differential geometry - 國立臺灣大學dragon/lecture notes/dg-fall-2012-ntu-20150901.pdf ·...

Differential Geometry

Chin-Lung Wang

ABSTRACT. This is the preliminary version of my course notes inthe fall term of 2006 at NCU and 2012 at NTU. The aim is to pro-vide basic concepts in differential geometry for first year graduatestudents as well as advanced undergraduate students.

Contents

Chapter 1. Differentiable Manifolds 51. The category of Ck manifolds 52. Cut offs and the partition of 1 83. Tangent spaces 124. Tangent maps 155. Sub-manifolds and the Whitney imbedding theorem 186. Submersions and Sard’s theorem 237. Vector fields, flows, Lie derivatives and the Frobenius

integrability theorem 278. Existence, Uniqueness and Smoothness Dependence of

ODE 339. Exercises 35

Chapter 2. Tensors and Differential Forms 411. The Tensor Algebra 412. The Exterior Algebra 453. Cartan’s operator d 464. Lie derivatives on tensors 485. Cartan’s homotopy formula 496. Integration on forms 527. Manifold with boundary and Stokes’ theorem 538. De Rham cohomology and the De Rham Theorem 569. Exercises 63

Chapter 3. Riemannian manifolds 691. Riemannian structure 692. Covariant Differentiation and Levi-Civita Connection 723. Geodesic, Exponential Map and Riemann Normal

Coordinate 763

4 CONTENTS

4. Riemann Curvature Tensor 825. Variation of Geodesics 846. Jacobi Fields 907. Space Forms 948. The Second Fundamental Form 969. Variation of Higher Dimensional Submanifolds 9810. Exercises 103

Chapter 4. Hodge Theorem 1151. Harmonic Forms 1152. Hodge Decomposition Theorem 1163. Bochner Principle 1194. Fourier Transform 1215. Sobolev spaces 1226. Elliptic Operators and Garding’s Inequality 1257. Proof of Compactness and Regularity Theorem 1288. Exercises 130

Chapter 5. Basic Lie Theory 1371. Categories of Lie groups and Lie algebras 1372. Exponential map 1413. Adjoint representation 1434. Differential geometry on Lie groups 1475. Homogeneous spaces 1506. Symmetric spaces 1547. Curvature for symmetric spaces 1598. Topology of Lie groups and symmetric spaces 1619. Exercises 164

Index 166

Chapter 1

DIFFERENTIABLE MANIFOLDS

1. The category of Ck manifolds

Definition 1.1. A (topological) manifold M is a topological space whichis (1) locally Euclidean (2) Hausdorff and (3) Second countable.

Here are some explanations of these concepts:(1) M is locally Euclidean if for each point p ∈ M there is a open

neighborhood U 3 p which is homeomorphic to an open set in Rd

for some d ∈N. Let

ϕ : U → ϕ(U) ⊂ Rd

be such a homeomorphism. The components xi : U → R of ϕ arecalled the coordinate functions and the pair (U, ϕ) is called a (coor-dinate) chart of M at p. It is customary to identify φ with the (columnvector) coordinate function

x = (x1, · · · , xd)t.

(2) M is Hausdorff if for any p 6= q in M there are neighborhoodU 3 p, V 3 q such that U ∩V = ∅.

(3) M is second countable if there is countable basis for its topology.Recall that a basis is a collection of open subset such that any openset can be written as a (possibly infinite) union of certain constituentsfrom that collection.

Exercise 1.1. Show that Rd (with the standard Euclidean topology)is a manifold by finding an explicit countable basis.

It is not a priori clear why condition (3) should be there. A possi-ble reason goes as follows: If a topological space M is Hausdorff and

5

6 1. DIFFERENTIABLE MANIFOLDS

second countable, then any subset S ⊂ M with the induced topol-ogy is also Hausdorff and second countable. In particular any lo-cally Euclidean subset in Rd is a manifold. Conversely we will provelater that any manifold as defined above is indeed a subspace in Rd

(the Whitney Imbedding Theorem). Hence the abstract definition ofmanifolds does not really lead to anything outside Euclidean spaces.

Given a manifold M and two charts (Ua, φa) in Rda and (Ub, φb)

in Rdb with Ua ∩Ub 6= ∅, we form the coordinate transition function

φab := φa φ−1b : φb(Ua ∩Ub)→ φb(Ua ∩Ub)

which is a homeomorphism. It is intuitively clear that we shouldhave da = db, which will be the dimension of M. However, the onlyknown proofs are by no means elementary, except in one case:

Exercise 1.2. Let Rd1 ∼= Rd2 (homeomorphic). If d1 = 1 show thatd2 = 1. Investigate the case d1 = 2 and reduce the problem to theJordan Curve Theorem.

The general case will be outlined later (c.f. Exercise 1.19) by meansof certain approximation theorems and ideas in homotopy theory.

In this course we are mainly interested in differentiable manifoldsinstead of general topological manifolds. We call a collection of charts(Ua, φa)a∈A a Ck atlas of M if (1) the transition functions φab’s areall Ck mappings for some fixed k ∈N∪ ∞ and (2)

⋃a∈A Ua = M.

Exercise 1.3. For a manifold with a Ck atlas, k ≥ 1, show that thedimension d = dim M is well defined on each connected componentof M.

When a manifold M is equi-dimensional of dimension d, we usu-ally denote it by Md, if no confusion with the cartesian product M×· · · ×M is likely to occur.

Given a Ck atlas (Ua, φa)a∈A on M, a chart (U, φ) is Ck relatedto it if both the transition functions φ φ−1

a and φ−1a φ are Ck for all

a ∈ A. It is convenient to add all Ck related charts into a given atlas.

Exercise 1.4. Show that the enlarged collection of charts Uα, φα)α∈Aalso forms a Ck atlas. Moreover, it is a maximal atlas in the sense thatany chart which is Ck related to it is already contained in it.

1. THE CATEGORY OF Ck MANIFOLDS 7

Definition 1.2. A Ck (differentiable) structure on M is a maximal atlasof Ck charts. A Ck manifold is a manifold together with a Ck struc-ture. When a Ck manifold is given, the term charts of it will alwaysmean Ck charts.

Formally the case k = 0 is simply a topological manifold. Fromthe definition it is an immediately question whether it is possible toselect from all charts a sub-collection which defines a C1 structure oreven a Ck structure for higher k. These are important and highly non-trivial problems in manifold theory. There are C0 manifolds whichadmit no C1 structures. In later chapters we will address on someof these questions. For the moment, we will only remark that (1)A famous theorem Whitney says that any C1 manifold indeed ad-mits (contains) C∞ structures, (2) The C2 condition is the minimumrequirement to define the notion of curvature, a concept introducedby Gauss and Riemann which lead to the birth of modern differen-tial geometry, and will be vastly studied in this course. Thus in thiscourse, differentiable manifolds will always mean C∞ manifolds.

A function f : Md → R is Ck at p ∈ M if f x−1 is Ck at x(p) ∈ Rd

for one chart (U, x) which contains p. Since

f x−1β = f x−1

α (xα x−1β ),

by the definition of Ck structure the notion of Ck is independent ofthe choice of charts. Denote by Ck(U) the space of functions that areCk at all points in U.

Likewise a function f : Mm → Nn between two Ck manifolds iscalled Ck if

y f x−1 : x( f−1(V) ∩U) ⊂ Rm → y(V) ⊂ Rn

is Ck for any choice of charts (U, x) on M and (V, y) on N. It isenough to check it for any two special atlas. Denote by Ck(M, N) thespace of all such Ck functions.

A mapping f : M → N between two Ck manifolds is a diffeomor-phism if f−1 is well defined and both f and f−1 are Ck. This is thenotion of isomorphisms in the category of Ck manifolds.


Exercise 1.5. For any Ck manifold Md and p ∈ M, show that thereare charts with x(U) = B0(r), the open ball of radius r in Rd, as wellas charts with x(U) = Rd.

Exercise 1.6. Consider M = R with one chart given by (R, φ) whereφ(t) = t3. Show that this defines a C∞ structure on M. Is M diffeo-morphic to R with the standard C∞ structure (R, id)?

There could be many Ck structures on a manifold, but it is hardto find non-diffeomorphic ones. The set of equivalence classes ofdifferentiable structures up to diffeomorphism is a delicate object forstudy, which again will be briefly discussed in later chapters.

2. Cut offs and the partition of 1

Are there any C∞ functions on a C∞ manifold M besides the con-stants? For each charts (U, x) the coordinate functions xi’s are bydefinition C∞ on U but it may not be possible to extend xi to a C∞

function on M.One of the basic principles in differential geometry is try to (1)

compute things locally via differential calculus and (2) find a wayto patch local information together to get global results. This sectionestablishes the existence of partition of unity which is the simplest toolin this regard.

Recall that a topological space M is paracompact if every opencover Uαα∈A of it has a locally finite open refinement Vββ∈B, inthe sense that

(1) Local finiteness: for each p ∈ M, there is a neighborhood U 3 psuch that Vβ ∩U = ∅ except possibly for a finite number of Uβ’s.

(2) Refinement: There is a map ρ : B→ A such that Vβ ⊂ Uρ(β) forall β ∈ B. The map ρ may not be injective nor surjective.

A manifold is more than paracompact. In fact we have an easybut important

Lemma 1.3. Let M be a locally compact topological space which is Haus-dorff and second countable (e.g. a manifold), then M is σ compact. Namely,

2. CUT OFFS AND THE PARTITION OF 1 9

there is a countable sequence of increasing open sets Gii∈N with Gi com-pact, Gi ⊂ Gi+1 and M =

⋃∞i=1 Gi.

PROOF. Let Wii∈N be any given countable basis.

Exercise 1.7. Show that by removing those Wi with noncompact clo-sure Wi we still get a basis. (Notice that the Hausdorff condition isneeded.)

Thus we may assume that Wi is compact for all i.Let G1 = W1. The set Gi is constructed inductively: Suppose that

Gi is constructed. Since Gi is compact and covered by Wj’s, there is asmallest j(i) ∈N so that

Gi ⊂W1 ∪ · · · ∪Wj(i).

We then define Gi+1 = W1 ∪ · · · ∪Wj(i). It remains to show that Gi+1

is compact. This follows from

Gi+1 ⊂ W1 ∪ · · · ∪ Wj(i)

since a closed set in a (finite union of) compact set is compact.

Lemma 1.4. Let M =⋃∞

i=1 Gi be σ compact. Then every open coverUαα∈A has a countable locally finite refinement Vjj∈N with Vj beingcompact.

PROOF. For each i ∈N, consider the open annulus Si := Gi+1\Gi−2.(We put Gi = ∅ for i ≤ 0.) Then Gi\Gi−1 is compact and containedin Si. It is covered by Uα ∩ Siα∈A hence is covered by a finite num-ber of them. By putting together all these finite open sets we get acountable sequence Vjj∈N. Each Vj is of the form Uα ∩ Si, so

Vj = Uα ∩ Si ⊂ Si ⊂ Gi+1

is closed in a compact set. Hence Vj is itself compact.Finally, Vj is locally finite since if p ∈ Si then only those Vj’s

constructed from Si−1, Si and Si+1 may possibly intersect Si nontriv-ially.


Now we discuss cut off (or bump) functions . Let

f (t) =

e−1/t for t > 0,

0 for t ≤ 0.

Exercise 1.8. Show that f ∈ C∞(R) and f (n)(0) = 0 for all n ∈N.

The function

g(t) =f (t)

f (t) + f (1− t)=

1

1 + e1t−

11−t

is then C∞ and non-decreasing with g(t) = 0 for t ≤ 0 and g(t) = 1for t ≥ 1.

The function h(t) = g(2 + t)g(2 − t) is a cut off function withh = 1 on [−1, 1] and h = 0 outside (−2, 2). For a higher dimensionalversion we consider

ψ(x1, · · · , xd) =d

∏i=1

h(xi) ∈ C∞(Rd).

Then ψ = 1 on [−1, 1]d and ψ = 0 outside (−2, 2)d. Alternatively wemay consider the radially symmetric function

ψ(x) = h(|x|) ∈ C∞(R)

which has ψ|B0(1) = 1 and ψ|Rd\B0(2) = 0.In general for a continuous function f on a topological space M

its support is defined to be

supp f = p ∈ M | f (p) 6= 0.

For a closed set B ⊂ M, a cut off function for B is a non-negativecontinuous function f such that supp f = B. The functions ψ aboveare special C∞ cut off functions of standard cube and closed balls.Definition 1.5. Given an open cover Uαα∈A of a Ck manifold M, apartition of unity subordinate to Uα is a countable collection of Ck

functions ψjj∈N on M such that(1) 0 ≤ ψj ≤ 1 for all j.(2) supp ψj is a locally finite (closed) refinement of Uα.(3) ∑j∈N ψj(p) = 1 for all p ∈ M.

2. CUT OFFS AND THE PARTITION OF 1 11

There will be no convergence issue in (3) since by (2) the sum willbe a finite sum over a neighborhood of any point p.

Theorem 1.6 (Existence of Partition of Unity). Let M be a Ck manifoldwith Uαα∈A an open cover. Then there is a Ck partition of unity ψij∈N

subordinate to Uα with supp ψj being compact. Without the compactsupport requirement we may label the partition of unity by the same set Awith ψα 6≡ 0 for at most a countable subset of A.

PROOF. Let M =⋃∞

i=1 Gi as given by Lemma 1.3. We will modifythe proof of Lemma 1.4 to construct ψj.

For each i ∈N, Gi\Gi−1 is compact and contained in Si = Gi+1\Gi−2.For each p ∈ Gi\Gi−1, let (Wp, x) be a chart at p such that Wp ⊂Uα ∩ Si for some α ∈ A and x(Wp) = B0(3). Let Vp = x−1(B0(2)).Define a Ck cut off function for Vp ⊂Wp ⊂ Uα by

ψp =

ψ x on Wp,

0 on M\Wp.

There is a finite subcover of the open cover Vp of Gi\Gi−1. Byputting together all such finite open sets for all i ∈ N, we get thedesired locally finite refinement Vjj∈N as in Lemma 1.4.

Let ψj be the corresponding cut off function for Vj. For each p ∈M, there is a (finite number of) ψj with ψj(p) 6= 0, hence we maydefine

ψj =ψj

∑i ψi∈ C∞(M),

which clearly satisfies ∑j ψj = 1 with supp ψj = supp ψj = Vj beingcompact.

For the last statement, for each α ∈ A, we may simply let

ψα = ∑Vj⊂Uα

ψj.

Here ψα ≡ 0 if no such i exists. The proof is complete.

Exercise 1.9. Investigate the theorem for the case when M = R withthe open cover given by a single set U = M = R.


Exercise 1.10. Let A (resp. U) be a closed (resp. open) set in a Ck

manifold M with A ⊂ U. Show that there exists f ∈ Ck(M) suchthat f |A ≡ 1 and f |M\U ≡ 0. Is that possible to make supp f = A?

3. Tangent spaces

It is a priori not obvious how to generalized the concept of tan-gent vectors to manifolds. In fact this is a challenge problem for C0

manifolds. We will give two definitions of it for Ck manifolds withk ∈N∪ ∞.

Let p ∈ Rd and X ∈ Rd be a vector. For a C1 function f definednear p, the directional derivative

X f := DX f (p) =ddt

f (p + tX)∣∣t=0 = lim

t→0

f (p + tX)− f (p)t

is defined, which is a derivation (first order differential operator) inthe sense that

(1) Linearity: X(a f + bg) = aX f + bXg and(2) Lebnitz rule: X( f g) = (X f )g(p) + f (p)Xg.

Conversely, it is interesting to see whether a derivation deter-mines a vector. We will see shortly that this is indeed the case forderivations on C∞ functions.

For a Ck manifold, denote by Ckp the space of germs of Ck functions

at p. It consists of functions which are defined on some neighbor-hood of p and two functions f , g are identified if f |U = g|U for someU 3 p.Definition 1.7. Let M be a Ck manifold and p ∈ M. The Zariski tan-gent space DpM is the vector space which consists of all derivationsX : Ck

p → R.

For any chart (U, x) at p, partial derivatives ∂/∂xi|p are examplesof tangent vectors: For f ∈ Ck

p,

∂

∂xi

∣∣∣p

f :=∂( f x−1)

∂xi (x(p)).

The following lemma is a useful substitute of the Taylor expan-sion especially for functions that has only limited differentiability.

3. TANGENT SPACES 13

Lemma 1.8. Let f ∈ Ck(B0(r)). Then

f (x1, · · · , xd) =d

∑i=1

xig(x1, · · · , xd)

with gi ∈ Ck−1(B0(r)) and gi(0) = ∂ f /∂xi(0).

PROOF. By the Fundamental Theorem of Calculus,

f (x)− f (0) =∫ 1

0

ddt

f (tx) dt =d

∑i=1

∫ 1

0

∂ f∂xi (tx)xi dt

=d

∑i=1

xi∫ 1

0

∂ f∂xi (tx) dt =

d

∑i=1

xig(x),

where gi(x) =∫ 1

0

∂ f∂xi (tx) dt ∈ Ck−1(B0(r)) and g(0) =

∂ f∂xi (0) as ex-

pected.

Theorem 1.9. For a C∞ manifold with (U, x) a chart at p, the partialderivatives form a basis of DpM. Indeed for any X ∈ DpM,

X =d

∑i=1

X(xi)∂

∂xi

∣∣∣p.

PROOF. Since X(1) = X(1 · 1) = X(1) · 1 + 1 · X(1) = 2X(1),we have X(1) = 0, hence X(a) = aX(1) = 0 for any constant a.For simplicity of notations we assume that x(p) = 0. Then for anyf ∈ C∞

p ,

X f = X( f − f (p)) = X(∑i xigi)

= ∑i X(xi)gi(p) + xi(p)X(gi)

= ∑i X(xi)∂( f x−1)

∂xi (0)

= ∑i X(xi)∂

∂xi

∣∣∣p

f .

This proves the theorem.


The proof (and the theorem) fails for Ck manifolds if k < ∞ be-cause then gi ∈ Ck−1 only and the second line about the Lebnitz ruledoes not make sense. This can be analyzed in purely algebraic terms:

Proposition 1.10. For any Ck manifold M, k ∈ 0, 1, 2, · · · , ∞,

DpM ∼= (mp/m2p)∗.

Where mp = f ∈ Ckp | f (p) = 0 is the maximal ideal at p and m2

pconsists of all finite sum of products ∑ figi with fi, gi ∈ mp.

PROOF. Let X ∈ DpM. X is by definition a linear functional X :mp → R. To show that X induces a map mp/m2

p → R we need toverify that X|m2

p= 0. But this follows from the Lebnitz rule readily:

For fi, gi ∈ mp,

X(∑i figi) = ∑i X( fi)gi(p) + fi(p)Xgi(p) = 0.

Conversely given ψ : mp/m2p → R we claim that Xψ f := ψ( f −

f (p)) defines a derivation Xψ on Ckp. Indeed,

Xψ( f g) = ψ( f g− f (p)g(p))

= ψ(( f − f (p))(g− g(p)) + ( f − f (p))g(p) + f (p)(g− g(p))

= (Xψ f )g(p) + f (p)Xψg

where we use the fact that ( f − f (p))(g− g(p)) ∈ m2p.

Exercise 1.11. Let M be a Ck manifold. Show that

dim DpM = dim mp/m2p =

dim M if k = ∞,

∞ if k < ∞.

(Hint: For k = 1, study functions f (x) = (x1)a for 1 < a < 2.)

Let (U, x) and (V, y) be two charts at p, then for any f ∈ C∞p ,

∂

∂xi

∣∣∣p

f =∂( f x−1)

∂xi (x(p)) =∂( f y−1 y x−1)

∂xi (x(p))

= ∑j

∂( f y−1)

∂yj (y(p))∂yj

∂xi (x(p)) = ∑j

∂yj

∂xi (x(p))∂

∂yj

∣∣∣p

f .

4. TANGENT MAPS 15

Thus two vector representations ∑i ai ∂

∂xi

∣∣∣p

and ∑j bi ∂

∂yj

∣∣∣p

on two

charts correspond to the same vector precisely when their coefficientssatisfy the transformation rule

bj = ∑i

ai ∂yj

∂xi (x(p)).

This condition only requires C1 structure, so we make the

Definition 1.11. The tangent space TpM for a Ck manifold M withk ≥ 1 consists of compatible systems of vectors in each chart at pwhich satisfy the transformation rule.

4. Tangent maps

From now on we work in the C∞ category. In particular TpM =

DpM and tangent vectors are precisely derivations. Given f : M →N a C∞ map and p ∈ M, we define the tangent map d fp : TpM →Tf (p)N via

(d fpX)h = X(h f ),

where X ∈ TpM and h ∈ C∞f (p).

The tangent map is indeed the generalization of derivative of amap in Calculus and is also denoted by D fp, D f (p), d f (p), f∗p andperhaps the most commonly used f ′(p). It is the linearization (firstorder approximation) of the original map.

As in Calculus, d fp is linear and satisfies the chain rule. Namelyfor g : N → S be another C∞ map we have

d(g f )p = dg f (p) d fp.

The proof in Calculus is a bit tricky, but the proof now is completelyformal: For X ∈ TpM and h ∈ C∞

g f (p),

(d(g f )pX)h = X(h g f ) = (d fpX)(h g) = dg f (p)(d fpX)h.

Exercise 1.12. Show that in charts (U, x) at p ∈ M and (V, y) at

f (p) ∈ N, d fp is represented by d fx(p) with Jacobian matrix[

∂ f j

∂xi

]


where f = y f x−1 : x(U ∩ f−1(V))→ Rdim N. Namely

d fp

( ∂

∂xi

∣∣∣p

)= ∑

j

∂ f j

∂xi (x(p))∂

∂yj

∣∣∣p.

Two special cases with one of the manifolds being R are particu-larly interesting.

Example 1.12. The first is the total differential d f of a function f : M→R. Let y be the coordinate of R, then d fpX = a∂/∂y|p for some a.By plug in h = y in the definition of d fp we get a = X f . Since thereis only one basis ∂/∂y|p for TpR, following the usual convention wedenote a vector in R simply by its coefficient. Also we drop the pointp if no confusion is likely to occur. Then

d f (X) = X f .

Each coordinate xi is a C∞ function at p and we compute

dxi( ∂

∂xj

)=

∂xi

∂xj = δij.

That is, the differentials dxi’s form a dual basis of the cotangent space

T∗p M := (TpM)∗ = Hom(TpM, R)

with respect to the basis ∂/∂xi’s. Moreover,

d f = ∑i

∂( f x−1)

∂xi dxi.

This follows by looking at the values of both sides on the basis vec-tors ∂/∂xi’s.

Example 1.13. The second example is the tangent vector

γ′(t) = dγt

( ∂

∂t

)of a parameterized C1 curve γ : (a, b) ⊂ R→ M with parameter t ∈(a, b). Here, following the usual convention, we identify γ′(t) ≡ dγt

as its image vector since there is one basis vector ∂/∂t on Tt(a, b).In local chart (U, x), the curve is represented by

t 7→ γ(t) := x γ(t) = (x1(t), · · · , xm(t))t

4. TANGENT MAPS 17

and

γ′(t) = ∑i(xi)′(t)

∂

∂xi

∣∣∣γ(t)

.

The way d fp approximates f is best explained through the in-verse/implicit function theorem. We start with the simplest notions,namely the injectivity and surjectivity of f .Definition 1.14. A map f ∈ C∞(M, N) is an immersion at p ∈ M if thelinear map d fp is injective, it is a submersion at p if d fp is surjective.

Lemma 1.15. Let f : Mm → Nn be C∞.

(1) If f is an immersion at p (so m ≤ n), then there are charts (U, x)at p and (V, y) at f (p) such that f |U is represented by yi =

f i(x) = xi for i = 1, . . . , m and yi = 0 for i ≥ m + 1. That is, Uis a coordinate slice of V.

(2) If f is a submersion at p (so m ≥ n), then there are charts (U, x)at p and (V, y) at f (p) such that f |U is represented by yi =

f i(x) = xi for i = 1, . . . , n. That is, f is a coordinate projectionfrom U to V.

PROOF. For (1), start with any charts such that f (U) ⊂ V. Sinced fx(p) is injective, it has rank m. By reordering of coordinates yi’s’

we may assume that the first m× m square matrix[

∂ f i

∂xj (x(p))]m

i,j=1is invertible. Denote by

y =

[y1

y2

]= f (x) =

[f 1(x)f 2(x)

]under Rn = Rm ×Rn−m, then by the inverse function theorem y1 =

f 1(x) is invertible over some x(p) ∈W ⊂ x(U). By using (x−1(W), y1 =

f 1 x) as a new chart at p and let g = f 2 ( f 1)−1, the map f becomesa graph of g:

y =

[y1

g(y1)

].

By a simple change of coordinates

z =

[z1

z2

]=

[y1

y2 − g(y1)

]


near f (p) ∈ N, we get the desired coordinate charts.For (2), again we start with any charts with f (U) ⊂ V. Since

d fx(p) is surjective, it has rank n. By reordering of coordinates xi’s

we may assume that the first n × n square matrix[

∂ f i

∂xj (x(p))]n

i,j=1

is invertible. Denote by x = (x1, x2)t under Rm = Rn ×Rm−n andconsider the map F : x(U)→ y(V)×Rm−n defined by[ y

x2

]:= F(x1, x2) =

[f (x1, x2)

x2

].

Since

dFx(p) =

[D1 f D2 f

0 idm−n

]is invertible, the inverse G = F−1 exists over a smaller neighborhoodW 3 x(p).

The result follows by using (y, x2)t as the new coordinate systemat p.

Exercise 1.13. Show that f ∈ C∞(M, N) can be locally representedby

f (x) = (x1, · · · , xk, 0, · · · , 0)t

for some k ≤ m if and only if that d fp has constant rank k for allp ∈ M.

5. Sub-manifolds and the Whitney imbedding theorem

There is a well defined notion of sub-objects in a reasonably givencategory.

Definition 1.16. For a manifold N, a topological subspace M ⊂ N isa submanifold if there is an atlas (Uα, xα)α∈A on N such that therestriction

(Uα ∩M, xα|Uα∩M)α∈A

also form an atlas on M. This definition applies to any Ck manifolds.

5. SUB-MANIFOLDS AND THE WHITNEY IMBEDDING THEOREM 19

Let f : M → N be an immersion. By lemma 1.15 (1), for anyp ∈ M there is a chart U 3 p so that f |U is injective and f (U) isa submanifold of N. However, f may not be injective globally, e.g.parameterized plane curves with self-intersections.

Even if f is an injective immersion, the image may not be a mani-fold at all!Example 1.17. Consider the plane curve in polar coordinates r =

sin 2θ with θ ∈ (0, π). The parametrization γ : (0, π)→ R2 given by

(x(θ), y(θ)) = (r cos θ, r sin θ) = (sin 2θ cos θ, sin 2θ sin θ)

is an injective immersion of (0, π) into R2. But the point (0, 0) ∈γ((0, π)) does not have any locally Euclidean neighborhood, whenthe image γ((0, π)) is equipped with the subspace topology in R2.

Even if the image is a manifold, it may not be equipped with theinduced subspace topology:

Exercise 1.14. Let a ∈ R\Q and consider the map

f : R→ S1 × S1 : t 7→ (eit, eiat)

where we identify S1 as a subset in C. Show that f is an injectiveimmersion and f (R) is dense in S1 × S1.

Definition 1.18. A C∞ map f : M → N is an imbedding if it is an in-jective immersion which induces a homeomorphism f : M ∼→ f (M)

with f (M) ⊂ N being equipped with the subpace topology.

Lemma 1.19. If f : M → N is a C∞ imbedding then f (M) ⊂ N is a C∞

submanifold and f : M→ f (M) is a diffeomorphism.

PROOF. The condition that f being a homeomorphism means thatfor U a open neighborhood at p, there are open set V ⊂ N at f (p)such that f−1(V) = U and U is homeomorphic to f (U) = V ∩ f (M)

under f . By lemma 1.15 (1) we may select U to be a coordinate sliceof V and hence f (M) is a C∞ submanifold of N. f : M → f (M) is adiffeomorphism since they have identically the same atlas.

A continuous map f : M → N between topological spaces iscalled closed if the image of a closed set is closed. It is clear that an


injective continuous closed map induces a homeomorphism onto itsimage, so an injective closed immersion is an imbedding.

Similarly f is open if it send open sets to open sets. An injec-tive open immersion is also an imbedding. However, an imbeddingneeds not be closed nor open. E.g. an interval (a, b) along the x-axisin R2.

A continuous map is proper if the inverse image of a compact setis compact.

Exercise 1.15. Let f ∈ C(M, N) with N being Hausdorff. Show that:(1) If M is compact then f is proper as well as closed. (2) If M, N aremanifolds and f is proper then it is also closed.

Thus for compact domain manifolds there is no serious topo-logical issues to concern; the notion of immersions and imbeddings(= injective immersions here) are precise and convenient. For non-compact domain manifolds, extra information on f are usually cru-cial.

The Whitney imbedding theorem says that any manifold is noth-ing more than an imbedded submanifold in the Euclidean space. Be-fore we prove this fundamental result we need the

Definition 1.20. A set A ⊂ Rd has measure zero if for any ε > 0 thereis a countable cover by balls Bi with ∑i |Bi| < ε. A set A ⊂ Md in aCk manifold (k ≥ 1) has measure zero if for any chart (U, x) the setx(A ∩U) has measure zero in Rd.

By the standard diagonal argument we see that a countable unionof measure zero sets also has measure zero. Also it is trivial thatmeasure zero sets can not contain open sets. To see that the laterdefinition makes sense we need “measure zero” to be independentof choice of coordinates. Indeed, more is true:

Exercise 1.16. (1) If f : U → Rd is C1 and A ⊂ U ⊂ Rd has measurezero then f (A) also has measure zero. (2) If f : Mm → Nn is C1 andm < n, then f (M) has measure zero, in particular it is not surjective.

5. SUB-MANIFOLDS AND THE WHITNEY IMBEDDING THEOREM 21

Theorem 1.21 (Whitney Imbedding Theorem, 1936). Every C∞ mani-fold Md admits a C∞ closed imbedding into R2d+1 and a C∞ closed immer-sion in R2d.

PROOF. Here we will only give the proof for the (simpler) casethat M is compact.

Step 1: Construct an imbedding f : M → RN for some largeN ∈N. (This step requires only the C1 structure.)

For any p ∈ M, consider a chart (Up, xp) with xp(Up) = B0(2).The open cover U′p := x−1

p (B0(1))p∈M admits a finite subcoverindexed by 1, . . . , k. Consider cut off functions ψik

i=1 with ψi ≡ 1on U′i and supp ψi ⊂ Ui. Define a C∞ map

f :=k

∏i=1

(ψixpi , ψi)

≡ (ψ1xp1 , ψ1, · · · , ψkxpk , ψk) : M→ (Rd+1)k = Rk(d+1).

To see that f is an immersion, we notice that

d fp = (d(ψ1xp1), dψ1, · · · , d(ψkxpk), dψk) : TpM→ Tf (p)Rk(d+1).

Let p ∈ U′i . Since ψi|U′i ≡ 1, we get d(ψixpi)p = d(xpi)p. This is theidentification map TpM ∼= Txpi (p)R

d hence in particular that d fp isinjective.

To see that f is injective, given p 6= p′ ∈ M, if there is an i suchthat p, p′ ∈ U′i , then the component ψixpi = xpi gives different coor-dinates for p and p′. Otherwise p ∈ U′i and p′ 6∈ U′i for some i andthen ψi(p) = 1 > ψi(p′).

Step 2: Reduction of imbedding dimension N to 2d + 1.This step works for any given closed imbedding f : M → RN

(we need only the C2 condition on M and f , and M may be non-compact). The idea is find a direction v ∈ SN−1 and compose fwith the projection map πv : RN → v⊥ ∼= RN−1. The new mapfv := πv f is closed since projection maps are clearly closed mapsand composition of closed maps are again closed.

To have fv being injective it is equivalent to require that for anyp 6= q ∈ M, the vector

−−−−−→f (p) f (q) is not parallel to v. More precisely,


let ∆ : M→ M×M be the diagonal map ∆(p) = (p, p) and considerthe secant map

σ : M×M\∆(M)→ SN−1/±1 =: RPN−1

defined by

σ((p, q)) = ± f (p)− f (q)|| f (p)− f (q)|| .

Since σ is a C∞ map from a 2d manifold to an N − 1 manifold, if2d < N − 1 (that is, N > 2d + 1) then σ can not be surjective; in factIm σ has measure zero. Thus fv is injective if we select v 6∈ im σ.

To have fv being an immersion it is equivalent to require thatd(πv)x = πv is injective on Tx f (M) for all x ∈ f (M). Without loss ofgenerality we identify M as its image f (M) in RN. Then the tangentbundle TM ⊂ M×RN and the unit sphere bundle

S(TM) ⊂ M× SN−1

as a C∞ manifold of dimension d + (d− 1) = 2d− 1 is defined. Themap

T : S(TM)→ SN−1/±1 : (p, v) 7→ ±v

is a C∞ map from a 2d− 1 manifold to an N− 1 manifold. If 2d− 1 <

N − 1 (that is, N > 2d) then im T has measure zero and so T is notsurjective. Thus fv is an immersion if we select v 6∈ imT.

When N > 2d + 1, im σ ∪ im T ⊂ SN−1 also has measure zerohence the desired projection direction v can be selected. This com-pletes the proof.

Exercise 1.17. For a Ck manifold M, show that TM is a Ck−1 mani-fold by constructing an atlas on it and computing the transition func-tions.

A similar idea leads to applications to homotopy theory:

Theorem 1.22. If f : Sk → Sn is continuous with k < n, then f ishomotopic to a constant map. That is, πk(Sn) = 0 for k < n.

6. SUBMERSIONS AND SARD’S THEOREM 23

PROOF. It is trivial if f is not onto since Sn \ p ∼= Rn is con-tractible. This is indeed the case if f is C1 by Exercise 1.16.

Now the idea is simply to approximate f by a C1 (in fact C∞)function f : Sk → Sn within a δ-error with δ < π.

Exercise 1.18. Prove the C∞ approximation for f ∈ C(Sk, Sn) withinany δ > 0. In fact show that the C1 approximation is always possiblefor any f ∈ C(M, N) where M, N are both compact C1 manifolds.

With this done, then for each x ∈ Sk, the two vectors f (x), f (x)span a two dimensional plane Vx ⊂ Rn+1 and there is a uniquehomotopy F(x, t) from f (x) to f (x) through the shorter great cir-cle Vx ∩ Sn. F(x, t) is clearly continuous, hence f is homotopic to f ,which is C1 and hence homotopic to a constant map.

Exercise 1.19 (Invariance of dimension). As a corollary, show thatRn ∼= Rm (homeomorphic)⇐⇒ n = m.

6. Submersions and Sard’s theorem

After discussing submanifolds induced from immersions, we nowconsider those induced from submersions, i.e. d f is surjective.

N

Mf−1(q2)

f−1(q1)

q2

q1

Definition 1.23. Let f : M → N be smooth. A point q ∈ N is calleda regular value of f if d fp is surjective for all p ∈ f−1(q). Otherwiseq is called a singular value . When q is a regular value, the preimagef−1(q) is a, possibly non-connected, submanifold of M and f can beparametrized as a coordinate projection locally at p by Lemma 1.15.


Any point p ∈ f−1(q), i.e. d fp surjective, is referred as a regular point.

Definition 1.24. A point p ∈ M is called a critical point of f if p is notregular, i.e. d fp is not surjective. If N = R, this means d fp = 0. Wedenote by C( f ) the set of all critical points.

Intuitively, the map f establishes a kind of nice local fiber spacestructure on M outside the singular values f (C( f )). Thus it is impor-tant to know more about the properties of f (C( f )).

Theorem 1.25 (Sard’s Theorem). f (C( f )) has measure 0 in N.

PROOF. Only have to prove the case of charts (U, x) with boundedU.

Consider f : U ⊂ Rm → Rn. Let

Ci = x ∈ U | Dα f (x) = 0, ∀ α, |α| ≤ i

and C = C( f ). So C ⊃ C1 ⊃ C2 ⊃ · · · .This proof consists three steps:

(1) f (C \ C1) has measure 0.(2) f (Ci \ Ci+1) has measure 0.(3) f (Ck) has measure 0 for some k large enough.

And we use induction on the dimensions m + n.If m + n = 1, C = C1 and f (C) consist of only one point. The

theorem is trivial in this case. So we assume that m + n ≥ 2.Let p ∈ C \C1, say ∂ f 1(p)/∂x1 6= 0. Through a coordinate change

h : U → Rm,x 7→ ( f 1(x), x2, · · · , xm)t,

we have

dhp =

∂1 f 1(p) ∗

0 Im−1

which has non-zero Jacobian. By the Inverse Function Theorem,there is a neighborhood V ⊂ U such that h−1 exists on V := h(V).

6. SUBMERSIONS AND SARD’S THEOREM 25

U

Vp

h(p)

V

f

hh−1

Rm

f = f h−1

Let f = f h−1 : V → Rm. We have f 1(t, · · · ) = t and

q ∈ C( f ) ∩ V ⇐⇒ h−1(q) ∈ C( f ) ∩V.

Then f : V → Rn splits into

ft : (t ×Rm−1) ∩ V → t ×Rn−1,

and

d f =

1 0

∗ d ft

, q = (t, r) ∈ C( f ) ⇐⇒ r ∈ C( ft).

By the induction hypothesis, C( ft) has measure zero in the hy-perplane t ×Rn−1. Since C( f ) \ C1 ⊂

⋃t C( ft), by Fubini theorem

in Lebesgue integral,

| f (C( f ) \ C1)| ≤∫

t

∣∣ ft(C( ft))∣∣ dt = 0.

Remark 1.26. This argument does not really need the full power ofLebesgue theory. We only need the theory of measure 0 for the proof.

Consider the same argument on all p ∈ C \ C1. Since C \ C1 canbe covered by countable union of such (C( f ) \ C1)’s, we concludethat f (C \ C1) has measure 0.


Secondly, for any p ∈ Ci \ Ci+1, we may assume Dα f (p) = 0 forall |α| ≤ i but Dβ f 1(p) 6= 0 for some β = α + (1, 0, · · · , 0).

Write f (α)(x) = Dα f 1(x). Again by changing coordinates,

h : U → Rm,x 7→ ( f α(x), x2, · · · , xm)t,

dhp =

∂1 f (α)(p) ∗

0 Im−1

is invertible and there is a neighborhood V ⊂ U such that h−1 exists.Also, h(Ci ∩V) ⊂ 0 ×Rm−1. Let f = f h−1. We have

h−1(q) ∈ Ci( f )⇐⇒ q ∈ Ci( f ) = Ci( f0)

where f0 : (0 ×Rm−1)⋂

h(V)→ Rn.By induction, f (C( f0)) has measure 0, and so does f (Ci \ Ci+1)

by the countable covering argument as before.

Thirdly, we claim that f (Ck) has measure 0 for large k through avolume estimate. By the Taylor expansion, there exists a constant Awhich depends only on k, m and n such that for p ∈ Ck,

f (p + h) = f (p) + R(p, h), |R(p, h)| ≤ A|h|k+1

for all |h| < δ. Let vn = vol(B0(1)). In each such ball Bp(δ) we have

| f (Ck ∩ Bp(δ))| ≤ vn Anδ(k+1)n

Now we cover Ck by a finite number of such balls. We need at most(2d/δ)m balls with d = diam U. Pick k satisfying k + 1 > m/n, then

| f (Ck)| ≤ vn An(2d)mδ(k+1)n−m.

Since A and d are independent of δ, by taking δ → 0 we get thatf (Ck) has measure 0. The proof is complete.

Remark 1.27. Sard’s Theorem holds for Ck maps with k > maxm−n, 0.

7. VECTOR FIELDS AND FROBENIUS THEOREM 27

7. Vector fields, flows, Lie derivatives and the Frobenius integrabil-ity theorem

Definition 1.28. Let U ⊂ M be open. X ∈ T(U) is called a vector field iffor each p ∈ U, we assign a vector X(p) ∈ TpM at p. We say X ∈ C∞

if X f ∈ C∞(U) for all f ∈ C∞(U).

XTM

M•p

•Xp

?

π

We’ve known that TM is a smooth manifold where charts andcoordinate functions on TM are of the form

(π−1(U), x1, · · · , xm, dx1, · · · , dxm).

So we also have equivalent definitions for C∞ vector fields:

Proposition 1.29. The followings are equivalent:

(1) X( f ) ∈ C∞(U) for any f ∈ C∞(U);

(2) X = ∑ ai(∂/∂xi), ai ∈ C∞(U);

(3) X : U → TM is C∞.

In Euclidean space, for a Ck vector field F : U ⊂ Rm → Rm wecan find a solution to the initial value problem

∂x∂t

= F(x(t)), x = (x1(t), · · · , xm(t))

x(0, x0) = x0

.

This is the existence and uniqueness theorems of ODE.

Theorem 1.30. There exists a unique solution x(t, x0) which is Ck in botht and x0. (In fact, Ck+1 in t.)

We also have a manifold version:


Theorem 1.31. Let X be a C∞ vector field on M. For any p ∈ M, thereexists (ap, bp) ⊂ R

⋃±∞ and γp : (ap, bp)→ M such that γp(0) = p,γ′p(t) = X(γp(t)) and γp is the maximal among curves with this property.

The solution γp is called an integral curve of X.Furthermore, for any fixed t ∈ R, it is useful to consider the flow

φt(p) := γp(t)

on the domainDt := p ∈ M | t ∈ (ap, bp).

Clearly, φt is a diffeomorphism from Dt to D−t. (We remark that ifM is compact, the interval of existence (ap, bp) can be extended to(−∞, ∞) and φt will be a diffeomorphism from M to itself.) Andφtt∈R defines a one-parameter group of (local) diffeomorphisms:

• φt φs = φt+s

• φ−1t = φ−t

It is fundamental and natural to ask: How to take derivatives of avector field? Notice that it does not make sense to compare the twovectors Xp and Xq when p 6= q. In Rd, we take derivatives by movingf (γ(t)) back to γ(0) and calculate

vp( f ) = limt→0

f (γ(t))− f (γ(0))t

along any C1 curve γ with γ(0) = p and γ′(0) = v.On a manifold, one solution is the Lie derivative.

Definition 1.32. Given two vector fields V, X. Let φt be the flow gen-erated by V. The Lie derivative is defined by

LV X(p) := limt→0

1t

((φ−t)∗Xφt(p) − Xp

)=

ddt∣∣t=0(φ−t)∗Xφt(p)

Mγ

•p • γ(t)

Xp

Xγ(t)

I(φ−t)∗


We shall see shortly that the Lie derivative corresponds to an-other simple binary operation on vector fields, the Lie bracket.

Definition 1.33. Let X, Y be C∞ vector fields on U. The Lie bracket isdefined by [X, Y] := XY−YX, i.e. [X, Y]p f = Xp(Y( f ))−Yp(X( f ))

It seems that [X, Y] is a second order operator. In fact, it is indeedan operation on T(U). That is,

Proposition 1.34. [X, Y] is still a C∞ vector field on U.

PROOF. Let X = ∑ ai ∂

∂xi , Y = ∑j

bj ∂

∂xj and f ∈ C∞. By Lebnitz

rule,

[X, Y]p f = ∑ij

ai ∂

∂xi

(bj ∂

∂xj f)− bj ∂

∂xj

(ai ∂

∂xi f)

= ∑ij

ai ∂bj

∂xi∂ f∂xj + aibj ∂

∂xi∂ f∂xj − bj ∂ai

∂xi∂ f∂xj − bjai ∂

∂xj∂ f∂xi

= ∑ij

(ai ∂bj

∂xi∂

∂xj − bj ∂ai

∂xi∂

∂xj

)f

The Lie bracket has the following properties:

Proposition 1.35. Let X, Y be C∞ vector fields

(1) (anti-symmetry) [X, Y] = −[Y, X]

(2) (Jacobi identity) [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 0

Now we can prove the important formula:

Theorem 1.36 (Lie). (LVW)p = [V, W]p


PROOF. Let V generate φt and W generate ψs. Choose a test func-tion h ∈ C∞

p . By definition,

(LVW)ph

=ddt

∣∣∣t=0

(φ−t)∗Wφt(p)h =ddt

∣∣∣t=0

Wφt(p)h φ−t

=ddt

dds

h φ−t ψs(φt(p))∣∣∣t=0,s=0

=dds

ddt

h φ−t ψs φt(p)∣∣∣t=0,s=0

=dds

(dh(−Vφ−tψsφt(p)

)+ dh φ−t∗ ψs∗

(Vφt(p)

)) ∣∣∣t=0,s=0

=dds

(−dh(Vψs(p)) + d(h ψs)(Vp)

) ∣∣∣s=0

= − dds

∣∣∣s=0

Vh(ψs(p)) + Vp

(dds

∣∣∣s=0

h ψs(p))

= −Wp(Vh) + Vp(Wh) = [V, W]ph.

We may extend these concepts to some higher dimensional cases.

Definition 1.37. A k-dimensional distribution D on a manifold M is achoice of a k-dimensional subspace D(p) of TpM for each p ∈ M. Wesay D is C∞ if for every p ∈ M, there is a neighborhood U such thatD is spanned by k C∞ vector fields X1, X2, · · · , Xk on U.

The question in higher dimension becomes: For a given distribu-tion D near p, does there exists a k dimensional submanifold S withp ∈ S such that TqS = D(q) for all q ∈ S?

Clearly a necessary condition for this integrability is the follow-ing:

Definition 1.38. D is called involutive if [X, Y] ∈ D for all X, Y ∈ D .Here a vector field X ∈ D means Xp ∈ D(p) for all p.

This turns out to be sufficient at least locally:

Theorem 1.39 (Frobenius Integrability). If D is involutive, then for allp ∈ M there exists a maximal integral manifold Sp ⊂ M passing throughp such that TqSp = D(q) for all q ∈ Sp.


As in the case of integral curves, Sp ⊂ M needs not be closed.That is Sp may not be with the induced topology.

The proof is based on two lemmas:

Exercise 1.20. Let f : S→ M be C∞ and X, Y ∈ C∞(TS). If f∗X = X′,f∗Y = Y′ on M, then f∗([X, Y]) = [ f∗X, f∗Y].

For f∗X = X′, we call X and X′ are f -related.

Lemma 1.40. If X1, · · · , Xk are k vector fields near a point p ∈ M thatspans D and [Xi, Xj] = 0 for each i, j, then locally near p there exists acoordinate submanifold S 3 p such that Xi = ∂/∂xi, i = 1, 2, · · · , k.

That is, there exists a coordinate system (U, x) at p such that S can belocally represented as xj = 0 | j = k + 1, · · · , m

PROOF. Assume [V, W] = 0. Let V generate φ and W generate ψ.We claim that φt ψs(p) = ψs φt(p).

t

s−t−s

pM

φt(p)t

s

ψs φt

ψs φt(p)

Let c(t, s) = ψ−s φ−t ψs φt. Since LVW = 0 implies φ−t∗W =

W, directly differentiation shows that

∂

∂sc(t, s) = −Wψ−sφ−tψsφt(p) + ψ−s∗ φ−t∗(Wψsφt(p))

= −Wc(t,s) + Wc(t,s) = 0

Similarly, ∂c(t, s)/∂t = 0. So c(t, s) = p for all well-defined t, s.Now, suppose Xi generates φi for i = 1, 2, . . . , k near p. We define

c(t1, t2, . . . , tk) := φ1t1 φ2

t2 · · · φktk .


We can check its tangent vector can be generated by X1, . . . , Xk:

c∗(∂/∂t1) =∂

∂t1 φ1t1 · · · φk

tk = X1

∣∣∣c(t1,...,tk)

and since every φiti

commutes with each other,

c∗(∂/∂ti) =∂

∂ti φiti φ1

t1 · · · φiti · · · φk

tk = Xi

∣∣∣c(t1,...,tk)

.

Let S := c(t1, . . . , tk) | tj ∈ Ij ∀ j where Ij is chosen to be small

enough such that every φjtj is well-defined for tj ∈ Ij. We can see

that c∗ is non-vanishing and then c is an immersion into M. ByLemma 1.15, there exists a chart (U, x) near p such that xi(c(t1, . . . , tk)) =

ti for i = 1, . . . , k and xi(c(t1, . . . , tk)) = 0 for i = k + 1, . . . , m. AndS ∩U is a coordinate slice xj = 0 | j = k + 1, . . . , m on M.

PROOF OF FROBENIUS THEOREM. Since the distribution D is a kdimensional subspace, we can project it into an Euclidean space:

TpM

p

R⟨

∂∂x1 , . . . , ∂

∂xk

⟩S

D

??π∗

Then there exists a U 3 p such that

D ∼= R

(∂

∂x1 , . . . ,∂

∂xk

), ∀ q ∈ U.

By lifting each Xi ∈ D from ∂/∂xi, i.e. π∗(Xi) = ∂/∂xi. Hence

0 =

[∂

∂xi ,∂

∂xj

]= [π∗(Xi), π∗(Xj)] = π∗[Xi, Xj]⇐⇒ [Xi, Xj] = 0.

8. EXISTENCE, UNIQUENESS AND SMOOTHNESS DEPENDENCE OF ODE 33

Applying Lemma, there exists an integral manifold S near p suchthat Tq(S) = D(q) for all q ∈ U.

8. Existence, Uniqueness and Smoothness Dependence of ODE

When given such an ODE: X′(t) = F(X(t)), F ∈ C1(O, Rn) a vector field.

X(0) = x0,

by the existence and uniqueness theorems of solutions to ODE sys-tems, there exists a unique flow φ(t, x0) satisfying the equation fort ∈ J, an interval of existence. There are two proofs of this theo-rem, one of which going through a classical method by estimates(c.f. Hirsch and Smale, Differential Equations, Dynamical Systems,and an Introduction to Chaos) and the other provided by the inversefunction theorem on Banach space. In this section, we will assumethe theorems.

It is obviously that ∂∂t φ(t, x0) exists. So, our goal is to discuss: how

smooth the solution φ(t, x0) depends on its initial value x0?Suppose there are two solutions X(t) and X(t) with the given

initial data x0 and x0 + z0. The key point is to estimate ‖X(t)− X(t)‖in terms of x0 and z0. We consider the variational equation:

(*)

U′(t) = A(t)U(t),

U(0) = z0

where A(t) = F′(X(t)), which is C0 depending on t. The idea is that,when z0 is small, X(t) + U(t) should approximate X(t) with initialx0 + z0. In fact, this comes from the intuition that if F is C2, thenthe solution to the variational problem is just the first order term ofX(t)’s Taylor expansion in z0.


Proposition 1.41. Let U(t, ξ) be the flow of (*), i.e. U(0, ξ) = ξ, x0 + ξ ∈O, and Y(t, ξ) be the flow of X′(t) = F(X(t)) with Y(0, ξ) = ξ. Then

lim‖ξ‖→0

‖Y(t, ξ)− X(t)−U(t, ξ)‖‖ξ‖ = 0

uniformly on an interval of existence J.

Assuming the proposition, the theorem follows immediately.

Theorem 1.42. The flow φ(t, x0) of the ODE system X′(t) = F(X(t)),X(0) = x0 is of Ck as well as F ∈ Ck.

PROOF. By the proposition,

φ(t, x0 + ξ)− φ(t, x0) = Y(t, ξ)− X(t) = U(t, ξ) + o(|ξ|).

Note that from solving the linear system U′(t, ξ) = A(t)U(t, ξ) withU(0, ξ) = ξ, we can see that U(t, ξ) = eA(t)ξ is linear in ξ. HenceD2φ(t, x0)ξ = U(t, ξ) and φ(t, x) is of C1 in x.

Back to the variational equation (*), we will get

ddt(D2φ(t, x0)) = F′(φ(t, x0))D2φ(t, x0)

with D2φ(0, x0) = idRn . Then by induction, F ∈ Ck will imply thatφ(t, x) is Ck in x.

PROOF OF PROPOSITION. We can write the differential equationsinto integrals as:

X(t) = x0 +∫ t

0F(X(s))ds,

Y(t, ξ) = x0 + ξ +∫ t

0F(Y(s, ξ))ds,

U(t, ξ) = ξ +∫ t

0F′(X(s))U(s, ξ)ds.

By Taylor expansion, we have such an estimate:

‖Y(t, ξ)− X(t)−U(t, ξ)‖ ≤∫ t

0‖F(Y(s, ξ))− F(X(s))− F′(X(s))U(s, ξ)‖ds

≤∫ t

0‖F′(X(s))‖‖Y(s, ξ)− X(s)−U(s, ξ)‖+ ‖R(X(s), Y(s, ξ)− X(s))‖ds

9. EXERCISES 35

where R is the first order remainder term.We use the Gronwall’s inequality to deal with the iteration of dif-

ference appearing in the integral.

Exercise 1.21 (Gronwall’s inequality, an easy version). If u ∈ C1[0, d],u > 0 and u satisfies u(t) ≤ c +

∫ t0 Ku(s)ds for some positive con-

stants c, K. Then u ≤ ceKt on [0, d].

Therefore, the constant c = |ξ| can be taken to be small and

‖Y(t, ξ)− X(t)‖ ≤ ‖ξ‖+∫ t

0‖F(Y(s, ξ))− F(X(s))‖ds

≤ ‖ξ‖+∫ t

0‖F′‖‖Y(s, ξ)− X(s)‖ds

Choose K large such that ‖F′‖ < K on a small neighborhood. ByGronwall’s inequality, ‖Y(t, ξ)− X(t)‖ ≤ ‖ξ‖eKt. So, for any ε > 0,we can choose ξ small such that the remainder term

‖R(X(s), Y(s, ξ)− X(s))‖ ≤ ε‖Y(s, ξ)− X(s)‖.

Denote g(t) = ‖Y(t, ξ)− X(t)−U(t, ξ)‖ and rewrite

g(t) ≤∫ t

0Kg(s) + ε‖ξ‖eKsds ≤ ε‖ξ‖C +

∫ t

0Kg(s)ds

for some bounded constant C depending on F and the existece inter-val J. By Gronwall’s inequality again, g(t) ≤ ε‖ξ‖CeKt and hence

‖Y(t, ξ)− X(t)−U(t, ξ)‖‖ξ‖ ≤ εC

which is uniformly in t.

9. Exercises

1. Show that Rd (with the standard Euclidean topology) is a mani-fold by finding an explicit countable basis.

2. Let Rd1 ∼= Rd2 (homeomorphic). If d1 = 1 show that d2 = 1.Investigate the case d1 = 2 and reduce the problem to the JordanCurve Theorem.


3. For a manifold with a Ck atlas, k ≥ 1, show that the dimensiond = dim M is well defined on each connected component of M.

4. Show that the enlarged collection of charts Uα, φα)α∈A also formsa Ck atlas. Moreover, it is a maximal atlas in the sense that anychart which is Ck related to it is already contained in it.

5. For any Ck manifold Md and p ∈ M, show that there are chartswith x(U) = B0(r), the open ball of radius r in Rd, as well ascharts with x(U) = Rd.

6. Consider M = R with one chart given by (R, φ) where φ(t) = t3.Show that this defines a C∞ structure on M. Is M diffeomorphicto R with the standard C∞ structure (R, id)?

7. Show that by removing those Wi with noncompact closure Wi westill get a basis. (Notice that the Hausdorff condition is needed.)

8. Show that f ∈ C∞(R) and f (n)(0) = 0 for all n ∈N.9. Investigate the theorem for the case when M = R with the open

cover given by a single set U = M = R.10. Let A (resp. U) be a closed (resp. open) set in a Ck manifold M

with A ⊂ U. Show that there exists f ∈ Ck(M) such that f |A ≡ 1and f |M\U ≡ 0. Is that possible to make supp f = A?

11. Let M be a Ck manifold. Show that

dim DpM = dim mp/m2p =

dim M if k = ∞,

∞ if k < ∞.

(Hint: For k = 1, study functions f (x) = (x1)a for 1 < a < 2.)12. Show that in charts (U, x) at p ∈ M and (V, y) at f (p) ∈ N, d fp

is represented by d fx(p) with Jacobian matrix[

∂ f j

∂xi

]where f =

y f x−1 : x(U ∩ f−1(V))→ Rdim N. Namely

d fp

( ∂

∂xi

∣∣∣p

)= ∑

j

∂ f j

∂xi (x(p))∂

∂yj

∣∣∣p.

13. Show that f ∈ C∞(M, N) can be locally represented by

f (x) = (x1, · · · , xk, 0, · · · , 0)t

9. EXERCISES 37

for some k ≤ m if and only if that d fp has constant rank k for allp ∈ M.

14. Let a ∈ R\Q and consider the map

f : R→ S1 × S1 : t 7→ (eit, eiat)

where we identify S1 as a subset in C. Show that f is an injectiveimmersion and f (R) is dense in S1 × S1.

15. Let f ∈ C(M, N) with N being Hausdorff. Show that: (1) If Mis compact then f is proper as well as closed. (2) If M, N aremanifolds and f is proper then it is also closed.

16. (1) If f : U → Rd is C1 and A ⊂ U ⊂ Rd has measure zero thenf (A) also has measure zero. (2) If f : Mm → Nn is C1 and m < n,then f (M) has measure zero, in particular it is not surjective.

17. For a Ck manifold M, show that TM is a Ck−1 manifold by con-structing an atlas on it and computing the transition functions.

18. Prove the C∞ approximation for f ∈ C(Sk, Sn) within any δ > 0.In fact show that the C1 approximation is always possible for anyf ∈ C(M, N) where M, N are both compact C1 manifolds.

19. As a corollary, show that Rn ∼= Rm (homeomorphic)⇐⇒ n = m.20. Let f : S → M be C∞ and X, Y ∈ C∞(TS). If f∗X = X′, f∗Y = Y′

on M, then f∗([X, Y]) = [ f∗X, f∗Y].21. If u ∈ C1[0, d], u > 0 and u satisfies u(t) ≤ c +

∫ t0 Ku(s)ds for

some positive constants c, K. Then u ≤ ceKt on [0, d].22. (Warner Ch.1 #10) Let M be a compact manifold of dimension n,

and let f : M → Rn be C∞. Prove that f cannot everywhere benon-singular.

23. (Warner Ch.1 #3) Let Uα be an open cover of a manifold M.Prove that there exists a refinement Vα such that Vα ⊂ Uα foreach α.

24. (Warner Ch.1 #9) Let f : R2 → R be defined by

f (x, y) = x3 + xy + y3 + 1.

For which points p = (0, 0), p = (13 , 1

3), p = (−13 ,−1

3) is f−1( f (p))

an imbedded submanifold in R?


25. (Warner Ch.1 #16) Let N ⊂ M be a submanifold. Let γ : (a, b) →M be a C∞ curve such that γ(a, b) ⊂ N. Show that it is not neces-sarily true that γ(t) ∈ Nγ(t) for each t ∈ (a, b).

26. (Warner Ch.1 #17) Prove that any C∞ vector field on a compactmanifold is complete.

27. (Warner Ch.1 #18) Prove that a C∞ map f : R2 → R1 cannot beone-to-one.

28. (Warner Ch.1 #23) A Riemannian structure on a differentiable man-ifold M is a smooth choice of a positive definite inner product〈 , 〉m on each tangent space Mm, smooth in the sense that when-ever X and Y are C∞ vector fields on M, then 〈X, Y〉 is a C∞ func-tion on M. Prove that there exists a Riemannian structure on everydifferentiable manifold. You will need to use a partition of unityargument. A Riemannian manifold is a differentiable manifold to-gether with a Riemannian structure.

29. (Warner Ch.1 #6) Prove that if ψ : M→ N is C∞, one-to-one, onto,and everywhere non-singular, then ψ is a diffeomorphism.

30. (Warner Ch.1 #19) Supply the details of the equivalence of theFrobenius theorem 1.39 and the classical version:

Remark 1.43 (classical Frobenius theorem). Let U and V be opensets in Rm and Rn respectively. We use coordinates r1, . . . , rm onRm and s1, . . . , sn on Rn. Let

b : U ×V → M(n, m)

be a C∞ map of U × V into the set of all n×m real matrices, andlet (r0, s0) ∈ U ×V. If

∂biβ

∂rγ−

∂biγ

∂rβ+

n

∑j=1

(∂biβ

∂sjbjγ −

∂biγ

∂sjbjβ

)= 0

(i = 1, . . . , n; γ, β = 1, . . . , m)

on U ×V, then there exist neighborhoods U0 of r0 in U and V0 ofs0 in V and a unique C∞ map

α : U0 ×V0 → V

9. EXERCISES 39

such that if

αs(r) = α(r, s) (s ∈ V0, r ∈ U0)

thenαs(r0) = s, dαs

∣∣r = b(r, α(r, s))

for all (r, s) ∈ U0 ×V0.31. (Warner Ch.1 #20) Let ϕ : N → M be C∞, and let X be a C∞

vector field on N. Suppose that dϕ(X(p)) = dϕ(X(q)) wheneverϕ(p) = ϕ(q). Is there a smooth vector field Y on M which is ϕ-related to X?

Chapter 2

TENSORS AND DIFFERENTIAL FORMS

The idea of a tensor is a natural generalization of a vector field.There are various types of tensor fields and differential forms asso-ciated with the tangent spaces and cotangent spaces:

TpM T∗p M

V =⟨

∂∂xi

⟩V∗ = Hom(V, R) =

⟨dxi∣∣

p

⟩.

In general, tensors are just geometrical objects because the relation-ships between them must be independent of the choices of coordi-nate systems.

In this section, we shall first develop some facts from multilinearalgebra and then apply these concepts to manifolds.

1. The Tensor Algebra

There are several aspects to define tensors and the tensor prod-ucts. Here we try to introduce them just in an intuitive way.

For given vector spaces V, W, say

V = 〈v1, v2, . . . , vm〉 , W = 〈w1, w2, . . . , wn〉 ,

we define V⊗W as a vector space spanned by vi⊗wj’s, i.e. V⊗W =⟨vi ⊗ wj

∣∣1 ≤ i ≤ m, 1 ≤ j ≤ m⟩, where ⊗ is bilinear over R. We can

see that V ⊗W has dimension mn and the definition is independentof the choices of basis. Also, as a consequnce, V ⊗W is canonicallyisomorphic to W ⊗V and (V ⊗W)⊗U is canonically isomorphic toV ⊗ (W ⊗U).

41

42 2. TENSORS AND DIFFERENTIAL FORMS

In particular, we consider the dual space V∗ and the tensor prod-uct

V ⊗ · · · ⊗V ⊗V∗ ⊗ · · · ⊗V∗ = V⊗r ⊗V∗⊗s =: ⊗r,sV = V(r,s),

which is called (r, s)-tensors of V and denoted as Tr,s. The first rcomponents are called the contravariant part and the last s ones arecalled the covariant part.

Since we can recognize a vector by its duality, we may also iden-tify the (r, s)-tensors with multilinear functions:

Tr,s(V) = (⊗rV)⊗ (⊗sV) ∼= L(Vs; Vr) ∼= L(V∗s, Vr; R).

Example 2.1. Here are some examples of tensors.

(1) T ∈ ⊗1,2V can be identified with T(x, y) ∈ L(V, V; V) by

V ⊗V∗ ⊗V∗ L(V, V; V)

T = ∑i,j,k aijkvi ⊗ v∗j ⊗ v∗k ↔ ∑i aijkvi = T(vi, vk)

(2) ⊗0,2(V) = V∗ ⊗V∗.e.g. the metric tensor g = ∑i,j gijdxi ⊗ dxj.

(3) ⊗1,3(V) = V ⊗V∗ ⊗V∗ ⊗V∗.

e.g. the curvature tensor: R = ∑i,j,k,l Rijkl

∂

∂xi ⊗ dxj ⊗ dxk ⊗dxl.

Particularly, the symmetric and alternating tensors are usuallyseen in differential geometry. A symmetric tensor T ∈ Symk(V∗) ⊂⊗k(V∗) satisfies

T(· · · , vi, · · · , vj, · · · ) = T(· · · , vj, · · · , vi, · · · ) ∀ i, j.

For example, the metric tensor g is indeed a symmetric 2-tensor. Letv = ∑n

i=1 aiei,

g(v, v) =n

∑i,j=1

g(ei, ej)aiaj ∼ g =n

∑i,j=1

gije∗i ⊗ e∗j .

Note that q(v) = g(v, v) is a quadratic form. Conversely, we havethe polarization formula for the homogeneous polynomial of degree

1. THE TENSOR ALGEBRA 43

2:g(v, w) =

12(q(v + w)− q(v)− q(w)).

Exercise 2.1. Find the polarization formula of T ∈ Symk(V∗) forgeneral k.

On the other hand, an alternating tensor T ∈ Λk(V) ⊂ ⊗pV∗ =

⊗0,p(V) is a skew-symmetric form satfisfies

T(· · · , vi, · · · , vj, · · · ) = (−1)T(· · · , vj, · · · , vi, · · · ).

By linearity, this implies that T(· · · , v, · · · , v, · · · ) = 0. Moreover,suppose vi = ∑j aj

iej.

T(v1, · · · , vk) = T(· · · , ∑j

ajiej, · · · )

= ∑p∈Pn

k

ap(1)1 ap(2)

2 · · · ap(k)k T(ep(1), · · · , ep(k))

= ∑c∈Cn

k

∑σ∈Sk(c)

aσ(c(1))1 · · · aσ(c(k))

k T(eσ(c(1)), · · · , eσ(c(k)))

= ∑

c∈Cnk

T(ec(1), · · · , ec(k))det Vc

where Vc means the k× k minor of the matrix [v1 · · · vk] with respectto c:

[v1 v2 · · · vn] =

a11 · · · a1

k... . . . ...

an1 · · · an

k

.

In summary, for V = 〈e1, · · · , en〉, V∗ = 〈e∗1 , · · · , e∗n〉, the tensorT ∈ Tr,s(V) can be decomposed into such a linear combination:

T = ∑I,J

Ti1 i2 ··· irj1 j2 ··· js (ei1 ⊗ · · · ⊗ eir)⊗ (e∗1 ⊗ · · · ⊗ e∗s )

= ∑I,J

T IJ eI ⊗ eJ .

Now we go back to manifolds: V = TpM, V∗ = T∗p M. On a localchart (U, x), the tangent bundle TU has frame ∂

∂x1 , · · · , ∂∂xn and the

cotangent bundle has a dual-frame dx1, · · · , dxn.


Definition 2.2. We call a tensor field T ∈ Tr,s(U) (actually, it meansTr,s(TU)) is C∞ if and only if the coefficients T I

J (x) ∈ C∞(U) where

T = ∑i,j

Ti1i2···irj1 j2···js

∂

∂xi1⊗ · · · ⊗ ∂

∂xir⊗ dxj1 ⊗ · · · ⊗ dxjs .

We have mentioned that the relationships between tensors areindependent of the choice of coordinate systems.

Uα

Uβ

φ

x

y

(x1, x2, . . . , xn)(y1, y2, . . . , yn)

y x−1

Let (Uα, x), (Uβ, y) be two charts and φαβ = y x−1 be the coordi-nate transformation. Now we can see the changes on basis and dualbasis:

yi = φiαβ(x1, x2, · · · , xn)⇒

[∂yi

∂xj

]= φ′αβ

∂

∂yi = ∑j

∂xj

∂yi∂

∂xj , dyi = ∑j

∂yi

∂xj dxj

So the tangent vectors (the contravariant part) vary with the inversetransformation and the dual vectors (the covariant part) vary withthe transformation. For a r, s tensor field T under the transition(U, x)→ (V, y), we have

T = ∑I,J

Ti1,··· ,irj1,··· ,js

∂

∂xi1⊗ · · · ⊗ ∂

∂xir⊗ dxj1 ⊗ · · · ⊗ dxjs

= ∑I′,J′

T I′J′

∂

∂yi′1⊗ · · · ⊗ ∂

∂yi′r⊗ dyj′1 ⊗ · · · ⊗ dyj′s

2. THE EXTERIOR ALGEBRA 45

where

T I′J′ = ∑

I,JTi1,··· ,ir

j1,··· ,js∂yi′1

∂xi1· · · ∂yi′r

∂xi′r

∂xj1

∂yj′1· · · ∂xjs

∂yj′s.

More generally, for a smooth mapping f : M → N, tensors on Mand N can be connected by the differential map and pull-back map:

Mf

// N

TpMf∗ // Tf (p)N

induce// T∗f (p)Nf ∗

// T∗p M

w // f ∗(w)(v) = w( f∗v)

Hence we can compute for T = ∑j1,...,js hj1,...,js dyj1 ⊗ · · · ⊗ dyjs ,

f ∗(T) = ∑j1,...,js

hj1,...,js f d(yj1 f )⊗ · · · ⊗ d(yjs f ).

2. The Exterior Algebra

Now we go back to think about the alternating tensors. Consid-ering the ideal

I(V) := 〈· · · ⊗ a⊗ · · · ⊗ a⊗ · · · 〉 ⊂ ⊗V = ⊗∞r=0 (⊗r(V))

and taking the quotient

Λ(V) := ⊗V/

I(V).

Definition 2.3 (exterior algebra and exterior product). The natural ho-momorphism

⊗V → Λ(V)

α 7→ α

induces the multiplication (on the algebra Λ(V)) which satisfies

(a ∧ b) ∧ c = a⊗ b ∧ c = a⊗ (b⊗ c) = a ∧ (b ∧ c).

We call Λ(V) an exteror algebra and the multiplication ∧ the exteriorproduct (wedge product) on Λ(V).


We can see that Λ(V) = ⊕∞p=0Λp(V) is a Z2-graded ring where

Λp(V) = R〈ei1 ∧ · · · ∧ eip | i1 < · · · < ip〉 ∼= RCn

p .Note that the alternating p-tensors can also be constructed by

anti-symmetrization of the p-tensors ei1 ⊗ · · · ⊗ eip . For an example,we can define a 2-form by

a ∨ b :=a⊗ b− b⊗ a,

a ∨ b ∨ c :=a⊗ b⊗ c− a⊗ c⊗ b− b⊗ a⊗ c

+ b⊗ c⊗ a + c⊗ a⊗ b− c⊗ b⊗ a.

The two constructions are equivalent. We have such a vector spaceisomorphism

∨(V) −→ Λ(V)

a ∨ b 7−→ 12 a ∨ b = 1

2 a⊗ b− b⊗ a = a⊗ b = a ∧ b(a ∨ b) ∨ c 7−→ 1

2(a ∨ b) ∨ c = 12 a ∨ b ∧ c = a ∧ b ∧ c

.

By extending the mapping linearly, we have ∨(V) ∼= Λ(V).

3. Cartan’s operator d

Now focus on the alternating tensors on manifolds. We denoteAr(M) := C∞(Λr(T∗M)) = C∞alternating r-forms.

For any f ∈ C∞(M), we can see the differential d f as a smoothmapping from TM into R. So d f can be considered as a 1-form,d f : M→ Λ∗(M), which is called the exterior derivative of a 0-form f .

Another example comes from the Green’s theorem: for (P, Q) aC1 vector field and D a regular domain with smooth boundary ∂D,we have

∫∂D Pdx + Qdy =

∫D(Qx − Py)dxdy. The integration of 2-

form is related to a 1-form.So we have such an important differential operator d which is

unique in this sense:

Theorem 2.4 (Cartan’s exterior derivative, d). There exists a uniquedifferential operator d : Ap(M)→ Ap+1(M) such that

(1) d f (X) = X f , for any f ∈ C∞(M).(2) d2 = 0.(3) (Lebnitz rule) d(ω ∧ η) = dω ∧ η + (−1)pω ∧ dη.

3. CARTAN’S OPERATOR d 47

Actually, Cartan write down the exact formula, see 2.7. However,to show this formula, since dω only depends on only v1, . . . , vp, wemust argue that this formula is independent of choices to the exten-sions.

Now we just show the existence and uniqueness of d by consid-ering on a local coordinate.

PROOF. Choose a local chart (U, x) near p. Let ω = f IdxI =

fi1,i2,...,ip dxi1 ∧ dxi2 ∧ · · · ∧ dxip . By Lebnitz rule and d2 = 0, we havedω = d f I ∧ dxI , which is unique. Now we still need to check theconsistency in another coordinate system. Let (V, y) be another localchart near p. Write ω into y coordinates.

ω = ∑ fi1,...,ip

∂xi1

∂yα1· · · ∂xip

∂yαpdyα1 ∧ · · · ∧ dyαp = ∑

AfAdyA.

Take the exterior derivative in y coordinates.

dω =∑A

d fA ∧ dyA = ∑A

d

(fi1,...,ip

∂xi1

∂yα1· · · ∂xip

∂yαp

)∧ dyA

=∑A

d fi1,...,ip

(∂xi1

∂yα1· · · ∂xip

∂yαp

)∧ dyA

+ fi1,...,ip d

(∂xi1

∂yα1· · · ∂xip

∂yαp

)∧ dyA

= d f I ∧ dxI + f I d2xI = d f I ∧ dxI .

Recall for any f : M→ N, we have the induced pull-back map

f ∗ : A∗(N) −→ A∗(M)

f ∗(w)(v1, v2, . . . , vp) 7−→ w( f∗(v1), f∗(v2), . . . , f∗(vp))

and f ∗(ω ∧ η) = f ∗(ω)∧ f ∗(η). The operator d is commutative withthe pull-back map.

Theorem 2.5. Given a smooth map f : M → N. f ∗ is a ring homomor-phism and d f ∗(ω) = f ∗(dω).


PROOF.

d f ∗(gdxI) =d(g f ∧i d(xi f )) = d(g f ) ∧(∧i d(xi f )

)= f ∗(dg ∧ dxI) = f ∗

(d(gdxI)

).

4. Lie derivatives on tensors

By extending the Lebnitz rule, we can apply derivatives on usualtensors. For example, along the flow generated by a vector field, wecan take the Lie derivative of a tensor: Let T ∈ Tr,s(M) and X ∈C∞(TM). Let X generate the flow φt.

p

φt(p)

φ∗t

TpM Tφt(p)Mφt∗

φ∗t = φ−t∗

T(φt(p))

T(p)

⊗

⊗

LXT(p) := limt→0

1t(φ∗t T(φt(p))− T(p)

)=

ddt

∣∣∣t=0

φ∗t T(φt(p)).

Exercise 2.2. Check the Lebnitz rule of Lie derivative: LX(T ⊗ S) =LXT ⊗ S + T ⊗ LXS.

By Lebnitz rule, for any tensor T ∈ Tr,s, where

T = ∑ Ti1,...,irj1,...,js (x)

∂

∂xi1⊗· · ·⊗ ∂

∂xir⊗ dxj1⊗· · ·⊗ dxjs = ∑ T I

J∂

∂X I ⊗ dx J ,

we can compute the Lie derivative of T:

LXT = ∑(LXT IJ )

∂

∂xI ⊗ dx J + T IJ (LX

∂

∂xI )⊗ dx J + T IJ

∂

∂xI ⊗ (LXdx J).

5. CARTAN’S HOMOTOPY FORMULA 49

5. Cartan’s homotopy formula

There is an intrinsic way to write down the Lie derivative. Whenconsidering the duality of forms and vectors, (ω, Y) = ω(Y), wehave

Exercise 2.3. LX(ω, Y) = (LXω, Y) + (ω, LXY).

Furthermore, we can derive this formula:

Theorem 2.6 (Cartan’s homopoty formula).

LX = ιXd + dιX.

Here the operator ιX : Ap(M) → Ap−1(M) is the interior product de-fined by

ιXω(v1, . . . , vp−1) = ω(X, v1, . . . , vp−1).

Exercise 2.4. Show that ιX(ω∧ η) = ιX(ω)∧ η +(−1)deg ωω∧ ιX(η).

PROOF. By Lebnitz rule of LX, ι and d, it is enough to prove thecase ω ∈ A1(M). Notice that LX commutes with d:

LXdω =ddt

∣∣∣t=0

φ∗t (dω) = d(

ddt

∣∣∣t=0

φ∗t (ω)

)= dLXω.

Consider the more simple case: ω = f dh.

LX( f dh) = (LX f )dh + f (LXdh)

= X( f )dh + f d(LXh)

= d f (X)dh + f d(dh(X))

= d f (X)dh + d( f dh(X))− d f dh(X).

And

(dιX + ιXd) f dh = d( f dh(X)) + ιXd f ∧ dh

= d( f dh(X)) + d f (X)dh− d f dh(X)

= LX( f dh).

In the last section, we’ve mentioned that there is also an intrinsicexpression for exterior derivative.


Theorem 2.7 (Cartan’s intrinsic formula).

dω(v0, v1, . . . , vp) =p

∑i=0

(−1)ivi(ω(v0, . . . , vi, . . . , vn))

+ ∑0≤i≤j≤p

(−1)i+jw([vi, vj], v0, . . . , vi, . . . , vj, . . . , vp)

where v0, . . . , vp are locally extended vector fields of v0, . . . , vp.

Before proving it, we introduce a useful argument to check whethera quantity is tensorial or not.

Lemma 2.8 (Fundamental Theorem of Tensor Calculus). Let F : C∞(TM)×· · · × C∞(TM) → C∞(TM) be a r-linear map on vector fields. ThatF comes from a tensor (or F is pointwisely defined), if and only if, F isfunctional-linear in each component.

Exercise 2.5. Prove the fundamental theorem of tensor calculus.

PROOF OF THE INTRINSIC FORMULA. Clearly, LHS is a tensor. De-note the RHS as R(v0, . . . , vp). Let g be a smooth function. We checkif R is functional-linear, i.e. R(gv0, . . . , vp) = gR(v0, . . . , vp). Noticethat we can’t take g out as a coefficient directly, because g may bedifferentiated by some vectors.

R(gv0, . . . , vp) =gR(v0, . . . , vp) +p

∑i=1

(−1)ivi(g)ω(v0, . . . , vi, . . . , vn)

−p

∑j=1

(−1)jω((vjg)v0, . . . , vj, . . . , vp)

We can see that vjg is just a scalar and then the last two terms can becancelled. Hence R is a tensor.

Since both LHS and RHS are tensorial, it doesn’t matter how weextend the vector fields, and we may check the formula in any coor-dinate system. Consider a local coordinate (U, x) and v0, . . . , vp = ∂

∂x0 , . . . , ∂∂xp . So [vi, vj] = 0 for each i, j.

5. CARTAN’S HOMOTOPY FORMULA 51

ω ∈ Ap(M). W.L.O.G, we may assume ω = f dx1 ∧ dx2 ∧ · · · ∧dxp. Hence

LHS = dω(v0, . . . , vp)

=n−1

∑i=0

∂

∂xi f dxi ∧ · · · ∧ dxp(v0, . . . , vp)

=∂

∂x0 f dx0 ∧ · · · ∧ dxp(v0, . . . , vp)

= v0 f = RHS.

One of applications of the formula is reformation of Frobeniusintegrability theorem, in so-called differential ideals.

Definition 2.9. An ideal I ⊂ A(M) (ideal under ∧) is called a differ-ential ideal if it is closed under exterior differentiation, i.e. dI ⊂ I .

Let E be a p dimensional distribution. There is a subspace ofdifferential forms that dual to E

E IE := ω ∈ A(M)∣∣ ω∣∣E = 0

which vanish on E and is an ideal.

Proposition 2.10. E is involutive, i.e. [E, E] ⊂ E, if and only if dIE ⊂ IE.

PROOF. The key point of the proof is that IE can be generated by1-forms. Consider on a local chart U and E

∣∣U = f1v1 + · · · + cpvp

for v1, . . . , vp independent vector fields at each q ∈ U. Complete itinto a basis of TU: v1, . . . , vp, vp+1, . . . , vn and take the dual basisθ1, . . . , θp, θp+1, . . . , θn. Then

IE∣∣U = A(U) ∧ θp+1 + A(U) ∧ θp+1 + · · ·+ A(U) ∧ θn

= 〈θp+1, . . . , θn〉.

Let ω ∈ θp+1, . . . , θn. By Cartan’s intrinsic formula,

dω(X, Y) = X(ω(Y))−Y(ω(X)) + ω([X, Y])

= ω([X, Y])


since ω vanishes on E. Hence dω∣∣E = 0 if and only if [X, Y] ∈ E for

every X, Y ∈ E.

We have such a result directly by original Frobenius theorem.

Corollary 2.11. If a differential ideal I is locally generated by 1-forms,then there exists an integral submanifold Mx ∀ x ∈ M, I

∣∣Mx

= 0.

Exercise 2.6. Determine which one of the following is integrable. Ifso, integrate it.

(1) I = 〈θ〉 with θ = (xy− zx2)dx + dy in R3.(2) θ = yzdx + xzdy− dz in R3.(3) θ = dy− zdx in R3.

6. Integration on forms

Let ω = f dx1 ∧ · · · ∧ dxn ∈ Anc (R

n) (with compact support). Wemay identify the integration of the form with the Riemann integral:∫

Rnω :=

∫Rn

f dx1dx2 · · · dxn.

Let y be another coordinate, this is well-defined only if the JacobianJ := det( ∂xi

∂yj ) > 0 by observing from the formula in Calculus∫Rn

f dx1dx2 · · · dxn =∫

Rnf |J| dy1dy2 · · · dyn.

Definition 2.12 (orientable manifold). An atlas is orientable if it is pos-sible to pick an atlas such that det(φ′uv) > 0 for any two coordinateu, v. Such an atlas is also called an orientation.

Note that an orientable manifold always has two orientations.

Fact 2.13. Mn is orientable ⇐⇒ ∃ ω ∈ An(M) such that ω 6= 0 ev-erywhere on M.

PROOF. Let ω be a non-vanishing top form on M. On a localchart (U, x), write ω = f dx1 ∧ · · · ∧ dxn. Since ω 6= 0, f > 0or f < 0 everywhere. If f > 0, we pick (U, x). Otherwise, pick

7. MANIFOLD WITH BOUNDARY AND STOKES’ THEOREM 53

(U, (x2, x1, x3, . . . , xn)). Hence by changing coordinate and use ω 6=0, we can check det(φ′uv) > 0 for every coordinate u, v.

Conversely, if M is oriented by the atlas (U, x) and say UφU→

Rn. Let ωU = φ∗U(dx1 ∧ · · · ∧ dxn) on U and (PU, U)’s be a P.O.U. Setω = ∑U PUωU. Then ω is non-vanishing everywhere.

Such an ω in the fact is called a volume form/maesure/density.Now assume M is oriented with an atlas. Let η ∈ An

c (M). Wecan define the integration when fixing the orientation:∫

Mη := ∑

U

∫U

PUη = ∑U

∫Rn

(φ−1U )∗(PUη).

Since all det φ′uv > 0, the integration is well-defined.

Remark 2.14. The integration is independent of the choices of P.O.U.

PROOF. Given two P.O.U., (ρi, Ui) and (τj, Vj).

∑i

∫Ui

ρiη = ∑i

∫M

ρi(∑j

τjη) = ∑i,j

∫M

τj(ρiη) = ∑j

∫M

τjη.

We want to introduce a similar result as fundamental theorem ofcalculus in Cartan’s differential forms. Recall the Stokes’ theorem:for ω ∈ An−1(M), ∫

Mdω =

∫∂M

ω.

But what is ∂M? We need to develop the notations of manifold withboundary.

7. Manifold with boundary and Stokes’ theorem

Definition 2.15 (manifold with boundary). M is a manifold with bound-ary if M is Hausdorff and of second countable, and there exists anopen cover (U, x) such that

x : U → Rn

or x : U →Hn := (x1, x2, . . . , xn) | xn ≥ 0

is a homeomorphism onto an open set.


And M is called Ck if and only if y x−1 : x(U ∩V) → y(U ∩V)

is Ck for all (U, x), (V, y).

UV

y(V)

x(U)x

y φVU

M

Remark 2.16. If we restrict the chart transition on the boundary,

y x−1∣∣x(U∩V)∩∂Hn

is also a Ck diffeomorphism. Hence ∂M is also a Ck manifold withdim ∂M = n− 1.

If M is oriented, then we can get the induced orientation on ∂M.Let x, y be two coordinates near a point on the boundary.

y x−1(x1, . . . , xn−1, 0) = (y1, . . . , yn−1, 0).

Since M is oriented,

J∣∣∣∂M

= det

∣∣∣∣∣∣∣∣∣∣∣

∂y1

∂x1 · · · ∂y1

∂xn−1 ∗...

... ∗∂yn−1

∂x1 · · · ∂yn−1

∂xn−1 ∗0 · · · 0 ∂yn

∂xn

∣∣∣∣∣∣∣∣∣∣∣> 0

and∂yn

∂xn

∣∣∣∂M

> 0 =⇒ det(y x−1∣∣Hn) > 0.

So ∂M is an oriented manifold.

Theorem 2.17 (Stokes’ theorem). Let M be an oriented manifold withboundary ∂M. Let ω ∈ An−1

c (M). Then∫M

dω = (−1)n−1∫

∂Mω.

7. MANIFOLD WITH BOUNDARY AND STOKES’ THEOREM 55

PROOF. We assume the existence of P.O.U. of manifold with bound-ary. Let (Uα, ρα) be a P.O.U. of M. If this holds for Rn, then∫

Mdω = ∑

α

∫Uα

ραdω

= ∑α

∫Uα

d(ραω)− (dρα)ω

= ∑α

(−1)n∫

∂Uα

ραω−∫

Md(1)ω

= (−1)n−1∫

∂Mω.

It remains to prove this for Rn. There are 2 cases: on Rn and Hn.Write ω = ∑n

i=1 fi(x)dx1 ∧ · · · dxi · · · ∧ dxn. On Rn,∫Rn

dω =∫

Rn ∑i

∂ fi

∂xi (−1)i−1dx1 ∧ · · · ∧ dxn

= ∑i(−1)i−1

∫Rn

∂ fi

∂xi dx1 · · · dxn

(Fubini) = ∑i(−1)i−1

∫Rn−1

(∫R

∂ fi

∂xi dxi)

dx1 · · · dxi · · · dxn

= ∑i(−1)i−1

∫Rn−1

(∂ fi

∂xi

∣∣∣∣∞−∞

)dx1 · · · dxi · · · dxn

= 0 = (−1)n−1∫

∂Rnω

since each fi is of compact support. On Hn,∫Hn

dω = ∑i

∫Hn

∂ fi

∂xi (−1)i−1dx1 ∧ · · · ∧ dxn

= (−1)n−1∫

Rn−1

(fn

∣∣∣∞0

)dx1 · · · dxn−1

= (−1)n−1∫

∂Hnω.

For oriented M, let ∂M be oriented by (−1)n−1dx1 ∧ · · · ∧ dxn−1.Then we can drop the sign in Stokes’ theorem. That’s why we chooseouter normal vector in the divergence theorem for 3-dim cases.


8. De Rham cohomology and the De Rham Theorem

Since d2 = 0, we can easily see the complex structure

· · · −→ Ap−1(M)dp−1−→ Ap(M)

dp−→ Ap+1(M) −→ · · · .

It is naturally to ask whether the quotient space

HpdR(M) =

ker dp

Im dp−1

is non-trivial or not.Definition 2.18. The p-th de Rham cohomology group of M is the quo-tient space of the real vector space of closed p-forms modulo thesubspace of exact p-forms.

α ∈ Ap(M) is a closed p-form if dα = 0 and is an exact p-form ifthere exists a β such that α = dβ.

We also define the compactly supported de Rham cohomologyby

Hpc (M) :=

ker dp∣∣

Apc (M)

Im dp−1∣∣

Ap−1c (M)

.

The non-trivial cohomology groups are important invariants andcan help us to classify the spaces.

Theorem 2.19 (Poincare Lemma). If U is contractible, then HpdR(U) =

0, ∀ p ≥ 1.

Remark 2.20. Let M be connected. For p = 0, f ∈ A0(M) and d f = 0⇒ f is constant. So H0

dR(M) ∼= R; H0c (M) ∼= R if M is compact, ∼= 0

if M is non-compact.The result of Poincare lemma is a kind of homopoty invariance.For f0, f1 : M → N two C∞ maps, we say that f0 ∼ f1 is C∞

homotopic if ∃ F : M× [0, 1] → N which is C∞ such that F(x, 0) =

f0(x) and F(x, 1) = f1(x).

Theorem 2.21 (homotopy invariance).

f ∗0 ≡ f ∗1 : HpdR(N)→ Hp

dR(M).

8. DE RHAM COHOMOLOGY AND THE DE RHAM THEOREM 57

Definition 2.22. f ∗ : HpdR(N)→ HdR(M) is defined by such a diagram

Ap−1(N) //

f ∗

Ap(N) //

f ∗

Ap+1(N)

f ∗

Ap−1(M) // Ap(M) // Ap+1(M)

PROOF OF POINCARE LEMMA. Let M = N = U. Let f0 be theidentity map and f1 be a constant map.

Mid //

Mconstoo

x 7−→ xx 7−→ c

We have id∗ = c∗ = 0 =⇒ HpdR(M) = 0.

PROOF OF HOMOTOPY INVARIANCE. Let dω = 0. Need to showthat f ∗1 ω − f ∗0 ω = dη for some η. Recall the homotopy formula:LX = ιXd + dιX. Let X = ∂

∂t = ∂t.

L∂/∂tF∗ω = ι∂t dF∗ω + dι∂t F∗ω = dι∂t F

∗ω.

Since dF∗ω = F∗dω = 0. Write F∗ω = α + dt ∧ β. Then

L∂t F∗ω = dM×I ι∂t(α + dt ∧ β) = dMβ + dt ∧ ∂β

∂t= ∂tα− dt ∧ ∂tβ.

Therefore ∂tα = dMβ.By fundamental theorem of calculus,

f ∗1 ω− f ∗0 ω = α(1)− α(0) =∫ 1

0∂tαdt =

∫ 1

0dMβdt = dM

(∫ 1

0βdt)

.

Exercise 2.7.

(1) M compact oriented manifold (without boundary) of dimen-sion n. Show that Hn

dR(M) ∼= R.(2) Prove Brouwer fixed point theorem:

(a) @ C∞ map f : Bn → Sn−1 with f∣∣Sn−1 = idSn−1 .

(b) For any smooth g : Bn → Bn, ∃ x ∈ Bn such that g(x) =x.


Let M be a C∞ compact manifold. As in physics, we often connectthe concepts of fields and particles. Now we associate the topologicalsingular cohomology with the de Rham cohomology via an integralmap:

HpdR(M) ∫∼ // Hp(M; R) ≡ Hp(M; R)∗

[ω] //∫

ω (·) : [σ] 7−→∫

σω

where ω ∈ Ap(M) and σ ∈ Sp(M) is a p-chain. The map∫

is well-defined by Stokes’ theorem.

We recall one of the definitions of singular homology.

Definition 2.23. For any topological space M, a singular p-cube is acontinuous map σ : [0, 1]p → M. The singular homology Sp(M) is afree abelian group generated by all singular p-cubes, i.e.

σ ∈ Sp(M), σ =N

∑i=1

aiσi

for σi’s singular p-cubes and ai’s scalars in a commutative ring suchas R, Z.

x1

x2

x3Ip−1(1,1)

Like in de Rham cohomology, we have a boundary map in thesingular homology:

∂p : Sp(M) −→ Sp−1(M)

Ip 7−→ ∂Ip

where

∂Ip =p

∑i=1

1

∑α=0

(−1)α+i Ip−1(i,α)


and Ip−1(i,α) is one of the two faces (α = 0, 1) that perpendicular to the

i-th axis:

Ip−1(i,α) : Ip−1 −→ Ip

(t1, t2, . . . , tp−1) 7−→ (t1, . . . , ti−1, α, ti, tp−1).

Similar to d2 = 0, the boundary map has the same property:Fact 2.24. ∂ ∂ = 0.

It’s clearly to see this holds on a cube [0, 1]p:

∂

∂

Now for a singular p-cube σ,

∂σ =p

∑i=1

1

∑α=0

(−1)i+ασ(i,α)

=p

∑i=1

1

∑α=0

(−1)i+ασ Ip−1(i,α)

= σ ∂Ip.

So ∂2σ = 0.

Exercise 2.8. Check the sign of boundary agree with our previousdefinition of induced orientation on boundary. And also check ∂2 =

0.

Also, we can observe the sequence

· · · // Sp+1(M)∂p+1

// Sp(M)∂p

// Sp−1(M) // · · · .


The homology group is defined as the quotient

Hp(M; Z) =ker ∂p

Im∂p+1.

The definition of cohomology is taking the dual:

S∗p+1(M) S∗p(M)∂∗oo S∗p−1(M)

∂∗oo

where ∂∗ ∂∗ = 0 and

Hp(M; Z) =ker ∂∗

Im∂∗.

We have Hp(M; F) ∼= Hp(M; F)∗.Recall in the Poincare lemma (homotopy invariance):

Mf1 //

f0

// N , f1 ∼ f2 =⇒ f1∗ = f2∗.

Let f : M→ N. This induces a map between the homology group

f# : Sp(M) −→ Sp(N) : σ 7−→ f#(σ) = f σ.

f

σ

f#(σ) = f σ

Ip

M N

By definition, we have ∂ f# = f#(∂) and then f∗ : Hp(M) →Hp(N).


Proposition 2.25. ∃ T : Sp(M) → Sp+1(N) such that f1# − f2# =

±∂T ± T∂.

Sp+1(M)∂ //

f1#f0#

Sp(M)∂ //

f1#f0#

T

yy

Sp−1(M)

f1#f0#

T

yy

Sp+1(N)∂ // Sp(N)

∂ // Sp−1(N)

.

f1#M N

f0#

σ

f1#(σ)

f0#(σ)T∂

∂

T

Construction of T: let F : M × [0, 1] → N, F(x, 0) = f0(x),F(x, 1) = f1(x). Define T(σ)(x) = F(σ(x), t).

Exercise 2.9. Determine the sign in the homotopy formula in T.

Corollary 2.26. Hp(U, Z) = 0 if U is contractable for all p ≥ 1.

Now back to the de Rham map. The deRham map is well-defined,i.e.

∫(ω) is independent of the choice of ω in [ω] ∈ Hp

dR(M). Weneed to explain the definition of integration over chains: for σ : Ip →M, ∫

σω :=

∫Ip

σ∗ω


which is defined by the Riemann integral. Then extend the inte-gration linearly to all combinations σ = ∑N

i=1 aiσi. Choose anotherω + dη ∈ Hp

dR(M). This well-definition comes from the Stokes’ the-orem:

∫σ dη =

∫∂σ η = 0.

Lemma 2.27 (Mayer-Vietoris sequence). U, V open sets in M, then ∃an exact sequence

0 // Ap(U ∪V) // Ap(U)⊕ Ap(V) // // Ap(U ∩V) // 0

α // (α∣∣U, α

∣∣V)

(ξ, η) // ξ∣∣U∩V − η

∣∣U∩V

.

Why exact? Choose a P.O.U. with φU, φV for U, V such thatsuppφU ⊂ U, suppφV ⊂ V, φU + φV = 1 on U ∩ V. Let (ξ, η) =

(φVγ,−φUγ). Then φVγ∣∣U∩V − (−φUγ)

∣∣U∩V = γ.

Fact 2.28. Both of i, j commute with d.

So there exists a long exact sequence:

HpdR(U ∪V)

i∗ // HpdR(U)⊕ Hp

dR(V)j∗ // Hp

dR(U ∩V)δ // Hp+1

dR (U ∪V)

γ // d(φV) ∧ γ

.

(δ is called the connecting map)

(φVγ,−φUγ)U

γoo

0 // Ap(U ∪V) //

d

Ap(U)⊕ Ap(V) //

d

Ap(U ∩V)

d

0 // Ap+1(U ∪V) // Ap+1(U)⊕ Ap+1(V) // Ap(U ∩V)

(dφV) ∧ γ (d(φVγ),−d(φUγ))oo

9. EXERCISES 63

Hp−1(U ∪V) //

∫

Hp(U ∪V) //

∫

Hp(U)⊕ Hp(V) //

∫

Hp(U ∩V) //

∫

Hp+1(U ∪V)∫

Hp−1dR (U ∪V)

δ // HpdR(U ∪V)

i∗ // HpdR(U)⊕ Hp

dR(V)j∗ // Hp

dR(U ∩V)δ // Hp+1

dR (U ∪V)

Exercise 2.10. Prove that the Mayer-Vietoris sequence is an exactsequence and check all diagrams commute.

PROOF OF DE RHAM THEOREM. It is true for 1 ball U. By induc-tion on the number of convex open cover. If it is true for U, V, whyis it true for U ∩V.

Fact 2.29. On a Riemannian manifold (M, g), there exists convexneighborhoods.

U1, . . . , Un, Un+1 convex =⇒ (U1 ∪ · · · ∪ Un) ∩ Un+1 = (U1 ∩Un+1) ∪ · · · ∪ (Un ∩Un+1).

9. Exercises

1. Find the polarization formula of T ∈ Symk(V∗) for general k.2. Check the Lebnitz rule of Lie derivative: LX(T ⊗ S) = LXT ⊗ S +

T ⊗ LXS.3. LX(ω, Y) = (LXω, Y) + (ω, LXY).4. Show that ιX(ω ∧ η) = ιX(ω) ∧ η + (−1)deg ωω ∧ ιX(η).5. Prove the fundamental theorem of tensor calculus.6. Determine which one of the following is integrable. If so, integrate

it.(a) I = 〈θ〉 with θ = (xy− zx2)dx + dy in R3.(b) θ = yzdx + xzdy− dz in R3.(c) θ = dy− zdx in R3.

7. (a) M compact oriented manifold (without boundary) of dimen-sion n. Show that Hn

dR(M) ∼= R.(b) Prove Brouwer fixed point theorem:

(i) @ C∞ map f : Bn → Sn−1 with f∣∣Sn−1 = idSn−1 .

(ii) For any smooth g : Bn → Bn, ∃ x ∈ Bn such that g(x) =x.


8. Check the sign of boundary agree with our previous definition ofinduced orientation on boundary. And also check ∂2 = 0.

9. Determine the sign in the homotopy formula in T.10. Prove that the Mayer-Vietoris sequence is an exact sequence and

check all diagrams commute.11. (Warner Ch.2 #2)

(a) Show that homogeneous tensor are generally not decompos-able.

(b) Show that if dim V ≤ 3, then every homogeneous element inΛ(V) is decomposable.

(c) Let dim V > 3. Give an example of an indecomposable ho-mogeneous element of Λ(V).

(d) Let α be a differential form. Is α ∧ α ≡ 0?12. (Warner Ch.2 #4) Derive the following three formulae.

f ∧α g(v1, . . . , vp+q)

= ∑p,q shuffles

(sgn π) f (vπ(1), . . . , vπ(p))g(vπ(p+1), . . . , vπ(p+q)),

f ∧β g(v1, . . . , vp+q)

=1

(p + q)! ∑π∈Sp+q

(sgn π) f (vπ(1), . . . , vπ(p))g(vπ(p+1), . . . , vπ(p+q)),

f ∧α g =(p + q)!

p!q!f ∧β g.

Here a permutation π ∈ Sp+q is called a “p, q shuffle” if π(1) <

π(2) < · · · < π(p) and π(p + 1) < · · · < π(p + q).13. (Warner Ch.2 #13) Let V be an n-dimensional real inner product

space. We extend the inner product from V to all of Λ(V) by set-ting the inner product of elements which are homogeneous of dif-ferent degrees equal to zero, and by setting

〈w1 ∧ · · · ∧ wp, v1 ∧ · · · ∧ vq〉 = det〈wi, vj〉

and then extending bilinearly to all of Λp(V). Prove that if e1, . . . , en

is an orthonormal basis of V, then the corresponding basis

eΦ = ei1 ∧ ei2 ∧ · · · ∧ eir , i1 < i2 < · · · < ir

9. EXERCISES 65

where Φ = i1, · · · , ir runs over all subsets of 1, . . . , n, is anorthonormal basis for Λ(V).

Since Λn(V) is one-dimensional, Λn(V) \ 0 has two compo-nents. An orientation on V is a choice of a component of Λn(V) \0. If V is an oriented inner product space, then there is a lineartransformation

∗ : Λ(V)→ Λ(V),

called star, which is well-defined by the requirement that for anyorthonormal basis e1, . . . , en of V (in particular, for any re-orderingof a given basis),

∗ (1) = ±e1 ∧ · · · ∧ en, ∗(e1 ∧ · · · ∧ en) = ±1,

∗ (e1 ∧ · · · ∧ ep) = ±ep+1 ∧ · · · ∧ en,

where one takes “+” if e1∧ · · · ∧ en lies in the component of Λn(V) \0 determined by the orientation and “−” otherwise. Observethat

∗ : Λp(V)→ Λn−p(V).

Prove that on Λp(V),

∗∗ = (−1)p(n−p).

Also prove that for arbitrary v, w ∈ Λp(V), their inner product isgiven by

〈v, w〉 = ∗(w ∧ ∗v) = ∗(v ∧ ∗w).

14. (Warner Ch.2 #7) Let w ∈ Ap(M) and let X, Y0, . . . , Yp be C∞ vec-tor fields on M. Show that

LY0

(w(Y1, . . . , Yp)

)=(LY0w)(Y1, . . . , Yp)

+p

∑i=1

w(Y1, . . . Yi−1, LY0Yi, Yi+1, . . . , Yp)

and use the case p = 1 to derive

LX = ι(X) d + d ι(X)

on A∗(M) for p = 1, and then for all p.


15. (Warner Ch.2 #11) Let I be an ideal of forms on M locally gener-ated by r independent 1-forms. Say I is generated by w1, . . . , wr

on U. Then the condition that I be a differential ideal is equiva-lent to each of(a) dwi = ∑j wij∧wj for some 1-forms wij (for each such (U, w1, . . . , wr)).(b) If w = w1 ∧ · · · ∧ wr, then dw = α ∧ w for some 1-forms α (for

each such (U, w1, . . . , wr)).16. (Warner Ch.4 #1) A d-dimensional manifold X for which there ex-

ists an immersion f : X → Rd+1 is orientable if and only if thereis a smooth nowhere-vanishing normal vector field along (X, f ).

17. (Warner Ch.4 #2) Prove that the real projective space Pn is ori-entable if and only if n is odd.(Hint: Observe that the antipodal map on the n-sphere Sn is orientation-preserving if and only if n is odd.)

18. (Warner Ch.4 #3) Carry out in detail the proof of the existenceof local orthonormal frame fields on a Riemannian manifold (cf.Exercise 1.28 and the next chapter).

19. (Warner Ch.4 #6) Let w be the volume form of an oriented Rie-mannian manifold of dimension n. Let X1, . . . , Xn and Y1, . . . , Yn

be vector fields on M. Prove that

w(X1, . . . , Xn) · w(Y1, . . . , Yn) = det〈Xi, Yj〉.

Prove also that

w(X1, . . . , Xn)w = X1 ∧ · · · ∧ Xn,

where Xi is the 1-form dual (via the Riemannian structure) to thevector field Xi.

20. (Warner Ch.4 #17) Using de Rham cohomology, prove that thetorus T2 is not diffeomorphic with the 2-sphere S2.

21. (Warner Ch.4 #18)(a) Prove that every closed 1-form in the open shell

1 <

(3

∑i=1

r2i

)1/2

< 2

in R3 is exact.

9. EXERCISES 67

(b) Find a 2-form in the above shell that is closed but not exact.(c) Prove that the above shell is not diffeomorphic with the open

unit ball in R3.22. (a) M compact orientable manifold of dimension n, show that

HndR(M) ∼= R.

(b) Prove that Brouwer fixed point theorem:(i) @ C∞ map f : Bn → Sn−1 with f

∣∣Sn−1 = idSn−1

(ii) For any C∞ map g : Bn → Bn, ∃ x such that g(x) = x.23. (Warnar ch.4 #4) Prove the divergence theorem. First assume M is

oriented and use Stokes’ theorem together with the identity∫∂D∗V =

∫∂D〈V,~n〉.

The easiest way to see the identity is to choose a local orientedorthonormal frame field e1, . . . , en on a neighborhood of a point of∂D, such that at points of ∂D, e1 is the outer unit normal vectorand e2, . . . , en form an oriented basis of the tangent space to ∂D.Then express ∗V and 〈V,~n〉 in terms of this local frame field andits dual coframe field w1, . . . , wn. Finally, show that the theoremholds for a regular domain D in a Riemannian manifold M whichis not necessarily orientable.

24. (Warnar ch.4 #16)(a) Prove that every closed 1-form on S2 is exact.(b) Let

σ =r1dr2 ∧ dr3 − r2dr1 ∧ dr3 + r3dr1 ∧ dr2

(r21 + r2

2 + r23)

3/2

in R3 \ 0. Prove that σ is closed.(c) Evaluated

∫S2 σ. How does this show that σ is not exact?

(d) Let

α =r1dr1 + r2dr2 + · · ·+ rndrn

(r21 + r2

2 + · · ·+ r2n)

n/2

in Rn \ 0. Find ∗α and prove that ∗α is closed.(e) Evaluate

∫Sn−1 ∗α. Is ∗α exact?

Chapter 3

RIEMANNIAN MANIFOLDS

1. Riemannian structure

Definition 3.1. (M, g) is a Riemannian manifold if M is a C∞ manifoldand g ∈ C∞(M, Sym2(T∗M)), where gp is positive definite for everyp ∈ M.

Such a g is called a Riemannian metrix. Locally we can write g =

∑i,j gijdxi ⊗ dxj, gij = g( ∂∂xi ,

∂∂xj ).

For any given differentiable manifold M, we can define a metricthrough a P.O.U. Uα, ρα: For M = ∪αUα, we can put arbitrarymetric gα on Uα (e.g. the Euclidean inner product). Then define g =

∑α ραgα. Since each gα is a metric, it is easily to see that g is also ametric.

On the other hand, a major question in Riemannian geometry isthe isometric embedding problem: For any (M, g), doese there exists anembedding f such that f : M → RN where g = f ∗(gRN

) for someN?

This result is proved by Nash after 1954 (1954: C1 embedding;1956: Ck embedding). But it will not be mentioned here.

Now with an inner product, we want to define the concept ofvolume on manifolds: ∫

Mf :=

∫M

f ω

where ω is a top form which coincide with the metric. On TpM, lete1, e2, . . . , en be an orthonormal basis. Then we can choose a C∞ dualorthonormal frame (1-form) on TM

∣∣U, φn, . . . , φn. ω can be simply

chosen as φ1 ∧ · · · ∧ φn.

69

70 3. RIEMANNIAN MANIFOLDS

Exercise 3.1. On a local coordinate chart, if ω = hdx1 ∧ · · · ∧ dxn,show that h =

√det(gij). In general, show that v1, . . . , vk ⊂ Rn, the

k-dim’l volume of

[v1, . . . , vk] =

k

∑i=1

tivi

∣∣∣ 0 ≤ ti ≤ 1

is√

det(VtV) where

V =

| | · · · |v1 v2 · · · vk

| | · · · |

.

On an inner product space (V, 〈 , 〉), we can define the Hodgeoperator: for e1, . . . , en O.N.B.,

∗ : Λp(V) // Λn−p(V)

e1 ∧ · · · ∧ en // ep+1 ∧ · · · ∧ en

.

We require that for α = ei1 ∧ · · · ∧ eip ∈ Λp(V), ∗α ∈ Λn−p(V) withα ∧ ∗α = e1 ∧ · · · en, then extend it linearly.

For general p-vectors α, β ∈ Λp(V), α = ∑|I|=p aIei1 ∧ · · · ∧ eip ,β = ∑|I|=p bIei1 ∧ · · · ∧ eip ,

α ∧ (∗β) = ∑|I|,|J|=p

aIbJ(ei1 ∧ · · · ∧ eip) ∧ ∗(ej1 ∧ · · · ejp)

= 〈α, β〉e1 ∧ · · · ∧ en

where 〈α, β〉 is the usual induced inner product on Λp(V).

Exercise 3.2. Check that ∗2 = (−1)p(n−p).

Recall in the Exercise 3.1, for v = [v1 · · · vp], w = [w1 · · ·wp], wecan also define the inner product 〈v, w〉 := det(wtv).

Define ∗ : Ap(M) → An−p(M) pointwisely on T∗p (M). Thestokes’ theorem can be associated with the usual divergence theo-rem in R3 via this operator.∫

ΩdivF dV =

∫∂Ω

F · n dA.

1. RIEMANNIAN STRUCTURE 71

F = (P, Q, R) vector field in R3. We can write the formula into dif-ferential forms.∫

ΩdivF dV =

∫Ω(Px + Qy + Rz)dx ∧ dy ∧ dz.∫

∂ΩF · n dA =

∫∂Ω

Pdy ∧ dz + Qdz ∧ dx + Rdx ∧ dy.

n

The divergence theorem follows from the general Stokes’ theo-rem.

For general (M, g), the vector fields are associated with the 1-forms by

V ∼ // V∗

v // v : v(w) := 〈v, w〉φ φ ∈ V∗oo

(We may also write φ as φ# or φ[). Note that one of the advantage forus to considering forms is: it is well-defined to add or wedge formsof any degree but we can’t do the same thing on vector fields.

Similarly, for F = P ∂∂x + Q ∂

∂y + R ∂∂z , F = Pdx + Qdy + Rdz and

grad f = d f ,

divF = ∗d ∗ F,

curlF = dF.

As an example for higher dimension, in R3+1, the Maxwell equa-tion can be written as divF = 0

curlF = 0.

By the way, Λ2(R4) ∼= C6.


Exercise 3.3. Verify the original Stokes theorem in R3∫Ω∇× F · n dA =

∫∂Ω

F · dr.

2. Covariant Differentiation and Levi-Civita Connection

After introducing the metric structure on a manifold, we mayfurther consider how to do differentiation on it.

For example, on Rn, let F be a vector-valued function and v a unitvector. The directional derivative at some point p is defined by

DvF := limt→0

1t(F(p + tv)− F(p)) .

Also recall in chapter 1, we define the Lie derivative: whose differ-ential is given by a flow generated by the vector field v.

So, what we really want is the notion of “parallel translation”. Theconcepts of parallel translation is developed by Christoffel, Levi-Civita, Ricii, Koszul, . . . for many years.

Anyway, no matter how we translate the vectors, the differentia-tion should follow linearity and Lebnitz rule at least.

Definition 3.2 (covariant differentiation). A covariant differentiation isan operator

∇X : C∞(TM)→ TpM, ∀ X ∈ TpM, p ∈ M

such that

(1) it is R-linear,(2) satisfies Lebnitz rule: ∇X( f Y) = X( f )Y + f∇XY,(3) and it is functional linear in X.

In other words,

∇ : C∞(TM)× C∞(TM) // C∞(TM)

(X, Y) // ∇XY

which is functional linear in X. Or equivalently, ∇ : C∞(TM) →A1(TM) = A1(M)⊗ C∞(TM).

Example 3.3. (1) M = Rn, ∇ = D.

2. COVARIANT DIFFERENTIATION AND LEVI-CIVITA CONNECTION 73

(2) Let M → Rn be a hypersurface and ∇ = D. ∇XY :=(DXY)T, the tangential part to TpM of the usual derivativein Rn.

M

p

TpM(DXY)T

(DXY)N

X

DXY

On a manifold, a covariant differentiation can be constructed bypatching the induced local constructions. Let Mm = ∪αUα and (Uα, φα)

be a P.O.U., Uα → Rm.Given ∇α any covariant differentiation in Uα. Is ∇ := ∑α fα∇α

a covariant differentiation operator? For an easier case, ∇ = h∇.∇X( f Y) = h(∇X f Y) = hX( f )Y + h f∇XY = hX( f )Y + f ∇XY. Leb-nitz rule implies that h must be constant 1.

For ∇ = ∑ fα∇α,

∇X( f Y) = ∑α

fαX( f )Y + f ∇XY = X( f )Y + f ∇XY

provided by ∑α fα = 1. So ∇ is a covariant differentiation if and onlyif it is an affine linear combination, i.e. ∑α fα = 1.

In general, we call ∇ an affine structure on M. And in particular,for a P.O.U., ∇ = ∑α ϕα∇α is definitely a covariant differentiationoperator.

The definition of covariant differentiations is independent of Rie-mannian structure. Now we want to introduce a covariant differen-tiation which is compatible with the Riemannian metric. It is calleda Levi-Civita connection.

Levi-Civita observed that


(1) in case of Mm → Rn,

X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉 = 〈DXY, Z〉+ 〈Y, DXZ〉.

Notice that Z ∈ TpM, 〈(DXY)N, Z〉 = 0.(2) Also, the torsion-free condition:

∇XY−∇YX = (XY−YX)T = [X, Y]T = [X, Y].

i.e., ∇XY−∇YX− [X, Y] = 0.

Exercise 3.4. M → Rn. Show the induced connection is torsion-free.

Theorem 3.4 (Fundamental Theorem of Riemannian Geometry (Lev-i-Civita)). Given (M, g), ∃ ∇ = ∇L.C. which is metrical (compatiblewith g) and torsion-free.

PROOF. We assume the existence first. Given any 3 vector fieldsX, Y, Z. By compatibility and torsion-free condition,

X〈Y, Z〉 = 〈∇XY, Z〉+ 〈Y,∇XZ〉,Y〈X, Z〉 = 〈∇YZ, X〉+ 〈Z,∇YX〉

= 〈∇YZ, X〉+ 〈Z,∇XY + [Y, X]〉Z〈X, Y〉 = 〈∇ZX, Y〉+ 〈X,∇ZY〉

= 〈∇XZ + [Z, X], Y〉+ 〈X,∇YZ + [Z, Y]〉

⇒ 〈∇XY, Z〉 =12

(X〈Y, Z〉+ Y〈Z, X〉 − Z〈X, Y〉

+ 〈Z, [X, Y]〉+ 〈Y, [Z, X]〉 − 〈X, [Y, Z]〉)

.

Then get uniqueness.

Exercise 3.5. Prove the existence of Levi-Civita connection by check-ing this really defines a covariant differentiation.

In local coordinate (x1, . . . , xm), we define the Christoffel symbolΓk

ij. Let ∂i := ∂∂xi and

∇∂i ∂j =:m

∑k=1

Γkij∂k.

2. COVARIANT DIFFERENTIATION AND LEVI-CIVITA CONNECTION 75

So

∇∂i( f j∂j) = (∂i f j)∂j + f j ∑k

Γkij∂k.

We only need to calculate Γkij to know the differentiation. Denote

〈∂i, ∂j〉 = gij. By using the formula in Levi-Civita’s theorem,

∑`

Γ`ijg`k =

12(∂igjk + ∂jgki − ∂kgij

).

We use the convention (gij) = (gij)−1, i.e. ∑q gpqgqr = δ

pr . So we

have the formula for Christoffel symbol:

(3.1) Γkij =

12

gk`(∂igjk + ∂jgki − ∂kgij).

We’ve talked about that the way we take derivatives depends onparallel translations.

γ(t)

F(t)

M

Now given a C∞ curve γ : [0, 1] → M.γ′(t) = dγ

(∂∂t)≡ ∂

∂t = ∂t. Let F bea C∞ vector field along γ. Write DF

dt =

∇∂t F = ∇tF as the covariant differentia-tion along γ.

Definition 3.5. We say that F is parallel along γ if ∇∂t F = 0.

Let F = ∑ f i∂i, γ(r) = (x1(t), . . . , xn(t)).

∇∂t F =∂ f i

∂t∂i + f i∇∂t∂i

=∂ f i

∂t∂i + f i xjΓk

ij∂k

=(∂ f i

∂t+ f k xjΓi

kj)∂i.

This is a first order ODE in t. By existence and uniqueness theoremof ODE, we can show that for an affine structure, the parallel trans-lation always exists and determined by the initial vector.

Especially, for ∇ = ∇L.C., parallel translation preserves innerproducts. Let V1, V2 be two parallel vector field along γ.

γ〈V1, V2〉 = 〈∇γV1, V2〉+ 〈V1,∇γV2〉 = 0.


3. Geodesic, Exponential Map and Riemann Normal Coordinate

Definition 3.6. γ is a geodesic if ∇γγ = 0.

Example 3.7. For Mm → Rn,

∇γγ = (Dγγ)T = (D∂t γ)T = r′′(t)T.

A geodesic is by definition a curve parametrized by a parametert which is proportional to its arclength. Let γ be a curve on (M, g).If ∇γγ = 0,

γ〈γ, γ〉 = 2〈∇γγ, γ〉 = 0.

So |γ| is constant. And the arclength s(t) =∫ t

a |γ|dt. In particular,when γ is parametrized by arclength s, |γ′(t)| = 1 ∀ t.

Similarly, we can write down the geodesic equation in local coor-dinates. γ = xi∂k.

This is a second order, nonlinear ODE.

∇γγ = ∑k

(xk + ∑

i,jΓk

ij xi xj)

∂k = 0.

Still by the existence and uniqueness theorem, γ exists and is uniquelydetermined by γ(t0) and γ′(t0) in a local neighborhood.

The local existence ensures us to find the geodesics in any direc-tion.Definition 3.8 (exponential map). For p ∈ M, the exponential mapexpp : TpM → M is defined by expp(tv) = γ(t), where γ is a ge-odesic satisfying γ(0) = p, γ′(0) = v.

pv

expp(v)M

TpM

By existence of ODE, the exponential map expp is well-defined onsome neighborhood U 3 0. But it may not be extended to infinity:Example 3.9. Consider M = R2 \ 0. The geodesics are lines on theR2 plane. It is clearly that the geodesics can’t across the origin.

3. GEODESIC, EXPONENTIAL MAP AND RIEMANN NORMAL COORDINATE 77

R2

p

vO

Fact 3.10. expp is a local diffeomorphism near 0 ∈ TpM The smooth-ness follows from smooth dependence of ODE on initial value.

By inverse function theorem, we only need to compute

d(expp)0 : T0(TpM)(∼= TpM)→ TpM.

By direct calculation,

d(expp)0(v) :=ddt

∣∣∣t=0

expp(tv) = v.

Hence d(expp)0 = id.An easy application of the exponential map is Riemainnian normal

coordinate. Let e1, . . . , em be an orthonormal basis of TpM. In theexistence neighborhood of exponential map, we can define the newcoordinate function

(x1, x2, . . . , xm) 7→ expp(x1e1 + x2e2 + · · ·+ xmem).

M

TpMp e1

e2

Why is this good?

Fact 3.11. Under this coordinate, Γkij(p) = 0 ∀ i, j, k.

PROOF. Fix h = (h1, h2, . . . , hm) . Let x(t) = (x1(t), x2(t), . . . , xm(t)) =(th1, th2, . . . , thm). Consider γ(t) = expp(th). Recall the geodesic


equation:

xk(t) + ∑i,j

Γkij x

i(t)xj(t) = 0.

Since x(t) is linear in t, ∑i,j Γkij x

i(t)xj(t) = 0. Let t = 0. Then we have

∑i,j

Γki,j(p)hihj = 0 ∀ k.

Since this is true for arbitrary h in existence neighborhood, we con-clude that Γk

i,j(p) = 0.

Here’s an application of the normal coordinates. We can showthe local minimality of geodesics.

Consider the polar coordinate on the tangent space: (r, θ1, . . . , θm−1).Denote grr = g(∂r, ∂r), grα = g(∂r, ∂θα), gαβ = g(∂θα , ∂θβ).

Let p, q be two points connected by a geodesic γ. Suppose p, q arejointed by another curve Γ.

p

v

M

TpMB

exp(B)q

gG

exp

Lemma 3.12 (Gauss lemma). grα = 0 ∀ α.

Hence

(ds)2 = (dr)2 + ∑α,β

gαβdθα ⊗ dθβ.


Then the arc length of Γ

|Γ| =∫

ds =∫ √√√√(dr

dt

)2

+ ∑α,β

gαβdθα

dt⊗ dθβ

dtdt

≥∫ ∣∣∣∣dr

dt

∣∣∣∣ dt = `

since g is a positive definite symmetric tensor.

PROOF OF GAUSS LEMMA. We prove this form:

〈d(expp)v(v), d(expp)v(w)〉q = 〈v, w〉p.

for w ∈ Tv(TpM) ∼= TpM.

p

v

M

TpM

B

qγ

w

d(expp)v(v)

d(expp)v(w)TqM

Let w = w1 + w2 where w1 ⊥ v, w2 = λv. Then for the w2 part,

〈d(expp)vv, d(expp)vw2〉 = λ〈d(expp)vv, d(expp)vv〉 = λ〈v, v〉.

This is followed by definition,

d(expp)vv =ddt∣∣t=0 expp(v + tv) =

ddt∣∣t=0γ(t + 1) = γ′(1)

and since |γ′(t)| is constant.For the w1 part, we need to show

〈d(expp)vv, d(expp)vw1〉=0 0 = λ〈v, w1〉.

Consider a parametrized surface in M, X(t, s) := expp(tv(s)), wherev(0) = v, v′(0) = w1 and |v(s)| is constant.


γ

p

M

TpM

qw2

w1vv(s)

Xt =∂X

∂t= X∗

( ∂

∂t

), Xs =

∂X

∂s= X∗

( ∂

∂s

).

Then

〈d(expp)vv, d(expp)vw1〉q = 〈Xt, Xs〉(1, 0).

By definition of geodesic,

∂

∂t〈Xt, Xs〉 = 〈∇∂tXt, Xs〉+ 〈Xt,∇∂tXs〉

= 〈0, Xs〉+ 〈Xt,∇∂sXt + [Xt, Xs]

= 〈Xt,∇∂sXt〉 =12

∂s〈Xt, Xt〉 = 0.

since 〈Xt, Xt〉 = |v(s)|2 is constant.

Exercise 3.6. Prove Gauss lemma using the fact Γαrr = 0 and then

∂rgrα = 0 ∀ α.

At the end of this section, recall in de Rham theorem, we ap-ply the existence of convex neighborhood in Fact 2.29. Now we canprove this fact.

Bp(δ)

pq

q′? Given p ∈ M. Fix δ > 0 such that

∀ q ∈ Bp(δ), expq is a diffeomor-phism on some Bq(rq). Simply letr0 = min

q∈Bp(δ0)rq > 0.

As in the picture, any q, q′ may not be jointed by a geodesic insidethe neighborhood.


Now we see (M, g) as a metric space (M, d) where d(p, q) =

infγ joins p,q L(γ). Reset δ = r0/4. Let W := Bp(r)/4). Now forall q ∈ W, expq is a diffeomorphism on W since d(q, q′) ≤ d(q, p) +d(p, q′) < r0/2 for all q, q′ ∈ W. So there at least exists a minimalgeodesic γ connecting q and q′ with γ ⊂ Bq(r0/2) ⊂ Bp(r0).

Want to prove γ ⊂W = Bp(r0/4).

r(t)γ(t)

p

qq′

Recall the geodesic equation: xk(t) + Γkij x

i(t)xj(t) = 0. γ(t) =

(x1(t), . . . , xm(t)). For the second order equation, we will apply themaximum principle. Define the distance function r(t) := d(p, γ(t)).

r2 =m

∑k=1

xkxk, rr′ =m

∑k=1

xk xk.

(r′)2 + rr′′ = ∑k

xkxk + (xk)2

= −∑k

Γkij x

i xjxk + ∑k(xk)2.

Suppose r reach the max value at t = t0. If γ(t0) = q or q′, the done.If not, we have r′(t0) = 0, r′′(t0) < 0.

Hence LHS ≤ 0 and when r0 is chosen small enough, RHS > 0leads a contradiction.


4. Riemann Curvature Tensor

R(X, Y) detects the non-commutativity of covariant differentia-tion, that is:

R(X, Y)Z := ∇X∇YZ−∇Y∇XZ−∇[X,Y]Z

≡([∇X,∇Y]−∇[X,Y]

)Z.

for X, Y, Z ∈ TpM and X, Y, Z any vector fields that extend X, Y, Z.The last term in definition is a correction to make it functional linear.We can see that R is a (1, 3) tensor. With local coordinates, say

R(∂i, ∂j)∂` =: ∑ Rkìj∂k.

In convenience, we drop all summation symbols. By computations,

R(∂i, ∂j)∂` = ∇i∇j∂` −∇j∇i∂` −∇[i,j]∂`

= ∇i(Γsj`∂s)−∇j(Γt

i`∂t)

= (∂iΓkj`)∂k + Γs

j`Γkis∂k − (∂jΓk

i`)∂k − Γti`Γ

kjt∂k

= (∂iΓkj` − ∂jΓk

i` + Γsj`Γ

kis − Γs

i`Γkjs)∂k.

Also, we define

Rkìj := gkmRmìj = 〈R(∂i, ∂j)∂`, ∂k〉.

The convention of tensor contraction is very common to see. Forexample, we apply it to covariant derivatives.

Example 3.13. Let Z = f p∂p be a vector field. We denote

∇i( f p∂p) =: f p;i ∂p, ∇i∇j( f p∂p) =: f p

;ji∂p.

So

( f p;ji − f p

;ij)∂p = R(∂i, ∂j) f p∂p = f qRpqij∂p,

and in other words, we can change the order of covariant differenti-ation via the commutation formula

f p;ij = f p

;ij + f qRpqij.

4. RIEMANN CURVATURE TENSOR 83

Exercise 3.7. Extend the covariant derivative to all tensors. Andprove the commutation formula:

Ti1···irj1···js;ji = Ti1···ir

j1···js;ij + Tqi2···irj1···js Ri1

qij + Ti1qi3···irj1···js Ri2

qji + · · ·

+ Ti1···irpj2···js Rp

j1 ji + · · ·+ Ti1···irj1···js−1 p Rp

js ji.

There are many symmetries on Riemann’s curvature tensor. Clearlyby definition, we have Rk

ìj = −Rk`ji since R(X, Y) = −R(Y, X).

Proposition 3.14 (Symmetries on R).

(0) Rkìj = −Rk

`ji.(1) Torsion free: the first Bianchi identity.

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

i.e. Rk[ìj] := Rk

ìj + Rkij` + Rk

jì = 0.(2) Metrical: skew symmetry in Rkìj = −R`kij.

PROOF. For (1), since these are all tensors, we may check this inany basis:

R(∂i, ∂j)∂k + R(∂j, ∂k)∂i + R(∂k, ∂i)∂j

= ∇i∇j∂k −∇j∇i∂k +∇j∇k∂i −∇k∇j∂i +∇k∇i∂j −∇i∇k∂j

= 0

since ∇i∂j −∇j∂i = [∂i, ∂j] = 0.For (2), that is, we want to show: 〈R(X, Y)Z, W〉+ 〈Z, R(X, Y)W〉 =

0. Let X = ∂i, Y = ∂j.

〈∇i∇jZ, W〉 − 〈∇j∇iZ, W〉= ∂i〈∂jZ, W〉 − 〈∇jZ,∇iW〉 − ∂j〈∇iZ, W〉+ 〈∇iZ,∇jW〉= ∂i∂j〈Z, W〉 − ∂i〈Z,∇jW〉 − 〈∇jZ,∇iW〉− ∂j∂i〈Z, W〉+ ∂j〈Z,∇iW〉+ 〈∇iZ,∇jW〉.

Now by exchanging Z and W, we’ll get a (−1) sign.

Exercise 3.8. From these symmetries, we can get some further re-sults:


(1) The symmetries (3.14) implies Rk`ij = Rijk`.(2) Second Bianchi identity:

0 = Rij[k`;m] := Rijk`;m + Rij`m;k + Rijmk;`.

Furthermore, we introduce the Ricci curvature tensor:

Ric := Rijdxi ⊗ dxj ∈ Sym2(T∗), Rij = ∑`

R`i`j = ∑ gk`Rki`j,

and the scalar curvature:

S := gijRij.

These quantities are related to Einstein field equation.

Rij −S2

gij = Ti j

where Ti j is called the energy momentum tensor in space.When the space is vacuum, we take Tij ≡ 0. In this case,

Rij −S2

gij = 0 ⇒ gij(Rij −S2

gij) = S− S2

m⇒ S, Rij ≡ 0.

For uniform case, Tij = Mgij for some constant M > 0 and we’llhave the equation

Rij = λgij.

Exercise 3.9. For n ≥ 3, if Rij = λgij, then λ is constant. (hint: useBianchi identity.)

5. Variation of Geodesics

Let γ : [a, b] → M be (piecewise) C∞. We consider the one pa-rameter variation:

F : [a, b]× (−ε, ε)→ M.

Let T = F∗ ∂∂t be the tangent vector field and V = F∗ ∂

∂s be the varia-tion field for Fs(t) := F(s, t). F0(t) = γ. We assume the vector fieldsare F-related here, that is, [V, T] = [F∗ ∂

∂s , F∗ ∂∂t ] = F∗[ ∂

∂s , ∂∂t ]. (If not, it

still work. May refer the book by Cheeger and Ebin.)

5. VARIATION OF GEODESICS 85

M

t

sε ∂s

∂t

γ

−ε

Fs

The length of Fs:

L(s) =∫ b

a〈T, T〉1/2(t, s) dt.

In this section, we can establish the minimality of arclength of geodesicsfrom the variation method.

L′(s) =∂

∂s

∫ b

a〈T, T〉1/2 dt =

∫ b

aV〈T, T〉1/2 dt

=∫ b

a

2〈∇VT, T〉2〈T, T〉1/2 dt =

∫ b

a

〈∇TV, T〉〈T, T〉1/2 dt (torsion free)

=∫ b

a

T〈V, T〉 −V〈∇TT〉〈T, T〉1/2 dt.

May assume γ is parametrized by its arclength, i.e. 〈T, T〉(t, 0) = 1.Then

(3.2) L′(0) = 〈V, T〉∣∣∣ba−∫ b

a〈V,∇TT〉dt.

This is called the first variation formula.

Fact 3.15. γ is a “critical point” for any variation F such that F(a, s) =P, F(b, s) = Q with P, Q ∈ M fixed if and only if γ is a geodesic, i.e.∇TT = 0.

Note that P, Q are fixed, so V(a) = V(b) = 0.

Remark 3.16. Instead of the end-fixed condition, we may consider〈V, T〉(t, 0) = 0 for all t ∈ [a, b], which is called “normal variation”.


We still assume γ is parametrized by arclength. Differentiate itagain

L′′(s) =∂

∂s

∫ b

a

〈∇TV, T〉〈T, T〉 dt

=∫ b

a

V〈∇TV, T〉〈T, T〉1/2 −

〈∇TV, T〉2〈T, T〉1/2 dt.

For the first term, we can change the order of differentiation by thecurvature tensor:

V〈∇TV, T〉 = 〈∇V∇TV, T〉+ 〈∇TV,∇VT〉

= 〈∇T∇VV, T〉+ 〈R(V, T)V, T〉+ ‖∇TV‖2

= T〈∇VV, T〉 − 〈∇VV,∇TT〉 − 〈R(V, T)T, V〉+ ‖∇TV‖2

Also,〈∇TV, T〉 = T〈V, T〉 − 〈V,∇TT〉.

Now assume that γ is a geodesic. So we get this second variationformula:(3.3)

L′′(0) = 〈∇VV, T〉∣∣∣ba+∫ b

a‖∇TV‖2 − 〈R(V, T)T, V〉 − (T〈V, T〉)2 dt.

From 〈∇TV,∇TV〉 = T〈V,∇TV〉 − 〈V,∇T∇TV〉 and

〈V,∇TV〉 = 〈V,∇VT〉 = V〈V, T〉 − 〈∇VV, T〉,

we have the second form of this formula:(3.4)

L′′(0) = V〈V, T〉∣∣∣ba−∫ b

a〈∇T∇TV + R(V, T)T, V〉+ (T〈V, T〉)2 dt.

Now we want to make some applications by these formulas.Recall: R(X, Y, Z, W) := 〈R(X, Y)W, Z〉 and

Ric(U, V) := ∑i

R(U, ei, V, ei).

Definition 3.17 (sectional curvature). For a two-dimensional plane Ein the tangent space, the sectional curvature is defined by

K(X, Y) =R(X, Y, X, Y)‖X ∧Y‖


for any independent pair of vectors X, Y in E.

Exercise 3.10. Show that K(X, Y) is the Gauss curvature of surfaceX(u, v) = expp(uX + vY). (hint: may assume that ‖X‖ = ‖Y‖ =

1, 〈X, Y〉 = 0)

Exercise 3.11 (Polarization formula). Let K(X, Y) = R(X, Y, X, Y).Then

〈R(X, Y)Z, W〉 = K(X + W, Y + Z)− K(X + W, Y)− K(X + W, Z)

− K(X, Y + Z)− K(W, Y + Z) + K(X, Z) + K(W, Y)

− K(Y + W, X + Z) + K(Y + W, X) + K(Y + W, Z)

+ K(Y, X + Z) + K(W, X + Z) + K(Y, Z)− K(W, X).

Definition 3.18. (M, g) has constant sectional curvature at p if K(X, Y) =k(p) for all X, Y ∈ TpM independent vectors.

Exercise 3.12. (M, g) has constant sectional curvature at p if andonly if

〈R(X, Y)W, Z〉 = K(p)(〈X, Z〉〈Y, W〉 − 〈X, W〉〈Y, Z〉),

that is

Rijk` = K(p)(gikgj` − gi`gjk).

Definition 3.19. A Riemannian manifold (M, g) is complete if (M, d)is complete where d(p, q) = inf L(γ) among all γ joins p, q.

Theorem 3.20 (Hopf-Rinow-deRham). The following are equivalent:

(1) (M, g) is complete.(2) expp is defined on all TpM for all p ∈ M.(3) expp is defined on all TpM for one p.(4) Any closed bounded set is compact.

We omit the proof here.

Corollary 3.21. That (M, g) is complete implies it is geodesic convex,which means any two points can be joined by a minimal geodesic.


Theorem 3.22 (Bonnet-Myer). Let (Mm, g) be complete with Ric ≥(m − 1)K > 0. Then M is compact with diamM ≤ π√

K. Note that

diamM := supp,q∈M d(p, q).

PROOF. If there exist p, q ∈ M with d(p, q) > π√K

. Let γ be ashortest geodesic joining p, q, say γ(0) = p, γ(`) = q. Take e1 = T =

γ′. Pick ei(0)i=2,...,m an O.N.B. of TpM and let ei(t) be the paralleltranslation of ei(0) along γ.

Now let Vi = sin(πt/`)ei. Construct the variation

Fi(t, s) = expγ(t)(sVi).

Since Fi is an endpoint-fixed normal variation,

0 ≤ ∑i>1

L′′i (0) = ∑i

∫ `

0‖∇TVi‖2 − 〈R(Vi, T)T, Vi〉dt.

= ∑i

∫ `

0‖π

`cos

π

`tei‖2 − sin2 π

`t〈R(ei, e1)e1, ei〉 dt

=≤ m− 12

((

π

`)2 − K

)< 0.

a contradiction since ` > π/√

K.

Another application of second variation formula is this theorem.

Theorem 3.23 (Synge). If Mm is compact with sectional curvature K > 0,then π1(M) is finite. Moreover,

(1) If m is even and M is orientable, then π1(M) = 1 ∼= Z2.(2) If m is odd, M must be orientable.

PROOF. For (1), if π1(M) 6= 1, then there exists a [γ] ∈ π1(M)

which can be represented by a curve γ with shortest length. (Henceγ is a smooth geodesic.) Here we leave the existence of γ as an exer-cise:` := infΓ∼γ L(Γ). Take a minimizing sequence Γ1, Γ2, . . . such thatΓi → γ in a subsequence.


Γiγ

M

p

γ

v(0)

v(`)

TpMM

Let e1(t) := T(t) = γ′(t), t ∈ [0, `] and e2(0), . . . , em(0) be O.N.B.of (e1(0))⊥ in TpM. By parallel translations, we can get orthonormalframes ei(t) along γ.

Now for any v = v(0) ∈ TpM, applying parallel translation alongγ on v(0), we can get v(`) ∈ TpM which in general may not be iden-tical with v(0). But this map is a holonomy transformation and wecan write

v = v(0) 7→ v(`) = Av

for some linear transformation on TpM. And in this case, we canonly consider the induced transformation

A : (e1(0))⊥ → (e1(0))⊥ ∼= Rm−1.

Since m − 1 is odd and M is orientable, the linear transformationA ∈ SO(m − 1) and must has 1 as its eigenvalue. So we can getAv = v for some v ∈ (e1(0))⊥. Using V = v as the variation field, bysecond variation formula, we have

0 ≤ L′′(0) =∫ `

0‖∇TV‖2 − 〈R(V, T)T, V〉 dt

=∫ `

00− K(V, T) dt < 0,

which is a contradiction.We leave (2) as an exercise.

Exercise 3.13. Complete the proofs of other parts of Synge theorem:Let Mm be compact and with positive sectional curvature K.

(1) For any homotopy class of closed curves [γ], there exists aminimal geodesic γ0 ∈ [γ].


(2) For m is odd, M is orientable.

6. Jacobi Fields

In the other form of second variation formula:

L′′(0) = V〈V, T〉∣∣∣ba−∫ b

a〈∇2

TV + R(V, T)T, V〉+ (T〈V, T〉)2 dt.

In case of a normal variation, we found that ∇2TV + R(V, T)T is an

important quantity to effect this value.We define that for a geodesic γ, T := γ′, a vector field along γ is

Jacobi if ∇2TV + R(V, T)T = 0. We call this Jacobi equation.

The Jacobi equation is a second order ODE. So V can be uniquelydetermined by V(0), V′(0) = ∇TV

∣∣t=0, and the solution space has

dimension 2m.

Theorem 3.24. J is Jacobi if and only if J is a variation vector field ofgeodesics, i.e.

J = F∗∂

∂s

∣∣∣s=0

for some F(t, s) which is a geodesic in t for any s.

PROOF. Assume F(t, s) is a family of geodesics for all s.

J

γ

F(t, s)

s = 0

Let T = F∗ ∂∂t , V = F∗ ∂

∂s . Then

∇2TV = ∇T∇TV = ∇T∇VT

= ∇V∇TT + R(T, V)T.

So we get the Jacobi equation since ∇TT = 0 for all s.

6. JACOBI FIELDS 91

p = c(0)

γ

J(0)

c(s)

Conversely, let c(s) be a curve with c(o) = p, c′(0) = J(0). LetT, J be the parallel translation of T(0), J′(0). Define

F(t, s) := expc(s)(t(T + s J′)),

which is clearly a variation of geodesics with F(0, s) = c(s). Now itsuffice to check: V = F∗ ∂

∂s = J, i.e. V(0) = J(0), V′(0) = J′(0).

V(0) =∂F∂s

∣∣∣(s,t)=(0,0)

(t=0)=

dds

expc(s)(0)∣∣∣s=0

= c′(0) = J(0),

V′(0) = ∇TV∣∣∣(s,t)=(0,0)

= ∇VT∣∣∣(s,t)=(0,0)

= ∇V

[(d expc(s))t(T+s J′)

∂

∂t(t(T + s J′)

)] ∣∣∣(s,t)=(0,0)

(t=0)= ∇ ∂

∂sId(T + s J′)

∣∣∣s=0

= J′(0).

Now we consider an example. Since S2 is complete, exp is well-defined on all TNS2.

N

S

J

J(π) = 0

(d expN)`v

`v T`vTNS2

We find that J(π) = 0 for any J with J(0) = 0.


Definition 3.25. For P, Q ∈ γ, we say Q is a conjugate point of Pif Q is a singular point of expP, i.e. if Q = expP(`v), (d expP)`v :T`v(TPM)→ TQM has a non-trivial kernel .

Pthe set of conjugate points

So for any vector w in such a kernel, we will construct a Jacobifield with J(0) = 0 with J(`) related to w.

Let F(t, s) := expp(t(v + s w` )), then

∂F∂s

∣∣∣s=0,t=`

= (d expp)`v(w).

So if J = ∂F∂s

∣∣s=0 is Jacobi with J(0) = 0, J(`) = (d expp)`v(w) for

general w. Hence w is in the kernel if and only if J(`) = 0 and J′(0) =w/`. And that is to say, J(`), J′(0) can determine each other if andonly if (d expp)`v is invertible.

Corollary 3.26. Q conjugates to P if and only if there exists a Jacobi fieldJ 6= 0 with J(P) = J(Q) = 0. Furthermore, there’s a correspondencebetween the Jacobi field and kernel:

J Jacobi,J(P) = 0 = J(Q)

↔ ker((expp)∗)`v

J 7→ w := J′(0)`.

Also, if Q is conjugate to P, then P is conjugate to Q.For more applications, may see Milnor’s Morse theory. Now we

can see some applications.

Theorem 3.27 (Cartan-Hadamard theorem). If M is complete with K <

0 and simply connected, then ∀p ∈ M, expp : TpM → M is a coveringmap. In particular, the universal cover M ∼= Rm (topologically).

Lemma 3.28. If M complete with K ≤ 0, then ∀ p ∈ M, γ geodesicthrough p, there doesn’t exist a conjugate point of p along γ.

6. JACOBI FIELDS 93

PROOF. Suppose there exists a conjugate point. And let J be aJacobi field with J(0) = J(`) = 0. By Jacobi equation,

0 = −∫ `

0〈∇2

T J + R(J, T)T, J〉 dt

= −∫ `

0T〈∇T J, J〉 − ‖∇T J‖2 + 〈R(J, T)T, J〉 dt

= 〈∇T J, J〉∣∣∣`0+∫ `

0‖∇T J‖2 − 〈R(J, T)T, J〉 dt.

Since J(0) = J(`) = 0 and K ≤ 0, ∇T J ≡ 0 along γ. But this impliesJ ≡ 0, a contradiction.

Lemma 3.29. If X is complete, f : X Y is a local isometry, then f is acovering map.

We leave this lemma as an exercise.

Exercise 3.14. Prove Lemma (3.29) and show by example that it failsif X is not complete.

PROOF OF CARTAN-HADAMARD THEOREM. By Lemma 3.28, exppis a local diffeomorphism on the whole TpM. Hence

( , ) := exp∗p〈 , 〉

is a Riemann metric on TpM ∼= Rm which makes expp a local isome-try.

By Hopf-Rinow theorem, (TpM, ( , )) is complete since exp0 :T0(TpM)→ TpM is an isometry.

By Lemma 3.29, expp is a covering map.

Example 3.30 (Space forms). (1) Sn, K > 01.(2) Rn/Γ, Γ ∼= Zn. K ≡ 0.(3) Hn/Γ, Γ: discrete group of isometries. K < 0.

Hn := x ∈ Rn | ‖x‖ < 1, g = ds2 =4‖dx‖2

1− ‖x‖2 .

Exercise 3.15. Show that Hn complete, K ≡ −1; Sn, K ≡ 1.


The second application is that we can see how curvatures are re-lated to construct isometries.

If there exists p ∈ X, Φ : TpX → TqY is an isometry of the innerproduct space, then Φ induces a local diffeomorphism:

y = ϕ(x) := expq Φ exp−1p

for v ∈ TxX where x ∈ γ. Let Pγ be the parallel translation along γ.Define gγ(v) = Pγ Φ P−1

γ (v).

Φ

pq

X Yϕ

x

v

γ γ

Theorem 3.31 (Cartan). If R(gγ(v), gγ(w))gγ(z) = gγ(R(v, w)z) forall v, w, z ∈ TxX, x in a neighborhood of p, then ϕ is a local isometry.

PROOF. Let J be the Jacobi field along γ. J(0) = 0, J(`) = v. Sinced exp is local diffeomorphic, that means there’re no conjugate points.Then J := gγ(J) is also a Jacobi field on γ.

Also, J′(0) = ΦJ′(0) by chain rule, since J(0) = 0, so

J(`) = (d expp)`γ′(0) J′(0),

J(`) = (d expq)`γ′(0) J′(0)

= (d expq)`γ′(0) Φ (d expp)−1`γ′(0) J(`)

= d(expq Φ exp−1p )J(`)

= (dϕ)x J(`).

Note that dΦ = Φ by linearity.

7. Space Forms

As model spaces, we often take manifolds with constant sectionalcurvature into account. In particular, those complete and simply

7. SPACE FORMS 95

connected Riemannian manifolds with constant sectional curvaturesare called space forms.

Now we will show that for each K, say K ∈ −1, 0, 1, the spaceforms are isometric:

Theorem 3.32. (1) K ≡ −1⇐⇒ M ∼= Hm.(2) K ≡ 0⇐⇒ M ∼= Rm.(3) K ≡ 1⇐⇒ M ∼= Sm.

PROOF. For (1)(2), K ≤ 0. We compare Hm, Rm with M 3 p.

Step 1: Construct any linear isometry: T0Hm L→ TpM.Step 2: Matching geodesics. As in Cartan’s theorem,

ϕ = expp L exp−10 : U 3 0→ M.

This map is well-defined by completeness.

0

Hm

p

M

In case of K ≡ −1, exp0 : T0Hm → Hm is a covering map. Sowe get ϕ : Hm → M as a diffeomorphism. Since for all X, Y, Z, W,〈R(X, Y)Z, W〉 ≡ −1. We may apply Cartan’s theorem: ϕ is a localisometry.

We leave the positive case as an exercise.

Exercise 3.16. M complete, simply connected with K ≡ 1. Show thatM ∼= Sm.

So if M has constant sectional curvature, M ∼= space form/Γwhere Γ is a discrete group of isometries, Γ ∼= π1(M).

For example, we can talk about hyperbolic geometry.


Hm

ds = 4|dz|2(1−|z|2)2 ds = dx2+dy2

y2

Exercise 3.17. For Poincare upper half plane, show that all geodesicare circles that perpendicular to the real line. Bonus: use this to provethat the isometry group is exactly SL(2, R).

All isometries between these hyperbolic spaces are precisely

SL(2, R) :=

[a bc d

]∣∣∣∣∣ ad− bc = 1

,

[a bc d

](z) =

az + bcz + d

.

8. The Second Fundamental Form

Now we consider all these properties on submanifolds. Let

f : Mm → Mm+k

where (M, g) with connection∇LC, (M, g) with∇LC are Riemannian

manifolds.

M → M

P

NpTpM

We consider the Riemannian structure on M is induced from M:

g(X, Y) = g(X, Y), ∀ X, Y ∈ TpM, ∀ p ∈ M,

8. THE SECOND FUNDAMENTAL FORM 97

and

∇LC = (∇LC)T

where T means the tangential projection.We can decompose the tangent space TpM = TpM⊕ Np.In general, for X, Y, Z, W ∈ TpM,

〈R(X, Y)Z, W〉 6= 〈R(X, Y)Z, W〉.

For example, Sm → Rm+1. But the sectional curvature K ≡ 1 on theball while K ≡ 0 in Rm.

We define difference of the covariant differentiation as the secondfundamental form.

Definition 3.33 (second fundamental form).

I(X, Y) ≡ B(X, Y) := (∇XY)N

where Y is a vector field extension of Y and N is the perpendicularpart to TpM of the vector.

We can see that B is a symmetric bilinear form since

B(X, Y)− B(Y, X) = (∇XY−∇YX)N = [X, Y]N = 0.

Also we define the mean curvature vector to be the trace of B,−→H =

H = tr(B):

H(p) =m

∑i=1

B(ei, ej) for e1, . . . , em ONB of TpM,

=m

∑i,j=1

gijB(∂i, ∂j) in any coordinate charts.


So we can get the relation of R and R.

〈R(X, Y)W, Z〉= 〈∇X∇YW, Z〉 − 〈∇Y∇XW, Z〉 − 〈∇[X,Y]W, Z〉

= 〈∇X∇YW, Z〉 − 〈∇Y∇XW, Z〉 − 〈∇[X,Y]W, Z〉 since Z ∈ TpM

= X〈∇YW, Z〉 − 〈∇YW,∇XZ〉 −Y〈∇XW, Z〉+ 〈∇XW,∇YZ〉 − 〈∇[X,Y]W, Z〉

= X〈∇YW, Z〉 − 〈∇YW,∇XZ〉+ 〈B(Y, W),∇XZ〉 −Y〈∇XW, Z〉

+ 〈∇XW,∇YZ〉 − 〈B(X, W),∇YZ〉 − 〈∇[X,Y]W, Z〉

= 〈R(X, Y)W, Z〉+ 〈B(X, Z), B(Y, W)〉 − 〈B(X, W), B(Y, Z)〉.

This is called Gauss equation.

Example 3.34. Let Mm → Rm+1 be a hypersurface. Choose a localcoordinate (x1, . . . , xm). Denote ∂i = Xi =

∂∂xi and

ìj = B(∂i, ∂j) = (∇∂i ∂j)N =

(∂

∂xi Xj

)N= XN

ji .

Now for M = Smr → Rm+1, we claim that ìj = − 1

R gij: note that thenormal vector is N = X/r, so

ìj = Xij · N = (Xi · N)j − (Xi · Nj) = −Xi ·Xj

r= −1

rgij.

In this case, we can write down the Gauss equation for balls:

Rijk` =1r2 (gikgj` − gi`gjk).

9. Variation of Higher Dimensional Submanifolds

Reference:

• Lawson: Lectures on minimal submanifolds.• Spivak: A Comprehensive Introduction to Differential Ge-

ometry, (V)

For an immersion ı : Mn → (Nn+p, g) and considering the in-duced metric on (M, g), we take variation of Mn in N:

F : (−ε, ε)×M → N,

9. VARIATION OF HIGHER DIMENSIONAL SUBMANIFOLDS 99

which is C∞ with F0 = ı, Fs : M → N. Let η := F∗ ∂∂s be the variation

vector field.Recall the n-dimensional area functional:

A(s) =∫

MsdA =

∫M

√g(s)dx1 ∧ · · · ∧ dxn

where g = det(gij) and g = F∗g.Take derivative:

A′(s) =∫

M

∂

∂s

√g(s) dx1 ∧ · · · ∧ dxn.

By the reductive formula for determinant, denoting Gij as the cofac-tor to gij,

g′(s) = g′ijGij = g′ijgijg.

Hence

A′(s) =∫

M

g′ijgijg

2√

gdx1 ∧ · · · ∧ dxn =

12

∫M

g′ijgij dA(s).

Let p ∈ M. Pick Riemann normal coordinate (x1, . . . , xn) (of M)at p. Then

g′ij(0) =∂

∂s〈∂i, ∂j〉

∣∣∣s=0

= 〈∇η∂i, ∂j〉+ 〈∂i,∇η∂j〉.

Since

〈∇∂i η, ∂j〉 = 〈∇∂i η, ∂j〉 = ∂i〈η, ∂j〉 − 〈η,∇∂i ∂j〉= ∂i〈η, ∂j〉 − 〈η,∇∂i ∂j + B(∂i, ∂j)〉= ∂i〈η, ∂j〉 − 〈η, B(∂i, ∂j)〉

(and 〈∂i,∇η∂j〉 is the same by symmetry), we get the first variationformula:

A′(0) =∫

Mgij(∂i〈η, ∂j〉 − 〈η, B(∂i, ∂j)

)dA

=∫

MdivηT − 〈η,

−→H 〉 dA

=∫

∂M〈ηT, ν〉 dσ−

∫M〈η,−→H 〉 dA.

Definition 3.35. M is called a minimal submanifold (immersion) in N ifand only if

−→H = 0.


Now our problem is: Is a minimal submanifold “locally minimal”in the usual sense? Even in R3, given any piecewise smooth curva Γ,is there an area minimizing surface S with ∂S = Γ?

Differentiate again:

A′′(s) =12

∫M

∂

∂s(g′ijg

ij√q)dx1 ∧ · · · ∧ dxn

=∫

M

( 12

g′′ijgij√g︸︷︷︸

(1)

+12

g′ij(gij)′√

g︸︷︷︸(2)

+12

g′ijgij(√

g)′︸︷︷︸(3)

)dx1 ∧ · · · ∧ dxn.

At the point p, s = 0, pick the Riemann normal coordinate and thengij(p) = δij, ∇∂i ∂j = 0.

(1) = η〈∇∂i ∂i, ∂i〉 = 〈∇η∇η∂i, ∂i〉+ 〈∇η∂i,∇η∂i〉

= 〈∇η∇∂i η, ∂i〉+ ‖∇∂i η‖2

= 〈R(η, ∂i)η, ∂i〉+ 〈∇∂i∇ηη, ∂i〉+ ‖∇∂i ηT‖2 + ‖∇∂i η

N‖2

= −R(η, ∂i, η, ∂i) + ∂i〈∇ηη, ∂i〉 − 〈∇ηη,∇∂i ∂i〉+ ‖∇∂i ηT‖2 + ‖∇∂i η

N‖2.

For (2), g−1g = id, (g−1)′g + g−1g′ = 0, (g−1)′ = −g−1g′g−1.

(2) = −12 ∑

i,jg′ij(gikg′k`g`j) = −1

2

n

∑i,j(g′ij)

2

= −12 ∑

i,j

(〈∇η∂i, ∂j〉+ 〈∂i,∇η∂j

)2

And for (3),

(3) =14

(∑i,j

g′ijgij

)2

=(

∑〈∇η∂i, ∂i〉)2

Put together, and we get

9. VARIATION OF HIGHER DIMENSIONAL SUBMANIFOLDS 101

Theorem 3.36 (Second variation formula).

A′′(0) =∫

∂M〈∇ηη, ν〉 −

∫M

(〈∇ηη,

−→H 〉+ ∑

i〈R(∂i, η)η, ∂i〉

)+∫

M∑i,j〈∇∂i η, ∂j〉2 + ∑

i‖∇∂i η

N‖2

− 12

∫M

n

∑i,j=1

(〈∇η∂i, ∂j〉+ 〈∂i,∇η∂j〉

)2

+∫

M

(∑

i〈∇η∂i, ∂i〉

)2.

For i : M → N (minimal), F normal variation (η ⊥ TM), we maylet η = f µ where µ is the unit normal vector. In this case,

〈∇η∂i, ∂j〉 = 〈∇∂i η, ∂j〉 = ∂i〈η, ∂j〉 − 〈η,∇∂i ∂j〉= −〈η, B(∂i, ∂j)〉,

and the calculation for 〈∂i,∇η∂j〉 is the same. So

−12

∫M

n

∑i,j=1

(〈∇η∂i, ∂j〉+ 〈∂i,∇η∂j〉

)2= −2

∫M

n

∑i,j=1| f |2〈µ, B(∂i, ∂j)〉2.

Combining with the term ∑〈∇∂i η, ∂j〉2, we get

−∫

M| f |2‖〈B, µ〉‖2.

Also, ∑i〈∇η∂i, ∂i〉 = ∑i〈∇∂i η, ∂i〉 = − f ∑i〈∇∂i ∂i, µ〉 = − f 〈−→H , µ〉.For ∑i ‖∇

N∂i

η‖2, if the dimension of vertical part is 1, we may sayη = f en+1 (µ = en+1) and

∇∂i η = ∇∂i( f en+1) = (∂i f )en+1 + f∇∂i en+1.

Since 〈∇∂i en+1, en+1〉 = 12 ∂i〈en+1, en+1〉 = 0, f∇∂i en+1 ∈ TM. So

∇N∂i

η = (∂i f )en+1, ∑i‖∇N

∂iη‖2 = ‖d f ‖2 = ‖∇ f ‖2.

We have the corollary:


Corollary 3.37. For ∂M fixed normal variation of a minimal immersionM# N,

A′′(0) =∫

M‖∇ f ‖2 −

(Ric(−→n ,−→n ) + ‖B‖2

)f 2,

where −→n = en+1.

As an application, we introduce the result which is a lemma usedin the proof of positive mass theorem.

Theorem 3.38 (Schoen - Yau). If M is a compact oriented 3-fold (3-dimensional manifold) with the scalar curvature R > 0, then 6 ∃ M # Nsuch that M is compact orientable surface with g(M) ≥ 1 and stable(A′′(0) ≥ 0).

PROOF. If M is stable, i.e. A′′(0) ≥ 0 and∫M

(Ric(−→n ,−→n ) + ‖B‖2) f 2 ≤

∫M|∇ f |2, ∀ f .

Let f ≡ 1. We apply this lemma:

Lemma 3.39. Ric(−→n ,−→n ) = 12 R − K − 1

2‖B‖2 where K is the Gausscurvature on M.

With this lemma, we have∫M

12

R +12‖B‖2 −

∫M

K ≤ 0

⇒ 2πχ(M) = 4π(1− g(M)) ≤ 0

a contradiction.

PROOF OF LEMMA. Pick Riemann normal coordinate at p.

R = gijgk`Rikj`

= 2(R1212 + R1313 + R2323)

= 2(Ric(e3, e3) + R1212).

Rijk` = Rijk`+ hikhj`− hi`hjk and K = R1212 = R1212 + h11h22− h12h21.Also,

B =

[h11 h12

h21 h22

], tr(B) = h11 + h22 = 0.

10. EXERCISES 103

‖B‖2 = h211 + h2

12 + h221 + h2

22 = 2h211 + 2h2

12 and h11h22 − h12h21 =

−(h211 + h2

12) =12‖B‖2. Hence we get the formula.

10. Exercises

1. On a local coordinate chart, if ω = hdx1 ∧ · · · ∧ dxn, show thath =

√det(gij). In general, show that v1, . . . , vk ⊂ Rn, the k-dim’l

volume of

[v1, . . . , vk] =

k

∑i=1

tivi

∣∣∣ 0 ≤ ti ≤ 1

is√

det(VtV) where

V =

| | · · · |v1 v2 · · · vk

| | · · · |

.

2. Check that ∗2 = (−1)p(n−p).3. Verify the original Stokes theorem in R3∫

Ω∇× F · n dA =

∫∂Ω

F · dr.

4. M → Rn. Show the induced connection is torsion-free.5. Prove the existence of Levi-Civita connection by checking this re-

ally defines a covariant differentiation.6. Prove Gauss lemma using the fact Γα

rr = 0 and then ∂rgrα = 0 ∀ α.7. Extend the covariant derivative to all tensors. And prove the com-

mutation formula:

Ti1···irj1···js;ji = Ti1···ir

j1···js;ij + Tqi2···irj1···js Ri1

qij + Ti1qi3···irj1···js Ri2

qji + · · ·

+ Ti1···irpj2···js Rp

j1 ji + · · ·+ Ti1···irj1···js−1 p Rp

js ji.

8. From these symmetries, we can get some further results:(a) The symmetries (3.14) implies Rk`ij = Rijk`.(b) Second Bianchi identity:

0 = Rij[k`;m] := Rijk`;m + Rij`m;k + Rijmk;`.

9. For n ≥ 3, if Rij = λgij, then λ is constant. (hint: use Bianchiidentity.)


10. Show that K(X, Y) is the Gauss curvature of surface X(u, v) =

expp(uX + vY). (hint: may assume that ‖X‖ = ‖Y‖ = 1, 〈X, Y〉 =0)

11. Let K(X, Y) = R(X, Y, X, Y). Then

〈R(X, Y)Z, W〉 = K(X + W, Y + Z)− K(X + W, Y)− K(X + W, Z)

− K(X, Y + Z)− K(W, Y + Z) + K(X, Z) + K(W, Y)

− K(Y + W, X + Z) + K(Y + W, X) + K(Y + W, Z)

+ K(Y, X + Z) + K(W, X + Z) + K(Y, Z)− K(W, X).

12. (M, g) has constant sectional curvature at p if and only if

〈R(X, Y)W, Z〉 = K(p)(〈X, Z〉〈Y, W〉 − 〈X, W〉〈Y, Z〉),

that is

Rijk` = K(p)(gikgj` − gi`gjk).

13. Complete the proofs of other parts of Synge theorem: Let Mm becompact and with positive sectional curvature K.(a) For any homotopy class of closed curves [γ], there exists a

minimal geodesic γ0 ∈ [γ].(b) For m is odd, M is orientable.

14. Prove Lemma (3.29) and show by example that it fails if X is notcomplete.

15. Show that Hn complete, K ≡ −1; Sn, K ≡ 1.16. M complete, simply connected with K ≡ 1. Show that M ∼= Sm.17. For Poincare upper half plane, show that all geodesic are circles

that perpendicular to the real line. Bonus: use this to prove thatthe isometry group is exactly SL(2, R).

18. Determine all the geodesics on the sphere Sn.19. (Do Carmo Ch.2 #8) Consider the upper half-plane

R2+ = (x, y) ∈ R2 | y > 0

with the metric given by g11 = g22 = 1y2 , g12 = 0 (metric of Lo-

batchevski’s non-euclidean geometry).(a) Show that the Christoffel symbols of the Riemannian connec-

tion are: Γ111 = Γ2

12 = Γ122 = 0, Γ2

11 = 1y , Γ1

12 = Γ222 = − 1

y .

10. EXERCISES 105

(b) Let v0 = (0, 1) be a tangent vector at point (0, 1) of R2+ (v0

is a unit vector on the y-axis with origin at (0, 1)). Let v(t)be the parallel transport of v0 along the curve x = t, y = 1.Show that v(t) makes an angle t with the direction of the y-axis, measured in the clockwise sense.

20. (Do Carmo Ch.3 #5) Let M be a Riemannian manifold and X ∈X (M) (the set of all vector field of class C∞ on M). Let p ∈ Mand let U ⊂ M be a neighborhood of p. Let ϕ : (−ε, ε)×U → Mbe a differentiable mapping such that for any q ∈ U the curvet → ϕ(t, q) is a trajectory of X passing through q at t = 0 (U andϕ are given by the fundamental theorem for ordinary differentialequations). X is called a Killing field (or an infinitesimal isometry) if,for each t0 ∈ (−ε, ε), the mapping ϕ(t0, ·) : U ⊂ M → M is anisometry. Prove that(a) A vector field v on Rn may be seen as a map v : Rn → Rn; we

say that the field is linear if v is a linear map. A linear field onRn, defined by a matrix A, is a Killing field if and only if A isanti-symmetric.

(b) Let X be a Killing field on M, p ∈ M, and let U be a normalneighborhood of p on M. Assume that p is a unique pointof U that satisfies X(p) = 0. Then, in U, X is tangent to thegeodesic spheres centered at p.

(c) Let X be a differentiable vector field on M and let f : M →N be an isometry. Let Y be a vector field on N defined byY( f (p)) = d fp(X(p)), p ∈ M. Then Y is a Killing field if andonly if X is also a Killing vector field.

(d) X is Killing⇔ 〈∇YX, Z〉+ 〈∇ZX, Y〉 = 0 for all Y, Z ∈ X (M)

(the equation above is called the Killing equation).(e) Let X be a Killing field on M with X(q) 6= 0, q ∈ M. Then there

exists a system of coordinates (x1, . . . , xn) in a neighborhoodof q, so that the coefficients gij of the metric in this systemcoordinates do not depend on xn.

21. (Do Carmo Ch.4 #4) Let M be a Riemannian manifold with thefollowing property: given any two points p, q ∈ M, the paralleltransport from p to q does not depend on the curve that joins p to


q. Prove that the curvature of M is identically zero, that is, for allX, Y, Z ∈ X (M), R(X, Y)Z = 0.

22. Define covariant derivatives on differential forms.23. (Do Carmo Ch.4 #8)(Schur’s theorem) Let Mn be a connected Rie-

mannian manifold with n ≥ 3. Suppose that M is isotropic, that is,for each p ∈ M, the sectional curvature K(p, σ) does not dependon σ ⊂ TpM. Prove that M has constant sectional curvature, thatis, K(p, σ) also does not depend on p.

24. (Do Carmo Ch.4 #10) (Einstein manifolds) A Riemannian man-ifold Mn is called an Einstein manifold if, for all X, Y ∈ X (M),Ric(X, Y) = λ〈X, Y〉, where λ : M → R is a real valued funciton.Prove that:(a) If Mn connected and Einstein , with n ≥ 3, then λ is constant

on M.(b) If M3 is a connected Einstein manifold then M3 has constant

sectional curvature.25. Show that the sectional curvature is the Gauss curvature.26. (Do Carmo Ch.4 #6) Let M be a Riemannian manifold. M is a

locally symmetric space if ∇R = 0, where R is the curvature tensorof M.(a) Let M be a locally symmetric space and let γ : [0, `) → M be

a geodesic of M. Let X, Y, Z be parallel vector fields along γ.Prove that R(X, Y)Z is a parallel field along γ.

(b) Prove that if M is locally symmetric, connected, and has di-mension two, then M has constant sectional curvature.

(c) Prove that if M has constant (sectional) curvature, then M is alocally symmetric space.

27. (Do Carmo Ch.5 #5) (Jacobi fields and conjugate points on locallysymmetric spaces) Let γ : [0, ∞) → M be a geodesic in a locallysymmetric space M and let v = γ′(0) be its velocity at p = γ(0).Define a linear transformation Kv : TpM→ TpM by

Kv(x) = R(v, x)v, x ∈ TpM.

(a) Prove that Kv is self-adjoint.

10. EXERCISES 107

(b) Choose an orthonormal basis e1, . . . , en of TpM that diago-nalizes Kv, that is,

Kv(ei) = λiei, i = 1, . . . , n.

Extend the ei to fields along γ by parallel transport. Showthat, for all t,

Kγ′(t)(ei(t)) = λiei(t).

where λi does not depend on t.(c) Let J(t) = ∑i xi(t)ei(t) be a Jacobi field along γ. Show that

the Jacobi equation is equivalent to the system

d2xi

dt2 + λixi = 0, i = 1, . . . , n.

(d) Show that the conjugate points of p along γ are given byγ(πk/

√λi), where k is a positive integer and λi is a positive

eigenvalue of Kv.28. (Do Carmo Ch.7 #1) If M, N are Riemannian manifolds such that

the inclusion i : M → N is an isometric immersion, show by anexample that the strict inequality of metrics dM > dN can occur.

29. (Do Carmo Ch.7 #4) Consider the universal covering

π : M→ R2 \ (0, 0)

of the Euclidean plane minus the origin. Introduce the coveringmetric on M. (Note: For M a covering space of M, it is possibleto give the covering space a Riemannian structure such that thecovering map π : M→ M is a local isometry.) Show that M is notcomplete and not extendible, and that the Hopf-Rinow theorem isnot true for M (this shows that the definition of non-extendibility,though natural, is not a satisfactory one).

30. (Do Carmo Ch.7 #5) A divergent curve in a Riemannian manifoldM is a differentiable mapping α : [0, ∞) → M such that for anycompact set K ⊂ M there exists t0 ∈ (0, ∞) with α(t) /∈ K for allt > t0 (that is, α “escapes” every compact set in M). Define the


length of a divergent curve by

limt→∞

∫ t

0|α′(t)| dt.

Prove that M is complete if and only if the length of any divergentcurve is unbounded.

31. (Do Carmo Ch.7 #6) A geodesic γ : [0, ∞) → M in a Riemann-ian manifold M is called a ray starting from γ(0) if it minimizesthe distance between γ(0) and γ(s), for any s ∈ (0, ∞). Assumethat M is complete, non-compact, and let p ∈ M. Show that Mcontains a ray starting form p.

32. (Do Carmo Ch.7 #12) A Riemannian manifold is said to be homoge-neous if given p, q ∈ M there exists an isometry of M which takesp into q. Prove that any homogeneous manifold is complete.

33. (Do Carmo Ch.8 #1) Consider, on a neighborhood in R2, n > 2 themetric

gij =δij

F2

where F 6= 0 is a function of (x1, . . . , xn) ∈ Rn. Denote by Fi =∂F∂xi

,

Fij =∂2F

∂xi∂xj, etc.

(a) Show that a necessary and sufficient condition for the metricto have constant curvature K is Fij = 0, i 6= j

F(Fjj + Fii) = K + ∑ni=1(Fi)

2.

(b) Use (a) to prove that the metric gij has constant curvature K ifand only if

F = G1(x1) + G2(x2) + · · ·+ Gn(xn),

where

Gi(xi) = ax2i + bixi + ci

andn

∑i=1

(4cia− b2i ) = K.

10. EXERCISES 109

(c) Put a = K/4, bi = 0, ci = 1/n and obtain the formula ofRiemann

gij =δij(

1 + K4 ∑ x2

i)2

for a metric gij of constant curvature K. If K < 0, the metric gij

is defined in a ball of radius of√

4−K .

(d) If K > 0, the metric is defined on all of Rn. Show that such ametric on Rn is not complete.

34. (Do Carmo Ch.6 #1) Let M1 and M2 be Riemannian manifolds,and consider the product M1 ×M2, with the product metric. Let∇1 be the Riemannian connection of M1 and let ∇2 be the Rie-mannian connection of M2.(a) Show that the Riemannian connection∇ of M1×M2 is given

by ∇Y1+Y2(X1 + X2) = ∇1Y1

X1 +∇2Y2

X2, for X1, Y1 ∈ X (M1),X2, Y2 ∈ X (M2).

(b) For every p ∈ M1, the set (M2)p = (p, q) ∈ M1 × M2; q ∈M2 is a submanifold of M1 × M2, naturally diffeomorphicto M2. Prove that (M2)p is a totally geodesic submanifold ofM1 ×M2.

(c) Let σ(x, y) ⊂ T(p,q)(M1 ×M2) be a plane such that x ∈ TpM1

and y ∈ TqM2. Show that K(σ) = 0.35. (Do Carmo Ch.6 #2) Show that x : R2 → R4 given by

x(θ, ϕ) =1√2(cos θ, sin θ, cos ϕ, sin ϕ), (θ, ϕ) ∈ R2

is an immersion of R2 into the unit sphere S3(1) ⊂ R4, whoseimage x(R2) is a torus T2 with sectional curvature zero in the in-duced metric.

36. (Do Carmo Ch.6 #4) Let N1 ⊂ M1, N2 ⊂ M2 be totally geodesicsubmanifolds of the Riemannian manifolds M1 and M2 respec-tively. Prove that N1× N2 is a totally geodesic submanifold of theproduct M1 ×M2 with the product metric.

37. (Do Carmo Ch.6 #5) Prove that the sectional curvature of the Rie-mannian manifold S2 × S2 with the product metric, where S2 is


the unit sphere in R3, is non-negative. Find a totally geodesic, flattorus, T2, embedded in S2 × S2.

38. (Do Carmo Ch.6 #8) (The Clifford torus) Consider the immersionx : R2 → R4 given in the above exercise 35.(a) Show that the vectors

e1 = (− sin θ, cos θ, 0, 0), e2 = (0, 0,− sin ϕ, cos ϕ)

form an orthonormal basis of the tangent space, and that thevectors

n1 =1√2(cos θ, sin θ, cos ϕ, sin ϕ),

n2 =1√2(− cos θ,− sin θ, cos ϕ, sin ϕ)

form an orthonormal basis of the normal space.(b) Use the fact that

〈Snk(ei), ej〉 = −〈∇ei nk, ej〉 = 〈∇ei ej, nk〉,

where ∇ is the covariant derivative (that is, the usual deriv-ative) of R4, and i, j, k = 1, 2, to establish that the matrices ofSn1 and Sn2 with respect to the basis e1, e2 are

Sn1 =

[−1 00 −1

], Sn2 =

[1 00 −1

](c) From the above exercise 35, x is an immersion of the torus

T2 into S3(1) (the Clifford torus). Show that x is a minimalimmersion.

39. (Do Carmo Ch.6 #11) Let f : Mn+1 → R be a differentiable func-tion. Define the Hessian, Hess f of f at p ∈ M as the linear operator

Hess f : TpM→ TpM, (Hess f )Y = ∇Y grad f , Y ∈ TpM,

where ∇ is the Riemannian connection of M. Let a be a regularvalue of f and let Mn ⊂ Mn+1 be the hypersuperface in M definedby M = p ∈ M | f (p) = a. Prove that:(a) The Laplacian ∆ f is given by

∆ f = trace(Hess f ).

10. EXERCISES 111

(b) If X, Y ∈ X (M), then

〈(Hess f )Y, X〉 = 〈Y, (Hess f )X〉.

Conclude that Hess f is self-adjoint, hence determines a sym-metric bilinear form on TpM, p ∈ M, given by (Hess f )(X, Y) =〈(Hess f )X, Y〉, X, Y ∈ TpM.

(c) The mean curvature H of M ⊂ M is given by

nH = −div(

grad f|grad f |

).

(d) Observe that every embedded hypersurface Mn ⊂ Mn+1 islocally the inverse image of a regular value. Conclude from(c) that the mean curvature H of such a hypersuperface isgiven by

H = − 1n

divH,

where N is an appropriate local extension of the unit normalvector field on Mn ⊂ Mn+1.

40. (Do Carmo Ch.8 #8) (Riemannian submersions) A differentiablemapping f : Mn+k → Mn is called a submersion if f is surjective,and for all p ∈ M, d f p : TpM → Tf ( p)M has rank n. In this case,for all p ∈ M, the fiber f−1(p) = Fp is a submanifold of M anda tangent vector of M, tangent to some Fp, p ∈ M, is called avertical vector of the submersion. If, in addition, M and M haveRiemannian metrics, the submersion f is said to be Riemannian if,for all p ∈ M, d fp : TpM → Tf (p)M preserves lengths of vectorsorthogonal to Fp. Show that:(a) If M1 × M2 is the Riemannian product, then the natural pro-

jections πi : M1 ×M2 → Mi, i = 1, 2 are Riemannian submer-sions.

(b) Let the tangent bundle TM be given the Riemannian metricas:

〈V, W〉(p,v) = 〈dπ(V), dπ(W)〉p + 〈Dvdt

(0),Dwds

(0)〉p

for (p, v) ∈ TM, V, W tangent vectors at (p, v) in TM whereV = α′(0), W = β′(0) for curves α, β chosen such that α(t) =


(p(t), v(t)), β(t) = (q(s), w(s)), p(0) = q(0) = 0, v(0) =

w(0) = v (cf. Do Carmo Ch.3 #2). Show that the projectionπ : TM→ M is a Riemannian submersion.

41. (Do Carmo Ch.8 #9) (Conneciton of a Riemannian submersion)Let f : M → M be a Riemannian submersion. A vector x ∈ TpMis horizontal if it is orthogonal to the fiber. The tangent space TpMthen admits a decomposition TpM = (TpM)h ⊕ (TpM)v, where(TpM)h and (TpM)v denote the subspaces of horizontal and ver-tical vectors, respectively. If X ∈ X (M), the horizontal lift X of Xis the horizontal field defined by d f p(X( p)) = X( f (p)).(a) Show that X is differentiable.(b) Let ∇ and ∇ be the Riemannian connections of M and M

respectively. Show that

∇XY = ∇XY +12[X, Y]v, X, Y ∈ X (M),

where Zv is the vertical component of Z.(c) [X, Y]v( p) depends only on X( p) and Y( p).

42. (Do Carmo Ch.8 #10) (Curvature of a Riemannian submersion)Let f : M → M be a Riemannian submersion. Let X, Y, Z, W ∈X (M), X, Y, Z, W be their horizontal lifts, and let R and R be thecurvature tensors of M and M respectively. Prove that:(a)⟨

R(X, Y)Z, W⟩= 〈R(X, Y)Z, W〉 − 1

4⟨[X, Z]v, [Y, W]v

⟩+

14⟨[Y, Z]v, [X, W]v

⟩− 1

2⟨[Z, W]v, [X, Y]v

⟩(b) K(σ) = K(σ) + 3

4

∣∣[X, Y]v∣∣2 ≥ K(σ), where σ is the plane gen-

erated by the orthonormal vectors X, Y ∈ X (M) and σ is theplane generated by X, Y.

43. (Do Carmo Ch.8 #11) (The complex projective space) Let

Cn+1 \ 0 = (z0, . . . , zn) = Z 6= 0 | zj = xj + iyj, j = 0, . . . , n

be the set of all non-zero (n + 1)-tuples of complex numbers zj.Define equivalence relation on Cn+1 \ 0: (z0, . . . , zn) ∼ W =

(w0, . . . , wn) if zj = λwj, λ ∈ C, λ 6= 0. The equivalence class

10. EXERCISES 113

of Z will be denoted by [Z] (the complex line passing through theorigin and through Z). The set of such classes is called, by analogywith the real case, the complex projective space Pn(C) of complexdimension n.(a) Show that Pn(C) has a differentiable structure of a manifold

of real dimension 2n and that P1(C) is diffeomorphic to S2.(b) Let (Z, W) = z0w0 + · · ·+ znwn be the hermitian product on

Cn+1, where the bar denotes complex conjugation. IdentifyCn+1 ≈ R2n+2 by putting zj = xj + iyj = (xj, yj). Show that

S2n+1 = N ∈ Cn+1 ≈ R2n+2 | (N, N) = 1

is the unit sphere in R2n+2.(c) Show that the equivalence relation ∼ induces on S2n+1 the

following equivalence relation: Z ∼W if eiθZ = W. Establishthat there exists a differentiable map (the Hopf fibering) f :S2n+1 → Pn(C) such that

f−1([Z])

=eiθ N ∈ S2n+1 | N ∈ [Z] ∩ S2n+1, 0 ≤ θ ≤ 2π

=[Z] ∩ S2n+1.

(d) Show that f is a submersion.44. (Do Carmo Ch.8 #12) (Curvature of the complex projective space)

Define a Riemannian metric on Cn+1 \ 0 in the following way:If Z ∈ Cn+1 \ 0 and V, W ∈ TZ(C

n+1 \ 0),

〈V, W〉Z =Real(V, W)

(Z, Z).

Observe that the metric 〈 , 〉 restricted to S2n+1 ⊂ Cn+1 \ 0 co-incides with the metric induced from R2n+2.(a) Show that, for all 0 ≤ θ ≤ 2π, eiθ : S2n+1 → S2n+1 is an isom-

etry, and that, therefore, it is possible to define a Riemannianmetric on Pn(C) in such a way that the submersion f is Rie-mannian.


(b) Show that, in this metric, the sectional curvature of Pn(C) isgiven by

K(σ) = 1 + 3 cos2 ϕ,

where σ is generated by the orthonormal pair X, Y, cos ϕ =

〈X, iY〉, and X, Y are the horizontal lifts of X and Y, respec-tively. In particular, 1 ≤ K(σ) ≤ 4.

Chapter 4

HODGE THEOREM

1. Harmonic Forms

Let Ap(M) be differential p-forms on a compact closed manifoldMn with a given Riemannian metric g and an inner product

(α, β) :=∫

M〈α, β〉dV =

∫M

α ∧ ∗β,

where ∗ : Λp → Λn−p, ∗2 = (−1)(n−p)p. We get a pre-Hilbert space.Now the question is that: for [α] ∈ Hp

dR(M), can we pick a (or thebest) representative of [α]? e.g. infβ ‖α + dβ‖.

Denote d∗ as the adjoint of d : Ap(M)→ Ap+1(M).

(α, d∗β) = (dα, β) =∫

Mdα ∧ ∗β

=∫

Md(α ∧ ∗β)− (−1)pα ∧ d(∗β)

= (−1)np+1∫

Mα ∧ ∗(∗d ∗ β)

So d∗ = (−1)np+1 ∗ d∗.Let dα = 0.

(α + tdβ, α + tdβ) = (α, α) + 2(α, dβ)t + (dβ, dβ)t2 ≥ (α, α)

⇐⇒ 0 = (α, dβ) = (d∗α, β), ∀ β ⇐⇒ d∗α = 0.

Hence we have dα = 0, d∗α = 0⇔ ‖α‖ is minimal.Here we introduce the Laplace operator

∆ = (d + d∗)2 = dd∗ + d∗d.

We can see that

(∆α, α) = (dd∗α + d∗dα, α) = (d∗α, d∗α) + (dα, dα) = 0,115

116 4. HODGE THEOREM

and then ∆α = 0⇔ dα = 0, d∗α = 0.Clearly, this operator ∆ is self-adjoint. And the truly crucial prop-

erty we need is that ∆ is elliptic.So the general question becomes: how to solve ∆α = β? when is

it solvable? The answer goes:Let H = α | ∆α = 0. ∆α = β is solvable if and only if β ∈H⊥.The “if part” is easier. For any γ ∈H,

(β, γ) = (∆α, γ) = (α, ∆γ) = 0 ⇒ β ∈H⊥.

For example, let β ∈ A0(M) (a function). If ∆α = β is solvable,then β ∈H⊥ = R⊥, i.e.

∫M β dV = 0.

Conversely, the “only if part” comes from a standard formulationin fundamental analysis.

Let L : H → H be a bounded linear operator on some pre-HilbertspaceH. Consider the equation Lw = β. For any φ ∈ H,

(w, L∗φ) = (Lw, φ) = (β, φ).

We can see that any solution w defines a linear functional ` such that

`(L∗φ) = (β, φ) ∀ φ ∈ H.

Such a ` is called a weak solution for Lw = β.

Remark 4.1. If H is a Hilbert space (i.e. complete), then by Rieszrepresentation theorem, such ` will come from some α ∈ H.

2. Hodge Decomposition Theorem

Here we recall two theorems in standard analysis.

Theorem 4.2 (Regularity theorem). For Lw = β elliptic linear PDE,any weak solution is automatically smooth.

Theorem 4.3 (Compactness theorem). For a sequence αn ∈ H, if ‖αn‖ ≤C, ‖Lαn‖ ≤ C are bounded, then αn has a Cauchy subsequence (w.r.t.L2 norm).

2. HODGE DECOMPOSITION THEOREM 117

Both of the two theorems are implied by the Garding inequality.For f ∈ H,

‖ f ‖s+d ≤ C(‖L f ‖s + ‖ f ‖s),

where d is the order of L, and the norms are Sobolev norms.Now back to the case L = ∆, H = Ap(M) = Ap. Let H = Hp :=

α ∈ Ap(M) | ∆α = 0 be the set of harmonic p-forms.By assuming the two theorems, we have the theorem.

Theorem 4.4 (Hodge decomposition theorem).

(1) dim H < ∞, and(2) Ap = Hp ⊕ ∆Ap, i.e. Im∆ = H⊥.

PROOF. Firstly, we can infer that dim H < ∞ by compactness. Ifdim H = ∞, we may select

u1, u2, . . . ∈H, ‖ui‖ = 1, ui ⊥ uj ∀ i 6= j

satisfying ∆ui = 0. But it’s clear that ui has no Cauchy subse-quence.

Also by compactness, H ⊂ H is a closed subspace. We have thedecomposition Ap = H⊕H⊥.

Hence we can pick w1, w2, . . . , wl, an O.N.B. of H such that

α = β +l

∑i=1

(α, wi)wi = β + H(α)

where β ∈H⊥ and H(α) is the harmonic projection of α.We still need to show H⊥ = ∆Ap.It’s clear that H⊥ ⊃ ∆Ap since

(∆α, γ) = (α, ∆γ) = 0, ∀ γ ∈H.

The other side comes from the following claim:

Claim 4.5. There exists c > 0 such that ‖β‖ ≤ c‖∆β‖, ∀ β ∈H⊥.

If the claim holds, let α ∈H⊥. Define a linear functional ` by

`(∆∗φ) := (α, φ), on∆Ap ⊂ Ap.


Check that ` is bounded: let φ1 = φ− H(φ).

‖`(∆φ)‖ = ‖`(∆φ1)‖ = ‖(α, φ1)‖ ≤ ‖α‖ · ‖φ1‖(claim) ≤ c‖α‖ · ‖∆φ1‖ = (c‖α‖) · ‖∆φ‖.

By Hahn-Banach theorem, ` can be extended to a bounded linearfunction on Ap(M), i.e. ` is a weak solution to ∆w = α.

Then by regularity theorem, there exists a smooth w ∈ Ap(M)

such that ∆w = α. So finally H⊥ ⊂ ∆Ap.

Now we prove the claim.

PROOF OF CLAIM. Suppose the contrary, there exists a sequenceof β j ∈H⊥, ‖β j‖ = 1, ‖∆β j‖ → 0.

By compactness theorem, we may assume β j itself is a Cauchy se-quence. Define `(ψ) = limj→∞(β j, ψ) for ψ ∈ Ap which is bounded.And

`(∆ψ) = limj→∞

(β j, ∆ψ) = limj→∞

(∆β j, ψ) = 0,

i.e. ` is a weak solution of ∆β = 0.By regularity theorem, there exists a β ∈ Ap(M) such that `(ψ) =

(β, ψ) and ∆β = 0, i.e. β ∈H.But ‖β j‖ = 1 and β j ∈ H⊥ imply that ‖β‖ = 1 6= 0, β ∈ H⊥,

which leads a contradiction.

So far, these works for any elliptic L. In particular for ∆, we cango further. Since Ap = H⊕ (∆Ap), we define the Green operator

G :=

0 on H

∆−1 on H⊥, i.e. I = H + ∆G

where H : Ap → H is the harmonic projection. Hence Gα is theunique solution in H⊥ for ∆w = α− H(α).

Exercise 4.1.

(1) G will commute with any operator T with T∆ = ∆T.(2) G is a bounded, self-adjoint, compact operator.

3. BOCHNER PRINCIPLE 119

Notice that ∆ = dd∗ + d∗d, d∆ = ∆d, d∗∆ = ∆d∗, and then[d, G] = [d∗, G] = [∆, G] = 0. So

Ap = H⊕ Im∆ = H⊕ Imd⊕ Imd∗.

We can rewrite the identity I = H + ∆G as

α = H(α) + (dd∗ + d∗d)Gα

= H(α) + d(d∗Gα) + d∗(dGα).

So firstly we have the following result.

Proposition 4.6. If dα = 0, α = H(α) + d(d∗Gα).

Notice that the harmonic representation is unique. (e.g. α1− α2 =

dβ, H(α1)− H(α2) = H(dβ) = 0)There are some applications.

Proposition 4.7 (Poincare duality).

HpdR(M) ∼= Hn−1

dR (M)∗.

PROOF. Since ∗∆ = ∆∗, ∗2 = (−1)p(n−p), by using harmonicforms,

Hp ∗∼= Hn−p.

3. Bochner Principle

Proposition 4.8 (Bochner formula). Let M be compact.

∆ = −tr∇2 −∑i,j

ηi ∧ ι(ej)R(ei, ej),

for ej an O.N.B., ηi a dual basis. And ∇2x,y = ∇x∇y −∇∇xy.

Exercise 4.2. Prove the Bochner formula 4.8.

Corollary 4.9.

(1) If Ric > 0 (In fact, Ric ≥ 0, and > 0 at one point), b1 =

h1(M) = 0.


(2) If Ric ≥ 0, h1(M) ≤ n = dim(M) and if Ric > 0 at some point,then h1(M) = 0. The equality holds if and only if M is a flattorus, i.e. M ∼= Rn/Γ, Γ ∼= Zn.

PROOF. For (i), let θ ∈ A1(M). Say θ = ∑i aiηi where ηi is the

dual frame to the local frame ei at some p.At p, we have

∇2ei,ei

= ∇ei∇ei −∇∇ei ei .

So

0 = 〈∆θ, θ〉 −∑i〈∇ei∇ei θ, θ〉︸︷︷︸

(1)

− 〈∑ ηi ∧ ι(ej)(∇i∇j −∇j∇i)ηk, η`〉aka`︸︷︷︸

(2)

.

By direct calculation,

(1) = ∑i

ei〈∇ei θ, θ〉 − |∇ei θ|2 = −1

2∆|θ|2 − |∇θ|2.

(2) = ∑〈Rkjijη

i, η`〉aka` = Ric(θ, θ).

Therefore

0 = 〈∆θ, θ〉 = 12

∆|θ|2 + |∇θ|2 + Ric(θ, θ).

• If θ 6= 0, let |θ(p)| > 0 be maximal. Then ∆|θ|2 ≥ 0 bysecond derivative test. But this contradicts to Ric(θ, θ) > 0.• In case of Ric ≥ 0 and > 0 at one point.

0 = −12

∫M

∆|θ|2 +∫

M|∇θ|2 +

∫M

Ric(θ, θ).

By Stokes’ theorem, since ∂M = ∅, we have

0 =∫

M|∇θ|2 +

∫M

Ric(θ, θ)⇒ ∇θ ≡ 0.

i.e. θ is parallel and then θ is constant. And this contradictsto Ric > 0 at one point.

For (ii), again Ric ≥ 0 implies θ 6= 0 is parallel, determined bysome θp ∈ T∗p (M).

Hence h1 ≤ dim T∗p (M) = dim M.

4. FOURIER TRANSFORM 121

Moreover, if the equality holds, h1 = dim M = m, then the uni-versal cover M −→ M has m parallel 1-forms. Therefore it has mparallel vector fields.

Hence M ∼= Rm.

4. Fourier Transform

Here we define some notions in this section:

(1) x = (x1, . . . , xm) ∈ Rm.(2) x · y = x1y1 + x2y2 + · · ·+ xmym, |x| = (x · x)1/2.(3) α = (α1, . . . , αm) multi-index, αi ∈N∪ 0, |α| = ∑m

i=1 αi.(4) xα = xα1

1 xα22 · · · x

αmm .

(5) dα = ∂|α|

∂xα11 ···∂xαm

m, Dα = (−i)|α|dα.

To prove the regularity theorem and compactness theorem, weneed to investigate the concepts of weak derivatives, which we in-troduce here is defined through Fourier transformations.

Definition 4.10 (Fourier transformation). For ξ ∈ Rm, x ∈ Rm, letf ∈ C∞

0 (Rm, C) be a smooth function with compact support. TheFourier transformation of f is defined as

f (ξ) = (2π)−m/2∫

Rmeix·ξ f (x) dx.

Also recall the convolution of two functions

f ∗ g :=∫

Rmf (x− y)g(y) dy =

∫Rm

f (z)g(x− z) dz.

On the purpose of doing operations on weak functions, in manytimes we can use convolutions to construct smooth approximations.

Let f ∈ C∞0 (Rm) with

∫Rm f = 1, f ≥ 0, f (0) 6= 0. For u ∈

R+, define fu(x) := 1um f

(xu

). We can see that

∫Rm fu = 1. Such

a sequence fu is called a “δ-function”, and we can write it as δ0 =

limu→0

fu.

Proposition 4.11. For g ∈ C0, fu ∗ g(x)→ g(x) as u→ 0+ is a smoothapproximation, i.e., δ0 = limu→0 fu is the identity element for ∗.


We also have inverse Fourier transforms defined as in the follow-ing.

Definition 4.12 (Inverse Fourier Transform).

g(x) :=∫

Rmeiξ·xg(ξ) dξ.

Now we list the basic properties and correspondence with con-volutions in the following theorem.

Theorem 4.13 (Basic Properties of Fourier Transform).

(1) ˇf = f .

(2) Dαξ f (ξ) = xα f (x), Dα

x f (x) = ξα f (ξ).

(3) f ∗ g = f g, f ∗ g = f · g.

(4) Fourier transform is an isometry, i.e., L2(Rm)· // L2(Rm)·

oo .

5. Sobolev spaces

By observing the formula Dαx f = ξα f , we notice that derivatives

of f will corresponds multiplications of f in the frequency space.This formula enlightens the idea of weak derivatives.

Let s ∈ Rm, which is regarded as the order of L2-derivative. De-fine the s-norm of a L2-function f by

| f |2s :=∫

Rm(1 + |ξ|2)s| f (ξ)|2 dξ,

and the inner product (·, ·)s by

( f , g)s :=∫

Rm(1 + |ξ|2)s f (ξ)g(ξ) dξ.

We can easily check that C∞0 (Rm) → L2(Rm).

Let the Sobolev space Hs(Rm) be the completion of C∞0 (Rm) in

L2(Rm) (w.r.t the s-norm). Then again by the formula Dαx f = ξα f ,

Dα : Hs ∩ C∞0 −→ Hs−|α|,

|Dαx f |2s−|α| =

∫Rm

(1 + |ξ|2)s−|α||ξα f (ξ)|2 dξ.

5. SOBOLEV SPACES 123

The core concept here is, while we define the (weak) derivativeby multiplication in frequency space, how does f ∈ Ck actually if flies in some Sobolev space Hs, i.e. ξα f (ξ) lies in L2?

Theorem 4.14 (key lemmas).

(1) (Sobolev lemma) If s > k + m/2 and f ∈ Hs, then f ∈ Ck and

| f |∞,k ≤ C| f |s.

(2) (Rellich lemma) If s > t, Hs → Ht is a compact imbedding.(3) (interpolation) Let s > t > u. ∀ ε > 0, ∃ C(ε) such that

| f |t ≤ ε| f |s + C(ε)| f |u

for all f ∈ C∞0 .

PROOF.

(1) Let k = 0. First we consider f ∈ C∞0 (Rm).

| f (x)| =∣∣∣∣∫

Rmeixξ f (ξ) dξ

∣∣∣∣=

∣∣∣∣∫Rm

[eixξ f (ξ)(1 + |ξ|2)s/2

](1 + |ξ|2)−s/2 dξ

∣∣∣∣≤ | f |s

∫Rm

1(1 + |ξ|2)s dξ (Holder’s inequality)

≤ C| f |s

for some C depending on s > m/2.

Now for any f ∈ Hs, we can choose a sequence f jHs−→ f .

By above result,

| fi − f j|∞ ≤ C| fi − f j|s

infers that fi forms a Cauchy sequence in C0 and then f ∈C0. So

| f |∞ ≤ | f − fi|∞ + | fi|∞ ≤ C(| f − fi|s + | fi|s)

and then | f |∞ ≤ C| f |s by taking i→ ∞.


For k > 0, apply the same argument to Dα f , |α| = k.

|Dα f |∞ ≤ C|Dα f |s ≤ C| f |s+k,

⇒ | f |∞,k ≤ C| f |s+k.

Hence Hs ⊂ Ck.(2) Now consider K ⊂cpt Rm, fn ∈ Hs with supp( fn) ⊂ K,| fn|s ≤ C.

Let g ∈ C∞0 (Rm) with g ≡ 1 on K. Then g · fn = fn,

fn = g ∗ fn.

∂j fn = ∂j(g ∗ fn) = (∂j g) ∗ fn.

|∂j fn(ξ)| ≤∫

K|∂j g(ξ − ζ) fn(ζ)| dζ

≤ | fn|s∫

K

|∂j g(ξ − ζ)|2

(1 + |ξ|2)s dζ

≤ Ch(ξ) ≤ C1

where h(ξ) is a continuous function in ξ.Again apply the same process to fn(ξ), and we also have

| fn(ξ)| ≤ C1.Hence fn is a uniformly bounded and equicontinuous

sequence on any compact subset. By Arzela-Ascoli theo-rem and a diagonal argument, there exists convergent sub-sequences of fn (still denoted as fn) uniformly on each com-pact subset.

Now we want to show that fn converges in Ht for s > t.

| f j − fk|2t =∫

Rm| f j − fk|2(1 + |ξ|2)t dξ

We decompose the domain into two parts and have the fol-lowing estimates.

6. ELLIPTIC OPERATORS AND GARDING’S INEQUALITY 125

For |ξ| ≥ r, (1 + |ξ|2)t ≤ (1 + r2)t−s(1 + |ξ|2)s and∫|ξ|≥r| f j − fk|2(1 + |ξ|2)t dξ

≤ (1 + r2)t−s∫

Rm| f j − fk|(1 + |ξ|2)s dξ

≤ 2C(1 + r2)t−s.

For any ε > 0, we can choose r large such that 2C(1+ r2)t−s <

ε/2.On the other hand, |ξ| ≤ r is a compact set. We can

pick j, k large such that∫|ξ|≤r| f j − fk|2(1 + |ξ|2)t dξ < ε/2.

Hence | f j − fk|2t < ε for j, k large enough.(3) For s > t > u, there exists Cε such that

(1 + |ξ|2)t ≤ ε(1 + |ξ|2)s + Cε(1 + |ξ|2)u.

Then we get the interpolation inequality immediately.

6. Elliptic Operators and Garding’s Inequality

Now we consider:

(E, k)

a vector bundle of rank k, e.g. E = Λp(T∗M), k = (mk ).

M

f : a section

XX

compact manifold,

and C∞(M, E) is the vector space of all C∞ sections.Definition 4.15. We say L : C∞(M, E) → C∞(M, E) is a linear differ-ential operator of order d ∈ N, if locally we can regard L as a mapbetween locally bundles:

C∞(U, E∣∣U)

L−→ C∞(U, E∣∣U),

L f = ∑α

Aα(x)Dα f , x ∈ U,


where U ⊂ Rm, E∣∣U∼= U ×Rk, f = ( f 1, · · · , f k) and Aα being k× k

matrices.In particular, we usually write

L = ∑|α|=d

Aα(x)Dα + ∑|α|<d

Aα(x)Dα

where ∑|α|=d Aα 6= 0.

• L is elliptic over U if

p(x, ξ) = ∑|α|=d

Aα(x)ξα (x ∈ U, ξ 6= 0)

is an invertible matrix. p is called the symbol of L.• L is strictly elliptic over S if∣∣∣∣∣∑α

Aα(x)ξα · v∣∣∣∣∣ ≥ C|ξ||α||v|, ∀ x ∈ S, ξ 6= 0

for some C independent of ξ.In particular, ellipticity implies uniform ellipticity over

any compact subsets.

Now we want to reduce to the Euclidean case.Pick a local trivialization (Ui, φi, ψi) where Ui are charts on M, φi

is a P.O.U. and ψ is the trivialization of E∣∣Ui

.

∼Ui ×Rk

ψ

f

MUi

For any f ∈ C∞(M, E), f = ∑i f iσi, i.e. φi( f ) = ( f 1, · · · , f k). Andfor s ∈ R,

| f |s = ∑i|ψi(φi · f )|s.

6. ELLIPTIC OPERATORS AND GARDING’S INEQUALITY 127

Then we can define Hs(M, E) to be the completion of C∞(M, E) withrespect to the norm | · |s.

Denote H0(M, E) = L2(M, E).Remark 4.16.⋃s∈R

Hs =: H−∞ ⊃ · · · ⊃ Hs ⊃s<t· · ·⊃ Hs ⊃ · · · ⊃ H∞ :=

⋂s∈R

Hs = C∞.

Note that ∪sHs = C∞ follows from the Sobolev lemma.

Exercise 4.3.

(1) Show that all | · |s are independent of choice of (Ui, φi, ψi) aslong as φi, ψi (and their inverse) have bounded derivation(i.e. norm equivalence).

(2) All the key lemmas still hold.

To ensure the regularity, we need to show that the weak solu-tion belongs to some Hs for higher order s actually and use the keylemmas. This estimate comes from the following theorem.

Theorem 4.17 (Garding inequality). P is an elliptic operator of order don (M, E). There exists C > 0 such that

| f |s+d ≤ C (|P f |s + | f |s) ,

for all f ∈ Hs+d.

PROOF. Under some local chart (Ui, φi, ψi), we rewrite P = P0 +

P1 + P2 where

P0 = ∑|α|=d

Aα(0)Dα, P1 = ∑|α|<d

Aα(x)Dα, P2 = ∑|α|=d

(Aα(x)−Aα(0))Dα

for some p ∈ Ui, p↔ x = 0.Locally, by uniform ellipticity at p,

|P0 f |2s =∫(1 + |ξ|2)s|P0 f (ξ)|2 dξ

=∫(1 + |ξ|2)s

∣∣∣∣∣∑α=d

Aα(0)ξα f (ξ)

∣∣∣∣∣2

dξ

≥ C′∫(1 + |ξ|2)s|ξ|2d| f (ξ)|2 dξ.


(|P0 f |s + | f |s)2 ≥∫(1 + C′′|ξ|d)2(1 + |ξ|2)s| f (ξ)|2 dξ

≥∫

C(1 + |ξ|2)d(1 + |ξ|2)s| f (ξ)|2 dξ ≥ C21 | f |2s+d.

We can easily get the estimates

|P1 f |s ≤ C2| f |s+d−1

|P2 f |s ≤C1

2| f |s+d + C2| f |s+d−1

by choosing U small enough a priori.

Exercise 4.4. Show that in the proof of this inequality 4.17, there ex-ists open neighborhood U 3 p independent of f such that the aboveestimate for P2 f holds. Notice that P2 f (0) = 0.

Hence

|P f |s + | f |s ≥ |P0 f |s − |P1 f |s − |P2 f |s + | f |s

≥ C1

2| f |s+d − 2C2| f |s+d−1.

⇒ | f |s+d ≤ C(|P f |s + | f |s + | f |s+d−1).

By interpolation inequality, (ε = 1/2C)

| f |s+d−1 ≤1

2C| f |s+d + C(ε)| f |s.

Finally we have12| f |s+d ≤ C|P f |s + C| f |s.

7. Proof of Compactness and Regularity Theorem

Theorem 4.18 (Compactness theorem). For a sequence αn ∈ H, if‖αn‖ ≤ C, ‖Lαn‖ ≤ C are bounded, then αn has a Cauchy subsequence(w.r.t. L2 norm).

PROOF. Since un ∈ C∞ on compact M, un, Lun is uniformly boundedin L2-norm. By Garding inequality,

|un|2 ≤ C(|Lun|0 + |un|0) ≤ C.

7. PROOF OF COMPACTNESS AND REGULARITY THEOREM 129

By Rellich lemma, un ∈ H2 and un has a Cauchy subsequence inH0.

Theorem 4.19 (Regularity theorem). L is an elliptic operator of d ∈ N.If Lu = v, v ∈ Ht, u ∈ H−∞ := ∪s∈RHs, then u ∈ Ht+d.

PROOF. Suppose u ∈ Hs for some s ∈ R. It’s enough to showthat if

Lu ∈ Hs−d+1 ⇒ u ∈ Hs+1.

The key here is to show that the difference quotient

uh(x) :=u(x + h)− u(x)

|h| ∈ Hs

for any vector h.

u(·+ h)(ξ) =∫

e−iξxu(x + h) dx

= eihxi∫

e−i(x+h)ξu(x + h) d(x + h)

= eihξ u(ξ).

⇒ uh(ξ) =eihξ − 1|h| h(ξ) =

(ih · ξ|h| + o(|h|)

)u.

So if uh(x) ∈ Hs, then for h 6= 0 small,

⇒∫(1 + |ξ|2)s|ξ|2|u(ξ)|2 dξ < ∞

since u ∈ Hs.So we get

u ∈ Hs, |uh|s ≤ C ∀ h 6= 0⇐⇒ u ∈ Hs+1.

Now we need to get estimate for |uh|s.

Lemma 4.20.L(uh) = (Lu)h − Lh(Thu)

where Thu(x) = u(x + h) denotes the translation operator, and Lh =

∑α Ahα(x)Dα.


PROOF OF LEMMA.

Aα(x)Dα(u(x + h)− u(x))

= Aα(x + h)Dαu(x + h)− Aα(x)Dαu(x)

− (Aα(x + h)− Aα(x))Dαu(x + h).

Back to the proof. Notice that Lu ∈ Hs−d+1 infers that |(Lu)h|s−d

is uniformly bounded for any small h. And then

|uh|s ≤ C(|L(uh)|s−d + |uh|s−d)

≤ C(|(Lu)h|s−d + |Lh(Thu)|s−d + |uh|s−d)

≤ C(|Lu|s−d+1 + |u|s−d+1).

So |uh|s ≤ C for any small h and hence |u|s+1 ≤ C.

8. Exercises

1. (a) G will commute with any operator T with T∆ = ∆T.(b) G is a bounded, self-adjoint, compact operator.

2. Prove the Bochner formula 4.8.3. (a) Show that all | · |s are independent of choice of (Ui, φi, ψi) as

long as φi, ψi (and their inverse) have bounded derivation (i.e.norm equivalence).

(b) All the key lemmas still hold.4. Show that in the proof of this inequality 4.17, there exists open

neighborhood U 3 p independent of f such that the above esti-mate for P2 f holds. Notice that P2 f (0) = 0.

5. (Warner Ch.6 #6) Derive explicit formulas for d, ∗, δ and ∆ in Eu-clidean space. In particular, show that if

α = ∑i1<···<ip

αIdxi1 ∧ · · · ∧ dxip ,

then

∆α = (−1) ∑i1<···<ip

(n

∑i=1

∂2αI

∂x2i

)dxi1 ∧ · · · ∧ dxip .

8. EXERCISES 131

6. (Warner Ch.6 #7) Let ϕ belong to the C∞ periodic functions P onthe plane. Prove that

∥∥∥∥ ∂2ϕ

∂x∂y

∥∥∥∥ ≤ 12‖∆ϕ‖

7. (Warner Ch.6 #8) The Rellich lemma 4.14 says that the naturalinjection i : Ht → Hs for s < t is a compact operator; that is, ittakes bounded sequences into sequences with convergent subse-quences. An analogous example of this phenomenon is the fol-lowing. Let C denote the Banach space of periodic continuousfunctions on the real line, say with period 2π, and with norm thesup-norm ‖ · ‖∞. Let C1 be the subset of C consisting of functionswith continuous first derivative. As a norm for C1 we take

‖ f ‖ = ‖ f ‖∞ +

∥∥∥∥d fdx

∥∥∥∥∞

.

Use Arzela-Ascoli theorem to prove that the natural injection i :C1 → C is a compact operator.

8. (Warner Ch.6 #9) We shall consider a number of elliptic equationsof the form Lu = f on the real line. In each case, f will be smoothand periodic of period 1, and we look for solution u also periodicof period 1. This restriction to periodic functions makes this inessence a problem on a compact space, the unit circle. We let u′ =du/dx, etc.(a) u′ = f . This is the simplest example of an elliptic operator

which exhibits all of the essential ingredients of the theory.What is the formal adjoint of this differential operator? Showthat there is a solution u (periodic) if and only if f is orthogo-nal to the kernel of this adjoint.

(b) u′ − u = f . What is the kernel (in the periodic functions) inthe case? What are the necessary and sufficient conditions onf for there to exist a periodic solution?

(c) u′′ = f . Show that this operator is formally self-adjoint. Showthat there is a periodic solution if and only if f is orthogonal


to the kernel; and using the fact that∫ x

0

(∫ t

0f (s) ds

)dt =

∫ x

0f (s)

(∫ x

0dt)

ds,

show that the unique solution orthogonal to the kernel is

u(x) =∫ x

0t(x− 1) f (t) dt +

∫ 1

xx(t− 1) f (t) dt

− 12

∫ 1

0t(t− 1) f (t) dt.

This explicitly exhibits the Green’s operator for this case.(d) u′′ + 4π2u = f . Show that this operator is formally self-

adjoint. What is the kernel? Derive an explicit formula forthe solution u, and show that u is periodic if and only if f isorthogonal to the kernel.

9. (Warner Ch.6 #12) Let α and β be n-forms on a compact orientedmanifold Mn such that

∫M α =

∫M β. Prove that α and β differ by

an exact form.10. (Warner Ch.6 #13) Show that the compactness theorem cannot be

strengthened to the assertion of the existence of a subsequencewhich is convergent in Ap(M).

11. (Warner Ch.6 #16) (The Eigenvalues of the Laplacian) This is anextended exercise in which the fundamental properties of the eigen-functions and eigenvalues of the Laplacian are developed. Proofsfor the more difficult parts are outlined, and in some cases aregiven nearly in full.

Consider the Laplace-Beltrami operator ∆ acting on the p-formsAp(M) for some fixed p. A real number λ corresponding to whichthere exists a not identically zero p-form u such that ∆u = λu iscalled an eigenvalue of ∆. If λ is an eigenvalue, then any p-form usuch that ∆u = λu is called an eigenfunction of ∆ corresponding tothe eigenvalue λ. The eigenfunctions corresponding to a fixed λ

form a subspace of Ap(M) called the eigenspace of the eigenvalueλ.(a) Prove that the eigenvalues of ∆ are non-negative.(b) Prove that the eigenspaces of ∆ are finite dimensional.

8. EXERCISES 133

(c) Prove that the eigenvalues have no finite accumulation point.(d) Prove that eigenfunctions corresponding to distinct eigenval-

ues are orthogonal.(e) (Existence) In order for the above statements to have sub-

stance we must prove that there exists eigenvalues of ∆. Firstof all, zero is an eigenvalue if and only if there are non-trivialharmonic p-forms on M, and the corresponding eigenspaceis precisely the space Hp of harmonic forms. We shall nowestablish that ∆ has a positive eigenvalue – in fact, a wholesequence of eigenvalues diverging to +∞. Consider ∆ to berestricted to (Hp)⊥. Then we have ∆ : (Hp)⊥ → (Hp)⊥,and also we have the Green’s operator G : (Hp)⊥ → (Hp)⊥,and ∆Gα = α, G∆α = α for all α ∈ (Hp)⊥. Observe that theeigenvalues of G

∣∣(Hp)⊥ are the reciprocals of the eigenvaluesof ∆

∣∣(Hp)⊥. Let

η = sup‖ϕ‖=1,ϕ∈(Hp)⊥

‖Gϕ‖.

Then η > 0 and ‖Gϕ‖ ≤ η‖ϕ‖ for every ϕ ∈ (Hp)⊥. We shallprove that 1/η is an eigenvalue of ∆. Let ϕi ∈ (Hp)⊥ be amaximizing sequence for η; that is, ‖ϕj‖ = 1 and ‖Gϕj‖ → η.First, we claim that ‖G2ϕj − η2ϕj‖ → 0, for

‖G2ϕj − η2ϕj‖2 =‖G2ϕj‖2 − 2η2〈G2ϕj, ϕj〉+ η4

≤η2‖Gϕj‖2 − 2η2‖Gϕj‖2 + η4 → 0.

Second, we claim that ‖Gϕj − ηϕj‖ → 0. For if we let ψj =

Gϕj − ηϕj, then

0← 〈ψj, G2ϕj − η2ϕj〉=〈ψj, Gψj + ηψj〉

=〈ψj, Gψj〉+ η‖ψj‖2 ≥ η‖ψj‖2,

where we have used the fact that 〈ψj, Gψj〉 ≥ 0 (check why?)Now there is a subsequence of the ϕj, call it ϕj, such thatGϕj is Cauchy. Define a linear functional ` on Ap(M) by


setting

`(β) = limj→∞

η〈Gϕj, β〉, β ∈ Ap(M).

Prove that ` is a non-trivial weak solution of

(∆− 1/η)u = 0.

From this and the fact that ∆− 1/η is elliptic, conclude thatλ = 1/η is an eigenvalue of ∆.

(f) (Existence of Other Eigenvalues) Suppose that we have eigen-values λ1 ≤ λ2 ≤ · · · ≤ λn and corresponding eigenfunc-tions u1, u2, . . . , un (orthonormalized) for ∆

∣∣(Hp)⊥. Let Rn bethe subspace of (Hp)⊥ spanned by u1, . . . , un. Observe thatG and ∆ map (Hp ⊕ Rn)⊥ into itself, then define

ηn+1 = sup‖ϕ‖=1, ϕ∈(Hp⊕Rn)⊥

‖Gϕ‖

and proceed as in part (e) to establish that λn+1 = 1/ηn+1 isan eigenvalue of ∆. Clearly λn+1 ≥ λn.

(g) (L2 Completeness) Let λ1 ≤ λ2 ≤ · · · be the eigenvaluesof ∆ on Ap(M), where each eigenvalue is included as manytimes as the dimension of its eigenspace, with a correspond-ing orthonormalized sequence of eigenfunctions ui. Letα ∈ Ap(M). Then

limn→∞

∥∥∥∥∥α−n

∑i=1〈α, ui〉ui

∥∥∥∥∥ = 0.

For the proof, let k be the dimension of Hp. Then there existsβ ∈ (Hp)⊥ such that Gβ = α−∑n

i=1〈α, ui〉ui. It follows that∥∥∥∥∥α−n

∑i=1〈α, ui〉

∥∥∥∥∥ =

∥∥∥∥∥G

(β−

n

∑i=k+1

〈β, ui〉βi

)∥∥∥∥∥for n > k. But, by the definition of λn+1,∥∥∥∥∥G

(β−

n

∑i=k+1

〈β, ui〉ui

)∥∥∥∥∥ ≤ 1λn+1

∥∥∥∥∥β−n

∑i=k+1

〈β, ui〉ui

∥∥∥∥∥≤ 1

λn+1‖β‖ → 0, as n→ ∞.

8. EXERCISES 135

(h) (Uniform Completeness) The uniform norm ‖α‖∞ is definedon Ap(M) by

‖α‖∞ = supm∈M

(∗(α ∧ ∗α)(m))1/2 .

From the Sobolev inequality 4.14 and the compactness of M,one can conclude that there exists a large enough integer kand a constant c > 0 such that

‖α‖∞ ≤ c‖(1 + ∆)kα‖

for every α ∈ Ap(M).Let α ∈ Ap(M), and let Pn(α) = ∑n

i=1〈α, ui〉ui, where we arecontinuing with the notation of (g). Now ∆Pn = Pn∆, so that

‖α− Pn(α)‖∞ ≤c‖(1 + ∆)k[α− Pn(α)]‖=‖ϕ− Pn ϕ‖ → 0,

where ϕ = (1 + ∆)kα.12. (Warner Ch.6 #17) We define the operator ∆2 : Ap(M) → Ap(M)

by ∆2α = ∆(∆α). Discuss the solvability of ∆2α = β.13. (Warner Ch.6 #18) Consider the operator

L =n

∑i,j=1

aij∂2

∂xi∂xj+

n

∑i=1

bi∂

∂xi+ c

acting on C2(Rn). Show that there is no loss of generality in as-suming that aij = aji, and prove that L is elliptic at a point x if andonly if the matrix (aij(x)) is positive (or negative) definite.

In particular, show that the wave equation

u =∂2u∂x2 −

∂2u∂y2 = f

is not elliptic, and give an example where

u = f ∈ C∞, but u /∈ C∞.

14. (Warner Ch.6 #19) Consider ∆ : Ap(M)→ Ap(M). Prove that if λ

is the minimum eigenvalue of ∆, and if c > −λ, then (∆+ c)α = β

can be solved for every β ∈ Ap(M).


15. (Warner Ch.6 #22) Let ϕn, for n = 1, 2, . . . be a periodic C∞ func-tion on the plane which agrees with log log( 1

r+1/n ) for 0 ≤ r ≤ 12 ,

where r =√

x2 + y2. Show that there is no constant c > 0 suchthat

‖ϕn‖∞ ≤ c‖ϕn‖1, for all n.

This shows, in the case n = 2, that the restriction t ≥ [n/2] + 1(s > k + m/2 in this lecture note) in the Sobolev lemma 4.14 isessential.

Chapter 5

BASIC LIE THEORY

1. Categories of Lie groups and Lie algebras

A C∞ manifold G is a Lie group if G has a group structure and thegroup law

G× G → G; (g, h) 7→ gh−1

is C∞. For g ∈ G, we denote the left multiplication map h 7→ gh byLg and right multiplication h 7→ hg by Rg. We have the induced mapon tangent spaces:

dLg : ThG → TghG; dRg : ThG → ThgG.

A vector field X ∈ C∞(TG) is left invariant if Xgh = dLgXh for allg, h ∈ G. The Lie algebra g = Lie G of G is the vector space of all leftinvariant vector fields (l.i.v.f.) under bracket operation. Namely, asdifferential operators, for f ∈ C∞(G):

[X, Y] f := X(Y f )−Y(X f ).

Since a l.i.v.f. X is determined by its value Xe at the identity e ∈ G,we identify

g ∼= TeG.

Abstractly, a vector space L over a field F with an F-bilinear map[ , ] : L× L→ L is called a Lie algebra (over F) if [x, y] = −[y, x] and

[[x, y], z] + [[y, z], x] + [[z, x], y] = 0. (Jacobi identity).

It is clear that the bracket of vector fields has this property.

Example 5.1. Consider the general linear group

G = GL(n, R) = g ∈ Mn×n(R) | det g 6= 0 .137

138 5. BASIC LIE THEORY

From Cramer’s rule, we see that g 7→ g−1 is C∞ hence that G is a Liegroup.

As an open subset of Mn×n(R) ∼= Rn2, we have TeG = Mn×n(R).

The matrix algebra has a natural Lie algebra structure gl(n, R) de-fined by

[A, B] := AB− BA.

Theorem 5.2. gl(n, R) coincides with Lie G.

PROOF. From (gh(t))′ = gh′(t), we see that (Lg)∗A = gA forg ∈ G, A ∈ TeG. Thus if A is the l.i.v.f. with Ae = A, then Ag =

gA. Let G → Rn2with coordinates (xij)

ni,j=1 being the entries of the

corresponding matrix g. Then a tangent vector A = (aij) ∈ TeG andA are equivalent to

A = ∑i,j

aij∂

∂xij

∣∣∣∣e

and Ag = ∑i,j(gA)ij

∂

∂xij

∣∣∣∣g

respectively. From

∑m

∂

∂xij(xkmbml) = ∑

mδkiδmjbml = δkibjl,

we compute

[A, B]e = ∑i,j,k,l

(aij

∂

∂xij

((gB)kl

) ∂

∂xkl− bij

∂

∂xij

((gA)kl

) ∂

∂xkl

)∣∣∣∣g=e

= ∑i,j,l

aijbjl∂

∂xil

∣∣∣∣e− bijajl

∂

∂xil

∣∣∣∣e= ∑

i,l(AB− BA)il

∂

∂xil

∣∣∣∣e.

This corresponds to AB− BA precisely.

A Lie subgroup H < G is itself a Lie group such that H is both asubgroup and an immersion. We allow H ⊂ G to be not closed.Example 5.3. Subgroups of matrix groups are the main sources of Liegroups.

(i) Let SL(n, R) = g ∈ GL(n, R) | det g = 1 . Consider asmooth curve t 7→ g(t) with g(0) = e and det g(t) = 1.Then we compute tr g′(0) = 0. So its Lie algebra is given bysl(n, R) = A ∈ Mn×n(R) | tr A = 0 .

1. CATEGORIES OF LIE GROUPS AND LIE ALGEBRAS 139

(ii) Let O(n, R) = g ∈ GL(n, R) | gTg = e . Consider a smoothcurve t 7→ g(t) with g(0) = e and g(t)Tg(t) = e. Then wecompute g′(0)T + g′(0) = 0. So its Lie algebra is given byo(n, R) = A ∈ Mn×n(R) | AT = −A .

(iii) Let SO(n, R) = g ∈ O(n, R) | det g = 1 . It is clear thatO(n, R) has two connected components and SO(n, R) is theidentity component, so so(n, R) = o(n, R).

(iv) Sp(2n, R) = g ∈ M2n×2n(R) | gT Jg = J , where

J =

(0 In

−In 0

).

Its Lie algebra sp(2n, R) is given by A ∈ M2n×2n(R) | AT J =−JA .

(v) We have similar Lie groups GL(n, C), SL(n, C), O(n, C), SO(n, C)

and Sp(n, C). Indeed they are defined by algebraic equa-tions with integer coefficient, so they can take values in anyfield. The corresponding Lie algebras gl(n, C), sl(n, C), so(n, C)

and sp(n, C) are complex Lie algebras.(vi) Let U(n) = g ∈ GL(n, C) | g∗g = e . Consider a smooth

curve t 7→ g(t) with g(0) = e and g(t)∗g(t) = e. Then wecompute g′(0)∗ + g′(0) = 0. So its Lie algebra is given byu(n) = A ∈ Mn×n(C) | A∗ = −A . Notice that u(n) is areal Lie algebra.

(vii) SU(n) = g ∈ U(n) | det g = 1 . su(n) = sl(n, C) ∩ u(n).

All these subgroups can be realized as the subgroup preservingcertain additional structure. For “S”, g preserves volume. For “O”,g preserves the Euclidean inner product. For “Sp”, g preserves thenon-degenerate symplectic form

xT Jy = (x1yn+1 − xn+1y1) + · · ·+ (xny2n − x2nyn).

And for “U”, g preserves the Hermitian inner product.sl(n, C), so(2n, C), sp(n, C) and so(2n + 1, C) are known as classi-

cal complex semi-simple Lie algebras of type An, Bn, Cn and Dn respec-tively. (To be explained later).


Theorem 5.4. Given a Lie group G. There is an one to one correspondencebetween connected Lie subgroups of G and Lie subalgebras of g.

PROOF. This follows from the Frobenius Theorem in differentialgeometry. For a basis Xi of a Lie subalgebra h of g, we defined asubspace distributionHg which is spanned by Xig for all g ∈ G. Thedistribution H =

⋃g∈GHg is integrable. Indeed, for any two C∞

vector fields V = ∑ fiXi and W = ∑ giXi, we compute

[V, W] = ∑ figj[Xi, Xj] + ∑ fi(Xigj)Xj −∑ gj(Xj fi)Xi ∈ H.

We then take H to be the maximal integral submanifold passingthrough e ∈ G.

To check that H is a group, let g ∈ H. The map Lg maps themanifold H to gH. The left invariance says that dLgHh = Hgh, hencegH is also an integral submanifold. Now H and gH both contain theelement g, hence the maximality (uniqueness) implies that H = gH.This implies that H is closed under multiplication and also g−1 ∈ H(since e ∈ H). So H is a subgroup of G.

Finally, H is a Lie groups simply because the map H × H → Hsending (g, h) to gh−1 is the restriction of the given C∞ map G×G →G.

Remark 5.5. For any Lie group G, the tangent bundle TG is a trivialvector bundle with global frame given by any basis of g.

More generally, a Lie group homomorphism ρ : G → H is a C∞ mapwhich is also a group homomorphism. The tangent map dρ : TG →TH is compatible with l.i.v.f.’s. To see this, ρ(gg′) = ρ(g)ρ(g′) meansρ Lg = Lρ(g) ρ, so

dρ dLg = dLρ(g) dρ.

Thus dρ : g → h. dρ is indeed a Lie algebra homomorphism in thesense that

dρ[X, Y] = [dρ(X), dρ(Y)],

which is easily verified from the definitions.

2. EXPONENTIAL MAP 141

2. Exponential map

We call a nontrivial Lie group homomorphism R → G a one pa-rameter subgroup, even though it may not be injective. The exponentialmap links Lie algebras with Lie groups through the consideration ofall one parameter subgroups. Before treating the abstract setting, welook at the case for matrix groups.

Example 5.6. For A ∈ Mn×n(C), t ∈ C, we define the absolutelyconvergent series

etA = 1 + tA +t2

2!A2 + · · ·+ tk

k!Ak + · · · .

It is easily checked that if AB = BA then eAeB = eA+B. Hence eA hasinverse e−A and so eA ∈ GL(n, C). Moreover γ(t) = etA is the oneparameter subgroup with

γ′(t) = etA A = dLγ(t)A = Aγ(t).

That is, etA is the integral curve of the l.i.v.f. determined by A ∈gl(n, C).

The discussion works for C being replaces by R. Also if we takeA be in a Lie subalgebra, the eA lies in the corresponding Lie sub-group. This follows from the previous theorem. But we can also seehow it works explicitly: For example,

tr A = 0 =⇒ det eA = etr A = 1.

Also

A∗ = −A =⇒ (eA)∗eA = eA∗eA = e−AeA = In.

Now we turn to a general Lie group G. Let X ∈ g. Since RX < g

is a one dimensional Lie subalgebra, by the previous theorem its inte-gral curve is a one dimensional subgroup H. By taking the universalcover R → H if necessary, we get a one parameter subgroup whichwe denote by t 7→ exp tX. We shall give a direct proof of this withstronger conclusions.


Let φt be the flow generated by X. That is, φt(g) is the curve withφ0(g) = g and

ddt

φt(g) = Xφt(g).

Theorem 5.7. The range of t is R for all g ∈ G. Moreover, φt : G → G is aone-parameter group of diffeomorphisms as right translations φt = Rφt(e).

PROOF. Consider the curve gφt(e). Since gφ0(e) = g and

ddt(

gφt(e))= dLg

(dLφt(e)Xe

)= dLgφt(e)Xe

= Xgφt(e),

we conclude that φt(g) = gφt(e) = Rφt(e)g.By substituting g = φs(e) we find φs(e)φt(e) = φt(φs(e)) =

φt+s(e). This shows that for g = e, the range of t can be extended toall R and φt(e) is a one parameter subgroup. The theorem is provedby using the relation φt(g) = gφt(e) again.

Now we define the exponential map

exp : g→ G

by exp tX = φt(e) where φt is the flow generated by X. Since

(d exp)0(X) =ddt

∣∣∣∣t=0

exp tX = X,

we get (d exp)0 = Idg and exp is invertible near 0 ∈ g.

Corollary 5.8. If H < G is a Lie subgroup, then H is generated by exp h.

However, exp is not necessarily surjective, hence exp g is not nec-essarily a group.

Exercise 5.1. Let X ∈ sl(2, R) and d =√|det X|. Then

(i) eX = (cosh d)I2 +1d (sinh d)X if det X < 0.

(ii) eX = (cos d)I2 +1d (sin d)X if det X > 0.

(iii) eX = I2 + X if det X = 0.

3. ADJOINT REPRESENTATION 143

Let ga =

(a 00 a−1

)∈ SL(2, R). Then ga lies in a unique one pa-

rameter subgroup if a > 0. ga lies in infinitely many one parametersubgroup if a = −1. If a 6= −1 and a < 0, then ga 6∈ exp sl(2, R).

3. Adjoint representation

3.0.1. Three adjoints Ig, Adg and adX. For g ∈ G, let Ig : G → G bethe inner automorphism Ig(h) = LgRg−1(h) = Rg−1 Lg(h) = ghg−1.Since Ig(e) = e, we get its differential

Adg := dIg : g→ g

as a Lie algebra automorphism. From dIgg′ = d(Ig Ig′) = dIg dIg′ ,we get the adjoint representations

Ad : G → Aut g

and

ad := d(Ad) : g→ End g.

For G a matrix group, g is a matrix Lie algebra and it is clear thatAdg(Y) = gYg−1. For g(t) a curve with g(0) = e and g′(0) = X wethen compute

adX(Y) = (g(t)Yg(t)−1)′(0) = XY−YX = [X, Y].

This property holds true in general:

Theorem 5.9. For X, Y ∈ g,

adXY = [X, Y].


PROOF. Let f ∈ C∞(G) and φ, ψ be the flows generated by X, Y.Then

(adXY) f =ddt

∣∣∣∣t=0

(Adexp tXY) f

=ddt

∣∣∣∣t=0

dds

∣∣∣∣s=0

f (Iexp tX(exp sY))

=dds

∣∣∣∣s=0

ddt

∣∣∣∣t=0

f (exp tX · exp sY · exp(−tX))

=dds

ddt(

f φ−t ψs φt(e))(0, 0)

=dds

d f (−Xψs(e)) + d( f ψs)Xe

∣∣∣∣s=0

= − dds

∣∣∣∣s=0

Xψs(e) f + Xe

(dds

∣∣∣∣s=0

f ψs

)= − d

ds

∣∣∣∣s=0

(X f ) ψs(e) + XeY f

= −YeX f + XeY f = [X, Y]e f .

Remark 5.10. Readers with experience in differential geometry mayobserve that the proof is identical with the one for Lie derivative LXY =

[X, Y]. Indeed,

adXY =ddt

∣∣∣∣t=0

(Adexp tXY)

=ddt

∣∣∣∣t=0

dRexp(−tX)dLexp tXY

=ddt

∣∣∣∣t=0

dφ−tY = LXY

by the left invariance of Y and the definition of LXY.

It is harder to get explicit formula for Adg in the abstract setting.We have such a formula in two special cases, both are based on the

3. ADJOINT REPRESENTATION 145

commutative diagram

Gρ//

OO

exp

HOOexp

gdρ// h

To see this, simply notice that ρ exp tX and exp dρ(tX) are both oneparameter subgroups in H with the same tangent vector dρ(X) att = 0.

By applying the diagram to ρ = Ig, we get:

exp(AdgX) = g(exp X)g−1.

(For matrix groups this is obvious).By applying the diagram to H = Aut g, ρ = Ad and g = exp X,

we get

Adexp XY = eadXY.

With these preparation, we give some applications of the adjointrepresentation:

3.0.2. Center of a Lie group. A Lie algebra is called abelian if [X, Y] =0 for all X, Y.

Proposition 5.11. Let G be a connected Lie group, then Center(G) =

Ker Ad. In particular, G is abelian if and only if g is abelian.

PROOF. If g is in the center, then for all t ∈ R and X ∈ g,

exp tX = g(exp tX)g−1 = exp Adg tX = exp tAdg X.

Hence X = Adg X for all X. That is, Adg = idg.Conversely, g ∈ Ker Ad implies that exp X = g(exp X)g−1. Hence

g commutes with all elements in a neighborhood of e in G. By theconnectedness of G we conclude that g commutes with every ele-ments in G.

Corollary 5.12. [X, Y] = 0 implies that exp X · exp Y = exp(X + Y).


PROOF. Let h be the two dimensional abelian Lie subalgebra of gspanned by X and Y. Consider the Lie group H generated by exp h.The proposition show that H is abelian and so the curve γ(t) =

exp tX · exp tY is an one parameter subgroup. Since γ′(0) = X + Y,we conclude that exp tX · exp tY = exp t(X + Y).

Corollary 5.13. If G is a connected Lie groups with trivial center, then

Ad : G → Aut g = GL(g)

is a faithful representation. In particular, G is a matrix subgroup.

3.0.3. Normal Lie subgroups. A subspace h of g is a Lie ideal if[h, g] ⊂ h. In this case we denote by hC g. It is clear that h is at leasta subalgebra.

Proposition 5.14. Let H < G be a connected Lie subgroup of a connectedLie group. Then

HC G ⇐⇒ h := Lie H C g.

PROOF. Let g = exp X with X ∈ g and Y ∈ h,If h is a Lie ideal of g, then

g(exp Y)g−1 = exp Adg Y

= exp(eadXY)

= exp[(

I + adX +12!

ad2X + · · ·

)Y]

∈ exp h ⊂ H.

Since H is generated by h, this proves that H is normal.Conversely, if H is normal, then the above computation shows

that

γ(t) := exp(eadtXY) ∈ H.

Hence h 3 γ′(0) = adXY = [X, Y] and h is a Lie ideal.

3.1. Fundamental correspondences.

4. DIFFERENTIAL GEOMETRY ON LIE GROUPS 147

3.1.1. Equivalence of categories.

Theorem 5.15. Let G and H be connected Lie groups with Lie algebras gand h, If G is simply connected, then there is a one to one correspondence be-tween Lie group homomorphisms G → H and Lie algebra homomorphismsg→ h.

IDEA OF PROOF. The is proved by exploring the Frobenius theo-rem on the product group G×H in a manner similar to the subgroupcase.

Indeed a morphism ρ : G → H is equivalent to a subgroupΓ ⊂ G × H (graph of ρ) such that πG : G × H → G maps Γ ontoG bijectively.

The given map g → h gives rise to a Lie subalgebra of g⊕ h andby the subgroup case we have proved, the corresponding Lie sub-group exists. The remaining problem is to prove the bijectivity of Γonto G when G is simply connected.

Exercise 5.2. Complete the remaining problem of Theorem 5.15.

3.1.2. Ado’s imbedding theorem.

Theorem 5.16. Every (finite dimensional) Lie algebra can be regarded as aLie subalgebra of some gl(n, R). Hence every simpley connected Lie groupis a subgroup of GL(n, R). Moreover, every compact Lie group can beimbedded as a closed subgroup of some O(n, R).

For a proof, see Bourbaki.

4. Differential geometry on Lie groups

4.1. Levi-Civita connection. Any inner product 〈 , 〉e on TeG = g

uniquely determined a left invariant (Riemannian) metric on G byleft translations. Namely for v, w ∈ TgG,

〈v, w〉g := 〈dLg−1v, dLg−1w〉e.

A bi-invariant metric is a metric which is both left and right in-variant. We will shortly determine all Lie groups which admit bi-invariant metrics.


Proposition 5.17. (i) For any left invariant metric 〈 , 〉 on G, and X, Y ∈g, the Levi-Civita connection is given by

∇XY = 12([X, Y]− ad∗XY− ad∗YX).

(ii) If 〈 , 〉 is bi-invariant, then 〈adZX, Y〉+ 〈X, adZY〉 = 0 for X, Y, Z ∈g. In particular, ∇XY = 1

2 [X, Y].Moreover, R(X, Y)Z = −1

4 [[X, Y], Z] and R(X, Y, X, Y) = 14 |[X, Y]|2 ≥

0.

PROOF. Recall that the Levi-Civita connection is the unique firstorder differential operator ∇X : C∞(TM) → C∞(TM) with ∇XY −∇YX = [X, Y] (torsion free) and X 〈Y, Z〉 = 〈∇XY, Z〉 + 〈X,∇YZ〉(metrical). For any three vector fields X, Y, Z ∈ C∞(TM), a cycliccomputation leads to

2 〈∇XY, Z〉 = X 〈Y, Z〉+ Y 〈Z, X〉 − Z 〈X, Y〉− 〈Z, [Y, X]〉 − 〈Y, [X, Z]〉 − 〈X, [Y, Z]〉 .

If X, Y, Z ∈ g, all the inner products are constant in G. This leads to(i).

For (ii), the bi-invariance implies in particular that for X, Y, Z ∈ g,⟨Adexp tZX, Adexp tZY

⟩= 〈X, Y〉 .

Take differentiation at t = 0 leads to 〈adZX, Y〉+ 〈X, adZY〉 = 0. Inthe above formula, only the term − 〈Z, [Y, X]〉 is left, hence ∇XY =12 [X, Y].

By the definition of the Riemann curvature operator,

R(X, Y)Z = ∇X∇YZ−∇Y∇XZ−∇[X,Y]Z

= 14 [X, [Y, Z]]− 1

4 [Y, [X, Z]]− 12 [[X, Y], Z] = −1

4 [[X, Y], Z],

where the Jacobi identity is used to rewrite the second term. Finally,

R(X, Y, Z, W) := 〈R(X, Y)W, Z〉 = −14〈[[X, Y], W], Z〉

= 14〈[W, [X, Y]], Z〉 = −1

4〈[X, Y], [W, Z]〉 = 14〈[X, Y], [Z, W]〉,

where the adW invariance of 〈 , 〉 is used.

It is also straightforward to deduce from (i):

4. DIFFERENTIAL GEOMETRY ON LIE GROUPS 149

Corollary 5.18. For left invariant metrics,

R(X, Y, X, Y) = |ad∗XY + ad∗YX|2 − 〈ad∗XX, ad∗YY〉

− 34 |[X, Y]|2 − 1

2〈[[X, Y], Y], X〉 − 12〈[[Y, X], X], Y〉.

Exercise 5.3. Show the Corollary 5.18 by Proposition 5.17(i).

4.1.1. Lie groups with bi-invariant metrics.

Theorem 5.19. A connected Lie group G with a bi-invariant metric iscomplete, the exponential map is surjective and its one parameter subgroupscoincides with geodesics through e ∈ G.

PROOF. By Proposition 5.17, for any l.i.v.f. X, ∇XX = 12 [X, X] =

0. Hence one parameter subgroups are the same as geodesics throughe ∈ G. This implies that geodesics through e can be extended infin-itely, so G is complete by the Hopf-Rinow theorem. In particular, thetwo exponential maps exp and expe (in Riemannian geometry) coin-cide and are surjective.

Corollary 5.20. If G has a bi-invariant metric, then any Lie group immer-sion H → G is totally geodesic.

Corollary 5.21. There is no bi-invariant metrics on SL(2, R).

Exercise 5.4. When G is compact, the bi-invariant metrics alwaysexist. For example, for G ⊂ O(n, R) ⊂ Sn2−1(

√n), the Euclidean

metric 〈A, B 〉 = tr ABT is bi-invariant.

Example 5.22. The Euclidean metric on Rn is clearly bi-invariant.These examples turns out to be basically all the examples:

Theorem 5.23. A simply connected Lie group G which admits a bi-invariantmetric takes the form G = Rn × H for H compact and n ∈ Z≥0.

PROOF. Let zC g be the center, which is clearly an ideal. Thenh := z⊥ < g is also an ideal: For a ∈ z⊥, b ∈ g, and c ∈ z,

〈[b, a], c〉 = −〈a, [b, c]〉 = 0 =⇒ [b, a] ∈ z⊥.


(This holds true for any ideal z.) Since G is simply connected, thedecomposition g = z⊕ h leads to G = Z × H with Lie Z = z andLie H = h.

The center ZCG is simply connected and abelian, hence Z ∼= Rn

for some n. Let e1, . . . , eh ∈ h be an orthonormal basis. For any X ∈ h,the group H with the induced bi-invariant metric has Ricci curvature

Ric(X, X) = 14

h

∑i=1|[X, ei]|2 > 0.

By translation, this show that the Ricci curvature has a positive lowerbound on H. Hence by the theorem of Bonnet-Meyer H must becompact.

5. Homogeneous spaces

5.1. General homogeneous spaces. Let H < G be a closed Liesubgroup. Then the coset space G/H = gH | g ∈ G has a naturalC∞ manifold structure such that the projection map π : G → G/His C∞. G acts transitively on G/H by left translations. Also the sta-bilizer (also called isotropy subgroup) G[gH]

∼= H at each point [gH].Conversely, given a transitive C∞ action G×M→ M on a C∞ mani-fold M. Let H = Gm0 for some m0 ∈ M. Then G/H ∼= M. A space ofthe form G/H is called a homogeneous space. If HCG then G/H isa also Lie group.Example 5.24. Here are some standard examples:

(i) O(n) × Sn−1 → Sn−1 is transitive and O(n)en∼= O(n − 1).

So Sn−1 ∼= O(n)/O(n− 1).(ii) U(n)× S2n−1 → S2n−1 is transitive and U(n)en

∼= U(n− 1).So S2n−1 ∼= U(n)/U(n− 1). Similarly, S2n−1 ∼= SU(n)/SU(n−1). In particular, S1 ∼= U(1) and S3 ∼= SU(2) are Lie groups.

(iii) Real projective space: RPn−1 = Sn−1/±1. So

RPn−1 ∼= O(n)/O(n− 1)× ±1 ∼= SO(n)/O(n− 1).

(iv) Complex projective space: CPn−1 = (Cn\0)/C×. So

CPn−1 ∼= S2n−1/S1 ∼= U(n)/U(n− 1)×U(1) ∼= SU(n)/U(n− 1).

5. HOMOGENEOUS SPACES 151

(v) Stiefel manifold of k-frames: GL(n, R) × Vn,k → Vn,k is tran-sitive where Vn,k is the set of all k frames in Rn. For S =

e1, . . . , ek,

GS =

(I A0 B

)∈ GL(n, R)

.

So Vn,k∼= GL(n, R)/GS. For Vn,k the set of all orthonormal

k-frames,

Vn,k∼= O(n)/O(n− k) ∼= SO(n)/SO(n− k).

For complex Stiefel manifold VCn,k of k-frames in Cn,

VCn,k∼= U(n)/U(n− k) ∼= SU(n)/SU(n− k).

(vi) Grassmannian manifolds: Let Gn,k be the set of all k-dimensionalsubspaces in Rn, then Gn,k

∼= Vn.k/O(k) ∼= O(n)/O(n− k)×O(k) and dim Gn,k = k(n − k). Similarly for the complexGrassmannian

GCn,k∼= VC

n.k/U(k) ∼= U(n)/U(n− k)×U(k).

It is a complex manifold with dimC GCn,k = k(n− k). Grass-

mannians generalizes projective spaces. They are very im-portant for the study of vector bundles.

(vii) Poincare’s upper half plane: Let H = z ∈ C | Im z > 0 .SL(2, R) acts on H transitively by(

a bc d

)z =

az + bcz + d

,

The stabilizer at i is SO(2, R), so H ∼= SL(2, R)/SO(2, R). His non-compact and analytically isomorphic to the unit disk,an example of the bounded symmetric domains. The doublecoset space

Γ\H ∼= Γ\SL(2, R)/SO(2, R)


with Γ < SL(2, R) contains all Riemann surfaces of genusg ≥ 2 (uniformization theorem). If Γ < SL(2, Z) is an arith-metic subgroup, then it represents certain moduli spaces ofelliptic curves.

5.2. Riemannian homogeneous spaces. For further study, weneed notions and results from differential geometry. Let (M, ds2)

be a Riemannian manifold. That is, ds2 is a family of inner products〈 , 〉x on Tx M varying smoothly in x ∈ M. An isometry g : M→ M isa C∞ map such that g∗ds2 = ds2. Equivalently, 〈dg(v), dg(w)〉g(x) =

〈v, w〉x for all v, w ∈ Tx M. It is known that the full isometry group

G ≡ O(M, ds2) := g ∈ C∞(M, M) | g∗ds2 = ds2

is a Lie group. For each x ∈ M, Gx induces a linear representationρ : Gx → O(Tx M). Since an isometry maps geodesics to geodesics,ρ(h) determines h through the geodesic exponential map expx : U ⊂Tx M → M and thus ρ is injective. In particular, each isotropy groupGx is compact.

A connected Riemannian manifold (M, ds2) is Riemannian homo-geneous if for any two points x, y ∈ M, there exists an isometry g suchthat g(x) = y. In this case, we have a transitive action G×M → Mand M ∼= G/Gx. In particular, M is homogeneous with compactisotropy.

Proposition 5.25. A Riemannian homogeneous space is complete.

Exercise 5.5. Prove the Proposition 5.25

A natural question arises: When is a general homogeneous spaceM ∼= G/H Riemannian homogeneous? That is we are searching for met-rics on G/H such that G acts on it as isometries. Such a metric is calleda G-invariant metric, which may not always exist. Also there couldbe different ways to represent M as a group quotient. Thus we needto clarify these issues first.

5. HOMOGENEOUS SPACES 153

In considering the homogeneous structure we may assume thatG acts on G/H effectively in the sense that any g ∈ G\e acts non-trivially. Indeed,

g[kH] = [kH]⇐⇒ k−1gk ∈ H ⇐⇒ g ∈ kHk−1.

Hence g acts trivially if and only if g ∈ ⋂k∈G kHk−1 =: H0. It is clearthat H0 is the largest subgroup of H with H0C G. Thus

G/H ∼=G/H0

H/H0=: G1/H1

has an effective G1 action.Denote G → G/H by g 7→ g := gH. There is a natural identi-

fication Te G/H = g/h. Since AdH and adh act on g and leave thesubspace h invariant, we get the natural adjoint actions on g/h in-duced from π : g→ g/h.

Lemma 5.26. For h ∈ H, dLh ≡ Adh modulo h on Te G/H.

PROOF. Differentiate the equation h exp(tX)H = h exp(tX)h−1H.

Proposition 5.27. A G-invariant metric on the homogeneous space M =

G/H is equivalent to an inner product 〈 , 〉 on g/h ∼= TeM which isAdH-invariant. If H is connected, this is equivalent to “adh-invariance”:Namely, for A ∈ h, X, Y ∈ g/h,

〈adAX, Y〉+ 〈X, adAY〉 = 0.

PROOF. The necessity of AdH invariance on 〈 , 〉 follows from theabove lemma. To see its sufficiency, we simply define for v, w ∈Tg G/H

〈v, w〉g := 〈dLg−1v, dLg−1w〉.Then 〈v, w〉gh = 〈dLh−1dLg−1v, dLh−1dLg−1w〉 = 〈dLg−1v, dLg−1w〉 =〈v, w〉g. Hence the left invariant metric on G/H is well defined.

The remaining statement on adh is left as an exercise.

Exercise 5.6. Show the remaining statement on adh in Proposition 5.27.s


Theorem 5.28. Assume that G acts on M = G/H effectively. Then Madmits a G invariant metric if and only if AdH ⊂ GL(g) has compactclosure.

Moreover, G invariant metrics on G/H are precisely left invariant met-rics on G which is also H bi-invariant.

PROOF. (⇒) Write G/H = G∗/H∗ with G∗ = O(M, ds2), H∗ =G∗e . Then G → G∗, and hence g → g∗, is injective. We know thatim AdH∗ ⊂ GL(g∗) is compact since H∗ is. To realize it inside the or-thogonal group we simply pick an arbitrary inner product on g∗ andaverage it by this compact image so that the resulting inner product〈 , 〉∗ on g∗ is AdH∗-invariant. (This is the same procedure to con-struct bi-invariant metrics on a compact Lie group.) Let 〈 , 〉 = 〈 , 〉|∗g.Then it is clear that the image AdH ⊂ O(g, 〈 , 〉).

(⇐) If AdH has compact closure K ⊂ GL(g), starting with any in-ner product on g the averaging procedure over K again producesan AdH-invariant inner product 〈 , 〉 on g. Let p := h⊥ which isisomorphic onto g/h under π. It is clear that AdH(p) ⊂ p since〈AdH(h

⊥), h〉 = 〈h⊥, AdHh〉 = 0. Thus 〈 , 〉|p defines the desiredAdH-invariant inner product on g/h.

6. Symmetric spaces

6.1. Local and global symmetric spaces. A connected Riemann-ian manifold (M, ds2) is a symmetric space if for all x ∈ M there is anisometry sx : M → M such that x is an isolated fixed point of sx anddsx : Tx M → Tx M sends v → −v. It is locally symmetric if sx existsonly locally.

To construct local isometry, consider the map sx which reversesgeodesics γ with γ(0) = x:

sx(γ(t)) = γ(−t).

This coincides with sx when (M, ds2) is locally symmetric, becauselocal isometry maps geodesics to geodesics and geodesics are deter-mined by initial conditions γ(0) and γ′(0).

6. SYMMETRIC SPACES 155

Proposition 5.29. Symmetric spaces are Riemannian homogeneous.

PROOF. Let G = O(M, ds2). In particular it contains the sub-group generated by the symmetries sx, x ∈ M. We only need toshow that G acts on M transitively. For any x, y which are joined bya geodesic γ with γ(0) = x, γ(T) = y, let sz be the isometry withz = γ(T/2). Clearly sz(x) = y.

In general, x and y can be joined by a sequence of broken geodesicsγi. Then we take the isometry to be the composite of those szi ’s.

Proposition 5.30. In terms of curvature, (M, ds2) is locally symmetric ifand only if that ∇R = 0, that is the curvature tensor is parallel.

PROOF. Indeed, “⇒” is easy: For any tensor T of even degree,∇T is of odd degree. Since sx is a local isometry, we get

∇T = s∗x(∇T) = −∇T,

hence ∇T = 0. “⇐” is a consequence of the Cartan Theorem.

Corollary 5.31. Simply connected locally symmetric spaces are symmetric.

This follows from∇R = 0 and the Cartan-Ambrose-Hicks Theorem.

Theorem 5.32. A connected Lie group G with a bi-invariant metric, e.g.,for G compact times Euclidean, is a G× G symmetric space.

PROOF. Let G × G act on G by (g, h)α = gαh−1. Then G ∼=G× G/G, with the stabilizer at e ∈ G being the diagonal group iso-morphic to G. We claim that the map

sg : h 7→ gh−1g

defined the symmetry at g.We check this for se : h 7→ h−1 first. Indeed, near e ∈ G the map

se is given by exp X 7→ exp(−X). From this we see that se reversesone parameter subgroups and dse = −IdTeG.

To show that se is an isometry, consider any point g ∈ G anda vector v = dLgX ∈ TgG with X ∈ TeG. Then v = γ′(0) where


γ(t) = g exp tX. Then seγ(t) = exp(−tX)g−1 = Rg−1 exp(−tX).Hence

(dse)g v = (dse)g γ′(0) = −dRg−1 X.

With w = dLgY, we compute by using bi-invariance of the metricthat⟨

(dse)g v, (dse)g w⟩=⟨−dRg−1 X,−dRg−1Y

⟩= 〈X, Y〉 = 〈v, w〉 .

For general g ∈ G, sg = LgRgse is the composite of three isome-tries, hence sg is also an isometry.

It remains to check that (dsg)g = −IdTgG. As before let v =

dLgX ∈ TgG. γ(t) = g exp tX. Then sgγ(t) = g exp(−tX)g−1g =

g exp(−tX). Hence

(dsg)g v = (dsg)g γ′(0) = −dLgX = −v.

This completes the proof that G is symmetric.

6.2. Symmetric spaces via Lie algebras. When is a homogeneousspace M = G/H symmetric? This will be reduced to a problem onLie algebras. Recall that σ ∈ Aut G ia an involution if σ 6= IdG andσ2 = IdG.

Theorem 5.33. (Basic structure theorem for symmetric spaces).

(a) Let M = G/H be a symmetric space with G = O(M, ds2), then

σ : G → G; g 7→ σ(g) = sxgsx

is an involution of G and K = Gσ is a closed subgroup contain-ing H such that K = H. H contains no non-trivial normalsubgroup of G.

(b) Conversely, let G be a Lie group with an involution σ. Let K = Gσ

and fix a G-invariant metric 〈 , 〉 on M = G/K. Let σ be thediffeomorphism on M induced from σ. If 〈 , 〉 is σ-invariant thenM is symmetric.

(c) A simply connected Lie group G with an involution σ is equivalentto a Z2 graded decomposition g = h⊕ p in the sense that

[h, h] ⊂ h, [h, p] ⊂ p, [p, p] ⊂ h.

6. SYMMETRIC SPACES 157

Given σ, the subalgebra h and the subspace p are the±1 eigenspaceof dσ : g→ g respectively.

PROOF. For (a), σ is an involution since

σ(gh) = sxghsx = (sxgsx)(sxhsx) = σ(g)σ(h)

and σ2(g) = σ(sxgsx) = sx(sxgsx)sx = g. The group K ∩ H is bothopen (exercise) and closed in K, hence F = H. H contains nonon-trivial normal subgroup of G since otherwise the action of G onM is not effective.

For (b), 〈 , 〉 is σ-invariant means that σ is an isometry on M. Since(dσe)2 = idg, we have the ±1 eigenspace decomposition g = k⊕ p

and TeM ∼= p. So dσe = −idTe M. Thus se := σ is the symmetryat e. We noticed that a Riemannian homogeneous space which issymmetric at one point is then symmetric everywhere. Indeed, thesymmetry at g is given by

sg := Lg σ Lg−1 = Lg σ Lg−1 (mod K).

It is clear that sg is well defined, sg(g) = g, s2g = idM and sg is an

isometry. The property (dsg)g = −id can be easily checked as in theLie group case.

For (c), let v ∈ h and w ∈ p. Then

dσ[v, w] = [dσ(v), dσ(w)] = [+v,−w] = −[v, w].

Hence [h, p] ⊂ p. The proofs of the other two inclusions are similar.Conversely, given Z2 graded decomposition g = h ⊕ p, define

a Lie algebra morphism T : g → g with T|h = id and T|p = −id.Since G is simply connected, this gives rise to a Lie group morphismσ : G → G. Since d(σ2) = dσ dσ = T T = idg, we conclude thatσ2 = idG by the unique correspondence between morphisms.

So the problem on constructing symmetric spaces is reduced tofinding a Z2 decomposition g = h⊕ p with compatible inner prod-uct 〈 , 〉 on p. Combining with Proposition 5.27 and Theorem 5.28, the


corresponding metric on G/H can be constructed from a left invari-ant metric on G which is bi-invariant on H. Examples are providedby the semi-simple Lie groups.

6.3. Examples via semi-simple Lie groups. Let g be a Lie alge-bra over F = R or C. Define the Killing form

B(X, Y) = tr(ad X ad Y); B : g× g→ F.

It is the main source to provide adjoint invariant quadratic forms:

Lemma 5.34 (Exercise). B is ad-invariant: B(adZX, Y)+ B(X, adZY) =0.

We say that g is semi-simple if B is non-degenerate, g is simple if gis not abelian and g contains no proper Lie ideals.

Theorem 5.35. g is semi-simple if and only if g is a direct sum of simpleideals.

The proof for the “only if” part is similar to the proof of Theorem5.23 by using B in place of the bi-invariant metric. The “if” partfollows from the Killing-Cartan criterion which will not be presentedhere.

We say G is semi-simple (simple) if g is semi-simple (simple). ForG simple, every bi-invariant metric 〈 , 〉 is determined by its value ate and proportional to the Killing form.

Example 5.36. We give two main series of examples of symmetricspaces G/H that arise from semi-simple Lie groups G. Notice thatwe had seen that H may always be assumed to be compact.

(i) Type I: G is compact and B is negative definite. E.g.

SO(2n)/U(n), SO(p + q)/SO(p)× SO(q),

SU(2n)/SO(n), SU(p + q)/SU(p)×U(q),

Sp(n)/U(n), Sp(p + q)/Sp(p)× Sp(q).

These includes spheres, projective spaces and Grassmanni-ans.

7. CURVATURE FOR SYMMETRIC SPACES 159

(ii) Type II: G is non-compact and B is indefinite.In this case there is a maximal compact subalgebra h and

Z2 decomposition g = h ⊕ p (the Cartan decomposition).Moreover, B is negative definite on h and positive definite on p.E.g.

SO(p, q)/SO(p)× SO(q),

with respect to the indefinite inner product ∑pi=1 x2

i −∑p+qj=p+1 x2

j .For q = 1, we get the Poincare upper half space Hp.

Similarly

SU(p, q)/U(p)× SU(q),

with respect to the indefinite Hermitian inner product ∑pi=1 |zi|2−

∑p+qj=p+1 |zj|2. For q = 1, we get the unit ball in Cp. Other ex-

amples are SL(n, R)/SO(n, R) (for n = 2 we had seen thatthis gives Poincare upper half plane), SO(n, C)/SO(n, R),SL(n, C)/SU(n).

The main theory of Cartan says that any simply connected sym-metric space may be decomposed into a product of three factors

M = M0 ×M+ ×M−,

where M0 is a Euclidean space, M+ is of compact type and M− is ofnon-compact type. Both M+ and M− may be further decomposedinto irreducible factors and each factor can be constructed from cer-tain semi-simple Lie algebras in a way similar to the above examples.The details can be found in Helgason’s classic text.

7. Curvature for symmetric spaces

7.1. Riemannian submersion. A map f : (M, g) → (M, g) is aRiemannian submersion if it is a C∞ submersion f : M → M andd f : ThM → TM is an isometry, where TM = TvM ⊕ ThM is theorthogonal decomposition defined by: Tv

p M := ker d f p is the verticaltangent space which is also the tangent space of the fiber submanifoldMp := f−1(p) with p = f ( p), and Th

p M := (Tvp M)⊥ is the horizontal

tangent space.


If f ( p) = p and X ∈ TpM, then there is a unique horizontal liftX ∈ TpM such that d f pX = X. Under such a lifting, one may relatethe Levi-Civita connection and Riemannian curvature tensor on Min terms of those on M. This is particularly useful in dealing withRiemannian homogeneous spaces or symmetric spaces of the formG → M = G/H. The following simple relations, due to O’Neill, canbe found in most textbook in Riemannian geometry. The proofs areleft as exercises.

Theorem 5.37. Let f : M→ M be a Riemannian submersion. Then

(a) ∇XY = ∇XY + 12 [X, Y]v for any vector fields X, Y on M and any

lifts X, Y. The vertical component [X, Y]v is tensorial in X and Y.(b) For any X, Y, Z, W ∈ TpM,

R(X, Y, Z, W) = R(X, Y, Z, W) + 12〈[X, Y]v, [Z, W]v〉

+ 14〈[X, Z]v, [Y, W]v〉 − 1

4〈[X, W]v, [Y, Z]v〉.

(c) R(X, Y, X, Y) = R(X, Y, X, Y) + 34 |[X, Y]v|2.

Combining with the curvature formula for Lie groups, we mayachieve

Theorem 5.38. (a) Let G be a compact semi-simple Lie group withan involution σ (σ2 = id). Let g = h⊕ p be the ± eigenspacedecomposition. Then −B defines a bi-invariant metric on G andG/H is a symmetric space with curvature

R(X, Y, X, Y) = |[X, Y]|2.

(b) Let G be a non-compact semi-simple Lie group and g = h ⊕ p

be a Z2 decomposition as in Example 5.36 (ii). Then B|p definesan invariant metric on G/H and make it a symmetric space withcurvature

R(X, Y, X, Y) = −|[X, Y]|2.

PROOF. We only give the proof for (b). The proof for (a) is similarand easier.

8. TOPOLOGY OF LIE GROUPS AND SYMMETRIC SPACES 161

For the Riemannian submersion G → G/H with the left invariantmetric on G defined by

〈 , 〉|h = −B|h, 〈 , 〉|p = B|p,

we have Tve G = h and Th

e G = p. Let X, Y, Z ∈ T[H]G/H ∼= p. ByCorollary 5.18 and Theorem 5.37 (c), we get

R(X, Y, X, Y) = |ad∗XY + ad∗YX|2 − 〈ad∗XX, ad∗YY〉

− 34 |[X, Y]p|2 − 1

2〈[[X, Y], Y], X〉 − 12〈[[Y, X], X], Y〉.

Since [p, p] ⊂ h, [X, Y]p = 0. Also [X, Z] ∈ h implies that 〈ad∗XY, Z〉 =〈Y, [X, Z]〉 = 0, hence ad∗XY ∈ h. Now for T ∈ h, the left invariantmetric 〈 , 〉 on G is also right invariant under H (i.e. ad-invariance ofB) says that

〈[T, X], Y〉+ 〈X, [T, Y]〉 = 0,

which is equivalent to 〈ad∗XY+ ad∗YX, T〉 = 0, hence ad∗XY+ ad∗YX =

0. By setting X = Y we get also ad∗XX = 0. So only the last two termsremained in the curvature formula.

Since [[X, Y], Y], [[Y, X], X] ∈ p, we compute by ad-invariance ofB

R(X, Y, X, Y) = −12 B([[X, Y], Y], X)− 1

2 B([[Y, X], X], Y)

= 12 B([X, Y], [X, Y]) + 1

2 B([Y, X], [Y, X]).

Since [X, Y] ∈ h, this gives −|[X, Y]|2. The proof is complete.

8. Topology of Lie groups and symmetric spaces

For a group action G on a manifold M, a differential form ω ∈Λp(M) is an invariant form if g∗ω = ω for all g ∈ G.

Theorem 5.39. Let M = G/H be a symmetric space with compact G,Then

H∗(M, R) ∼= A∗inv(M) = H∗(M).

PROOF. By the de Rham theorem H∗(M, R) ∼= H∗dR(M, R), hencewe need to show that every invariant form is also closed and everyclosed differential form is equivalent to an unique invariant form.


Step 1. ω ∈ Apinv(M) ⇒ dω = 0. We show first that ω := s∗xω ∈

Apinv(M) for any x ∈ M. For this, recall sxg = σ(g)sx, where σ is the

involution. So

g∗ω = g∗s∗xω = s∗xσ(g)∗ω = s∗xω = ω.

From dsx = −Id on Tx M we get ω|x = (−1)pω|x. Together with theinvariance of ω and ω, this implies that ω = (−1)pω.

Now dω and dω are also invariant forms (since d commutes withg∗) and s∗xdω = d(s∗xω) = dω ∈ Ap+1

inv (M). So similarly dω =

(−1)p+1dω. But we also have dω = (−1)pdω, hence we concludedω = 0.

Step 2. dω = 0⇒ ω ∼ ω ∈ Apinv(M). We prove this step for any

homogeneous space with compact G. On a Lie group G, pick up anyleft invariant metric, its volume form give rise to invariant measuredµ which can be normalized to have total volume 1 if G is compact.

For any g ∈ G, g∗ω ∼ ω since the map g : M → M is homo-topic to identity. This holds true for any affine linear combination:∑ µig∗i ω ∼ ω with ∑ µi = 1. Taking limits (by definition of Riemannsum) we find that

ω :=∫

Gg∗ω dµg ∼ ω.

ω ∈ Λpinv(M) since for any h ∈ G,

h∗ω = h∗∫

Gg∗ω dµg =

∫G

h∗g∗ω h∗dµg =∫

G(gh)∗ω dµgh = ω.

Step 3. We show that an exact invariant form must be zero. Fixa G invariant metric on M. We recall the Hodge star operator ∗ :ΛpT∗x M → Λn−pT∗x M. The G invariance implies that ∗g∗ = g∗∗ forany g ∈ G. Hence ω ∈ Ap

inv(M) ⇒ ∗ω ∈ Apinv(M). In particular

d(∗ω) = 0 by step 1.For ω, η ∈ Ap(M),

〈ω, η〉 =∫

Mω ∧ ∗η

8. TOPOLOGY OF LIE GROUPS AND SYMMETRIC SPACES 163

is a inner product on Ap(M). Now for ω = dη an exact invariant pform,

〈ω, ω〉 =∫

Mdη ∧ (∗ω) =

∫M

d(η ∧ (∗ω))− (−1)p∫

Mη ∧ d(∗ω) = 0

by Stokes theorem and d(∗ω) = 0. Hence ω = 0 as desired.

Step 4. It remains to show that invariant forms are precisely har-monic forms. If ω ∈ Ap

inv(M), we have just seen that

dω = 0 and d∗ω = (−1)p(n−p) ∗d∗ω = 0.

Thus 4ω = (dd∗ + d∗d)ω = 0. If we assume the Hodge theoremwhich says that H∗dR(M, R) ∼= H∗(M), then harmonic forms must beinvariant forms.

We would like to give a direct proof: First notice that 4η = 0 ifand only if that dη = 0 and d∗η = 0, which is seen from the identity

〈4η, η〉 = ||dη||2 + ||d∗η||2.

Let 4ω = 0. For any X ∈ g, we compute by Cartan’s homotopyformula

LXω = ιXdω + dιXω = dιXω.

The invariance of the metric implies that4 commutes with LX, henceLXω is also harmonic. So 〈LXω, LXω〉 = 〈LX, dιXω〉 = 〈d∗LXω, ιXω〉 =0 and then LXω = 0. By definition of Lie derivatives this implies thatω is an invariant form. The proof is completed.

Corollary 5.40. For a connected compact Lie group G, viewed as a G× Gsymmetric space, the de Rham cohomology are given by bi-invariant forms,which are precisely harmonic forms:

H∗(G, R) ∼= A∗G× G-inv(G) ∼= Λ∗Ad-inv[g∗].

PROOF. Only the last equality requires explanation. The left in-variant forms are uniquely determine by their values at e ∈ G, so wehave

A∗left-inv(G) ∼= Λ∗[g∗].


A left invariant form is right invariant if and only if it is adjoint in-variant:

R∗g−1ω = R∗g−1 L∗g ω = I∗g ω.

The inner automorphism Ig induces the adjoint action Adg on g,hence on the dual space g∗. This gives the action I∗g on left invari-ant forms.

Corollary 5.41. Let G be a compact Lie group and Ω(X, Y, Z) := B([X, Y], Z)where B is the Killing form. If Ω 6= 0, say if G is semi-simple, thenH3(G) 6= 0.

PROOF. The skew symmetry of Ω in X, Y is obvious. In Z, thisis equivalent to adjoint invariance of B. The adjoint invariance of Ωfollows from the adjoint invariance of B and the Jacobi identity.

We conclude the discussion with two fundamental results in co-homology and homotopy theory of Lie groups without proof.

Theorem 5.42 (Hopf Theorem). For connected Lie group G, H∗(G) is afinitely generated free exterior algebra Λ[y1, . . . , yn], with yi being of odddegree. For example, H∗(U(n), R) ∼= Λ[y1, . . . , y2n−1], deg yi = i.

Theorem 5.43 (Bott Periodicity Theorem).

(i) Unitary case: πi−1(SU(2m)) ∼= πi+1(SU(2m)) for 1 ≤ i ≤ 2m.Hence for U := lim

−→U(m), we have πi−1(U) ∼= πi+1(U).

(ii) Orthogonal case: For O := lim−→

O(n), we have πi(O) ∼= πi+8(O).

The first eight values πi(O) for 0 ≤ i ≤ 7 are Z2, Z2, 0, Z, 0, 0, 0, Z

respectively.

9. Exercises

1. Let X ∈ sl(2, R) and d =√|det X|. Then

(i) eX = (cosh d)I2 +1d (sinh d)X if det X < 0.

(ii) eX = (cos d)I2 +1d (sin d)X if det X > 0.

(iii) eX = I2 + X if det X = 0.

9. EXERCISES 165

Let ga =

(a 00 a−1

)∈ SL(2, R). Then ga lies in a unique one

parameter subgroup if a > 0. ga lies in infinitely many one pa-rameter subgroup if a = −1. If a 6= −1 and a < 0, then ga 6∈exp sl(2, R).

2. Complete the remaining problem of Theorem 5.15.3. Show the Corollary 5.18 by Proposition 5.17(i).4. Prove the Proposition 5.255. Show the remaining statement on adh in Proposition 5.27.s6. Show that in K ∩ H in Theorem 5.33 is open.7. Show the Lemma .8. (Warner Ch.6 #20) (The Peter-Weyl Theorem) The representative

ring of a compact Lie group G is the ring generated over the com-plex numbers by the set of all continuous functions f for whichthere is a continuous homomorphism ρ : G → GL(n, C) for somen such that f = ρij for some choice of i and j. The Peter-Weyl the-orem states that representative ring is dense in the space of complex-valued continuous functions on G in the uniform norm. That is, if gis a complex valued continuous function on G, and if ε > 0 isgiven, then there is a function f in the representative ring suchthat | f (σ)− g(σ)| < ε for all σ ∈ G. We outline the proof of thistheorem which is based on the uniform completeness of the eigen-functions of the Laplacian. One can choose a Riemannian struc-ture on G such that each of the diffeomorphisms `σ for σ ∈ G (lefttranslation by σ) is an isometry (that is, 〈v, w〉τ = 〈d`σv, d`σw〉στ

for all τ ∈ G and all v, w ∈ Gτ). Since the C∞ functions aredense in the space of continuous functions in the uniform norm,and since by result of above exercise (Warner Ch.6 #16 (h)) the di-rect sum of the eigenspaces of the Laplacian is dense in the spaceof C∞ functions in the uniform norm, it suffices for the Peter-Weyl theorem to prove that each eigenfunction of the Laplacian∆ : C∞(G)→ C∞(G) belongs to the representative ring.

Now, G acts on the C∞ functions on G by

σ( f ) = f `σ for σ ∈ G.


Prove that since the `σ are isometries, this action commutes withthe Laplacian

∆( f `σ) = (∆ f ) `a, for σinG.

Let Vλ be the (finite dimensional) eigenspace associated with theeigenvalue λ of ∆ : C∞(G) → C∞(G). Prove that the action of Gleaves Vλ invariant. Then let ϕ1, . . . , ϕn be a basis of Vλ, and let

σ(ϕi) = ∑j

gji(σ)ϕj.

Then σ → gji(σ) is a homomorphism of G → GL(n, R). Provethat this homomorphism is continuous. Then observe that

ϕi(σ) = ϕi `σ(e) = ∑j

gji(σ)ϕj(e),

so that ϕi belongs to the representative ring.9. (cf. Do Carmo Ch.1 #7) When G is compact, the bi-invariant met-

rics always exist. For example, for G ⊂ O(n, R) ⊂ Sn2−1(√

n), theEuclidean metric 〈A, B 〉 = tr ABT is bi-invariant.

Index

closed map, 19critical point, 24cut off function, 10

diffeomorphism, 7differentiable

manifold, 6structure, 7

distribution, 30

existence and uniqueness theoremsof ODE, 27

Frobenius integrability, 30classical version, 38

Hausdorff space, 5

imbedding, 19immersion, 17involutive, 30

Lie bracket, 29Lie derivative, 28locally Euclidean, 5

manifold, 5

partition of unity, 10existence, 11

proper map, 20

regular point, 24

regular value, 23Riemannian manifold, 38, 69Riemannian structure, 38

Sard’s theorem, 24second countable, 5singular value, 23submersion, 17

tangent map, 15tangent space, 15

Zariski tangent space, 12

vector field, 27

Whitney imbedding theorem, 20

167

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