Differentiable Manifolds Notes

7/29/2019 Differentiable Manifolds Notes

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iv Contents

12. Connection 1-forms, curvature 2-forms and the Cartan

structure equations 6213. The Gauss-Bonnet formula 67

14. Index of a vector field 71

15. Stokess Theorem 73

16. Exercises on generalized geometry 74

Chapter 3. Geodesics 77

1. The first variation of arc length 77

2. Geodesics and the exponential map 82

Chapter 4. Introduction to Geometric Analysis 91

1. The Hessian and Laplacian of a function 91

2. Hypersurfaces in Riemannian manifolds 96

3. Distance functions 99

4. Volume 100

5. Covariant tensors and differentiable forms 101

Bibliography 107


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Preface

These are notes for my Fall 2011 Math 250A class, which uses do CarmosRiemannian Geometry book. Thanks to the students for helpful suggestions.

Initially I have strived for quantity instead of quality. Hopefully thenotes will continue to be edited and corrected in the future. So for themoment, errors, like pages, abound. As the chapters move forward, thequality of the notes follows the second law of thermal dynamics, namelydisorder increases.

Chapter 1. In 1 we define differentiable manifold and give the simplestexamples of Euclidean spaces and spheres.

In 2 we linearize a manifold at a point. The tangent space comprises

equivalence classes of curves in terms of how they act on functions as di-rectional derivative. Equivalently, tangent vectors are derivations, acting onfunctions linearly and satisfying the product rule.

In 3, given a map between manifolds, we can linearize it a point to getthe differential, which maps between tangent spaces. We introduce diffeo-morphisms and tangent bundle.

In 4 we state the result that the inverse image of a regular value is asubmanifold.

In 5 we consider vector fields, their Lie brackets, and 1-parametergroups of diffeomorphisms.

In 6 we define the Lie derivative and show that, acting on vector fields,it is the same as the Lie bracket.

In 7 we consider cotangent bundles and 1-forms.

v


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vi Preface

Chapter 2. In 1 we define Riemannian manifolds and give basic ex-

amples such as Euclidean space, the sphere, and hyperbolic space.In 2 we discuss the existence of Riemannian metrics and introduce

length and distance.

In 3 we discuss the geometry of Euclidean hypersurfaces, introducingthe second fundamental form.

In 4 we discuss Lie groups and left-invariant metrics.

In 5 we prove the existence of the Levi-Civita connection. This includesa prelude to tensors.

In 6 we discuss covariant differentiation along a curve.

In 7 we define the Riemann curvature tensor and discuss its symmetries.

In 8 we define the Ricci tensor.In 9 we discuss pull backs and Lie derivatives of tensors.

In 10 we discuss the covariant derivative of a tensor.

In 11 define the connection 1-forms and curvature 2-forms and derivethe Cartan structure equations.

In 12 we present some exercises on generalized geometry.

Chapter 3. In 1 we discuss the first variation of arc length; vectorfields along a map.

In 2 we discuss geodesics and the exponential map. Gauss lemma,HopfRinow, conjugate points, cut points, injectivity radius defintion. The

sphere as an example.Chapter 4. In 1 we discuss the Hessian and Laplacian of a function.

Bochner formula for |f|2. Divergence. Basic equations for Ricci solitons(quasi-Einstein metrics).

In 2 we discuss hypersurfaces in Riemannian manifolds, including levelsets and parametrized hypersurfaces.

In 3 we discuss distance functions and derive the Ricatti equation.

In 4 we discuss volume.

In 5 we discuss tensors and differential forms; exterior derivative.


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2 1. Differentiable Manifolds

for v = (v1, v2, . . . , vn) Rn. Note that for any q and v we have q+ tv

dom() for t sufficiently small. The linear transformation dq is representedby the matrix (

i

xj(q)), where =

1, 2, . . . , m

and where the entry

i

xj(q) is in the ith row and jth column.

Given an open set V Rn, a function f : V R is differentiable ifit is differentiable at each point of V. Let k be a nonnegative integer. Afunction f : V R is called Ck if it is continuously differentiable up to orderk, i.e., all of the partial derivatives of f of order k exist as functions on Vand they are all continuous on V. Smooth or C means Ck for all k 0.There are many other notions of regularity of functions such as Lipschitz,real analytic, etc.

1.2. The definition of differentiable manifold.A differentiable manifold is simply a locally Euclidean space in the dif-

ferentiable category.

Definition 1.1. An n-dimensional differentiable manifold atlas (orsimply atlas) on a set M is a family of injective maps

x : U M

of open sets U Rn into M for in a set such that

(1) x (U) = M,

(2) for each , with x (U) x (U) W = , x1 (W)

and x1 (W) are open subsets ofR

n

and the (bijective) transitionmaps x1 x : x

1 (W) x

1 (W) and x

1 x : x

1 (W)

x1 (W) are differentiable. (We say that (U, x) and (U, x) are(differentiably) compatible).

The map x : U M is called a parametrization or local coordi-nate chart.

The simplest example of an atlas on a set is Rn with the single mapidRn : R

n Rn.

Exercise 1.2. Given an atlas {(U, x)} on a set M, show that thereexists a unique atlas A on M such that

(1) {(U, x)} A,

(2) If B is any atlas such that {(U, x)} B, then B A.We callA amaximal atlas or a differentiable manifold struc-ture on M.

Hint: Define A to be the set of all (U, x) such that x : U Rn M iscompatible with each (U, x) , i.e., for each with x (U) x (U)


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1. Basics of differentiable manifolds 3

W = , x1 (W) and x

1 (W) are open subsets of Rn and the maps

x1 x : x1 (W) x1 (W) and x1 x : x1 (W) x1 (W) aredifferentiable.

Definition 1.3. Ann-dimensional differentiable manifold is a setMwith an n-dimensional maximal atlas A.

The definition of Ck manifold, for 0 k , is same as for differ-entiable manifold except that we require that the transition maps x1 xand x1 x are C

k.

Remark 1.4. The reader has probably noticed that given a pseudogroup of local transformations of Rn, we may define a manifold by simplyrequiring thatx1

x

, x1

x

(see Kobayashi and Nomizu [7]). In this

way we may define the notions of Lipschitz manifold, real analytic manifold,etc.

Euclidean space, as an n-dimensional C manifold, is Rn with theunique maximal C atlas containing the parametrization idRn : R

n Rn.

A differentiable manifold structure defines a topology on M.

Definition 1.5. A setA M is defined to beopeniffx1 (A x (U)) Rn

is open for all (U, x) A. The topology on M is the collection of open sets.

Exercise 1.6. Show that

(1) this definition of open set indeed defines a topology on M,(2) for each(U, x) A, x (U) M is open with respect to this topology

and x : U M is continuous.

Solution: (1)

(i) Clearly M is open since x1 ( x (U)) = Rn is open.Also, M is open since x1 (M x (U)) =U is open.

(ii) If{A}K is a collection of open sets in M, then for any parametriza-tion (U, x)

x1KA x (U) = x1

K(A x (U)) = Kx1 (A x (U))

is a union of open sets in Rn and hence open in Rn.

(iii) If{A}K is a finite collection of open sets, then for any parametriza-tion (U, x)

x1

K

A

x (U)

= x1

K

(A x (U))

=K

x1 (A x (U))


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2. Tangent spaces 5

2. Tangent spaces

Roughly speaking, the tangent space is the linearization of a manifold at apoint.

2.1. The notion of tangent space.

Let Dp be the set of functions on M that are differentiable at p. Let

Ck,locp be the set of functions f on M such that there exists a neighborhoodV of p on which f is Ck.

Motivated by the notion of directional derivative, the following is a stan-dard definition.

Definition 1.9. A derivation atp is a linear map X : D R that satisfies

the product rule:1

X(f g) = f(p) X(g) + g (p) X(f) .

The tangent space TpM at p is the set of derivations at p. We also callX(f) the directional derivative of f in the direction X.

A variant of this definition, which we shall also use, is to replace D byCk,locp , where 2 k .

Exercise 1.10. Show that TpM is a real vector space.

Let 1 : M R denote the constant function 1 (x) 1. Let X be a

tangent vector at some point p M, i.e., a derivation at p. Then by theproduct rule and since 12 = 1 we have

X(1) = X

12

= 1 X(1) + 1 X(1) = 2X(1) ,

which implies X(1) = 0. Hence, for any constant function const, we have

(1.2) X(const) = X(const 1) = const X(1) = 0.

Given a differentiable curve : (, ) M with (0) = p, we definethe corresponding derivation

(0) : D R

by

(1.3) (0)(f) = ddtt=0

(f ) (t) .

Exercise 1.11. Show that (0) is a derivation at p, so that (0) TpM.

1The derivation X being linear means that for any c R and f, g D,

X(cf+ g) = cX(f) + X(g) .


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Solution. The verification of linearity is elementary. As for the product

rule, we compute

(0)(f g) =d

dt

t=0

((f g) ) (t)

=d

dt

t=0

((f ) (t) (g ) (t))

= (f )(0)d

dt

t=0

((g ) (t)) + (g )(0)d

dt

t=0

((f ) (t))

= f(p) (0)(g) + g (p) (0)(f) .

Definition 1.12. We may also define the tangent space Tp

M to be the setof (0), where : (, ) M is a differentiable curve with (0) = p.

The following, which we will prove later, implies that the two definitionsof TpM are the same.

Theorem 1.13. Let M be a C2 manifold. If X is a derivation at p, thenthere exists a differentiable curve such that (0) = X.

2.2. The tangent space is n-dimensional.

Manifolds are defined by parametrizations, so it is natural to understandthe tangent space this way. Let x : U M be a parametrization withp x (U). Let ei = (0, . . . , 1

ith

, 0, . . . , 0) for 1 i n. {ei}ni=1 is called the

standard basis for Rn. Consider the Euclidean coordinate lines

(1.4) i : t x1 (p) + tei.

Pushing these curves forward by the parametrization x, we obtain the co-ordinate curves:

(1.5) ci (t) x (i (t)) = x x1 (p) + tei .

Since ci is differentiable and ci (0) = p, ci (0) : D R is a tangent vector at

p, called a coordinate tangent vector.

Theorem 1.14. Let M be a differentiable manifold. With Definition 1.12of the tangent space, {ci (0)}

ni=1 is a basis for TpM. In particular, TpM is

n-dimensional.


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2. Tangent spaces 7

Proof. We use the parametrization x to understand an arbitrary tangent

vector (0). We compute

(0)(f) =d

dt

t=0

((f ) (t))

=d

dt

t=0

(f x)

x1

(t)

= d (f x)x1(p)

d

dt

t=0

x1

(t)

,

where the Euclidean differential d (f x)x1(p) is defined as in (1.1). Note

that

x1

(0) = x1 (p).

Letd

dt

t=0

x1

(t) v =

v1, . . . , vn

Rn.

Then we may write

(0)(f) = d (f x)x1(p) (v)

for any f D and differentiable curve . So given a parametrization x,the derivation (0) depends only on d

dt

t=0

x1

(t)

= v Rn. Thisindicates that TpM should be n-dimensional. We now prove this rigorously.

Now we show that (0) may be written as a linear combination of thec

i

(0). Using v = ni=1 viei and the linearity of d (f x)x1(p), we computethat (0)(f) = d (f x)x1(p) (v)

=ni=1

vid (f x)x1(p) (ei)

=ni=1

vid

dt

t=0

f

x

x1 (p) + tei

=n

i=1vi

d

dt

t=0

(f ci) (t)

=ni=1

vici (0)(f) .

Since this true for each f D, we conclude that

(0) =ni=1

vici (0) ,


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where vi = ddt t=0 x

1

i

( (t)). This shows that {ci (0)}

ni=1 spans the

tangent space TpM in Definition 1.12.

To see that the ci (0) are linearly independent, suppose that ai R are

such thatni=1

aici (0) = 0.

Then choosing f =

x1j

, we obtain

0 =ni=1

aici (0)

x1j

.

Now

ci (0)

x1j

=d

dt

t=0

x1j

ci

(t)

=d

dt

t=0

ji t

= ji .

Hence 0 =n

i=1 aiji = a

j . This completes the proof of Theorem 1.14.

Exercise 1.15. Let x : U M be a parametrization, p x (U), andci (t) = x

x1 (p) + tei

. Show that the derivation ci (0)p at p satisfies

ci (0)p (f) = (f x)qix1 (p)

for any C function f defined in a neighborhood of p.

Solution. We compute

ci (0)p (f) =d

dt

t=0

(f ci) (t)

=d

dt

t=0

(f x)

x1 (p) + tei

=n

j=1(f x)

qj x1 (p) d

dt t=0 x1 (p) + teij

=(f x)

qi

x1 (p)

since ddtt=0

x1 (p) + tei

j= ji .

We may identify a vector v Rn with the tangent vector at a pointq Rn defined by the derivation (0), where (t) = q+ tv. In this way


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2. Tangent spaces 9

we have a natural vector space isomorphism between Rn and TqRn for any

q Rn.

2.3. Equivalence of the two definitions of tangent space.

Now we return to Definition 1.9 of tangent space as the set of all deriva-tions at a point. We prove that each derivation arises from a curve.

Proof of Theorem 1.13. Let M be a C2 manifold. Let X be a derivationat p, that is, X : Dp R is a linear function such that

X(f g) = f(p) X(g) + g (p) X(f) .

Let x : U M be a parametrization with x (0) = p. We shall use thefollowing fact (proved below):

If f C2,locp , then there exist constants ai R and functions gi C1,locp

with gi (p) = 0 such that

(1.6) f(q) = f(p) +n

i=1 ai x1

i

(q) +n

i=1 gi (q) x1

i

(q) .

Let Xi = X((x1)i) R. Since X is a derivation, we compute for f C2,locpthat

X(f) =ni=1

X

ai

x1i

+ gi

x1i

=ni=1

(ai + gi (p)) X

x1i

+

x1i

(p) X(gi)

=

ni=1

aiXx1i=

ni=1

aiXi,

where we used gi (p) = 0 and

x1i

(p) = 0.


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On the other hand, ni=1

Xici (0)

(f) =

ni=1

Xici (0)(f)

=ni=1

Xin

j=1

ajci (0)

x1j

=n

i,j=1

Xiajji

=n

i=1Xiai

since ci (0)

x1j

= ji . Since the above formulas are true for any f Dp,

we conclude that X =n

i=1 Xici (0) when acting on f C

2,locp , where

Xi = X

x1i

R. By Lemma 1.16 below,

X =ni=1

Xici (0)

acting on f Dp. Note that X = (0), where

(t) = x

t

X1, . . . , X n

.

Proof of (1.6): By precomposing with the map x, (1.6) is equivalent to:

h (v) = h (0) +ni=1

aivi +

ni=1

ki (v) vi

for v = v1, . . . , vn U Rn, where h = f x C1 and ki = gi x C1and where ki (0) = 0.We shall prove that if h : U R is C2, then there exist a C1 function

ki with ki (0) = 0 such that

(1.7) h (v) = h (0) +ni=1

h

xi(0) vi +

ni=1

ki (v) vi.


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3. Differential of a map and tangent bundle 11

Proof of (1.7): We have

h (v) h (0) =1

0

d

dth (tv) dt

=

10

ni=1

vih

xi(tv) dt

=ni=1

vih

xi(0) +

ni=1

vi1

0

h

xi(tv)

h

xi(0)

dt.

Define

gi (v)

1

0

h

xi(tv)

h

xi(0)

dt.

Since (v) hxi

(tv) hxi

(0) is C1 in a neighborhood of 0, we have gi

C1,loc0 and gi (0) = 0.

The one loose end we have left is to prove the following.

Lemma 1.16. Let M be a C2 manifold. If X and Y are derivations at psuch that X equals Y when restricted to C2,locp , then X = Y (on all of Dp).

3. Differential of a map and tangent bundle

3.1. Differential of a map.

Let M and N be differentiable manifolds. A continuous map : M N is differentiable at p M if for any parametrizations (U, x) of M and(V, y) of N we have that y1 x is (a vector-valued function of an opensubset of Euclidean space) differentiable at x1 (p).

Let : M N be a map which is differentiable at p. Its differentialat p is defined by

dp : TpM T(p)N,

where

(1.8) dp

(0) ( ) (0)

for any differentiable curve : (, ) M with (0) = p.

Exercise 1.17. Show that the differential of is well defined, i.e., if and are differentiable curves such that (0) = (0), then ( ) (0) =( ) (0).

The differential satisfies (composition rule)

d ( )p = d(p) dp.


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If and its inverse are differentiable, then

d 1(p)

= (dp)1 .

We leave the verification of these properties, which follow from their Eu-clidean counterparts, to the reader.

3.2. Diffeomorphisms.

Let Mn and Nn be C differentiable manifolds. A bijective map :M N is called a diffeomorphism if both and 1 are C. Givenp M, a local diffeomorphism at p is a diffeomorphism : U V,where U is a neighborhood of p in M and U is a neighborhood of (p) in

N.

Proposition 1.18 (Inverse function theorem). If a C map : M Nand p M are such that dp is an isomorphism, then there exist neighbor-hoods U of p and V of (p) such that |U : U V is a diffeomorphism,i.e., is a local diffeomorphism at p.

3.3. Tangent bundle.

Given a differentiable manifold Mn, we can bundle the tangent spacesTpM, p M, together to form the tangent bundle. As a point-set, it isdefined by the disjoint union

TM = pM

TpM.

Equivalently, we may define it as the point-set

TM = {(p, v) : p M, v TpM} .

We have the projection map : TM M defined by (p, v) = p. Notethat 1 (p) = TpM.

We shall now give the tangent bundle TM a manifold structure. Givena parametrization (U, x) ofM, i.e., x :U Rn x (U) M, at each pointp x (U) we have the basis of tangent vectors

xip= ci (0)p ,

where ci (t) = x

x1 (p) + tei

. The subscript p is used to denote that the

tangent vector xi

is at p.

The coordinate tangent vectors are the push forwards by the parametriza-tion of the standard basis vectors ofRn.


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3. Differential of a map and tangent bundle 13

Lemma 1.19. Let ei (0, . . . , 0, 1

ith

, 0, . . . , 0) for i = 1, . . . , n. Considering

U as a differentiable manifold (it is an open subset ofRn), we have

(1.9)

xi x(q)= (dx)q (ei) ,

where ei TqRn, for each q U and i.

Proof. Given q U and i, define the Euclidean coordinate line i : (, ) U by

i (t) q+ tei.

Then i (0) = ei and

ci (t) = x (i (t)) .

By the definition of differential (1.8),

(dx)q (ei) = (dx)q

i (0)

= (x i) (0) = ci (0)x(q) =

xi x(q).

By taking linear combinations of the formula in the above lemma, weobtain an expression for (dx)q.

Lemma 1.20.

(dx)q (u) =n

i=1 ui

xi

x(q)

.

Proof. Using the linearity of (dx)q, we compute

(dx)q (u) = (dx)q (ni=1

uiei)

=ni=1

ui (dx)q (ei)

=ni=1

ui

xi x(q)

with the last equality by Lemma 1.19.

Note that for each q U, (dx)q : TqRn = Rn Tx(q)M is a vector space

isomorphism.

Corresponding to (U, x) define a parametrization (V, y) of the tangentbundle TM by

y : VU Rn TM,


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where

y (q, u) (x (q) , (dx)q (u) )for q U and u Rn. First note that since x :U x (U) is a bijection andeach (dx)q is an isomorphism, y : U R

n 1 (x (U)) is a bijection.

Lemma 1.21. Let {(U, x)} be a Ck+1 differentiable manifold struc-

ture on M. Then the maximal extension of the collection of correspondingparametrizations {(V, y)} is a C

k differentiable manifold structure onTM.

Proof. Since M = x (U) and since y (V) =

1 (x (U)), wehave TM =

y (V). Next, we need to show that the transition maps

are differentiable. By definition, we have

y (q, u) = (x (q) , (dx)q (u) )

for q U and u Rn. Thus for any , such that x (U) x (U) =

, the transition function y1 y is given by

y1 y (q, u) = ((x1 x) (q) , (dx)

1x(q)

(dx)q (u) )

= ((x1 x) (q) , d(x1 x)q (u) )

since (dx)1 = d(x1 ) and d(x

1 x) = d(x

1 ) dx. Since M is C

k+1,

we have that x1 x Ck+1 and hence y1 y C

k.

Denote = x1 x, which maps an open subset ofRn diffeomorphicallyto an open subset ofRn. Then

y1 y (q, u) = ( (q) , dq (u) ).

Then the differential of the transition function y1 y is given by

d

y1 y v

w

=

dq (v)

dq (w) + (Hess )q (u, v)

(1.10)

=

dq 0

(Hess )q (u, ) dq

vw

.

Here (Hess )q

(u, v) d

dt t=0 dq+tv (u) is the Hessian (or second deriva-tive) of .A differentiable manifold N is orientable if N has a (not maximal)

differentiable structure {(V, y)} such that the differentials of the transitionmaps y1 y have positive determinant.

Exercise 1.22. Using formula (1.10), prove that the tangent bundle TMis orientable.


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5. Vector fields, Lie brackets and Lie derivatives 15

Solution (sketch). We have

det

d

y1 y

= det dq 0

(Hess )q (u, ) dq

= (det(dq))

2 > 0.

4. Critical and regular points of functions and level surfaces

Let Mm andNn be C manifolds and let : M N be a C map. We saythat p M is a critical point of if the differential dp : TpM T(p)Nis not surjective. In this case, (p) is called a critical value of . Notethat if m < n, then each point p M is a critical point of any map .

We say that q N is a regular value of if it is not a critical valueof . In particular, if q / (M), then q is a regular value of .

Proposition 1.23. Suppose m n. If q N is a regular value of , thenthe set

Q 1 (q) = {p M : (p) = q} M

has the following property. For each p Q there exists a neighborhood V ofp in M and a C homeomorphism

x : U Rmn Q V M

such that (dx)u : TuU = Rmn Tx(u)M is injective for each u U. We

call Q a regular submanifold of dimension m n.

Exercise 1.24. Prove that Q is an (m n)-dimensional manifold.

Define f : Rn+1 R by f(x) = |x|2. One sees that

dfx = fx =

f

x1,

f

x2, . . . ,

f

xn+1

is given by dfx = 2x. As a map, dfx is

dfx (v) =n+1i=1

f

xivi

for v =

v1, v2, . . . , vn+1

Rn+1. So x is a critical point of f if and only ifx = 0. We conclude that if r > 0, then f1

r2

= Sn (r), i.e., the sphere of

radius r, is a a regular submanifold of dimension n.

5. Vector fields, Lie brackets and Lie derivatives

5.1. Vector fields.

Much of the study of differential and Riemannian geometry is in the C

category. For convenience, we now let Mn be a C differentiable manifold,so that the transition maps x1 x are C

.


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Definition 1.25. A vector field X on M is an assignment to each point

p M a tangent vector Xp X(p) TpM.

Equivalently, if : TM M denotes the projection map, then a vectorfield is a map X : M TM satisfying X = idM, the identity map ofM.

If Mn is a C differentiable manifold and if 0, then we say that Xis a C vector field if with respect to any C parametrization (U, x) wehave that each Xi :U R, defined by

Xi (p) Xp((x1)i),

is a C function for i = 1, . . . , n.

Exercise 1.26. Show that for any p x (U),

X(p) =ni=1

Xi (p)

xip.

Let f : U Rn R be a C function on an open set. The gradientvector field of f is

f = grad f =

f

x1, . . . ,

f

xn

.

For example, if U = Rn and f(x) = |x|2 =

x1

2

+

x2

2

+ + (xn)2, then

f = 2x.If X =

X1, X2

is a vector field on R2, define the vector field J(X) by

J(X) =

X2, X1

.

The underlying transformation J : R2 R2 is called an almost complexstructure. Note that X, J(X) = 0. Also,

J

|x|2

= J

2

x1, x2

= 2

x2, x1

.

If X =

X1, . . . , X n

is a C vector field on U Rn, then its diver-gence is the function defined by

div X =ni=1

Xixi

.

Note that if f : U Rn R is a C function, then its Laplacian is givenby

f =ni=1

xi(f)i =

ni=1

2f

(xi)2.


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Exercise 1.27. Compute |x|r onRn (possibly minus the origin) forr R.

Does r = 2 n for n 3 give you a special answer.

5.2. Lie brackets of vector fields.

Definition 1.28. The Lie bracket of two C vector fields X and Y isdefined, as a derivation at each point in M, by

(1.11) [X, Y] (f) X(Y (f)) Y (X(f))

for C any function f : M R.

We need to show that the Lie bracket is well defined, namely that it isa derivation at each point.

Let C

(M) denote the set of C

functions on M. Note that for avector field X and a function f, we have X(f) is a function. At a pointp M it is easy to check that (exercise)

[X, Y]p : C (M) R

is an R-linear function, i.e.,

[X, Y]p (cf + g) = c [X, Y]p (f) + [X, Y]p (g) .

Thus, to show that [X, Y]p is a derivation at p it is enough to verify the

product rule. We compute for any f, g C (M) and p M,2

[X, Y]p (f g) = Xp (Y (f g)) Yp (X(f g))

= Xp (f Y (g) + gY (f)) Yp (f X(g) + gX(f))

= f(p) Xp (Y (g)) + g (p) Xp (Y (f)) + Xp (f) Yp (g) + Xp (g) Yp (f)

f(p) Yp (X(g)) g (p) Yp (X(f)) Yp (f) Xp (g) Yp (g) Xp (f)

= f(p) (Xp (Y (g)) Yp (X(g))) g (p) (Xp (Y (f)) Yp (X(f)))

= f(p) [X, Y]p (g) g (p) [X, Y]p (f) .

Since [X, Y]p is R-linear on C (M) and since [X, Y]p (f g) = f(p) [X, Y]p (g)

g (p) [X, Y]p (f), we conclude that [X, Y]p is a derivation at p and that the

Lie bracket [X, Y] is a vector field on M.

It is clear that

[Y, X] = [X, Y] .Exercise 1.29. Let U be an open subset ofRn. Show that for vector fieldsV and W on U we have

[V, W] = V (W) W(V) .

2To simplify the way the expressions look, you may omit the parentheses and the subscripts

p in the notation.


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Exercise 1.33. Prove that for any C vector fields X and Y and C

functions g and h,(1.12) [gX,hY] = g (Xh) Y h (Y g) X+ gh [X, Y] .

Solution. Acting as a derivation on a C function f : M R, wehave

[gX,hY] f = gX(hY f) hY (gX f)

= g (X h Y f + h X(Y f)) h Y g X f hg Y (Xf)

= g (Xh) Y f h (Y g) Xf + gh [X, Y] f.

Let X =

i X

i xi

and Y =

j Y

j xj

. Using (1.12) and the fact that

xi , xj = 0, we compute[X, Y] = [

i

Xi

xi,j

Yj

xj]

=i,j

Xi

xiYj

xj Yj

xjXi

xi+ XiYj

xi,

xj

=i,j

Xi

Yj

xi

xj Yj

Xi

xj

xi

,

that is,

(1.13) [X, Y] =

ni,j=1

XiYjxi YiXjxi xj .The Jacobi identity says that for C vector fields X, Y, and Z, we

have

(1.14) [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0.

We may see this as follows. Acting as derivations on a function f : M R,we have

[X, [Y, Z]] (f) = X([Y, Z] (f)) [Y, Z] (X(f))

= X(Y (Z(f)) Z(Y (f))) Y (Z(X(f))) + Z(Y (X(f)))

= X Y Z f X Z Y f Y Z X f + ZY Xf ,where in the last line we omitted the () in the notation for the sake of brevity.By cyclically permuting the X,Y,Z, we obtain

[Z, [X, Y]] f = Z X Y f Z Y X f X Y Z f + Y X Z f

and

[Y, [Z, X]] = Y Z X f Y X Z f Z X Y f + XZY f .


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Summing these three equations, we obtain 6 pairs of cancelling terms (each

permutation of XY Z appears with both a plus and a minus. So we get theJacobi identity.

A real (or complex) Lie algebra is a real (or complex) vector space gwith a binary operation [ , ] : g g g, called the Lie bracket, with theproperties that for any c R (or C) and u,v,w g,

(1) (alternating) [v, u] = [u, v],

(2) (bilinear) [u + cv,w] = [u, w] + c [v, w] ,

(3) (Jacobi identity) [[u, v] , w] + [[v, w] , u] + [[w, u] , v] = 0.

The set (M) of C vector fields on M with the Lie bracket definedby (1.11) is a Lie algebra.

For a differentiable curve : (, ) M, let (t) be its correspondingderivation at (t), defined by

(t) (f) = (f ) (t)

for f C (M).

Lemma 1.34. Let M be a differentiable manifold and let X be a differen-tiable vector field on M.

Then for each p M there exists > 0, a neighborhood U of p, and adifferentiable map

: (, ) U M

such that for each q U the differentiable curve q : (, ) M definedby

q (t) (t, q) .

satisfies q (0) = q and

q (t) = X(q (t))

for t (, ).

Moreover q is unique in the sense that if : (, ) M is a differ-entiable curve such that (0) = q and

(t) = X( (t))

for t (, ), then = q. We callq an integral curve (or trajectory)to the vector field X.

The essence of the proof of this, which we omit, is the existence anduniqueness theorem for systems of ordinary differential equations (ODE).By pulling back the problem by x : U M from M to U Rn, we mayassume that M is an open subset ofRn. Given a differentiable vector field


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6. The Lie derivative on vector fields 21

X on an open set U Rn and given q U, there exists a unique solution

q to the system of ODE:dqdt

(t) = X(q (t)) ,

q (0) = q.

Moreover, this solution depends differentiably on q.

For each t (, ), define the map t : U M by t (q) (t, q). Itturns out that for each t (, ), t (U) is open and t : U t (U) is adiffeomorphism. {t} is called a local flow of X. We also say that {t} isthe 1-parameter (local) group of diffeomorphisms generated by X.

Exercise 1.35. Show that t+s = t s.

If : M N is a C diffeomorphism and if X is a C vector field onM, then d (X), defined by

d (X) (q) = d1(q)

X1(q)

is a C vector field on N. We call d (X) the push forward of X by .Some books denote this by (X).

When M is compact we obtain the following.

Lemma 1.36. Let M be a compact differentiable manifold and let X be adifferentiable vector field on M. Then there exists a differentiable map

: R M Msuch that for each q M the differentiable curve q : R M defined by

q (t) (t, q) .

satisfies q (0) = q and

q (t) = X(q (t))

for t R.

Assuming M is compact, for each t R, we define the differentiable mapt : M M by t (q) (t, q) (the local flow is now a global flow). Sincett = tt = t+(t) = 0 = idM, we have (t)

1 = t. So each t

is a diffeomorphism ofM. So t is a 1-parameter group of diffeomorphisms.

6. The Lie derivative on vector fields

6.1. Lie derivative is the Lie bracket.

The Lie bracket is related the concept of Lie derivative (which we nowconsider for vector fields and later consider for more general tensors).


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Let X be a C vector field on M. Given p M, let t : V M be

a local flow of X, where V is a neighborhood of p and t (, ) for some > 0. For q V, since t (q) = (t, q) = q (t) and

dqdt (t) = Xq(t), we

have

dtdt

(q) = Xt(q).

Let X and Y be C vector fields on M. The Lie derivative ofY withrespect to X is defined by:

(LXY)p d

dt

t=0

dt

Yt(p)

= limt0

1tdt Yt(p) d0 Y0(p)

= limt0

1

t

dt

Yt(p)

Yp

,

where t is a local flow of X. So the Lie derivative is a measure of how Ychanges under the action of the local flow of X.

On vector fields, the Lie derivative and Lie bracket are the same.

Lemma 1.37. LXY = [X, Y]; in other words, for each p M,

[X, Y]p = limt0

1

t dt Yt(p) Yp .Proof. Step 1. L

xi

xj

= 0. We first show that L xi

xj

= 0, which by

Lemma 1.31 implies that L xi

xj

= [ xi

, xj

]. Let (i)t be the correspond-

ing local flow of the vector field xi

on x (U). For p in its domain,

(1.15) (i)t (p) = x

x1 (p) + tei

,

which is the same as ci (t) in (1.5). Indeed, differentiating this in t, we havethe defining equation for the (i)t:

d (i)tdt

(p) = dxx1(p)+tei (ei) =

xi (i)t(p).

Then, given t, the differential of (i)t (as a function of p) is

(d (i)t)p = dxx1(p)+tei d

x1p

.


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So, regarding the expression in the definition of Lie derivative, we haved (i)t

(i)t(p)

xj (i)t(p)

=

dxx1((i)t(p))tei

d

x1

(i)t(p)

xj (i)t(p)

= dxx1((i)t(p))tei

(ej)

=

xj p

since x

x1 ((i)t (p)) tei

= p and since (1.9) implies d

x1q

( xi q

) =

ei. Thus

L xi

xj

p

= limt0

1td (i)t xj (i)t(p) xj p = 0.

Remark. So far we have only shown L xi

xj

= 0 =xi

, xj

. One

of the difficulties of the proving the general case this way is that it is notobvious how to obtain a formula for the local flow t for X in general. Wecan circumvent this difficulty.

Step 2. L x1

Y. Now consider L x1

Y (we take i = 1 for conve-

nience and we replace xj

by any vector field Y). Recall that (1)t (p) =

x x1 (p) + te1. Let

Yq =n

j=1

Yi (q)

xj q.

Note that from (1.13) we have

x1, Y

=j

Yj

x1

xj.

Then d (1)t

(1)t(p)

Y(1)t(p)

= dxx1((1)t(p))te1 d x1(1)t(p)Y(1)t(p)= dx

x1((1)t(p))te1

nj=1

Yj ((1)t (p)) ej

=

nj=1

Yj ((1)t (p))

xj p.


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Proof of (1.17). Let (U, x) be a parametrization with p x (U),

x1 (p) = 0, and (dx)0 (e1) = Xp (we may obtain this by precomposingan arbitrary parametrization containing p in its image with an appropriateaffine transformation ofRn). Let {t} be the local flow of X. Then thereexists > 0 such that the map

x : U (, )

x =

x2, . . . , xn

Rn1 : |x| <

M

given by

x

t, x2, . . . , xn t

x

0, x2, . . . , xn

is a well defined parametrization. Since dx0 = dx0, that x is a parametriza-

tion for some > 0 sufficiently small follows from the inverse function the-orem (Proposition 1.18). Note that x

0, x2, . . . , xn

= x

0, x2, . . . , xn

.

Then for any q x (U),

(x)1 q= dx

(x)1(q)(e1)

=d

dt

t=(x)1(q)1

t

x

0, x1 (q)2 , . . . , x1 (q)n

= Xq.

6.2. Push forward of the Lie bracket.

Let : N M be a diffeomorphism and let X and Y be vector fieldson N. Then, acting on any function f : M R, we have

d ([X, Y]) (f) = (XY Y X) (f )

= X((d (Y) (f)) ) Y ((d (X) (f)) )

= d (X) (d (Y) (f)) d (Y) (d (X) (f))

= [d (X) , d (Y)] (f) .

Hence we have the diffeomorphism invariance of the Lie bracket:

(1.18) d ([X, Y]) = [d (X) , d (Y)] .

Using this, we now give another proof of the Jacobi identity (1.14). GivenX,Y ,Z (M), let {t} be the 1-parameter local group of diffeomorphisms


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generated by X. Then, using Lemma 1.37, we compute that

[X, [Y, Z]]p = (LX [Y, Z])p(1.19)

=d

dt

t=0

dt

[Y, Z]t(p)

=

d

dt

t=0

[dt (Y) , dt (Z)]p

=

d

dt

t=0

dt (Y) , Z

p

+

Y,

d

dt

t=0

dt (Z)

p

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

Thus

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

Exercise 1.38. Show that for 1-parameter family of vector fields Y (t) andZ(t),

d

dt[Y (t) , Z(t)] =

dY

dt(t) , Z(t)

+

Y (t) ,

dZ

dt(t)

.

This justifies the fourth equality in (1.19).

Solution. Given f C (M), let g (t) = Y (t) (f) and h (t) =Z(t) (f). Then

d

dt[Y (t) , Z(t)] (f) =

d

dt(Y (t) (h (t)) Z(t) (g (t)))

= Y (t) (h (t)) + Y (t) h (t) Z (t) (g (t)) Z(t) g (t)= Y (t) (Z(t) (f)) + Y (t)

Z (t) (f)

Z (t) (Y (t) (f)) Z(t)

Y (t) (f)

=

Y (t) , Z(t)

(f) +

Y (t) , Z (t)

(f) .

7. Cotangent spaces, cotangent bundles and 1-forms

7.1. Cotangent space and cotangent bundle.

Let Mn be a C manifold. Given p M, the cotangent space at p is

Tp M (TpM) = { | : TpM R is linear} .

That is Tp M is the dual vector space of TpM. Analogous to the tangentbundle, we define the cotangent bundle to be the disjoint union

TM =

pM

Tp M,

which is equivalent to

TM =

(p, v) : p M, v Tp M

.


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7. Cotangent spaces, cotangent bundles and 1-forms 27

We have the projection map : TM M defined by (p, ) = p. Note

that 1 (p) = Tp M.

7.2. 1-forms and dual bases.

A C 1-form (or, less commonly, cotangent vector field) on Mis a C section of TM, that is, a C map : M TM satisfying = idM. The notion of 1-form is dual to the notion of vector fieldin that TM is replaced by TM. We may also consider a 1-form as afunction : TM R such that |TpM : TpM R is linear for each p M.

We denote the set of 1-forms by C(TM).

Given a parametrization (U, x), define the 1-forms dxj, for j = 1, . . . , n,

on x (U) by

(1.20) dxj

xip

= ji

at each point p x (U). Namely,

dxjnj=1

is the basis of Tp M dual to the

basis { xi

}ni=1 of TpM.

Exercise 1.39. Show, for eachj = 1, . . . , n, thatdxj Tp M is the same as

the differential d(x1)j : TpM R, where the functions (x1)j : x (U) R

are defined by x1 = ((x1)1, . . . , (x1)n).

Exercise 1.40. Show that if is a C 1-form on M and (U, x) is aparametrization of M, then on x (U) we have

=ni=1

idxi,

where i (xi

) : x (U) R. The functions i are called the compo-nents of with respect to the parametrization (U, x).

Exercise 1.41. Show that the cotangent bundle TM is a 2n-dimensionalC manifold.

7.3. Lie derivative of a 1-form.

Let Mn be a C manifold. Let X be a C vector field on M and letf : M R be a C function. The Lie derivative of f with respect to Xis defined by:

(LXf) (p) d

dt

t=0

(f t) (p) ,


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where {t} is the local flow of X. We compute that

(LXf) (p) =ddtt=0

(f p) (t)

= p (0)(f)

= Xp (f) ,

where p is the integral curve to X with p (0) = p. So, acting on functions,the Lie derivative is simply the directional derivative.

Let : N M be a C map. Given a C 1-form on M, its pullback by to N is defined by

(1.21) () (u) = (d (u))

for u TN. Then

is a C

1-form on N.Let X be a C vector field on M and let be a C 1-form on M. The

Lie derivative of with respect to X is defined by:

(LX)p d

dt

t=0

t

t(p)

(1.22)

= limt0

1

t

t

t(p)

p

.

Exercise 1.42. Show that if(U, x) is a parametrization ofM, then for eachi and j,

L xi

dxj = 0

in x (U).Exercise 1.43 (Lie derivative product rule). Show that ifX and Y areC

vector fields on M and is a C 1-form on M, then3

(1.23) X( (Y)) = (LX) (Y) + ([X, Y]) .

For example, as components with respect to (U, x),

(L xi

)j =jxi

.

3Note that X( (Y)) = LX ( (Y)) since (Y) is a function.


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Chapter 2

Riemannian Manifolds

The concept of Riemannian manifold is a generalization of Euclidean spacewith the Euclidean geometry (inner product).

1. Introduction to Riemannian metrics

1.1. Euclidean geometry.

Euclidean geometry is determined by the standard Euclidean inner prod-uct ,

Rnon Rn. The length of a vector X Rn is defined by

|X|

X, X

Rn

and the angle between two nonzero vectors X and Y is defined by

(X, Y) cos1

X, YRn

|X| |Y|

.

More generally, an inner product , on a real vector space V isdefined to satisfy:

v, u = u, v ,

u1 + cu2, v = u1, v + c u2, v ,

u, u 0,

u, u = 0 if and only ifu = 0

for any c R and where the us and vs are in V. Of course, u, v1 + cv2 =u, v1 + c u, v2 follows by symmetry. That is,

, : V V R

is a symmetric, bilinear, positive-definite form.

29


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30 2. Riemannian Manifolds

1.2. Definition of Riemannian metric and isometry.

Since each tangent space of a differentiable manifold is a (finite-dimen-sional) real vector space, we may extend the infinitesimal (at each point)notion of geometry to manifolds.

Definition 2.1. A Riemannian metric on a differentiable manifold Mn

is an assignment to each p M an inner productgp , p onTpM. WhenM is a C differentiable manifold, we say that g is C if it depends on pin a C way.

More precisely, let (U, x) be a parametrization of M and define thecorresponding coefficients

gij : x (U) R

by

gij g(

xi,

xj)

for i, j = 1, . . . , n. We say that the Riemannian metric g is C if for anyparametrization (U, x) the corresponding coefficients gij of the metric areC functions.

Equivalently, g is C iff for any C vector fields X, Y on M, the func-tion p gp (X(p) , Y (p)) is C

.

Definition 2.2. A C Riemannian manifold is a pair (Mn, g), where

Mn

is a C

differentiable manifold and g is a C

Riemannian metric.

The induced Riemannian metric on the graph of a Euclidean function isgiven by the following.

Exercise 2.3. Let U Rn be an open set and let f : U R be a C

function. Define

x : U Rn+1 = Rn R

by

x (q) = (q, f(q)) .

Compute dxq (ei) for q U, and compute

gij (q) dxq (ei) , dxq (ej) ,

where , is the standard Euclidean inner product onRn+1.

Solution. For x

q1, . . . , q n

=

q1, . . . , q n, f

q1, . . . , q n

we have

dxq (ei) =x

qi(q) =

0, . . . , 1, . . . , 0,

f

qi(q)

= ei +

f

qi(q) en+1,


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(b) linear maps A satisfying Av,Aw = v, w for all v, w Rn, i.e.,

orthogonal transformations. The group of orthogonal transfor-mations is denoted by O (n,R).

(1) Graph. Let U Rn be an open set and let f : U R be a C

function. Define the C manifold

Mn = {(q, f(q)) : q U } Rn+1.

Recall that we have the global parametrization x : U M defined by

x (q) = (q, f(q)) .

The coordinate vectors may identified with

xi x(q) = ei +

f

qi (q) Rn+1

.

The Euclidean inner product , Rn+1

induces a C Riemannian metric gon M defined by

g (u, v) = u, vRn+1

for u, v TpM, where p is any point in M. Exercise 2.3 asks you to computegij .

(2) The unit sphere Sn =

x Rn+1 : |x| = 1

Rn+1. Recall thatfor p Sn we may identify its tangent space with

TpSn = p

x Rn+1 : x, p = 0

.

As a hypersurface in Rn+1, Sn has the induced Riemannian metric defined,like above, by

gSn (u, v) = u, vRn+1

for u, v TpSn, where p Sn.

The isometry group of Sn is O (n + 1,R).

Recall that given Sn, we have the parametrization

x : = Rn Sn {}

defined, as the inverse of stereographic projection, by

x (q) =

2q+ |q|2 1 |q|2 + 1 .

Let = N = (0, . . . , 0, 1) be the north pole and identify N = Rn {0}with Rn by (q, 0) q for q Rn.

Exercise 2.7. Compute the Riemannian metric g onRn defined by

gq (u, v) = gSn

(dxN)q (u) , (dxN)q (v)


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1. Introduction to Riemannian metrics 33

for u, v TqRn = Rn. How does g (u, v) compare with the Euclidean

metric u, vRn? We remark that, by definition, (Rn, g) is isometric to(Sn {N} , gSn).

(2) Hyperbolic space. There are a number of models of hyperbolicspace. All of the following examples are isometric to each other.

(a) Disk model. Let

D = {x Rn : |x| < 1} .

The hyperbolic disk metric gD on D is defined by: for x D,

(gD)x (u, v) = 4 u, vRn

1 |x|22 .

(b) Half-space model. Let

H = {x Rn : xn > 0} .

The hyperbolic half-space metric gH on H is defined by: for x H,

(gH)x (u, v) =u, v

Rn

x2n

(c) Hyperboloid model. Let

L =

x Rn+1 : x21 + + x2n x

2n+1 = 1, xn+1 > 0

.

The hyperbolic half-space metric gH on H is defined by: for x L,

(gL)x (u, v) = u, vMn+1 u1v1 + + unvn un+1vn+1.

Exercise 2.8. Show that for each x L, (gL)x is positive definite on TxL.So (L, gL) is a Riemannian manifold.

Exercise 2.9. Show that the map F : (H, gH) (D, gD) given by

F

x1, . . . , xn

4 |x|2

|x|2 + 4xn + 4,

4x1

|x|2 + 4xn + 4, . . . ,

4xn1

|x|2 + 4xn + 4

is well defined and an isometry.


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Remark. Note that

Fx1, . . . , xn = |x|2 + 42 16 (xn)2|x|2 + 4xn + 4

=

|x|2 4xn + 4|x|2 + 4xn + 4

=

n1i=1 (x

i)2 + (xn 2)2n1i=1 (x

i)2 + (xn + 2)2.

Exercise 2.10. Following up on the above exercise, find explicit isometriesbetween the different models of hyperbolic space. Hint: see the expositorypaper [2].

2. Properties of Riemannian metrics

Given a differentiable manifold, we would like to know that there actuallyexists a Riemannian metric on it. Fortunately this is generally true.

2.1. Existence of Riemannian metrics.

The existence of a Riemannian metric on a differentiable manifold isgiven by Proposition 2.10 on p. 43 of Do Carmo (see also Proposition 11.26on p. 284 of Lee [8]). The idea of the proof is to paste together Euclideanmetrics, defined via parametrizations, using a partition of unity subordinateto the covering by parametrizations.

Theorem 2.11 (Existence of Riemannian metrics). If Mn is a connected,Hausdorff C manifold admitting a countable number of parametrizationswhose coordinate neighborhoods cover M, then there exists a C Riemann-ian metric g on M.

Fact. Under the above hypotheses, there exists a countable number ofparametrizations {(U, x)}A such that {x (U)}A is a locally finitecover of M (i.e., for each p M there exists a neighborhood V of p suchthat only a finite number of the x (U) intersect V) and there exists a C

partition of unity {}A subordinate to the cover {x (U)}A of M,i.e.,

(1) : M [0, 1] is a C function with support in x (U) for each A,

(2)

A (p) = 1 (this is a finite sum) at each point p M.


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2. Properties of Riemannian metrics 35

Proof of Theorem 2.11. Let {(U, x)}A and {}A be as above.

For each A, define the Riemannian metric g on x (U) by

g (u, v) =

d

x1p

(u) , d

x1p

(v)Rn

for u, v TpM, where p x (U).1 We extend g to a symmetric bilinear

form on each TpM for p M by setting it to be zero outside of x (U).Then g

is a C symmetric bilinear form on each TpM for p M. Notethat g

is positive definite wherever = 0 (recall that supp() x (U)). Define

g A

g,

which at each point ofM is a finite sum. g is a C

symmetric bilinear formon each TpM for p M. Moreover, for any p M there exists A suchthat (p) > 0 and p x (U). Then for any u TpM {0} we have

g (u, u) =A

g (u, u) g

(u, u) > 0.

Of course, gp (0, 0) = 0 for any p M. Thus g is a C Riemannian metricon M.

2.2. Lengths of paths and distance.

Let (Mn, g) be a Riemannian manifold. Given a piecewise C1 path : [a, b] M, its length is defined by

(2.2) L ()

ba

(u) du.This length is invariant under reparametrization of the path . That is, if : [c, d] [a, b] is a C1 diffeomorphism, then

L ( ) = L () .

Indeed, by the chain rule and the change of variables formula,

L ( ) = b

a ddu ( ) (u) du=

ba

( (u)) (u) du= L () .

1Note that g

xi

, x

j

= ij .


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We say that a C1 path (u) is parametrized by arc length if | (u)| 1.

Assuming | (u)| = 0 for all u [a, b], we may reparametrize by arc lengthby defining

s = 1 (u)

ua

(u) du,so that ds

du(u) = | (u)|. Then dds ( ) (s)

= ( (s)) duds (s)

= (u) 1

dsdu

(u)

= 1.

The distance function d : M M [0, ) is defined by

(2.3) d (x, y) inf

L () ,

where the infimum is taken over all piecewise C paths : [0, 1] Mn

with (0) = x and (1) = y.

3. Hypersurfaces of Euclidean space

Euclidean hypersurfaces provide basic and concrete examples of Riemannianmanifolds.

Let Mn Rn+1 be a regular (Euclidean) hypersurface, i.e., for eachp M there exists a neighborhood V ofp in M and a C homeomorphism

x : U Rn M V Rn+1

such that dxu : TuU = Rn Tx(u)R

n+1 = Rn+1 is injective for each u U.

Exercise 2.12. Show that there is a unique C manifold structure on Mthat contains each (U, x) such thatx : U M V is a C homeomorphismand dxu is injective for u U.

The induced Riemannian metric I on M, also called the first fun-damental form, is defined by

I (U, V) U, VRn+1for U, V TpM, where p M. This is simply the restriction of the Eu-clidean inner product to the tangent spaces TpM. It is easy to see that I isa C Riemannian metric.

We may naturally identify TpM with an n-dimensional subspace ofRn+1.

In particular, at each p M there are exactly two unit vectors perpendicularto TpM, called unit normals. If M is orientable, then there exactly two


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3. Hypersurfaces of Euclidean space 37

choices of C unit normal vector fields defined at all points of M. Let

: M Rn+1 be such a choice. We have(1) || 1 on M,

(2) p, U = 0 for each U TpM, p M,

(3) i : M R is C for each i = 1, . . . , n + 1.

Note that if , : (, ) M are differentiable curves with (0) = (0) in the Euclidean sense considering , : (, ) Rn+1, then for anyC function f : M R we have (f ) (0) = (f ) (0), i.e., (0) = (0) as derivations on M. Of course, the same holds true for Rn+1-valuedfunctions on M such as the normal : M Rn+1.

Given p M, the second fundamental form at p

II : TpM TpM R

is defined by

II (U, V) U() , V

for U, V TpM.

First we note the following elementary fact. If : (, ) Rn is adifferentiable curve and if V and W are vector fields defined on the imagecurve (, ) such that V ( (t)) and W ( (t)) are differentiable, then

d

dtV ( (t)) , W( (t)) =

d

dtV ( (t)) , W ( (t))

(2.4)

+V ( (t)) , ddt

W ( (t)) .Now extend V to a vector field on M defined in a neighborhood U M

of p. Since , V 0 in U, at p we have by (2.4)

0 = U, V = U() , V + , U(V) ,

i.e., the second fundamental form may be rewritten as

(2.5) II (U, V) = , U(V) .

Next extend U on M to a neighborhood of p. We compute at p that

II (U, V) II (V, U) = , U(V) + , V (U)= , [U, V]

= 0,

where we used Exercise 1.29 and [U, V]p TpM is the Lie bracket of U andV as vector fields on M. Hence the second fundamental form is symmetric:

II (U, V) = II (V, U) .


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We may also describe the second fundamental form in terms of the

Weingarten mapL : TpM TpM

defined by

L (U) = U() .

The reason L (TpM) TpM is because

L (U) , = U() , =1

2U(, ) = 0.

By definition,

II (U, V) = L (U) , V .

Since II is symmetric, the linear map L is self-adjoint, i.e., L (U) , V =U, L (V).

The eigenvalues of the Weingarten map, which we denote by

1, . . . , n,

are called the principal curvatures. Note that since L is self-adjoint, thereexists an orthonormal basis of eigenvectors {ei}

ni=1 for TpM such that

L (ei) = iei,

where the i are real. Note that the principal curvatures are associated to achoice of unit normal vector field . If we choose the opposite unit normal,then all the signs of the i change.

The mean curvature (depends on the choice of ) at a point p M is

the sum of the principal curvatures (i.e., the trace of the Weingarten map)

(2.6) H(p) 1 + + n = tr (L) =ni=1

II(ei, ei) ,

where {ei} is any orthonormal basis for TpM.

The Gauss curvature is the product of the principal curvatures (i.e.,the determinant of the Weingarten map)

(2.7) K 1 n = det L =detII

detI.

We now calculate using a parametrization x : U M V of our hy-

persurface, which is a C

homeomorphism such that each dxu is injective.Now the coordinate tangent vectors are

xi= ci (0) = ei (x)

for i = 1, . . . , n. The components of the induced Riemannian metric (firstfundamental form) on M are given in M V by

Iij = ei (x) , ej (x)Rn+1 .


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4. Lie groups 39

We may pull back the Riemannian metric I on M by x to a Riemannian

metric g on U, defined by(2.8) g (U, V) I (dx (U) , dx (V)) = dx (U) , dx (V)

Rn+1

for U, V TxU = Rn. Then

gij (p) gp (ei, ej) = Iij (x (p))

for p U.

The coefficients of the second fundamental form are

IIij = II (ei (x) , ej (x)) = ei (x) () , ej (x)Rn+1 ,

where ei (x) () is the the tangent vector ei (x) to M acting on as anRn+1-valued function on M. Note that ei (x) () = ei ( x). By (2.5) and

ei (x) (ej (x)) = ei (ej (x)), we may rewrite this asIIij = , ei (ej (x))Rn+1 .

More explicitly, i.e., keeping track at which point each quantity is, we have

IIij (x (p)) =

(ei)p (x)

x(p)

() , (ej)p (x)

Rn+1

=

(x (p)) , (ei)p

(ej)p (x)

Rn+1

.

We may pull back the second fundamental form II to U by defining

h (U, V) II (dx (U) , dx (V))

for U, V TxU = Rn. Then

hij (p) hp (ei, ej) = IIij (x (p))

for p U.

Exercise 2.13. Compute the second fundamental form, Weingarten map,mean curvature, and Gauss curvature for the graph in Exercise 2.3.

4. Lie groups

A Lie group G is a set G with:

(1) a binary operation, called multiplication, : G G G such that(G, ) is a group,

(2) a C differentiable manifold structure on G

such that the map : G G G

defined by (x, y) = x y1

is C, where G G has the product manifold differentiable structure.


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Define the right and left multiplication maps

Rx : G G and Lx : G G

byRx (y) = yx and Lx (y) = xy.

It is easy to see that for any x G, Rx is a diffeomorphism with inverseRx1 and that Lx is a diffeomorphism with inverse Lx1.

A C vector field X on G is left-invariant if

dLx (Xy) = Xxy

for each x, y G. Right-invariant means

dRx (Xy) = Xyx

for each x, y G.Let e denote the identity element ofG. Given a tangent vector Xe TeG,

we define a corresponding left-invariant vector field (LIVF) on G by

Xx = dLx (Xe)

for each x G (exercise: check that X is left-invariant). In this way wemay identify the set of left-invariant vector fields on G with TeG.

A subspace h of a Lie algebra g is called a Lie subalgebra if it is closedunder the Lie bracket operation.

Exercise. Show that ifX and Y are LIVF on a Lie group, then [X, Y]is a LIVF.

Because of this, LIVF is a Lie subalgebra of the set of all vector fieldson G. In other words, TeG is a Lie algebra with Lie bracket defined by

[Xe, Ye] [X, Y]e .

Let , e be an inner product on TeG. Define a corresponding Riemann-ian metric , on G by

(2.9) u, vx = (dLx1)x (u) , (dLx1)x (v)e

for any x G and u, v TxG. Note that if u, v TeG, then

(dLx)e (u) , (dLx)e (v)x = u, ve .

If a Riemannian metric g on G satisfies (2.9), then we say that g is a left-invariant metric.

One similarly defines right-invariant metric on a Lie group.

Examples:

Let F be R or C.

1. The general linear group GL (n,F) is the set of invertible n nmatrices with entries in F, i.e., {A : det A = 0} , with matrix multiplication.


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5. Levi-Civita connection 41

2. The orthogonal group O (n,F) is the setA GL (n,F) : AT A = A AT = I ,where T denotes transpose and I denotes identity. O (n,F) is a subgroup ofGL (n,F) . We have

O (n,F) = {A GL (n,F) : Av,Aw = v, w for all v, w Fn} .

3. The special orthogonal group SO (n,F) is the set

O (n,F) {A : det A = 1} .

4. The unitary group U (n) is the subgroup of GL (n,C) given by

A : AT A = A AT = I .Define the Hermitian inner product on Cn by (v, w) =

i v

iwi. Then

U (n) = {A GL (n,C) : (Av,Aw) = (v, w) for all v, w Cn} .

For further discussion of Lie groups, see Lee [8], pp. 3740, pp. 93100,pp. 194199, and Chapter 9.

5. Levi-Civita connection

Tangent vectors act on C functions and we call this action directionaldifferentiation by analogy with Euclidean space. The Lie derivative is a

derivative acting on vector fields and it is equal to the Lie bracket. However,it is not analogous to the directional derivative. On a Riemannian manifoldwe do naturally have such a notion, called covariant differentiation.

5.1. Existence of the Levi-Civita connection.

Let (Mn, g) be a Riemannian manifold. Let (M) denote the set ofC

vector fields. The Levi-Civita connection (or Riemannian covariantderivative) is the unique map

: (M) (M) (M) ,

denoted by (X, Y) = XY, satisfying: for each X,Y ,Z (M) and

f C

(M),(1) fX+YZ = fXZ+ YZ,

(2) X (Y + Z) = XY + XZ,

(3) X (f Y) = fXY + X(f) Y,

(4) XY YX = [X, Y],

(5) XY, Z = XY, Z + Y, XZ.


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A map satisfying only (1)(3) is called an affine connection. Property

(4) is called symmetric or torsion-free; property (5) is called compatiblewith the metric.

We need to show:

Theorem 2.14. The Levi-Civita connection exists, is unique, and satisfies

XY, Z =1

2{XY, Z + Y Z, X ZX, Y(2.10)

+ [X, Y] , Z + [Z, X] , Y [Y, Z] , X}.

Proof. (Uniqueness) If a map satisfies properties (1)(5), then by (5)

XY, Z = XY, Z + Y, XZ ,

Y Z, X = YZ, X + Z, YX ,ZX, Y = ZX, Y + X, ZY .

Adding the first two equations and subtracting the third equation yields

XY, Z + Y Z, X ZX, Y

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

ZX, Y + YZ, X X, ZY ,

which by (4) equals

2 XY, Z [X, Y] , Z [Z, X] , Y + [Y, Z] , X .

Hence we obtain

(2.11) 2 XY, Z = T (X,Y ,Z) ,

where the map T : (M) (M) (M) C (M) is defined by

T(X,Y ,Z) XY, Z + Y Z, X ZX, Y(2.12)

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

This is equivalent to formula (2.10) for . Since this formula uniquelydetermines , if the Levi-Civita connection exists, then it is unique.

(Existence) It is easy to see that T(X,Y ,Z) is R-linear in each compo-nent. We compute that (see Exercise 2.16 below)

(2.13) T(X,Y ,f Z) = f T(X,Y ,Z) = T(f X,Y ,Z) ,

i.e., T (X,Y ,Z) is C (M)-linear in the first and third components. Hence,by Lemma 2.18 below,

(2.14) T (X,Y ,Z) (p) M(Xp, Y , Z p) R

depends only on Xp, Zp TpM and Y (M). Given Xp TpM andY (M), define

LXp,Y : TpM R


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by

(2.15) LXp,Y(Zp) M(Xp, Y , Z p) .

Then LXp,Y is an R-linear function. By (2.69), this implies that there existsa unique tangent vector LXp,Y TpM such that

(2.16) LXp,Y(Zp) =

LXp,Y, Zp

for each Zp TpM. Define : (M) (M) (M) by

(2.17) (XY) (p) 1

2LXp,Y,

which is easily seen to be well-defined. Note that by (2.17), (2.16), (2.15)

and (2.14), we have

(XY) (p) , Zp

1

2LXp,Y, Zp

=

1

2LXp,Y(Zp)

=1

2M(Xp, Y , Z p)

=1

2T(X,Y ,Z) (p) ,

which is (2.11), a formula we derived assuming the existence of an affine

connection satisfying properties (1)(5). We now show, in fact, that XYsatisfies the five properties of the Levi-Civita connection.

Since T (X,Y ,Z) is C (M)-linear in the first component, we have thatLXp,Y is a linear in Xp, which implies that (XY) (p) depends linearly onXp. I.e., property (1) holds. Similarly, we see that (XY) (p) is R-linear inY, so that property (2) holds.

For each X,Y ,Z (M), we have

2 (XY) (p) (YX) (p) , Zp = LXp,Y(Zp) LYp,X (Zp)

= T (X,Y ,Z) (p) T(Y ,X,Z) (p)

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

[Y, X] , Z [Z, Y] , X + [X, Z] , Y) (p)

= 2 [X, Y] , Z (p)

by (2.12). Since Z is arbitrary, this implies that for each X, Y (M),

(XY) (p) (YX) (p) = [X, Y] (p) ,

i.e., property (4) holds.


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Finally,

2 (XY, Z + Y, XZ)

= T(X,Y ,Z) + T(X,Z,Y )

= XY, Z + Y Z, X ZX, Y + XZ, Y + ZY, X Y X, Z

+ [X, Y] , Z + [Z, X] , Y [Y, Z] , X

+ [X, Z] , Y + [Y, X] , Z [Z, Y] , X

= 2XY, Z ,

which verifies property (5). Hence defined by (2.17) satisfies properties(1)(5).

Remark 2.15. By (2.17), the Levi-Civita connection is equivalent to maps : TpM (M) TpM for all p M.

Exercise 2.16. Verify (2.13).

Exercise 2.17. Let : (Nn, h) (Mn, g) be an isometry. Let g and h

denote the Levi-Civita connections of g and h, respectively. Prove that

(2.18) hXY = (d)1

gd(X)

d (Y)

.

5.2. A prelude to tensors.

Now we prove a result we used above and which is related to the conceptof tensor, to be discussed later in more detail.

Lemma 2.18. Suppose that T : (M) (M) satisfies

(1) T(X+ Y) = T(X) + T (Y),

(2) T(f X) = f T(X).Then for any X (M) we have T (X)p depends only on Xp. Inother words, for any p M and any two C vector fieldsX1, X2 (M) with X1 (p) = X2 (p), we have T(X1)p = T (X2)p.

When the map T satisfies (1) and (2), we call it a tensor.

We first need the following.

Lemma 2.19. If X1, X2 (M) are such that X1 = X2 in a neighborhoodV of a point p M, then T(X1)p = T(X2)p.

Proof. Let f : M R be a smooth function with f(p) = 0 and supp (f) V. Then f X1 = f X2 (M). Hence T (f X1) = T(f X2) (M) and, inparticular, f(p) T(X1)p = f(p) T (X2)p TpM. The lemma follows sincef(p) = 0.


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Proof of Lemma 2.18. Suppose p M and X (M) are such that

Xp = 0. The lemma will follow from showing that

(2.19) T(X)p = 0.

Let (U, x) be a parametrization ofM with p x (U). Then the vector fields{ xi

}ni=1 are a basis for TqM for each q x (U). Then in x (U) we have

X =ni=1

Xi

xi,

where Xi C (x (U)) and Xi (p) = 0. Let : M R be a smoothfunction with (p) = 1 in a neighborhood of p and supp () x (U). Thenby Lemma 2.19,

T(X)p = T

ni=1

Xi

xi

p

= (p)ni=1

Xi (p) T

xi

p

= 0.

The lemma indeed follows since ifX1, X2 (M) with X1 (p) = X2 (p),then X1 X2 (M) satisfies (X1 X2)p = 0, which by (2.19) impliesthat T (X1 X2)p = 0. Thus T(X1)p = T (X2)p.

Exercise 2.20. LetD, D : (M) (M) (M) be affine connections.Show that their difference D D : (M) (M) (M) is a tensor.The main point is to show that for X,Y ,Z (M) and f C (M),

(2.20) DX(f Y) DX (f Y) = fDXY DXY .Since we also have

DfXY DfXY = f

DXY DXY

,

we say that D D is a tensor.

Exercise 2.21. Let(U, x) be a parametrization of M and let

gk

denotethe inverse matrix of (gij), i.e.,

(2.21)n

k=1gikgjk =

ij.

Show that if V is a tangent vector, then

(2.22) V =n

k,=1

gk

V,

x

xk.

Hint. We just need to show that the inner products of the lhs and rhsof (2.22) with any

xiare equal.


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5.3. The Christoffel symbols.

For differential geometry calculations, it is useful to consider the Levi-Civita connection with respect to the coordinate vectors {

xi}. In particu-

lar, we define the Christoffel symbols kij by

(2.23) xi

xj=

nk=1

kij

xk.

We think of the Christoffel symbols as being the components of the Levi-Civita connection with respect to a parametrization. From (2.10) and[ xi

, xj

] = 0, we obtain

xi

xj

,

x =

1

2

xi

gj +

xj

gi

x

gij .Hence, by (2.22) we have the fundamental formula for the Christoffel sym-bols:

Lemma 2.22.

(2.24) kij =1

2

n=1

gk

xigj +

xjgi

xgij

,

where

gk

is the inverse of the metric component matrix as in (2.21).

Let U Rn be an open set and let f : U R+ be a C positivefunction. Define the Riemannian metric

g = f2gRn .

With respect to the identity parametrization x = idU, we have

gij = f2ij .

Thus (2.24) yields

(2.25) kij =1

f

f

xijk +

f

xjik

f

xkij

.

5.4. Covariant differentiation of 1-forms.

The covariant derivative of a 1-form on M is defined by

(2.26) (X) (Y) X( (Y)) (XY)

for all X, Y TpM, p M. The motivation for this definition is that isessentially the product rule, i.e.,

X( (Y)) = (X) (Y) + (XY) .

Exercise 2.23. Show that if is a 1-form, then X() = (X).


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6. Covariant differentiation along a path 47

Solution. We compute

X(), Y = X, Y , XY

= X( (Y)) (XY)

= (X) (Y)

= (X), Y .

Exercise 2.24. Show that if and are 1-forms, then for any vector X,

(2.27) X, = X, + , X .

Solution. Let g1() (M). We have

X, = X,

= X(), + , X(

)

= (X), + , (X)

= X, + , X ,

where we used X() = (X).

6. Covariant differentiation along a path and parallel

translation

In this section we discuss covariant differentiation along a curve. Later we

shall discuss the more general notion of covariant differentiation along amap, which, in the case of a map from a 2-dimensional rectangle, is usefulfor computing variation of arc length formulas for families of curves.

Let Mn be a C manifold and let D : (M) (M) (M) bean affine connection. Let c : I M be a C curve, where I R is aninterval. By a C vector field along c we mean a C curve V : I TMwith V = c, i.e., V (t) Tc(t)M. Let (c) denote the set of C

vectorfields along c.

Lemma 2.25. There exists a unique map Ddt

: (c) (c), called covari-ant differentiation alongc, with the properties that: for anyV, W (c)and C function f : I R,

(i) Ddt (V + W) =DdtV +

DdtW,

(ii) Ddt

(f V) = fDdt

V + dfdt

V,

(iii) For any V (M) we have Ddt

(V c) = DcV.

Proof. (Uniqueness) Suppose that Ddt : (c) (c) satisfies (i)(iii). Lett I and (U, x) be a parametrization with c (t) x (U). Given V (c),


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we may write

(2.28) V (t) =ni=1

Vi (t)

xi c(t),

where Vi : I R is a C function. By (i) and (ii), we have

(2.29)D

dtV =

ni=1

Vi (t)D

dt

xi c

+

ni=1

dVi

dt(t)

xi c.

Hence to prove uniqueness, we only need to show that Ddt

xi

c

is uniquely

determined. By (ii), for any W (M) and p M, Ddt

W(p) depends onlyon W in any given neighborhood of p. By (iii), we have

(2.30)

D

dt xi c (t) = Dc(t) xi .By (2.29), (2.28) and (2.30), D

dtV is uniquely determined.

(Existence) Given t I and a parametrization (U, x), define

(2.31)

D

dtV

(t)

ni=1

Vi (t) Dc(t)

xi+

ni=1

dVi

dt(t)

xi c(t),

where Vi (t) is defined by (2.28). Clearly (i) holds. We have

D

dt(f V) =

ni=1

f Vi (t) Dc(t)

xi+

ni=1

d(f V)i

dt(t)

xi c

= fDdt

V +ni=1

dfdt

(t) Vi (t) xi

c

= fD

dtV +

df

dtV,

which is (ii). Suppose V (M) and define V V c. We compute

Dc(t)V = Dc(t)

ni=1

Vi

xi

(2.32)

=n

i=1

c (t)

Vi

xi c(t)+ Vi (c (t)) Dc(t)

xi

=ni=1

dVidt

(t)

xi c(t)+ Vi (t) Dc(t)

xi

=

D

dtV

(t) ,

so that (iii) holds. To prove existence, we just need to show that the defini-tion (2.31) of

Ddt

V

(t) is independent of the parametrization. With (U, x)


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6. Covariant differentiation along a path 51

parametrization with c (t) x (U). By (2.32), we haveDdt

V, W

+

V,D

dtW

=ni=1

dVidt

xi

, W

+ Vi

Dc

xi, W

+n

j=1

dWj

dt

xj, V

+ Wj

Dc

xj, V

=n

i,j=1

dVi

dtWjgij + V

iWj

Dc

xi,

xj

+n

i,j=1

dWj

dtVigji + W

jVi

Dc

xj,

xi

=n

i,j=1

ddt

ViWj

gij + V

iWjc (gij)

=d

dtV, W

with the penultimate inequality usingDc

xi,

xj

+

Dc

xj,

xi

= c

xj,

xi

,

which in turn follows from (2.35).

As a consequence, along a curve the inner products of parallel vector

fields are constant.

Corollary 2.29. Suppose thatD is an affine connection which is compatiblewith the metric. If V and W are parallel vector fields along c, then |V (t)|2

and V (t) , W (t) are constant. In particular, if{ei,0}ni=1 is an orthonormal

frame at c (t0), where t0 I, then the unique parallel vector fields ei alongc with ei (t0) = ei,0, 1 i n, satisfy

ei (t) , ej (t) = ij

for t I and 1 i, j n.

Let (Mn, g) be a Riemannian manifold and let be its Levi-Civita

connection. We say that a path c : I M is a geodesic if its unit tangentvector field is parallel along C, that is,

(2.37)

dt

c

|c|

= 0,

where dt denotes the Levi-Civita covariant derivative along c. The motiva-tion for this definition will be apparent when we discuss the first variationof arc length of curves.


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We say that a geodesic has constant speed if |c (t)| is constant for

t I. We have that (2.37) and |c| being constant is equivalent to theconstant speed geodesic equation:

(2.38)

dtc = 0.

By (2.34) and (2.33), the constant speed geodesic equation, with respectto a parametrization, is:

(2.39)dxk

dt(t) +

ni,j=1

xi (t) xj (t) kji (c (t)) = 0,

where xj (t) = (x1)j c (t). If we define cj (t) = (x1)j (c (t)), then thismay be written as

d2ck

dt2(t) +

ni,j=1

dci

dt(t)

dcj

dt(t)

kji x

c1, . . . , cn

(t) = 0.

Let c (t)

x1 c

(t) =

c1 (t) , . . . , cn (t)

. Then this is a (nonlinear)system of ode of the form

(2.40)d2c

dt2(t) = F

dc

dt(t) , c (t)

,

where F : Rn U Rn is a C function.2 By the existence and uniquenessof solutions to systems of ode, we have that given q U and X Rn,there exists > 0 and a unique C curve c : (, ) U satisfying (2.40)and, equivalently, (2.39). Moreover, by the smooth dependence of solutionsof ode, c (t) is a C function of q, X, and t. This translates to the sameresults for the solution c (t) = x (c (t)) to (2.38).

7. The Riemann curvature tensor

The Lie bracket measures the noncommutativity of the directional derivative

acting on functions: [X, Y] f = X(Y (f)) Y (X(f)). By the torsion-freecondition (4) in the definition of the Levi-Civita connection, the commutatorof the first (Riemannian) covariant derivative is equal to the Lie bracket:XY YX = [X, Y]. The Riemann curvature tensor is the commutatorof the second covariant derivative acting on vector fields.

2Note that F is quadratic in dcdt

(t).


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7. The Riemann curvature tensor 53

7.1. Tensors and C (M)-linearity.

We may extend Lemma 2.18 to the following. Let k (M) (M) (M) (k-fold product) and let T : k (M) A, where A is C (M)or (M). We say that T is a C (M)-multilinear map if it is C (M)-linear in each component, i.e., for each 1 i k,

T(X1, . . . , X i1, Xi + Yi, Xi+1, . . . , X k)

= T(X1, . . . , X i1, Xi, Xi+1, . . . , X k)

+ T(X1, . . . , X i1, Yi, Xi+1, . . . , X k)

and

T (X1, . . . , X i1, f Xi, Xi+1, . . . , X k) = f T (X1, . . . , X i1, Xi, Xi+1, . . . , X k)

for any f C (M) and C vector fields X and Y.

Remark 2.30. The map T beingR-multilinear is the same definition withf C (M) replaced by f R.

Lemma 2.31. IfT is aC (M)-multilinear map, then for anyX1, . . . X k (M) we have thatT (X1, . . . , X k)p A depends only on(X1)p, . . . , (Xk)p TpM.

Exercise 2.32. Prove this lemma.

If T : k (M) C (M) is a C (M)-multilinear map, then it isequivalent to the map

T : kTM R

defined by

T

(X1)p , . . . , (Xk)p

T(X1, . . . , X k)p

for arbitrary extensions of (Xi)p TpM to Xi (M). Note that, by

Lemma 2.31, we have that T is well defined and for each p M, T

kTpM

:

k

TpM R is R-multilinear. For simplicity, we shall use the notation forboth T and T. We call T a covariant k-tensor.

Note that a 1-form is the same thing as a covariant 1-tensor. Indeed,recall that a 1-form may be considered as a function : TM R suchthat |TpM : TpM R is linear for each p M.

A Riemannian metric is the same as a covariant 2-tensor which is sym-metric and positive definite.


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7.2. Definition of Rm and its first properties.

More precisely, the Riemann curvature (3, 1)-tensor

Rm : (M) (M) (M) (M)

is defined by

(2.41) Rm (X, Y) Z X (YZ) Y(XZ) [X,Y]Z

for X , Y , Z (M). Note, for example, that YZ (M), so thatX (YZ) is well defined.

Clearly we have the antisymmetry:

(2.42) Rm (Y, X) Z = Rm (X, Y) Z.

Furthermore, one may easily check that Rm is R-multilinear. That is, forc R,

Rm (X1 + cX2, Y) Z = Rm (X1, Y) Z+ c Rm (X2, Y) Z,(2.43)

Rm (X, Y1 + cY2) Z = Rm (X, Y1) Z+ c Rm (X, Y2) Z,

Rm (X, Y) (Z1 + cZ2) = Rm (X, Y) Z1 + c Rm (X, Y) Z2,

where the Xs, Ys and Zs are vector fields.

Lemma 2.33. For any X,Y ,Z (M) and any C function f : M R,we have(2.44)

Rm (f X,Y ) Z = Rm (X,f Y ) Z = Rm (X, Y) (f Z) = fRm (X, Y) Z.

By Lemma 2.31, (Rm(X, Y) Z)p TpM depends only onXp, Yp, Zp TpM.

Proof. (1) We compute that

Rm (f X,Y ) Z = fX (YZ) Y(fXZ) [fX,Y]Z= fX (YZ) (fY(XZ) + Y (f) XZ) f[X,Y]Y(f)XZ

= fRm (X, Y) Z.

(2) Since Rm (X, Y) Z is antisymmetric in X and Y, we have

Rm (X,f Y ) Z = fRm (X, Y) Z.


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7. The Riemann curvature tensor 55

(3) We also compute

Rm (X, Y) f Z = X (Y(f Z)) Y(X (f Z)) [X,Y] (f Z)

= X ((Y f) Z+ fYZ) Y((Xf) Z+ fXZ)

([X, Y] f) Z f[X,Y]Z

= X(Y f) Z+ (Y f) XZ+ (Xf) YZ+ fX (YZ)

Y (Xf) Z (Xf) YZ (Y f) XZ fY(XZ)

([X, Y] f) Z f[X,Y]Z

= fRm (X, Y) Z,

where we used X(Y f)Y (Xf)[X, Y] f = 0 and cancelled terms to obtainthe last equality.

For eachp M, since (Rm (X, Y) Z)p TpM depends only on Xp, Yp, Zp TpM, there exists a well-defined map

Rmp : TpM TpM TpM TpM

given by

Rmp (Xp, Yp, Zp) (Rm(X, Y) Z)p ,

where X,Y ,Z (M) are any vector fields satisfying X(p) = Xp, Y (p) =Yp and Z(p) = Zp. Since Rm is C

(M)-multilinear, we obtain that Rm pis R-multilinear. Since 3 copies ofTpM are mapped into 1 copy ofTpM, wesay that Rm is a tensor of type (3, 1) or that Rm is a (3, 1)-tensor.

Exercise 2.34. Let : (Nn, h) (Mn, g) be an isometry. Let Rm g andRmh denote the Riemann curvature tensors of g and h, respectively. Provethat

(2.45) Rm h (X, Y) Z = (d)1 (Rm g (d (X) , d (Y)) d (Z)) .

It is convenient to define the second covariant derivative of a vectorfield by

2 : (M) (M) (M) (M) ,

where

(2.46) 2 (X,Y ,Z) 2X,YZ X (YZ) XYZ.

Using XY YX = [X, Y], we see that Rm is the commutator of thesecond covariant derivative:

(2.47) Rm (X, Y) Z = 2X,YZ 2Y,XZ.

We shall later see why this definition is natural from the point of view oftensors. One computes that for f C (M),

(2.48) 2fX,YZ = 2X,fYZ = f

2X,YZ.


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By Lemma 2.31, 2 (X,Y ,Z)p depends only on Xp, Yp TpM and Z

(M). We also compute that2X,Y(f Z) = f

2X,YZ+ Y (f) XZ+ X(f) YZ(2.49)

((XY) f) Z+ X(Y (f)) Z.

Note that (2.44) also follows from (2.48), (2.49), and (2.47).

Exercise 2.35. Prove (2.48) and derive Rm (X, Y) (f Z) = fRm (X, Y) Zfrom (2.49).

7.3. Algebraic identities of the curvature and the first Bianchiidentity.

The Riemann curvature covariant 4-tensor (or (4, 0)-tensor) isdefined by

Rm (X,Y ,Z,W) Rm (X, Y) Z, W .

Some basic symmetries of the Riemann curvature tensor are(2.50)Rm (X,Y ,Z,W) = Rm (Y ,X,Z,W) = Rm (X,Y ,W,Z) = Rm (Z,W,X,Y) .

The first equality in (2.50) is (2.42). We prove the second and third equalitiesbelow (see (2.51) and (2.52)).

Besides (2.50) there are additional symmetries that the Riemann curva-ture tensor satisfies.

Proposition 2.36. The Riemann curvature tensor satisfies the first Bianchiidentity, i.e.,

Rm (X, Y) Z+ Rm (Y, Z) X+ Rm (Z, X) Y = 0.

Proof. To see the first Bianchi identity, we simply sum the following threeformulas (by definition)

Rm (X, Y) Z = X (YZ) Y(XZ) [X,Y]Z,

Rm (Y, Z) X = Y(ZX) Z(YX) [Y,Z]X,

Rm (Z, X) Y = Z(XY) X (ZY) [Z,X]Y,

while applying the torsion-free property VW WV = [V, W], to obtain

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

[Y,Z]X [Z,X]Y [X,Y]Z

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

= 0

by the Jacobi identity (1.14).


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for i,j,k = 1, . . . , n. The components of Rm as a (4, 0)-tensor are

Rijk

Rm

xi,

xj

xk,

x

=n

m=1

gmRmijk .

In a parametrization, (2.50) may be written as

Rijk = Rjik = Rijk = Rkij .

Lemma 2.38 (Components of the Riemann curvature (3, 1)-tensor). Wehave

(2.53) Rijk = ijk jik +

np=1

pjkip pikjp ,where i

xi

.

Proof. Using (2.41), (2.23), and [i, j] = 0, we compute

Rijk = Rm (i, j) k

= i

jk

j (ik)

= i

n=1

jk

j

n=1

ik

= ijk j

ik +

np=1

pjkip pikjp=

ijk jik + np=1

pjk

ip

pik

jp

.

In a parametrization, the first Bianchi identity is

(2.54) Rijk + Rjki + Rkij = 0,

8. The sectional curvature

If P TxM is a 2-plane, then the sectional curvature of P is defined by

(2.55) K(P) Rm (e1, e2) e2, e1 ,

where {e1, e2} is an orthonormal basis of P; this definition is independentof the choice of such a basis.


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10. Pull backs and Lie derivatives of tensors 59

Exercise 2.39. Show that if X and Y are any two vectors spanning P, then

(2.56) K(P) = Rm (X, Y) Y, X|X|2 |Y|2 X, Y2

.

9. The Ricci tensor

The Ricci tensor is the covariant 2-tensor defined by

Rc (X, Y) =ni=1

Rm (X, ei) ei, Y .

Exercise 2.40. Show that Rc is symmetric.

We may also consider Rc as the linear map

Rc : TpM TpM

defined by

(2.57) Rc (X, Y) Rc (X) , Y

for each X, Y TpM.

Exercise 2.41 (Geometric interpretation of tracing). Show that the traceof a symmetric 2-tensor is given by the following formula:

Trace g () =1

n

Sn1

(V, V) d (V) ,

where Sn1

is the unit (n 1)-sphere, nn its volume, and d its volumeform. From this, show for any unit vectorU that 1n1 Rc (U, U) is the aver-age of the sectional curvatures of planes containing the vector U. Similarly,1n

R (p) is the average of Rc (U, U) over all unit vectors U Sn1 TpM.

Solution. There exists an orthonormal basis {ei}ni=1 such that =n

i=1 iei e

i . Furthermore, Trace g () =

ni=1 i and

1

n

Sn1

V, ei2 d (V) = 1.

See also [5].

10. Pull backs and Lie derivatives of tensors

Let T be a C covariant k-tensor on a C manifold Mn. Generalizing thenotion of the pull back of a 1-form, i.e., a covariant 1-tensor, we define thepull back of T by a C map : N M as

(T) (X1, . . . , X k) T (d (X1) , . . . , d (Xk)) .

This generalizes definition (1.21), which is the case where k = 1.


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For example, for a Euclidean hypersurface Mn Rn+1 parametrized by

a map x : U M, definition (2.8) says thatg x(I),

i.e., the Riemannian metric g on U is defined to be the pull back by x of theRiemannian metric I on M.

More generally, given a Riemannian manifold (Mn, g) and a C immer-sion : N M, i.e., for each q N we have dq is injective, the pullback metric on N is defined by

(g) (X, Y) = g (d (X) , d (Y)) .

Thus definition (2.1) of : (Nn, h) (Mn, g) being an isometry is the

same as

g = h.Directly generalizing (1.22), we may now define the Lie derivative of a

covariant k-tensor T with respect to a C vector field X as:

(LXT)p d

dt

t=0

t

Tt(p)

= limt0

1

t

t

Tt(p)

Tp

.

Note that ift

Tt(p)

= Tp for p M and all t such that t (p) is defined,then LXT = 0. For example, ifX is such that t are (local) isometries, thenLXg = 0.

Definition 2.42. We say that X is a Killing vector field if LXg = 0.

The following is the product rule for the Lie derivative.

Proposition 2.43.(2.58)

(LXT) (Y1, . . . , Y k) = X(T(Y1, . . . , Y k)) k

i=1T (Y1, . . . , [X, Yi] , . . . , Y k) ,

or equivalently,

LX (T(Y1, . . . , Y k)) = (LXT) (Y1, . . . , Y k) +k

i=1T (Y1, . . . , LXYi, . . . , Y k) .

10.1. Pulled back bundles and affine connections.

Let : Nn Mm be a C map. The pulled back tangent bundleTM is defined by

(2.59) TM =xN

T(x)M.

Equivalently, we may define it as the point-set

(2.60) TM =

(x, v) : x N, v T(x)M

.


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11. Covariant derivative of a tensor 61

We have the projection map : TM N defined by (x, v) = x. Note

that 1 (x) = T(x)M.

Exercise 2.44. Show that TM is a C manifold of dimension n + m.

11. Covariant derivative of a tensor

The covariant derivative of a covariant k-tensor T is defined by the prod-uct rule:(2.61)

(XT) (Y1, . . . , Y k) X(T (Y1, . . . , Y k)) k

i=1T (Y1, . . . , XYi, . . . , Y k) .

If XT = 0 for each X (M), then we say that T is parallel.

Exercise 2.45

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