2Vector fields
III Differential Geometry
2.1 The tangent bundle
Recall that we had the notion of a tangent vector. If we have a curve
γ
:
I → M
,
then we would like to think that the derivative
˙γ
“varies smoothly” with time.
However, we cannot really do that yet, since for different
t
, the value of
˙γ
lies in
different vector spaces, and there isn’t a way of comparing them.
More generally, given a “vector field”
f
:
p 7→ v
p
∈ T
p
M
for each
p ∈ M
, how
do we ask if this is a smooth function?
One way to solve this is to pick local coordinates
x
1
, · · · , x
n
on
U ⊆ M
. We
can then write
v
p
=
X
i
α
i
(p)
∂
∂x
i
p
.
Since
α
i
(
p
)
∈ R
, we can say
v
p
varies smoothly if the functions
α
i
(
p
) are smooth.
We then proceed to check that this does not depend on coordinates etc.
However, there is a more direct approach. We simply turn
T M =
[
p∈M
T
p
M
into a manifold. There is then a natural map
π
:
T M → M
sending
v
p
∈ T
p
M
to
p
for each
p ∈ M
, and this is smooth. We can then define the smoothness of
f using the usual notion of smoothness of maps between manifolds.
Assuming that we have successfully constructed a sensible
T M
, we can define:
Definition
(Vector field)
.
A vector field on some
U ⊆ M
is a smooth map
X : U → T M such that for all p ∈ U, we have
X(p) ∈ T
p
M.
In other words, we have π ◦ X = id.
Definition
(
Vect
(
U
))
.
Let
Vect
(
U
) denote the set of all vector fields on
U
. Let
X, Y ∈ Vect(U), and f ∈ C
∞
(U). Then we can define
(X + Y )(p) = X(p) + Y (p), (fX)(p) = f (p)X(p).
Then we have X + Y, fX ∈ Vect(U). So Vect(U) is a C
∞
(U) module.
Moreover, if V ⊆ U ⊆ M and X ∈ Vect(U ), then X|
V
∈ Vect(V ).
Conversely, if
{V
i
}
is a cover of
U
, and
X
i
∈ Vect
(
V
i
) are such that they
agree on intersections, then they patch together to give an element of
Vect
(
U
).
So we say that Vect is a sheaf of C
∞
(M) modules.
Now we properly define the manifold structure on the tangent bundle.
Definition (Tangent bundle). Let M be a manifold, and
T M =
[
p∈M
T
p
M.
There is a natural projection map π : T M → M sending v
p
∈ T
p
M to p.
Let
x
1
, · · · , x
n
be coordinates on a chart (
U, ϕ
). Then for any
p ∈ U
and
v
p
∈ T
p
M, there are some α
1
, · · · , α
n
∈ R such that
v
p
=
n
X
i=1
α
i
∂
∂x
i
p
.
This gives a bijection
π
−1
(U) → ϕ(U) × R
n
v
p
7→ (x
1
(p), · · · , x
n
(p), α
1
, · · · , α
n
),
These charts make
T M
into a manifold of dimension 2
dim M
, called the tangent
bundle of M.
Lemma. The charts actually make T M into a manifold.
Proof. If (V, ξ) is another chart on M with coordinates y
1
, · · · , y
n
, then
∂
∂x
i
p
=
n
X
j=1
∂y
j
∂x
i
(p)
∂
∂y
j
p
.
So we have
˜
ξ ◦ ˜ϕ
−1
: ϕ(U ∩ V ) × R
n
→ ξ(U ∩ V ) × R
n
given by
˜
ξ ◦ ˜ϕ
−1
(x
1
, · · · , x
n
, α
1
, · · · , α
n
) =
y
1
, · · · , y
n
,
n
X
i=1
α
i
∂y
1
∂x
i
, · · · ,
n
X
i=1
α
i
∂y
n
∂x
i
!
,
and is smooth (and in fact fiberwise linear).
It is easy to check that
T M
is Hausdorff and second countable as
M
is.
There are a few remarks to make about this.
(i) The projection map π : T M → M is smooth.
(ii) If U ⊆ M is open, recall that
Vect(U) = {smooth X : U → T M | X(p) ∈ T
p
M for all p ∈ U }.
We write
X
p
for
X
(
p
). Now suppose further that
U
is a coordinate chart,
then we can write any function
X
:
U → T M
such that
X
p
∈ T
p
M
(uniquely) as
X
p
=
n
X
i=1
α
i
(p)
∂
∂x
i
p
Then X is smooth iff all α
i
are smooth.
(iii)
If
F ∈ C
∞
(
M, N
), then D
F
:
T M → T N
given by D
F
(
v
p
) = D
F |
p
(
v
p
) is
smooth. This is nice, since we can easily talk about higher derivatives, by
taking the derivative of the derivative map.
(iv)
If
F ∈ C
∞
(
M, N
) and
X
is a vector field on
M
, then we cannot obtain a
vector field on
N
by D
F
(
X
), since
F
might not be injective. If
F
(
p
1
) =
F (p
2
), we need not have DF (X(p
1
)) = DF (X(p
2
)).
However, there is a weaker notion of being F -related.
Definition
(
F
-related)
.
Let
M, N
be manifolds, and
X ∈ Vect
(
M
),
Y ∈
Vect(N) and F ∈ C
∞
(M, N ). We say they are F -related if
Y
q
= DF |
p
(X
p
)
for all
p ∈ M
and
F
(
p
) =
q
. In other words, if the following diagram commutes:
T M T N
M N
DF
X
F
Y
.
So what does
Vect
(
M
) look like? Recall that a vector is defined to be a
derivation. So perhaps a vector field is also a derivation of some sort.
Definition
(
Der
(
C
∞
(
M
)))
.
Let
Der
(
C
∞
(
M
)) be the set of all
R
-linear maps
X : C
∞
(M) → C
∞
(M) that satisfy
X (fg) = f X (g) + X (f )g.
This is an R-vector space, and in fact a C
∞
(M) module.
Given X ∈ Vect(M), we get a derivation X ∈ Der(C
∞
(M)) by setting
X (f)(p) = X
p
(f).
It is an exercise to show that
X
(
f
) is smooth and satisfies the Leibniz rule.
Similar to the case of vectors, we want to show that all derivations come from
vector fields.
Lemma. The map X 7→ X is an R-linear isomorphism
Γ : Vect(M) → Der(C
∞
(M)).
Proof. Suppose that α is a derivation. If p ∈ M , we define
X
p
(f) = α(f)(p)
for all f ∈ C
∞
(M). This is certainly a linear map, and we have
X
p
(fg) = α(fg)(p) = (fα(g) + gα(f))(p) = f(p)X
p
(g) + g(p)X
p
(f).
So
X
p
∈ T
p
M
. We just need to check that the map
M → T M
sending
p 7→ X
p
is smooth. Locally on M, we have coordinates x
1
, · · · , x
n
, and we can write
X
p
=
n
X
i=1
α
i
(p)
∂
∂x
i
p
.
We want to show that α
i
: U → R are smooth.
We pick a bump function
ϕ
that is identically 1 near
p
, with
supp ϕ ⊆ U
.
Consider the function ϕx
j
∈ C
∞
(M). We can then consider
α(ϕx
j
)(p) = X
p
(ϕx
j
).
As
ϕx
j
is just
x
j
near
p
, by properties of derivations, we know this is just equal
to α
j
. So we have
α(ϕx
j
) = α
j
.
So α
j
is smooth.
From now on, we confuse
X
and
X
, i.e. we think of any
X ∈ Vect
(
M
) as a
derivation of C
∞
(M).
Note that the product of two vector fields (i.e. the composition of derivations)
is not a vector field. We can compute
XY (fg) = X(Y (fg))
= X(f Y (g) + gY (f ))
= X(f )Y (g) + f XY (g) + X(g)Y (f) + gXY (f).
So this is not a derivation, because we have the cross terms
X
(
f
)
Y
(
g
). However,
what we do have is that XY − Y X is a derivation.
Definition
(Lie bracket)
.
If
X, Y ∈ Vect
(
M
), then the Lie bracket [
X, Y
] is
(the vector field corresponding to) the derivation XY − Y X ∈ Vect(M).
So Vect(M) becomes what is known as a Lie algebra.
Definition
(Lie algebra)
.
A Lie algebra is a vector space
V
with a bracket
[ · , · ] : V × V → V such that
(i) [ · , · ] is bilinear.
(ii) [ · , · ] is antisymmetric, i.e. [X, Y ] = −[Y, X].
(iii) The Jacobi identity holds
[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0.
It is a (painful) exercise to show that the Lie bracket does satisfy the Jacobi
identity.
The definition of the Lie algebra might seem a bit weird. Later it will come
up in many different guises and hopefully it might become more clear.