8Connections
III Differential Geometry
8.4 Curvature
The final topic to discuss is the curvature of a connection. We all know that
R
n
is flat, while S
n
is curved. How do we make this precise?
We can consider doing some parallel transports on
S
n
along the loop coun-
terclockwise:
We see that after the parallel transport around the loop, we get a different vector.
It is in this sense that the connection on S
2
is not flat.
Thus, what we want is the following notion:
Definition
(Parallel vector field)
.
We say a vector field
V ∈ Vect
(
M
) is parallel
if V is parallel along any curve in M.
Example.
In
R
2
, we can pick
ξ ∈ T
0
R
2
∼
=
R
2
. Then setting
V
(
p
) =
ξ ∈ T
p
R
2
∼
=
R
2
gives a parallel vector field with V (0) = ξ.
However, we cannot find a non-trivial parallel vector field on S
2
.
This motivates the question — given a manifold
M
and a
ξ ∈ T
p
M
non-zero,
does there exist a parallel vector field
V
on some neighbourhood of
p
with
V (p) = ξ?
Naively, we would try to construct it as follows. Say
dim M
= 2 with
coordinates
x, y
. Put
p
= (0
,
0). Then we can transport
ξ
along the line
{y
= 0
}
to define
V
(
x,
0). Then for each
α
, we parallel transport
V
(
α,
0) along
{x
=
α}
.
So this determines V (x, y).
Now if we want this to work, then
V
has to be parallel along any curve, and
in particular for lines
{y
=
β}
for
β 6
= 0. If we stare at it long enough, we figure
out a necessary condition is
∇
∂
∂x
i
∇
∂
∂x
j
= ∇
∂
∂x
j
∇
∂
∂x
i
.
So the failure of these to commute tells us the curvature. This definition in fact
works for any vector bundle.
The actual definition we will state will be slightly funny, but we will soon
show afterwards that this is what we think it is.
Definition
(Curvature)
.
The curvature of a connection d
E
: Ω
0
(
E
)
→
Ω
1
(
E
) is
the map
F
E
= d
E
◦ d
E
: Ω
0
(E) → Ω
2
(E).
Lemma. F
E
is a tensor. In particular, F
E
∈ Ω
2
(End(E)).
Proof.
We have to show that
F
E
is linear over
C
∞
(
M
). We let
f ∈ C
∞
(
M
) and
s ∈ Ω
0
(E). Then we have
F
E
(fs) = d
E
d
E
(fs)
= d
E
(df ⊗ s + fd
E
s)
= d
2
f ⊗ s − df ∧ d
E
s + df ∧ d
E
s + fd
2
E
s
= fF
E
(s)
How do we think about this? Given X, Y ∈ Vect(M), consider
F
E
(X, Y ) : Ω
0
(E) → Ω
0
(E)
F
E
(X, Y )(s) = (F
E
(s))(X, Y )
Lemma. We have
F
E
(X, Y )(s) = ∇
X
∇
Y
s − ∇
Y
∇
X
s − ∇
[X,Y ]
s.
In other words, we have
F
E
(X, Y ) = [∇
X
, ∇
Y
] − ∇
[X,Y ]
.
This is what we were talking about, except that we have an extra term
∇
[X,Y ]
,
which vanishes in our previous motivating case, since
∂
∂x
i
and
∂
∂x
j
commute in
general.
Proof. We claim that if µ ∈ Ω
1
(E), then we have
(d
E
µ)(X, Y ) = ∇
X
(µ(Y )) − ∇
Y
(µ(X)) − µ([X, Y ]).
To see this, we let µ = ω ⊗ s, where ω ∈ Ω
1
(M) and s ∈ Ω
0
(E). Then we have
d
E
µ = dω ⊗ s − ω ∧ d
E
s.
So we know
(d
E
µ)(X, Y ) = dω(X, Y ) ⊗ s − (ω ∧ d
E
s)(X, Y )
By a result in the example sheet, this is equal to
= (Xω(Y ) − Y ω(X) − ω([X, Y ])) ⊗ s
− ω(X)∇
Y
(s) + ω(Y )∇
X
(s)
= Xω(Y ) ⊗ s + ω(Y )∇
X
s
− (Y ω(X) ⊗ s + ω(X)∇
Y
s) − ω([X, Y ]) ⊗ s
Then the claim follows, since
µ([X, Y ]) = ω([X, Y ]) ⊗ s
∇
X
(µ(Y )) = ∇
X
(ω(Y )s)
= Xω(Y ) ⊗ s + ω(Y )∇
X
s.
Now to prove the lemma, we have
(F
E
s)(X, Y ) = d
E
(d
E
s)(X, Y )
= ∇
X
((d
E
s)(Y )) − ∇
Y
((d
E
s)(X)) − (d
E
s)([X, Y ])
= ∇
X
∇
Y
s − ∇
Y
∇
X
s − ∇
[X,Y ]
s.
Definition (Flat connection). A connection d
E
is flat if F
E
= 0.
Specializing to the Riemannian case, we have
Definition
(Curvature of metric)
.
Let (
M, g
) be a Riemannian manifold with
metric
g
. The curvature of
g
is the curvature of the Levi-Civita connection,
denoted by
F
g
∈ Ω
2
(End(T M)) = Ω
0
(Λ
2
T
∗
M ⊗ T M ⊗ T
∗
M).
Definition (Flat metric). A Riemannian manifold (M, g) is flat if F
g
= 0.
Since flatness is a local property, it is clear that if a manifold is locally
isometric to
R
n
, then it is flat. What we want to prove is the converse — if
you are flat, then you are locally isometric to
R
n
. For completeness, let’s define
what an isometry is.
Definition
(Isometry)
.
Let (
M, g
) and (
N, g
0
) be Riemannian manifolds. We
say G ∈ C
∞
(M, N) is an isometry if G is a diffeomorphism and G
∗
g
0
= g, i.e.
DG|
p
: T
p
M → T
G(p)
N
is an isometry for all p ∈ M .
Definition
(Locally isometric)
.
A manifold
M
is locally isometric to
N
if for
all
p ∈ M
, there is a neighbourhood
U
of
p
and a
V ⊆ N
and an isometry
G : U → V .
Example.
The flat torus obtained by the quotient of
R
2
by
Z
2
is locally isometric
to R
2
, but is not diffeomorphic (since it is not even homeomorphic).
Our goal is to prove the following result.
Theorem.
Let
M
be a manifold with Riemannian metric
g
.Then
M
is flat iff
it is locally isometric to R
n
.
One direction is obvious. Since flatness is a local property, we know that if
M is locally isometric to R
n
, then it is flat.
To prove the remaining of the theorem, we need some preparation.
Proposition.
Let
dim M
=
n
and
U ⊆ M
open. Let
V
1
, · · · , V
n
∈ Vect
(
U
) be
such that
(i)
For all
p ∈ U
, we know
V
1
(
p
)
, · · · , V
n
(
p
) is a basis for
T
p
M
, i.e. the
V
i
are
a frame.
(ii) [V
i
, V
j
] = 0, i.e. the V
i
form a frame that pairwise commutes.
Then for all
p ∈ U
, there exists coordinates
x
1
, · · · , x
n
on a chart
p ∈ U
p
such
that
V
i
=
∂
∂x
i
.
Suppose that
g
is a Riemannian metric on
M
and the
V
i
are orthonormal in
T
p
M. Then the map defined above is an isometry.
Proof.
We fix
p ∈ U
. Let Θ
i
be the flow of
V
i
. From example sheet 2, we know
that since the Lie brackets vanish, the Θ
i
commute.
Recall that (Θ
i
)
t
(
q
) =
γ
(
t
), where
γ
is the maximal integral curve of
V
i
through q. Consider
α(t
1
, · · · , t
n
) = (Θ
n
)
t
n
◦ (Θ
n−1
)
t
n−1
◦ · · · ◦ (Θ
1
)
t
1
(p).
Now since each of Θ
i
is defined on some small neighbourhood of
p
, so if we just
move a bit in each direction, we know that
α
will be defined for (
t
0
, · · · , t
n
)
∈
B = {|t
i
| < ε} for some small ε.
Our next claim is that
Dα
∂
∂t
i
= V
i
whenever this is defined. Indeed, for t ∈ B and f ∈ C
∞
(M, R). Then we have
Dα
∂
∂t
i
t
(f) =
∂
∂t
i
t
f(α(t
1
, · · · , t
n
))
=
∂
∂t
i
t
f((Θ
i
)
t
◦ (Θ
n
)
t
n
◦ · · · ◦
\
(Θ
i
)
t
i
◦ · · · ◦ (Θ
1
)
t
1
(p))
= V
i
|
α(t)
(f).
So done. In particular, we have
Dα|
0
∂
∂t
i
0
= V
i
(p),
and this is a basis for
T
p
M
. So D
α|
0
:
T
0
R
n
→ T
p
M
is an isomorphism. By the
inverse function theorem, this is a local diffeomorphism, and in this chart, the
claim tells us that
V
i
=
∂
∂x
i
.
The second part with a Riemannian metric is clear.
We now actually prove the theorem
Proof of theorem.
Let (
M, g
) be a flat manifold. We fix
p ∈ M
. We let
x
1
, · · · , x
n
be coordinates centered at
p
1
, say defined for
|x
i
| <
1. We need to
construct orthonormal vector fields. To do this, we pick an orthonormal basis at
a point, and parallel transport it around.
We let
e
1
, · · · , e
n
be an orthonormal basis for
T
p
M
. We construct vector
fields
E
1
, · · · , E
n
∈ Vect
(
U
) by parallel transport. We first parallel transport
along (
x
1
,
0
, · · · ,
0) which defines
E
i
(
x
1
,
0
, · · · ,
0), then parallel transport along
the
x
2
direction to cover all
E
i
(
x
1
, x
2
,
0
, · · · ,
0) etc, until we define on all of
U
.
By construction, we have
∇
k
E
i
= 0 (∗)
on {x
k+1
= · · · = x
n
= 0}.
We will show that the
{E
i
}
are orthonormal and [
E
i
, E
j
] = 0 for all
i, j
. We
claim that each E
i
is parallel, i.e. for any curve γ, we have
D
γ
E
i
= 0.
It is sufficient to prove that
∇
j
E
i
= 0
for all i, j.
By induction on k, we show
∇
j
E
i
= 0
for
j ≤ k
on
{x
k+1
=
· · ·
=
x
n
= 0
}
. The statement for
k
= 1 is already given
by (∗). We assume the statement for k, so
∇
j
E
i
= 0 (A)
for
j ≤ k
and
{x
k+1
=
· · ·
=
x
n
= 0
}
. For
j
=
k
+ 1, we know that
∇
k+1
E
i
= 0
on
{x
k+2
=
· · ·
=
x
n
= 0
}
by (
∗
). So the only problem we have is for
j
=
k
and
{x
k+2
= · · · = x
n
= 0}.
By flatness of the Levi-Civita connection, we have
[∇
k+1
, ∇
k
] = ∇
[∂
k+1
,∂
k
]
= 0.
So we know
∇
k+1
∇
k
E
i
= ∇
k
∇
k+1
E
i
= 0 (B)
on
{x
k+2
=
· · ·
=
x
n
= 0
}
. Now at
x
k+1
= 0 , we know
∇
k
E
i
vanishes. So it
follows from parallel transport that ∇
k
E
i
vanishes on {x
k+2
= · · · = x
n
= 0}.
As the Levi-Civita connection is compatible with
g
, we know that parallel
transport is an isometry. So the inner product product
g
(
E
i
, E
j
) =
g
(
e
i
, e
j
) =
δ
ij
.
So this gives an orthonormal frame at all points.
Finally, since the torsion vanishes, we know
[E
i
, E
j
] = ∇
E
i
E
j
− ∇
E
j
E
i
= 0,
as the E
i
are parallel. So we are done by the proposition.
What does the curvature mean when it is non-zero? There are many answers
to this, and we will only give one.
Definition
(Holonomy)
.
Consider a piecewise smooth curve
γ
: [0
,
1]
→ M
with
γ
(0) =
γ
(1) =
p
. Say we have a linear connection
∇
. Then we have a notion of
parallel transport along γ.
The holonomy of ∇ around γ: is the map
H : T
p
M → T
p
M
given by
H(ξ) = V (1),
where V is the parallel transport of ξ along γ.
Example.
If
∇
is compatible with a Riemannian metric
g
, then
H
is an isometry.
Example.
Consider
R
n
with the usual connection. Then if
ξ ∈ T
0
R
n
, then
H(ξ) = ξ for any such path γ. So the holonomy is trivial.
Example.
Say (
M, g
) is flat, and
p ∈ M
. We have seen that there exists a
neighbourhood of
p
such that (
U, g|
U
) is isometric to
R
n
. So if
γ
([0
,
1])
∈ U
,
then H = id.
The curvature measures the extent to which this does not happen. Suppose
we have coordinates x
1
, · · · , x
n
on some (M, g). Consider γ as follows:
(s, 0, · · · , 0)
(s, t, · · · , 0)(0, t, · · · , 0)
0
Then we can Taylor expand to find
H = id +F
g
∂
∂x
1
p
,
∂
∂x
2
p
!
st + O(s
2
t, st
2
).