold, and we have virtually completed the construction.The third step is to construct the slice neighborhood.

As a first step, we produce coordinate neighborhoods:

(1) (O2, ) a coordinate neighborhood in N centered at x0(2) (U2, ) a coordinate neighborhood in M centered at f(x0)

with f1(U2

N) = O2

and ( f)x0

(T Nx0

) = Rn {0}

Take O1 N a coordinate neighborhood in N centered at x0, and U1 M such thatf1(U1 N) = O1. Such a U1 exists since f is a homeomorphism to its image, and f(N)has the subspace topology. Take (U2, ) a coordinate neighborhood ofM centered at f(x0)with U2 U1. Let O2 f

1(U2), x0 O2. Then (O2, ) is a coordinate neighborhoodof N centered at x0. Let v1, , vn span ( f)x0(T Nx0) R

m, and let v1, , vm be abasis of Rm. Let H : Rm Rm be the isomorphism H(

mi=1 aivi) = (a1, , am). Then

(U2, H) is a coordinate neighborhood in M centered at f(x0) and (Hf)x0(T Nx0) =Rn {0}. Let = H and the coordinate neighborhoods are constructed.

The second step is to cut down the neighborhood of f(x0) so that f(N) looks likethe graph of a function. This step requires the inverse function theorem. We producecoordinate neighborhoods:

(1) (O3, ) a coordinate neighborhood in N centered at x0(2) (U3, ) a coordinate neighborhood in M centered at f(x0), : U3 W3 W2

Rn Rmn

(3) a C function g : W3 W2such that (f(N) U4) is the graph of g.

Let W2 Rn and W4 R

mn be open sets such that W4 W2 is a neighborhood of0 (U2) R

m. Now define U4 = 1(W4 W2) and O4 = f

1(U4). Then O4 O2,(O4, ) is a chart centered at x0, and U4 U2, (U4, ) is a chart centered at f(x0). Let

p1 : Rm

Rn

be the projection onto the first n coordinates and p2 : Rm

Rmn

be theprojection onto the last m n coordinates. The function p1 f 1 maps the open

set (O4) to W4. Since ( f)x0(T Nx0) = Rn {0}, D(p1 f

1)(0) has rank n,


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CHAPTER 6 IMMERSIONS AND EMBEDDINGS 3

i.e., it is an isomorphism. By the Inverse Function Theorem, there is a neighborhood V of

0 Rn

, V (O4) and a neighborhood W3 of 0 Rn

, W3 W4 such thatp1 f

1 : V W3

is a diffeomorphism. Let O3 = 1(V) and U3 =

1(W3 W2). Then (U3, ) is acoordinate chart centered at f(x0), : U3 W3 W2, and (O3, ) is a coordinate chartcentered at x0. Let g be the composition

W3(p1f

1)1

1(O3)(p2f

1) W2

The function g is the composition of two C functions. We now observe that the graphof g is (f(N) U3). The points in (f(N) U3) are f

1((O3)). If x (O3),

then its coordinates in W3 W2 is (p1 f 1(x), p2 f 1)(x)) which agreeswith the graph of g. The second step is established.

The third step is to produce the slice neighborhood. Take W1 an open set with compactclosure and W1 W3. Let be such that 0 < < max{|g(x) y | x W1, y R

n \ W2}.Let V1 W1 W2 be the open set {(x, y) W1 W2 | |g(x) y| < }. Let : V1 W1 B(0) by (x, y) = (x, y g(x)). The map is a diffeomorphims with inverse(x, y) (x, y + g(x)). The image of the graph ofg under is W1 {0}. Let U =

1(V1),then (U, ) is the slice neighborhood: y = g(x) if and only if (x, y) = (x, 0).

It remains to show that f is a diffeomorphism. Since f is a homeomorphism to itsimage, f has a continuous inverse. We need to see that f is smooth as is its inverse. We

use Proposition 2.18***. Given x N, there is a chart of f(N) about x that arises froma slice chart about x in M, Proposition 3.3***. Let (U, ), : U W1 W2 be theslice chart and (U f(N), p1 ) the chart for f(N). The map f is a diffeomorphism if

p1 f 1 and its inverse are C in a neighborhood of (x) and (f(x)), respectively.

Now, since (U, ) is a slice neighborhood,

p1 f 1 = f 1.

The derivative D( f 1)(x) has rank n since and are diffeomorphisms, and f hasrank n. By the Inverse Function Theorem, p1 f

1 is C on a neighborhood of(f(x)). By Proposition 2.18***, f and f1 are smooth functions.

Some authors use the terminology that the image of a manifold under an immersion isa submanifold, but this usage is less common. Furthermore it requires the immersion inthe definition. We will use the term immersed submanifold.

Definition. Suppose N and M are manifolds and f : N M is an immersion. Then(N, f) is an immersed submanifold.

This terminology is suggested by Exercise 1.***

Proposition 6.5***. Suppose N and M are manifolds and f : N M is a one-to-oneimmersion. If N is compact, then f is an embedding.

Proof. We just need to show that f is a homeomorphism to its image. It is a one-to-onecontinuous map from a compact space to a Hausdorff space. By a standard result in generaltopology, f is a homeomorphism.


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4 CHAPTER 6 IMMERSIONS AND EMBEDDINGS

ExercisesExercise 1***. Suppose that f : N M is a one-to-one immersion. Show that for everyx N there is a neighborhood U of x such that f |U: U M is an embedding. Showthat the result holds even if f is not one-to-one.

The next exercise is a difficult and interesting exercise.

Exercise 2***. Every compact n-manifold embeds in RN for some N.

This result is true without the hypothesis of compactness.The dimension N can be taken to be 2n. That every n-manifold embedds in R2n is a

result by H. Whitney. It is also interesting to note that every compact n-manifold immersesin R2n(n) where (n) is the number of ones in the dyadic expansion of n. This resultwas proven by Ralph Cohen. The connection to the dyadic expansion and that this resultis the best possible arose in work by William S. Massey.

Exercise 3***. Let f : RP2 R3 by f([x , y , z]) = (xy,xz,yz). Show that f is awell-defined smooth function. Is f one-to-one? Is f an immersion?

Let g : RP2 R4 byg([x , y , z]) = (xy,xz,yz,x4). Is g an embedding or an immersion?


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CHAPTER 7

VECTOR BUNDLES

We next begin addressing the question: how do we assemble the tangent spaces at variouspoints of a manifold into a coherent whole? In order to guide the decision, consider thecase of U Rn an open subset. We reflect on two aspects.

The first aspect is that the total derivative of a C function should change in a C

manner from point to point. Consider the C

map f : U Rm

. Each point x U givesa linear map fx = Df(x) : Rn Rm the total derivative which is represented by anm n matrix, the Jacobian matrix of Df(x). The Jacobian matrix is a matrix of C

functions in x. While for each x U there is a linear map

Df(x) : T Ux = Rn Rm = TRmf(x)

these fit together to give a C map on the product

U Rn Rm Rm

(x, v) (f(x), Df(x)(v)).

The second aspect is that we wish to define vector fields. A vector field is a choice oftangent vector at each point. For an open subset U of Rn, a vector field is just given by afunction g : U Rn (as the reader probably learned in Advanced Calculus). In order tokeep track of the tail, we write the vector field as

V : U U Rn

x (x, g(x)).

Any C

function g gives a vector field. The complication on a manifold M is that thevector with tail at x M must be in the vector space T Mx and these vector spaces changewith x. In this chapter, we study the required concepts to assemble the tangent spaces ofa manifold into a coherent whole and construct the tangent bundle. The tangent bundleis an example of an object called a vector bundle.

Definition 7.1***. Suppose Mn is a manifold. A real vector bundle over M consists ofa topological space E, a continuous map : E M and a real vector space V (calledthe fiber) such that for each m M, 1(m) is a vector space isomorphic toV, and thereexists an open neighborhood U of m, and a homeomorphism

U : 1

(U) U V

copyright c2002

Typeset by AMS-TEX

1


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2 CHAPTER 7 VECTOR BUNDLES

such that 1U

(m, ) : {m} V 1(m) is a linear isomorphism.

The bundle is smooth ifE is a smooth manifold, is smooth, andU is a diffeomorphism.

In these notes, all vector bundles will be smooth. We may denote a vector bundle by : E M

differential geometry (2006) - jimmie lawson

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