L4: Lie derivative

A symplectic form is a closed non-degenerate

2-form. What does it look like? Pointwise, a

non-degenerate skew-form ω ∈ Λ2V on a vector

space V is one for which

(i) ω(u, v) = −ω(v, u) and

(ii) ω(u, v) = 0∀v ⇒ u = 0.

Example: ω0 =∑ni=1 dxi∧dyi on R 2n (a 2-form

with constant co-efficients). Define a matrix,

with respect to the basis x1, . . . , xn, y1, . . . , yn,

J0 =

(

0 In−In 0

)

. The group Sp2n(R ) com-

prises linear maps A for which A∗ω0 = ω0, i.e.

ω(Av,Au) = ω(v, u), which says AtJ0A = J0.

Lemma: if V, ω is a symplectic vector space

there is a basis u1 . . . un, v1 . . . vn of V such that

ω|〈ui〉 ≡ 0, ω|〈vj〉 ≡ 0 and ω(ui, vj) = δij. Such

a basis is called a symplectic basis.

Before giving the proof, note if U ⊂ (V, ω),

there is a symplectic orthogonal complement

U⊥ω = v ∈ V |ω(v, u) = 0 ∀u ∈ U

U is a symplectic subspace if U ∩ U⊥ω = 0,

which means ω|U is non-degenerate.

Proof of Lemma: We induct on dim(V ). By

definition ω is non-degenerate, so given u1 ∈ V ,

there is v ∈ V such that ω(u1, v) 6= 0; let v1 =

v/|ω(u1, v)| so ω(u1, v1) = 1. Then U = 〈u1, v1〉

is a symplectic subspace of V , so (by induction)

there is a symplectic basis u2, . . . , un, v2, . . . , vnfor U⊥ω, and this completes the proof.

Corollary: a symplectic vector space has even

dimension. ω is non-degenerate if and only if

ωn 6= 0 ∈ ΛnV .

The second statement is true in the standard

model (R 2n, ω0). If ω is degenerate, ω(u, v) =

0∀v, extend u to a basis u = u1, u2, . . . u2n of

V and note ωn(u1, . . . , u2n) = 0.

Corollary: A two-dimensional manifold is sym-

plectic if and only if it is orientable. The sphere

S2k is not symplectic if k > 1.

Proof: if dim(M) = 2, a non-degenerate 2-

form is a volume form, and any such is ob-

viously closed. For the second statement, if

M2k, ω is symplectic, the 2-form defines a class

[ω] ∈ H2dR(M). The k-th cup-product of this

class with itself is pointwise at TxM the non-

zero form ωkx, i.e. ωk is a volume form, hence

non-zero in cohomology. So [ω]k 6= 0 ∈ H2kdR(M)

(which is just R ), so all the even cohomologygroups H2i(M) have non-zero dimension.

We can easily build more examples. For in-

stance, products of symplectic manifolds are

symplectic: (M,ω) × (N, σ) = (M × N,ω ⊕ σ).

The last notation is shorthand for π∗Mω+π∗Nσ.

So the family Σg1× · · ·×Σgk gives (closed) ex-

amples in all possible dimensions.

At the end of the last lecture, we said a func-

tion on a symplectic manifold defines a flow

f 7→ df 7→ (µω)−1(df) = Xf = X 7→ φXt t∈(−δ,δ).

Being a canonical construction, we expect the

flow to preserve the symplectic structure. For-

mally, we prove such statements using the Lie

derivative. If X ∈ Γ(TM) is a vector field with

flow φt, defined at least locally for small time,

and α ∈ Ω∗(M) is some differential form, then

LX(α) =d

dt(φ∗tα)|t=0

is the Lie derivative of α with respect to X, and

measures the way α changes along the flow.

This is actually a much more natural differen-

tial operator than either covariant differenti-

ation with respect to a connexion, or exterior

differentiation of forms themselves: and LX(α)

is the same kind of object as α itself. We can

also differentiate covariant tensors (like vec-

tor fields) as well as contravariant tensors like

forms: just replace φ∗t by φt∗.

The interior product ι is defined by

〈ιXω, u〉 = 〈ω,X ∧ u〉

for X ∈ Γ(TM), u ∈ Λp(TM), ω ∈ Ωp+1(M).Note ιXω = µω(X) in our previous notation.The following formulae are often useful:

(i) LX(f) = X · f for f ∈ C∞(M)(ii) LX(Y ) = [X,Y ] for X,Y in Γ(TM)(iii) LX(dα) = d(LXα) for α ∈ Ω

∗(M)(iv) LX = ιX d+ d ιX [Cartan’s formula](v) dθ(X,Y ) = X · θ(Y )− Y · θ(X)− θ([X,Y ])for vector fields X,Y and θ ∈ Ω1(M)

(i) and (iii) are immediate from the definitions.

For (iv), note ιX is an antiderivation:

ιX(u ∧ v) = ιXu ∧ v+ (−1)deg(u)u ∧ ιXv

So LX and ιXd+ dιX are both derivations pre-serving degree & both commute with d. Henceit suffices to check the identity on functions.

This reduces to ιX(df) = df(X) = X · f . Weshall not prove (v). Recall [X,Y ] = 0 iff theflows generated by X and Y commute, whichwould certainly imply that LX(Y ) = 0.

To prove (ii), first drink a large glass of grappa.

Then we compute, by showing the two sides

take the same value on any function f , at any

point m. Suppose X integrates to φt and Y

integrates to ψs. Now (LXY )(f)|m equals

d

dt

(

(dφ−tYφt(m))|t=0

)

f =d

dt(Yφt(m) · (fφ−t))t=0

Define H(t, u) = f φ−t ψu φt (m)

Define K(t, u, s) = f φs ψu φt (m) both de-

fined in a neighbourhood of the origin. Then

check:

(i) ∂2H|(t,0) = Yφt(m) · (f φ−t)

(ii) ∂12(H)|(0,0) = LXY (f)|m

(iii) ∂12(H)|(0,0) = ∂12(K)|(0,0,0)−∂23(K)|(0,0,0)

(iv) ∂2K|(t,0,0) = (Y · f)φt(m)

(v) ∂12(K)|(0,0,0) = (X · (Y · f))m, similarly:

(vi) ∂23(K)|(0,0,0) = (Y · (X · f))m

So LXY = XY − Y X = [X,Y ], as required. If

you’re not convinced, you didn’t drink enough

grappa.

Lemma: The Hamiltonian flow of a function

on a symplectic manifold preserves the sym-

plectic form, i.e. is through symplectomor-

phisms.

Proof: Let f : M → R be smooth, with asso-

ciated vector field X = (µω)−1(df), with ω the

symplectic form. Cartan’s formula says

LXω = ιXdω+ d(ιXω)

but dω = 0 as symplectic forms are closed;

and ιXω = µω(X) = df so d(ιXω) = d2f =

0. Hence LXω = 0, which exactly says ω is

preserved by the flow of the vector field.

Example: solving Hamilton’s equations

∂H

∂p= q;

∂H

∂q= −p

is exactly following the flow of XH = (∂H∂p ,−

∂H∂q ),

which satisfies ιXHω0 = dH. Hence, classical

mechanical flows are through (not necessarily

linear!) symplectomorphisms of R 2n.