Riemannian Geometry

Amin Saied

D. Lotay

Contents

1

Chapter 1

Introduction

1.1 History

Riemann (1854) invented Riemannian geometry during a lecture whilst lectured by Gauss!

Lost of applications in analysis, number theory, general relativity and geometry. It’s evenused in financial mathematics.

Fundamental objets of Riemannian geometry: Geodesics and Curvature.

Objective of the course is to understand these and the relationships between them.

Key application: “Local conditions” ⇒ “global constraints”. What that means is Geome-try near each point ⇒ topological conditions.

Riemannian geometry consists of concrete notions. Course strategy will be to understand thebasics of Riemannian geometry using as few abstract definitions as possible.

Essentially the only abstract idea we will need is “Manifold”.

A manifold M is a topological space such that for all p ∈ M there exists open U containingp such that U is homeomorphic (a homeomorphism is a continuous bijection with continuousinverse) to an open set Euclidean space.

Manifolds sound boring but can have very interesting topology and geometry. In fact, manifoldsare the natural object on which to do geometry.

Examples: Rn and any open set in Rn, curves and surfaces. We are going to focus on Rie-mannian manifolds. Here we can measure lengths and angles between (tangent) vectors.

Simplest non-trivial examples are surfaces in R3.

On Riemannian manifolds we can define our fundamental objects:geodesics - These are curves which locally minamise distance between two points. (i.e. in Rnthe shortest distance between two points is a straight line. On a sphere it is a curve)

And curvature - how a manifold deviates from flat Rn Euclidean space near each point.

2

1.2 Syllabus

1. Surfaces in R3 - A comprehensive review of the theory with plenty of examples, but willomit details of proofs. This will culminate in Gauss Theorem Eqrigium (Latin for Gauss’remarkable theorem).

2. Manifolds - A quick review

3. Riemannian manifolds - The real material starts

4. Geodesics - The first variation formula

5. Curvature - Will define various types of curvature (Richi curvature is fundamental in Gen.Relativity)

6. Riemannian Submanifolds - Can relate back to surfaces in R3

7. Jaconi Fields (relate geodesics and curvature)

8. Completeness (Hopf-Rinow Theorem and Hadamard Theorem)

9. Constant Curvature - We will classify the complete spaces with constant curvature (Hy-perbolic Space)

10. Second Variation Formula and Applications - Myers’ Theorem and Synge’s Theorem

1.3 Multivariable Calculus

V ⊆ Rn open. F : V → Rm, (F1, . . . , Fm).Let x1, . . . , xn be indeterminants in Rn

Definition 1.1. F is smooth at p ∈ V (i.e. infinitely differentiable) if all partial derivativesof all orders of Fi exist and are continuous. Then F is smooth if it is smooth at every point p.

E.g. F : R2 → R is smooth if ∂k+lF∂xk

exists and is cts for all k, l.

Definition 1.2. Let F : V → Rm, where V ⊂ Rn and F is smooth. The differential of F at

q ∈ V is the linear map dFq : Rn → Rm with m× n matrix

(∂Fi∂xj

(q)

).

Examples

1. F : R→ R⇒ dFq = multiplication by F ′(q)

2. F : R2 → R⇒

dFq(u, v) =

(∂F

∂x(q),

∂F

∂y(q)

)(uv

)=

⟨(∂F

∂x,∂F

∂y

), (u, v)

⟩

3

3. F : R3 → R⇒

dFq(p) =

⟨(∂F

∂x,∂F

∂y,∂F

∂z

)︸ ︷︷ ︸

grad F

, (u, v, w)

⟩

4. F (x, y, z) = x2 + y2 + z2

Matrix is dFq =

2x2y2z

∣∣∣∣∣∣q

= 2q. Hence dFq(p) = 2〈q, p〉.

5. F (x, y, z) = (√x2 + y2 − α)2 + z2 Not smooth at (0, 0, z).

dF(x,y,z) = grad F

= 2(xr

(r − α),y

r(r − α), z

)Where r =

√x2 + y2 > 0.

6. Remark F (q + h) = F (q) + dFq(h) + o(|h|)

Let A ∈Mn(R) i.e. an n× n matrix.

F (A) = ATA

F (A+ h) = (A+ h)T (A+ h) = ATA(=F (A)) +ATh+ hTA+ hTh(=0(|h|))

Comparing with the remark then we have that

dFA(h) = ATh+ hTA

Lemma1.3. (Chain Rule)

d(F ·G)q = dFG(q) · dGq (1.1)

(F ·G)′(q) = F ′(G(q))G′(q)

Theorem 1.4. (Inverse Funtion Theorem)Let F : U → Rm where U ⊂ Rn open, F smooth.

Suppose there exists p ∈ U : dFq : Rn → Rn is invertable.Then ∃ V containing p and W containing F (p) such that F : V → W is a diffeomorphism (asmooth bijection with smooth inverse).

Moreover dF−1F (q) = (dFq)

−1 ∀ q ∈ V .

4

Theorem 1.5. (Implicit Function Theorem)Let F : U → Rn where U ⊆ Rn+m open, F smooth.

Suppose ∃ p ∈ U : dFp : Rn+m → Rn is surjective.

Let c = F (p) then ∃ open V ⊆ Rn, open W ∈ Rm such that p ∈ V × W , and a smoothmap G : V →W such that F−1(c) ∩ (V ×W ) = {(q,G(q)) : q ∈ V }

5

Chapter 2

Surfaces in R3

How to rigorously define what we might mean by a surface in R3.

Examples

1. f : U → R where U ⊂ R2. Then consider the graph of f , denote Γf .Γf = {(x, y, z) : z = f(x, y)}.

2. However, we can’t get everything we want as a surface in this way, as graphs of functions,for instance the sphere, S2.

There are certain things that we want to rule out, so let’s motivate our definition with a fewnon-examples and try to see what makes them bad. Our starting point for our definition willbe that, for each point on S our surface, we can construct a map from an open set U to V ∩ Sfor an open set V about our point, such that the map satisfies certain conditions.

Non-Examples

1. We don’t want surfaces that intersect themselves e.g.

6

To ensure this we put condition in the the map must be a bijection. If not we can getsomething like above where the two points shown in U get mapped into the same pointin V ∩ S.

2. We don’t want surfaces that nearly touch but don’t. Say then that map must havecontinuous inverse.

3. Consider the curve given by x(t) = (t2, t3)

This is a smooth bijection with a continuous inverse, but we don’t want this either. Theproblem with this is that it’s derivative at t = 0 is 0.

Definition 2.1. S ⊆ R3 is called a (regular) surface if ∀ p ∈ S ∃ open V ⊆ R3, p ∈ V, ∃open U ⊆ R2, ∃ smooth function x : U → V ∩ S such that the following two conditions hold:

1. x is a smooth bijection with continuous inverse

2. ∀ q ∈ U : dxq is injective.

Let’s have a look at what’s going on here:

7

Examples

1. Plane:S = {(x, y, 0) : x, y ∈ R2}. Is S a surface? Check against the definition.

∀ p ∈ S (write p = (x, y, 0)) ∃ V ⊆ R3, p ∈ V choose V = R3

U = R2

x(u, v) = (u, v, 0)

Jacobian is

1 00 10 0

which is injective (rank 2). Everything else clear.

8

2. Cylinder:S = {(x, y, z) : x2 + y2 = 1} Many approaches e.g. use cylindrical polars U = {(θ, z) :θ ∈ (−π, π), z ∈ R}

x(θ, z) = (cos θ, sin θ, z)

In this case we can take V to be {(x, y, z) : x ≥ 0, y 6= 0}

3. x(u, v) = {(v cosu, v sinu, u) : u, v ∈ R} – HelicoidHere we are defining the surface using the definition i.e. by actually stating the map. Thiscan make life a bit easier.

4. x(u, v) = {(cosh v · cosu, cosh v · sinu, v) : u, v ∈ R} – CatenoidAgain, this has been defined by a function from an open set U = R2 to an open setV = R3, so we can immediately see the definition set up.

Both 3. and 4. are examples of minimal surfaces. That is, any slight perturbation willincrease the area of the surface.

9

2.1 Level Sets

We would like to have a theorem which made it a bit easier to get surfaces.

Goal: To have a criterion for when a level set, e.g. F (x, y, z) = 0, is a surface.

e.g. sphere F (x, y, z) = x2 + y2 + z2 − 1

Proposition 2.3. F : U → R smooth, U ⊂ R3 open, c ∈ F (u). If dFp : R3 → R is non-zerofor all p ∈ F−1(c), then F−1(c) is a surface.

We call this set the level set F−1(c) = {(x, y, z) : F (x, y, z) = c}.

Examples

1. F (x, y, z) = z2

F−1(c) =

{two planes if c > 0one plane if c = 0

dF(x,y,z) = 0. So we see that proposition 2.3 doesn’t apply here. But it is still a sur-face. That is, proposition 2.3 is a sufficient, but not a necessary condition.

2. F (x, y, z) =x2

A2+y2

B2+z2

C2

The level sets are ellipsoids (or a point if c = 0).

Can see that the level set of 0 is not a surface, and indeed, the derivative here is 0 at theorigin.

3. F (x, y, z) = x2 + y2 − z2

10

dF(0,0,0) = (∂F

∂x,∂F

∂y,∂F

∂z)

∣∣∣∣(0,0,0)

= (2x, 2y, 2z)|(0,0,0) = 0

hence by proposition 2.3 F−1(0) is not a surface.

4. F (x, y, z) = (√x2 + y2 − 2)2 + z2.

Can see a rotational symmetry (as can swap x and y and get same equation). To under-stand the curve then look at a slice y = 0, and then rotate this to get the surface. Belowwe plot values of a slice of F−1(c) for c = 0, 1/2, 2, 4.

• c = 0 – We just have 2 points, at ±2. Rotation will give a circle. My question: Isthis is a surface? I don’t think so but the where is the derivative 0? The derivativeis, (

2(√x2 + y2)((x2 + y2) · x) , 2(

√x2 + y2)((x2 + y2) · y) , 2z

)• 0 < c < 4 – Here rotation gives a torus (see below), which is a surface.

• c ≥ 4 – Have the circles intersecting, so there is a ‘sharp edge’, as it were uponrotation. Not a surface.

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Remark

• Proposition 2.3 is a sufficient, but not a necessary condition(e.g. F (x, y, z) = z2)

• Constants c satisfying the condition are called regular values of F

• (U, x) is called a coordinate chart

• x(U) = V ∩ S is called a coordinate neighborhood of p

• On V ∩S we have curvilinear coordinates given by mapping the (u, v)-coordinate linesto S using x

To prove proposition 2.3 we need a lemma:

Lemma 2.2. U ⊆ R2 open set, and f : U → R is a smooth function, then the graph Γf ⊆ R3

is a surface.

Proof. Define x(u, v) := (u, v, f(u, v)) ∈ R3.

NTS conditions from definition 2.1 are satisfied. x is clearly smooth, as f is a smooth map. Itis clear that x is a bijection too. The inverse map sends a point (x, y, z) ∈ Γf to (x, y) ∈ U andR3 → R2 by (x, y, z) 7→ (x, y) is a linear map, and therefore is continuous.

Finally NTS the differential is injective:

Let q = (u, v). Compute dxq, dxq =

∂u∂u

∂u∂v

∂v∂u

∂v∂v

∂f∂u

∂f∂v

=

1 00 1fu fv

And this is injective (it has

rank 2).

Proposition 2.3. F : U → R smooth, U ⊂ R3 open, c ∈ F (u). If dFp : R3 → R is non-zerofor all p ∈ F−1(c), then F−1(c) is a surface.

We call this set the level set F−1(c) = {(x, y, z) : F (x, y, z) = c}.

Remark Proof of proposition 2.3 amounts to deducing the implicit function theorem fromthe inverse function theorem.

12

Proof of 2.3Idea is to show that locally F−1(c) is the graph of some function.(Useful special case to keep in mind: F (x, y, z) = f(x, y)− z for some given function f)

We’re assuming that dFp 6= 0 for all p ∈ F−1(c). Therefore we can choose coordinates(x, y, z) ∈ R3 so that ∂F

∂z

∣∣p

doesn’t vanish. (And in fact ∂F∂z

∣∣p6= 0 for all q sufficiently close to

p)

Define a new function G(x, y, z) := (x, y, f(x, y, z))

Claim:The Inverse Function Theorem applies to this function i.e. the Jacobian dGp is invertible.

pf:

dGp =

∂x∂x

∂x∂y

∂x∂z

∂y∂x

∂y∂y

∂y∂z

∂z∂x

∂z∂y

∂z∂z

=

1 0 00 1 0Fx Fy Fz

⇒ det(dGp) = Fz 6= 0

Therefore it is invertible.

So the inverse function theorem applies i.e. there exists nhds U containing p, V containingG(p), and a smooth function H : V → U such that H ·G = IdU and G ·H = IdV .

It’s clear that H (u, v, w)︸ ︷︷ ︸coords on V

= (u, v, h(u, v, w)) where h is some function. (As G doesn’t touch

the first two coordinates so neither can it’s inverse)

How to construct f such that F−1(c) = Γf locally near p.

Claim:h(x, y, c) = z for all (x, y, z) ∈ F−1(c) ∩ U

Follows from F (x, y, z) = c and fact that H ·G = IdU

Homework Run through the proof with the special case stated at the beginning.

2.2 Tangent Spaces to Surfaces in R3

S ⊂ R3 surface, p ∈ S

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Definition 2.7. A vector X ∈ R3 is called a tangent vector to the surface at p if X = α′(0)for some curve α : (−ε, ε)→ R3 for which α(−ε, ε) ⊆ S and α(0) = p.

The tangent space, Tp := space of all such vectors X.

Remark Definition 2.7 makes sense for any set S ⊂ R3. TpS is always going to be a cone(i.e. dilation invarient). The idea is TpS is closed under addition (i.e. is a vector space/a linersubspace of R3) iff the set S is smooth at p.

Here p isnot a smoothpoint, sotangent vectorsdo not form avector space,as they are notclosed underaddition.

Proposition 2.8. If S is a smooth surface, p ∈ S, x : U → V ∩ S any coordinate chart for aS near p, then the set TpS = dxq(R2) where x(q) = p.

Note: This is evidently a two dimensional subspace of R3. It is the space spanned by thecolumns of the Jacobian matrix of x.

Proposition 2.9. If F : U → R, (where U ⊆ R3 an open set) is smooth, and if c ∈ F (U) is aregular value, then Tp

[F−1(c)

]= ker dFp = (gradpF )⊥ for all p ∈ F−1(c)

Definition 2.4. f : S → Rn is smooth at p iff f · x : U → Rn is smooth in the usual sense.

f : S → S is smooth at p iff y−1 · f · x is smooth in the usual sense.

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Definition 2.10.f : s→ S smooth, p ∈ S. Then we can define the derivative / differential dfp : TpS → Tf(p)Sa linear map, as follows

X ∈ TpS then by definition X = α′(0). Consider f · α in S.

(f · α)(0) = f(p). Then define dfp(X) := (f · α)′(0) ∈ Tf(p)S

DiffeomorphismsMostly interested in diffeomorphisms and local diffeomorphisms.

Examples

1. f : R3 → R3, f(x, y, z) = (Ax,By,Cz), S2 = {x2 + y2 + z2 = 1} = S

f(S2) = ellipsoid = { x2

A2+y2

B2+z2

C2= 1} = S

This is a diffeomorphism. It preserves the topology.

2. T 2 = {(√x2 + y2 − 2)2 + z2 = 1}, the torus.

f : R2 → T 2 by (u, v) 7→ ((2 + cos v) cosu, (2 + cos v) sinu, sin v)

15

Each squaregets mappedinto the torus,so this mapis surjective,but clearly notinjective.

This is a local diffeomorphism. This is a doubly periodic map,

Proposition 2.11.f : S → S a smooth map, say dfp : TpS → Tf(p)S is invertible for all p ∈ S, then f is a localdiffeomorphism (i.e. for all p ∈ S ∃ nhds U containing p, V containing f(p), f : U → V issmoothly invertible). In case f itself is bijective, we call f a diffeomorphism.

2.3 The Gauss Map

Curvature of a surface.

The spherehas varyingcurvature asthe normalvector changesas p is varied.The planehas constantcurvature asthe normalvectors are thesame as p isvaried.

Orientability A complication can arise with certain surfaces. It is best explained with anexample.

The Mobius strip:

16

As p varies the normal vector moves around the strip, but when you return the your originalpoint the normal vector can change sign, as demonstrated above (badly!).

Surfaces on which this can happen are called non-orientable surfaces.

Definition 2.12. The normal space to S at p

NpS := (TpS)⊥

Example Computing normal vectors

1. (U, x) a chart. distinguished tangent vectors∂x

∂u,∂x

∂v∈ TpS

unit normal vector ν(p) =∂x∂u ×

∂x∂v

|∂x∂u ×∂x∂v |

LOCAL

Note: Careful, swapping u and v let’s ν(p) pick up a minus sign.

2. F : R3 → R smoth function, c ∈ R regular value of F (i.e. (gradF )(p) 6= 0 for allp ∈ F−1(c))

S := F−1(c) GLOBAL

TpS = ker dfp = (gradpF )⊥

unit normal vector ν(p) =gradpF

|gradpF |

Note: Mobius strip is not the level set of a smooth function.

Definition 2.16. A Gauss Map for S is a smooth map

S2 is unitvectors

ν : S → S2

such that ν(p) ∈ NpS for all p ∈ S. 1

1So the normal vector at a point

17

Remark

1. If ν is a Gauss map then so is −ν

2. Gauss maps always exist locally (see Example 1 above)

3. On regular level sets Gauss maps exist globally (see Example 2 above)

Local

ν(p) = ±∂x∂u ×

∂x∂v

|∂x∂u ×∂x∂v |

(?)

Global If S = F−1(c) have a global Gauss map

ν =grad F

|grad F |

Proposition 2.14. Gauss maps exist globally iff the surface is “orientable”.

Definition 2.13. S orientable: ⇐⇒ can cover S by coordinate charts (Uα, xα), xα : Uα →S,U ⊆ R2 such that for all pairs (α, β) for which xα(Uα)∩xβ(Uβ) 6= ∅ the Jacobian determinantdet d(x−1

β · xα) > 0 wherever defined.

Example

1. f : U → R, U ⊆ R2

⇒ graph Γf ⊆ R3 is orientable

I only need 1 chart to cover.

2. A surface that can be covered by 2 charts whose intersection is connected is orientable.

(Pf: Swap coordinates in one of the charts if necessary e.g. sphere, cylinder)

18

Idea of Proof of 2.14.The expression (?) changes by the sign of the Jacobian under a change of coordinates

(u, v) 7→ (v, u) Jacobian

(0 11 0

) det(−1)

2.4 Curvature

Key Idea Consider differential of the Gauss map

ν : S → S2

dνp : TpS → Tν(p)S2 = ν(p)⊥ = TpS

Definition 2.22.dνp : TpS → TpS is the shape operator or the Weingasten map

Associated bilinear form: Second Fundamental Form:

II : TpS × TpS → R(X,Y ) 7→ 〈dνp(X), Y 〉

Examples

1. Plane: ν is a constant vector ⇒ dν = 0

2. Unit Sphere:

ν : S2 → S2

ν = Id

dνp = Id : TpS2 → TpS

2

19

3. Cylinder: x2 + y2 = 1

Gauss map: (x, y, z) 7→ (−y, x, 0).

In terms of the basis {v1, v2} for TpS the shape operator

dνp =

(1 00 0

)Question How to compute shape operator in terms of coordinate chart (U, x)?

Basis for the tangent space:

(∂x

∂u,∂x

∂v

)Compute the matrix of the second fundamental form

II

(∂x

∂u,∂x

∂v

)=

⟨dν

(∂x

∂u

),∂x

∂u

⟩=

∂

∂u〈u, ∂x

∂u〉︸ ︷︷ ︸

zero everywhere

− 〈ν, ∂2x

∂u2〉

in terms of

{∂x

∂u,∂x

∂v

}

II = −

⟨∂2x

∂u2, ν

⟩ ⟨∂2x

∂u∂v, ν

⟩⟨∂2x

∂v∂u, ν

⟩ ⟨∂2x

∂v2, ν

⟩

Proposition 2.17. II is a symmetric bilinear form (i.e. the shape operator dνp : TpS → TpSis self-adjoint).

In particular, dνp is diagonisable.

20

Definition 2.23. The eigenvalues of dνp are called principle curvatures k1(p), k2(p)

k1k2 = det (dνp) – “Gaussian curvature” K

k1 + k2 = tr (dνp) – “mean curvature” H

Example

1. Plane: k1 = k2 = 0 K = H = 0

2. Sphere: k1 = k2 = 1 K = 1, H = 2

3. Cylinder k1 = 1, k2 = 0 K = 0, H = 1

Remark H is orientation dependent

K is orientation independent.

H = the gradient of the are functional. H = 0 ⇐⇒ stationary point for the area func-tional (minimal surface).

K is intrinsic to the surface. If two surfaces have the same K then locally they have thesame ‘kind of minimal distance’.

Remark In matrix form

II =

II

(∂x

∂u,∂x

∂u

)II

(∂x

∂u,∂x

∂v

)

II

(∂x

∂v,∂x

∂u

)II

(∂x

∂v,∂x

∂v

)

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Chapter 3

Manifolds

Definition 3.1. A (Hausdorff paracompact) topological space M is a (smooth) n-dimensionalmanifold if there exists {(Ua, xa) : a ∈ A} where Ua ⊆ Rn are open, xa : Ua → xa(Ua) ⊆M isa homeomorphism such that

1.⋃a∈A

xa(Ua) = M

2. Whenever Vab := xa(Ua) ∩ xb(Ub) 6= ∅, the map

x−1b · xa : x−1

a (Vab) ⊆ Rn → x−1b (Vab) ⊆ Rn

is a diffeomorphism.

Call {(Ua, xa)} an atlas for M and (Ua, xa) a coordinate chart for M .

Examples

1. Rn

2. Open sets in Rn

Remark

1. Basically Hausdorff and paracompact is the same as a metric space.

2. It can be useful to make weaker topological assumptions or ask for transition maps notto be smooth.

3. Do not need the third condition in Do Carmo where he says it must be maximal.

Proposition 3.2. Surfaces in R3 are 2-dimensional manifolds.

Proof R3 is a metric space ⇒ surfaces S in R3 are metric spaces. Coordinate charts on Sgive a potential atlas. By definition of a smoth map on S we have that x−1

b · xa is smooth andinvertable with smooth inverse (i.e. a diffeo). QED.

Proposition 3.3. Let f : U ⊆ Rn → Rn be smooth, then the graph Γf of f in Rn+m is ann-dimensional manifold.

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Proof One chart U and x(p) = (p, f(p))

Proposition 3.4. Let F : U ⊆ Rn+m → Rm be smooth, and c ∈ F (U) such that dFp :Rn+m → Rn is surjective for all p ∈ F−1(c) (i.e. c is a regular value of F ) then F−1(c) is ann-dimensional manifold.

Proof The implicit function theorem (Thm 1.5.) ⇒ the for every p ∈ F−1(c) there existsa nhd of p in F−1(c) which is the graph of a smooth function from Rn → Rn. So by propo-sition 3.3. F−1(c) is a manifold in a nhd of each point, and the transition maps we get arereparametrisations of graphs of smooth functions, and so are diffeomorphisms. QED.

Examples

1. F :n+1→ R by F (x1, . . . , xn+1) = x21 + x2

2 + · · ·x2n+1

Then dFp = 2p 6= 0 for p 6= 0 ⇒ 1 is a regular value of F ⇒ F−1(1) is an n-dimensionalmanifold called Sn, the n-sphere.

2. F : Mn(R)→Mn(R) by F (A) = ATA

F (A)T = F (A) so F : Mn(R)→ Symn(R) Mn(R) ∼=Rn2

, Symn(R) ∼=

R12n(n+1)

dFB(A) = BTA+ATB

If B ∈ F−1(I) i.e. BTB = I and C ∈ Symn(R)

dFB(12BC) = BT (1

2BC) + (12BC)TB = 1

2(BTB)C + 12C

TBTB = C

⇒ I is a regular value of F .

Proposition 3.4. ⇒ F−1(I) = {A ∈ Mn(R)|ATA = I} = O(n) is 12n(n − 1) dimen-

sional manifold.

A ∈ O(n)⇒ det(A)2 = det(ATA) = 1⇒ det(A) = ±1

det : O(n) → {−1, 1} is continuous ⇒ det−1(1) = {A ∈ O(n) : det(A) = 1} = SO(n)special orthogonal group is an 1

2n(n− 1) dimensional manifold.

Remark O(n) and SO(n) are Lie groups, which roughly speaking are groups which are alsomanifolds.

Theorem 3.5. Let M be an n-dimensional manifold and let G be a group. Suppose that forall x ∈ G there exists a diffeomorphism φx : M →M such that,

1. φe = id where e ∈ G is the identity

2. φxy = φx · φy ∀ x, y ∈ G

3. For all p ∈M there exists open U containing p such that U ∩φx(U) = ∅ for all x ∈ G\{e}.

We say that G acts properly discontinuously on M .

23

Define an equivalence relation on M by p ∼ q iff p = φx(q) for some x ∈ G:

Then G /∼ = M /G with the quotient topology is an n-dimensional manifold.

Proof Choose an atlas {Ua, xa} for M such that xa(Ua) ∩ φx(xa(Ua)) = ∅ for every x 6= e.Then atlas for M /G is {(Ua, π · xa)} where π : M →M /G is the natural projection. QED.

Examples Z2 = {−1, 1} with φ−1 = −id and Zn acts on Rn as φ(k1,...,kn)(x1, . . . , xn) =(x1 + k1, . . . , xn + kn)

1. Z2 acts properly discontinuously on Sn ⇒ Sn/Z2 is an n-dimensional manifold called

RPn, real projective n-space.

2. Cyylinder C = {(x, y, z) : x2 + y2 = 1, |z| < 1}

C/Z2 is a 2-dimensinal manifold called the Mobius band.

3. T2 /Z2 is a 2-dimensional manifold called the Klien Bottle

4. Rn /Zn is an n-dimensional manifold called Tn, the n-torus.

Definition 3.6. A map f : M → N where M,N are manifolds, is smooth at p ∈ M if thereexists coordinate charts (U, x) at p and (V, y) at f(p), with f(U) ⊆ V , such that y−1 ·x : U → Vis smooth. f is smooth if smooth at every p.

Remark (2) in the definition of manifold precisely ensures that this definition makes sense.

Definition 3.7. f : M → N is a diffeomorphism if f is a smooth bijection with smooth inverse.

f : M → N is a local diffeomorphism at p ∈ M if there exists open U containing p andopen V containing f(p) such that f : U → V is a diffeomorphism.

24

Proposition 3.8. The projection map π : M →M /G (where G acts properly discontinuouslyon M) is a local diffeomorphism.

Corollary 3.9. There exists a local diffeomorphism from Sn to RPn

Definition 3.10. M a manifold is orientable if there exists an atlas {(Ua, xa), a ∈ A} suchthat det d(x−1 · x)q > 0 for all q ∈ x−1

a (Vab).

An orientation on M is a choice of such an atlas.

Examples

1. Sn is orientable

2. Tn is orientable

3. Mobius band and Klien bottle are non-orientable

4. RPn is orientable ⇐⇒ n is odd

Let M be an n-dimensional manifold.

Definition 3.11. Let α : (−ε, ε) → M be a curve with α(0) = p, let f : U ⊆ M → R be asmooth function on U an open set containing p.

f ·α : (−ε, ε)→ R smooth at 0⇒ define α′(0) : f → (f ·α)′(0) to be the tangent vector to α at 0.

Let (U, x) be a coordinate chart and write

x−1 · α(t) = (a1(t), . . . , an(t)) ∈ Rn

(f · α)′(0) =d

dtf(a1(t), . . . , an(t))|t=0

=n∑i=1

a′i(0)∂f

∂xi

∣∣∣∣p

⇒ α′(0) =n∑i=1

a′i(0)∂

∂xi

∣∣∣∣p

in local coordinates.

Moreover, if I take

{∂i =

∂

∂xi

}as a basis then I can view α′(0) as (a′1(0), . . . , a′n(0)) ∈ Rn

Definition 3.12. X is a tangent vector to M at p if there exists α : (−ε, ε)→M with α(0) = pand α′(0) = X.

Let TpM be the set of all tangent vectors to M at p.

Definition 3.13. Let f : M → N be a map between manifolds, smooth at p ∈ M . X ∈TpM ⇒ ∃ curve α : (−ε, ε)→M such that α(0) = p and α′(0) = X ⇒ curve f ·α : (−ε, ε)→ N ,so (f · α)′(0) ∈ Tf(p)N .

Define dfp(X) = (f · α)′(0)

We have a linear map dfp : TpM → Tf(p)N called the differential of f at p.

25

Remarks

1. If (U, x) is a coordinate chart on M at p with x(q) = p ⇒ dxq : Rn → TpM is anisomorphism.

2. If {(Ua, xa)} is an orientation on M , can define an orientation on each TpM as follows:

{∂i} a basis for TpM , {Xi} another basis for TpM ⇒ Xi = Σjaij∂j so say that {Xi}is positively oriented if det(aij) > 0.

This is well defined precisely because det d(x−1b · xa)|q > 0.

3. We say that two orientations on M are the same if they induce the same orientation onevery tangent space.

Definition 3.13. f : M → N is a diffeomorphism ⇒ dfp : TpM → Tf(p)N is an isomorphismfor all p ∈M .

If M,N are oriented we say that f is orientation preserving if det dfp > 0 for all p ∈ M(where detdfp is calculated using positively oriented bases).

Examplef(p) = −p for p ∈ Rn then dfp = −id so det dfp = (−1)n therefore f is orientation preservingiff n is even.

Proposition 3.15 If f : M → N is smooth and dfp is an isomorphism, then f is a localdiffeomorphism at p.

Proof Apply the inverse function theorem on coordinate charts. QED.

We want to think about relating manifold theory tot eh theory of surfaces in R3.

Definition 3.16.f : M → N is an immersion if f is smooth and dfp is injective for all p ∈M .f : M → N is an embedding if it is an immersion and a homeomorphism onto its image f(M).

If if is an immersion/embedding we call f(M) an immersed/embedded submanifold on N .If we just say submanifold then assume embedded.

Proposition 3.17. A surface in R3 is a submanifold of R3.

Proof x is an embedding for the coordinate chart (U, x) ⇒ the inclusion map from S to R3

is an embedding. QED.

Proposition 3.18 F : Rn+m → Rm smooth, c a regular value of F then F−1(c) is an n-dimensional submanifold of Rn+m.

Moreover, TpF−1(c) = ker dFp.

26

Definition 3.19. TM = {(p,X) : p ∈ M,X ∈ TpM} is the tangent bundle of M . If

(x1, . . . , xn) are coordinates on M then

(x1, . . . , xn,

∂

∂x1, . . . ,

∂

∂xn

)are coordinates on TM

therefore TM is a 2n-dimensional manifold.

Examples

1. Let S = {(x, y, z) : x2 + y2 − 1}. Define f : S × R2 → TS by,

f(p, (u, v)) = (p, u(y∂x − x∂y) + v∂z)

Then f is a diffeomorphism, so TS ∼= S × R2.

2. For S2 we get TS2 = {(p,X) : p ∈ S2, X ∈ {p}⊥}

I can identify TS2 with the space of oriented affine lines in R3 : (p,X)→ line through Xin dirextion of p.

Here TS2 � S2 × R2.

Definition 3.20. A vector field X on M is an assignment of X(p) ∈ TpM for all p ∈M suchthat the map p→ (p,X(p)) ∈ TM is smooth.

Remark

1. Vector fields X on M give a way of differentiating functions on M (since X(p) ∈ TpMdifferentiates functions at p0.

2. Vector field is a choice of “arrows” (vectors in Rn) tangent to M at each point.

Examples

1. Cylinder {(x, y, z) : x2 + y2 = 1}

X = y∂x − x∂y, Y = ∂z

2. On S2, X is well defined and non-zero everywhere except at the north and south poles.

(Hairy Ball Theorem: There does not exist a nowhere vanishing vector field on S2)

27

Definition 3.21 Let α : (−ε, ε)→ M be a curve on a manifold M . X is a vector field alongα if X assigns X(α(t)) ∈ Tα(t)M for each t such that t 7→ (α(t), X(α(t)) ∈ TM is smooth.

A particular case is given by α′, the velocity field of α, where α′(t) is the tangent vector to α at t.

X,Y vector fields. Say X = Σiai∂i, Y = Σjbj∂j Consider the composition XY = Σi,jai∂ibj∂j .This may not be a vector field.

Lemma 3.22. X,Y vectore fields ⇒ ∃ vector fields Z on M such that Zf = (XY − Y X)ffor all smooth functions f : M → R.

Proof Let {(Ua, xa)} be an atlas for M . Suppose xa(Ua) for some a,

X =∑

i ai∂i, Y =∑

j bj∂j (ai, bj : xa(Ua)→ R)

(XY − Y X)f = (∑i,j

(ai∂ibj − bj∂jai)f (?)

Suppose Z exists. Then Z is given by (?) on xa(Ua)⇒ Z is unique.

Now, instead, define Z on each xa(Ua) by (?). By uniqueness, if Vab 6= ∅ then Z|xa(Ua)(p) =Z|xb(Ub)(p) for all p ∈ Vab therefore Z is well defined. QED.

Definition 3.23. X,Y vector fields then define the Lie Bracket,

[X,Y ] = XY − Y X

is a vector field.

Proposition 2.34. (Properties of the Lie Bracket)X,Y, Z vector fields, a, b function on M .

1. [X,Y ] = −[Y,X] (skew symmetric)

2. [X + Y,Z] = [X,Z] + [Y, Z] (linear)

3. [aX, bY ] = ab[X,Y ] + aX(b)Y − bY (a)X (where X(b), Y (a) are functions)

4. [[X,Y ], Z] + [[Y,Z], X] + [[Z,X], Y ] = 0 (Jacobi Identity)

Definition 3.25. If X a vector field, then a curve α : (−ε, ε)→M is an intergral curve ofX if α′(t) = X(α(t))

Definition 3.26. X a vector field, p ∈M ⇒ ∃ open U containing p and smooth map

ΦX : (−ε, ε)× U →M (for any ε > 0)

such that αq(t) = ΦX(t, q) is the unique integral curve of X with Xq(0) = q ∈ U .

Let φXt (q) = ΦX(t, q)∀q ∈ U (φXt : U →M)

Then {φXt : t ∈ (−ε, ε)} is called the flow of X. The φXt are diffeomorphisms onto theirimage. (Notice that φX0 is the identity)

28

Chapter 4

Riemmanian Manifolds

Definition 4.1. A Riemannian metric g on M is an assignment of an inner product gp oneach tangent space TpM such that if X,Y are vector fields on an open set U ⊆M then g(X,Y )is a smooth function on U .

p 7→ gp(X(p), Y (p))

Then I call (M, g) a Riemannian manifold

Fact: Every manifold admits a Riemannian metric.

Examples

1. Usual metric on Rn (standard metric on each tangent space)

2. First fundamental form of a surface in R3

I(X,Y ) = 〈X,Y 〉R3 for X,Y vector fields on the surfaces.

3. (M, gM ), (N, gN ) (M ×N, gM×N ) the product Riemannian geometry.

4. A group G acting properly discontinuously on M a Riemannian metric on M /G , thendefine g by:

gp(X,Y ) = hπ(p)(dπp(X), dπp(Y )) (Covering metric)

(Recall π : M →M /G)

Definition 4.2 f : (M, g)→ (N,h) is an isometry if f is a diffeomorphism and gp(X,Y ) =hf(p)(dfp(X), dfp(Y )), for all X,Y ∈ TpM , for all p ∈M .

f(M, g) → (N,h) is a local isometry at p ∈ M if there exists open U containing p, openV containing f(p) such that f : U → V is an isomtery.

f is an isomtery if f is f is a local isomtery everywhere.

Remark

1. (U, x) coordinate chart on (M, g) then g(∂i, ∂j) = gij smooth functions on x(U) (where∂i, ∂j vector fields on x(U)).

Therefore can now view g locally as a symmetric n× n matrix of smooth functions (gij)

29

2. If there exists coordinate charts (U, x) for (M, q) and (U, y) for (N,h) such that Gij = hijon U then y · x−1 : x(U)→ y(U) is an isometry.

Examples

1. G acting purely discontinuously on M , h Riemannian metric on M /G ⇒ define a Rie-mannian metric g on M by

gp(X,Y ) = hπ(p)(dπp(X), dπp(Y )) (4.1)

for p ∈M and where π : M →M /G . g is called the covering metric.

This construction will work whenever I have a covering map π : M → (N,h).

If instead I start with a metric g on M such that the diffeomorphisms φx are isometriesfor all x ∈ G, then I can use (4.1) to define an induced quotient metric on M /G .

2. Suppose f : (M, g) → (N,h) is an immersion. Then we can consider the metric g on Mby,

gp(X,Y ) = hf(p)(dfp(X), dfp(Y ))

I say that f is an isometric immersion if g = g.

Theorem 4.3. The Fundamental Theorem of Riemannian GeometryLet (M, g) be a Riemannian manifold. Given vector fields X,Y defined on open sets U ⊆M, ∃!way of defining a vector field ∇XY on U such that, if a, b are functions on U , and Z is a vectorfield on U , then

1. ∇aX+bY Z = a∇XZ + b∇Y Z

2. ∇X(Y + Z) = ∇XY +∇XZ

3. ∇X(aY ) = a∇XY +X(a)Y

4. X(g(Y,Z)) = g(∇XY, Z) + g(Y,∇XZ)

5. ∇XY −∇YX = [X,Y ]

Call ∇XY the Levi-Civita Connection of g.

Remark Condition 1,2,3 say that ∇ is an immersion. 4 says that ∇ is compatable with themetric, 5 says that ∇ is torsion-free or symmetric.

Proof Suppose there exists ∇ satisfying 1, 2, 3, 4, 5. Then apply 4 three times, i.e.

X(g(Y,Z)) = g(∇XY,Z) + g(Y,∇XZ)

Y (g(Z,X)) = g(∇Y Z,X) + g(Z,∇YX)

Z(g(X,Y )) = g(∇ZX,Y ) + g(X,∇ZY )

Using 5 get,

X(g(Y, Z)) + Y (g(Z,X))− Z(g(X,Y )) =

2g(∇XY, Z) + g(X, [Y,Z])− g(Y, [Z,X])− g(Z, [X,Y ])

30

(e.g. have g(X, [Y,Z]) = g(X,∇Y Z −∇ZY ) = g(X,∇Y Z)− g(X,∇ZY ). If you bash these outit should add up to what we want)

⇒ g(∇XY,Z) =1

2{X(g(Y,Z)) + Y (g(Z,X))− Z(g(X,Y ))

− g(X, [Y, Z]) + g(Y, [Z,X]) + g(Z, [X,Y ])} (4.2)

Now the RHS of this is independent of ∇, so implies that ∇ is unique. Now define ∇ by (4.2)instead, and recover conditions 1 through 5. Condition 1, 2, 3 are easy but messy. Won’t dothem. For 4 observe that the last 5 terms in (4.2) are anti-symmetric in Y, Z.

⇒ g(∇XY,Z) + g(Y,∇XZ) =1

2X(g(Y, Z)) +

1

2X(g(Z, Y ))

= X(g(Y,Z))

For 5 I show that the first 5 terms in (4.2) are symmetric in X,Y

⇒ g(∇XY,Z)− g(∇YX,Z) =1

2g(Z, [X,Y ])− 1

2g(Z, [Y,X])

= g([X,Y ], Z)

�

Remark (4.2) is well defined if Y is defined along an integral curve of X.

Definition 4.4. A topological group G is a Lie Group if it is a manifold such that the maps

G×G→ G

(x, y) 7→ xy

and

G→ G

x 7→ x−1

are smooth.

Examples

1. n× n matrices Mn(R) form a group under addition thought of as Rn2as a manifold with

addition/subtraction smooth maps.

2. GL(n,R) = {n × n matrices over R : det 6= 0}. This is open in Rn2, a Lie group of

dimension n2.

3. SL(n,R) = {A ∈Mn(R) : det(A) = 1}. A Lie group of dimension n2 − 1.

4. O(n) and SO(n) are Lie groups of dimension 12n(n− 1).

Definition 4.5. Left and Right AdjoinDefine Lx, Rx by Lx(y) = xy and Rx(y) = yx for all x, y ∈ G. We say that a metric g on G isleft invariant if

gy(X,Y ) = gLx(y)(d(Lx)yX, d(Lx)yY )

for all x, y ∈ G,X, Y ∈ TyG. Similarly can define a right invariant metric on G.

31

Remark We say that a metric g on G is bi-invariant iff g is left and right invariant.

Example If I take any inner product ge on TeG, then I define a Riemannian metric g on Gby

gx(X,Y ) = ge(d(Lx−1)xX, d(Lx−1)xY )

with X,Y ∈ TxG, x ∈ G. Then of course g is left invariant.

Remarks

1. TG = G× TeG

2. G is not guarenteed to have a bi-invariant metric, but it will have one if G is compact.

3. The example defines a bi-invariant metric iff ge(X, [Y,Z]) = ge([X,Y ], Z) for all X,Y, Z ∈TeG.

Definition 4.6. X is a left invariant vector field on G if d(Lx)y(X(y)) = X(xy) for allx, y ∈ G.

Given X(e) ∈ TeG, can define a left invariant vector field X by,

X(x) := d(Lx)e(X(e))

therefore ∃ a one-to-one correspondence between TeG and left invariant vector fields in G.

Lemma 4.7. If X,Y are left invariant vector fields in G, then [X,Y ] is also a left invariantvector field in G.

Definition 4.8. Lemma 4.7. ⇒ one can define a bracket operation [·, ·] on TeG by,

[Xe, Ye] := [X,Y ]e

We call g = TeG with this bracket the Lie Algebra of G.

Examples

1. g(GL(n,R)) = Mn(R) with commutator as the Lie bracket.

2. g(SL(n,R)) = {A ∈Mn(R) : tr(A) = 0}.

c.f. tr([A,B]) = tr(AB)− tr(BA) = 0

(M, g) a Riemannian manifold. (U, x) a coordinate chart. Vector fields E1, . . . , En on x(U)given by Ei = ∂

∂xi

X,Y vector fields on x(U)⇒ X =∑

i aiEi, Y =∑

j bjEj Using proper-ties 1, 2, 3, 4, 5from defintionof ∇

32

∇XY = ∇∑i aiEi

∑j

bjEj

=∑i

ai∇Ei(∑j

bjEj)

=∑i,j

aibj∇EiEj + aiEi(bj)Ej

∇EiEj =∑k

ΓkijEk

⇒ ∇XY =∑k

(X(bk) +∑i,j

Γkijaibj)Ek

Therefore ∇ is defined locally by Γkij

Definition 4.9. The functions Γkij are the Christoffel symbols of ∇ (or of g) in the coordi-nate nhd x(U).

Remark Γkij depends on the choice of (U, x).

Proposition 4.10. If g = (gij) on x(U), and g−1 = (gij) on x(U) then

Γkij = Γkji (4.3)

and,

Γkij =1

2

∑l

gkl(∂igjl + ∂jgil − ∂lgij) (4.4)

Remark Γkij = Γkji justifies the notion of symmetric connection.

Proof of 4.10.

∇EiEj −∇EjEi = [Ei, Ej ] = 0

⇐⇒∑k

(Γkij − Γkji)Ek = 0

⇐⇒ Γkij = Γkji

Proves (4.3). To prove (4.4),

g(∇EiEj , El) =∑m

g(ΓmijEm, El)

=∑m

Γmij gml

=1

2

{Ei(g(Ej , El)) + Ej(g(Ei, El)− El(g(Ei, Ej)))

}=

1

2

(∂igjl + ∂jgil − ∂lgij

)Multiply by gkl for result.

�

33

Examples

1. For the cylinder, x(u, v) = (cosu, sinu, v)⇒ xu and xv are orthonormal⇒ g =

(1 00 1

)⇒

Γkij = 0

2. For S2, x(u, v) = (cosu sin v, sinu sin v, cos v), u ∈ (0, 2π), v ∈ (0, π).

E1 = xu, E2 = xv are orthogonal, but g(E1, E1) = sin2 v and g(E2, E2) = 1

Using (4.4) we can calculate the following

Γ111 = 0,Γ1

22 = 0,Γ112 = cot v

Γ211 = − sin v cos v,Γ2

22 = 0,Γ212 = 0

For example

Γ112 =

1

2

∑l

g1l(∂1g2l + ∂2g1l − ∂lg12)

=1

2g11(∂1g21 + ∂2g11 − ∂1g12) +

1

2g12(∂1g22 + ∂2g12 − ∂2g12)

where g11 = sin2 v, g22 = 1, g12 = g21 = 0 as E1, E2 are orthogonal. Hence,

Γ112 =

1

2g11(∂2g11) =

1

2g11(

∂

∂vsin2 v) =

1

2g11(2 cos v sin v)

This somehow boils down to cot v. Find out how.

∇E1E1 = − sin v cos vE2

∇E2E2 = 0

∇E1E2 = cot vE1 Away from theequator E2begins to curve‘up’, or at leastdifferently. E1is unchangedaway from theequator.

34

Definition 4.11. Given a curve α in M and a vector field X along α then the covariantderivative of X along α is

DX

Dt= ∇α′X (4.5)

In x(U) I can write α(t) = (x1(t), . . . , xn(t)) and X =∑

i ai(t)Ei.

α′(t) =∑

i x′i(t)Ei = d

dt

DX

Dt=∑k

(a′k +∑i,j

Γkijaix′j)Ek (4.6)

If X = α′, then,

Dα′

Dt=∑k

(x′′k +∑i,j

Γkijx′ix′j)Ek (4.7)

Example For T 2 in R3, take x(u, v) = ((2 + sin v) cosu, (2 + sin v) sinu, cos v),

Let α(t) = (s cos t, 2 sin t, 1) be a curve on T 2.

E1 = xu, E2 = xv

Notice that E1|(t,0) = α′(t). Take X(t) = (0, 0, 1) ∈ Tα(t)R3, g(X,Ei) = 0

⇒ E1(g(X,Ei))︸ ︷︷ ︸=0

= g(∇E1X,Ei) + g(X,∇E1Ei)︸ ︷︷ ︸=0

Therefore ∇E1X is normal to T 2.

g(X,X) = 1⇒ g(∇E1X,X) = 0⇒ ∇E1X = DXDt = 0

Definition 4.12. A vector field X along a curve α is called parallel if DXDt = 0

Theorem 4.13. Let p, q ∈ M,α : [0, L] → M is a curve such that α(0) = p and α(L) = q,and let X0 ∈ TpM

∃! parallel vector field X along α such that X(p) = X0

The map τα : TpM → TqM given by τα(X0) = X(q) is an isomorphism called the paralleltransport along α.

35

Proof STP this in the case where α([0, L]) ⊆ x(U) since using compactness of [0, L] we cancover [0, L] with a finite number of intersecting open intervals such that the curve segmentsare contained in the coordinate nhds, and uniqueness implies we get a well defined vector fieldalong all of α.

X is parallel along all of α iff the right had side of (4.6) is zero. This is a linear first order ordi- Second con-dition fromdefinition 4.11.nary differential equation in n unknowns (a1, . . . , an) with initial conditions (a1(0), . . . , an(0)) =

X0 ⇒ ∃! solution.

τα is an isomorphism because we can use the backward curve β(t) = α(L− t), i.e. (τ)−1 = τβ

�

Remark This allows us to connect two tangent spaces.

36

Chapter 5

Geodesics

Definition 5.1. A curve γ in M is a geodesic if γ′ is parrallel along γ.

That is, a curve γ : I →M is a geodesic ifD

dtγ′ = 0 for all t ∈ I.

If γ is a geodesic thend

dt〈γ′, γ′〉 = 2〈D

dtγ′, γ′〉 = 0 (do Carmo)

d

dtg(γ, γ′) = 2g(∇γ′γ′, γ′) = 0 (lectures)

In other words, |γ′| = c, a constant. We say γ is normalised if |γ′| = 1.

In local cordinates, γ(t) = (x1(t), . . . , xn(t)) we have

x′′k +∑i,j

Γkijx′ix′j = 0

are the geodesic equations

Examples

1. Rn : Γkij = 0⇒ geodesics satisfy x′′i = 0 i.e. straight lines

2. Cylinder x(u, v) = (cosu, sinu, v)

Γkij = 0⇒ geodesics are helices (cos(at+ b), sin(at+ b), ct+ d)

3. S2: x(u, v) = (cosu sin v, sinu sin v, cos v)

Recall from previous example that only non-zero Γkij are Γ112 = cot v and Γ2

11 = − sin v cos v.Hence we get the geodesic equations

u′′ + 2 cot v u′ v′ = 0v′′ − sin v cos v u′ u′ = 0

v = constant ⇒ u not a constant.v = π

2 ⇒ u′′ = 0

Hence geodesic with v a constant is the equator.

37

Remark Motivation for studying geodesics: We want to know how to go from one point toanother on a Riemannian manifold in a ‘distinguished’ way. In Rn could take the straight linebetween two points. On a Riemannian manifold, we travel along s geodesic.

Proposition 5.2. ∃! vector field G on TM where integral curves are (γ, γ′) for geodesics γ inM .

G is called a geodesic field and it’s flow is called geodesic flow.

Proof If (x1, . . . , xn, y1, . . . , yn) with yi = ∂∂xi

are local coordinates on TM , then (γ, γ′) satisfy

x′k = yk

y′k = −∑i,j

Γkijyiyj (†)

If (p,X) ∈ TM ∃! solutions to (†) with γ(0) = p and γ′(0) = X defined from t ∈ (−ε, ε)say ⇒ uniqueness ofG if it exists. Use (†) to define G locally. Uniqueness implies G definedglobally.

�

38

Theorem 5.3. Let p ∈ M . There exists open U containing p, ε > 0 and a smooth mapΓ : (−2, 2)× V →M where

V = {(q,X) : q ∈ U,X ∈ TqM s.t. |X| < ε}

such that γ(q,X)(t) = Γ(t, q,X) curve is the unique geodesic in M with γ(0) = q, γ′(0) = X.

Lemma 5.4. If γ geodesic defined on −δ, δ) with γ(0) = q, γ′(0) = X then the geodesic γwith γ(0) = q, γ′(0) = sX is defined on (− δ

s ,δs) for s > 0, and γ(t) = γ(st).

Proof Let α(t) = γ(st) for s > 0.

Therefore α is defined on (− δs ,

δs), α(0) = γ(0) = q

α′(0) = sγ′(0) = sX and

∇α′α′ = ∇sγ′sγ′ = s2∇γ′γ′ = 0

Therefore α geodesic. Uniqueness ⇒ α = γ

�

Proof of 5.3. 5.2. ⇒ geodesic flow Φ is defined on (−δ, δ)×V for some δ > 0, ε > 0 and openU containing p.

5.4. ⇒ define Γ(t, q,X) = Φ( δ2 t, q,δ2X).

�

Definition 5.5. Use the notation of Theorem 5.3.

Define exp : V →M by exp(q,X) = γ(q,X)(1).

This is called the exponential map. We usually restrict to

expp : Bε(0) ⊆ TpM →M

given by expp(X) = γ(p,X)(1).

Remark expp(sX) = γ(p,sX)(1) = γ(p,X)(s). So

39

Proposition 5.6. ∃ ε > 0 such that expp : Bε(0) ⊆ TpM → M is a diffeomorphism onto it’simage.

Proof STP d(expp)0 is an isomorphism, since this implies expp is a local diffeomorphism, andthen we are done. (since I can shrink the bit I’m looking at at it will be a diffo onto it’s image).

We may identify T0(TpM) with TpM . So, if X ∈ T0(TpM) consider

d(expp)o(X) =d

dt{expp(tX)}|t=0

=d

dt(γ(p,X)(t))|t = 0

= γ′(p,X)(0) = X

Hence d(expp)0 : To(TpM)→ TpM is the identity, so an isomorphism.

�

Theorem 5.7. Given p ∈ M there exists open W containing p, and δ > 0 such that ∀ p ∈W, expq : Bδ(0) ⊆ TpM →M is a diffeomorphism onto it’s image.

Proof Let U, V, ε be as in Theorem 5.3.

Define F : V →M ×M sending (q,X) 7→ (q, expq(X))

Then, WRT to the natural coordinates on the product on M ×M

dF |(p,0) =

(I I0 I

)(in block notation, by proof of prop 5.6.)

Therefore dF(p,0) is an isomorphism ⇒ F is a local diffeomorphism at (p, 0).

Therefore ∃ δ ∈ (0, ε) and open U ⊆ U , open W ∈M ×M with (p, p) ∈ W such that if

V = {(q,X) : q ∈ U ,X ∈ TqM s.t. |X| < δ} ⊆ V

then F : V → W is a diffeo.

Choose open W containing p such that W ×W ⊆ W

If q ∈W ⇒ F ({q} ×Bδ(0)) ⊇ {q} ×W so W ⊆ expq(Bδ(0)).

�

40

Definition 5.8. A piecewise smooth curve (pw-curve) α : [0, 1]→M is a continuous curvesuch that ∃ a partition 0 = t0 < t1 < · · · < tk < tk+1 = L with α|(ti,ti+1) smooth ∀ i. Thepoints α(ti) are the vertices of α and the angle between α′(ti−) and α′(ti+) is called thevertex angle.

Definition 5.9. The length of a piecewise linear curve α : [0, L]→M is

L(α) =

∫ L

0|α′(t)|dt

α is minimising if L(α) ≤ L(β)∀ pw-linear β : [0, L]→M with β(0) = α(0) and β(L) = α(L)

Definition 5.10. p ∈ M open, V containing p is a normal neighborhood of p if ∃ openU ⊆ TpM such that expp : U → V is a diffeomorphism.

If ε > 0 such that the closed ball Bε(0) ⊆ U then we define

Bε(p) := expp(Bε(0))

to be the geodesic ball of radius ε about p. Moreover we define the boundry Sε(p) = ∂(Bε(p))to be the geodesic sphere of radius r about p.

(Prop 5.6 ⇒ normal nhds exist)

We say open W ⊆M is a totally normal nhd if W is normal for all q ∈W .

(Theorem 5.7. ⇒ existance of totally normal nhds exist)

Proposition 5.11. Geodesics γ : [0, L] → M contained in Bε(p) where p = γ(0) and ε > 0,are minimising.

Moreover, if α : [0, L] → M is a pw-curve with the same endpoints, and L(α) = L(γ) thenthey have the same image i.e. α([0, L]) = γ([0, L])

Lemma 5.12. (Symmetry)Let A, V ⊆ R2 with V open, A connected, V ⊆ A ⊆ V with ∂A a pw-curve with vertex angles6= π.Let f : A→M be smooth, and (u, v)-coordinates on A. Then

D

du

∂f

∂v=D

dv

∂f

∂u

whereD

du,D

dvare covarient derivatives along curves v = constant, u = constant.

41

Proof (U, x) a coordinate chart at p ∈ f(A)⇒ x−1 ◦ f(u, v) = (x1(u, v), . . . , xn(u, v)). There-fore

D

du

∂f

∂v=D

du

∑i

∂xi∂v

∂

∂xi=∑i

∂2xi∂u∂v

∂

∂xi+∑i,j

∂xi∂v

∂xj∂u∇ ∂

∂xj

∂

∂xi

This is symmetric in u, v since ∇ ∂∂xj

∂∂xi

is symmetric in i, j.

�

Lemma 5.13. (Gauss Lemma)Let p ∈M,X ∈ TpM such that expp(X) is well defined. If Y ∈ TX(TpM) = TpM

1 then

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

Proof Write Y = Y T + Y ⊥ where Y T ∈ span{X} and Y ⊥ ∈ {X}⊥.

g(d(expp)X(X), d(expp)X(Y T )) = g(X,Y T )

by definition of of expp since Y T = λX. So it is enough to show that

I = g(d(expp)X(X), d(expp)X(Y ⊥)) = 0

if Y ⊥ 6= 0.

∃ ε > 0 and a curve X(t) for t ∈ (−ε, ε) in TpM such that X(0) = X and X ′(0) = Y ⊥

with |X(t)| = constant, and expp(sX(t)) is well-defined for s ∈ [0, 1]

Let f(s, t) = expp(sX(t)), for s ∈ [0, 1], t ∈ (−ε, ε)

∂f

∂s= d(expp)sX(t)(X(t))

∂f

∂t= d(expp)sX(t)(sX

′(t))

Therefore I = g(∂f∂s (1, 0), ∂f∂t (1, 0))

Curves with t fixed (s 7→ f(s, t)) are geodesics and symmetry theorem ⇒

∂

∂sg(∂f

∂s,∂f

∂t) = g(

D

dt

∂f

∂s,∂f

∂t) + g(

∂f

∂s,D

dt

∂f

∂t)

= g(∂f

∂s,D

dt

∂f

∂s)

=1

2

∂

∂tg(∂f

∂s,∂f

∂s)

1this is an obvious equality

42

which is proportional to 12∂∂t |X(t)|2 = 0. Therefore

g(∂f

∂s,∂f

∂t)(s, 0) = g(

∂f

∂s,∂f

∂t)(1, 0) ∀ s = I

∂f

∂t(s, 0) = d(expp)sX(sY ⊥)

= s2d(expp)X(Y ⊥)→ 0 (as s→ 0)

⇒ I = 0

Proof of 5.11. Suppose α is a pw-curve with same endpoints as γ. Suppose that α([0, L]) isnot contained within the geodesic ball ⇒ let T ∈ [0, L] be the least such that α(T ) ∈ Sε(p) ⇒L(α) ≥ L(α|[0,T ]) ≥ ε > L(γ)

So suppose α([0, L]) ⊆ Bε(p). Suppose WLOG α(t) 6= p. α(t) = expp(r(t)X(t)) for t > 0,where r : (0, L]→ R+ piecewise smooth, and X(t) a curve in TpM with |X(t)| = 1.

Write α(t) = f(r(t), t). Hence α′(t) = r′(t) dfdr + dfdt (chain rule).

Now use Gauss lemma: g( dfdr ,dfdt ) = g(X,X ′) = 0 since |X| = 1, constant. And g( dfdr ,

dfdr ) =

g(X,X) = 1.

Therefore |α′(t)|2 = |r′|2 + |dfdt |2 ≥ |r′|2

Let δ > 0, ∫ L

δ|α′(t)|dt ≥

∫ L

δ|r′(t)|dt

≥∫ L

δr′(t)dt

= r(L)− r(δ)

Let δ → 0⇒ L(α) ≥ r(L) = 2L(γ) (notice, s 7→ expp(sX(L)) is a geodesic, which is γ).

Moreover, L(α) = L(γ) only if dfdt = 0 and |r′| = r′ ⇒ α(t) = expp(r(t)X(L)) (with r′ > 0) ⇒ α

is a monotonic reparametrisation of γ.

�

Proposition 5.14. γ : [0, L] → M pw-curve with |γ′| constant and minimsing ⇒ γ is ageodesic.

Proof Let t ∈ [0, L], W a totally normal nhd of γ(t)⇒ ∃ closed interval I ⊆ (0, L such thatt ∈ I, I0 6= 0 and γ(I) ⊆W,γ|I pw-curve and joins two points p, q in a geodsic ball.

Proposition 5.11. ⇒ L(γ|I) is length of radial geodesic (s 7→ expp(sX)) for p to q ⇒ γ|I isa geodesic in I ⇒ γ is a geodesic at t.

2maybe should be r(L)− r(0)

43

5.1 Variation

Definition 5.15. Let α : [0, L] → M be a pw-curve. A variation of α is a cts mappingf : (−ε, ε) × [0, L] → M such that f(0, t) = α(t) ∀ t ∈ [0, L] and there exists some partition0 = t0 ≤ t1 ≤ · · · ≤ tk ≤ tk+1 = L with f |(−ε,ε)×[ti,ti+1] is smooth for all i.

A variation is proper if f(s, 0) = α(0) and f(s, L) = α(L) ∀ s.

We say the variation field of f is the vector field Vf (t) =∂f

∂s(0, t) along α.

Proposition 5.16. Given a piecewise smooth vector field V along a pw-curve α, there existsa variation f of α such that V = Vf .Moreover, if V (0) = V (L) = 0 we can choose f proper.

Proof For every t ∈ [0, L] let Wt = {expα(t)X : |X| < δt} be a totally normal nhd of α(t).

Thus {Wt : t ∈ [0, L]} is an open cover for α([0, L]) compact. Therefore ∃ finite subcover{Wt1 , . . . ,Wtk} and δ = mini{δti}.

Let ε ∈ (0, δmaxt∈[0,L] |V (t)|) and let f(s, t) = expα(t)(sV (t)) for s ∈ (−ε, ε) and t ∈ [0, L].

Vf (t) =∂f

∂s(0, t) = d(expα(t))0(V (t)) = V (t)

If V (0) = V (L) = 0 then f(s, 0) = α(0) and f(s, L) = α(L) and this is proper.

�

Definition 5.17. Energy of a pw-curve α : [0, L]→M is

E(α) =

∫ L

0|α′(t)|2dt

44

Energy of a variation f of α is

Ef (s) =

∫ L

0

∣∣∣∣∂f∂t (s, t)

∣∣∣∣2 dtLemma 5.18. L(α)2 ≤ L · E(α) with equaity iff |α′| is constant. L is just

a number,whereas L(α)is a function.

∫ L

01 = L

Proof (Cauchy-Scwartz):(∫ L

0a(t)b(t)dt

)2

≤∫ L

0a(t)2dt

∫ L

0b(t)2dt

with equality iff b = λ for some constant λ. Let a = 1 and b = |α′|.

�

Lemma 5.19. Let p, q ∈M,γ : [0, L]→M be a minimising geodesic from p to q. ∀ pw-curvesα : [0, L]→M with α(0) = p, α(L) = q we have E(γ) ≤ E(α) with equality iff α is a minimisinggeodesic.

Proof Lemma 5.18 ⇒ LE(γ) = L(γ)2 ≤ L(α)2 ≤ LE(α) with equality iff |α′| is constant and L is just a num-ber

L(α) = L(γ). Proposition 5.14. ⇒ result.

�

Theorem 5.20. (First Variation Formula)Let α : [0, L]→M be a pw-curve, and let f be a variation of α. Then

1

2E′f (0) = −

∫g

(Vf ,

D

Dtα′)dt−

k∑i=1

g(Vf (ti), α′(ti+)− α′(ti−))

− g(Vf (0), α′(0)) + g(Vf (L), α′(L))

Remark E′f (0) = 0 if α is a geodesic and f is proper.

Proof of 5.20.

d

ds

∫ ti+1

ti

g

(∂f

∂t,∂f

∂t

)dt = 2

∫ ti+1

ti

g

(D

Ds

∂f

∂t,∂f

∂t

)dt

= 2

∫ ti+1

ti

g

(D

Dt

∂f

∂s,∂f

∂t

)dt

= 2

∫ ti+1

ti

d

dtg

(∂f

∂s,∂f

∂t

)dt− 2

∫ ti+1

ti

g

(∂f

∂s,D

Dt

∂f

∂t

)dt

⇒ 1

2E′f (s) =

k∑i=0

[g

(∂f

∂s,∂f

∂t

)]ti+1

ti

−∫ L

0g

(∂f

∂s,D

Dt

∂f

∂t

)dt

Set s = 0

�

45

Corollary 5.21. A pw-curve α : [0, L] → M a geodesic ⇐⇒ ∀ proper variations f ofα,E′f (0) = 0.

Proof ⇒: α a geodesic,

∴ α smooth and DDtα

′ = 0⇒ F ′f (0) = 0∀ proper f .

⇐: E′f (0) = 0 ∀ proper variations f ,

Let h : [0, L] → R≥0 such that h(t) > 0 for t 6= ti for all i and h(ti) = 0. Let V (t) =h(t)Ddtα

′ ⇒ V (0) = V (L) = 0⇒ ∃ proper variation f of α such that V = Vf (prop 5.16).

Theorem 5.20. ⇒ 0 = E′f (0) = −∫ L

0 h(t)| ddtα′|2dt⇒ α(ti,ti+1) geodesic ∀ i.

Let V (t) be a piecewise smooh vector field along α such that

V (0) = V (L) = 0

andV (ti) = α′(ti+)− α′(ti−) (i ∈ {1, . . . , k})

There exists proper variation f such that V = Vf ⇒ 0 = E′f(0) = −

∑ki=1 |α′(ti+)−α′(ti−)|2 ⇒

α is continuously differentiable on [0, L]. Uniqueness of solutions to ODEs ⇒ α is a geodesic.

�

46

Chapter 6

Curvature

Let (M, g) be a Riemannian n-dimensional manifold.

Definition 6.1. Riemannian curvature R of g is given by,

R(X,Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z

for vector fields X,Y, Z in an open set U in M .

Example M = Rn then [X,Y ] = 0 ∀ X,Y and

∇X∇Y = ∇Y∇X ∀ X,Y

∴ R = 0⇒M is flat (or zero curvature)

Remark(U, x) coordinate charts, vector fields ∂i = ∂

∂xi(hence [∂i, ∂j ] = 0), and,

R(∂i, ∂j)∂k = (∇∂i∇∂j −∇∂j∇∂i)∂k =∑l

Rlijk∂l

We can calculate,

Rlijk = ∂iΓjkl − ∂jΓlik +∑m

ΓlimΓmjk =∑m

ΓljkΓmki

Example

1. Γkij = 0 for Rn and Γkij preserved by local isometries ⇒ Tn = Rn /Zn has R = 0 as well.

2. Cylinder in R3 has Γkij = 0⇒ cylinder is flat.

3. For S2 we have coordinates x(u, v)→ S2, E1 = xu, E2 = xv

⇒ R(E1, E2)E2 = E1

∇E1E1 = − sin v cos vE2,∇E2E2 = 0 and ∇E1E2 = ∇E2E1 = cot vE1

Proposition 6.2. R(·, ·) is bilinear, R(X,Y ) is linear.

Proof Exersize. (see his notes)

47

Remark Consider g(R(X,Y )Z,W ). This is allowed because R(X,Y )Z is a vector field. Inlocal coordinates

g(R(∂i, ∂j)∂k, ∂l) =∑m

Rlijkgml =: Rijkl

Proposition 6.3. (Symmetries of R)Let X,Y, Z,W e vector fields on an open set in M .

1. g(R(X,Y )Z,W ) = −g(R(Y,X)Z,W )

Equivalently, Rijkl = −Rjikl

2. g(R(X,Y )Z,W ) = −g(R(X,Y )W,Z)

Equivalently, Rijkl = −Rijlk

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

Equivalently, Rijkl = Rklij

4. (Bianchi Identity)g(R(X,Y )Z,W ) + g(R(Y, Z)X,W ) + g(Z,X)Y,W ) = 0

Equivalently, Rijkl +Rjkil +Rkijl = 0

Definition 6.4. Let σ = Span{X,Y } ⊆ TpM be a 2-dimensional subspace. The sectionalcurvature of σ (at p) is given by

K(σ) = K(X,Y ) =g(R(X,Y )Y,X)

|X|2|Y |2 − g(X,Y )2∈ R

Lemma 6.5. K(σ) is independent of the choice of basis for σ.

Proof Let σ = Span{X,Y }. Any other basis {Z,W} for sigma can be written as Z =aX + bY,W = cX + dY for some a, b, c, d ∈ R such that ad− bc 6= 0.

g(R(Z,W )W,Z) = (ad− bc)2g(R(X,Y )Y,X)

by proposition 6.3.

|Z|2|W |2−gp(Z,W )2 = (a2|X|2+2abg(X,Y )+b2|Y |2)(c2|X|2+2cdg(X,Y )+d2|Y |2)−(ac|X|2+(ad+ bc)g(X,Y ) + bd|Y |2)2 = (ad− bc)2(|X|2|Y |2 − g(X,Y )2)

This is just algebra, but it shows that the choice of basis is irrelevant. This is good.

�

Lemma 6.6 Let R be an operator with the same properties as R in Propositions 6.2. and6.3. . Suppose that ∀ p ∈M and ∀ 2-dimensional subspaces σ = Span{X,Y } ⊆ TpM we havethat

K(σ) =gp(R(X,Y )Y,X)

|X|2p|Y |2p − gp(X,Y )2= K(σ)

Then R = R.

48

Proof On problems 4 (with hints).

Lemma 6.7. M has constant sectional curvature K ⇐⇒ Constantsectionalcurvaturemeans that forany choice ofσ we get thesame value Kg(R(X,Y )Z,W ) = K(g(X,W )g(Y, Z)− g(X,Z)g(Y,W ))

Proof (⇐)

K(X,Y )(|X|2|Y |2 − g(X,Y )2) = g(R(X,Y )Y,X)

= K(g(X,X)g(Y, Y )− g(X,Y )g(X,Y ))

Hence K(X,Y ) = K constant.

(⇒) Define R by

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

∴ Lemma 6.6. ⇒ R = R

�

Example In S2,

K(E1, E2) =g(R(E1, E2)E2, E1)

|E1|2=|E1|2

|E1|2= 1

Definition 6.8. The Rici curvature, Ric of M is given by

Ric(X,Y ) = Tr(X → R(X,Y )Z)

where X,Y, Z are vector fields on open V ⊆M .

If {E1, . . . , En} are an orthonormal basis for vector fields on V then

Ric(Y,Z) =

n∑i=1

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

n∑i=1

g(R(Y,Ei)Ei, Z)

We define Ric(Y ) by Ric(Y ) is a vec-tor field

g(Ric(Y ), Z) = Ric(Y,Z)

If |Y | = 1, we say Ric(Y, Y ) is the Rici curvature in the direction of Y .

Coordinate chart (U, x) on M , ∂1, . . . , ∂n} the usual basis for vector fields on x(U), then we candefine

Rij = Ric(∂i, ∂j) =∑j

g(R(∂i, Ek)Ek, ∂j)

=∑k,l

Rikljgkl

49

Definition 6.9. The Scalar curvature S of M , is given by a map fromvector fields tovector fields,so can take it’strace

S = Tr(X 7→ Ric(X))

for vector fields X defined on open V ⊆M

S =

n∑i=1

g(R(Ei, Ej)Ej , Ei)

=∑i,j

g(R(Ei, Ej)Ej , Ei)

In local coordinates

S =∑i,j,k,l

Rijklgijgkl

=∑i,j

Rijgij

Remark S = 0 is a weak condition. In general it does not imply that M is flat. We haveweakened the condition ever and ever again, we have taken traces everywhere we can andsummed lots of the information out.

Examples

1. S2

x(u, v) = (cosu sin v, sinu sin v, cos v), v ∈ (0, π), ∂1 = xu, ∂2 = xv

|∂1|2 = sin2 v |∂2|2 = 1 g(∂1, ∂2) = 0

Let E1 = ∂1|∂1| , E2 = ∂2. Then

Ric(∂1, ∂1) =2∑i=1

g(R(∂1, Ei)Ei, ∂1)

= g(R(∂1, E2)E2, ∂1)

= g(R(∂1, ∂2)∂2, ∂1) = sin2 v

Ric(∂1, ∂2) = 0

Ric(∂2, ∂2) = 1

Therefore we can get a matrix,

(Rij) =

(sin2 v 0

0 1

)= (gij)

Moreover, S =∑

i,j Ri,jgi,j = Tr

(1 00 1

)= 2. For s2, Ric = g, S = 2.

2. (M, g) is Einstein if Ric = λg for some constant λ. Ric = λg are Einstein’s vacuum fieldequations from General Relativity.

Remark: λ = 0 is especially interesting (Ricci-flat).

50

Proposition 6.10. Recall the notation of Lemma 5.12 (the symmtery lemma).In partcular we have a smooth map f : A ⊆ R2 → M with coordinates (u, v) in A. X isa vector field along f if X assigns X(f(u, v)) ∈ Tf(u,v)M ∀ (u, v) ∈ A such that(u, v) 7→f(u, v), X(f(u, v)) ∈ TM is smooth.

Then (D

Du

D

Dv− D

Dv

D

Du

)X = R(∂uf, ∂vf)X

Proof By definition of R since[DDu ,

DDv

]= 0.

�

6.1 Tensors in Riemannian Geometry

Remark We only need covariant tensors because of g

Definition 6.11. T is a (covariant) tensor of order r if T is a multi-linear map taking r(local) vector fields on M to a local smooth functions on M .

If (U, x) is a coordinate chart on M , basis {∂1, . . . , ∂n} for vector fields on x(U), then

T (∂i1 , . . . , ∂in) = Ti1...in

are the components of T WRT {∂1, . . . , ∂n}.

Examples

1. Let X be a vector field on M . X defines a tensor of order 1 by

ωX(Y ) = g(X,Y )

with components ai if X =∑n

i=1 ai∂i.

2. The metric g is a tensor of order 2 with components gij .

3. The Riemann curavtue R defines a tensor of order 4 by different R’s, itis poor nota-tion, but that isthe way it is

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

with componants Rijkl.

4. The Ricci curvature Ric defines a tensor of order 2 with components Rij

5. The operator ∇ given by

∇(X,Y, Z) = g(∇XY,Z)

is NOT a tensor.

51

Definition 6.12. The covariant derivative of a tensor T of order R is given by

∇T (X1, . . . , Xr, Y ) = Y (T (X1, . . . , Xr))−r∑j=1

T (X1, . . . , Xj−1∇YXj , Yj+1, . . . , Xr)

which is a tensor of order r+1.

We define (∇Y T )(X1, . . . , Xr) = ∇T (X1, . . . , Xr, Y )

Examples

1. For vector fields X,Y, Z

(∇Y ωX)(Z) = ∇ωX (Z, Y )

= Y (ωX(Z))− ωX(∇Y Z)

= Y (g(X,Z))− g(X,∇Y Z)

= g(∇YX,Z)

= ω∇YX(Z)

2.

∇g(X,Y, Z) = Z(g(X,Y ))− g(∇ZX,Y )− g(X,∇Z , Y )

= 0

Therefore ∇g = 0 ⇐⇒ g is a parallel tensor.

Remark

d(expp)sX(sY ⊥) = s d(expp)sX(Y ⊥)

Hence d(expp)sX(Y ⊥)→ d(expp)0(Y ⊥) = Y ⊥ as s→ 0.

dfp =(∂fi∂xj

(p))

matrix of smooth functions at p. If f : M → N smooth then

df : TM → TN

(p,X) 7→ (f(p), dfp(X))

is smooth.

Geodesic frame: Xi(t), Xj(t) parallel along α(t), then g(Xi, Xj) is constant along α.

d

dtx(Xi, Xj) = g(∇α′Xi, Xj) + g(Xi,∇α′Xj) = 0

(∇α′Xi = ∇α′Xj = 0). i.e. parallel transport map is a metric isomorphism. Geodesic coordi-nates / normal coordinates.

52

Chapter 7

Riemannian Submanifolds

i : (Mn, gM )→ (Nn+m, g)

isometric embedding (but can assume just immersion mainly). For p ∈M we write

TpN = TpM ⊕NpM

where NpM := (TpM)⊥. Therefore if X ∈ TpN , there exists unique XT ∈ TpM,X⊥ ∈ NpMsuch that X = XT +X⊥

Definition 7.1 X is a vector field along M if it assigns X(p) ∈ TpN ∀ p ∈ M such thatp→ (p,X(p)) is smooth.

We can uniquely write X = XT + X⊥ where XT (p) ∈ TpM and X⊥(p) ∈ NpM for all psuch that p→ (p,XT (p)) and p→ (p,X⊥(p)) are smooth.

We call XT a tangent vector field and X⊥ a normal vector field.

Remark Let ∇N ,∇M be the Levi-Civita connections of N and M respectively.

If X,Y vector fields on M then∇MX Y = (∇NX Y )T

where X, Y are any extension of X,Y to local vector fields on N . Since X, Y are arbitrry, wewill simply write X = X, Y = Y .

Definition 7.2. The second fundamental form B(X,Y ) of M is

B(X,Y ) = ∇NXY −∇MX Y = (∇NXY )⊥

for X,Y local tangent vector fields on M so B(X,Y )(p) ∈ NpM ∀ p ∈M .

Proposition 7.3. B(X,Y ) is bilinear and symmetric.

Proof Bilinearity follows from fact that ∇N ,∇M are bilinear.

B(X,Y )−B(Y,X) = ∇NXY −∇MX Y −∇NY X +∇MY X= [X,Y ] + [Y,X] = 0

�

53

Definition 7.4. For a tangent vector field X and a normal vector field ξ on open U in M , wedefine the shape operator Sξ by

G(Sξ(X), Y ) = g(B(X,Y ), ξ)

for all tangent vector fields on U . B(X,Y ) symmetric ⇒ Sξ is self adjoint.

Remark Notice that

G(Sξ(X), Y ) = g(B(X,Y ), ξ)

= g(∇NXY, ξ)= X(g(Y, ξ))− g(Y,∇Xξ)= −g((∇Xξ)T , Y ) (g(Y, ξ) = 0)

Proposition 7.5. Sξ(X) = −(∇Xξ)T

Definition 7.6. The normal connection ∇⊥ on M is given by

∇⊥Xξ = (∇NXξ)⊥ = ∇NXξ − (∇NXξ)T = ∇NXξ + Sξ(X)

for local tangent vector fields X and normal vector fields ξ on M .

We define the normal curvature R⊥ on M by

R⊥(X,Y )ξ = (∇⊥X∇⊥Y = ∇⊥Y∇⊥X −∇⊥[X,Y ])ξ

Theorem 7.7. (The Fundamental Equations)Let X,Y, Z,W be local tangent vector fields, and let ξ, ζ be local normal vector fields on M .

(Gauss)

g(RN (X,Y )Z,W ) = g(RM (X,Y )Z,W ) + g(B(X,Z), B(Y,W ))− g(B(X,W ), B(Y, Z))

(Codazzi)

g(RN (X,Y )Z, ξ) = g((∇NXB)(Y,Z), ξ)− g((∇NY B)(X,Z), ξ)

where (∇NXB)(Y, Z) = ∇⊥XB(Y,Z) = B∇(MX Y,Z)−B(Y,∇MX Z).

(Ricci)

g(RN (X,Y )ξ, ζ) = g(R⊥(X,Y )ξ, ζ) + g([Sξ, Sζ ]X,Y )

where [Sξ, Sζ ] = Sξ ◦ Sζ − Sζ ◦ Sξ

Corollary 7.8. If N has constant section curvature K then

1. K = KM (X,Y ) +|B(X,Y )|2 − g(B(X,X), B(Y, Y ))

|X|2|Y |2 − g(X,Y )2

2. (∇NXB)(Y,Z) = (∇NY B)(X,Z)

3. g(R⊥(X,Y )ξ, ζ) = −g([Sξ, Sζ ]X,Y )

54

Proof of Theorem 7.7.

RN (X,Y )Z = (∇NX∇NY −∇NY ∇NX −∇N[X,Y ])Z

= ∇NX(∇MY Z +B(Y,Z))−∇NY (∇MZ +B(X,Z))−∇M[X,Y ]Z −B([X,Y ], Z)

= (∇MX∇MY Z −∇MY ∇MX Z −∇M[X,Y ]Z) +B(X,∇MY Z)− SB(Y,Z)X +∇⊥XB(Y, Z)

−B(Y,∇MX Z) + SB(X,Z)Y −∇⊥YB(X,Z)−B([X,Y ], Z)

Therefore g(RN (X,Y )Z,W ) = g(RM (X,Y )Z,W )− g(SB(Y,Z)X,W ) + g(SB(X,Z)Y,W ). Whichimplies (Gauss) by definition of the shape operator.

g(RN (X,Y )Z, ξ) = g(B(X,∇MY Z), ξ) + g(∇⊥XB(Y,Z), ξ)− g(B(Y,∇MX Z), ξ)

− g(∇⊥YB(X,Z), ξ)− g(B([X,Y ], Z), ξ)

= g((∇NXB)(Y, Z), ξ)− g((∇NY B)(X,Z), ξ)

where [X,Y ] = ∇MX Y −∇MY X. This implies (Codazzi).

RN (X,Y )ξ = ∇NX∇NY ξ −∇NY∇NXξ −∇N[X,Y ]ξ

= ∇NX(−Sξ(Y ) +∇⊥Y ξ)−∇NY (−Sξ(Y ) +∇⊥Y ξ) + Sξ([X,Y ])−∇⊥[X,Y ]ξ

= −∇MX Sξ(Y )−B(Sξ(Y ), X)− S∇⊥Y ξ(X) +∇MY Sξ(X)+

B(Sξ(X), Y ) + S∇⊥Xξ(Y ) + Sξ([X,Y ]) +∇⊥X∇⊥Y ξ −∇⊥Y∇⊥Xξ −∇⊥[X,Y ]ξ

Therefore g(RN (X,Y )ξ, ζ) = g(R⊥(X,Y )ξ, ζ)+g(B(X,Sξ(Y ), ζ)−b(B(Sξ(X), Y, ζ) = g(R⊥(X,Y )ξ, ζ)+g(Sζ ◦ Sξ(Y ), X) − g(Sζ ◦ Sξ(X), Y ). Can pull the S’s across since as self-adjoint (i.e. g(Sζ ◦Sξ(Y ), X) = g(Sξ(Y ), Sζ(X)) = g(Y, Sξ ◦ Sζ(X)))

�

7.1 Hypersurfaces

Definition 7.9. An n-dimensional submanifold of an n + 1 dimensional manifold is called ahypersurface.

Assume M is an oriented hypersurface in oriented Riemannian (N, g).

Definition 7.10. Let p ∈ M,ν ∈ NpM such that |ν| = 1. Then Sν : TpM → TpM is selfadjoint so there exists (positive) orthonormal basis of eigenvectors {E1, . . . , En}. Determine νuniquely by requiring {E1, . . . , En} is a (positive) orthonormal basis from TpN ⇒ smooth mapν : M → TN such that ν(p) ∈ NpM ∀ p ∈M and |ν(p)| = 1 ∀ p.

If SνEi = λiEi we call λi the principle curvature at p, and Ei the principle directions at p.

We call KM (p) = λ1 · · ·λn the Gaussian curvature at p andHM (p) = 1

n(λ1 + · · ·+ λn) the mean curvature at p.

Definition 7.11. If N = Rn+1 then ν : M → Sn by ν(p) ∈ NpM ⊆ TpRn+1 = Rn+1. We callν the Gauss map and we can view dνp : TpM → TpM (as TpM is parallel to Tν(p)S

n).

55

Lemma 7.12. Tf N = Rn+1, ν is the Gauss map of M , then dνp = −Sν(p).

Proof If X ∈ TpM ∃ curve α such that α(0) = p, α′(0) = X. Therefore

dνp(X) =d

dt(ν ◦ α(t))|t=0 = (∇Xν)T = −Sν(p)(X)

(since ∇X 〈ν, ν〉︸ ︷︷ ︸=1

= 2〈ν,∇Xν〉 = 0)

�

Remark Kn(p) = det(−dνp) and HM (p) = − 1n tr(dνp).

Gauss equations ⇒ KM (X,Y ) − KN (X,Y ) = g(B(X,X), B(Y, Y )) − |B(X,Y )|2 if X,Y or-thogonal and unit length.

M hypersurface ⇒ Sν(Ei) = λiEi ⇒

g(B(Ei, Ej), ν) = g(SνEimEj) = λiδij

∴ if i 6= j then KM (Ei, Ej)−KN (Ei, Ej) = λiλj .

In particular if N = R3 then M is a surace in R3 and KM (E1, E2) = λ1λ2 = KM the Gaussiancurvature ⇒ Gauss Theorem Egregium.

Proposition 7.13. The sectional curvatures of Sn are all 1.

Proof Take ν(p) = −p. Lemma 7.12⇒ Sν = Id⇒ eigenvalues of Sν are all 1 asKSn(Ei, Ej) =1 for i 6= j.

�

7.2 Totally Geodesic and Minimal Submanifolds

Let ι : (Mn, gM )→ (Nn+m, g) be an isometric embedding (or immersion).

Definition 7.14. M is geodesic at p if B = 0 at p. M is totally geodesic if B = 0.

Proposition 7.15. M is geodicic at p iff every geodesic in M starting at p is geodesic in Nat p.

Proof Let γ be a geodesic in M with γ(0) = p, and let X = γ′.

∇MXX = 0 ⇐⇒ (∇MXX)T = 0

Therefore B(X,X) = (∇NXX)⊥ = 0 at t = 0 iff ∇NXX = 0 at t = 0.

Since every Y ∈ TpM can be realised as tangent vector to a geodesic at 0 starting at p weare done.

�

56

Remark Let V be a normal nhd of p ∈ N and let V = expp U , for U ⊆ TpN . Let σ be a2-plane in TpN . Then M = expp(σ ∩ U) is a 2-dim submanifold of N with p ∈M .

Poposition 7.15 ⇒M is geodesic at p⇒ B = 0 at p⇒ KM(σ) = KN

(σ) by Gauss equations.

KM (σ) is the Gaussian curvature of M at p. Therefore the sectional curvature KN (σ) ofN at p is the Gaussian curvature at p of a small surface in N containing p.

Definition 7.16. Given a local orthogonal basis ξ1, . . . , ξm of normal vector fields on M , wedefine the mean curvature vector by

H =1

n

m∑j=1

tr(Sξj )ξj

We say M is minimal if H = 0 i.e. tr(Sξ) = 0 ∀ ξ ∈ NpM, ∀ p ∈M .

Remark Minimal submanifolds are stationary points for the volume functional as can belocally minimising (but can also be maximising by the nature of the second derivative test).

57

Chapter 8

Jacobi Fields

Definition 8.1. Given a geodesic γ, the equation for a vector field J along γ given by

D2

Ds2J(s) +R(J(s), γ′(s))γ′(s) = 0

is called the Jacobi equation and it’s solution J is called a Jacobi field.

Consider f(s, t) = expp(sX(t)) for a curveX(t) in TpM withX(0) = X andX ′(0) = Y, s ∈ [0, 1].

d(expp)sX =∂f

∂t(s, 0) = J(s)

measures how geodesics spread from geodesic s→ expp(sX).

0 =D

Dt

(D

Ds

(∂f

∂s

))=

D

Ds

D

Dt

(∂f

∂s

)−R

(∂f

∂s,∂f

∂t

)∂f

∂s(Prop 6.10)

=D2

Ds2

(∂f

∂t

)+R

(∂f

∂t,∂f

∂s

)∂f

∂s(Lemma 5.12)

Set t = 0⇒ Jacobi equation. Therefore J is a Jacobi field.

58

Example γ′ is a Jacobi field everywhere non zero but zero derivation. sγ′ is a Jacobi fieldwith every non zero derivation γ′

Remark J is determined by J(0) and J ′(0) uniquely. Now J(0) ∈ Rn and J ′(0) ∈ Rn, sothere exist 2n linearly independent Jacobi fields along a geodesic.

Example Suppose M has constant sectional curvature K. Then R(J, γ′)γ′ = KJ if {J, γ′}everywhere orthonormal. Therefore the Jacobi equation in this situation (orthonormal) is

J ′′ +KJ = 0

Let X be a unit parallel vector field along γ such that g(X, γ′) = 0 then J with J(0) = 0 andJ ′(0) = X(0) is given by

K = λ2 > 0 ⇒ λ−1 sin(λs)X(s)

K = 0 ⇒ sX(s)

K = λ2 < 0 ⇒ λ−1 sinh(λs)(X(s)

Proposition 8.2 Let γ : [0, L] → M be a geodesic, γ(0) = p, J Jacobi field along γ withJ(0) = 0. Let X(t) be a curve in TpM such that X(0) = Lγ′(0) and X ′(0) = J ′(0) (whereX ′(0) ∈ TLγ′(0)(TpM) = T0(TpM) 3 J ′(0)). If

f(s, t) = expp(s

LX(t))

and

J(s) =∂f

∂t(s, 0)

then J = J .

Proof

D

DsJ(s) =

D

Ds(d(expp)sγ′(0)(sJ

′(0)))

= d(expp)sγ′(0)(J′(0)) + s

D

Dsd(expp)sγ′(0)(J

′(0))

(8.1)

⇒ J ′(0) = d(expp)0(J ′(0)) = J ′(0) and J(0) = 0 = J(0). J Jacobi field ⇒ by uniqueness thatJ = J

�

Corollary 8.3. If J(0) = 0 then J(s) = d(expγ(0))sγ′(0)(sJ′(0)).

Proposition 8.4. If J(s) = d(expp)sX(sY ) where γ(0) = p, γ′(0) = X,Y ∈ TX(TpM) suchthat |Y | = 1. Then Taylors expan-

sion

|J(s)|2 = s2 =1

3g(R(X,Y )Y,X)s4 + o(s4)

59

Proof Observe that J(0) = 0⇒ |J(0)|2 = 0. First term

g(J, J)′(0) = 2g(J, J ′)(0) = 0

g(J, J)′′(0) = 2g(J ′, J ′)(0) + 2g(J, J ′′)(0)

= 2|J ′(0)|2 = 2|Y |2 = 2

g(J, J)′′′(0) = 6g(J ′, J ′′)(0) + 2g(J, J ′′′)(0)

= −6g(J ′, R(J, γ′)γ′)(0) (by Jacobi eq)

g(J, J)(iv)(0) = 6g(J ′′, J ′′)90) + 8g(J ′, J ′′′)(0) + 2g(J, J (iv)(0)

= −8g(J ′,D

DsR(J, γ′)γ′)(0)

Observe that

g(D

DsR(J, γ′)γ′, Z) =

d

dsg(R(J, γ′)γ′, Z)− g(R(J, γ′)γ′, Z ′)

=d

dsg(R(Z, γ′)γ′, J)− g(R(J, γ′)γ′, Z ′)

= g(D

DsR(Z, γ′)γ′, J) + g(R(Z, γ′)γ′, J ′)− g(R(J, γ′)γ′, Z ′)

= g(R(Z, γ′)γ′, J ′)(0) (at s = 0)

g(J, J)(iv)(0) = −8g(R(J ′, γ′)γ′, J ′)(0)

= −8g(R(Y,X)X,Y )

Taylors theorem ⇒ |J(s)|2 = 22!s

2 − 84!g(R(X,Y )Y,X)s4 + o(s4).

�

Curve X(t) ∈ TpM with |X(t) = 1| ∀ t,X ′(0) = Y and |Y | = 1. Rays sX(t) deviate from sX(0)with velocity |sX ′(0)| = s.

Moreover, geodesic expp(sX(t)) deviate from expp(sX(0)) with velocity |J(s)|. Therefore ifK < 0 geodesics in M spread more than rays in TpM and vice versa if K > 0.

60

Definition 8.4. Let γ : [0, L] → M be a geodesic. γ(T ) is conjugate to γ(0) (along γ) ifT ∈ (0, L] and ∃ Jacobi field J 6= 0 along γ such that J(0) = J(T ) = 0.

The multiplicity of γ(T ), mult(γ(T )) = max number of such linearly independent Jacobifields.

Remark ∃ at most n lin. ind. Jacobi fileds along γ with J(0) = 0 defined by J ′(0). MoreoverJ : s 7→ sγ′(0) satisfies J(0) = 0 but J 6= − for s > 0⇒ mult(γ(T )) ≤ n− 1.

Example For Sn, γ(0) and γ(π) are conjugate points and mult(γ(π)) = n− 1.

Definition 8.5. Conjugate locus of p = { first conjugate points ∀ geodesics from p} = C(p)

Proposition 8.6. γ(T ) is conjugate to γ(0) ⇐⇒ Tγ′(0) is a critical point of expγ(0). More-over, mult(γ(T )) = dim ker d(expγ(0))Tγ′(0)

Proof Suppose J 6= 0 is a Jabobi field along γ with J(0) = J(T ) = 0.

J(s) = d(expγ(0))sγ′(0)(sJ′(0))

⇒ 0 = J(T ) = d(expγ(0))Tγ′(0)(TJ′(0)︸ ︷︷ ︸6=0

)

61

⇐⇒ Tγ′(0) is a critical point.

mult(γ(T )) = k ⇐⇒ ∃ lin. ind. fields J1, . . . , Jk such that

Ji(0) = Ji(T ) = 0, Ji 6= 0

⇐⇒ ∃ J ′1(0), . . . , J ′k(0) lin. ind. in Tγ′(0)(Tγ(0)M).

But {J ′1(0), . . . , J ′k(0)} = ker(d(expγ(0))Tγ′(0))

�

Proposition 8.7. ∃ Jacobi fields along γ ⇒

g(J, γ′)(s) = sg(J ′, γ′)(0) + g(J, γ′)(0)

Proof

d

dsg(J ′, γ′) = g(J ′′, γ′) + g(J ′,

D

Dsγ′︸ ︷︷ ︸

=0

) (since γ geodesic)

= −g(R(J, γ′)γ′, γ′) = 0 (by symmetry of R)

⇒ g(J ′, γ′)(s) = g(J ′, γ′)(0)

d

dsg(J, γ′) = g(J ′, γ′) + g(J,

D

Dsγ′︸ ︷︷ ︸

=0

)

= g(J ′, γ′)(0)

⇒ g(J, γ′)(s) = sg(J ′, γ′)(0) + g(J, γ′)(0)

�

Corollary 8.8.

J(0) = J(L) = 0⇒ g(J, γ′) = 0

Jacobi field along γ : [0, L]→M .

Corollary 8.9.

J(0) = 0⇒ g(J, γ′)(0) = 0

and

g(J ′, γ′)(0) = 0 ⇐⇒ g(J, γ′) = 0

Therefore dim{ Jacobi fields J along γ : J(0) = 0, g(J, γ′) = 0} = n− 1

Proposition 8.10. If X0 ∈ Tγ(0)M and XL ∈ Tγ(L)M where γ : [0, L] → M is geodesic,and γ(0) and γ(L) are not conjugate, then there exists a unique Jacobi field along γ such thatJ(0) = X0 and J(L) = XL.

62

Proof Let J = { Jacobi fields J along γ : J(0) = 0}. Let

F : J → Tγ(L)M

J 7→ J(L)

γ(0) and γ(L) not conjugate ⇒ F is injective.

dimJ = n = dimTγ(L)M ⇒ F is an isomorphism.

∴ ∃! J0 ∈ J such that F (J0) = J0(L) = XL

Similarly ∃! JL Jacobi field such that JL(L) = 0 and JL(0) = X0 ⇒ J = J0 + JL is whatwe want.

�

Corollary 8.11. Let {J1, . . . , Jn−1} be a basis for { Jacobi fields J along γ : J(0) = 0 andg(J ′, γ′)(0) = 0}. If γ(s) /∈ C(γ(0)) then {J1(T ), . . . , Jn−1(T )} is a basis for {γ′(T )}⊥ ⊆ Tγ(T )M

63

Chapter 9

Completeness

Assume (M, g) connected n-dimensional Riemannian manifold with n ≥ 2.

Definition 9.1. M is (geodesically) complete if all geodesics are defined for all time t i.e.expp(X) is defined for all X ∈ TpM ∀ p ∈M .

Definition 9.2. If p, q ∈M , define

d(p, q) = inf{L(α) : α pw − curve from p to q}

Proposition 9.3. (M,d) is a metric space.

Proof Normal nhds contain geodesic balls and vice versa so manifold topology of M andmetric topology are the same.

Clearly d(p, p) = 0, d(p, q) = d(q, p) and if α, β are pw-curves from p to q and q to r then

d(p, r) ≤ L(α) + L(β) ≤ d(p, q) + d(q, r) (triangle inequality)

If p 6= q, M Hausdorff ⇒ ∃ ε > 0 such that q /∈ Bε(p) ⇒ for any pw-curve α from p to q,L(α) ≥ ε > 0⇒ d(p, q) > 0

�

Now we state the big theorem of this chapter, and one of the more complicated theorems toprove in the course.

Theorem 9.4. (Hopf-Rinov)The following are equivalent:

1. expp is defined on all of TpM for some p ∈M

2. Closed bounded subsets of M are compact (Hiene-Borel)

3. (M,d) is a complete metric space

4. M is geodesically complete

Moreover, if M is complete then for all points p, q in M there exists a geodesic γ from p to qwhich realises the least distance i.e. d(p, q) = L(γ).

64

Corollary 9.5.M compact ⇒ M complete.

Corollary 9.6.A closed (connected) submanifold N of complete M is complete.

Corollary 9.7.R2\{0, 0} is not complete.

p q δ γ ε R x0 y0 γ(s0) α d(γ(s), q) = R− s

Proof of Hopf-Rinov1. ⇒ 2.WTS ∀ q ∈ M ∃ geodesic γ : [0, 1] → M such that γ(0) = p and γ(1) = q and d(p, q) = L(γ).

Let q ∈M and let R = d(p, q). Let δ > 0 be such that Bδ(p) is defined and Sδ(p) = ∂Bδ(p).

The map x 7→ d(x, q) is continuous⇒ d(x, q) is a minimum for x = x0 ∈ Sδ(p) say. ∴ X ∈ TpMwith |X| = 1 such that x0 = expp(δX). Let γ(s) = expp(sX) ∀ s ∈ R.

Let A = {s ∈ [0, R] : d(γ(s), q) = R− s}.

Claim: A = [0, R]Pf: 0 ∈ A⇒ A 6= ∅. A is closed because s 7→ d(γ(s), q) is continuous. STP now that A is open. A set open,

closed andnon-emptyis the wholething.Let s0 ∈ A, s0 < R. Let ε > 0 be such that Bε(γ(s0)) and Sε(γ(s0)) are defined. As be-

fore ∃ y0 ∈ Sε(γ(s0)) such that d(y0, q) is a minimum over Sε(γ(s0)).

s0 ∈ A⇒ R− s0 = d(γ(s0), q)

= ε+ miny∈Sε(γ(s0))

d(y, q)

= ε+ d(y0, q)

⇒ d(y0, q) = R− (s0 + ε)

If y0 = γ(s0 + ε) then s0 + ε ∈ A, so A open. Now,

d(p, y0) ≥ d(p, q)− d(q, y0)

= R− (R− (s0 + ε))

= s0 + ε

The pw-curve α from p to γ(s0) (following γ) and from γ(s0) to y0 following radial geodesicfrom γ(s0), has L(α) = s0 + ε⇒ d(p, y0) = s0 + ε⇒ α is a geodesic⇒ (by uniqueness) α = γ (when they are both defined). /pf of claim

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Shown ∀ q ∈M ∃ geodesic γ : [0, 1]→M such that γ(0) = p and γ(1) = q and d(p, q) = L(γ).

If K ⊆M is closed and bounded then

K ⊆ BdR(p) for curve R > 0

⊆ expp(BR′(0)) for some R′ > 0 (by claim)

expp continuous ⇒ expp(BR′(0)) is compact ⇒ K compact.

2. ⇒ 3.Let (pn) be a Cauchy sequence in (M,d). Then (pn) is bounded ⇒ K = {pn} is closed andbounded ⇒ K compact ⇒ (pn) has a convergent subsequence ⇒ (pn) converges (using theCauchy property).

3. ⇒ 4.Suppose M is not geodesically complete. There exists normalised geodesic γ(s) defined fors < s0 bot not for s = s0. Let (sn) be strictly increasing sequence in [0, so) convergingto s0 for which γ(s0) is defined ⇒ (sn) is Cauchy ⇒ (γ(sn)) is Cauchy in (M,d) sinced(γ(sn), γ(sm)) = |sn − sm|.

(M,d) is complete ⇒ γ(sn)→ p0 as n→∞

Let W be a totally normal nhd of p0 such that ∃ δ > 0 with expq : Bδ(0) → M . Let N besufficiently large but if n,m > N then γ(sn) ∈W ∀ n > N and d(γ(sn), γ(sm)) < δ ∀ n,m > N .

Let n,m > N . Then there exists unique geodesic α : [0, 1] → W such that α(0) = γ(sn)and α(1) = γ(sm). Necessarily α and γ coincide where they are both defined.

Since expγ(sn) is a diffeomorphism in Bδ(0) and its image contains W , we have that α ex-tends γ beyond s0 giving a contradiction.

4. ⇒ 1.Obvious

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Theorem 9.8. (Hadamard)M complete, simply connected, K ≤ 0⇒ expp : TpM →M is a diffeomorphism ∀ p ∈M

Lemma 9.9. M complete, K ≤ 0 ⇒ C(p) = ∅ ∀ p ∈ M ⇒ expp : TpM → M is a surjectivelocal diffeomorphism for all p ∈M .

Proof Let p ∈M and let γ : [0,∞)→M be a geodesic such that γ(0) = p. Let J be a Jacobifield along γ such that J 6= 0 and J(0) = 0 (so J ′(0) 6= 0).

∴ g(J, J)′(0) = 2g(J, J ′)(0) = 0g(J, J)′′(0) = 2|J ′(0)|2 > 0⇒ g(J, J)′(t) > 0 ∀ t > 0 sufficiently small.

Jacobi equation ⇒

g(J, J)′′ = 2g(J ′, J ′) + 2g(J ′′, J)

= 2|J ′|2 − 2g(R(J, γ′)γ′, J)

= 2|J ′|2 − 2K(J, γ′)(|J |2|γ′|2 − g(J, γ′)2)

≥ 0

∴ g(J, J)′ is increasing ⇒ g(J, J)′(t) > 0 ∀ t > 0⇒ g(J, J) is strictly increasing for t > 0g(J, J)(0) = 0⇒ g(J, J)(t) > 0 ∀ t > 0⇒ C(p) = ∅

Applying prop 8.6. and theorem 9.4. gives the result.

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Lemma 9.10. M complete, F : M → N surjective local diffeomorphism such that |dfp| ≥1 ∀ p ∈M ⇒ f covering map.(|dfp(X) ≥ |X| ∀ X ∈ TpM ∀ p ∈M)

Proof Fact: f covering map ⇐⇒ f has the curve-lifting property (i.e. given a curve α inN, ∀ p ∈ f−1(α(0)) ∃ curve β in M such that β(0) = p and f ◦ β = α.

f surjective local diffeomorphism ⇒ ∃ ε > 0 and curve β : [0, ε] → M such that β(0) = pand f ◦ β(t) = α(t) ∀ t ∈ [0, ε]

Let A := {T ∈ [0, 1] : ∃ β : [0, T ]→M s.t. β(0) = p and f ◦ β|[0,T ] = α|[0,T ]}

Shown that [0, ε) ⊆ A

f local diffeomorphism ⇒ A is half-open on the right. So suppose, for a contradiction, thatA = [0, t0) for some t0 < 1.

∃ increasing sequence (tn) in A such that tn → t0 as n→∞

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d(β(tn), β(0)) ≤∫ tn

0|β′(t)|dt

≤∫ tn

0|dfβ(t)(β

′(t))|dt (by hypothesis)

=

∫ tn

0|α′(t)|dt

≤∫ t0

0|α′(t)|dt = L(α|[0,to])

Therefore (β(Tn)) is a bounded sequence in M .

M complete ⇒ β(tn)→ q ∈M as n→∞.

Let V 3 q be open s.t. f |V V → f(V ) is a diffeomorphism.

f ◦ β(tn) = α(tn)→ α(t0) n→∞

Therefore α(t0) ∈ f(V )

Continuity of α⇒ ∃ δ > 0 s.t. α[(t0 − δ, t0 + δ)] ⊆ f(V ) ∃ N ∈ Z+ s.t. tN ∈ (t0 − δ, t0 + δ)⇒β(t0 − δ, tN ]) ⊆ V

f |V diffeomorphism⇒ ∃ curve β : (t0−δ, t0+δ)→ V s.t. f ◦β(t) = α(t) for all t ∈ (t0−δ, t0+δ)

Moreover f ◦ β = f ◦ β on [t0 − δ, tN ]⇒ β = β in [t0 − δ, tN ]⇒ β extends β to [δ, t0 + δ)⇒ A ⊇ [0, t0 + δ) which contradicts our initial assumption that to ≤ 1

Therefore A = [0, 1]

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Proof of Theorem 9.8.Lemma 9.9. ⇒ expp : TpM →M is a surjective local diffeomorphism.

Define a Riemannian metric h on TpM such that expp is a local isometry i.e.

hX(Y, Z) = gexpp(X)(d(expp)X(Y ), d(expp)X(Z))

Geodesics in TpM through 0 are straight lines (by definition of expp). so by theorem 9.4., h iscomplete.

Lemma 9.10 ⇒ expp is a covering map.

TpM and M simply connected ⇒ expp is a diffeomorphism.

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Remark We actually proved something stronger. We proved that if ∃ p ∈M complete simplyconnected s.t. C(p) = ∅ then expp : TpM →M is a diffeomorphism.

Corollary 9.11. M complete, K ≤ 0⇒ universal cover M is diffeomorphic to Rn

Corollary 9.12. M complete, simply connected, K ≤ 0⇒M non-compact.

Example

1. T 2 has complete flat metric but T 2 not diffeomorphic to Rn ⇒ need simply connected inTheorem 9.8.

2. S2 (or Sn for n > 2) cannot have a metric with K ≤ 0 since it is compact and simplyconnected.

3. RPn cannot have a metric with K ≤ 0 since the universal cover is Sn

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Chapter 10

Constant Curvature

Theorem 10.1. (Cartan)Let (M, gM ) and (N, gN ) be (n-dimensional) Riemannian manifolds.

1. Let p ∈ M, q ∈ N, ι : TpM → TqN be an isometry, let expp and expq be the exponentialmaps.

2. Let V be a normal nhd of p such that expq is defined on U = ι ◦ exp−1p (V )

3. Let f = expq ◦ ι ◦ exp−1p : V → expq(U)

4. For r ∈ V let γr : [0, L] → M be the (unique) normalised geodesic s.t. γr(0) = p andγr(L) = r and let γr : [0, L]→ N be the (unique) normalised geodesic s.t. γr(0) = q and(γr)′(0) = ι((γr)′(0))

5. For r ∈ V and s ∈ [0, L] let P rs : TpM → Tγr(s) be the parallel transport along γr and let

P rs : TqN → Tγr(s)N be the parallel transport along γr

6. For r ∈ V and s ∈ [0, L], let ϕrs = P rs ◦ ι ◦ (P rs )−1 : Tγr(s)M → Tγr(s)N . Notice thatϕrL : TrM → Tf(r)N .

If

gM (RM (X,Y )Z,W ) = gN (RN (ϕrL(X), ϕrL(Y ))ϕrL(Z), ϕrL(W ))

for all X,Y, Z,W ∈ TrM ∀ r ∈ V then f : V → f(V ) is a local isometry and dfp = ι.

We can paraphrase this as “the curvature determines the metric”.

Proof Let r ∈ V,X ∈ TrM , J the unique Jacobi field along γr such that J(0) = 0 andJ(L) = X (prop 8.10). Let {E1, . . . , En} be an orthonormal basis of TpM such that En(γr)(0)and let Ei(s) = P rx (Ei) (orthonormal basis for Tγr(s)M ∀ s ∈ [0, L])

We can write J(s) =∑n

i=1 yi(s)Ei(s)

Jacobi equation ⇒

γ′′ +

n∑j=1

gM (RM (Ei, En)En, Ei)yi = 0 (?)

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Let J(s) = ϕrs(J(s)) for s ∈ [0, L], a vector field along γr.

Let Ei(s) = ϕrs(Ei(s)) for s ∈ [0, L]. {Ei} form an orthonormal basis. Hence,

J(s) =n∑i=1

yi(s)Ei(s)

(?) and hypothesis ⇒

y′′i +n∑j=1

gN (RN (Ei, En)En, Ei)yi = 0

so J is a Jacobi field with J(0) = 0.

ϕrs is an isometry ⇒ |J(s)| = |J(s)|.

WTS: J(L) = dfr(X) = dfr(J(L)), then dfr is an isometry and we are done.

J ′(0) = ι(J ′(0)) and J(s) = d(expp)s(γr)′(0)(sJ′(0)) and J(s) = d(expq)s(γr)′(0)(sJ

′(0))

Therefore,

J(s) = d(expq)s(γr)′(0)(sJ′(0))

= d(expq)L(γr)′(0)(Lι(J′(0)))

= d(expq)L(γr)′(0) ◦ ι ◦ (d(expp)L(γr)′(0))−1(J(L))

= dfr(J(L))

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Lemma 10.2. F,G : M → N local isometries from M connected with F (p) = G(p) anddFp = dGp for some p ∈M , then F = G.

Proof Let q ∈ M . There exists curve α : [0, 1] → M such that α(0) = p and α(1) = w. LetA = {t ∈ [0, 1] : F (α(t)) = G(α(t)) and dFα(t) = dGα(t)}.

0 ∈ A, A clearly closed.

Let V be a normal nhd of p such that F,G : V → F (V ) = G(V ) are isometries. Thenf = F−1 ◦G : V → V is an isometry such that f(p) = p and dfp = id. If r ∈ V the ∃! X ∈ TpMs.t. expp(X) = r.

Therefore f(r) = f(expp(X)) = expf(p)(dfp(X)) (since f is an isometry)= expp(X) = r⇒ f |y is the identity F = G on V .

Therefore sup A > 0 since ∃ T ∈ (0, 1) such that α(t) ∈ V ∀ t ∈ [0, T ].

If sup A = T ∈ A < 1 then we repeat the above argument for α(T ) to get a contradiction.Therefore A = [0, 1]

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Corollary 10.3. If M,N have the same constant curvature p ∈M, q ∈ N and ι : TpM → TqNis an isometry, then there exists normal nhds V 3 p,W 3 q and a unique isometry f : V → Ws.t. f(p) = q and dfp = ι.

Theorem 10.4. Let Hn = {(x1, . . . , xn) ∈ Rn : xn > 0} and define a Riemannian metric g by

gij =δijx2n

In coordinates (x1, . . . , xn).

Then hyperbolic space (Hn, g) is a connected, simply connected, n-dimensional Riemannianmanifold which is complete and has constant sectional curvature −1.

Proposition 10.5. Straight lines and semi-circles with centre on xn = 0 which are perpen-dicular to the hyperplane xn = 0 are the geodesics of Hn.

M constant sectional curvature K. Can always scale such that K ∈ {−1, 0, 1}.

Theorem 10.6. Let M be complete, simply connected, n-dimensional Riemannian manifoldwith constant sectional curvature K ∈ {−1, 0, 1}. Then M is isometric to Hn,Rn or Sn.

Remark For general K < 0 we write Hn(K) for hyperbolic space with constant sectionalcurvature K. Also, for general K ≥ 0 we write Sn(K) for sphere with constant sectionalcurvature.

Proof Suppose K = −1 or 0 and let N be either Hn or Rn.

Let p ∈ M, q ∈ N and isometry ι : TpM → TqN . Then Hadamard theorem ⇒ F =expq ◦ι ◦ exp−1

p : M → N is well defined.

Cartin’s theorem (10.1) ⇒ F is a local isometry. Lemma 9.10 ⇒ F is a covering map. M,Nsimply connected ⇒ F is a diffeomorphism ⇒ F is an isometry.

Now suppose K = 1, Let p ∈M, q ∈ Sn and ι : TqSn → TpM isometry. Let F = expp ◦ι◦exp−1p :

Sn\{−q} →M well-defined.

Cartin’s theorem ⇒ F is a local isometry.

Take r ∈ Sn\{q,−q}. Let s = F (r), and

J = dFr : TrSn → TsM

Define G = exps ◦J ◦ exp−1r : Sn\{−r} →M

Let N = Sn\{−q, r} connected (n ≥ 2).

r ∈ N,F (r) = s = G(r) and dFr = J = dGr

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Lemma 10.2 ⇒ F = G on N . Let

H =

{F (t) t 6= −qG(t) t 6= −r

Then H : Sn →M is a well-defined surjective local isometry. As before, H is an isometry.

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Proposition 10.7.Let M be complete with constant sectional curvature, and let M be the universal cover of Mwith the covering metric. Then there exists some subgroup Γ of the group of isometries of M

s.t. M is isometric to M /Γ with the induced metric.

Remark Γ acting properly discontinuously on M .

Proof Let ξ : M →M be the coving map, and Γ be the group of covering transformations.

Then ξ(p) = ξ(q) ⇐⇒ ∃ ϕ ∈ Γ s.t. ϕ(p) = q ⇐⇒ π(p) = π(q) where π : M → M /Γis projection.

Therefore there exists bijections f : M → M /Γ such that f ◦ ξ = π

ξ, π are local isometries. Therefore f is a local isometry. But f is a bijection, therefore itsan isometry.

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Remark Problem sheet 5 ⇐⇒ M complete ⇐⇒ M complete ⇐⇒ M /Γ complete.

So M is isometric to one of

Hn /Γ , Rn /Γ , Sn /Γ

for some Γ.

Proposition 10.8. M complete 2m-dimensional and constant sectional curvature 1. Then Mis isometric to S2m or RP2m

Proof O(2m+ 1) is the isometry group of S2n so let Γ ⊆ O(2m+ 1) act properly discontinu-ously on S2n.

Let φ ∈ Γ

detφ = 1⇒ (since (2m+1) is odd dimension) 1 is an eigenvalue of φ⇒ φ has a fixed point

⇒ φ =

(1 0

0 A

)where A ∈ SO(2m)

⇒ φ fixes S2m i.e. φ = S2m

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(Since Γ acts properly discontinuously)

detφ = −1⇒ detφ2 = 1⇒ φ2 = id on S2m

⇒ φ = ±id on S2m

Therefore S2m/Γ = S2m or RP2m. Result by Theorem 10.6 and Prop 10.7

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Example If M is a surface with genus > 1 then there exists a complete metric on M withconstant sectional curvature −1. Therefore M = H2

/Γ, for some subgroup Γ of the isometriesof the hyperbolic plane. (lots of complex analysis used in proof of this, so not going to prove it)

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