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.