3Hamiltonian vector fields

III Symplectic Geometry



3.1 Hamiltonian vector fields
Definition
(Hamiltonian vector field)
.
Let (
M, ω
) be a symplectic manifold. If
H C
(
M
), then since
˜ω
:
T M T
M
is an isomorphism, there is a unique
vector field X
H
on M such that
ι
X
H
ω = dH.
We call X
H
the Hamiltonian vector field with Hamiltonian function H.
Suppose
X
H
is complete (e.g. when
M
is compact). This means we can
integrate X
H
, i.e. solve
ρ
t
t
(p) = X
H
(ρ
t
(p)), ρ
0
(p) = p.
These flow have some nice properties.
Proposition.
If
X
H
is a Hamiltonian vector field with flow
ρ
t
, then
ρ
t
ω
=
ω
.
In other words, each ρ
t
is a symplectomorphism.
Proof. It suffices to show that
t
ρ
t
ω = 0. We have
d
dt
(ρ
t
ω) = ρ
t
(L
X
H
ω) = ρ
t
(dι
X
H
ω + ι
X
H
dω) = ρ
t
(ddH) = 0.
Thus, every function
H
gives us a one-parameter subgroup of symplectomor-
phisms.
Proposition. ρ
t
preserves H, i.e. ρ
t
H = H.
Proof.
d
dt
ρ
t
H = ρ
t
(L
X
H
H) = ρ
t
(ι
X
H
dH) = ρ
t
(ι
X
H
ι
X
H
ω) = 0.
So the flow lines of our vector field are contained in level sets of H.
Example. Take (S
2
, ω = dθ dh). Take
H(h, θ) = h
to be the height function. Then X
H
solves
ι
X
H
(dθ dh) = dh.
So
X
H
=
θ
, ρ
t
(h, θ) = (h, θ + t).
As expected, the flow preserves height and the area form.
We have seen that Hamiltonian vector fields are symplectic:
Definition
(Symplectic vector field)
.
A vector field
X
on (
M, ω
) is a symplectic
vector field if L
X
ω = 0.
Observe that
L
X
ω = ι
X
dω + dι
X
ω = dι
X
ω.
So
X
is symplectic iff
ι
X
ω
is closed, and is Hamiltonian if it is exact. Thus,
locally, every symplectic vector field is Hamiltonian and globally, the obstruction
lies in H
1
dR
(M).
Example.
Take (
T
2
, ω
= d
θ
1
d
θ
2
). Then
X
i
=
θ
i
are symplectic but not
Hamiltonian, since ι
X
i
ω = dθ
2i
is closed but not exact.
Proposition.
Let
X, Y
be symplectic vector fields on (
M, ω
). Then [
X, Y
] is
Hamiltonian.
Recall that if
X, Y
are vector fields on
M
and
f C
(
M
), then their Lie
bracket is given by
[X, Y ]f = (XY Y X)f.
This makes χ(M), the space of vector fields on M, a Lie algebra.
In order to prove the proposition, we need the following identity:
Exercise. ι
[X,Y ]
α = [L
X
, ι
Y
]α = [ι
X
, L
Y
]α.
Proof of proposition.
We need to check that
ι
[X,Y ]
ω
is exact. By the exercise,
this is
ι
[X,Y ]
ω = L
X
ι
Y
ω ι
Y
L
X
ω = d(ι
X
ι
Y
ω) + ι
X
dι
Y
ω + ι
Y
dι
X
ω ι
Y
ι
Y
dω.
Since
X, Y
are symplectic, we know d
ι
Y
ω
= d
ι
X
ω
= 0, and the last term always
vanishes. So this is exact, and
ω
(
Y, X
) is a Hamiltonian function for [
X, Y
].
Definition
(Poisson bracket)
.
Let
f, g C
(
M
). We then define the Poisson
bracket {f, g} by
{f, g} = ω(X
f
, X
g
).
This is defined so that
X
{f,g}
= [X
f
, X
g
].
Exercise.
The Poisson bracket satisfies the Jacobi identity, and also the Leibniz
rule
{f, gh} = g{f, h} + {f, g}h.
Thus, if (
M, ω
) is symplectic, then (
C
(
M
)
, {·, ·}
) is a Poisson algebra.
This means it is a commutative, associative algebra with a Lie bracket that
satisfies the Leibniz rule.
Further, the map
C
(
M
)
χ
(
M
) sending
H 7→ X
H
is a Lie algebra
(anti-)homomorphism.
Proposition. {f, g} = 0 iff f is constant along integral curves of X
g
.
Proof.
L
X
g
f = ι
X
g
df = ι
X
g
ι
X
f
ω = ω(X
f
, X
g
) = {f, g} = 0.
Example. If M = R
2n
and ω = ω
0
=
P
dx
j
dy
j
, and f C
(R
2n
), then
X
f
=
X
i
f
y
i
x
i
f
x
i
y
i
.
If ρ
0
(p) = p, then ρ
t
(p) = (x(t), y(t)) is an integral curve for X
f
iff
dx
i
dt
=
f
y
i
,
y
i
t
=
f
x
i
.
In classical mechanics, these are known as Hamilton equations.