BookII.pdf

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Manifolds and Differential Geometry: Vol II

Jeffrey M. Lee


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ii


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Contents

0.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

1 Comparison Theorems 3

1.1 Rauchs Comparison Theorem3

1.1.1 Bishops Volume Comparison Theorem . . . . . . . . . . . 51.1.2 Comparison Theorems in semi-Riemannian manifolds . . 5

2 Submanifolds in Semi-Riemannian Spaces 72.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Curves in Submanifolds . . . . . . . . . . . . . . . . . . . . . . . 122.3 Hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Lie groups II 15

3.1 Lie Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.1.1 Proper Lie Group Actions . . . . . . . . . . . . . . . . . . 183.1.2 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . 263.2.1 Reductive case . . . . . . . . . . . . . . . . . . . . . . . . 35

4 Killing Fields and Symmetric Spaces 41

5 Fiber Bundles II 455.1 Principal and Associated Bundles . . . . . . . . . . . . . . . . . . 455.2 Degrees of locality . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6 Connections and Gauge Theory 57

7 Geometric Analysis 597.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.1.1 L2, Lp, L . . . . . . . . . . . . . . . . . . . . . . . . . . 597.1.2 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 607.1.3 Elliptic Regularity . . . . . . . . . . . . . . . . . . . . . . 617.1.4 Star Operator II . . . . . . . . . . . . . . . . . . . . . . . 61

7.2 The Laplace Operator . . . . . . . . . . . . . . . . . . . . . . . . 62

iii


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iv CONTENTS

7.3 Spectral Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 667.4 Hodge Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677.5 Dirac Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

7.5.1 Clifford Algebras . . . . . . . . . . . . . . . . . . . . . . . 687.5.2 The Clifford group and spinor group . . . . . . . . . . . . 73

7.6 The Structure of Clifford Algebras . . . . . . . . . . . . . . . . . 737.6.1 Gamma Matrices . . . . . . . . . . . . . . . . . . . . . . . 74

7.7 Clifford Algebra Structure and Representation . . . . . . . . . . 757.7.1 Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . 757.7.2 Hyperbolic Spaces And Witt Decomposition . . . . . . . . 767.7.3 Witts Decomposition and Clifford Algebras . . . . . . . . 777.7.4 The Chirality operator . . . . . . . . . . . . . . . . . . . 787.7.5 Spin Bundles and Spin-c Bundles . . . . . . . . . . . . . . 797.7.6 Harmonic Spinors . . . . . . . . . . . . . . . . . . . . . . 79

8 Classical Mechanics 818.1 Particle motion and Lagrangian Systems . . . . . . . . . . . . . . 81

8.1.1 Basic Variational Formalism for a Lagrangian . . . . . . . 828.1.2 Two examples of a Lagrangian . . . . . . . . . . . . . . . 85

8.2 Symmetry, Conservation and Noethers Theorem . . . . . . . . . 858.2.1 Lagrangians with symmetries. . . . . . . . . . . . . . . . . 878.2.2 Lie Groups and Left Invariants Lagrangians . . . . . . . . 88

8.3 The Hamiltonian Formalism . . . . . . . . . . . . . . . . . . . . . 88

9 Complex Manifolds 919.1 Some complex linear algebra . . . . . . . . . . . . . . . . . . . . 91

9.2 Complex structure . . . . . . . . . . . . . . . . . . . . . . . . . . 949.3 Complex Tangent Structures . . . . . . . . . . . . . . . . . . . . 979.4 The holomorphic tangent map. . . . . . . . . . . . . . . . . . . . 989.5 Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 989.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1009.7 The holomorphic inverse and implicit functions theorems. . . . . 100

10 Symplectic Geometry 10310.1 Symplectic Linear Algebra . . . . . . . . . . . . . . . . . . . . . . 10310.2 Canonical Form (Linear case) . . . . . . . . . . . . . . . . . . . . 10510.3 Symplectic manifolds . . . . . . . . . . . . . . . . . . . . . . . . . 10510.4 Complex Structure and Kahler Manifolds . . . . . . . . . . . . . 10710.5 Symplectic musical isomorphisms . . . . . . . . . . . . . . . . . 110

10.6 Darbouxs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 11010.7 Poisson Brackets and Hamiltonian vector fields . . . . . . . . . . 11210.8 Configuration space and Phase space . . . . . . . . . . . . . . . 11510.9 Transfer of symplectic structure to the Tangent bundle . . . . . . 11610.10Coadjoint Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . 11810.11The Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

10.11.1 The configuration in R3N . . . . . . . . . . . . . . . . . . 120


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0.1. PREFACE v

10.11.2Modelling the rigid body on SO ( 3 ) . . . . . . . . . . . . . 1 2 010.11.3The trivial bundle picture . . . . . . . . . . . . . . . . . . 121

10.12The momentum map and Hamiltonian actions . . . . . . . . . . . 121

11 Poisson Geometry 12511.1 Poisson Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 125

0.1 Preface


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0.1. PREFACE 1

Hi


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2 CONTENTS


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Chapter 1

Comparison Theorems

1.1 Rauchs Comparison Theorem

In this section we deal strictly with Riemannian manifolds. Recall that for a(semi-) Riemannian manifold M, the sectional curvature KM(P) of a 2planeP TpM is

R (e1 e2) , e1 e2for any orthonormal pair e1, e2 that span P.

Lemma 1.1 IfYis a vector field along a curve : [a, b] Mthen ifY(k) is theparallel transport of

k

tY(a) along we have the following Taylor expansion:

Y(t) =mk=0

Y(k)(t)

k! (t a)k +O(|t a|m+1)

Proof. Exercise.

Definition 1.2 IfM, g andN, h are Riemannian manifolds andM : [a, b] M and N : [a, b] Nare unit speed geodesics defined on the same interval[a, b] then we say that KM KN along the pair (M, N) if KM(QM(t))KN(PN(t))for every allt [a, b]and every pair of2-planesQM(t) TM(t)M,PN(t) TN(t)N.

We develop some notation to be used in the proof of Rauchs theorem. LetMbe a given Riemannian manifold. IfY is a vector field along a unit speedgeodesic M such thatY(a) = 0 then let

IMs (Y, Y) :=IM|[a,s](Y, Y) =

sa

tY(t), tY(t) + RM,YM, Y(t)dt.

IfY is an orthogonal Jacobi field then IMs (Y, Y) = tY, Y(s) by theorem??.

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4 CHAPTER 1. COMPARISON THEOREMS

Theorem 1.3 (Rauch) Let M, g and N, h be Riemannian manifolds of thesame dimension and let M : [a, b]

M and N : [a, b]

N unit speed

geodesics defined on the same interval [a, b]. Let JM and JN be Jacobi fieldsalongM andN respectively and orthogonal to their respective curves. Supposethat the following four conditions hold:

(i) JM(a) = JN(a) and neither ofJM(t) orJN(t) is zero fort (a, b](ii)

tJM(a)= tJN(a)(iii) L(M) = dist(M(a), M(b))(iv) KM KN along the pair(M, N)

ThenJM(t) JN(t) for allt [a, b].

Proof. Let fM be defined by fM(s) := JM(s) and hM by hM(s) :=IMs (JM, JM)/JM(s)2 fors (a, b]. DefinefN andhNanalogously. We havefM(s) = 2I

Ms (J

M, JM) and fM/fM = 2hM

and the analogous equalities for fN and hN. Ifc (a, b) then

ln(JM(s)2) = ln(JM(c)2) + 2 s

c

hM(s)ds

with the analogous equation for N. Thus

ln JM(s)

2

JN(s)2 = ln

JM(c)

2

JN(c)2 + 2

s

c

[hM(s)

hN(s

)]ds

From the assumptions (i) and (ii) and the Taylor expansions for JM andJN wehave

limca

JM(c)2JN(c)2

= 0

and so

ln

JM(s)2JN(s)2

= 2 lim

ca

sc

[hM(s) hN(s)]ds

If we can show that hM(s) hN(s) 0 fors (a, b] then the result will follow.So fixs0

(a, b] letZM(s) :=JM(s)/JM(r)and ZN(s) :=JN(s)/JN(r).We now define a parameterized families of sub-tangent spaces along M by

WM(s) : = M(s) TM(s)M and similarly for WN(s). We can choose a

linear isometry Lr : WN(r)WM(r) such that Lr(JN(r)) =JM(r). We nowwant to extend Lr to a family of linear isometries Ls : WN(s) WM(s). Wedo this using parallel transport by

Ls:= P(M)sr Lr P(N)rs.


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1.1. RAUCHS COMPARISON THEOREM 5

Define a vector field Y along M byY(s) := Ls(JM(s)). Check that

Y(a) =JM(a)

Y(r) =JM(r)

Y2 = JN2tY2 =

tJN2The last equality is a result of exercise ?? where in the notation of that exercise(t) := P(M)rt Y(t). Since (iii) holds there can be no conjugates alongM uptor and so IMr is positive definite. NowY JM is orthogonal to the geodesicM and so by corollary ??we have IMr (J

M, JM) IMr (Y, Y) and in fact

IMr

(JM, JM)

IMr

(Y, Y) = r

a tY

2 +RM(M, Y, M, Y)

ra

tY2 +RN(N, JN, N, JN) (by (iv))

=INr (JN, JN)

Recalling the definition ofY we obtain

IMr (JM, JM)/

JM(r)2 INr (JN, JN)/JN(r)2and so hM(r) hN(r) 0 butr was arbitrary and so we are done.

1.1.1 Bishops Volume Comparison Theorem

under construction

1.1.2 Comparison Theorems in semi-Riemannian mani-folds

under construction


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6 CHAPTER 1. COMPARISON THEOREMS


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Chapter 2

Submanifolds inSemi-Riemannian Spaces

2.1 Definitions

Let M be a d dimensional submanifold of a semi-Riemannian manifold M ofdimension n where d < n. The metric g(., .) =., . on Mrestricts to tensoron M which we denote by h. Since h is a restriction of g we shall also usethe notation., . for h. If the restriction h of is nondegenerate on each spaceTpM then h is a metric tensor on Mand we say that M is a semi-Riemanniansubmanifold ofM. If M is Riemannian then this nondegeneracy condition isautomatic and the metric h is automatically Riemannian. More generally, if

: M M , g is an immersion we can consider the pull-back tensor g definedby

g(X, Y) = g(T X , T Y).Ifg is nondegenerate on each tangent space then it is a metric onMcalled thepull-back metric and we call a semi-Riemannian immersion. IfM is alreadyendowed with a metric gM then if

g = gM then we say that : (M, gM)M , g

is an isometric immersion. Of course, if g is a metric at all, as it

always is if

M, g

is Riemannian, then the map : (M, g) M , g is anisometric immersion. Since every immersion restricts locally to an embeddingwe may, for many purposes, assume that Mis a submanifold and that is justthe inclusion map.

Definition 2.1 Let M , g be a Lorentz manifold. A submanifold M M issaid to be spacelike (resp. timelike, lightlike) ifTpM TpM is spacelike (resp.timelike, lightlike).

There is an obvious bundle on Mwhich is the restriction ofT M toM. Thisis the bundle T M

M

=pMTpM. Each tangent spaceTpM decomposes as

TpM =TpM (TpM)

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8 CHAPTER 2. SUBMANIFOLDS IN SEMI-RIEMANNIAN SPACES

where (TpM)

={v TpM :v, w = 0 for all w TpM}. Then T M =

p(TpM) its natural structure as a smooth vector bundle called the normalbundle to M inM. The smooth sections of the normal bundle will be denotedby

T M

or X(M). Now the orthogonal decomposition above is globalizes

asT M

M

=T M T MA vector field on Mis always the restriction of some (not unique) vector fieldon a neighborhood ofM. The same is true of any not necessarily tangent vectorfield alongM. The set of all vector fields along Mwill be denoted by X(M)

M

.

Since any function on Mis also the restriction of some function on Mwe mayconsider X(M) as a submodule of X(M)

M

. If X X(M) then we denoteits restriction to M by X

M

or sometimes just X. Notice that X(M) is a

submodule of X(M)M. We have two projection maps : TpM NpM andtan : TpM TpMwhich in turn give module projections nor : X(M)MX(M) and : X(M)

M

X(M). The reader should contemplate the followingdiagrams:

C(M) restr C(M) = C(M)

X(M)

restr X(M)M

tan X(M)

andC(M)

restr C(M) = C(M) X(M)

restr X(M)

M

nor X(M).

Now we also have an exact sequence of modules

0 X(M) X(M)M

tan X(M) 0which is in fact a split exact sequence since we also have

0 X(M) nor X(M)M

X(M) 0

The extension map X(M)M

X(M) is not canonical but in the presence ofa connection it is almost so: IfU(M) is the open tubular neighborhood ofMgiven, for sufficiently small by

U(M) = {p M : dist(p, M)< }

then we can use the following trick to extend any X X(M) to X(U(M))|M.First choose a smooth frame field E1,...,En defined along M so that Ei X(M)

M

. We may arrange if need to have the first d of these tangent to M.Now parallel translate each frame radially outward a distance to obtain asmooth frame field E1,...,En on U(M).

Now we shall obtain a sort of splitting of the Levi-Civita connection of Malong the submanifold M. The reader should recognize a familiar theme here


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2.1. DEFINITIONS 9

especially if elementary surface theory is fresh in his or her mind. First we noticethat the Levi-Civita connection

on Mrestrict nicely to a connection on the

bundle T MM M. The reader should be sure to realize that the space ofsections of this bundle is exactly X(M)

M

and so the restricted connection is a

map M

:X(M) X(M)M

X(M)M

. The point is that ifX X(M) andW X(M)

M then XW doesnt seem to be defined since X and W are not

elements ofX(M). But we may extendXandWto elements ofX(M) and thenrestrict again to get an element of X(M)

M

.Then recalling the local propertiesof a connection we see that the result does not depend on the extension.

Exercise 2.2 Use local coordinates to prove the claimed independence on theextension.

We shall write simply in place of

Msince the context make it clear when

the later is meant. Thus XW :=

X

W where

X and

W are any extensionsofX and W respectively.

Clearly we have XY1, Y2 = XY1, Y2+Y1, XY2 and so is a metricconnection on T M

M

. For a fixed X, Y X(M) we have the decomposition ofXYinto tangent and normal parts. Similarly, for V X(M) we can considerthe decomposition ofXVinto tangent and normal parts. Thus we have

XY = (XY)tan + (XY)XV = (XV)tan + (XV)

We make the following definitions:

XY := (XY)tan for all X, Y X(M)

b12(X, Y) := (

XY)

for all X, YX

(M)b21(X, V) := (XV)tan for all X X(M),V X(M)

XV := (XV) for all X X(M), V X(M)

Now ifX, Y X(M),V X(M) then 0 = Y, Vand so0 = XY, V

= XY, V + Y, XV= XY , V + Y, XVtan= b12(X, Y), V + Y, b21(X, V).

It follows that b12(X, Y), V = Y, b21(X, V). Now we from this thatb12(X, Y)is not only C(M) linear in Xbut also in Y. Thusb12 is tensorial and so foreach fixed p M, b12(Xp, Yp) is a well defined element ofTpM for each fixedXp, Yp TpM. Also, for any X1, X2 X(M) we have

b12(X1, X2) b12(X2, X1)=

X1X2 X2X1= ([X1, X2])

= 0.


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So b12 is symmetric. The classical notation forb12 is IIand the form is calledthe second fundamental tensor or the second fundamental form. For TpMwe define the linear map B() :=b12(, ). With this in mind we caneasily deduce the following facts which we list as a theorem:

1.XY := (XY)tan defines a connection onMwhich is identical with theLevi-Civita connection for the induced metric onM.

2. (XV) := XVdefines a metric connection on the vector bundle T M.3. (XY) := b12(X, Y ) defines a symmetric C(M)-bilinear form with

values in X(M).

4. (XV)tan := b21(X, V ) defines a symmetric C(M)-bilinear form withvalues in X(M).

Corollary 2.3 b21 is tensorial and so for each fixed p M, b21(Xp, Yp) isa well defined element of TpM

for each fixed Xp, Yp TpM and we have abilinear formb21: TpM TpM TpM.Corollary 2.4 The map b21(, ) : TpM TpM is equal toBt : TpMTpM

.

Writing any Y X(M)M

as Y = (Ytan, Y) we can write the map X :X(M)

M

X(M)M

as a matrix of operators: X BXBtX X

Next we define the shape operator which is also called the Weingarten map.

There is not a perfect agreement on sign conventions; some authors shapeoperator is the negative of the shape operator as defined by other others. Weshall defineS+ andS which only differ in sign. This way we can handle bothconventions simultaneously and the reader can see where the sign conventiondoes or does not make a difference in any given formula.

Definition 2.5 Let p M. For each unit vector u normal to M at p wehave a map called the () shape operator Su associated to u. defined bySu(v) :=

vUtan whereUis any unit normal field defined nearp such thatU(p) =u.

The shape operators{Su }u a unit normal contain essentially the same infor-mation as the second fundamental tensor II = b12. This is because for any

X, Y X(M) and U X(M) we haveSUX, Y =

XUtan , Y = U, XY= U, XY = U, b12(X, Y)= U, II(X, Y).

Note:U, b12(X, Y) is tensorial in U, X and Y . Of course , SUX= SUX.


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2.1. DEFINITIONS 11

Theorem 2.6 LetMbe a semi-Riemannian submanifold of M. We have theGauss equation

RVWX, Y = RVWX, Y II(V, X), II(W, Y) + II(V, Y), II(W, X)

Proof. Since this is clearly a tensor equation we may assume that [V, W] =0. With this assumption we have we haveRVWX, Y = (V W) (W V) where(V W) = VWX, Y

VWX, Y = VWX, Y + V(II(W, X)), Y= VWX, Y + V(II(W, X)), Y= VWX, Y +VII(W, X), Y II(W, X), VY=

V

WX, Y

+V

II(W, X), Y

II(W, X),

VY

SinceII(W, X), VY= II(W, X), VY = II(W, X), II(V, Y)

we have (V W) = VWX, YII(W, X), II(V, Y). Interchanging the rolesofV andW and subtracting we get the desired conclusion.

Another formula that follows easily from the Gauss equation is the followingformula (also called the Gauss formula):

K(v w) = K(v w) +II(v, v), II(w, w) II(v, w), II(v, w)v, vw, w v, w2Exercise 2.7 Prove the last formula (second version of the Gauss equation).

From this last version of the Gauss equation we can show that a sphereSn(r) of radius r in Rn+1 has constant sectional curvature 1/r2 for n > 1. If(ui) is the standard coordinates onRn+1 then the position vector field in Rn+1

is r =n+1

i=1 uii. Recalling that the standard (flat connection)D on Rn+1 is

just the Lie derivative we see that DXr=n+1i=1 Xuii = X. Now using the

usual identifications, the unit vector field U = r/r is the outward unit vectorfield on Sn(r). We have

II(X, Y), U = DXY, U=

1

r DXY, r

=

1

r Y, DXr

= 1

rY, X = 1

rX, Y.

Now letting M be Rn+1 and M be Sn(r) and using the fact that the Gausscurvature ofRn+1 is identically zero, the Gauss equation gives K= 1/r.

The second fundamental form contains information about how the semi-Riemannian submanifold Mbends about in M. First we need a definition:


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Definition 2.8 LetM be semi-Riemannian submanifold of M andN a semi-Riemannian submanifold ofN. A pair isometry : ( M, M)

(N , N) consists

of an isometry : M Nsuch that(M) =Nand such that |M :M Nis an isometry.

Proposition 2.9 A pair isometry : ( M , M)( N , N) preserves the secondfundamental tensor:

Tp II(v, w) =I I(Tp v, Tp w)

for allv, w TpMand allp M.

Proof. Let p Mand extend v, w TpM to smooth vector fields V andW. Since an isometry respects the Levi-Civita connections we have VW =VW. Now since is a pair isometry we have Tp(TpM) T(p)NandTp(TpM

) T(p)N. This means that : X(M)M X(N)Npre-serves normal and tangential components (X(M)) X(N) and (X(M)) X(N). We have

Tp II(v, w) = II(V, W)((p))=

VW ( (p))=

VW

( (p))

=VW ( (p))

=I I(V, W)((p))

=I I(V, W)((p))=I I(Tp v, Tp w)

The following example is simple but conceptually very important.

Example 2.10 LetMbe the strip 2 dimensional strip{(x,y, 0) : < x < }considered as submanifold ofR3 (with the canonical Riemannian metric). LetN be the subset ofR3 given by{(x,y, 1 x2) :1 < x < 1}. Exercise:Show thatMis isometric toM. Show that there is no pair isometry(R3, M) (R3, N).

2.2 Curves in SubmanifoldsIf : I M is a curve in Mand M is a semi-Riemannian submanifold ofMthen we have tY = tY +II(, Y) for any vector field Y along. IfY isa vector field in X(M)

M

or in X(M) then Y is a vector field along . Inthis case we shall still write tY = tY+ II(, Y) rather than t(Y ) =t(Y ) +II(, Y ).


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2.2. CURVES IN SUBMANIFOLDS 13

Recall that is a vector field along . We also have := twhich in thiscontext will be called the extrinsic acceleration (or acceleration in M. By

definition we have tY = tY. The intrinsic acceleration (accelerationin M) ist. Thus we have

= t+II(, ).

From this definitions we can immediately see the truth of the following

Proposition 2.11 If: I M is a curve inMandM is a semi-Riemanniansubmanifold of M thenis a geodesic inM if and only if(t) is normal to Mfor everyt I.

Exercise 2.12 A constant speed parameterization of a great circle inSn(r) isa geodesic. Every geodesic inSn(r) is of this form.

Definition 2.13 A semi-Riemannian manifoldM Mis called totally geodesicif every geodesic inM is a geodesic inM.

Theorem 2.14 For a semi-Riemannian manifoldM M the following condi-tions are equivalent

i) M is totally geodesic

ii) II

0

iii) For all v T M the M geodesic v with initial velocity v is such thatv[0, ] M for >0 sufficiently small.

iv) For any curve : I M, parallel translation along induced by inM is equal to parallel translation along induced by inM.

Proof. (i)=(iii) follows from the uniqueness of geodesics with a given ini-tial velocity.

iii)=(ii); Let v T M. Applying 2.11 tov we see that II(v, v) = 0. Since vwas arbitrary we conclude that II

0.

(ii)=(iv); Suppose vTpM. IfV is a parallel vector field with respect tothat is defined nearpsuch thatV(p) =v. Then tV = tV+II(, V) = 0+0for any with (0) =p so that Vis a parallel vector field with respect to .

(iv)=(i); Assume (iv). Ifis a geodesic in M then is parallel along with respect to. Then by assumption is parallel along with respect to. Thus is also a M geodesic.


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2.3 Hypersurfaces

If the codimension of M in M is equal to 1 then we say that M is a hyper-surface. If M is a semi-Riemannian hypersurface in M andu, u > 0 forevery u (TpM) we call M a positive hypersurface. Ifu, u < 0 for everyu (TpM) we callMa negative hypersurface. Of course, ifM is Riemannianthen every hypersurface in M is positive. The sign ofM, in M denoted sgn Mis sgnu, u.

Exercise 2.15 Suppose that c is a regular value of f C(M) then M =f1(c) is a semi-Riemannian hypersurface ifdf, df > 0 on all of M or ifdf, df 0 we have that f1(r2) is a semi-Riemannian hypersurface inRn+1, with sign and unit normal U = r/r. We shall divide these hyper-

surfaces into two classes according to sign.

Definition 2.16 Forn > 1 and0 v n, we define

Sn(r) = {x Rn+1, : x, x=r2.

Sn(r) is called the pseudo-sphere of index.

Definition 2.17 Forn > 1 and0 n, we define

Hn(r) = {x Rn+1(+1),+1 : x, x= r2.

Hn(r) is called the pseudo-hyperbolic space of radiusr and index.


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Chapter 3

Lie groups II

3.1 Lie Group Actions

The basic definitions for group actions were given earlier in definition ?? and??. As before we give most of our definitions and results for left actions and askthe reader to notice that analogous statements can be made for right actions.

Definition 3.1 Letl : G M Mbe a left action whereG is a Lie group andM a smooth manifold. If l is a smooth map then we say that l is a (smooth)Lie group action.

As before, we also use either of the notations gp or lg(p) for l(g, p). For

right actions r : M G M we write pg = rg(p) = r(p, g). A right actioncorresponds to a left action by the rule gp := pg1. Recall that for pM theorbit ofp is denoted Gp and we call the action transitive ifGp = M.

Definition 3.2 Let l be a Lie group action as above. For a fixed p M theisotropy groupofp is defined to be

Gp := {g G: gp= p}The isotropy group ofp is also called thestabilizerofp.

Exercise 3.3 Show thatGp is a closed subset and abstract subgroup ofG. Thismeans thatGp is a closed Lie subgroup.

Recalling the definition of a free action, it is easy to see that an action isfree if and only if the isotropy subgroup of every point is the trivial subgroupconsisting of the identity element alone.

Definition 3.4 Suppose that we have Lie group action of G on M. If N isa subset of M and gx x for all x N then we say that N is an invariantsubset. IfNis also a submanifold then it is called an invariant submanifold.

15


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16 CHAPTER 3. LIE GROUPS II

In this definition we include the possibility that N is an open submanifold.If N is an invariant subset of N then it is easy to set that gN = N wheregN = lg(N) for any g. Furthermore, if N is a submanifold then the actionrestricts to a Lie group action G N N.

If G is zero dimensional then by definition it is just a group with discretetopology and we recover the definition of discrete group action. We have alreadyseen several examples of discrete group actions and now we list a few examplesof more general Lie group actions.

Example 3.5 In case M = Rn then the Lie group GL(n,R) acts onRn bymatrix multiplication. Similarly, GL(n,C)acts onCn. More abstractly, GL(V)acts on the vector spaceV. This action is smooth sinceAx depends smoothly(polynomially) on the components ofA and on the components ofx Rn.

Example 3.6 Any Lie subgroup ofGL(n,R) acts onRn also by matrix multi-plication. For example, O(n,R) acts onRn. For everyx Rn the orbit ofx isthe sphere of radiusx. This is trivially true ifx = 0. In general, ifx = 0then,gx =x for any g O(n,R). On the other hand, if x, y Rn andx =y = r then letx := x/r andy := y/r. Extend to orthonormal bases(x = e1,...,en) and (y = f1,...,fn). Then there exists an orthogonal matrix Ssuch thatSei = fi fori= 1,...,n. In particular, Sx=y and so Sx = y.Exercise 3.7 From the last example we can restrict the action ofO(n,R) to atransitive action onSn1. NowS O(n,R)also acts onRn and by restriction onSn1. From the last example we know that O(n,R) acts transitively onSn1.Show that the same is true forSO(n,R) as long asn > 1.

A Lie group acts on itself in an obvious way:

Definition 3.8 For a Lie group G and a fixed elementg G, the maps Lg :G G andRg :G G are defined by

Lgx= gx forx GRgx= xg forx G

and are called left translationand right translation(byg) respectively.

The mapsGG Ggiven by (g, x) Lgxand (g, x) Rgxare Lie groupactions.

Example 3.9 If H is a Lie subgroup of a Lie group G then we can considerLh for anyh Hand thereby obtain an action ofH onG.

Recall that a subgroupHof a groupGis called a normal subgroup ifgkg1 Kfor any k Hand all g G. In other word, H is normal ifgH g1 H forall all g Gand it is easy to see that in this case we always have gHg1 =H.


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3.1. LIE GROUP ACTIONS 17

Example 3.10 If H is a normal Lie subgroup of G, then G acts on H byconjugation:

Cgh= ghg1

Suppose now that a Lie group G acts on smooth manifolds M and N. Forsimplicity we take both actions to be left action which we denote by l and respectively. A map : M Nsuch that lg =g for all g G, is saidto be an equivariant map (equivariant with respect to the given actions). Thismeans that for all g the following diagram commutes:

M N

lg gM

N

If is also a diffeomorphism then we have an equivalenceof Lie group actions.

Example 3.11 If : G H is a Lie group homomorphism then we can definean action ofG onH by(g, g1) = g(h) = L(g)h. We leave it to the reader toverify that this is indeed a Lie group action. In this situation is equivariantwith respect to the actions andL (left translation).

Example 3.12 Let Tn = S1 S1 be the ntorus where we identify S1with the complex numbers of unit modulus. Fix k = (k1,...,kn) Rn.ThenRacts onRn byk(t, x) =t x := x +tk. On the other hand,R acts on Tn byt (z1,...,zn) = (eitk1z1,...,eitknzn). The map Rn Tn given by(x1,...,xn) (eix

1

,...,eixn

) is equivariant with respect to these actions.

Theorem 3.13 (Equivariant Rank Theorem) Suppose thatf :M N issmooth and that a Lie group G acts on both M and Nwith the action on Mbeing transitive. If f is equivariant then it has constant rank. In particular,each level set of f is a closed regular submanifold.

Proof. Let the actions on M and Nbe denoted by l and respectively asbefore. Pick any two points p1, p2 M. SinceG acts transitively on M there isag withlgp1= p2. By hypothesis, we have the following commutative diagram

M f N

lg gM

f

Nwhich, upon application of the tangent functor gives the commutative diagram

M Tp1f N

Tp1 lg Tf(p1)gM

Tp2f N


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Since the maps Tp1 lg and Tf(p1)g are linear isomorphisms we see that Tp1fmust have the same rank as T

p1f. Since p

1 and p

2were arbitrary we see that

the rank off is constant on M.There are several corollaries of this neat theorem. For example, we know

that O(n,R) is the level set f1(I) where f : GL(n,R) gl(n,R) =M(n,R)is given by f(A) =ATA. The group O(n,R) acts on itself via left translationand we also let O(n,R) act on gl(n,R) by Q A := QTAQ (adjoint action).One checks easily that fis equivariant with respect to these actions and sincethe first action (left translation) is certainly transitive we see that O(n,R) isa closed regular submanifold ofGL(n,R). It follows from proposition ?? thatO(n,R) is a closed Lie subgroup of GL(n,R). Similar arguments apply forU(n,C) GL(n,C) and other linear Lie groups. In fact we have the followinggeneral corollary to Theorem 3.13 above.

Corollary 3.14 If : G H is a Lie group homomorphism then the kernelKer(h) is a closed Lie subgroup ofG.

Proof. Let G act on itself and on H as in example 3.11. Then is equiv-ariant and 1(e) = Ker (h) is a closed Lie subgroup by Theorem 3.13 andProposition ??.

We also have use for the

Corollary 3.15 Letl : G M Mbe a Lie group action andGp the isotropysubgroup of somep M. ThenGp is a closed Lie subgroup ofG.

Proof. The orbit map p : G M given by p(g) = gp is equivariantwith respect to left translation on G and the given action on M. Thus by

the equivariant rank theorem, Gp is a regular submanifold of G an then byProposition ?? it is a closed Lie subgroup.

3.1.1 Proper Lie Group Actions

Definition 3.16 Letl : G M Mbe a smooth (or merely continuous) groupaction. If the map P :G M M Mgiven by(g, p) (lgp, p) is proper wesay that the action is aproper action.

It is important to notice that a proper action is notdefined to be an actionsuch that the defining map l : G M M is proper.

We now give a useful characterization of a proper action. For any subset

K M, let g K := {gx : x K}.Proposition 3.17 Let l : G M M be a smooth (or merely continuous)group action. Then l is a proper action if and only if the set

GK := {g G: (g K) K= }

is compact wheneverKis compact.


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Proof. We follow the proof from [?];

Suppose thatl is proper so that the map Pis a proper map. Let G be thefirst factor projection G M G. Then

GK= {g : there exists a x K such that gx K}= {g : there exists a x M such that P(g, x) K K}=G(P

1(K K))

and so GK is compact.Next we assume that GK is compact for all compact K. IfC M M is

compact then letting K=1(C) 2(C) where 1 and 2 are first and secondfactor projectionsM M Mrespectively we have

P1(C)

P1(K

K) {

(g, x) : gp

K} GK K.

SinceP1(C) is a closed subset of the compact set GK Kit is compact. Thismeans thatPis proper sinceCwas an arbitrary compact subset ofM M.

Using this proposition, one can show that definition ?? for discrete actionsis consistent with definition 3.16 above.

Proposition 3.18 IfG is compact then any smooth action l: G M M isproper.

Proof. Let K M M be compact. We find compactC M such thatK

C

Cas in the proof of proposition 3.17.

Claim: P1(K) is compact. Indeed,

P1(K) P1(C C) = cCP1(C {c})= cC{(g, p) : (gp,p) C {c}}= cC{(g, c) : gp C} cC(G {c}) =G C

Thus P1(K) is a closed subset of the compact set G C and hence iscompact.

Exercise 3.19 Prove the followingi) Ifl : G

M

M is a proper action andH

Gis a closed subgroup then

the restricted actionH M M is proper.ii) IfN is an invariant submanifold for a proper actionl : G M M then

the restricted actionG N N is also proper.

Let us now consider a Lie group action l : GM M that is bothproper and free. The map orbit map at p is the map p : G M given byp(g) =g p. It is easily seen to be smooth and its image is obviously G p. In


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fact, if the action is free then each orbit map is injective. Also,p is equivariantwith respect to the left action ofG on itself and the action l :

p(gx) = (gx) p= g (x p)=g p(x)

for all x, gG. It now follows from Theorem 3.13 (the equivariant rank theo-rem) thatp has constant rank and since it is injective it must be an immersion.Not only that, but it is a proper map. Indeed, for any compact K M the set1p (K) is a closed subset of the set GK{p} and since the later set is compactby Theorem 3.17,1p (K) is compact. Now by exercise ?? obtain the result thatp is an embedding and each orbit is a regular submanifold ofM.

It will be very convenient to have charts on Mwhich fit the action ofG ina nice way. See figure 3.1.

Definition 3.20 Let M be ann-manifold andG a Lie group of dimensionk.If l: G M M is a Lie group action then an action-adapted chartonMis a chart(U, x) such that

i) x(U) is a product open setV1 V1 Rk Rnk = Rn

ii) if an orbit has nonempty intersection withU then that intersection has theform

{xk+1 =c1,....,xn =cnk}for some constantsc1,...,cnk.

Theorem 3.21 If l : G M M is a free and proper Lie group action thenfor everyp M there is an action-adapted chart centered atp.

Proof. Let p M be given. SinceG p is a regular submanifold we maychoose a regular submanifold chart (W, y) centered at p so that (G p) W isexactly given by yk+1 =... = yn = 0 in W. Let Sbe the complementary slicein W given by y1 = ... = yk = 0. Note that S is a regular submanifold. Thetangent space TpM decomposes as

TpM=Tp(G p) TpSLet : G S Mbe the restriction of the action l to the set G S. Also, letip :G

G

Sbe the insertion map g

(g, p) and let je :S

G

Sbe the

insertion map s (e, s). These insertion maps are embeddings and we havep = ipand alsoje = whereis the inclusionS M. NowTep(TeG) =Tp(G p) since p is an embedding. On the other hand,T p = T T ip andso the image ofT(e,p)must containTp(G p). Similarly, from the composition je = we see that the image ofT(e,p) must contain TpS. It follows thatT(e,p) : T(e,p)(G S) TpM is surjective and since T(e,p)(G S) and TpMhave the same dimension it is also injective.


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G

p

p.

Figure 3.1: Action-adapted chart

By the inverse function theorem, there is neighborhoodO of (e, p) such that| O is a diffeomorphism. By shrinking O further if necessary we may assumethat (O) W. We may also arrange that O has the form of a productO = A B for A open in G and B open in S. In fact, we can assume thatthere are diffeomorphisms : Ik A and : Ink B where Ik and Inkare the open cubes in Rk and Rnk given respectively by Ik = (1, 1)k andInk = (1, 1)nk and where (e) = 0 Rk and (p) = 0 Rnk. LetU := (A B). The map ( ) : Ik Ink U is a diffeomorphismand so its inverse is a chart. We must make one more adjustment. We mustshow that B can be chosen small enough that the intersection of each orbitwith B is either empty or a single point. If this were not true then there wouldbe a sequence of open sets Bi with compact closure (and with correspondingdiffeomorphismsi :I

k Bi as above) such that for every i there is a pair ofdistinct points pi, p

i Bi with gipi = pi for some sequence{gi} G. Since

manifolds are first countable and normal, we may assume that the sequence

{Bi} is a nested neighborhood basis which means that Bi+1 Bi for all iand for each neighborhood V of p, we have Bi V for large enough i. Thisforces both pi, and p

i = gipi to converge to p. From this we see that the

set K ={(gipi, pi), (p, p)} M M is compact. Recall that by definitionthe map P : (g, x) (gx,x) is proper. Since (gi, pi) = P1(gipi, pi) we seethat{(gi, pi)} is a subset of the compact set P1(K). Thus after passing toa subsequence we have that (gi, pi) converges to (g, p) for some g and hence


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p

A

G

B

S

S

ip

je

gi g and gipi gp. But this means we havegp = lim

igipi = lim

ipi = p

and since the action is free we conclude that g = e. But this is impossible sinceit would mean that for large enough i we would have gi

A and in turn this

would imply that

(gi, pi) = lgi(pi) = pi = le(p

i) = (e, p

i)

contradicting the injectivity of on A B. Thus after shrinking B we mayassume that the intersection of each orbit with B is either empty or a singlepoint. We leave it to the reader to check that with x:= ( ( ))1 :UIk Ink Rn we obtain a chart (U, x) with the desired properties.

For the next lemma we continue with the convention that I is the interval(1, 1).Lemma 3.22 Let x := ( ( ))1 : U Ik Ink = In Rn be anaction-adapted chart map obtained as in the proof of Theorem 3.21 above. Then

given anyp1 U, there exists a diffeomorphism : In

In

such that x isan action-adapted chart centered atp1.

Proof. Clearly all we need to do is show that for any a In there is adiffeomorphism : In In such that(a) = 0. Let ai be theith componentofa. Leti : I Ibe defined by

i := t(ai)


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wheretc(x) :=x cand : (1, 1) R is the useful diffeomorphism : x tan(

2x). The diffeomorphism we want is now (x) = (

1(x1),...,

1(xn)).

3.1.2 Quotients

If l : G M M is a Lie group action, then there is a natural equivalencerelation on M whereby the equivalence classes are exactly the orbits of theaction. The quotient space space (or orbit space) is denotedG\Mand we havethe quotient map : M G\M. We put the quotient topology on G\M sothatA G\Mis open if and only if1 (A) is open in M. The quotient mapis also open. Indeed, let U Mbe open. We want to show that (U) is openand for this it suffices to show that 1 ((U)) is open. But1 ((U)) is theunionglg(U) and this is open since each lg(U) is open.

Lemma 3.23 G\M is a Hausdorff space if the set :={(gp,p) : g G,p M} is a closed subset ofM M.

Proof. Letp, q G\M with (p) =p and (q) =q. Ifp =qthen p and qare not in the same orbit. This means that (p, q) / and so there must be aproduct open setUVsuch that (p, q) UV andUV disjoint from . Thismeans that (U) and (V) are disjoint neighborhoods ofp and qrespectively

Proposition 3.24 If l: G M M is a free and proper action thenG\M isHausdorff and paracompact.

Proof. To show that G

\M is Hausdorff we use the previous lemma. We

must show that is closed. But =P(G M) is closed sinceP is proper.To show that G\Mis paracompact it suffices to show that each connected

component ofG\Mis second countable. This reduces the situation to the casewhere G\M is connected. In this case we can see that if{Ui} is a countablebasis for the topology on M then{ (Ui)} is a countable basis for the topologyonG\M.

We will shortly show that if the action is free and proper then G\M has asmooth structure which makes the quotient map : M G\M a submersion.Before coming to this let us note that if such a smooth structure exists then it isunique. Indeed, if (G\M)A isG\Mwith a smooth structure given by maximalatlas A and similarly for (G\M)B for another atlas B then we have the followingcommutative diagram:

M

(G\M)A id (G\M)B

Sinceis a surjective submersion, Proposition ?? applies to show that (G\M)A id(G\M)B is smooth as is its inverse. This means thatA = B.


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Theorem 3.25 If l : G M M is a free and proper Lie group action thenthere is a unique smooth structure on the quotientG

\Msuch that

(i) the induced topology is the quotient topology and henceG\M is a smoothmanifold,

(ii) the projection: M G\M is a submersion,(iii) dim(G\M) = dim(M) dim(G).

Proof.Let dim(M) =n and dim(G) = k. We have already show that G\Mis a paracompact Hausdorff space. All that is left is to exhibit an atlas such thatthe charts are homeomorphisms with respect to this quotient topology. Let qG\Mand choosepwith(p) =q. Let (U, x) be an action-adapted chart centeredatp and constructed exactly as in Theorem 3.21. Let (U) =V

G

\Mand let

B be the slice x1 = = xk = 0. By construction |B : B V is a bijectionand in fact it is easy to check that |B is a homeomorphism and := ( |B)1is the corresponding local section. Consider the map y= 2 x where 2 isthe second factor projection2: RkRnk Rnk. This is a homeomorphismsince (2 x)|B is a homeomorphsim and 2 x = (2 x)|B . We nowhave chart (V, y).

Given two such charts (V, y) and (V , y) we must show that y1 y1 issmooth. The (V, y) and (V , y) are constructed from associated action-adaptedcharts (U, x) and (U , x) on M. Letq VV. As in the proof of Lemma 3.22 wemay find diffeomorphisms and so that (U, x) and (U , x) are action-adapted charts centered at points p1 1(q) and p2 1(q) respectively.Correspondingly, the charts (V, y) and (V , y) are modified to charts (V, y) and

(V , y) centered at qwhere

y :=2 x y :=2 x

One checks y y1 = 2 and similarly for y y1. From this it followsthat the overlap map y1

y will be smooth if and only if y1 y1 is smooth.

Thus we have reduced to the case where (U, x) and (U , x) are centered at p11(q) and p2 1(q) respectively. This entails that both (V, y) and (V , y)are centered at q V V. Now if we choose a g G such that lg(p1) = p2then by composing with the diffeomorphismlg we can reduce further to the casewherep1= p2. Here we use the fact that lg takes the set of orbits to the set oforbits in a bijective manner and the special nature of our action-adapted charts

with respect to these orbits. In this case the overlap map x x1 must have theform (a, b) (f(a, b), g(b)) for some smooth functions f and g. It follows thaty1 y1 has the form b g(b).

Similar results hold for right actions. In fact, some of the most importantexamples of proper actions are usually presented as right actions (the rightaction associated to a principal bundle). In fact, we shall see situations wherethere is both a right and a left action in play.


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Example 3.26 Consider S2n1 as the subset ofCn given by S2n1 ={Cn :

|| = 1

}. Here = (z1,...,zn) and

||= zizi. Now we let S1 act onS2n1 by (a, ) a = (az1,...,azn). This action is free and proper. The

quotient is the complex projective spaceCPn1.

S2n1

CPn1

These maps (one for each n) are called the Hopf maps. In this context S1 isusually denoted byU(1).In the sequel we will be considering the similar right actionSn U(1) Sn.In this case we think ofCn+1 as consisting of column vector and the action isgiven by(, a) a. Of course, sinceU(1) is abelian this makes essentially nodifference but in the next example we consider the quaternionic analogue wherekeeping track of order is important.

The quaternionic projective HPn1 space is defined by analogy with CPn1.The elements ofHPn1 are 1-dimensional subspaces of the right H-vector spaceHn. Let us refer to these as H-lines. Each of these are of real dimension 4.Each element ofHn\{0} determines an H-line and the H-line determined by(1,...,n)t will be the same as that determined (1,..., n)t if and only if thereis a nonzero element a Hso that (1,..., n)t = (1,...,n)ta= (1a,...,na)t.This defines an equivalence relation on Hn\{0} and thus we may also think ofHPn1 as (Hn\{0}) /. The element ofHPn1 determined by (1,...,n)t isdenoted by [1,...,n]. Notice that the subset{ Hn : || = 1} is S4n1. Justas for the complex projective spaces we observe that all such H-lines containpoints ofS4n1 and two points , S4n1 determine the same H-line if andonly if = a for some a with|a| = 1. Thus we can think ofHPn1 as aquotient of S4n1. When viewed in this way, we also denote the equivalenceclass of = (1,...,n)t S4n1 by [] = [1,...,n]. The equivalence classesare clearly the orbits of an action as described in the following example.

Example 3.27 ConsiderS4n1 considered as the subset ofHn given byS4n1 ={ Hn :|| = 1}. Here = (1,...,n)t and|| =ii. Now we define aright action ofU(1,H) onS4n1 by (, a) a = (1a,...,na)t. This actionis free and proper. The quotient is the quaternionic projective spaceHPn1 andwe have the quotient map denoted by

S4n1

HPn1

This map is also referred to as an Hopf map. Recall thatZ2 ={1, 1}acts on Sn1 = Rn on the right (or left) by multiplication and the action isa (discrete) proper and free action with quotientRPn1 and so the above twoexamples generalize this.


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For completeness we describe an atlas for HPn1. View HPn1 as thequotientS4n1/

described above. Let

Uk := {[] S4n1 Hn :k = 0}

and definek :Uk Hn1= R4n1 by

k([]) = (i([1,...,n]) = (111 ,...,1,...,n1n )

where as for the real and complex cases the caret symbol indicates that wehave omitted the 1 in the i-th slot so as to obtain an element ofHn1. Noticethat we insist that the 1i in this expression multiply from the right. Thegeneral pattern for the overlap maps become clear from the example 3 12 .Here have

3 12 (y1, y3,...,yn) = 3([y1, 1, y3,...,yn])=

y1y

13 , y

13 , y4y

13 ,...,yny

13

In the special casen = 1, we have an atlas of just two charts {(U1, 1), (U2, 2)}and in close analogy with the complex case we have U1 U2 = H\{0} and1 12 (y) = y1 =2 11 (y) fory H\{0}.

Exercise 3.28 Show that by identifyingH withR4 and modifying the stereo-graphic charts onS3 R4 we can obtain an atlas forS3 with overlap maps ofthe same form as forHP1 given above. Use this to show thatHP1=S3.

Combining the last exercise with previous results we have

RP1=S1CP1=S2HP1=S3

3.2 Homogeneous Spaces

LetHbe a closed Lie subgroup of a Lie group G. The we have arightaction ofH on G given by right multiplication r : G H G. The orbits of this rightaction are exactly the left cosets of the quotient G/H. The action is clearly freeand we would like to show that it is also proper. Since we are now talking about

a right action and G is the manifold on which we are acting, we need to showthat the mapPright : G H GGgiven by (p, h) (p, ph) is a proper map.The characterization of proper action becomes the condition that

HK := {h H : (K h) K= }

is compact whenever K is compact. To this end let Kbe any compact subsetofG. It will suffice to show that HK is sequentially compact and to this end let


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3.2. HOMOGENEOUS SPACES 27

{hi}iZ+ be a sequence inHK. Then there must be a sequences {ai} and {bi} inKsuch that a

ihi= b

i. SinceK is compact and hence sequentially compact, we

can pass to subsequences{ai(j)}jZ+ and{bi(j)}jZ+ so that limj ai(j) =aand limj bi(j) = b. Here i i(j) is a monotonic map on positive integers;Z+ Z+. This means that limj hi(j) = limj a1i(j)bi(j) = a1b. Thusthe original sequence{hi} is shown to have a convergent subsequence and weconclude that the right action is proper. Using Theorem 3.25 (or its analoguefor right actions) we obtain

Proposition 3.29 LetHbe a closed Lie subgroup of a Lie group G theni) the right actionG H G is free and properii) the orbit space is the left coset space G/H and this has a unique smoothmanifold structure such that the quotient map : G G/H is a surjection.Furthermore,dim(G/H) = dim(G)

dim(H).

IfKis a normal Lie subgroup ofG then the quotient is a group with multi-plication given by [g1][g2] = (g1K)(g2K) = g1g2K. The normality ofKis whatmakes this definition well defined. In this case, we may ask whether G/K isa Lie group. If K is closed then we know from the considerations above thatG/K is a smooth manifold and that the quotient map is smooth. In fact, wehave the following

Proposition 3.30 (Quotient Lie Groups) If K is closed normal subgroupof a Lie group G then G/K is a Lie group and the quotient map G G/Kis a Lie group homomorphism. Furthermore, ifh : G H is a surjective Liegroup homomorphism thenker(h) is a closed normal subgroup and the induced

maph: G/ ker(h) His a Lie group isomorphism.Proof. We have already observed thatG/Kis a smooth manifold and that

the quotient map is smooth. After taking into account what we know fromstandard group theory the only thing we need to prove for the first part isthat the multiplication and inversion in the quotient are smooth. it is an easyexercise using corollary ?? to show that both of these maps are smooth.

Consider a Lie group homomorphism h as in the hypothesis of the proposi-tion. It is standard that ker(h) is a normal subgroup and it is clearly closed. It

is also easy to verify fact that the inducedh map is an isomorphism. One canthen use Corollary ?? to show that the induced maph is smooth.

If a group G acts transitively on M (on the right or left) then M is calleda homogeneous space with respect to that action. Of course it is possiblethat a single group G may act on Min more than one way and so Mmay be ahomogeneous space in more than one way. We will give a few concrete examplesshortly but we already have an abstract example on hand.

Theorem 3.31 If H is a closed Lie subgroup of a Lie group G then the mapG G/H G given by l : (g, g1H) gg1H is a transitive Lie group action.ThusG/H is a homogeneous space with respect to this action.


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Proof.The fact thatl is well defined follows since ifg1H=g2H g12 g1 H

and so gg2

H = gg2

g12

g1

H = gg1

H. We already know that G/H is a smoothmanifold and : G G/H is a surjective submersion. We can form anothersubmersionidG: GG GG/Hmaking the following diagram commute:

G G idG

G

G G/H G/H

Here the upper horizontal map is group multiplication and the lower hori-zontal map is the action l. Since the diagonal map is smooth, it follows fromProposition?? thatl is smooth. We see that l is transitive by observing that ifg1H, g2H G/H then

lg2g11 (g1H) =g2H

It turns out that up to appropriate equivalence, the examples of the abovetype account for all homogeneous spaces. Before proving this let us look at someconcrete examples.

Example 3.32 Let M = Rn and G = Euc(n,R) the group of Euclidean mo-tions. We realizeEuc(n,R) as a matrix group

Euc(n,R) =

1 0v Q

: v Rn andQ O(n)

The action ofEuc(n,R) onRn is given by the rule 1 0

v Q

x= Qx +v

wherex is written as a column vector. Notice that this action is not given by amatrix multiplication but one can use the trick of representing the pointsx ofRn

by the (n+ 1) 1 column vectors

1x

and then we have

1 0v Q

1x

=

1Qx +v

. The action is easily seen to be transitive.

Example 3.33 As in the previous example we takeM =Rn but this time thegroup acting is the affine group Af f(n,R) which we realize as a matrix group:

Af f(n,R) = 1 0

v A

: v Rn andA GL(n,R)The action is

1 0v A

x= Ax +v

and this is again a transitive action.


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Comparing these first two examples we see that we have made Rn into ahomogeneous space in two different ways. It is sometime desirable to give dif-ferent names and/or notations for Rn to distinguish how we are acting on thespace. In the first example we might denoteEn (Euclidean space) and in thesecond case byAn and refer to it as affine space. Note that, roughly speakingthe action by E uc(n,R) preserves all metric properties of figures such as curvesdefined inEn. On the other hand,Af f(n,R) always sends lines to lines, planesto planes etc.

Example 3.34 Let M = H :={z C : Im z > 0}. This is the upper halfcomplex plane. The group acting onH will be Sl(2,R) and the action is givenby

a bc d

z= az+b

cz+d.

This action is transitive.

Example 3.35 We have already seen in Example 3.7 that bothO(n)andSO(n)act transitively on the sphereSn1 Rn so Sn1 is a homogeneous space in atleast two (slightly) different ways. Also, both SU(n) andU(n) act transitivelyS2n1 Cn.Example 3.36 Let Vn,k denote the set of all k-frames forR

n where by a k-frame we mean an ordered set of k linearly independent vectors. Thus annframe is just an ordered basis forRn. This set can easily be given a smoothmanifold structure. This manifold is called the (real) Stiefel manifold of k-

frames. The Lie group GL(n,R) acts (smoothly) on Vn,k by g (e1,...,ek) =(ge1,...,gek). To see that this action is transitive let(e1,...,ek)and(f1,...,fk)betwo k-frames. Extend each to nframes (e1,...,ek,...,en) and(f1,...,fk,...,fn)since we consider elements ofRn as column vectors these two nframes can beviewed as invertible n n matrices E and F. The if we let g := EF1 thengE= F which entailsg (e1,...,ek) = (ge1,...,gek) = (f1,...,fk).Example 3.37 Let Vn,k denote the set of all orthonormal k-frames for Rn

where by an orthonormalk-frame we mean an ordered set ofk orthonormal vec-tors. Thus an orthonormalnframe is just an orthonormal basis forRn. Thisset can easily be given a smooth manifold structure and is called the Stiefelmanifold of orthonormalk-frames. The group O(n,R)acts transitively onVn,k for reasons similar to those given in the last example.

Theorem 3.38 Let M be a homogeneous space via the transitive action l :G M Mand let Gp the isotropy subgroup of a point p M. Recall thatG acts on G/Gp. If G/Gp is second countable (in particular if G is secondcountable), then there is an equivariant diffeomorphism : G/Gp M suchthat(gGp) =g p.

Proof.We want to defineby the rule(gGp) =g pbut we must show thatthis is well defined. This is a standard group theory argument; ifg1Gp =g2Gp


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theng11 g2 Gp so that

g11 g2

p= p or g1 p= g2 p. This map is a surjective

by the transitivity of the action l. It is also injective since if(g1

Gp

) = (g2

Gp

)theng1 p= g2 por g11 g2 pwhich by definition means thatg11 g2 Gp andtheng1Gp = g2Gp. Notice that the following diagram commutes:

G

p

G/Gp

M

From Corollary ?? we see that is smooth.To show that is a diffeomorphism it suffices to show that the rank of

is equal to dim Mor in other words that is a submersion. Since(gg1Gp) =(gg1) p = g(g1Gp) the map is equivariant and so has constant rank. ByLemma ?? is a submersion and hence in the present case a diffeomorphism.

Without the technical assumption on second countability, the proof showsthat we still have that : G/Gp M is a smooth equivariant bijection.Exercise 3.39 Show that if instead of the hypothesis of second countability inthe last theorem we assume instead that p has full rank at the identity then: G/Gp M is a diffeomorphism.

Letl : G M Mbe a left action and fix p0 M. Denote the projectiononto cosets by and also write rp0 : g gp0. Then we have the followingequivalence of maps

G = G

rp0

G/H = MExercise 3.40 LetG act onM as above. LetH1 :=Gp1 (isotropy ofp1) andH2 := Gp2 (isotropy of p2) where p2 = gp1 for some g G. Show that thereis a natural Lie group isomorphisms H1= H2 and a natural diffeomorphismG/H1=G/H2 which is an equivalence of actions.

We now look again at some of our examples of homogeneous spaces andapply the above theorem.

Example 3.41 Consider again Example 3.32. The isotropy group of the origininRn is the subgroup consisting of matrices of the form

1 00 Q where Q O(n). This group is clearly isomorphic to O(n,R) and so by theabove theorem we have an equivariant diffeomorphism

Rn= Euc(n,R)O(n,R)


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Example 3.42 Consider again Example 3.33. The isotropy group of the origininRn is the subgroup consisting of matrices of the form

1 00 A

whereA Gl(n,R). This group is clearly isomorphic to Gl(n,R) and so by theabove theorem we have an equivariant diffeomorphism

Rn= Af f(n,R)Gl(n,R)

It is important to realize that there is an implied action onRn which is differentfrom that in the previous example.

Example 3.43 Now consider the action ofSl(2,R) on the upper half complex

plane as in Example 3.34. Let us determine the isotropy subgroup for the pointi=

1 . IfA= a b

c d

is in this subgroup then

ai +b

ci+d=i

so that bc ad = 1 and bd+ ac = 0. Thus A SO(2,R) (which is isomor-phic as a Lie group to the circle S1 = U(1,C)). Thus we have an equivariantdiffeomorphism

H= C+= Sl(2,R)SO(2,R)

Example 3.44 From example 3.35 we obtain

Sn1= O(n)O(n 1)

Sn1= SO(n)SO(n 1)

S2n1= U(n)U(n 1)

S2n1= SU(n)SU(n 1)

Example 3.45 Let (e1,..., en) be the standard basis forRn. Under the actionofGL(n,R)onVn,k given in Example 3.36, the isotropy group of the point whichis thek-planee= (ek+1,..., en) is the subgroup ofGl(n,R) of the form

A 00 id

forA Gl(n k,R)

We identify this group withGl(n k,R) and then we obtain

Vn,k= Gl(n,R)

Gl(n k,R)


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Example 3.46 A similar analysis leads to an equivariant diffeomorphism

Vn,k= O(n,R)O(n k,R)

where Vn,k is the Stiefel manifold of orthonormal k-planes of Example 3.37.Notice that takingk= 1 we recover Example 3.41.

Exercise 3.47 Show that ifk < n then we haveVn,k= SO(n,R)SO(nk,R) .

Next we introduce a couple of standard results concerning connectivity.

Proposition 3.48 LetG be a Lie group acting freely and properly on a smoothmanifoldM. Let the action be a left (reps. right) action. If bothG andM\G(resp. M/G) are connected thenM is connected.

Proof. Assume for concreteness that the action is a left action and thatG and M\G are connected. Suppose by way of contradiction that M is notconnected. Then there are disjoint open setU and V whose union is M. EachorbitG pis the image of the connected space G under the orbit map g g pand so is connected. This means that each orbit must be contained in one andonly one ofU andV. Now since the quotient map is an open map, (U) and(V) are open and from what we have just observed they must be disjoint and(U) (V) =M\G. This contradicts the assumption thatM\Gis connected.

Corollary 3.49 LetHbe a closed Lie subgroup ofG. Then if bothHandG/H

are connected thenG is connected.

Corollary 3.50 For eachn 1 then groupsSO(n), SU(n) andU(n) are con-nected while the group O(n) has exactly two components.

Proof. SO(1) and SU(1) are both connected since they each contain onlyone element. U(1) is the circle and so it too is connected. We use induction.Suppose that SO(k), SU(k) are connected for 1 k n 1. We show thatthis implies thatSO(n), SU(n) andU(n) are connected. From example 3.44 weknow thatSn1 =SO(n)/SO(n 1). SinceSn1 andSO(n 1) are connected(the second one by the induction hypothesis) we see that SO(n) is connected.The same argument works for S U(n) and U(n).

Every element ofO(n) has either determinant 1 or

1. The subsetSO(n)

O(n) since it is exactly{g O(n) : det g= 1}. If we fix an element a0 withdet a0 =1 then SO(n) and a0SO(n) are disjoint, open and both connectedsinceg a0g is a diffeomorphism which maps the first to the second. It is easyto show that S O(n) a0SO(n) = O(n).

We close this chapter by relating the notion of a Lie group action with thatof a Lie group representation. We given just a few basic definitions, some ofwhich will be used in the next chapter.


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Definition 3.51 A representation of a Lie group G in a finite dimensionalvector spaceV is a left Lie group action : G

V

Vsuch that for eachg

G

the map g :v (g, v) is linear. Thus a representation is a linear action.

The mapG Gl(V) given byg (g) := g is a Lie group homomorphismand will be denoted by the same letter as the action so that (g)v:= (g, v).In fact, given a Lie group homomorphism : G Gl(V) we obtain a linearaction by letting (g, v) :=(g)v. Thus a representation is basically the samething as a Lie group homomorphism into Gl(V) and is often defined as such. Thekernel of the action is the kernel of the associated homomorphism. A faithfulrepresentationis one that acts effectively and this means that the associatedhomomorphism has trivial kernel.

Exercise 3.52 Show that if: G V V is a map such thatg :v (g, v)is linear for allg , then is smooth if and only ifg :G Gl(V)is smooth foreveryg G. (AssumeV is finite dimensional as usual).

We have already seen one important example of a Lie group representation.Namely, the adjoint representation. The adjoint representation came from firstconsidering the action of G on itself given by conjugation which leaves theidentity element fixed. The idea can be generalized:

Theorem 3.53 Let l: G M Mbe a (left) Lie group action. Suppose thatp0 M is a fixed point of the action (lg(p0) = p0 for allg). The map

l(p0) :G Gl(Tp0M)

given byl(p0)(g) :=Tp0 lg

is a Lie group representation.

Proof. Since

l(p0)(g1g2) = Tp0(lg1g2) = Tp0(lg1 lg2)=Tp0 lg1 Tp0 lg2 =l(p0)(g1)l(p0)(g2)

we see that l(p0) is a homomorphism. We must show that l(p0) is smooth. ByExercise 3.52 this implies that the map G Tp0M Tp0M given by (g, v)Tp0 lg

v is smooth. It will be enough to show thatg

(Tp0 lg

v) is smooth

for any v Tp0M and any Tp0M. This will follow if we can show that forfixed v0 Tp0M, the map G T M given by g Tp0 lg v0 is smooth. Thismap is a composition

G T G T M=T(G M) T l T M

where the first map isg (0g, v0) which is clearly smooth.


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Definition 3.54 For a Lie group actionl : G M Mwith fixed pointp0 therepresentationl(p0) from the last theorem is called theisotropy representationfor the fixed point.

Now let us consider a transitive Lie group action l : G M M and apoint p0. For notational convenience, denote the isotropy subgroupGp0 byH.ThenHacts on M by restriction. We denote this action by : H M M

: (h, p) h for h H=Gp0Now notice that p0 is a fixed point of this action and so we have an isotropyrepresentation(p0) :H Tp0M Tp0M. On the other hand, we have anotheractionC :H GG where Ch :GG is given by ghgh1 for hH.The Lie differential ofCh is the adjoint map Adh : g g. The map Ch fixesHandAdhfixesh. Thus the mapAdh : g gdescends to a map

Adh : g/h g/h.

We are going to show that there is a natural isomorphism Tp0

M=

g/h suchthat for each h H the following diagram commutes:Adh : g/h g/h

(p0)h : Tp0M Tp0M

(3.1)

One way to state the meaning of this result is to say that h Adh is a repre-sentation ofHon the vector space g/hwhich is equivalent to the linear isotropyrepresentation. The isomorphism Tp0M

= g/h is given in the following verynatural way: Let h and consider Te() Tp0M. We have

Te() = d

dt

t=0

(exp t) = 0

since exp t h for all t. Thus h ker(Te). On the other hand, dim h =dim H= dim(ker(Te)) so in fact h= ker(Te) and we obtain an isomorphismg/h =Tp0M induced Te. Let us now see why the diagram 3.1 commutes. Letus take a scenic root to the conclusion since it allows us to see the big picturea bit better. First the following diagram clearly commutes:

G Ch G

G/H

Lh G/HElementwise we have

g Ch hgh1

gH Lh hgHUsing our equivariant diffeomorphism : G/H M we obtain an equivalentcommutative diagram

G Ch G

rp0 rp0 M

h M


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Applying these maps to exp t for g we have

exp t h(exp t)h1rp0 rp0

(exp t)p0 h(exp t)p0.

Applying the tangent functor (looking at the differential) we get the commuta-tive diagram

g Adh g

Terp0 Terp0

Tp0M

(p0)

h

Tp0M

and taking quotients, this gives the desired commutative diagram 3.1.

3.2.1 Reductive case

Definition 3.55 IfMis a homogeneous space with group G andH=Gp0 thenwe say that it is areductive homogeneous space if there exist a subspacem gwhich is complementary to h such thatAdh(m) m for allh H.

In general, we do notclaim that m is a Lie subalgebra ofg. The conditionon m is that it be Ad(H)-invariant. Since m is complementary to h we havem =g/h. Thusm =Tp0M. However, the assumption that m is Ad(H)-invariantgives more. Namely,h Adh|mis a representation ofHin mand it is equivalentto the isotropy representation. For the reductive case, we have a commutativediagram valid for all h H:

Adh|m m m

Adh : g/h g/h

(p0)h : Tp0M Tp0M

where the map mg/h is the restriction of the quotient map gg/h to thesubspacem. It is a linear isomorphism. In the reductive case, each of the maps

h Adh|m, h Adh andh (p0)h are equivalent and can be considered as aversion the isotropy representation.


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Suppose that V is an F-vector space and letB ={v1,...,vn} be a basisfor V. Then denoting the matrix representative of g with respect toB by[g]B we obtain a homomorphism G Gl(n,F) given by g [g]B. Ingeneral, a Lie group homomorphism of a Lie groupG into Gl(n,F) is calleda matrix representation of G. Notice that any Lie subgroup G of Gl(V) actson V in the obvious was simply by employing the definition ofGl(V) as a setof linear transformation of V. We call this the standard action of the linearLie subgroup ofGl(V) on V and the corresponding homomorphism is just theinclusion map G Gl(n,F). Choosing a basis, the subgroup corresponds toa matrix group and the standard action becomes matrix multiplication on theleft ofFn where the latter is viewed as a space of column vectors. This actionof a matrix group on column vectors is also referred to as a standard action.

Given a representation ofG in a vector space V we have a representation ofG in the dual space V by defining (g) := (g1)t : V V. Herewe have by definition

(g1)tv, w

=

(g1)v, w

for all v, w V and where

., .: V VF is the natural bilinear pairing which defines the dual. Thisdual representation is also sometimes called the contragradient representation(especially whenF = R).

Now letV andW be representations of a lie groupG in F-vector spaces Vand W respectively. We can then form the direct product representationV

W

by V Wg :=Vg Wg for gG and where we have Vg Wg (v, w) =(Vgv,

Ww).One can also form the tensor product of representations. The definitions

and basic facts about tensor products are given in the more general context ofmodule theory in Appendix ??. Here we given a quick recounting of the notionof a tensor product of vector spaces and then we defined tensor products ofrepresentations. Given two vector spaces V1and V2over some field F. Consider


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the spaceCV1V2 consisting of all bilinear maps V1 V2 W where W variesover all F-vector spaces but V

1 and V

2are fixed. A morphism from, say

1 :

V1 V2 W1 to 2 : V1 V2 W2 is defined to be a map : W1 W2such that the diagram

W1

V1 V2

1

2

W2

commutes.

There exists a vector space TV1,V2 together with a bilinear map : V1V2 TV1,V2 that has the following universal property: For every bilinear map: V1 V2 W there is a unique linear map: TV1,V2 W such that thefollowing diagram commutes:

V1 V2

W

TV1,V2

If such a pair (TV1,V2 , ) exists with this property then it is unique up toisomorphism inCV1V2 . In other words, if : V1 V2TV1,V2 is anotherobject with this universal property then there is a linear isomorphism TV1,V2

=TV1,V2 such that the following diagram commutes:TV1,V2

=

V1 V2

TV1,V2We refer to any such universal object as a tensor productof V1 and V2. Wewill indicate the construction of a specific tensor product that we denote byV1 V2 with corresponding map : V1 V2 V1 V2. The idea is simple:We let V1 V2 be the set of all linear combinations of symbols of the form


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v1 v2 forv1 V1 and v2 V2, subject to the relations

(v1+v2) v= v1 v+v2 vv (v1+v2) =v v1+v v2

r (v1 v2) =rv1 v2= v1 rv2The mapis then simply : (v1, v2) v1 v2. Somewhat more pedantically,letF(V1 V2) denote the free vector space generated by the set V1 V2 (theelements of V1 V2 are treated as a basis for the space and so the free spacehas dimension equal to the cardinality of the set V1 V2). Next we define anequivalence relation F(V1 V2) generated by the relations

(av1, v2) a(v1, v2)(v1, av2) a(v1, v2)

(v+w, v2) (v, v2) + (w, v2)(v1, v+w) (v1, v) + (v1, w)

for v1, v V1, v2, w V2 and a F. Then we let V1 V2 :=F(V1 Vk)/ and denote the equivalence class of (v1, v2) byv1 v2.

Tensor products of several vector spaces at a time are constructed similarlyto be a universal space in a category of multilinear maps. We may also form thetensor products two at a time and then use the easily proved fact (V W) U =V (W U) which is then denoted by V W U. Again the reader isreferred to Appendix ?? for more about tensor products.

Elements of the form v1

v2 generate V1

V2 and in fact, if

{e1,...,er

} is

a basis for V1 and{f1,...,fs} is a basis for V2 then set{ei fj : 1 i r, 1 j s}

is a basis for V1 V2 which therefore has dimension rs= dim V1dim V2.One more observation: IfA: V1 W1 and B : V2 W2 are linear, then

we can define a linear map A B: V1 V2 W1 W2. It is enough to defineA B on elements of the form v1 v2 and then extend linearly:

A B(v1 v2) = Av1 Bv2Notice that ifAandBare invertible thenABis invertible with (AB)1(v1v2) = A

1v1 B1v2. Pick bases for V1 and V2 as above and bases{e1,...,er}and

{f1,...,f

s

}for W1 and W2 respectively. The for

V1

V2 we can write

=ijei fj using the Einstein summation convention. We haveA B() = A B(ijei fj)

=ijAei Bfj=ijAki e

k Bljfl

=ijAki Blj(e

k fl )


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so that the matrix ofA B is given by (A B)klij =Aki Blj .

Now letV

andW

be representations of a lie groupG in F-vector spaces Vand W respectively. We can form a representation ofG in the tensor productspace by letting

V W

g:= Vg Wg for all g G. There is a variation on

the tensor product that is useful when we have two groups involved. IfV is arepresentation of a lie groupG1 in F-vector space V and W is representation ofa lie group G2 in theF-vector space W, then we can form a representation of athe Lie groupG1 G2also called the tensor product representation and denotedV W as before. In this case the definition is V W

(g1,g2):= Vg1 Wg2 .

Of course if it happens that G1 = G2 then have an ambiguity since V W

could be a representation of G or of G G. One usually determines whichversion is meant from the context. Alternatively, one can use pairs to denoteactions so that an action : G V V is denoted (G, ). Then the twotensor product representations would be (G G,

V

W

) and (G, V

W

)respectively.


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Chapter 4

Killing Fields andSymmetric Spaces

Following the excellent treatment given by Lang [?], we will treat Killing fields ina more general setting than is traditional. We focus attention on pairs (M, )where M is a smooth manifold and is a (not necessarily metric) torsionfree connection. Now let (M, M) and (N, N) be two such pairs. For adiffeomorphism : M N the pull-back connection is defined so that

(N)XY =(MXY)

for X, Y

X(N) and X is defined as before by X = T 1

X

. An

isomorphism of pairs (M, M) (N, N) is a diffeomorphism : M Msuch that = . Automorphism of a pair (M, ) is defined in the obviousway. Because we will be dealing with the flows of vector fields which are notnecessarily complete we will have to deal with maps :: M M with domaina proper open subsetO M. In this case we say that is a local isomorphismif |1(O)=|O.

Definition 4.1 A vector fieldX X(M) is called aKilling fieldifXt isa local isomorphism of the pair(M, ) for all sufficiently smallt.

Definition 4.2 LetM, g be a semi-Riemannian manifold. A vector fieldXX(M) is called ag

Killing field ifX is a local isometry.

It is clear that ifg is the Levi-Civita connection for M, g then any isometryofM, g is an automorphism of (M, g) and any gKilling field is a gKillingfield. The reverse statements are not always true. Letting Killg(M) denote thegKilling fields and Kill(M) theKilling fields we have

Killg(M) Killg(M)

41


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42 CHAPTER 4. KILLING FIELDS AND SYMMETRIC SPACES

Lemma 4.3 If(M, ) is as above then for anyX, Y , Z X(M) we have

[X, ZY] = Y,Z RY,ZX+ [X,Z]Y + Z[X, Y]Proof. The proof is a straightforward calculation left to the reader.

Theorem 4.4 Let M be a smooth manifold and a torsion free connection.ForX X(M), the following three conditions are equivalent:

(i) X is aKilling field

(ii) [X, ZY] = [X,Z]Y + Z[X, Y] for allY, Z X(M)

(iii)Y,ZX=RY,ZX for allY, Z X(M).

Proof. The equivalence of (ii) and (iii) follows from the previous lemma.Lett :=Xt . IfXis Killing (so (i) is true) then locally we have

ddt

tY =tLXY =t [X, Y] for all Y X(M). We also have tX=X. One calculatesthat

d

dtt (ZY) =t [X, ZY] = [tX, tZY]

= [X, tZY]and on the other hand

d

dttZtY = t [X,Z]tY + tZ(t [X, Y]).

Settingt = 0 and comparing we get (ii).Now assume (ii). We would like to show that ttZtY =ZY. We

show that ddtttZtY = 0 for all sufficiently small t. The thing to notice

here is that since the difference of connections is tensorial(Y, Z) = ddtttZtY

is tensorial being the limit of the a difference quotient. Thus we can assumethat [X, Y] = [X, Z] = 0. Thus tZ=Z and

tY =Y . We now have

d

dtttZtY

= d

dtt(ZY) =t[X, ZY]

=t([X,Z]Y + Z[X, Y]) = 0

But since

t

tZ

tYis equal to ZY when t= 0, we have that

t

tZ

tY =ZY for all t which is (i).Clearly the notion of a Jacobi field makes sense in the context of a general

(torsion free) connection on M. Notice also that an automorphism of a pair(M, ) has the property that is a geodesic if and only if is a geodesic.Proposition 4.5 X is aKilling field if and only ifX is a Jacobi fieldalong for every geodesic.


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43

Proof. If X is Killing then (s, t) Xs ((t)) is a variation of throughgeodesics and sot

X

(t) =

s s=0 Xs ((t)) is a Jacobi field. The proof ofthe converse (Karcher) is as follows: Suppose thatX is such that its restrictionto any geodesic is a Jacobi field. Then for a geodesic, we have

2(X ) = R(, X )= X X= X

where we have used to denote not only a field along but also an extensionto a neighborhood with [, X] = 0. But

X= X [, X] = X= X X= ,X

and so2

(X ) =,X. Now let v, w TpM for pM. Then there is ageodesic with (0) =v +w and soR(v, X)v+R(w, X)w+R(v, X)w+R(w, X)v

=R(v+w, X) (v+w) = ,X= v+w,v+wX= v,vX+ w,wX+ v,wX+ w,vX

Now replace w withw and subtract (polarization) to getv,wX+ w,vX=R(v, X)w+R(w, X)v.

On the other hand,v,wX w,vX = R(v, w)Xand adding this equation tothe previous one we get 2v,wX=R(v, X)w R(X, w)v R(w, v)Xand thenby the Bianchi identity 2v,wX = 2R(v, w)X. The result now follows fromtheorem 4.4.

We now give equivalent conditions for X X(M) to be agKilling field fora semi-Riemannian (M, g). First we need a lemma.

Lemma 4.6 For any vector fieldsX , Y , Z X(M) we haveLXY, Z = LXY, Z + Y, LXZ + (LXg) (Y, Z)

= LXY, Z + Y, LXZ + XY, Z + Y, XZProof. This is again a straightforward calculation using for the second

equality

(LX

g) (Y, Z) =LX

Y, ZLX

Y, Z

Y,LX

Z= LXY, Z + XY YX, Z + Y, XZ ZX

= LXY, Z + XY YX, Z + Y, XZ ZX= LXY, ZYX, Z Y, ZX + XY, Z + Y, XZ

0 + XY, Z + Y, XZ


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44 CHAPTER 4. KILLING FIELDS AND SYMMETRIC SPACES

Theorem 4.7 Let(M, g)be semi-Riemannian. X X(M)is agKilling fieldif and only if any one of the following conditions hold:

(i)LXg= 0.

(ii)LXY, Z = LXY, Z + Y, LXZ for allY, Z X(M).

(iii) For eachpM, the map TpM TpM given byv vX is skewsym-metric for the inner product., .p.

Proof. Since ifXis Killing thenXt g= g and in general for a vector fieldXwe have

d

dtXt g=

Xt LXg

the equivalence of (i) with the statement that X is gKilling is immediate.If (ii) holds then by the previous lemma, (i) holds and conversely. The

equivalence of (ii) and (i) also follows from the lemma and is left to the reader.


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Chapter 5

Fiber Bundles II

5.1 Principal and Associated Bundles

Let : E Mbe a vector bundle with typical fiber V and for every p M letGl(V, Ep) denote the set of linear isomorphisms from V to Ep. If we choose afixed basis{ei}i=1,...,k for V then we can identify each frame u = (u1,...,uk) atpwith the element ofGl(V, Ep) given byu(v) :=

viui wherev =

viei. With

this identification, notice that if := is the local section coming from aVB-chart (U, ) as described above then we have

(p) = |Ep for p U

Now let L(E) := pMGl(V, Ep) (disjoint union).It will shortly be clear that L(E) is a smooth manifold and the total space ofa fiber bundle. Let: L(E)Mbe the projection map defined by (u) =pwhenu Gl(V, Ep). Now observe thatGl(V) acts on the right of the set L(E).Indeed, the action L(E) Gl(V)L(E) is given by r : (u, g)ug =u g.If we pick a fixed basis for V as above then we may view g as a matrix and anelement u Gl(V, Ep) as a frame u = (u1,...,uk). In this case we have

ug= (uigi1,...,uig

ik)

It is easy to see that the orbit of a frame at p is exactly the set 1(p) =

Gl(V, Ep) and that the action is free. For each VB-chart (U, ) for E let be the associated frame field. Definef : U Gl(V)1 (U) byf(p, g) =(p)g. It is easy to check that this is a bijection. Let: 1 (U) UGl(V)be the inverse of this map. We have= (, ) where is a uniquely determinedmap. Starting with a VB-atlas{(U, )} for E, we construct a family{ :1 (U) U Gl(V)} of trivializations which gives a fiber bundle atlas{(U,)} for L(E) and simultaneously induces the smooth structure.

45


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46 CHAPTER 5. FIBER BUNDLES II

Definition 5.1 Let : E M be a vector bundle with typical fiber V. Thefiber bundle (L(E),,M,Gl(V)) constructed above is called the linear framebundle of E and is usually denoted simply by L(E). The frame bundle forthe tangent bundle of a manifold M is often denoted by L(M) rather than byL(T M).

Notice that any VB-atlas for E induces an atlas on L(E) according to ourconsiderations above. We have 1 (p, g) =( (p)g)

=( |Ep g)(p, (p)g).

Thus the transition functions ofL(E) are given by the standard transition func-tions ofEacting by left multiplication on Gl(V). In other words, the bundleL(E) has a Gl(V)-structure where the action ofGl(V) on the typical fiber Vis the standard one and the cocycle corresponding to the bundle atlas for L(E)that we constructed from a VB-atlas{(U, )} for original vector bundle E isthe very cocycle coming from this atlas on E.

We need to make one more observation concerning the right action ofGl(V)on L(E). Take a VB-chart for E, say (U, ), and let us look again at the asso-

ciated chart (1(U),) for L(E). First, consider the trivial bundlepr1 :UGl(V) Uand define the obvious rightaction on the total space 1(U)=U Gl(V) by ((p, g1), g) (p, g1g) := (p, g1) g. Then this action is transitiveon the fibers of this trivial bundle which are exactly the orbits of the action. Ofcourse since Gl(V) acts on L(E) and preserves fibers it also acts by restriction

on 1(U).

Proposition 5.2 The bundle map: 1(U) UGl(V)is equivariant withrespect to the right actions described above.

Proof. We first look at the inverse.1(p, g1)g = (p)g1g=1(p, g1g)Now to see this from the point of view of rather than its inverse just letu 1(U) L(E) and let (p, g1) be the unique pair such thatu =1(p, g1).Then

(ug) = 1(p, g1)g=1(p, g1g)= (p, g1g) = (p, g1) g=(u) g

A section ofL(E) over an open set U in M is just a frame field over U. Aglobal frame field is a global section ofL(E) and clearly a global section existsif and only ifE is trivial.


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5.1. PRINCIPAL AND ASSOCIATED BUNDLES 47

The linear frame bundles associated to vector bundles are example of prin-cipal bundles. For a frame bundle F(E) the following things stand out: Thetypical fiber is diffeomorphic to the structure group Gl(V) and we constructedan atlas which showed that L(E) had a Gl(V)-structure where the action wasleft multiplication. Furthermore there is a rightGl(V) action on the total space

L(E) which has the fibers as orbits. The charts have the form (1(U),) andderive from charts onEandis equivariant in a sense that(ug) =(u)gwhereif(u) = (p, g1) then (p, g1)g := (p, g1g) by definition. These facts motivate theconcept of a principal bundle.

Definition 5.3 Let : P Mbe a smooth fiber bundle with typical fiber a LiegroupG. The bundle (P,,M,G) is called a principal Gbundle if there isa smooth free right action ofG onP such

(i) The action preserves fibers; (ug) = (u) for allu P andg G.(ii) For each p M there exists a bundle chart (U, ) with p U and such

that if= (, ) then

(ug) = (u)g

for allu P andg G.

If the group G is understood, then we may refer to (P,,M,G) simply as aprincipal bundle.

Charts of the form described in (ii) in the definition are called principalbundle charts and an atlas consisting of principal bundle charts is called a

principal bundle atlas. If(u1) = (u2) then (u1) = (u2)g where g :=(u1)(u2)

1 and so

(u1) = ((u1), (u1)) =

((u2), (u2)g) = ((u2g), (u2g))

=(u2g)

Since is bijective we see that u1= u2g and so we conclude that the fibers areactually the orbits of the right action.

Notice that if (, U) and ( , U) are overlapping principal bundle chartswith = (, ) and = (, ) then

(ug)(ug)

1

= (u)gg

1

(u)

1

= (u)(u)

1

so that the map u (u)(u)1 is constant on fibers. This means there isa smooth function g :U U G such that

g(p) = (u)(u)1 (5.1)

whereu is any element in the fiber at p.


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48 CHAPTER 5. FIBER BUNDLES II

Lemma 5.4 Let(, U) and( , U) be overlapping principal bundle charts.For eachp

U

U

|1(p)

|1(p)1

(g) =g(p)g

where theg are given as above.

Proof. Let

|1(p)1

(g) = u. Then g = (u) and so |1(p) |1(p)(g) = (u). On the other hand, u 1(p) and so

g(p)g = (u)(u)1g= (u)(u)

1(u)

= (u) = |1(p) |1(p)(g)

From this lemma we see that the structure group of a principal bundle is Gacting on itself by left translation. Conversely if (P,,M,G) is a fiber bundlewith aGatlas withG acting by left translation then (P,,M,G) is a principalbundle. To see this we only need to exhibit the free right action. Letu P andchoose a char

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