2Vector fields
III Differential Geometry
2.3 Lie derivative
We now want to look at the concept of a Lie derivative. If we have a function
f
defined on all of
M
, and we have a vector field
X
, then we might want to
ask what the derivative of
f
in the direction of
X
is at each point. If
f
is a
real-valued function, then this is by definition
X
(
f
). If
f
is more complicated,
then this wouldn’t work, but we can still differentiate things along
X
using the
flows.
Notation. Let F : M → M be a diffeomorphism, and g ∈ C
∞
(M). We write
F
∗
g = g ◦ F ∈ C
∞
(M).
We now define the Lie derivative of a function, i.e. the derivative of a function
f
in the direction of a vector field
X
. Of course, we can obtain this by just
applying X(f), but we want to make a definition that we can generalize.
Definition
(Lie derivative of a function)
.
Let
X
be a complete vector field, and
Θ be its flow. We define the Lie derivative of g along X by
L
X
(g) =
d
dt
t=0
Θ
∗
t
g.
Here this is defined pointwise, i.e. for all p ∈ M, we define
L
X
(g)(p) =
d
dt
t=0
Θ
∗
t
(g)(p).
Lemma. L
X
(g) = X(g). In particular, L
X
(g) ∈ C
∞
(M, R).
Proof.
L
X
(g)(p) =
d
dt
t=0
Θ
∗
t
(g)(p)
=
d
dt
t=0
g(Θ
t
(p))
= dg|
p
(X(p))
= X(g)(p).
So this is quite boring. However, we can do something more exciting by
differentiating vector fields.
Notation.
Let
Y ∈ Vect
(
M
), and
F
:
M → M
be a diffeomorphism. Then
DF
−1
|
F (p)
: T
F (p)
M → T
p
M. So we can write
F
∗
(Y )|
p
= DF
−1
|
F (p)
(Y
F (p)
) ∈ T
p
M.
Then F
∗
(Y ) ∈ Vect(M). If g ∈ C
∞
(M), then
F
∗
(Y )|
p
(g) = Y
F (p)
(g ◦ F
−1
).
Alternatively, we have
F
∗
(Y )|
p
(g ◦ F ) = Y
F (p)
(g).
Removing the p’s, we have
F
∗
(Y )(g ◦ F ) = (Y (g)) ◦ F.
Definition
(Lie derivative of a vector field)
.
Let
X ∈ Vect
(
M
) be complete,
and
Y ∈ Vect
(
M
) be a vector field. Then the Lie derivative is given pointwise
by
L
X
(Y ) =
d
dt
t=0
Θ
∗
t
(Y ).
Lemma. We have
L
X
Y = [X, Y ].
Proof. Let g ∈ C
∞
(M, R). Then we have
Θ
∗
t
(Y )(g ◦ Θ
t
) = Y (g) ◦ Θ
t
.
We now look at
Θ
∗
t
(Y )(g) − Y (g)
t
=
Θ
∗
t
(Y )(g) − Θ
∗
t
(Y )(g ◦ Θ
t
)
t
| {z }
α
t
+
Y (g) ◦ Θ
t
− Y (g)
t
| {z }
β
t
.
We have
lim
t→0
β
t
= L
X
(Y (g)) = XY (g)
by the previous lemma, and we have
lim
t→0
α
t
= lim
t→0
(Θ
∗
t
(Y ))
g − g ◦ Θ
t
t
= Y (−L
X
(g)) = −Y X(g).
Corollary. Let X, Y ∈ Vect(M) and f ∈ C
∞
(M, R). Then
(i) L
X
(fY ) = L
X
(f)Y + fL
X
Y = X(f)Y + fL
X
Y
(ii) L
X
Y = −L
Y
X
(iii) L
X
[Y, Z] = [L
X
Y, Z] + [Y, L
X
Z].
Proof. Immediate from the properties of the Lie bracket.