III Symplectic Geometry - Symplectic manifolds

1Symplectic manifolds

III Symplectic Geometry

1.3 Symplectomorphisms and Lagrangians

We now consider Lagrangian submanifolds. These are interesting for many

reasons, but in this section, we are going to use them to understand symplecto-

morphisms.

Definition

(Lagrangian submanifold)

Let (

M, ω

) be a symplectic manifold,

and

L ⊆ M

a submanifold. Then

is a Lagrangian submanifold if for all

p ∈ L

is a Lagrangian subspace of

. Equivalently, the restriction of

vanishes and dim L =

dim M.

Example.

If (

M, ω

) is a surface with area form

, and

is any 1-dimensional

submanifold, then

is Lagrangian, since any 1-dimensional subspace of

Lagrangian.

What do Lagrangian submanifolds of

∗

look like? Let

be a 1-form on

X, i.e. a section of T

∗

X, and let

= {(x, µ

) : x ∈ X} ⊆ T

∗

The map

X → T

∗

is an embedding, and

dim X

dim T

∗

So this is a good candidate for a Lagrangian submanifold. By definition, X

Lagrangian iff s

∗

ω = 0. But

∗

ω = s

∗

(dα) = d(s

∗

α) = dµ.

is Lagrangian iff

is closed. In particular, if

h ∈ C

∞

(

), then

a Lagrangian submanifold of

∗

. We call

the generating function of the

Lagrangian submanifold.

There are other examples of Lagrangian submanifolds of the cotangent bundle.

Let

be a submanifold of

, and

x ∈ X

. We define the conormal space of

x to be

∗

S = {ξ ∈ T

∗

X : ξ|

= 0}.

The conormal bundle of S is the union of all N

∗

Example. If S = {x}, then N

∗

S = T

∗

Example. If S = X, then N

∗

S = S is the zero section.

Note that both of these are

-dimensional submanifolds of

∗

, and this is

of course true in general.

So if we are looking for Lagrangian submanifolds, this is already a good start.

In the case of

, this is certainly a Lagrangian submanifold, and so is

S = {x}. Indeed,

Proposition.

Let

∗

and

∗

. Then

L → M

is a Lagrangian

submanifold.

Proof.

-dimensional, then

S → X

locally looks like

→ R

, and it is

clear in this case.

Our objective is to use Lagrangian submanifolds to understand the following

problem:

Problem.

Let (

, ω

) and (

, ω

) be 2

-dimensional symplectic manifolds.

If f : M

→ M

is a diffeomorphism, when is it a symplectomorphism?

Consider the 4

-dimensional manifold

× M

, with the product form

ω = pr

∗

+ pr

∗

. We have

dω = pr

∗

dω

+ pr

∗

dω

= 0,

and ω is non-degenerate since





∗

∧n

∧ pr

∗

∧n

6= 0.

is a symplectic form on

× M

. In fact, for

, λ

∈ R \ {

}

, the

combination

∗

+ λ

∗

is also symplectic. The case we are particularly interested in is

= 1 and

= −1, where we get the twisted product form.

˜ω = pr

∗

− pr

∗

Let T

be the graph of f , namely

= {(p

, f(p

)) : p

∈ M

} ⊆ M

× M

and let γ

: M

→ M

× M

be the natural embedding with image T

Proposition. f

is a symplectomorphism iff

is a Lagrangian submanifold of

× M

, ˜ω).

Proof.

The first condition is

∗

, and the second condition is

∗

˜ω

= 0,

and

∗

˜ω = γ

∗

= γ

∗

= ω

− f

∗

We are particularly interested in the case where

are cotangent bundles.

, X

are

-dimensional manifolds and

∗

, then we can naturally

identify

∗

× X

)

∼

∗

× T

∗

The canonical symplectic form on the left agrees with the canonical symplectic

form on the right. However, we want the twisted symplectic form instead. So we

define an involution

σ : M

× M

→ M

× M

that fixes M

and sends (x, ξ) 7→ (x, −ξ) on M

. Then σ

∗

ω = ˜ω.

Then, to find a symplectomorphism

→ M

, we can carry out the following

procedure:

(i) Start with L a Lagrangian submanifold of (M

× M

, ω).

(ii) Twist it to get L

= σ(L).

(iii) Check if L

is the graph of a diffeomorphism.

This doesn’t sound like a very helpful specification, until we remember that

we have a rather systematic way of producing Lagrangian submanifolds of

(

, ×M

, ω

). We pick a generating function

f ∈ C

∞

(

× X

). Then d

is a

closed form on X

× X

, and

L = X

= {(x, y, ∂

(x,y)

, ∂

(x,y)

) : (x, y) ∈ X

× X

}

is a Lagrangian submanifold of (M

× M

, ω).

The twist L

is then

= {(x, y, d

(x,y)

, −d

(x,y)

) : (x, y) ∈ X

× X

}

We then have to check if

is the graph of a diffeomorphism

→ M

. If so,

we win, and we call this the symplectomorphism generated by

. Equivalently,

we say f is the generating function for ϕ.

To check if L

is the graph of a diffeomorphism, for each (x, ξ) ∈ T

∗

, we

need to find (y, η) ∈ T

∗

such that

ξ = ∂

(x,y)

, η = −∂

(x,y)

In local coordinates {x

, ξ

} for T

∗

and {y

, η

} for T

∗

, we want to solve

∂f

∂x

(x, y)

= −

∂f

∂y

(x, y)

The only equation that requires solving is the first equation for

, and then the

second equation tells us what we should pick η

to be.

For the first equation to have a unique smoothly-varying solution for

locally, we must require

det



∂

∂y



∂f

∂x



i,j

6= 0.

This is a necessary condition for

to generate a symplectomorphism

. If this

is satisfied, then we should think a bit harder to solve the global problem of

whether ξ

is uniquely defined.

We can in fact do this quite explicitly:

Example. Let X

= X

= R

, and f : R

× R

→ R given by

(x, y) 7→ −

|x − y|

= −

i=1

− y

)

What, if any, symplectomorphism is generated by f? We have to solve

∂f

∂x

= y

− x

= −

∂f

∂y

= y

− x

So we find that

= x

+ ξ

= ξ

So we have

ϕ(x, ξ) = (x + ξ, ξ).

If we use the Euclidean inner product to identify

∗

∼

, and view

as a

vector, then ϕ is a free translation in R

That seems boring, but in fact this generalizes quite well.

Example.

Let

be a connected manifold, and

a Riemannian

metric on

. We then have the Riemannian distance

X × X → R

, and we

can define

f : X × X → R

(x, y) 7→ −

d(x, y)

Using the Riemannian metric to identify

∗

∼

T X

, we can check that

T X → T X is given by

ϕ(x, v) = (γ(1), ˙γ(1)),

where

is the geodesic with

(0) =

and

˙γ

(0) =

. Thus,

is the geodesic

flow.