8Connections
III Differential Geometry
8.3 Riemannian connections
Now suppose our manifold
M
has a Riemannian metric
g
. It is then natural to
ask if there is a “natural” connection compatible with g.
The requirement of “compatibility” in some sense says the product rule is
satisfied by the connection. Note that saying this does require the existence of a
metric, since we need one to talk about the product of two vectors.
Definition
(Metric connection)
.
A linear connection
∇
is compatible with
g
(or
is a metric connection) if for all X, Y, Z ∈ Vect(M),
∇
X
g(Y, Z) = g(∇
X
Y, Z) + g(Y, ∇
X
Z).
Note that the first term is just X(g(Y, Z)).
We should view this formula as specifying that the product rule holds.
We can alternatively formulate this in terms of the covariant derivative.
Lemma.
Let
∇
be a connection. Then
∇
is compatible with
g
if and only if for
all γ : I → M and V, W ∈ J(γ), we have
d
dt
g(V (t), W (t)) = g(D
t
V (t), W(t)) + g(V (t), D
t
W (t)). (∗)
Proof. Write it out explicitly in local coordinates.
We have some immediate corollaries, where the connection is always assumed
to be compatible.
Corollary.
If
V, W
are parallel along
γ
, then
g
(
V
(
t
)
, W
(
t
)) is constant with
respect to t.
Corollary. If γ is a geodesic, then | ˙γ| is constant.
Corollary. Parallel transport is an isometry.
In general, on a Riemannian manifold, there can be many metric conditions.
To ensure that it is actually unique, we need to introduce a new constraint,
known as the torsion. The definition itself is pretty confusing, but we will chat
about it afterwards to explain why this is a useful definition.
Definition
(Torsion of linear connection)
.
Let
∇
be a linear connection on
M
.
The torsion of ∇ is defined by
τ(X, Y ) = ∇
X
Y − ∇
Y
X − [X, Y ]
for X, Y ∈ Vect(M).
Definition
(Symmetric/torsion free connection)
.
A linear connection is sym-
metric or torsion-free if τ (X, Y ) = 0 for all X, Y .
Proposition. τ is a tensor of type (2, 1).
Proof. We have
τ(fX, Y ) = ∇
fX
Y − ∇
Y
(fX) − [fX, Y ]
= f∇
X
Y − Y (f)X − f∇
Y
X − f XY + Y (fX)
= f(∇
X
Y − ∇
Y
X − [X, Y ])
= fτ (X, Y ).
So it is linear.
We also have τ (X, Y ) = −τ(Y, X) by inspection.
What does being symmetric tell us? Consider the Christoffel symbols in
some coordinate system x
1
, · · · , x
n
. We then have
∂
∂x
i
,
∂
∂x
j
= 0.
So we have
τ
∂
∂x
i
,
∂
∂x
j
= ∇
i
∂
j
− ∇
j
∂
i
= Γ
k
ij
∂
k
− Γ
k
ji
∂
k
.
So we know a connection is symmetric iff the Christoffel symbol is symmetric,
i.e.
Γ
k
ij
= Γ
k
ji
.
Now the theorem is this:
Theorem.
Let
M
be a manifold with Riemannian metric
g
. Then there exists
a unique torsion-free linear connection ∇ compatible with g.
The actual proof is unenlightening.
Proof. In local coordinates, we write
g =
X
g
ij
dx
i
⊗ dx
j
.
Then the connection is explicitly given by
Γ
k
ij
=
1
2
g
k`
(∂
i
g
j`
+ ∂
j
g
i`
− ∂
`
g
ij
),
where g
k`
is the inverse of g
ij
.
We then check that it works.
Definition
(Riemannian/Levi-Civita connection)
.
The unique torsion-free met-
ric connection on a Riemannian manifold is called the Riemannian connection
or Levi-Civita connection.
Example.
Consider the really boring manifold
R
n
with the usual metric. We
also know that T R
n
∼
=
R
n
→ R
n
is trivial, so we can give a trivial connection
d
R
n
f
∂
∂x
i
= df ⊗
∂
∂x
i
.
In the ∇ notation, we have
∇
X
f
∂
∂x
i
= X(f )
∂
∂x
i
.
It is easy to see that this is a connection, and also that it is compatible with the
metric. So this is the Riemannian connection on R
n
.
This is not too exciting.
Example.
Suppose
φ
:
M ⊆ R
n
is an embedded submanifold. This gives us a
Riemannian metric on M by pulling back
g = φ
∗
g
R
n
on M.
We also get a connection on
M
as follows: suppose
X, Y ∈ Vect
(
M
). Locally,
we know X, Y extend to vector fields
˜
X,
˜
Y on R
n
. We set
∇
X
Y = π(
¯
∇
˜
X
˜
Y ),
where π is the orthogonal projection T
p
(R
n
) → T
p
M.
It is an exercise to check that this is a torsion-free metric connection on M.
It is a (difficult) theorem by Nash that every manifold can be embedded in
R
n
such that the metric is the induced metric. So all metrics arise this way.