2Riemann curvature

III Riemannian Geometry

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(

V

2

T

∗

M ⊗ T

∗

M ⊗ T M) ⊆ Γ(T

1,3

M).

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

T

∗

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 ]

− [∇

X

, ∇

Y

].

In local coordinates, we can write

R =

R

i

j,k`

dx

k

dx

`

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

∈ Ω

2

M

(End(T M)).

Then we have

R(X, Y )

i

j

= R

i

j,k`

X

k

Y

`

.

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

p

M. In local coordinates, we write this as

R

ij,k`

= g

iq

R

q

j,k`

.

The first thing we want to prove is that

R

ij,k`

enjoys some symmetries we might

not expect:

Proposition.

(i)

R

ij,k`

= −R

ij,`k

= −R

ji,k`

.

(ii) The first Bianchi identity:

R

i

j,k`

+ R

i

k,`j

+ R

i

`,jk

= 0.

(iii)

R

ij,k`

= R

k`,ij

.

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

,

4) tensor as

R

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

ij

∂x

k

= g(∇

k

∂

i

, ∂

j

) + g(∂

i

, ∇

k

∂

j

).

We take a partial derivative, say with respect to x

`

, to obtain

∂

2

g

ij

∂x

`

∂x

k

= g(∇

`

∇

k

∂

i

, ∂

j

)+g(∇

k

∂

i

, ∇

`

∂

j

)+g(∇

`

∂

i

, ∇

k

∂

j

)+g(∂

i

, ∇

`

∇

k

∂

j

).

Then we know

0 =

∂

2

g

∂x

`

∂x

k

−

∂

2

g

∂x

k

∂x

`

= g([∇

`

, ∇

k

]∂

i

, ∂

j

) + g(∂

i

, [∇

`

, ∇

k

]∂

j

).

But we know

R(∂

k

, ∂

`

) = ∇

[∂

k

,∂

`

]

− [∇

k

, ∇

`

] = −[∇

k

, ∇

`

].

Writing R

k`

= R(∂

k

, ∂

`

), we have

0 = g(R

k`

∂

i

, ∂

j

) + g(∂

i

, R

k`

∂

j

) = R

ji,k`

+ R

ij,k`

.

So we are done.

(ii) Recall

R

i

j,k`

= (R

k`

∂

j

)

i

= ([∇

`

, ∇

k

]∂

j

)

i

.

So we have

R

i

j,k`

+ R

i

k,`j

+ R

i

`,jk

= [(∇

`

∇

k

∂

j

− ∇

k

∇

`

∂

j

) + (∇

j

∇

`

∂

k

− ∇

`

∇

j

∂

k

) + (∇

k

∇

j

∂

`

− ∇

j

∇

k

∂

`

)]

i

.

We claim that

∇

`

∇

k

∂

j

− ∇

`

∇

j

∂

k

= 0.

Indeed, by definition, we have

(∇

k

∂

j

)

q

= Γ

q

kj

= Γ

q

jk

= (∇

j

∂

k

)

q

.

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

(iii) Consider the following octahedron:

R

ik,`j

= R

ki,j`

R

i`,jk

= R

`i,kj

R

j`,ki

= R

`j,ik

R

jk,i`

= R

kj,`i

R

ij,k`

= R

ji,`k

R

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

(

−

2)

R

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

R

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

R

ij,k`

is a symmetric

bilinear form on

V

2

T

∗

p

M.

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 | =

p

g(X, X)g(Y, Y ) − g(X, Y )

2

,

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 |

2

.

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

X

or

Y

, and is symmetric

in

X

and

Y

. 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

K

(

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,

K

:

Gr

(2

, T

p

M

)

→ R

. This is called the sectional

curvature (of g).

It turns out the sectional curvature determines the Riemann curvature tensor

completely!

Lemma.

Let

V

be a real vector space of dimension

≥

2. Suppose

R

0

, R

00

:

V

⊗4

→ R

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

for the Riemann curvature tensor. We define

K

0

, K

00

:

Gr

(2

, V

)

→ R

as in the

sectional curvature. If K

0

= K

00

, then R

0

= R

00

.

This is really just linear algebra.

Proof. For any X, Y, Z ∈ V , we know

R

0

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

00

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

Using linearity of

R

0

and

R

00

, and cancelling equal terms on both sides, we find

R

0

(Z, Y, X, Y ) + R

0

(X, Y, Z, Y ) = R

00

(Z, Y, X, Y ) + R

00

(X, Y, Z, Y ).

Now using the symmetry property of R

0

and R

00

, this implies

R

0

(X, Y, Z, Y ) = R

00

(X, Y, Z, Y ).

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

R

0

(X, Y, Z, T ) + R

0

(X, T, Z, Y ) = R

00

(X, Y, Z, Y ) + R

00

”(X, T, Z, Y ).

We then rearrange and use the symmetries to get

R

0

(X, Y, Z, T ) − R

00

(X, Y, Z, T ) = R

0

(Y, Z, X, T ) − R

00

(Y, Z, X, T ).

We notice this equation says

R

0

(

X, Y, Z, T

)

− R

00

(

X, Y, Z, T

) is invariant under

the cyclic permutation

X → Y → Z → X

. So by the first Bianchi identity, we

have

3(R

0

(X, Y, Z, T ) − R

00

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

So we must have R

0

= R

00

.

Corollary.

Let (

M, g

) be a manifold such that for all

p

, the function

K

p

:

Gr(2, T

p

M) → R is a constant map. Let

R

0

p

(X, Y, Z, T ) = g

p

(X, Z)g

p

(Y, T ) − g

p

(X, T )g

p

(Y, Z).

Then

R

p

= K

p

R

0

p

.

Here

K

p

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

K

p

is a smooth

function of p.

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

δ

ij

, then we have

R

ij,ij

= −R

ij,ji

= K

p

,

and all other entries all zero.

Of course, the converse also holds.

Proof.

We apply the previous lemma as follows: we define

R

0

=

K

p

R

0

p

and

R

00

=

R

p

. It is a straightforward inspection to see that this

R

0

does follow the

symmetry properties of

R

p

, and that they define the same sectional curvature.

So R

00

= R

0

. We know K

p

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

We can further show that if

dim M >

2, then

K

p

is in fact independent of

p

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

p

(X, Y ) = tr(v 7→ R

p

(X, v)Y ).

In terms of coordinates, we have

Ric

ij

= R

q

i,jq

= g

pq

R

pi,jq

,

where g

pq

denotes the inverse of g.

This

Ric

is a symmetric bilinear form on

T

p

M

. This can be determined by

the quadratic form

Ric(X) =

1

n − 1

Ric

p

(X, X).

The coefficient

1

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

g

is the trace of

Ric

respect to g. Explicitly, this is defined by

s = g

ij

Ric

ij

= g

ij

R

q

i,jq

= R

qi

iq

.

Sometimes a convention is to define the scalar curvature as

s

n(n−1)

instead.

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

Ric

p

= (n − 1)K

p

g

p

,

and

s(p) = n(n − 1)K

p

.

Low dimensions

If

n

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

g

is also known as

the first fundamental form, and it is usually written as

g = E du

2

+ 2F du dv + G dv

2

.

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

R

12,12

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

R

12,12

=

1

2

s(EG − F

2

).

Thus

s/

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

s

2

= K =

LN − M

2

EG − F

2

,

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

Gaussian curvature. Also,

R

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.