III Riemannian Geometry (Full)

Part III — Riemannian Geometry

Based on lectures by A. G. Kovalev

Notes taken by Dexter Chua

Lent 2017

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

This course is a possible natural sequel of the course Differential Geometry offered in

Michaelmas Term. We shall explore various techniques and results revealing intricate

and subtle relations between Riemannian metrics, curvature and topology. I hope to

cover much of the following:

A closer look at geodesics and curvature. Brief review from the Differential Geometry

course. Geo desic coordinates and Gauss’ lemma. Jacobi fields, completeness and

the Hopf–Rinow theorem. Variations of energy, Bonnet–Myers diameter theorem and

Synge’s theorem.

Hodge theory and Riemannian holonomy. The Hodge star and Laplace–Beltrami

op erator. The Hodge decomposition theorem (with the ‘geometry part’ of the proof).

Bo chner–Weitzenb¨ock formulae. Holonomy groups. Interplays with curvature and de

Rham cohomology.

Ricci curvature. Fundamental groups and Ricci curvature. The Cheeger–Gromoll

splitting theorem.

Pre-requisites

Manifolds, differential forms, vector fields. Basic concepts of Riemannian geometry

(curvature, geodesics etc.) and Lie groups. The course Differential Geometry offered in

Michaelmas Term is the ideal pre-requisite.

Contents

1 Basics of Riemannian manifolds

2 Riemann curvature

3 Geodesics

3.1 Definitions and basic properties

3.2 Jacobi fields

3.3 Further properties of geodesics

3.4 Completeness and the Hopf–Rinow theorem

3.5 Variations of arc length and energy

3.6 Applications

4 Hodge theory on Riemannian manifolds

4.1 Hodge star and operators

4.2 Hodge decomposition theorem

4.3 Divergence

4.4 Introduction to Bochner’s method

5 Riemannian holonomy groups

6 The Cheeger–Gromoll splitting theorem

1 Basics of Riemannian manifolds

Before we do anything, we lay out our conventions. Given a choice of local

coordinates {x

}, the coefficients X

for a vector field X are defined by

X =

∂

∂x

In general, for a tensor field X ∈ T M

⊗q

⊗ T

∗

⊗p

, we write

X =

...k

...`

∂

∂x

⊗ ··· ⊗

∂

∂x

⊗ dx

⊗ ··· ⊗ dx

and we often leave out the ⊗ signs.

For the sake of sanity, we will often use implicit summation convention, i.e.

whenever we write something of the form

ijk

i`jk

we mean

i,j

ijk

i`jk

We will use upper indices to denote contravariant components, and lower

indices for covariant components, as we have done above. Thus, we always sum

an upper index with a lower index, as this corresponds to applying a covector to

a vector.

We will index the basis elements oppositely, e.g. we write d

instead of d

for a basis element of

∗

, so that the indices in expressions of the form

seem to match up. Whenever we do not follow this convention, we will write out

summations explicitly.

We will also adopt the shorthands

∂

∂x

, ∇

= ∇

∂

With these conventions out of the way, we begin with a very brief summary

of some topics in the Michaelmas Differential Geometry course, starting from

the definition of a Riemannian metric.

Definition

(Riemannian metric)

Let

be a smooth manifold. A Riemannian

metric

is an inner product on the tangent bundle

T M

varying smoothly

with the fibers. Formally, this is a global section of

∗

M ⊗T

∗

that is fiberwise

symmetric and positive definite.

The pair (M, g) is called a Riemannian manifold.

On every coordinate neighbourhood with coordinates

= (

, ··· , x

), we

can write

g =

i,j=1

(x) dx

and we can find the coefficients g

= g



∂

∂x

∂

∂x



and are C

∞

functions.

Example.

The manifold

has a canonical metric given by the Euclidean

metric. In the usual coordinates, g is given by g

= δ

Does every manifold admit a metric? Recall

Theorem

(Whitney embedding theorem)

Every smooth manifold

admits

an embedding into

for some

. In other words,

is diffeomorphic to a

submanifold of R

. In fact, we can pick k such that k ≤ 2 dim M.

Using such an embedding, we can induce a Riemannian metric on

restricting the inner product from Euclidean space, since we have inclusions

M → T

∼

More generally,

Lemma.

Let (

N, h

) be a Riemannian manifold, and

M → N

is an immersion,

then the pullback g = F

∗

h defines a metric on M.

The condition of immersion is required for the pullback to be non-degenerate.

In Differential Geometry, if we do not have metrics, then we tend to consider

diffeomorphic spaces as being the same. With metrics, the natural notion of

isomorphism is

Definition

(Isometry)

Let (

M, g

) and (

N, h

) be Riemannian manifolds. We

say

M → N

is an isometry if it is a diffeomorphism and

∗

. In other

words, for any p ∈ M and u, v ∈ T

M, we need



(df)

u, (df )



= g(u, v).

Example.

Let

be a Lie group. Then for any

, we have translation maps

, R

: G → G given by

(y) = xy

(y) = yx

These maps are in fact diffeomorphisms of G.

We already know that

admits a Riemannian metric, but we might want

to ask something stronger — does there exist a left-invariant metric? In other

words, is there a metric such that each L

is an isometry?

Recall the following definition:

Definition

(Left-invariant vector field)

Let

be a Lie group, and

a vector

field. Then X is left invariant if for any x ∈ G, we have d(L

)X = X.

We had a rather general technique for producing left-invariant vector fields.

Given a Lie group

, we can define the Lie algebra

. Then we can

produce left-invariant vector fields by picking some X

∈ g, and then setting

= d(L

The resulting vector field is indeed smooth, as shown in the differential geometry

course.

Similarly, to construct a left-invariant metric, we can just pick a metric at

the identity and the propagating it around using left-translation. More explicitly,

given any inner product on h·, ·i on T

G, we can define g by

g(u, v) = h(dL

−1

)

u, (dL

−1

)

for all

x ∈ G

and

u, v ∈ T

. The argument for smoothness is similar to that

for vector fields.

Of course, everything works when we replace “left” with “right”. A Rie-

mannian metric is said to be bi-invariant if it is both left- and right-invariant.

These are harder to find, but it is a fact that every compact Lie group admits a

bi-invariant metric. The basic idea of the proof is to start from a left-invariant

metric, then integrate the metric along right translations of all group elements.

Here compactness is necessary for the result to be finite.

We will later see that we cannot drop the compactness condition. There are

non-compact Lie groups that do not admit bi-invariant metrics, such as

, R

Recall that in order to differentiate vectors, or even tensors on a manifold,

we needed a connection on the tangent bundle. There is a natural choice for the

connection when we are given a Riemannian metric.

Definition

(Levi-Civita connection)

Let (

M, g

) be a Riemannian manifold.

The Levi-Civita connection is the unique connection

∇

: Ω

(

T M

)

→

Ω

(

T M

)

on M satisfying

(i) Compatibility with metric:

Zg(X, Y ) = g(∇

X, Y ) + g(X, ∇

Y ),

(ii) Symmetry/torsion-free:

∇

Y − ∇

X = [X, Y ].

Definition

(Christoffel symbols)

In local coordaintes, the Christoffel symbols

are defined by

∇

∂

∂x

= Γ

∂

∂x

With a bit more imagination on what the symbols mean, we can write the

first property as

d(g(X, Y )) = g(∇X, Y ) + g(X, ∇Y ),

while the second property can be expressed in coordinate representation by

= Γ

The connection was defined on

T M

, but in fact, the connection allows us to

differentiate many more things, and not just tangent vectors.

Firstly, the connection

∇

induces a unique covariant derivative on

∗

, also

denoted ∇, defined uniquely by the relation

Xhα, Y i = h∇

α, Y i + hα, ∇

Y i

for any X, Y ∈ Vect(M) and α ∈ Ω

(M).

To extend this to a connection

∇

on tensor bundles

q,p

≡

(

T M

)

⊗q

⊗

∗

⊗p

for any p, q ≥ 0, we note the following general construction:

In general, suppose we have vector bundles

and

, and

∈

Γ(

) and

∈

Γ(

). If we have connections

∇

and

∇

and

respectively, then

we can define

∇

E⊗F

⊗ s

) = (∇

) ⊗ s

+ s

⊗ (∇

Since we already have a connection on

T M

and

∗

, this allows us to extend

the connection to all tensor bundles.

Given this machinery, recall that the Riemannian metric is formally a section

g ∈

Γ(

∗

M ⊗ T

∗

). Then the compatibility with the metric can be written in

the following even more compact form:

∇g = 0.

2 Riemann curvature

With all those definitions out of the way, we now start by studying the notion

of curvature. The definition of the curvature tensor might not seem intuitive

at first, but motivation was somewhat given in the III Differential Geometry

course, and we will not repeat that.

Definition

(Curvature)

Let (

M, g

) be a Riemannian manifold with Levi-Civita

connection ∇. The curvature 2-form is the section

R = −∇ ◦ ∇ ∈ Γ(

∗

M ⊗ T

∗

M ⊗ T M ) ⊆ Γ(T

1,3

M).

This can be thought of as a 2-form with values in

∗

M ⊗ T M

End

(

T M

Given any X, Y ∈ Vect(M), we have

R(X, Y ) ∈ Γ(End T M ).

The following formula is a straightforward, and also crucial computation:

Proposition.

R(X, Y ) = ∇

[X,Y ]

− [∇

, ∇

In local coordinates, we can write

R =



j,k`



i,j=1,...,dim M

∈ Ω

(End(T M )).

Then we have

R(X, Y )

= R

j,k`

The comma between j and k is purely for artistic reasons.

It is often slightly convenient to consider a different form of the Riemann

curvature tensor. Instead of having a tensor of type (1, 3), we have one of type

(0, 4) by

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

for X, Y, Z, T ∈ T

M. In local coordinates, we write this as

ij,k`

= g

j,k`

The first thing we want to prove is that

ij,k`

enjoys some symmetries we might

not expect:

Proposition.

(i)

ij,k`

= −R

ij,`k

= −R

ji,k`

(ii) The first Bianchi identity:

j,k`

+ R

k,`j

+ R

`,jk

= 0.

(iii)

ij,k`

= R

k`,ij

Note that the first Bianchi identity can also be written for the (0

4) tensor as

ij,k`

+ R

ik,`j

+ R

i`,jk

= 0.

Proof.

(i)

The first equality is obvious as coefficients of a 2-form. For the second

equality, we begin with the compatibility of the connection with the metric:

∂g

∂x

= g(∇

∂

, ∂

) + g(∂

, ∇

∂

We take a partial derivative, say with respect to x

, to obtain

∂

∂x

= g(∇

∇

∂

, ∂

)+g(∇

∂

, ∇

∂

)+g(∇

∂

, ∇

∂

)+g(∂

, ∇

∇

∂

Then we know

0 =

∂

∂x

−

∂

∂x

= g([∇

, ∇

]∂

, ∂

) + g(∂

, [∇

, ∇

]∂

But we know

R(∂

, ∂

) = ∇

[∂

,∂

]

− [∇

, ∇

] = −[∇

, ∇

Writing R

= R(∂

, ∂

), we have

0 = g(R

∂

, ∂

) + g(∂

, R

∂

) = R

ji,k`

+ R

ij,k`

So we are done.

(ii) Recall

j,k`

= (R

∂

)

= ([∇

, ∇

]∂

)

So we have

j,k`

+ R

k,`j

+ R

`,jk

= [(∇

∇

∂

− ∇

∇

∂

) + (∇

∇

∂

− ∇

∇

∂

) + (∇

∇

∂

− ∇

∇

∂

)]

We claim that

∇

∂

− ∇

∇

∂

= 0.

Indeed, by definition, we have

(∇

∂

)

= Γ

= (∇

∂

)

The other terms cancel similarly, and we get 0 as promised.

(iii) Consider the following octahedron:

ik,`j

= R

ki,j`

i`,jk

= R

`i,kj

j`,ki

= R

`j,ik

jk,i`

= R

kj,`i

ij,k`

= R

ji,`k

k`,ij

= R

`k,ji

The equalities on each vertex is given by (i). By the first Bianchi identity,

for each greyed triangle, the sum of the three vertices is zero.

Now looking at the upper half of the octahedron, adding the two greyed

triangles shows us the sum of the vertices in the horizontal square is

(

−

ij,k`

. Looking at the bottom half, we find that the sum of the

vertices in the horizontal square is (−2)R

k`,ij

. So we must have

ij,k`

= R

k`,ij

What exactly are the properties of the Levi-Civita connection that make

these equality works? The first equality of (i) did not require anything. The

second equality of (i) required the compatibility with the metric, and (ii) required

the symmetric property. The last one required both properties.

Note that we can express the last property as saying

ij,k`

is a symmetric

bilinear form on

∗

Sectional curvature

The full curvature tensor is rather scary. So it is convenient to obtain some

simpler quantities from it. Recall that if we had tangent vectors

X, Y

, then we

can form

|X ∧ Y | =

g(X, X)g(Y, Y ) − g(X, Y )

which is the area of the parallelogram spanned by X and Y . We now define

K(X, Y ) =

R(X, Y, X, Y )

|X ∧ Y |

Note that this is invariant under (non-zero) scaling of

, and is symmetric

and

. Finally, it is also invariant under the transformation (

X, Y

)

7→

(X + λY, Y ).

But it is an easy linear algebra fact that these transformations generate all

isomorphism from a two-dimensional vector space to itself. So

(

X, Y

) depends

only on the 2-plane spanned by X, Y . So we have in fact defined a function on

the Grassmannian of 2-planes,

, T

)

→ R

. This is called the sectional

curvature (of g).

It turns out the sectional curvature determines the Riemann curvature tensor

completely!

Lemma.

Let

be a real vector space of dimension

≥

2. Suppose

, R

⊗4

→ R

are both linear in each factor, and satisfies the symmetries we found

for the Riemann curvature tensor. We define

, K

, V

)

→ R

as in the

sectional curvature. If K

= K

, then R

= R

This is really just linear algebra.

Proof. For any X, Y, Z ∈ V , we know

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

(X + Z, Y, X + Z, Y ).

Using linearity of

and

, and cancelling equal terms on both sides, we find

(Z, Y, X, Y ) + R

(X, Y, Z, Y ) = R

(Z, Y, X, Y ) + R

(X, Y, Z, Y ).

Now using the symmetry property of R

and R

, this implies

(X, Y, Z, Y ) = R

(X, Y, Z, Y ).

Similarly, we replace Y with Y + T , and then we get

(X, Y, Z, T ) + R

(X, T, Z, Y ) = R

(X, Y, Z, Y ) + R

”(X, T, Z, Y ).

We then rearrange and use the symmetries to get

(X, Y, Z, T ) − R

(X, Y, Z, T ) = R

(Y, Z, X, T ) − R

(Y, Z, X, T ).

We notice this equation says

(

X, Y, Z, T

)

− R

(

X, Y, Z, T

) is invariant under

the cyclic permutation

X → Y → Z → X

. So by the first Bianchi identity, we

have

3(R

(X, Y, Z, T ) − R

(X, Y, Z, T )) = 0.

So we must have R

= R

Corollary.

Let (

M, g

) be a manifold such that for all

, the function

Gr(2, T

M) → R is a constant map. Let

(X, Y, Z, T ) = g

(X, Z)g

(Y, T ) − g

(X, T )g

(Y, Z).

Then

= K

Here

is just a real number, since it is constant. Moreover,

is a smooth

function of p.

Equivalently, in local coordinates, if the metric at a point is

, then we have

ij,ij

= −R

ij,ji

= K

and all other entries all zero.

Of course, the converse also holds.

Proof.

We apply the previous lemma as follows: we define

and

. It is a straightforward inspection to see that this

does follow the

symmetry properties of

, and that they define the same sectional curvature.

So R

= R

. We know K

is smooth in p as both g and R are smooth.

We can further show that if

dim M >

2, then

is in fact independent of

under the hypothesis of this function, and the proof requires a second Bianchi

identity. This can be found on the first example sheet.

Other curvatures

There are other quantities we can extract out of the curvature, which will later

be useful.

Definition (Ricci curvature). The Ricci curvature of g at p ∈ M is

Ric

(X, Y ) = tr(v 7→ R

(X, v)Y ).

In terms of coordinates, we have

Ric

= R

i,jq

= g

pi,jq

where g

denotes the inverse of g.

This

Ric

is a symmetric bilinear form on

. This can be determined by

the quadratic form

Ric(X) =

n − 1

Ric

(X, X).

The coefficient

n−1

is just a convention.

There are still two indices we can contract, and we can define

Definition

(Scalar curvature)

The scalar curvature of

is the trace of

Ric

respect to g. Explicitly, this is defined by

s = g

Ric

= g

i,jq

= R

Sometimes a convention is to define the scalar curvature as

n(n−1)

instead.

In the case of a constant sectional curvature tensor, we have

Ric

= (n − 1)K

and

s(p) = n(n − 1)K

Low dimensions

= 2, i.e. we have surfaces, then the Riemannian metric

is also known as

the first fundamental form, and it is usually written as

g = E du

+ 2F du dv + G dv

Up to the symmetries, the only non-zero component of the curvature tensor is

12,12

, and using the definition of the scalar curvature, we find

12,12

s(EG − F

Thus

2 is also the sectional curvature (there can only be one plane in the

tangent space, so the sectional curvature is just a number). One can further

check that

= K =

LN − M

EG − F

the Gaussian curvature. Thus, the full curvature tensor is determined by the

Gaussian curvature. Also,

12,21

is the determinant of the second fundamental

form.

If n = 3, one can check that R(g) is determined by the Ricci curvature.

3 Geodesics

3.1 Definitions and basic properties

We will eventually want to talk about geodesics. However, the setup we need to

write down the definition of geodesics can be done in a much more general way,

and we will do that.

The general setting is that we have a vector bundle π : E → M.

Definition

(Lift)

Let

E → M

be a vector bundle with typical fiber

Consider a curve

: (

−ε, ε

)

→ M

. A lift of

is a map

: (

−ε, ε

)

→ E

π ◦ γ

= γ, i.e. the following diagram commutes:

(−ε, ε) M

For

p ∈ M

, we write

−1

(

{p}

)

∼

for the fiber above

. We can think

as the space of some “information” at

. For example, if

T M

, then the

“information” is a tangent vector at

. In physics, the manifold

might represent

our universe, and a point in

might be the value of the electromagnetic field

at p.

Thus, given a path

, a lift corresponds to providing that piece of

“information” at each point along the curve. For example, if

T M

, then we

can canonically produce a lift of

, given by taking the derivative of

at each

point.

Locally, suppose we are in some coordinate neighbourhood

U ⊆ M

such that

E is trivial on U. After picking a trivialization, we can write our lift as

(t) = (γ(t), a(t))

for some function a : (−ε, ε) → V .

One thing we would want to do with such lifts is to differentiate them, and

see how it changes along the curve. When we have a section of

on the whole

(or even just an open neighbourhood), rather than just a lift along a

curve, the connection provides exactly the information needed to do so. It is not

immediately obvious that the connection also allows us to differentiate curves

along paths, but it does.

Proposition.

Let

: (

−ε, ε

)

→ M

be a curve. Then there is a uniquely

determined operation

∇

from the space of all lifts of

to itself, satisfying the

following conditions:

(i) For any c, d ∈ R and lifts ˜γ

, γ

of γ, we have.

∇

(cγ

+ d˜γ

) = c

∇γ

+ d

∇˜γ

(ii) For any lift γ

of γ and function f : (−ε, ε) → R, we have

∇

(fγ

) =

+ f

∇γ

(iii)

If there is a local section

and a local vector field

such that

(t) = s(γ(t)), ˙γ(t) = V (γ(t)),

then we have

∇γ

= (∇

s) ◦ γ.

Locally, this is given by



∇γ



= ˙a

+ Γ

˙x

The proof is straightforward — one just checks that the local formula works,

and the three properties force the operation to be locally given by that formula.

Definition

(Covariant derivative)

The uniquely defined operation in the propo-

sition above is called the covariant derivative.

In some sense, lifts that have vanishing covariant derivative are “constant”

along the map.

Definition

(Horizontal lift)

Let

∇

be a connection on

with Γ

(

) the

coefficients in a local trivialization. We say a lift γ

is horizontal if

∇γ

= 0.

Since this is a linear first-order ODE, we know that for a fixed

, given any

initial a(0) ∈ E

γ(0)

, there is a unique way to obtain a horizontal lift.

Definition

(Parallel transport)

Let

: [0

→ M

be a curve in

. Given any

∈ E

γ(0)

, the unique horizontal lift of

with

(0) = (

(0)

, a

) is called the

parallel transport of

along

(0). We sometimes also call

(1) the parallel

transport.

Of course, we want to use this general theory to talk about the case where

is a Riemannian manifold,

T M

and

∇

is the Levi-Civita connection of

. In this case, each curve

(

) has a canonical lift independent of the metric or

connection given simply by taking the derivative ˙γ(t).

Definition

(Geodesic)

A curve

(

) on a Riemannian manifold (

M, g

) is called

a geodesic curve if its canonical lift is horizontal with respect to the Levi-Civita

connection. In other words, we need

∇˙γ

= 0.

In local coordinates, we write this condition as

¨x

+ Γ

˙x

= 0.

This time, we obtain a second-order ODE. So a geodesic is uniquely specified

by the initial conditions

(0) and

˙x

(0). We will denote the resulting

geodesic as γ

(t, a), where t is the time coordinate as usual.

Since we have a non-linear ODE, existence is no longer guaranteed on all

time, but just for some interval (

−ε, ε

). Of course, we still have uniqueness of

solutions.

We now want to prove things about geodesics. To do so, we will need to apply

some properties of the covariant derivative we just defined. Since we are lazy,

we would like to reuse results we already know about the covariant derivative

for vector fields. The trick is to notice that locally, we can always extend

˙γ

to a

vector field.

Indeed, we work in some coordinate chart around

(0), and we wlog assume

˙γ(0) =

∂

∂x

By the inverse function theorem, we note that

(

) is invertible near 0, and we

can write

(

) for small

. Then in this neighbourhood of 0, we can view

as a function of x

instead of t. Then we can define the vector field

˙γ(x

, ··· , x

) = ˙γ(x

, x

), ··· , x

)).

By construction, this agrees with ˙γ along the curve.

Using this notation, the geodesic equation can be written as

∇

˙γ



γ(t)

= 0,

where the

∇

now refers to the covariant derivative of vector fields, i.e. the

connection itself.

Using this, a lot of the desired properties of geodesics immediately follow from

well-known properties of the covariant derivative. For example,

Proposition. If γ is a geodesic, then |˙γ(t)|

is constant.

Proof.

We use the extension

˙γ

around

(0), and stop writing the underlines.

Then we have

˙γ(g( ˙γ, ˙γ)) = g(∇

˙γ

˙γ, ˙γ) + g( ˙γ, ∇

˙γ

˙γ) = 0,

which is valid at each q = γ(t) on the curve. But at each q, we have

˙γ(g( ˙γ, ˙γ)) = ˙x

∂

∂x

g( ˙γ, ˙γ) =

|˙γ(t)|

by the chain rule. So we are done.

At this point, it might be healthy to look at some examples of geodesics.

Example.

with the Euclidean metric, we have Γ

= 0. So the geodesic

equation is

¨x

= 0.

So the geodesics are just straight lines.

Example.

On a sphere

with the usual metric induced by the standard

embedding S

→ R

n+1

. Then the geodesics are great circles.

To see this, we may wlog

and

, for a standard basis

}

n+1

. We can look at the map

ϕ : (x

, ··· , x

) 7→ (x

, x

, −x

, ··· , −x

and it is clearly an isometry of the sphere. Therefore it preserves the Riemannian

metric, and hence sends geodesics to geodesics. Since it also preserves

and

we know

(

) =

by uniqueness. So it must be contained in the great circle

lying on the plane spanned by e

and e

Lemma.

Let

p ∈ M

, and

a ∈ T

. As before, let

(

t, a

) be the geodesic with

γ(0) = p and ˙γ(0) = p. Then

(λt, a) = γ

(t, λa),

and in particular is a geodesic.

Proof. We apply the chain rule to get

γ(λt, a) = λ ˙γ(λt, a)

γ(λt, a) = λ

¨γ(λt, a).

(

λt, a

) satisfies the geodesic equations, and have initial velocity

λa

. Then

we are done by uniqueness of ODE solutions.

Thus, instead of considering

(

t, a

) for arbitrary

and

, we can just fix

= 1, and look at the different values of

, a

). By ODE theorems, we know

this depends smoothly on

, and is defined on some open neighbourhood of

0 ∈ T

Definition

(Exponential map)

Let (

M, g

) be a Riemannian manifold, and

p ∈ M . We define exp

exp

(a) = γ(1, a) ∈ M

for a ∈ T

M whenever this is defined.

We know this function has domain at least some open ball around 0

∈ T

and is smooth. Also, by construction, we have exp

(0) = p.

In fact, the exponential map gives us a chart around

locally, known as

geodesic local coordinates. To do so, it suffices to note the following rather trivial

proposition.

Proposition. We have

(d exp

)

= id

where we identify T

∼

M in the natural way.

All this is saying is if you go in the direction of

a ∈ T

, then you go in the

direction of a.

Proof.

(d exp

)

(v) =

exp

(tv) =

γ(1, tv) =

γ(t, v) = v.

Corollary. exp

maps an open ball

, δ

)

⊆ T

U ⊆ M

diffeomorphically

for some δ > 0.

Proof. By the inverse mapping theorem.

This tells us the inverse of the exponential map gives us a chart of

around

p. These coordinates are often known as geodesic local coordinates.

In these coordinates, the geodesics from p have the very simple form

γ(t, a) = ta

for all a ∈ T

M and t sufficiently small that this makes sense.

Corollary.

For any point

p ∈ M

, there exists a local coordinate chart around

such that

– The coordinates of p are (0, ··· , 0).

– In local coordinates, the metric at p is g

(p) = δ

– We have Γ

(p) = 0 .

Coordinates satisfying these properties are known as normal coordinates.

Proof.

The geodesic local coordinates satisfies these property, after identifying

isometrically with (

, eucl

). For the last property, we note that the

geodesic equations are given by

¨x

+ Γ

˙x

= 0.

But geodesics through the origin are given by straight lines. So we must have

= 0.

Such coordinates will be useful later on for explicit calculations, since when-

ever we want to verify a coordinate-independent equation (which is essentially

all equations we care about), we can check it at each point, and then use normal

coordinates at that point to simplify calculations.

We again identify (T

N, g(p))

∼

, eucl), and then we have a map

(r, v) ∈ (0, δ) × S

n−1

7→ exp

(rv) ∈ M

This chart is known as geodesic polar coordinates. For each fixed

, the image of

this map is called a geodesic sphere of geodesic radius

, written Σ

. This is an

embedded submanifold of M.

Note that in geodesic local coordinates, the metric at 0

∈ T

is given by

the Euclidean metric. However, the metric at other points can be complicated.

Fortunately, Gauss’ lemma says it is not too complicated.

Theorem

(Gauss’ lemma)

The geodesic spheres are perpendicular to their

radii. More precisely,

(

t, a

) meets every Σ

orthogonally, whenever this makes

sense. Thus we can write the metric in geodesic polars as

g = dr

+ h(r, v),

where for each r, we have

h(r, v) = g|

In matrix form, we have

g =







1 0 ··· 0

. h







The proof is not hard, but it involves a few subtle points.

Proof. We work in geodesic coordinates. It is clear that g(∂

, ∂

) = 1.

Consider an arbitrary vector field

(

) on

n−1

. This induces a vector

field on some neighbourhood B(0, δ) ⊆ T

M by

X(rv) = X(v).

Pick a direction

v ∈ T

, and consider the unit speed geodesic

in the direction

of v. We define

G(r) = g(

X(rv), ˙γ(r)) = g(

X, ˙γ(r)).

We begin by noticing that

∇

∂

X − ∇

∂

= [∂

X] = 0.

Also, we have

G(r) = g(∇

˙γ

X, ˙γ) + g(

X, ∇

˙γ

˙γ).

We know the second term vanishes, since

is a geodesic. Noting that

˙γ

∂

∂r

we know the first term is equal to

g(∇

∂

, ∂

) =



g(∇

∂

, ∂

) + g(∂

, ∇

∂

)



X(g(∂

, ∂

)) = 0,

since we know that g(∂

, ∂

) = 1 constantly.

Thus, we know

(

) is constant. But

(0) = 0 since the metric at 0 is the

Euclidean metric. So G vanishes everywhere, and so ∂

is perpendicular to Σ

Corollary. Let a, w ∈ T

M. Then

g((d exp

)

a, (d exp

)

w) = g(a, w)

whenever a lives in the domain of the geodesic local neighbourhood.

3.2 Jacobi fields

Fix a Riemannian manifold

. Let’s imagine that we have a “manifold” of all

smooth curves on

. Then this “manifold” has a “tangent space”. Morally,

given a curve

, a “tangent vector” at

in the space of curve should correspond

to providing a tangent vector (in M ) at each point along γ:

Since we are interested in the geodesics only, we consider the “submanifold” of

geodesics curves. What are the corresponding “tangent vectors” living in this

“submanifold”?

In rather more concrete terms, suppose

(

) =

(

t, s

) is a family of geodesics

indexed by

s ∈

(

−ε, ε

). What do we know about

∂f

∂s



s=0

, a vector field

along f

We begin by considering such families that fix the starting point

, s

), and

then derive some properties of

∂f

∂s

in these special cases. We will then define a

Jacobi field to be any vector field along a curve that satisfies these properties.

We will then prove that these are exactly the variations of geodesics.

Suppose

(

t, s

) is a family of geodesics such that

, s

) =

for all

. Then

in geodesics local coordinates, it must look like this:

For a fixed p, such a family is uniquely determined by a function

a(s) : (−ε, ε) → T

such that

f(t, s) = exp

(ta(s)).

The initial conditions of this variation can be given by a(0) = a and

˙a(0) = w ∈ T

∼

We would like to know the “variation field” of

(

) =

(

0) =

(

t, a

) this

induces. In other words, we want to find

∂f

∂s

(

0). This is not hard. It is just

given by

(d exp

)

(tw) =

∂f

∂s

(t, 0),

As before, to prove something about

, we want to make good use of the

properties of

∇

. Locally, we extend the vectors

∂f

∂s

and

∂f

∂t

to vector fields

∂

∂t

and

∂

∂s

. Then in this set up, we have

˙γ =

∂f

∂t

∂

∂t

Note that in

∂f

∂t

, we are differentiating

with respect to

, whereas the

∂

∂t

the far right is just a formal expressions.

By the geodesic equation, we have

0 =

∇

˙γ = ∇

∂

Therefore, using the definition of the curvature tensor R, we obtain

0 = ∇

∂

∇

∂

∂t

= ∇

∂

∇

∂

− R(∂

, ∂

)∂

= ∇

∂

∇

∂

+ R(∂

, ∂

)∂

We let this act on the function f. So we get

0 =

∇

∂f

∂t

+ R(∂

, ∂

)

∂f

∂t

We write

J(t) =

∂f

∂s

(t, 0),

which is a vector field along the geodesic γ. Using the fact that

∇

∂f

∂t

∇

∂f

∂s

we find that J must satisfy the ordinary differential equation

∇

J + R( ˙γ, J) ˙γ = 0.

This is a linear second-order ordinary differential equation.

Definition

(Jacobi field)

Let

: [0

, L

]

→ M

be a geodesic. A Jacobi field is a

vector field J along γ that is a solution of the Jacobi equation on [0, L]

∇

J + R( ˙γ, J) ˙γ = 0. (†)

We now embark on a rather technical journey to prove results about Jacobi

fields. Observe that ˙γ(t) and t ˙γ(t) both satisfy this equation, rather trivially.

Theorem.

Let

: [0

, L

]

→ N

be a geodesic in a Riemannian manifold (

M, g

Then

(i) For any u, v ∈ T

γ(0)

M, there is a unique Jacobi field J along Γ with

J(0) = u,

∇J

(0) = v.

J(0) = 0,

∇J

(0) = k ˙γ(0),

then

(

) =

kt ˙γ

(

). Moreover, if both

(0)

∇J

(0) are orthogonal to

˙γ

(0),

then J(t) is perpendicular to ˙γ(t) for all [0, L].

In particular, the vector space of all Jacobi fields along

have dimension

2n, where n = dim M .

The subspace of those Jacobi fields pointwise perpendicular to

˙γ

(

) has

dimensional 2(n − 1).

(ii) J

(

) is independent of the parametrization of

˙γ

(

). Explicitly, if

˜γ

(

) =

˜γ(λt), then

J with the same initial conditions as J is given by

J(˜γ(t)) = J(γ(λt)).

This is the kind of theorem whose statement is longer than the proof.

Proof.

(i)

Pick an orthonormal basis

, ··· , e

, where

(0). Then

parallel transports

(

)

}

via the Levi-Civita connection preserves the

inner product.

We take e

to be parallel to ˙γ(0). By definition, we have

(0) = e

∇X

= 0.

Now we can write

J =

i=1

Then taking g(X

, ·) of (†) , we find that

¨y

j=2

R( ˙γ, X

, ˙γ, X

= 0.

Then the claims of the theorem follow from the standard existence and

uniqueness of solutions of differential equations.

In particular, for the orthogonality part, we know that

(0) and

∇J

(0)

being perpendicular to

˙γ

is equivalent to

(0) =

˙y

(0) = 0, and then

Jacobi’s equation gives

¨y

(t) = 0.

(ii) This follows from uniqueness.

Our discussion of Jacobi fields so far has been rather theoretical. Now that

we have an explicit equation for the Jacobi field, we can actually produce some

of them. We will look at the case where we have constant sectional curvature.

Example.

Suppose the sectional curvature is constantly

K ∈ R

, for

dim M ≥

We wlog |˙γ| = 1. We let J along γ be a Jacobi field, normal to ˙γ.

Then for any vector field T along γ, we have

hR( ˙γ, J) ˙γ, T i = K(g( ˙γ, ˙γ)g(J, T ) − g( ˙γ, J)g( ˙γ, T )) = Kg(J, T ).

Since this is true for all T , we know

R( ˙γ, J) ˙γ = KJ.

Then the Jacobi equation becomes

∇

J + KJ = 0.

So we can immediately write down a collection of solutions

J(t) =











sin(t

√

(t) K > 0

(t) K = 0

sinh(t

√

−K)

√

−K

(t) K < 0

for i = 2, ··· , n, and this has initial conditions

J(0) = 0,

∇J

(0) = e

Note that these Jacobi fields vanishes at 0.

We can now deliver our promise, proving that Jacobi fields are precisely the

variations of geodesics.

Proposition.

Let

: [

a, b

]

→ M

be a geodesic, and

(

t, s

) a variation of

γ(t) = f(t, 0) such that f(t, s) = γ

(t) is a geodesic for all |s| small. Then

J(t) =

∂f

∂s

is a Jacobi field along ˙γ.

Conversely, every Jacobi field along

can be obtained this way for an

appropriate function f.

Proof.

The first part is just the exact computation as we had at the beginning of

the section, but for the benefit of the reader, we will reproduce the proof again.

∇

= ∇

∇

∂f

∂s

= ∇

∇

∂f

∂t

= ∇



∇

∂f

∂t



− R(∂

, ∂

) ˙γ

We notice that the first term vanishes, because

∇

∂f

∂t

= 0 by definition of geodesic.

So we find

∇

= −R( ˙γ, J) ˙γ,

which is the Jacobi equation.

The converse requires a bit more work. We will write

(0) for the covariant

derivative of

along

. Given a Jacobi field

along a geodesic

(

) for

t ∈

, L

we let ˜γ be another geodesic such that

˜γ(0) = γ(0),

˜γ(0) = J(0).

We take parallel vector fields X

, X

along ˜γ such that

(0) = ˙γ(0), X

(0) = J

(0).

We put X(s) = X

(s) + sX

(s). We put

f(t, s) = exp

˜γ(s)

(tX(s)).

In local coordinates, for each fixed s, we find

f(t, s) = ˜γ(s) + tX(s) + O(t

)

as t → 0. Then we define

(t) = f(t, s)

whenever this makes sense. This depends smoothly on

, and the previous

arguments say we get a Jacobi field

J(t) =

∂f

∂s

(t, 0)

We now want to check that

. Then we are done. To do so, we have to

check the initial conditions. We have

J(0) =

∂f

∂s

(0, 0) =

d˜γ

(0) = J(0),

and also

(0) =

∇

∂f

∂s

(0, 0) =

∇

∂f

∂t

(0, 0) =

∇X

(0) = X

(0) = J

(0).

So we have

J = J.

Corollary. Every Jacobi field J along a geodesic γ with J(0) = 0 is given by

J(t) = (d exp

)

t ˙γ(0)

(tJ

(0))

for all t ∈ [0, L].

This is just a reiteration of the fact that if we pull back to the geodesic local

coordinates, then the variation must look like this:

But this corollary is stronger, in the sense that it holds even if we get out of the

geodesic local coordinates (i.e. when exp

no longer gives a chart).

Proof.

Write

˙γ

(0) =

, and

(0) =

. By above, we can construct the variation

f(t, s) = exp

(t(a + sw)).

Then

(d exp

)

t(a+sw)

(tw) =

∂f

∂s

(t, s),

which is just an application of the chain rule. Putting

= 0 gives the result.

It can be shown that in the situation of the corollary, if

a ⊥ w

, and

|a|

|w| = 1, then

|J(t)| = t −

K(σ)t

+ o(t

)

as t → 0, where σ is the plane spanned by a and w.

3.3 Further properties of geodesics

We can now use Jacobi fields to prove interesting things. We now revisit the

Gauss lemma, and deduce a stronger version.

Lemma (Gauss’ lemma). Let a, w ∈ T

M, and

γ = γ

(t, a) = exp

(ta)

a geodesic. Then

γ(t)

((d exp

)

a, (d exp

)

w) = g

γ(0)

(a, w).

In particular,

is orthogonal to

exp

{v ∈ T

|v|

. Note that the latter

need not be a submanifold.

This is an improvement of the previous version, which required us to live in

the geodesic local coordinates.

Proof. We fix any r > 0, and consider the Jacobi field J satisfying

J(0) = 0, J

(0) =

Then by the corollary, we know the Jacobi field is

J(t) = (d exp

)





We may write

= λa + u,

with

a ⊥ u

. Then since Jacobi fields depend linearly on initial conditions, we

write

J(t) = λt ˙γ(t) + J

(t)

for a Jacobi field J

a normal vector field along γ. So we have

g(J(r), ˙γ(r)) = λr|˙γ(r)|

= g(w, a).

But we also have

g(w, a) = g(λar + u, a) = λr|a|

= λr|˙γ(0)|

= λr|˙γ(r)|

Now we use the fact that

J(r) = (d exp

)

and

˙γ(r) = (d exp

)

and we are done.

Corollary

(Local minimizing of length)

Let

a ∈ T

. We define

(

) =

and ψ(t) a piecewise C

curve in T

M for t ∈ [0, 1] such that

ψ(0) = 0, ψ(1) = a.

Then

length(exp

◦ψ) ≥ length(exp

◦ϕ) = |a|.

It is important to interpret this corollary precisely. It only applies to curves

with the same end point in

. If we have two curves in

whose end

points have the same image in

, then the result need not hold (the torus would

be a counterexample).

Proof.

We may of course assume that

never hits 0 again after

= 0. We write

ψ(t) = ρ(t)u(t),

where ρ(t) ≥ 0 and |u(t)| = 1. Then

= ρ

u + ρu

Then using the extended Gauss lemma, and the general fact that if

(

) is a unit

vector for all t, then u · u

(u · u)

= 0, we have



(exp

◦ψ)(t)



(d exp

)

ψ(t)

(t)



= ρ

(t)

+ 2g(ρ

(t)u(t), ρ(t)u

(t)) + ρ(t)

|(d exp

)

ψ(t)

(t)|

= ρ

(t)

+ ρ(t)

|(d exp

)

ψ(t)

(t)|

Thus we have

length(exp

◦ψ) ≥

(t) dt = ρ(1) − ρ(0) = |a|.

Notation. We write Ω(p, q) for the set of all piecewise C

curves from p to q.

We now wish to define a metric on M, in the sense of metric spaces.

Definition

(Distance)

Suppose

is connected, which is the same as it being

path connected. Let (p, q) ∈ M. We define

d(p, q) = inf

ξ∈Ω(p,q)

length(ξ),

where

To see this is indeed a metric, All axioms of a metric space are obvious, apart

from the non-negativity part.

Theorem.

Let

p ∈ M

, and let

be such that

exp

B(0,ε)

is a diffeomorphism

onto its image, and let U be the image. Then

–

For any

q ∈ U

, there is a unique geodesic

γ ∈

Ω(

p, q

) with



(

)

< ε

Moreover,



(

) =

(

p, q

), and is the unique curve that satisfies this property.

– For any point q ∈ M with d(p, q) < ε, we have q ∈ U.

–

q ∈ M

is any point,

γ ∈

Ω(

p, q

) has



(

) =

(

p, q

)

< ε

, then

is a

geodesic.

Proof.

Let

exp

(

). Then the path

(

) =

exp

(

) is a geodesic from

of length

|a|

r < ε

. This is clearly the only such geodesic, since

exp

B(0,ε)

is a diffeomorphism.

Given any other path ˜γ ∈ Ω(p, q), we want to show (˜γ) > (γ). We let

τ = sup

t ∈ [0, 1] : γ([0, t]) ⊆ exp

(B(0, r))

Note that if τ 6= 1, then we must have γ(τ) ∈ Σ

, the geodesic sphere of radius

, otherwise we can continue extending. On the other hand, if

= 1, then we

certainly have

(

)

∈

, since

(

) =

. Then by local minimizing of length,

we have

(˜γ) ≥ (˜γ

[0,τ]

) ≥ r.

Note that we can always lift

˜γ[0, τ]

to a curve from 0 to

, since

exp

a diffeomorphism in B(0, ε).

By looking at the proof of the local minimizing of length, and using the same

notation, we know that we have equality iff τ = 1 and

ρ(t)

|(d exp

)

ψ(t)

ψ(t)u

(t)|

= 0

for all

. Since d

exp

is regular, this requires

(

) = 0 for all

(since

(

)

= 0

when

t 6

= 0, or else we can remove the loop to get a shorter curve). This implies

˜γ lifts to a straight line in T

M, i.e. is a geodesic.

Now given any

q ∈ M

with

(

p, q

)

< ε

, we pick

∈

[

r, ε

) and a path

γ ∈ Ω(p, q) such that (γ) = r

. We again let

τ = sup

t ∈ [0, 1] : γ([0, t]) ⊆ exp

(B(0, r

))

τ 6

= 1, then we must have

(

)

∈

, but lifting to

, this contradicts the

local minimizing of length.

The last part is an immediate consequence of the previous two.

Corollary.

The distance

on a Riemannian manifold is a metric, and induces

the same topology on M as the C

∞

structure.

Definition

(Minimal geodesic)

A minimal geodesic is a curve

: [0

→ M

such that

d(γ(0), γ(1)) = (γ).

One would certainly want a minimal geodesic to be an actual geodesic. This

is an easy consequence of what we’ve got so far, using the observation that a

sub-curve of a minimizing geodesic is still minimizing.

Corollary.

Let

: [0

→ M

be a piecewise

minimal geodesic with constant

speed. Then γ is in fact a geodesic, and is in particular C

∞

Proof.

We wlog

is unit speed. Let

t ∈

1], and pick

ε >

0 such that

exp

B(0,ε)

is a diffeomorphism. Then by the theorem,

[t,t+

ε]

is a geodesic.

So γ is C

∞

on (t, t +

ε), and satisfies the geodesic equations there.

Since we can pick

continuously with respect to

by ODE theorems, any

t ∈ (0, 1) lies in one such neighbourhood. So γ is a geodesic.

While it is not true that geodesics are always minimal geodesics, this is locally

true:

Corollary. Let γ : [0, 1] ⊆ R → M be a C

curve with |˙γ| constant. Then this

is a geodesic iff it is locally a minimal geodesic, i.e. for any

t ∈

1), there exists

δ > 0 such that

d(γ(t), γ(t + δ)) = (γ|

[t,t+δ]

Proof.

This is just carefully applying the previous theorem without getting

confused.

To prove

⇒

, suppose

is a geodesic, and

t ∈

1). We wlog

is unit speed.

Then pick

and

as in the previous theorem, and pick

. Then

γ|

[t,t+δ]

is a geodesic with length

< ε

between

(

) and

(

), and hence must have

minimal length.

To prove the converse, we note that for each

, the hypothesis tells us

γ|

[t,t+δ]

is a minimizing geodesic, and hence a geodesic, but the previous corollary. By

continuity,

must satisfy the geodesic equation at

. Since

is arbitrary,

is a

geodesic.

There is another sense in which geodesics are locally length minimizing.

Instead of chopping up a path, we can say it is minimal “locally” in the space

Ω(p, q). To do so, we need to give Ω(p, q) a topology, and we pick the topology

of uniform convergence.

Theorem.

Let

(

) =

exp

(

) be a geodesic, for

t ∈

1]. Let

(1).

Assume

is a regular point for

exp

for all

t ∈

1]. Then there exists

a neighbourhood of

in Ω(

p, q

) such that for all

in this neighbourhood,

(ψ) ≥ (γ), with equality iff ψ = γ up to reparametrization.

Before we prove the result, we first look at why the two conditions are

necessary. To see the necessity of

being regular, we can consider the sphere

and two antipodal points:

Then while the geodesic between them does minimize distance, it does not do so

strictly.

We also do not guarantee global minimization of length. For example, we

can consider the torus

= R

This has a flat metric from

, and the derivative of the exponential map is

the “identity” on

at all points. So the geodesics are the straight lines in

Now consider any two

p, q ∈ T

, then there are infinitely many geodesics joining

them, but typically, only one of them would be the shortest.

Proof.

The idea of the proof is that if

is any curve close to

, then we can use

the regularity condition to lift the curve back up to

, and then apply our

previous result.

Write

(

) =

ta ∈ T

. Then by the regularity assumption, for all

t ∈

1],

we know

exp

is a diffeomorphism of some neighbourhood

(

) of

(

) =

at ∈

onto the image. By compactness, we can cover [0

1] by finitely many such

covers, say W (t

), ··· , W (t

). We write W

= W (t

), and we wlog assume

0 = t

< t

< ··· < t

= 1.

By cutting things up, we may assume

γ([t

, t

i+1

]) ⊆ W

We let

U =

[

exp

Again by compactness, there is some

ε <

0 such that for all

t ∈

[

, t

i+1

], we

have B(γ(t), ε) ⊆ W

Now consider any curve

of distance

away from

. Then

([

, t

i+1

])

⊆ W

So we can lift it up to

, and the end point of the lift is

. So we are done

by local minimization of length.

Note that the tricky part of doing the proof is to make sure the lift of

has

the same end point as

, which is why we needed to do it neighbourhood

by neighbourhood.

3.4 Completeness and the Hopf–Rinow theorem

There are some natural questions we can ask about geodesics. For example, we

might want to know if geodesics can be extended to exist for all time. We might

also be interested if distances can always be realized by geodesics. It turns out

these questions have the same answer.

Definition

(Geodesically complete)

We say a manifold (

M, g

) is geodesically

complete if each geodesic extends for all time. In other words, for all

p ∈ M

exp

is defined on all of T

Example. The upper half plane

= {(x, y) : y > 0}

under the induced Euclidean metric is not geodesically complete. However,

and R

are diffeomorphic but R

is geodesically complete.

The first theorem we will prove is the following:

Theorem.

Let (

M, g

) be geodesically complete. Then any two points can be

connected by a minimal geodesic.

In fact, we will prove something stronger — let

p ∈ M

, and suppose

exp

defined on all of

. Then for all

q ∈ M

, there is a minimal geodesic between

them.

To prove this, we need a lemma

Lemma. Let p, q ∈ M. Let

= {x ∈ M : d(x, p) = δ}.

Then for all sufficiently small δ, there exists p

∈ S

such that

d(p, p

) + d(p

, q) = d(p, q).

Proof.

For

δ >

0 small, we know

= Σ

is a geodesic sphere about

, and

is compact. Moreover,

(

·, q

) is a continuous function. So there exists some

∈ Σ

that minimizes d( ·, q).

Consider an arbitrary

γ ∈

Ω(

p, q

). For the sake of sanity, we assume

δ <

d(p, q). Then there is some t such that γ(t) ∈ Σ

, and

(γ) ≥ d(p, γ(t)) + d(γ(t), q) ≥ d(p, p

) + d(p

, q).

So we know

d(p, q) ≥ d(p, p

) + d(p

, p).

The triangle inequality gives the opposite direction. So we must have equality.

We can now prove the theorem.

Proof of theorem.

We know

exp

is defined on

. Let

q ∈ M

. Let

q ∈ M

We want a minimal geodesic in Ω(

p, q

). By the first lemma, there is some

δ >

and p

such that

d(p, p

) = δ, d(p, p

) + d(p

, q) = d(p, q).

Also, there is some v ∈ T

M such that exp

v = p

. We let

(t) = exp



|v|



We let

I = {t ∈ R : d(q, γ

(t)) + t = d(p, q)}.

Then we know

(i) δ ∈ I

(ii) I is closed by continuity.

Let

T = sup{I ∩ [0, d(p, q)]}.

Since

is closed, this is in fact a maximum. So

T ∈ I

. We claim that

(

p, q

If so, then γ

∈ Ω(p, q) is the desired minimal geodesic, and we are done.

Suppose this were not true. Then

T < d

(

p, q

). We apply the lemma to

˜p

(

), and

remains as before. Then we can find

ε >

0 and some

∈ M

with the property that

d(p

, q) = d(γ

(T ), q) − d(γ

(T ), p

)

= d(γ

(T ), q) − ε

= d(p, q) − T − ε

Hence we have

d(p, p

) ≥ d(p, q) − d(q, p

) = T + ε.

Let γ

be the radial (hence minimal) geodesic from γ

(T ) to p

. Now we know

(γ

[0,T ]

) + (γ

) = T + ε.

concatenated with

[0,T ]

is a length-minimizing geodesic from

and is hence a geodesic. So in fact

lies on

, say

(

) for some

. Then

s ∈ I

, which is a contradiction. So we must have

(

p, q

), and

hence

d(q, γ

(T )) + T = d(p, q),

hence d(q, γ

(T )) = 0, i.e. q = γ

(T ).

Corollary

(Hopf–Rinow theorem)

For a connected Riemannian manifold (

M, g

the following are equivalent:

(i) (M, g) is geodesically complete.

(ii) For all p ∈ M, exp

is defined on all T

(iii) For some p ∈ M, exp

is defined on all T

(iv) Every closed and bounded subset of (M, d) is compact.

(v) (M, d) is complete as a metric space.

Proof.

(i) and (ii) are equivalent by definition. (ii)

⇒

(iii) is clear, and we proved

(iii) ⇒ (i).

–

(iii)

⇒

(iv): Let

K ⊆ M

be closed and bounded. Then by boundedness,

is contained in

exp

(

B(0, R)

). Let

be the pre-image of

under

exp

Then it is a closed and bounded subset of

, hence compact. Then

the continuous image of a compact set, hence compact.

– (iv) ⇒ (v): This is a general topological fact.

–

(v)

⇒

(i): Let

(

) :

I → R

be a geodesic, where

I ⊆ R

. We wlog

|˙γ| ≡

Suppose

I 6

. We wlog

sup I

a < ∞

. Then

lim

t→a

(

) exist by

completeness, and hence

(

) exists. Since geodesics are locally defined

near

, we can pick a geodesic in the direction of

lim

t→a

(

). So we can

extend γ further, which is a contradiction.

3.5 Variations of arc length and energy

This section is mostly a huge computation. As we previously saw, geodesics are

locally length-minimizing, and we shall see that another quantity, namely the

energy is also a useful thing to consider, as minimizing the energy also forces

the parametrization to be constant speed.

To make good use of these properties of geodesics, it is helpful to compute

explicitly expressions for how length and energy change along variations. The

computations are largely uninteresting, but it will pay off.

Definition (Energy). The energy function E : Ω(p, q) → R is given by

E(γ) =

|˙γ|

dt,

where γ : [0, T] → M.

Recall that Ω(

p, q

) is defined as the space of piecewise

curves. Often, we

will make the simplifying assumption that all curves are in fact

. It doesn’t

really matter.

Note that the length of a curve is independent of parametrization. Thus,

if we are interested in critical points, then the critical points cannot possibly

be isolated, as we can just re-parametrize to get a nearby path with the same

length. On the other hand, the energy

does depend on parametrization. This

does have isolated critical points, which is technically very convenient.

Proposition.

Let

: [0

, T

]

→ M

be a path from

such that for all

γ ∈

Ω(

p, q

) with

: [0

, T

]

→ M

, we have

(

)

≥ E

(

). Then

must be a

geodesic.

Recall that we already had such a result for length instead of energy. The

proof is just the application of Cauchy-Schwartz.

Proof. By the Cauchy-Schwartz inequality, we have

|˙γ|

dt ≥

|˙γ(t)| dt

with equality iff |˙γ| is constant. In other words,

E(γ) ≥

(γ)

So we know that if

minimizes energy, then it must be constant speed. Now

given any

, if we just care about its length, then we may wlog it is constant

speed, and then

(γ) =

2E(γ)T ≥

2E(γ

)T = (γ

So γ

minimizes length, and thus γ

is a geodesic.

We shall consider smooth variations

(

t, s

) of

(

) =

(

0). We require

that

: [0

, T

]

(

−ε, ε

)

→ M

is smooth. Since we are mostly just interested

in what happens “near”

= 0, it is often convenient to just consider the

corresponding vector field along γ:

Y (t) =

∂H

∂s



s=0

= (dH)

(t,0)

∂

∂s

Conversely, given any such vector field

, we can generate a variation

that

gives rise to Y . For example, we can put

H(t, s) = exp

(t)

(sY (t)),

which is valid on some neighbourhood of [0

, T

]

× {

}

. If

(0) = 0 =

(

), then

we can choose H fixing end-points of γ

Theorem (First variation formula).

(i) For any variation H of γ, we have

E(γ

)



s=0

= g(Y (t), ˙γ(t))|

−



Y (t),

∇

˙γ(t)



dt. (∗)

(ii) The critical points, i.e. the γ such that

E(γ

)



s=0

for all (end-point fixing) variation H of γ, are geodesics.

(iii) If |˙γ

(t)| is constant for each fixed s ∈ (−ε, ε), and |˙γ(t)| ≡ 1, then

E(γ

)



s=0

(γ

)



s=0

(iv)

is a critical point of the length, then it must be a reparametrization of

a geodesic.

This is just some calculations.

Proof.

We will assume that we can treat

∂

∂s

and

∂

∂t

as vector fields on an

embedded submanifold, even though H is not necessarily a local embedding.

The result can be proved without this assumption, but will require more

technical work.

(i) We have

∂

∂s

g( ˙γ

(t), ˙γ

(t)) = g



∇

˙γ

(t), ˙γ

(t)



= g



∇

∂H

∂s

(t, s),

∂H

∂t

(t, s)



∂

∂t



∂H

∂s

∂H

∂t



− g



∂H

∂s

∇

∂H

∂t



Comparing with what we want to prove, we see that we get what we want

by integrating

dt, and then putting s = 0, and then noting that

∂H

∂s



s=0

= Y,

∂H

∂t



s=0

= ˙γ.

(ii) If γ is a geodesic, then

∇

˙γ(t) = 0.

So the integral on the right hand side of (

∗

) vanishes. Also, we have

Y (0) = 0 = Y (T ). So the RHS vanishes.

Conversely, suppose γ is a critical point for E. Then choose H with

Y (t) = f(t)

∇

˙γ(t)

for some f ∈ C

∞

[0, T ] such that f(0) = f(T ) = 0. Then we know

f(t)



∇

˙γ(t)



dt = 0,

and this is true for all f. So we know

∇

˙γ = 0.

(iii)

This is evident from the previous proposition. Indeed, we fix [0

, T

], then

for all H, we have

E(γ

) =

(γ

)

and so

E(γ

)



s=0

(γ

)

(γ

)



s=0

and when s = 0, the curve is parametrized by arc-length, so (γ

) = T .

(iv)

By reparametrization, we may wlog

|˙γ| ≡

1. Then

is a critical point for

, hence for E, hence a geodesic.

Often, we are interested in more than just whether the curve is a critical

point. We want to know if it maximizes or minimizes energy. Then we need

more than the “first derivative”. We need the “second derivative” as well.

Theorem

(Second variation formula)

Let

(

) : [0

, T

]

→ M

be a geodesic with

|˙γ| = 1. Let H(t, s) be a variation of γ. Let

Y (t, s) =

∂H

∂s

(t, s) = (dH)

(t,s)

∂

∂s

Then

(i) We have

E(γ

)



s=0

= g



∇Y

(t, 0), ˙γ





(|Y

− R(Y, ˙γ, Y, ˙γ)) dt.

(ii) Also

(γ

)



s=0

= g



∇Y

(t, 0), ˙γ(t)







− R(Y, ˙γ, Y, ˙γ) − g( ˙γ, Y

)



dt,

where R is the (4, 0) curvature tensor, and

(t) =

∇Y

(t, 0).

Putting

= Y − g(Y, ˙γ) ˙γ

for the normal component of Y , we can write this as

(γ

)



s=0

= g



∇Y

(t, 0), ˙γ(t)







− R(Y

, ˙γ, Y

, ˙γ)



dt.

Note that if we have fixed end points, then the first terms in the variation

formulae vanish.

Proof. We use

E(γ

) = g(Y (t, s), ˙γ

(t))|

t=T

t=0

−



Y (t, s),

∇

˙γ

(t)



dt.

Taking the derivative with respect to s again gives

E(γ

) = g



∇Y

, ˙γ





t=0

+ g



∇

˙γ





t=0

−



∇Y

∇

˙γ



+ g



∇

˙γ



dt.

We now use that

∇

˙γ

(t) =

∇

˙γ

(t) + R



∂H

∂s

∂H

∂t



˙γ



∇



Y (t, s) + R



∂H

∂s

∂H

∂t



˙γ

We now set s = 0, and then the above gives

E(γ

)



s=0

= g



∇Y

, ˙γ





+ g



∇˙γ





−



∇



+ R( ˙γ, Y, ˙γ, Y )

dt.

Finally, applying integration by parts, we can write

−



∇



dt = − g



∇





∇Y



dt.

Finally, noting that

∇

˙γ(s) =

∇

Y (t, s),

we find that

E(γ

)



s=0

= g



∇Y

, ˙γ







− R(Y, ˙γ, Y, ˙γ)



dt.

It remains to prove the second variation of length. We first differentiate

(γ

) =

g( ˙γ

, ˙γ

)

∂

∂s

g( ˙γ

, ˙γ

) dt.

Then the second derivative gives

(γ

)



s=0

∂

∂s

g( ˙γ

, ˙γ

)



s=0

−



∂

∂s

g( ˙γ

, ˙γ

)





s=0

dt,

where we used the fact that g( ˙γ, ˙γ) = 1.

We notice that the first term can be identified with the derivative of the

energy function. So we have

(γ

)



s=0

E(γ

)



s=0

−



˙γ

∇

˙γ





s=0



dt.

So the second part follows from the first.

3.6 Applications

This finally puts us in a position to prove something more interesting.

Synge’s theorem

We are first going to prove the following remarkable result relating curvature

and topology:

Theorem

(Synge’s theorem)

Every compact orientable Riemannian manifold

(

M, g

) such that

dim M

is even and has

(

)

0 for all planes at

p ∈ M

simply connected.

We can see that these conditions are indeed necessary. For example, we can

consider

/ ±

1 with the induced metric from

. Then this is compact

with positive sectional curvature, but it is not orientable. Indeed it is not simply

connected.

Similarly, if we take

, then this has odd dimension, and the theorem

breaks.

Finally, we do need strict inequality, e.g. the flat torus is not simply connected.

We first prove a technical lemma.

Lemma.

Let

be a compact manifold, and [

] a non-trivial homotopy class

of closed curves in M . Then there is a closed minimal geodesic in [α].

Proof.

Since

is compact, we can pick some

ε >

0 such that for all

p ∈ M

, the

map exp

B(0,p)

is a diffeomorphism.

Let



inf

γ∈[α]



(

). We know that

 >

0, otherwise, there exists a

with



(

)

< ε

. So

is contained in some geodesic coordinate neighbourhood, but

then α is contractible. So  must be positive.

Then we can find a sequence

∈

[

] with

: [0

→ M

|˙γ|

constant,

such that

lim

n→∞

(γ

) = .

Choose

0 = t

< t

< ··· < t

= 1

such that

i+1

− t

2

So it follows that

d(γ

), γ

i+1

)) < ε

for all

sufficiently large and all

. Then again, we can replace

i+1

]

by a

radial geodesic without affecting the limit lim (γ

Then we exploit the compactness of

(and the unit sphere) again, and pass

to a subsequence of

{γ

}

so that

(

)

, ˙γ

(

) are all convergent for every fixed

i as n → ∞. Then the curves converges to some

→ ˆγ ∈ [α],

given by joining the limits

lim

n→∞

(

). Then we know that the length

converges as well, and so we know

ˆγ

is minimal among curves in [

]. So

ˆγ

locally minimal, hence a geodesic. So we can take γ = ˆγ, and we are done.

Proof of Synge’s theorem.

Suppose

satisfies the hypothesis, but

(

)

{

}

So there is a path

with [

]

= 1, i.e. it cannot be contracted to a point. By the

lemma, we pick a representative γ of [α] that is a closed, minimal geodesic.

We now prove the theorem. We may wlog assume

|˙γ|

= 1, and

ranges in

[0, T ]. Consider a vector field X(t) for 0 ≤ t ≤ T along γ(t) such that

∇X

= 0, g(X(0), ˙γ(0)) = 0.

Note that since g is a geodesic, we know

g(X(t), ˙γ(t)) = 0,

for all

t ∈

, T

] as parallel transport preserves the inner product. So

(

)

⊥

˙γ(T ) = ˙γ(0) since we have a closed curve.

We consider the map

that sends

(0)

7→ X

(

). This is a linear isometry

of (

˙γ

(0))

⊥

with itself that preserves orientation. So we can think of

as a map

P ∈ SO(2n − 1),

where

dim M

= 2

. It is an easy linear algebra exercise to show that every

element of

n −

1) must have an eigenvector of eigenvalue 1. So we can find

v ∈ T

such that

v ⊥ ˙γ

(0) and

(

) =

. We take

(0) =

. Then we have

X(T ) = v.

Consider now a variation

(

t, s

) inducing this

(

). We may assume

|˙γ

constant. Then

(γ

s=0

= 0

is minimal. Moreover, since it is a minimum, the second derivative must be

positive, or at least non-negative. Is this actually the case?

We look at the second variation formula of length. Using the fact that the

loop is closed, the formula reduces to

(γ

)



s=0

= −

R(X, ˙γ, X, ˙γ) dt.

But we assumed the sectional curvature is positive. So the second variation is

negative! This is a contradiction.

Conjugate points

Recall that when a geodesic starts moving, for a short period of time, it is

length-minimizing. However, in general, if we keep on moving for a long time,

then we cease to be minimizing. It is useful to characterize when this happens.

As before, for a vector field J along a curve γ(t), we will write

∇J

Definition (Conjugate points). Let γ(t) be a geodesic. Then

p = γ(α), q = γ(β)

are conjugate points if there exists some non-trivial

such that

(

) = 0 =

(

It is easy to see that this does not depend on parametrization of the curve,

because Jacobi fields do not.

Proposition.

(i)

(

) =

exp

(

), and

exp

(

βa

) is conjugate to

, then

is a singular

value of exp.

(ii) Let J be as in the definition. Then J must be pointwise normal to ˙γ.

Proof.

(i)

We wlog [

α, β

] = [0

1]. So

(0) = 0 =

(1). We

˙γ

(0) and

(0).

Note that

a, w

are both non-zero, as Jacobi fields are determined by initial

conditions. Then q = exp

(a).

We have shown earlier that if J(0) = 0, then

J(t) = (d exp

)

(tw)

for all 0

≤ t ≤

1. So it follows (d

exp

)

(

) =

(1) = 0. So (d

exp

)

has

non-trivial kernel, and hence isn’t surjective.

(ii) We claim that any Jacobi field J along a geodesic γ satisfies

g(J(t), ˙γ(t)) = g(J

(0), ˙γ(0))t + g(J(0), ˙γ(0)).

To prove this, we note that by the definition of geodesic and Jacobi fields,

we have

g(J

, ˙γ) = g(J

, ˙γ(0)) = −g(R( ˙γ, J), ˙γ, ˙γ) = 0

by symmetries of R. So we have

g(J, ˙γ) = g(J

(t), ˙γ(t)) = g(J

(0), ˙γ(0)).

Now integrating gives the desired result.

This result tells us g(J(t), ˙γ(t)) is a linear function of t. But we have

g(J(0), ˙γ(0)) = g(J(1), ˙γ(1)) = 0.

So we know g(J(t), ˙γ(t)) is constantly zero.

From the proof, we see that for any Jacobi field with J(0) = 0, we have

g(J

(0), ˙γ(0)) = 0 ⇐⇒ g(J(t), ˙γ(t)) = constant.

This implies that the dimension of the normal Jacobi fields along

satisfying

J(0) = 0 is dim M − 1.

Example.

Consider

⊆ R

with the round metric, i.e. the “obvious”

metric induced from

. We claim that

= (0

1) and

= (0

0) are

conjugate points.

To construct a Jacobi field, instead of trying to mess with the Jacobi equation,

we construct a variation by geodesics. We let

f(t, s) =





cos s sin t

sin s sin t

cos t





We see that when

= 0, this is the great-circle in the (

x, z

)-plane. Then we have

a Jacobi field

J(t) =

∂f

∂s



s=0





sin t





This is then a Jacobi field that vanishes at N and S.

When we are at the conjugate point, then there are many adjacent curves

whose length is equal to ours. If we extend our geodesic beyond the conjugate

point, then it is no longer even locally minimal:

We can push the geodesic slightly over and the length will be shorter. On the

other hand, we proved that up to the conjugate point, the geodesic is always

locally minimal.

In turns out this phenomenon is generic:

Theorem.

Let

: [0

→ M

be a geodesic with

(0) =

(1) =

such that

is conjugate to some

(

) for some

∈

1). Then there is a piecewise smooth

variation of f(t, s) with f(t, 0) = γ(t) such that

f(0, s) = p, f(1, s) = q

and (f ( ·, s)) < (γ) whenever s 6= 0 is small.

The proof is a generalization of the example we had above. We know that up

to the conjugate point, we have a Jacobi filed that allows us to vary the geodesic

without increasing the length. We can then give it a slight “kick” and then the

length will decrease.

Proof.

By the hypothesis, there is a

(

) defined on

t ∈

1] and

∈

1) such

that

J(t) ⊥ ˙γ(t)

for all t, and J(0) = J(t

) = 0 and J 6≡ 0. Then J

) 6= 0.

We define a parallel vector field

along

(

) =

−J

(

). We pick

θ ∈ C

∞

[0, 1] such that θ(0) = θ(1) = 0 and θ(t

) = 1.

Finally, we define

Z = θZ

and for α ∈ R, we define

(t) =

(

J(t) + αZ(t) 0 ≤ t ≤ t

αZ(t) t

≤ t ≤ 1

We notice that this is not smooth at

, but is just continuous. We will postpone

the choice of α to a later time.

We know

(

) arises from a piecewise

∞

variation of

, say

(

t, s

). The

technical claim is that the second variation of length corresponding to

(

) is

negative for some α.

We denote by

(

X, Y

)

the symmetric bilinear form that gives rise to the

second variation of length with fixed end points. If we make the additional

assumption that

X, Y

are normal along

, then the formula simplifies, and

reduces to

I(X, Y )

(g(X

, Y

) − R(X, ˙γ, Y, ˙γ)) dt.

Then for H

(t, s), we have

(γ

)



s=0

= I

+ I

= I(J, J)

= 2αI(J, Z)

= α

I(Z, Z)

We look at each term separately.

We first claim that I

= 0. We note that

g(J, J

) = g(J

, J

) + g(J, J

and

(

J, J

) added to the curvature vanishes by the Jacobi equation. Then

by integrating by parts and applying the boundary condition, we see that

vanishes.

Also, by integrating by parts, we find

= 2αg(Z, J

Whence

(γ

)



s=0

= −2α|J

+ α

I(Z, Z)

Now if

α >

0 is very very small, then the linear term dominates, and this is

negative. Since the first variation vanishes (

is a geodesic), we know this is a

local maximum of length.

Note that we made a compromise in the theorem by doing a piecewise

∞

variation instead of a smooth one, but of course, we can fix this by making a

smooth approximation.

Bonnet–Myers diameter theorem

We are going to see yet another application of our previous hard work, which

may also be seen as an interplay between curvature topology. In case it isn’t

clear, all our manifolds are connected.

Definition (Diameter). The diameter of a Riemannian manifold (M, g) is

diam(M, g) = sup

p,q∈M

d(p, q).

Of course, this definition is valid for any metric space.

Example. Consider the sphere

n−1

(r) = {x ∈ R

: |x| = r},

with the induced “round” metric. Then

diam(S

n−1

(r)) = πr.

It is an exercise to check that

K ≡

We will also need the following notation:

Notation.

Let

be two symmetric bilinear forms on a real vector space. We

say h ≥

h if h −

h is non-negative definite.

h ∈

Γ(

∗

) are fields of symmetric bilinear forms, we write

h ≥

≥

for all p ∈ M.

The following will also be useful:

Definition

(Riemannian covering map)

Let (

M, g

) and (

M, ˜g

) be two Rieman-

nian manifolds, and

M → M

be a smooth covering map. We say

is a

Riemannian covering map if it is a local isometry. Alternatively,

∗

˜g

. We

say

M is a Riemannian cover of M.

Recall that if

is in fact a universal cover, i.e.

is simply connected, then

we can (non-canonically) identify π

(M) with f

−1

(p) for any point p ∈ M.

Definition

(Bonnet–Myers diameter theorem)

Let (

M, g

) be a complete

dimensional manifold with

Ric(g) ≥

n − 1

where r > 0 is some positive number. Then

diam(M, g) ≤ diam S

(r) = πr.

In particular, M is compact and π

(M) is finite.

Proof.

Consider any

L < diam

(

M, g

). Then by definition (and Hopf–Rinow),

we can find

p, q ∈ M

such that

(

p, q

) =

, and a minimal geodesic

γ ∈

Ω(

p, q

)

with (γ) = d(p, q). We parametrize γ : [0, L] → M so that |˙γ| = 1.

Now consider any vector field

along

such that

(

) = 0 =

(

). Since

Γ is a minimal geodesic, it is a critical point for



, and the second variation

(

Y, Y

)

[0,L]

is non-negative (recall that the second variation has fixed end points).

We extend

˙γ

(0) to an orthonormal basis of

, say

˙γ

(0) =

, e

, ··· , e

We further let X

be the unique vector field such that

= 0, X

(0) = e

In particular, X

(t) = ˙γ(t).

For i = 2, ··· , n, we put

(t) = sin



πt



(t).

Then after integrating by parts, we find that we have

I(Y

, Y

)

[0,L]

= −

g(Y

+ R( ˙γ, Y

, ˙γ) dt

Using the fact that X

is parallel, this can be written as

sin

πt



− R( ˙γ, X

, ˙γ, X

)



dt,

and since this is length minimizing, we know this is ≥ 0.

We note that we have R( ˙γ, X

, ˙γ, X

) = 0. So we have

i=2

R( ˙γ, X

, ˙γ, X

) = Ric( ˙γ, ˙γ).

So we know

i=2

I(Y

, Y

) =

sin

πt



(n − 1)

− Ric( ˙γ, ˙γ)



dt ≥ 0.

We also know that

Ric( ˙γ, ˙γ) ≥

n − 1

by hypothesis. So this implies that

≥

This tells us that

L ≤ πr.

Since L is any number less that diam(M, g), it follows that

diam(M, g) ≤ πr.

Since

is known to be complete, by Hopf-Rinow theorem, any closed bounded

subset is compact. But M itself is closed and bounded! So M is compact.

To understand the fundamental, group, we simply have to consider a universal

Riemannian cover

M → M

. We know such a topological universal covering

space must exist by general existence theorems. We can then pull back the

differential structure and metric along

, since

is a local homeomorphism.

So this gives a universal Riemannian cover of

. But they hypothesis of the

theorem is local, so it is also satisfied for

. So it is also compact. Since

−1

(

)

is a closed discrete subset of a compact space, it is finite, and we are done.

It is an easy exercise to show that the hypothesis on the Ricci curvature

cannot be weakened to just saying that the Ricci curvature is positive definite.

Hadamard–Cartan theorem

To prove the next result, we need to talk a bit more about coverings.

Proposition.

Let (

M, g

) and (

N, h

) be Riemannian manifolds, and suppose

is complete. Suppose there is a smooth surjection

M → N

that is a local

diffeomorphism. Moreover, suppose that for any

p ∈ M

and

v ∈ T

, we have

|df

(v)|

≥ |v|. Then f is a covering map.

Proof.

By general topology, it suffices to prove that for any smooth curve

: [0

→ N

, and any

q ∈ M

such that

(

) =

(0), there exists a lift of

starting from from q.

[0, 1] N

˜γ

From the hypothesis, we know that

˜γ

exists on [0

, ε

] for some “small”

We let

I = {0 ≤ ε ≤: ˜γ exists on [0, ε]}.

We immediately see this is non-empty, since it contains

. Moreover, it is not

difficult to see that

is open in [0

1], because

is a local diffeomorphism. So it

suffices to show that I is closed.

We let {t

}

∞

n=1

⊆ I be such that t

n+1

> t

for all n, and

lim

n→∞

= ε

Using Hopf-Rinow, either

{˜γ

(

)

}

is contained in some compact

, or it is

unbounded. We claim that unboundedness is impossible. We have

(γ) ≥ (γ|

[0,t

]

) =

|˙γ| dt

|df

˜γ(t)

˜γ(t)| dt

≥

˜γ| dt

= (˜γ|

[0,t

]

)

≥ d(˜γ(0), ˜γ(t

)).

So we know this is bounded. So by compactness, we can find some

such that

˜γ

(

)

→ x

 → ∞

. There exists an open

x ∈ V ⊆ M

such that

is a

diffeomorphism.

Since there are extensions of

˜γ

to each

, eventually we get an extension to

within

, and then we can just lift directly, and extend it to

. So

∈ I

. So

we are done.

Corollary.

Let

M → N

be a local isometry onto

, and

be complete.

Then f is a covering map.

Note that since

is (assumed to be) connected, we know

is necessarily

surjective. To see this, note that the completeness of

implies completeness of

(

), hence

(

) is closed in

, and since it is a local isometry, we know

in particular open. So the image is open and closed, hence f(M ) = N .

For a change, our next result will assume a negative curvature, instead of a

positive one!

Theorem

(Hadamard–Cartan theorem)

Let (

, g

) be a complete Riemannian

manifold such that the sectional curvature is always non-positive. Then for every

point

p ∈ M

, the map

exp

M → M

is a covering map. In particular, if

(M) = 0, then M is diffeomorphic to R

We will need one more technical lemma.

Lemma.

Let

(

) be a geodesic on (

M, g

) such that

K ≤

0 along

. Then

has no conjugate points.

Proof.

We write

(0) =

. Let

(

) be a Jacobi field along

, and

(0) = 0. We

claim that if J is not identically zero, then J does not vanish everywhere else.

We consider the function

f(t) = g(J(t), J(t)) = |J(t)|

Then f (0) = f

(0) = 0. Consider

(t) = g(J

(t), J(t)) + g(J

(t), J

(t)) = g(J

, J

) − R( ˙γ, J, ˙γ, J) ≥ 0.

So f is a convex function, and so we are done.

We can now prove the theorem.

Proof of theorem.

By the lemma, we know there are no conjugate points. So

we know

exp

is regular everywhere, hence a local diffeomorphism by inverse

function theorem. We can use this fact to pull back a metric from

such that

exp

is a local isometry. Since this is a local isometry, we know

geodesics are preserved. So geodesics originating from the origin in

are

straight lines, and the speed of the geodesics under the two metrics are the same.

So we know

is complete under this metric. Also, by Hopf–Rinow,

exp

surjective. So we are done.

4 Hodge theory on Riemannian manifolds

4.1 Hodge star and operators

Throughout this chapter, we will assume our manifolds are oriented, and write

of the dimension. We will write

ε ∈

Ω

(

) for a non-vanishing form defining

the orientation.

Given a coordinate patch

U ⊆ M

, we can use Gram-Schmidt to obtain a

positively-oriented orthonormal frame

, ··· , e

. This allows us to dualize and

obtain a basis ω

, ··· , ω

∈ Ω

(M), defined by

) = δ

Since these are linearly independent, we can multiply all of them together to

obtain a non-zero n-form

∧ ··· ∧ ω

= aε,

for some

a ∈ C

∞

(

a >

0. We can do this for any coordinate patches, and

the resulting

-form agrees on intersections. Indeed, given any other choice

, ··· , ω

, they must be related to the original

, ··· , ω

by an element

Φ ∈ SO(n). Then by linear algebra, we have

∧ ··· ∧ ω

= det(Φ) ω

∧ ··· ∧ ω

= ω

∧ ··· ∧ ω

So we can patch all these forms together to get a global

-form

∈

Ω

(

)

that gives the same orientation. This is a canonical such

-form, depending only

and the orientation chosen. This is called the (Riemannian) volume form

of (M, g).

Recall that the

are orthonormal with respect to the natural dual inner

product on

∗

. In general,

induces an inner product on

∗

for all

= 0

, ··· , n

, which is still denoted

. One way to describe this is to give an

orthonormal basis on each fiber, given by

{ω

∧ ··· ∧ ω

: 1 ≤ i

< ··· < i

≤ n}.

From this point of view, the volume form becomes a form of “unit length”.

We now come to the central definition of Hodge theory.

Definition

(Hodge star)

The Hodge star operator on (

, g

) is the linear map

 :

∗

M) →

n−p

∗

satisfying the property that for all α, β ∈

∗

M), we have

α ∧ β = hα, βi

Since g is non-degenerate, it follows that this is well-defined.

How do we actually compute this? Since we have vector spaces, it is natural

to consider what happens in a basis.

Proposition.

Suppose

, ··· , ω

is an orthonormal basis of

∗

. Then we

claim that

(ω

∧ ··· ∧ ω

) = ω

p+1

∧ ··· ∧ ω

We can check this by checking all basis vectors of

, and the result drops

out immediately. Since we can always relabel the numbers, this already tells us

how to compute the Hodge star of all other basis elements.

We can apply the Hodge star twice, which gives us a linear endomorphism

 :

∗

M →

∗

M. From the above, it follows that

Proposition.

The double Hodge star



(

∗

)

→

(

∗

) is equal to

(−1)

p(n−p)

In particular,

1 = ω

, ω

= 1.

Using the Hodge star, we can define a differential operator:

Definition

(Co-differential (

))

We define

: Ω

(

)

→

Ω

p−1

(

) for 0

≤ p ≤

dim M by

δ =

(

(−1)

n(p+1)+1

 d p 6= 0

0 p = 0

This is (sometimes) called the co-differential.

The funny powers of (−1) are chosen so that our future results work well.

We further define

Definition (Laplace–Beltrami operator ∆). The Laplace–Beltrami operator is

∆ = dδ + δd : Ω

(M) → Ω

(M).

This is also known as the (Hodge) Laplacian.

We quickly note that

Proposition.

∆ = ∆  .

Consider the spacial case of (

M, g

) = (

, eucl

), and

= 0. Then a straight-

forward calculation shows that

∆f = −

∂

∂x

− ··· −

∂

∂x

for each

f ∈ C

∞

(

) = Ω

(

). This is just the usual Laplacian, except there

is a negative sign. This is there for a good reason, but we shall not go into that.

More generally, metric

(or alternatively a coordinate

patch on any Riemannian manifold), we have

|g| dx

∧ ··· ∧ dx

where |g| is the determinant of g. Then we have

∆

f = −

|g|

∂

(

|g|g

∂

f) = −g

∂

f + lower order terms.

How can we think about this co-differential δ? One way to understand it is

that it is the “adjoint” to d.

Proposition. δ

is the formal adjoint of d. Explicitly, for any compactly sup-

ported α ∈ Ω

p−1

and β ∈ Ω

, then

hdα, βi

hα, δβi

We just say it is a formal adjoint, rather than a genuine adjoint, because

there is no obvious Banach space structure on Ω

(

), and we don’t want to go

into that. However, we can still define

Definition

(

inner product)

For

ξ, η ∈

Ω

(

), we define the

inner

product by

hhξ, ηii

hξ, ηi

where ξ, η ∈ Ω

(M).

Note that this may not be well-defined if the space is not compact.

Under this notation, we can write the proposition as

hhdα, βii

= hhα, δβii

Thus, we also say δ is the L

adjoint.

To prove this, we need to recall Stokes’ theorem. Since we don’t care about

manifolds with boundary in this course, we just have

dω = 0

for all forms ω.

Proof. We have

0 =

d(α ∧ β)

dα ∧ β +

(−1)

p−1

α ∧ d  β

hdα, βi

+ (−1)

p−1

(−1)

(n−p+1)(p−1)

α ∧   d  β

hdα, βi

+ (−1)

(n−p)(p−1)

α ∧   d  β

hdα, βi

−

α ∧ δβ

hdα, βi

−

hα, δβi

This result explains the funny signs we gave δ.

Corollary. ∆ is formally self-adjoint.

Similar to what we did in, say, IB Methods, we can define

Definition

(Harmonic forms)

A harmonic form is a

-form

such that ∆

= 0.

We write

= {α ∈ Ω

(M) : ∆α = 0}.

We have a further corollary of the proposition.

Corollary. Let M be compact. Then

∆α = 0 ⇔ dα = 0 and δα = 0.

We say α is closed and co-closed .

Proof. ⇐ is clear. For ⇒, suppose ∆α = 0. Then we have

0 = hhα, ∆αii = hhα, dδα + δdαii = kδαk

+ kdαk

Since the L

norm is non-degenerate, it follows that δα = dα = 0.

In particular, in degree 0, co-closed is automatic. Then for all

f ∈ C

∞

(

we have

∆f = 0 ⇔ df = 0.

In other words, harmonic functions on a compact manifold must be constant.

This is a good way to demonstrate that the compactness hypothesis is required,

as there are many non-trivial harmonic functions on R

, e.g. x.

Some of these things simplify if we know about the parity of our manifold. If

dim M = n = 2m, then  = (−1)

, and

δ = −  d

whenever

p 6

= 0. In particular, this applies to complex manifolds, say

∼

with the Hermitian metric. This is to be continued in sheet 3.

4.2 Hodge decomposition theorem

We now work towards proving the Hodge decomposition theorem. This is a very

important and far-reaching result.

Theorem

(Hodge decomposition theorem)

Let (

M, g

) be a compact oriented

Riemannian manifold. Then

– For all p = 0, ··· , dim M , we have dim H

< ∞.

– We have

Ω

(M) = H

⊕ ∆Ω

(M).

Moreover, the direct sum is orthogonal with respect to the

inner product.

We also formally set Ω

−1

(M) = 0.

As before, the compactness of M is essential, and cannot be dropped.

Corollary. We have orthogonal decompositions

Ω

(M) = H

⊕ dδΩ

(M) ⊕ δdΩ

(M)

= H

⊕ dΩ

p−1

(M) ⊕ δΩ

p+1

(M).

Proof. Now note that for an α, β, we have

hhdδα, δdβii

= hhddδα, dβii

= 0.

dδΩ

(M) ⊕ δdΩ

(M)

is an orthogonal direct sum that clearly contains ∆Ω

(

). But each component

is also orthogonal to harmonic forms, because harmonic forms are closed and

co-closed. So the first decomposition follows.

To obtain the final decomposition, we simply note that

dΩ

p−1

(M) = d(H

p−1

⊕ ∆Ω

p−1

(M)) = d(δdΩ

p−1

(M)) ⊆ dδΩ

(M).

On the other hand, we certainly have the other inclusion. So the two terms are

equal. The other term follows similarly.

This theorem has a rather remarkable corollary.

Corollary.

Let (

M, g

) be a compact oriented Riemannian manifold. Then for

all

α ∈ H

(

), there is a unique

α ∈ H

such that [

] =

. In other words,

the obvious map

→ H

(M)

is an isomorphism.

This is remarkable. On the left hand side, we have

, which is a completely

analytic thing, defined by the Laplacian. On the other hand, the right hand

sides involves the de Rham cohomology, which is just a topological, and in fact

homotopy invariant.

Proof.

To see uniqueness, suppose

, α

∈ H

are such that [

] = [

]

∈

(M). Then

− α

= dβ

for some

. But the left hand side and right hand side live in different parts of

the Hodge decomposition. So they must be individually zero. Alternatively, we

can compute

kdβk

= hhdβ, α

− α

= hhβ, δα

− δα

= 0

since harmonic forms are co-closed.

To prove existence, let α ∈ Ω

(M) be such that dα = 0. We write

α = α

+ dα

+ δα

∈ H

⊕ dΩ

p−1

(M) ⊕ δΩ

p+1

(M).

Applying d gives us

0 = dα

+ d

+ dδα

We know d

= 0 since

is harmonic, and d

= 0. So we must have d

δα

= 0.

hhδα

, δα

= hhα

, dδα

= 0.

So δα

= 0. So [α] = [α

] and α has a representative in H

We can also heuristically justify why this is true. Suppose we are given some

de Rham cohomology class a ∈ H

(M). We consider

= {ξ ∈ Ω

(M) : dξ = 0, [ξ] = a}.

This is an infinite dimensional affine space.

We now ask ourselves — which

α ∈ B

minimizes the

norm? We consider

the function

→ R

given by

(

) =

kαk

. Any minimizing

is an

extremum. So for any β ∈ Ω

p−1

(M), we have



t=0

F (α + tdβ) = 0.

In other words, we have

0 =



t=0

(kαk

+ 2thhα, dβii

+ t

kdβk

) = 2hhα, dβii

This is the same as saying

hhδα, βii

= 0.

So this implies

δα

= 0. But d

= 0 by assumption. So we find that

α ∈ H

. So

the result is at least believable.

The proof of the Hodge decomposition theorem involves some analysis, which

we are not bothered to do. Instead, we will just quote the appropriate results.

For convenience, we will use

h·, ·i

for the

inner product, and then

k · k

the L

norm.

The first theorem we quote is the following:

Theorem

(Compactness theorem)

If a sequence

∈

Ω

(

) satisfies

kα

k <

C and k∆α

k < C for all n, then α

contains a Cauchy subsequence.

This is almost like saying Ω

(

) is compact, but it isn’t, since it is not

complete. So the best thing we can say is that the subsequence is Cauchy.

Corollary. H

is finite-dimensional.

Proof.

Suppose not. Then by Gram–Schmidt, we can find an infinite orthonormal

sequence

such that

= 1 and

∆

= 0, and this certainly does not have

a Cauchy subsequence.

A large part of the proof is trying to solve the PDE

∆ω = α,

which we will need in order to carry out the decomposition. In analysis, one

useful idea is the notion of weak solutions. We notice that if

is a solution,

then for any ϕ ∈ Ω

(M), we have

hω, ∆ϕi = h∆ω, ϕi = hα, ϕi,

using that ∆ is self-adjoint. In other words, the linear form



hω, ·i

: Ω

(

)

→

R satisfies

(∆ϕ) = hα, ϕi.

Conversely, if

hω, ·i

satisfies this equation, then

must be a solution, since for

any β, we have

h∆ω, βi = hω, ∆βi = hα, βi.

Definition

(Weak solution)

A weak solution to the equation ∆

is a

linear functional  : Ω

(M) → R such that

(i) (∆ϕ) = hα, ϕi for all ϕ ∈ Ω

(M).

(ii)  is bounded, i.e. there is some C such that |(β)| < Ckβk for all β.

Now given a weak solution, we want to obtain a genuine solution. If Ω

(

)

were a Hilbert space, then we are immediately done by the Riesz representation

theorem, but it isn’t. Thus, we need a theorem that gives us what we want.

Theorem

(Regularity theorem)

Every weak solution of ∆

is of the form

(β) = hω, βi

for ω ∈ Ω

(M).

Thus, we have reduced the problem to finding weak solutions. There is one

final piece of analysis we need to quote. The definition of a weak solution only

cares about what



does to ∆Ω

(

). And it is easy to define what



should do

on ∆Ω

(M) — we simply define

(∆η) = hη, αi.

Of course, for this to work, it must be well-defined, but this is not necessarily

the case in general. We also have to check it is bounded. But suppose this

worked. Then the remaining job is to extend this to a bounded functional on all

of Ω

(

) in whatever way we like. This relies on the following (relatively easy)

theorem from analysis:

Theorem

(Hahn–Banach theorem)

Let

be a normed vector space, and

a subspace. We let

→ R

be a bounded linear functional. Then

extends

to a bounded linear functional L → R with the same bound.

We can now begin the proof.

Proof of Hodge decomposition theorem.

Since

is finite-dimensional, by basic

linear algebra, we can decompose

Ω

(M) = H

⊕ (H

)

⊥

Crucially, we know (H

)

⊥

is a closed subspace. What we want to show is that

)

⊥

= ∆Ω

(M).

One inclusion is easy. Suppose α ∈ H

and β ∈ Ω

(M). Then we have

hα, ∆βi = h∆α, βi = 0.

So we know that

∆Ω

(M) ⊆ (H

)

⊥

The other direction is the hard part. Suppose

α ∈

(

)

⊥

. We may assume

non-zero. Since our PDE is a linear one, we may wlog kαk = 1.

By the regularity theorem, it suffices to prove that ∆

has a weak

solution. We define  : ∆Ω

(M) → R as follows: for each η ∈ Ω

(M), we put

(∆η) = hη, αi.

We check this is well-defined. Suppose ∆

= ∆

. Then

η −ξ ∈ H

, and we have

hη, αi − hξ, αi = hη − ξ, αi = 0

since α ∈ (H

)

⊥

We next want to show the boundedness property. We now claim that there

exists a positive C > 0 such that

(∆η) ≤ Ck∆ηk

for all

η ∈

Ω

(

). To see this, we first note that by Cauchy–Schwartz, we have

|hα, ηi| ≤ kαk · kηk = kηk.

So it suffices to show that there is a C > 0 such that

kηk ≤ Ck∆ηk

for every η ∈ Ω

(M).

Suppose not. Then we can find a sequence

∈

(

)

⊥

such that

kη

= 1

and k∆η

k → 0.

But then

∆

is certainly bounded. So by the compactness theorem, we

may wlog

is Cauchy. Then for any

ψ ∈

Ω

(

), the sequence

hψ, η

is Cauchy,

by Cauchy–Schwartz, hence convergent.

We define a : Ω

(M) → R by

a(ψ) = lim

k→∞

hψ, η

Then we have

a(∆ψ) = lim

k→∞

hη

, ∆ψi = lim

k→∞

h∆η

, ψi = 0.

So we know that

is a weak solution of ∆

= 0. By the regularity theorem

again, we have

a(ψ) = hξ, ψi

for some ξ ∈ Ω

(M). Then ξ ∈ H

We claim that

→ ξ

. Let

ε >

0, and pick

such that

n, m > N

implies

kη

− η

k < ε. Then

kη

− ξk

= hη

− ξ, η

− ξi ≤ |hη

− ξ, η

− ξi| + εkη

− ξk.

Taking the limit as

m → ∞

, the first term vansihes, and this tells us

kη

−ξk ≤ ε

So η

→ ξ.

But this is bad. Since

∈

(

)

⊥

, and (

)

∞

is closed, we know

ξ ∈

(

)

⊥

But also by assumption, we have

ξ ∈ H

. So

= 0. But we also know

kξk = lim kη

k = 1, whcih is a contradiction. So  is bounded.

We then extend



to any bounded linear map on Ω

(

). Then we are

done.

That was a correct proof, but we just pulled a bunch of theorems out of

nowhere, and non-analysts might not be sufficiently convinced. We now look

at an explicit example, namely the torus, and sketch a direct proof of Hodge

decomposition. In this case, what we needed for the proof reduces to the fact

Fourier series and Green’s functions work, which is IB Methods.

Consider

πZ

)

, the

-torus with flat metric. This has local

coordinates (

, ··· , x

), induced from the Euclidean space. This is convenient

because

∗

is trivialized by

{

∧ ···

}

. Moreover, the Laplacian is

just given by

∆(α dx

∧ ··· ∧ dx

) = −

i=1

∂

∂x

∧ ··· ∧ dx

So to do Hodge decomposition, it suffices to consider the case

= 0, and we are

just looking at functions C

∞

), namely the 2π-periodic functions on R.

Here we will quote the fact that Fourier series work.

Fact.

Let

ϕ ∈ C

∞

(

). Then it can be (uniquely) represented by a convergent

Fourier series

ϕ(x) =

k∈Z

ik·x

where

and

are vectors, and

k · x

is the standard inner product, and this is

uniformly convergent in all derivatives. In fact, ϕ

can be given by

(2π)

ϕ(x)e

−ik·x

dx.

Consider the inner product

hϕ, ψi = (2π)

¯ϕ

on 

, and define the subspace

∞



(ϕ

) ∈ 

: ϕ

= o(|k|

) for all m ∈ Z



Then the map

F : C

∞

) → 

ϕ 7→ (ϕ

is an isometric bijection onto H

∞

So we have reduced our problem of working with functions on a torus to

working with these infinite series. This makes our calculations rather more

explicit.

The key property is that the Laplacian is given by

F(∆ϕ) = (−|k|

In some sense, F “diagonalizes” the Laplacian. It is now clear that

= {ϕ ∈ C

∞

) : ϕ

= 0 for all k 6= 0}

)

⊥

= {ϕ ∈ C

∞

) : ϕ

= 0}.

Moreover, since we can divide by

|k|

whenever

is non-zero, it follows that

)

⊥

= ∆C

∞

4.3 Divergence

In ordinary multi-variable calculus, we had the notion of the divergence. This

makes sense in general as well. Given any X ∈ Vect(M), we have

∇X ∈ Γ(T M ⊗ T

∗

M) = Γ(End T M).

Now we can take the trace of this, because the trace doesn’t depend on the

choice of the basis.

Definition (Divergence). The divergence of a vector field X ∈ Vect(M ) is

divX = tr(∇X).

It is not hard to see that this extends the familiar definition of the divergence.

Indeed, by definition of trace, for any local frame field {e

}, we have

divX =

i=1

g(∇

X, e

It is straightforward to prove from definition that

Proposition.

div(fX) = tr(∇(fX)) = fdivX + hdf, Xi.

The key result about divergence is the following:

Theorem.

Let

θ ∈

Ω

(

), and let

∈ Vect

(

) be such that

hθ, V i

g(X

, V ) for all V ∈ T M. Then

δθ = −divX

So the divergence isn’t actually a new operator. However, we have some

rather more explicit formulas for the divergence, and this helps us understand

better.

To prove this, we need a series of lemmas.

Lemma. In local coordinates, for any p-form ψ, we have

dψ =

k=1

∧ ∇

ψ.

Proof.

We fix a point

x ∈ M

, and we may wlog we work in normal coordinates

at x. By linearity and change of coordinates, we wlog

ψ = f dx

∧ ··· ∧ dx

Now the left hand side is just

dψ =

k=p+1

∂f

∂x

∧ dx

∧ ··· ∧ dx

But this is also what the RHS is, because ∇

= ∂

at p.

To prove this, we first need a lemma, which is also useful on its own right.

Definition

(Interior product)

Let

X ∈ Vect

(

). We define the interior

product i(X) : Ω

(M) → Ω

p−1

(M) by

(i(X)ψ)(Y

, ··· , Y

p−1

) = ψ(X, Y

, ··· , Y

p−1

This is sometimes written as i(X)ψ = Xyψ.

Lemma. We have

(divX) ω

= d(i(X) ω

for all X ∈ Vect(M).

Proof. Now by unwrapping the definition of i(X), we see that

∇

(i(X)ψ) = i(∇

X)ψ + i(X)∇

ψ.

From example sheet 3, we know that ∇ω

= 0. So it follows that

∇

(i(X) ω

) = i(∇

X) ω

Therefore we obtain

d(i(X)ω

)

k=1

∧ ∇

(i(X)ω

)

k=1

∧ i(∇

X)ω

k=1

∧ i(∇

X)(

|g|dx

∧ ··· ∧ dx

)

= dx

(∇

X) ω

= (divX) ω

Note that this requires us to think carefully how wedge products work (

(

)(

α∧β

)

is not just α(X)β, or else α ∧ β would not be anti-symmetric).

Corollary (Divergence theorem). For any vector field X, we have

div(X) ω

d(i(X) ω

) = 0.

We can now prove the theorem.

Theorem.

Let

θ ∈

Ω

(

), and let

∈ Vect

(

) be such that

hθ, V i

g(X

, V ) for all V ∈ T M. Then

δθ = −divX

Proof.

By the formal adjoint property of

, we know that for any

f ∈ C

∞

(

we have

g(df, θ) ω

fδθ ω

So we want to show that

g(df, θ) ω

= −

fdivX

But by the product rule, we have

div(fX

) ω

g(df, θ) ω

fdivX

So the result follows by the divergence theorem.

We can now use this to produce some really explicit formulae for what

is,

which will be very useful next section.

Corollary. If θ is a 1-form, and {e

} is a local orthonormal frame field, then

δθ = −

k=1

i(e

)∇

θ = −

k=1

h∇

θ, e

Proof. We note that

hθ, e

i = h∇

θ, e

i + hθ, ∇

g(X

, e

) = g(∇

, e

) + g(X

, ∇

By definition of X

, this implies that

h∇

θ, e

i = g(∇

, e

So we obtain

δθ = −divX

= −

i=1

g(∇

, e

) = −

k=1

h∇

θ, e

We will assume a version for 2-forms (the general result is again on the third

example sheet):

Proposition. If β ∈ Ω

(M), then

(δβ)(Y ) = −

k=1

(∇

β)(e

, Y ).

In other words,

δβ = −

k=1

i(e

)(∇

β).

4.4 Introduction to Bochner’s method

How can we apply the Hodge decomposition theorem? The Hodge decomposition

theorem tells us the de Rham cohomology group is the kernel of the Laplace–

Beltrami operator ∆. So if we want to show, say,

(

) = 0, then we want to

show that ∆α 6= 0 for all non-zero α ∈ Ω

(M). The strategy is to show that

hhα, ∆αii 6= 0

for all

α 6

= 0. Then we have shown that

(

) = 0. In fact, we will show that

this inner product is positive. To do so, the general strategy is to introduce an

operator T with adjoint T

∗

, and then write

∆ = T

∗

T + C

for some operator C. We will choose T cleverly such that C is very simple.

Now if we can find a manifold such that C is always positive, then since

hhT

∗

T α, σii = hhT α, T αii ≥ 0,

it follows that ∆ is always positive, and so H

(M) = 0.

Our choice of

will be the covariant derivative

∇

itself. We can formulate

this more generally. Suppose we have the following data:

– A Riemannian manifold M.

– A vector bundle E → M.

– An inner product h on E.

– A connection ∇ = ∇

: Ω

(E) → Ω

(E) on E.

We are eventually going to take

∗

, but we can still proceed in the

general setting for a while.

The formal adjoint (∇

)

∗

: Ω

(E) → Ω

(E) is defined by the relation

h∇α, βi

E,g

hα, ∇

∗

βi

for all

α ∈

Ω

(

) and

β ∈

Ω

(

). Since

is non-degenerate, this defines

∇

∗

uniquely.

Definition (Covariant Laplacian). The covariant Laplacian is

∇

∗

∇ : Γ(E) → Γ(E)

We are now going to focus on the case

∗

. It is helpful to have the

following explicit formula for ∇

∗

, which we shall leave as an exercise:

As mentioned, the objective is to understand ∆

−∇

∗

∇

. The theorem is that

this difference is given by the Ricci curvature.

This can’t be quite right, because the Ricci curvature is a bilinear form on

T M

, but ∆

−∇

∗

∇

is a linear endomorphism Ω

(

)

→

Ω

(

). Thus, we need

to define an alternative version of the Ricci curvature by “raising indices”. In

coordinates, we consider g

Ric

instead.

We can also define this

Ric

without resorting to coordinates. Recall that

given an

α ∈

Ω

(

), we defined

∈ Vect

(

) to be the unique field such that

α(z) = g(X

, Z)

for all Z ∈ Vect(M ). Then given α ∈ Ω

(M), we define Ric(α) ∈ Ω

(M) by

Ric(α)(X) = Ric(X, X

With this notation, the theorem is

Theorem (Bochner–Weitzenb¨ock formula). On an oriented Riemannian mani-

fold, we have

∆ = ∇

∗

∇ + Ric .

Before we move on to the proof of this formula, we first give an application.

Corollary. Let (M, g) be a compact connected oriented manifold. Then

– If Ric(g) > 0 at each point, then H

(M) = 0.

– If Ric(g) ≥ 0 at each point, then b

(M) = dim H

(M) ≤ n.

– If Ric(g) ≥ 0 at each point, and b

(M) = n, then g is flat.

Proof. By Bochner–Weitzenb¨ock, we have

hh∆α, αii = hh∇

∗

∇α, αii+

Ric(α, α) ω

= k∇αk

Ric(α, α) ω

–

Suppose

Ric >

0. If

α 6

= 0, then the RHS is strictly positive. So the

left-hand side is non-zero. So ∆α 6= 0. So H

∼

(M) = 0.

–

Suppose

is such that ∆

= 0. Then the above formula forces

∇α

= 0.

So if we know

(

) for some fixed

x ∈ M

, then we know the value of

everywhere by parallel transport. Thus

is determined by the initial

condition

(

), Thus there are

≤ n

dim T

∗

linearly independent such

α.

–

(

) =

, then we can pick a basis

, ··· , α

. Then as above,

these are parallel 1-forms. Then we can pick a dual basis

, ··· , X

∈

Vect

(

). We claim they are also parallel, i.e.

∇X

= 0. To prove this, we

note that

hα

, ∇X

i + h∇α

, X

i = ∇hα

, X

But

hα

, X

is constantly 0 or 1 depending on

and

, So the RHS vanishes.

Similarly, the second term on the left vanishes. Since the

span, we know

we must have ∇X

= 0.

Now we have

R(X

, X

= (∇

]

− [∇X

, ∇

])X

= 0,

Since this is valid for all

i, j, k

, we know

vanishes at each point. So we

are done.

Bochner–Weitzenb¨ock can be exploited in a number of similar situations.

In the third part of the theorem, we haven’t actually proved the optimal

statement. We can deduce more than the flatness of the metric, but requires

some slightly advanced topology. We will provide a sketch proof of the theorem,

making certain assertions about topology.

Proposition. In the case of (iii), M is in fact isometric to a flat torus.

Proof sketch. We fix p ∈ M and consider the map M → R

given by

x 7→





i=1,···,n

∈ R

where the

are as in the previous proof. The integral is taken along any path

from

, and this is not well-defined. But by Stokes’ theorem, and the fact

that dα

= 0, this only depends on the homotopy class of the path.

In fact,

depends only on

γ ∈ H

(

), which is finitely generated. Thus,

is a well-defined map to

R/λ

for some

= 0. Therefore we obtain

a map

M →

(

)

. Moreover, a bit of inspection shows this is a local

diffeomorphism. But since the spaces involved are compact, it follows by some

topology arguments that it must be a covering map. But again by compactness,

this is a finite covering map. So M must be a torus. So we are done.

We only proved this for 1-forms, but this is in fact fact valid for forms of any

degree. To do so, we consider E =

∗

M, and then we have a map

∇ : Ω

(E) → Ω

(E),

and this has a formal adjoint

∇

∗

: Ω

(E) → Ω

(E).

Now if α ∈ Ω

(M), then it can be shown that

∆α = ∇

∗

∇α + R(α),

where

is a linear map Ω

(

)

→

Ω

(

) depending on the curvature. Then

by the same proof, it follows that if

R >

0 at all points, then

(

) = 0 for all

k = 1, ··· , n − 1.

R ≥

0 only, which in particular is the case if the space is flat, then we have

(M) ≤





= dim

∗

and moreover ∆α = 0 iff ∇α = 0.

Proof of Bochner–Weitzenb¨ock

We now move on to actually prove Bochner–Weitzenb¨ock. We first produce an

explicit formula for ∇

∗

, and hence ∇

∗

∇.

Proposition.

Let

, ··· , e

be an orthonormal frame field, and

β ∈

Ω

(

∗

Then we have

∇

∗

β = −

i=1

i(e

)∇

β.

Proof. Let α ∈ Ω

∗

M). Then by definition, we have

h∇α, βi =

i=1

h∇

α, β(e

)i.

Consider the 1-form given by

θ(Y ) = hα, β(Y )i.

Then we have

divX

i=1

h∇

, e

i=1

∇

, e

i − hX

, ∇

i=1

∇

hα, β(e

)i − hα, β(∇

i=1

h∇

α, β(e

)i + hα, ∇

(β(e

))i − hα, β(∇

i=1

h∇

α, β(e

)i + hα, (∇

β)(e

)i.

So by the divergence theorem, we have

h∇α, βi ω

i=1

hα, (∇

β)(e

)i ω

So the result follows.

Corollary. For a local orthonormal frame field e

, ··· , e

, we have

∇

∗

∇α = −

i=1

∇

α.

We next want to figure out more explicit expressions for d

and

d. To make

our lives much easier, we will pick a normal frame field:

Definition

(Normal frame field)

A local orthonormal frame

}

field is normal

at p if further

∇e

= 0

for all k.

It is a fact that normal frame fields exist. From now on, we will fix a point

p ∈ M

, and assume that

}

is a normal orthonormal frame field at

. Thus, the

formulae we derive are only valid at

, but this is fine, because

was arbitrary.

The first term dδ is relatively straightforward.

Lemma. Let α ∈ Ω

(M), X ∈ Vect(M). Then

hdδα, Xi = −

i=1

h∇

∇

α, e

Proof.

hdδα, Xi = X(δα)

= −

i=1

Xh∇

α, e

= −

i=1

h∇

∇

α, e

This takes care of one half of ∆ for the other half, we need a bit more work.

Recall that we previously found a formula for

. We now re-express the formula

in terms of this local orthonormal frame field.

Lemma. For any 2-form β, we have

(δβ)(X) =

k=1

−e

(β(e

, X)) + β(e

, ∇

X).

Proof.

(δβ)(X) = −

k=1

(∇

β)(e

, X)

k=1

−e

(β(e

, X)) + β(∇

, X) + β(e

, ∇

k=1

−e

(β(e

, X)) + β(e

, ∇

X).

Since we want to understand

for

a 1-form, we want to find a decent

formula for dα.

Lemma. For any 1-form α and vector fields X, Y , we have

dα(X, Y ) = h∇

α, Y i − h∇

α, Xi.

Proof. Since the connection is torsion-free, we have

[X, Y ] = ∇

Y − ∇

So we obtain

dα(X, Y ) = Xhα, Y i − Y hα, Xi − hα, [X, Y ]i

= h∇

α, Y i − h∇

α, Xi.

Finally, we can put these together to get

Lemma. For any 1-form α and vector field X, we have

hδdα, Xi = −

k=1

h∇

∇

α, Xi +

k=1

h∇

∇

α, e

i −

k=1

h∇

∇

α, e

Proof.

hδdα, Xi =

k=1

− e

(dα(e

, X)) + dα(e

, ∇

k=1

− e

(h∇

α, Xi − h∇

α, e

+ h∇

α, ∇

Xi − h∇

∇

α, e

k=1

− h∇

∇

α, Xi − h∇

α, ∇

Xi + h∇

∇

α, e

+ h∇

α, ∇

Xi − h∇

∇

α, e

= −

k=1

h∇

∇

α, Xi +

k=1

h∇

∇

α, e

i −

k=1

h∇

∇

α, e

What does this get us? The first term on the right is exactly the

∇

∗

∇

term

we wanted. If we add dδα to this, then we get

k=1

h([∇

, ∇

] − ∇

∇

)α, e

We notice that

, X] = ∇

X − ∇

= ∇

So we can alternatively write the above as

k=1

h([∇

, ∇

] − ∇

,X]

)α, e

The differential operator on the left looks just like the Ricci curvature. Recall

that

R(X, Y ) = ∇

[X,Y ]

− [∇

, ∇

Lemma

(Ricci identity)

Let

be any Riemannian manifold, and

X, Y, Z ∈

Vect(M) and α ∈ Ω

(M). Then

h([∇

, ∇

] − ∇

[X,Y ]

)α, Zi = hα, R(X, Y )Zi.

Proof. We note that

h∇

[X,Y ]

α, Zi+hα, ∇

[X,Y ]

Zi = [X, Y ]hα, Zi = h[∇

, ∇

]α, Zi+hα, [∇

, ∇

]Zi.

The second equality follows from writing [

X, Y

] =

XY −Y X

. We then rearrange

and use that R(X, Y ) = ∇

[X,Y ]

− [∇

, ∇

Corollary. For any 1-form α and vector field X, we have

h∆α, Xi = h∇

∗

∇α, Xi + Ric(α)(X).

This is the theorem we wanted.

Proof. We have found that

h∆α, Xi = h∇

∗

∇α, Xi +

i=1

hα, R(e

, X)e

We have

i=1

hα, R(e

, X)e

i =

i=1

g(X

, R(e

, X)e

) = Ric(X

, X) = Ric(α)(X).

So we are done.

5 Riemannian holonomy groups

Again let

be a Riemannian manifold, which is always assumed to be connected.

Let

x ∈ M

, and consider a path

γ ∈

Ω(

x, y

: [0

→ M

. At the rather

beginning of the course, we saw that

gives us a parallel transport from

. Explicitly, given any

∈ T

, there exists a unique vector field

along γ with

∇X

= 0, X(0) = X

Definition

(Holonomy transformation)

The holonomy transformation

(

)

sends X

∈ T

M to X(1) ∈ T

We know that this map is invertible, and preserves the inner product. In

particular, if x = y, then P (γ) ∈ O(T

∼

O(n).

Definition (Holonomy group). The holonomy group of M at x ∈ M is

Hol

(M) = {P (γ) : γ ∈ Ω(x, x)} ⊆ O(T

M).

The group operation is given by composition of linear maps, which corresponds

to composition of paths.

We note that this group doesn’t really depend on the point

. Given any

other

y ∈ M

, we can pick a path

β ∈

Ω(

x, y

). Writing

instead of

(

), we

have a map

Hol

(M) Hol

(M)

◦ P

−1

∈ Hol

(M)

So we see that Hol

(M) and Hol

(M) are isomorphic. In fact, after picking an

isomorphism O(

)

∼

)

∼

), these subgroups are conjugate as

subgroups of O(n). We denote this class by Hol(M ).

Note that depending of what we want to emphasize, we write

Hol

(

M, g

), or

even Hol(g) instead.

Really,

Hol

(

) is a representation (up to conjugacy) induced by the standard

representation of O(n) on R

Proposition. If M is simply connected, then Hol

(M) is path connected.

Proof. Hol

(

) is the image of Ω(

x, x

) in O(

) under the map

, and this map

is continuous from the standard theory of ODE’s. Simply connected means

Ω(x, x) is path connected. So Hol

(M) is path connected.

It is convenient to consider the restricted holonomy group.

Definition (Restricted holonomy group). We define

Hol

(M) = {P (γ) : γ ∈ Ω(x, x) nullhomotopic}.

As before, this group is, up to conjugacy, independent of the choice of the

point in the manifold. We write this group as Hol

(M).

Of course, Hol

(M) ⊆ Hol(M ), and if π

(M) = 0, then they are equal.

Corollary. Hol

(M) ⊆ SO(n) .

Proof. Hol

(

) is connected, and thus lies in the connected component of the

identity in O(n).

Note that there is a natural action of

Hol

(

) and

Hol

(

) on

∗

M for all p, and more generally tensor products of T

Fact.

– Hol

(

) is the connected component of

Hol

(

) containing the identity

element.

– Hol

(

) is a Lie subgroup of

(

), i.e. it is a subgroup and an immersed

submanifold. Thus, the Lie algebra of

Hol

(

) is a Lie subalgebra of

so(n), which is just the skew-symmetric n × n matrices.

This is a consequence of Yamabe theorem, which says that a path-connected

subgroup of a Lie group is a Lie subgroup.

We will not prove these.

Proposition

(Fundamental principle of Riemannian holonomy)

Let (

M, g

) be

a Riemannian manifold, and fix

p, q ∈ Z

and

x ∈ M

. Then the following are

equivalent:

(i) There exists a (p, q)-tensor field α on M such that ∇α = 0.

(ii)

There exists an element

∈

(

)

⊗p

⊗

(

∗

)

⊗q

such that

is invariant

under the action of Hol

(M).

Proof.

To simplify notation, we consider only the case

= 0. The general case

works exactly the same way, with worse notation. For α ∈ (T

∗

, we have

(∇

α)(X

, ··· , X

) = X(α(X

, ··· , X

)) −

i=1

α(X

, ··· , ∇

, ··· , X

Now consider a loop

: [0

→ M

be a loop at

. We choose vector fields

along γ for i = 1, ··· , q such that

∇X

= 0.

We write

(γ(0)) = X

Now if ∇α = 0, then this tells us

∇α

, ··· , X

) = 0.

By our choice of

, we know that

(

, ··· , X

) is constant along

. So we

know

α(X

, ··· , X

) = α(P

), ··· , P

)).

So α is invariant under Hol

(M). Then we can take α

= α

Conversely, if we have such an

, then we can use parallel transport to

transfer it to everywhere in the manifold. Given any y ∈ M, we define α

, ··· , X

) = α

), ··· , P

)),

where

is any path from

. This does not depend on the choice of

precisely because α

is invariant under Hol

(M).

It remains to check that

∞

with

∇α

= 0, which is an easy exercise.

Example.

Let

be oriented. Then we have a volume form

. Since

∇ω

= 0,

we can take

. Here

= 0 and

. Also, its stabilizer is

(

So we know Hol(M ) ⊆ SO(n) if (and only if) M is oriented.

The “only if” part is not difficult, because we can use parallel transport to

transfer an orientation at a particular point to everywhere.

Example. Let x ∈ M , and suppose

Hol

(M) ⊆ U(n) = {g ∈ SO(2n) : gJ

−1

= J

where



0 I

−I 0



By looking at

, we obtain

J ∈

Γ(

End T M

) with

∇J

= 0 and

−

1. This is a well-known standard object in complex geometry, and such

a J is an instance of an almost complex structure on M.

Example. Recall (from the theorem on applications of Bochner–Weitzenb¨ock)

that a Riemannian manifold (

M, g

) is flat (i.e.

(

)

≡

1) iff around each point

x ∈ M

, there is a parallel basis of parallel vector fields. So we find that (

M, g

)

is flat iff Hol

(M, g) = {id}.

It is essential that we use

Hol

(

M, g

) rather than the full

Hol

(

M, g

). For

example, we can take the Klein bottle

with the flat metric. Then parallel transport along the closed loop γ has



1 0

0 −1



In fact, we can check that Hol(K) = Z

. Note that here K is non-orientable.

Since we know that

Hol

(

) is a Lie group, we can talk about its Lie algebra.

Definition

(Holonomy algebra)

The holonomy algebra

hol

(

) is the Lie algebra

of Hol(M ).

Thus hol(M) ≤ so(n) up to conjugation.

Now consider some open coordinate neighbourhood

U ⊆ M

with coordinates

, ··· , x

. As before, we write

∂

∂x

, ∇

= ∇

∂

The curvature may also be written in terms of coordinates

j,k`

, and we

also have

R(∂

, ∂

) = −[∇

, ∇

Thus, hol(M) contains



t=0

P (γ

where γ

is the square

√

By a direct computation, we find

P (γ

) = I + λtR(∂

, ∂

) + o(t).

Here

λ ∈ R

is some non-zero absolute constant that doesn’t depend on anything

(except convention).

Differentiating this with respect to

and taking the limit

t →

0, we deduce

that at for p ∈ U, we have

= (R

j,k`

)

∈

∗

M ⊗ hol

(M),

where we think

hol

(

)

⊆ End T

. Recall we also had the

ij,k`

version, and

because of the nice symmetric properties of R, we know

ij,k`

)

∈ S

hol

(M) ⊆

∗

M ⊗

∗

Strictly speaking, we should write

)

∈ S

hol

(M),

but we can use the metric to go between T

M and T

∗

So far, what we have been doing is rather tautological. But it turns out this

allows us to decompose the de Rham cohomology groups.

In general, consider an arbitrary Lie subgroup

G ⊆ GL

(

). There is a

standard representation

(

) on

, which restricts to a representation

(ρ, R

) of G. This induces a representation (ρ

∗

)) of G.

This representation is in general not irreducible. We decompose it into

irreducible components (ρ

, W

), so that

∗

) =

We can do this for bundles instead of just vector spaces. Consider a manifold

with a

-structure, i.e. there is an atlas of coordinate neighbourhoods where the

transition maps satisfy

∂x

∈ G

for all

. Then we can use this to split our bundle of

-forms into well-defined

vector sub-bundles with typical fibers W

∗

M =

We can furthermore say that every

-equivariant linear map

→ W

induces a morphism of vector bundles φ : Λ

→ Λ

Now suppose further that

Hol

(

)

≤ G ≤

). Then we can show that

parallel transport preserves this decomposition into sub-bundles. So

∇

restricts

to a well-defined connection on each Λ

Thus, if

ξ ∈

Γ(Λ

), then

∇ξ ∈

Γ(

∗

M ⊗

), and then we have

∇

∗

∇ξ ∈

Γ(Λ

But we know the covariant Laplacian is related to Laplace–Beltrami via the

curvature. We only do this for 1-forms for convenience of notation. Then if

ξ ∈ Ω

(M), then we have

∆ξ = ∇

∗

∇ξ + Ric(ξ).

We can check that

Ric

also preserves these subbundles. Then it follows that

∆ : Γ(Λ

) → Γ(Λ

) is well-defined.

Thus, we deduce

Theorem.

Let (

M, g

) be a connected and oriented Riemannian manifold, and

consider the decomposition of the bundle of

-forms into irreducible representa-

tions of the holonomy group,

∗

M =

In other words, each fiber (Λ

)

⊆

∗

is an irreducible representation of

Hol

(g). Then

(i) For all α ∈ Ω

(M) ≡ Γ(Λ

), we have ∆α ∈ Ω

(M).

(ii) If M is compact, then we have a decomposition

(M) =

i,dR

(M),

where

i,dR

(M) = {[α] : α ∈ Ω

(M), ∆α = 0}.

The dimensions of these groups are known as the refined Betti numbers.

We have only proved this for

= 1, but the same proof technique can be

used to do it for arbitrary k.

Our treatment is rather abstract so far. But for example, if we are dealing

with complex manifolds, then we know that

Hol

(

)

≤

). So this allows us

to have a canonical refinement of the de Rham cohomology, and this is known

as the Lefschetz decomposition.

6 The Cheeger–Gromoll splitting theorem

We will talk about the Cheeger–Gromoll splitting theorem. This is a hard

theorem, so we will not prove it. However, we will state it, and discuss a bit

about it. To state the theorem, we need some preparation.

Definition

(Ray)

Let (

M, g

) be a Riemannian manifold. A ray is a map

(

) : [0

, ∞

)

→ M

is a geodesic, and minimizes the distance between any

two points on the curve.

Definition

(Line)

A line is a map



(

) :

R → M

such that



(

) is a geodesic,

and minimizes the distance between any two points on the curve.

We have seen from the first example sheet that if

is a complete unbounded

manifold, then

has a ray from each point, i.e. for all

x ∈ M

, there exists a

ray r such that r(0) = x.

Definition

(Connected at infinity)

A complete manifold is said to be connected

at infinity if for all compact set

K ⊆ M

, there exists a compact

C ⊇ K

such

that for every two points

p, q ∈ M \C

, there exists a path

γ ∈

Ω(

p, q

) such that

γ(t) ∈ M \K for all t.

We say M is disconnected at infinity if it is not connected at infinity.

Note that if

is disconnected at infinity, then it must be unbounded, and

in particular non-compact.

Lemma. If M is disconnected at infinity, then M contains a line.

Proof.

Note that

is unbounded. Since

is disconnected at infinity, we can

find a compact subset

K ⊆ M

and sequences

, q

→ ∞

m → ∞

(to make

this precise, we can pick some fixed point

, and then require

(

x, p

)

, d

(

x, q

)

→

∞) such that every γ

∈ Ω(p

, q

) passes through K.

In particular, we pick

to be a minimal geodesic from

parametrized

by arc-length. Then

passes through

. By reparametrization, we may assume

(0) ∈ K.

Since

is compact, we can pass to a subsequence, and wlog

(0)

→ x ∈ K

and ˙γ

(0) → a ∈ T

M (suitably interpreted).

Then we claim the geodesic

x,a

(

) is the desired line. To see this, since

solutions to ODE’s depend smoothly on initial conditions, we can write the line

(t) = lim

m→∞

(t).

Then we know

d((s), (t)) = lim

m→∞

d(γ

(s), γ

(t)) = |s − t|.

So we are done.

Let’s look at some examples.

Example. The elliptic paraboloid

{z = x

+ y

} ⊆ R

with the induced metric does not contain a line. To prove this, we can show that

any geodesic that is not a meridian must intersect itself.

Example.

Any complete metric

n−1

× R

contains a line since it is

disconnected at ∞.

Theorem

(Cheeger–Gromoll line-splitting theorem (1971))

If (

M, g

) is a com-

plete connected Riemannian manifold containing a line, and has

Ric

(

)

≥

0 at

each point, then

is isometric to a Riemannian product (

N × R, g

+ d

) for

some metric g

on N .

We will not prove this, but just see what the theorem can be good for.

First of all, we can slightly improve the statement. After applying the

theorem, we can check again if

contains a line or not. We can keep on

proceeding and splitting lines off. Then we get

Corollary.

Let (

M, g

) be a complete connected Riemannian manifold with

Ric

(

)

≥

0. Then it is isometric to

X × R

for some

q ∈ N

and Riemannian

manifold X, where X is complete and does not contain any lines.

Note that if

is zero-dimensional, since we assume all our manifolds are

connected, then this implies

is flat. If

dim X

= 1, then

∼

(it can’t be a

line, because a line contains a line). So again M is flat.

Now suppose that in fact

Ric

(

) = 0. Then it is not difficult to see from the

definition of the Ricci curvature that

Ric

(

) = 0 as well. If we know

dim X ≤

then M has to be flat, since in dimensions ≤ 3, the Ricci curvature determines

the full curvature tensor.

We can say a bit more if we assume more about the manifold. Recall (from

example sheets) that a manifold is homogeneous if the group of isometries

acts transitively. In other words, for any

p, q ∈ M

, there exists an isometry

M → M

such that

(

) =

. This in particular implies the metric is complete.

It is not difficult to see that if

is homogeneous, then so is

. In this case,

must be compact. Suppose not. Then

is unbounded. We will obtain a line

on X.

By assumption, for all

= 1

, ···

, we can find

, q

with

(

, q

)

≥

Since

is complete, we can find a minimal geodesic

connecting these two

points, parametrized by arc length. By homogeneity, we may assume that the

midpoint

(0) is at a fixed point

. By passing to a subsequence, we wlog

˙γ

(0) converges to some

a ∈ T

(

. Then we use

as an initial condition for

our geodesic, and this will be a line.

A similar argument gives

Lemma.

Let (

M, g

) be a compact Riemannian manifold, and suppose its uni-

versal Riemannian cover (

M, ˜g) is non-compact. Then (

M, ˜g) contains a line.

Proof.

We first find a compact

K ⊆

such that

(

) =

. Since

must be

complete, it is unbounded. Choose

, q

, γ

like before. Then we can apply deck

transformations so that the midpoint lies inside

, and then use compactness of

K to find a subsequence so that the midpoint converges.

We do more applications.

Corollary.

Let (

M, g

) be a compact, connected manifold with

Ric

(

)

≥

0. Then

–

The universal Riemannian cover is isometric to the Riemannian product

X × R

, with X compact, π

(X) = 1 and Ric(g

) ≥ 0.

– If there is some p ∈ M such that Ric(g)

> 0, then π

(M) is finite.

–

Denote by

(

) the group of isometries

M →

. Then

(

) =

(

)

E(R

), where E(R

) is the group of rigid Euclidean motions,

y 7→ Ay + b

where b ∈ R

and A ∈ O(q).

– If

M is homogeneous, then so is X.

Proof.

– This is direct from Cheeger–Gromoll and the previous lemma.

–

If there is a point with strictly positive Ricci curvature, then the same is

true for the universal cover. So we cannot have any non-trivial splitting.

So by the previous part,

must be compact. By standard topology,

|π

(M)| = |π

−1

({p})|.

–

We use the fact that

(

) =

(

). Pick a

g ∈ I

(

). Then we know

takes lines to lines. Now use that all lines in

M ×R

are of the form

p ×R

with p ∈ X and R ⊆ R

an affine line. Then

g(p ×R) = p

× R,

for some

and possibly for some other copy of

. By taking unions, we

deduce that g(p × R

) = p

× R

. We write h(p) = p

. Then h ∈ I(X).

Now for any

X × a

with

a ∈ R

, we have

X × a ⊥ p × R

for all

p ∈ X

So we must have

g(X × a) = X × b

for some b ∈ R

. We write e(a) = b. Then

g(p, a) = (h(p), e(a)).

Since the metric of

and

are decoupled, it follows that

and

must

separately be isometries.

We can look at more examples.

Proposition.

Consider

× R

for

= 2 or 3. Then this does not admit any

Ricci-flat metric.

Proof.

Note that

× R

is disconnected at

∞

. So any metric contains a line.

Then by Cheeger–Gromoll,

splits as a Riemannian factor. So we obtain

Ric

= 0 on the

factor. Since we are in

= 2

3, we know

is flat, as the

Ricci curvature determines the full curvature. So

is the quotient of

by a

discrete group, and in particular π

) 6= 1. This is a contradiction.

Let

be a Lie group with a bi-invariant metric

. Suppose the center

(

)

is finite. Then the center of

is trivial (since it is the Lie algebra of

G/Z

(

which has trivial center). From sheet 2, we find that

Ric

(

)

0 implies

(

) is

finite. The converse is also true, but is harder. This is done on Q11 of sheet 3 —

if π

(G) is finite, then Z(G) is finite.