Symplectic Topology

Lecture Notes

ZWEISTEIN

Desu-Cartes

1

Contents

1 Analytic and Geometric Prerequisites 3

1.1 Diﬀerential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 The Calculus of Diﬀerential Forms . . . . . . . . . . . . . . . . . 7

1.3 Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Symplectic Structures 12

2.1 Symplectic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Symplectic Embeddings . . . . . . . . . . . . . . . . . . . . . . . 14

2

1 Analytic and Geometric Prerequisites

The goal of this lecture is to discuss symplectic embeddings. To do this, we do

quite some preparatory work.

1.1 Diﬀerential Forms

We quickly review the calculus of diﬀerential forms here. We will mainly work in

R

n

here although the formalism can be generalized to arbitrary smooth manifolds.

Accordingly, we start this section with a quick review of multilinear algebra.

Definition 1.1 (Dual basis). Let

{e

1

, . . . , e

n

}

denote the standard basis of

R

n

.

The algebraic dual space of

R

n

is the vector space of linear forms

α

:

R

n

→ R

. We

will usually denote it (

R

n

)

∗

. The dual basis of (

R

n

)

∗

usually written

{e

∗

1

, . . . , e

∗

n

}

is deﬁned to be such that

e

∗

i

(

e

j

) =

δ

ij

. In other words,

e

∗

i

takes a vector and

spits out its i-th coeﬃcient.

Definition 1.2. We say that a multilinear map

α

:

R

n

× · · · × R

n

is skew-

symmetric if

α(v

1

, . . . , v

i

, v

j

, . . . , v

n

) = −α(v

1

, . . . , v

j

, v

i

, . . . , v

n

).

In other words, exchanging elements ﬂips the sign.

Ideally, we want to put together multiple basis elements of (

R

n

)

∗

. We will do

so by considering elements of the form

e

∗

1

∧ e

∗

2

∧ · · · ∧ e

∗

n

. More speciﬁcally, we set

Definition 1.3 (Wedge product). Let

{e

∗

1

, . . . , e

∗

n

}

denote the dual basis of

(

R

n

)

∗

. Let

I

= (

i

1

, . . . , i

k

) be a subset of 1

, . . . , n

. The wedge product of two

bases elements is deﬁned as

e

∗

i

∧ e

∗

j

(v

1

, v

2

) := det

e

i

(v

1

) e

i

(v

2

)

e

j

(v

1

) e

j

(v

2

)

More generally, we get

e

∗

i

1

∧ · · · ∧ e

∗

i

k

(v

1

, . . . , v

k

) = det

e

∗

i

1

(v

1

) · · · e

∗

i

1

(v

k

)

.

.

.

.

.

.

e

∗

i

k

(v

1

) · · · e

∗

i

k

(v

k

)

In the case where

k

= 2, this has the very nice geometric interpretation of

sending two vectors to the signed area of the parallelogram delimited by the

vectors. All of this allows us to deﬁne the vector space that will interest us.

Definition 1.4 (Exterior algebra). The

k

-th exterior algebra of (

R

n

)

∗

is the

vector space of all skew-symmetric multilinear forms from

R

n

to

R

equipped

with the wedge product. We usually denote it

∧

k

(

R

n

)

∗

. The wedge product is

associative and deﬁned such that

α ∧ β

= (

−

1)

kl

β ∧ α

where

α

takes

k

vectors,

and β takes l vectors.

3

It turns out there is still one important construction we have yet to discuss.

We have discussed dual spaces but as it turns out, this is more than a simple

construction. Recall from linear algebra that linear maps are the morphisms in

the category Vect

k

of k-vector spaces.

Proposition 1.5.

The correspondence

D

:

V 7→ V

∗

is a contravariant functor

from the category Vect

k

to itself.

There are two important aspects to this proposition. The ﬁrst one is that the

correspondence is a functor. Interestingly enough, this means our maps have to be

sent somewhere as well. The other one is that the functor is contravariant. This

means that if we have a map

φ

:

V → W

, we need to ﬁnd a map

φ

∗

:

W

∗

→ V

∗

.

Luckily, there is a very obvious candidate: just take the transpose. Indeed, this is

the construction we’re looking for. Our functor thus does the following. It takes

a vector space

V

and sends it to

V

∗

, and takes a map

φ

:

V → W

and sends it

to

φ

∗

:

W

∗

→ V

∗

. This is but one example of duality in category theory. Notice

that applying our functor twice to

V

gives us

V

∗∗

which is canonically isomorphic

to

V

because our vector spaces are ﬁnite-dimensional. More concretely, we have

that (

φ

∗

`

) =

`

(

φ

) so it acts as pre-composition in a sense. We want to emulate

this for our vector space ∧

k

(R

n

)

∗

.

Definition 1.6 (Pull-back). Let

φ

:

R

n

→ R

m

be a linear map and

α ∈ ∧

k

(

R

n

)

∗

.

We deﬁne the pull-back of φ to be

φ

∗

(α)[v

1

, . . . , v

k

] = α (φ(v

1

), . . . , φ(v

k

)) .

We are now ready to deﬁne diﬀerential forms. We have restricted our attention

to

R

n

but the general theory is not much harder. For our purposes, it will be

enough to consider our manifolds as embedded in real space which explains the

choice of sticking with R

n

.

Definition 1.7 (Diﬀerential k-form). A diﬀerential k-form is a smooth map

α : R

n

→ ∧

k

(R

n

)

∗

.

We will usually write α

p

to mean α applied at the point p ∈ R

n

.

The set of all diﬀerential

k

-forms on

R

n

is denoted by Ω

k

(

R

n

). This is a real

inﬁnite-dimensional vector space. We also set Ω

0

(

R

n

) :=

C

∞

(

R

n

), the space of

smooth functions on R

n

.

Remark 1.8. Unpacking the deﬁnition, we’re saying that a diﬀerential

k

-form

takes

k

vectors in

R

n

and spits out a number in

R

. Furthermore, the form should

also be multilinear, skew-symmetric and smooth. This deﬁnition is problematic

however, because it’s not so clear what is meant for such a map to be smooth.

One way to understand it is via vector ﬁelds.

Definition 1.9 (Vector ﬁeld). Let

M

be a smooth manifold. A vector ﬁeld

X

on

M

is a smooth section of the tangent bundle of

M

. In other words, it is a

map that takes a point

p

and sends it to a vector in

T

p

M

, where

T

p

M

denotes

the tangent space of M at p.

4

Example 1.10. Let M = R

n

. A vector ﬁeld X = (X

1

, . . . , X

n

) takes a point p

and sends it to (

X

1

(

p

)

, . . . , X

n

(

p

)). Notice that here,

T

p

R

n

≡ R

n

. Eﬀectively,

this means we’re taking a point to a vector in its tangent space, but in

R

n

they

are the same thing. To simplify notation, we write X

i

(p) as (X

i

)

p

.

As we have seen, our diﬀerential

k

-forms can eat vectors and spit out numbers.

As vector ﬁelds take a point and spit out vectors, we can instead apply diﬀerential

k

-forms to vector ﬁelds. This gives us the following deﬁnition of smooth: A

k-form α is smooth if the function

p 7−→ α

p

((X

1

)

p

, . . . , (X

n

)

p

)

is smooth for all smooth vector ﬁelds. This resolves the issue and insures our

deﬁnition is well-posed. Lastly, we discuss bases. As we have said before, Ω

k

(

R

n

)

is an inﬁnite-dimensional vector space. When restricting to a speciﬁc tangent

space T

p

R

n

however, we can deﬁne a basis for our diﬀerential k-forms.

Definition 1.11. Let

I

= (

i

1

, . . . , i

k

) be a subset of

{

1

, . . . , n}

. We deﬁne the

basis elements of Ω

k

(R

n

) at p to be

(dx

i

1

∧ · · · ∧ dx

i

k

)

p

: T

p

R

n

× · · · × T

p

R

n

→ R

n

.

They can be deﬁned using the basis elements of (T

p

R

n

)

∗

, namely:

(dx

i

1

∧ · · · ∧ dx

i

k

)

p

(v

1

, . . . , v

k

) = det

e

∗

i

1

(v

1

) · · · e

∗

i

1

(v

k

)

.

.

.

.

.

.

e

∗

i

k

(v

1

) · · · e

∗

i

k

(v

k

)

In a sense, we deﬁned a global basis that restricts to our usual dual basis

when zooming in on a speciﬁc tangent space. If we look closely, we will notice

that

dx

i

is just the projection on the

i

-th coordinate which is exactly what we

wanted. For instance, if

X

= (

X

1

, X

2

) is a vector ﬁeld, we have

dx

1

(

X

) =

X

1

.

Another detail we need to mention is that coeﬃcients in Ω

k

(

R

n

)

∗

are not numbers

anymore, but smooth functions. The wedge product also naturally transfers and

we have

(α ∧ β)

p

= α

p

∧ β

p

.

Remark 1.12. At this point, you might be scratching your head. Why are we

doing this exactly? You might have noticed that our basis elements look awfully

familiar to the usual inﬁnitesimals from calculus. Indeed,

dx

looks just like the

element used to perform integration. We do write

R

b

a

f

(

x

)

dx

after all. This isn’t

a coincidence.

If you have taken a course on diﬀerential geometry, you know that you can

associate to every function

f

:

R

n

→ R

a linear map called the diﬀerential

df

p

(surprise, surprise) at a point

p

. This diﬀerential encodes the information of all

the directional derivatives at p. Indeed, we have

df

p

(v) = lim

h→0

f(p + hv) − f(p)

h

.

5

This map varies smoothly with the base point and is obviously linear, which

means that the map

df : R

n

→ ∧

1

(R

n

)

∗

, p 7→ df

p

is a diﬀerential 1-form. Accordingly, we can decompose it in the basis

dx

1

, . . . , dx

n

.

We know from calculus that

df

p

(v) = h∇f(p), vi =

∂f

∂x

1

(p)v + · · · +

∂f

∂x

n

v.

Eﬀectively, this means that

df =

∂f

∂x

1

dx

1

+ · · · +

∂f

∂x

n

dx

n

.

Exercise 1.13. Are all 1-forms diﬀerentials of functions? Set

α

:=

α

1

(

x, y

)

dx

+

α

2

(

x, y

)

dy

and investigate. (Remember, the sum of diﬀerential

k

-forms is also a

diﬀerential k-form as they form vector spaces).

In the linear case, we have deﬁned the pullback of a function. We can extend

this deﬁnition in the case of diﬀerential forms. As is common in diﬀerential

geometry, we go from the smooth case to the linear case by diﬀerentiating. We

denote the diﬀerential of a map

φ

:

R

m

→ R

n

(sometimes called the pushforward)

at a point

p

as

Dφ

(

p

). In the case of real vector spaces, this is just the Jacobian.

Definition 1.14. Let

α ∈

Ω

k

(

R

m

) be a diﬀerential

k

-form and

φ

:

R

n

→ R

m

a

smooth map. We deﬁne the pull-back of the form α via φ to be

(φ

∗

α)

p

(X

1

, . . . , X

k

) = α

φ(p)

(Dφ(p)X

1

, . . . , Dφ(p)X

k

) .

This gives us a new diﬀerential

k

-form. Notice that the point goes from

p

to

φ(p) because Dφ(p) : T

p

R

n

→ T

φ

(p)R

m

.

Exercise 1.15. Show that φ

∗

(α ∧ β) = φ

∗

α ∧ φ

∗

β.

Example 1.16. At this point, an example would probably be a good idea. Let

φ be the polar coordinate transformation, that sends

r

ϑ

7−→

r cos ϑ = x

r sin ϑ = y

. It is a classical fact from vector analysis that its diﬀerential is the matrix

Dφ(r, ϑ) =

cos ϑ −r sin ϑ

sin ϑ r cos ϑ

Using the formula where we write φ

∗

instead of Dφ for brevity, we get

φ

∗

(dx ∧ dy) = det φ

∗

(r, ϑ) · (dr ∧ dϑ)

= rdr ∧ dϑ

6

which should look familiar if you’ve ever used the change of variable formula from

vector analysis. In a sense, we’ve said that

dxdy

transform into

rdrdϑ

. This is

actually completely true formally, but since we haven’t deﬁned integration yet,

we can’t really talk about it. Another less tedious way of solving this would’ve

been to use the fact that the wedge product and the pull-back commute.

Exercise 1.17. Show the same result using the fact that

φ

∗

(

dx ∧ dy

) =

φ

∗

dx ∧

φ

∗

dy. Hint: Remember that dx means the projection on the x-coordinate.

1.2 The Calculus of Diﬀerential Forms

We have now ported every tool we had in the linear case over to the smooth

case. It is time to see how the smooth case changes. The most obvious diﬀerence

is that we have ways of diﬀerentiating

k

-diﬀerential forms. The insight to be

gained here is that diﬀerential calculus measures how our function changes when

we change our base point. This is why it makes no sense in the linear context,

as there is no dependence on a base point. We start with the exterior derivative,

that we construct from scratch. We have seen above that to each smooth function

f

, we can associate a 1-form

df

. This construction is important and we would

like to generalize it. For 0-forms (smooth functions), we deﬁne this to be the

exterior derivative, that is to say,

df :=

n

X

i=1

∂f

∂x

i

dx

i

Since we know how to diﬀerentiate smooth functions, and those are the coeﬃcients

of forms of higher degree (the degree being

k

for a

k

-diﬀerential form), we have

a very simple way of deﬁning d for k-diﬀerential forms.

Definition 1.18 (Exterior derivative). Let

α

=

P

I

f

I

dx

I

be a

k

-diﬀerential

form where

|I|

=

k

and

dx

I

=

dx

i

1

∧ · · · ∧ dx

i

k

. We deﬁne the exterior derivative

to be the map d : Ω

k

(R

n

) → Ω

k+1

(R

n

) that sends

X

I

f

I

dx

I

7−→

X

I

df

I

∧ dx

I

.

Example 1.19. Let

α

=

f

(

x, y, z

)

dx ∧ dy ∧ dz

. Then

dα

=

df ∧ dx ∧ dy ∧ dz

.

For instance, if

f

(

x

) is the identity function

f

(

x

) =

x

, then

df

(

x

) =

dx

, our

usual projection.

This derivative enjoys a few nice properties.

Proposition 1.20. The exterior derivative satisﬁes the following properties:

1.

We have that

d

2

=

d ◦ d

= 0. This means that applying the exterior

derivative to a form twice gives us zero.

2.

The exterior derivative is compatible with the wedge product:

d

(

α ∧ β

) =

dα ∧ β + (−1)

k

α ∧ dβ, for a k-form α and an `-form β.

7

3.

The exterior derivative commutes with the pull-back: If

φ

:

R

n

→ R

m

is a

smooth map and α ∈ Ω

k

(R

m

), then dφ

∗

α = φ

∗

dα.

Definition 1.21. We say that a diﬀerential

k

-form

α

is closed if

dα

= 0. If

furthermore

α

=

df

for some smooth function

f

which we call a primitive, we

say that α is exact.

Notice that since

d

2

= 0, we have that exact implies closed. Understanding

when closed implies exact is the impetus behind de Rham cohomology.

Exercise 1.22. Show that the 2-form

η

(x,y)

=

xdy−ydx

x

2

+y

2

is closed, but not exact.

Hint: Try pulling it back using polar coordinates like we did above. Can you

tell what this form represents?

An important point is that our way of diﬀerentiating diﬀerential

k

-forms

yields (

k

+ 1)-forms. Is there a way to diﬀerentiate forms in a way that preserves

the degree? The answer is yes, but it is of a completely diﬀerent nature. We will

generalize the idea behind the directional derivative of vector calculus. First, we

need to discuss ﬂow.

Definition 1.23. Let

X

be a set. We deﬁne a ﬂow on

X

to be an action of the

group (

R,

+) on

X

. This is a mapping

ϕ

t

(

x

) :

R × X → X

such that

φ

0

(

x

) =

x

and φ

s

◦ φ

t

= φ

s+t

.

As it turns out, we can always associate a ﬂow locally to a vector ﬁeld X.

Proposition 1.24.

Let

X

:

T R

n

→ R

n

be a vector ﬁeld. We can always

associate locally a ﬂow to X, that is to say, there exists a map

ϕ : (−ε, ε) × R

n

→ R

n

, (t, x) 7→ ϕ

t

(x)

such that

d

dt

t=0

ϕ

t

(x) = X(x), ϕ

0

(x) = x.

It is with this construction that we can deﬁne our other derivative.

Definition 1.25 (Lie derivative). For

X

any vector ﬁeld, we deﬁne

L

X

:

Ω

k

(R

n

) → Ω

k

(R

n

), the Lie derivative in the direction of X by setting

L

X

α =

d

dt

t=0

ϕ

∗

t

α.

While this formula is perfectly okay that way, there is a simpler way to

compute the Lie derivative.

Definition 1.26 (Interior product). Let

X

be a vector ﬁeld on

R

n

and

α

a

k

-diﬀerential form. The interior product is the map

ι

X

that sends

α

to the

k − 1-form deﬁned by the property that

(ι

X

α)(X

1

, . . . , X

n

) = α(X, X

1

, . . . , X

n

).

8

What this does is contract the diﬀerential form

α

and the vector ﬁeld

X

.

This eﬀectively ﬁlls the ﬁrst spot in the

k

-diﬀerential form

α

with

X

. Using this

product, we arrive at a really surprising and interesting result.

Theorem 1.27 (Cartan’s magic formula). We have

L

X

= ι

X

◦ d + d ◦ ι

X

.

Proof.

First, it is easy to see that this holds for functions and exact one-forms.

Indeed,

L

X

(

f

) =

ι

X

df

and

d

(

df

) = 0. Now notice that

k

-diﬀerential forms are

nothing but sums of wedges of exact 1-forms multiplied by smooth functions.

If we can show that the formula is compatible with the wedge product, we’re

done. To this end, deﬁne

D

X

:=

ι

X

◦ d

+

d ◦ ι

X

. First, it is clear that

L

X

(

α ∧ β

) =

L

X

(

α

)

∧ β

+

α ∧ L

X

(

β

) from the deﬁnition of the Lie derivative

and the fact that the wedge product is bilinear. Now notice that our operator

D

X

satisﬁes the same property. Suppose that

α

is a

k

-diﬀerential form. This

holds true because

d

(

α ∧ β

) = (

dα

)

∧ β

+ (

−

1)

k

α ∧

(

dβ

) and we also have

ι

X

(

α∧β

) = (

ι

X

α

)

∧β

+(

−

1)

k

α∧

(

ι

X

β

). This in turn shows that (

L

X

−D

X

)

α

= 0

by decomposing in elementary summands of the form

fdx

i

1

∧ · · · ∧ dx

i

k

. This

completes the proof.

Having gone over ways to diﬀerentiate

k

-diﬀerential forms, we now look for

ways to integrate them. We already know how to integrate smooth functions,

but in a sense, we even know a little more. Let

γ

be a smooth curve in

R

n

. We

have seen in calculus how to integrate over such curves. There is a very natural

way to generalize this. As a small disclaimer, we’ll assume in this chapter that

all our forms are have compact support to avoid any technicalities.

Definition 1.28 (Integration of 1-forms). Let

γ

:

[a, b] → R

n

be a smooth curve

in R

n

and α ∈ Ω

1

(R

n

). We deﬁne the integral of α along γ to be

Z

γ

α :=

Z

b

a

α

γ(t)

˙

γ(t)

dt.

The fundamental theorem of calculus says that the integral of an exact 1-form

df is equal to the diﬀerence of fs evaluated at its end points.

Exercise 1.29. What does the integral of the one form

y dx

represent? Does it

look familiar? What’s diﬀerent?

Suppose now that

α

is a

k

-dimensional form. It is important to realize that

we will only be able to integrate

α

over

k

-dimensional submanifolds. The simplest

case is that of an open set.

Definition 1.30 (Integration of

k

-dimensional forms). Let

U ⊆ R

k

be an open

set and let

α

=

f

(

x

1

, . . . , x

k

)

dx

1

∧· · · ∧ dx

k

be a

k

-diﬀerential form. The integral

of α is the oriented usual integral in R

k

, that is to say:

Z

U

α :=

Z

U

f(x

1

, . . . , x

k

) dx

1

, . . . , dx

k

.

9

The only real diﬀerence is that by skew-symmetry, exchanging two of the

dx

i

ﬂips the sign of the integral. For instance,

Z

U

g(x, y)dx ∧ dy = −

Z

U

g(x, y)dy ∧ dx.

Extending the integral to submanifolds is not hard now that we know how

to do this.

Definition 1.31 (Integration over submanifolds). Let

ϕ

U

:

U → V ⊆ M

be

a chart for a

k

-dimensional submanifold

M

. Let

α ∈

Ω

k

(

U

). We deﬁne the

integral over V to be

Z

V

α =

Z

U

ϕ

∗

V

α.

Notice how we have transferred the problem of integrating in a manifold to

a simple integral in

R

k

via the pull-back of the form. Chaining those integrals

together via a partition of unity lets us integrate over the entire manifold

M

.

Lastly, we wish to discuss the cornerstone of vector calculus, the celebrated

theorem of Stokes. Most so-called fundamental theorems in calculus such as the

FTC, Green-Riemann and so on are just applications of this theorem. Here it is

in its full glory.

Theorem 1.32.

Let (

M, ∂M

) be an oriented (

k −

1)-dimensional manifold with

boundary and let α ∈ Ω

(k−1)

(M) (with compact support). Then,

Z

M

α =

Z

∂M

dα.

Exercise 1.33. Derive the fundamental theorem of calculus from this formula.

Hint: Take

α

=

f

(

x

) and

M

=

[a, b]

. What about the divergence theorem and

Green-Riemann?

1.3 Cohomology

In a ﬁrst course on homology, we usually discuss multiple ﬂavors of homology

functors. Usually, we discuss (semi-)simplicial homology ﬁrst and extend our

simplices to get singular homology. Another particular kind of homology that’s

very nice for computations is cellular homology. We have a few ways of computing

actual examples:

1. H

cell

∗

(X), the cellular homology groups,

2. The long exact sequence,

3. The Mayer-Vietoris sequence,

4. K¨unneth’s Formula.

10

Perhaps the reader is not familiar with the last one which states that

H

k

(

X ×

Y, K

) =

P

i+j=k

H

i

(

X, K

)

⊗ H

j

(

Y, K

), where

K

is a ﬁeld. The usual homology

construction uses chains. Cohomology is the dual notion: what happens if we

consider cochains instead? It is often the case that cohomology is a more natural

tool that homology. Indeed, cohomology has a few advantages to homology that

we will discuss now. Suppose we have a chain:

· · · C

k−1

C

k

C

k+1

· · ·

How do we make the arrows go the other direction? Well one idea is to apply

a Hom functor. Indeed, we have

· · · C

k−1

C

k

C

k+1

· · ·

where C

k

:= Hom(C

k

, R) where R is a ring.

Exercise 1.34. Show that if R = K a ﬁeld, then H

i

(X, K) ≡ H

i

(X, K).

We write

d

i

for the induced homomorphism between Hom-sets. One important

aspect of (singular) cohomology is that there is a striking link to analysis. As

we have seen earlier, the exterior derivative squares to zero. This means that we

can deﬁne a cohomology on the spaces of diﬀerential forms.

Definition 1.35 (De Rham cohomology). Let

d

k

: Ω

k

(

M

)

→

Ω

k+1

(

M

) denote

the exterior derivative. The cochain complex

· · · Ω

k−1

Ω

k

Ω

k+1

· · ·

d

k−1

d

k

induces cohomology groups H

k

dR

(M) := ker(d

i

)/im(d

i−1

).

We usually denote the cycle corresponding to

ω

by

[ω]

. So what do those

equivalence classes look like? We say that two forms

α

and

β

in Ω

k

(

M

) are

cohomologous if they diﬀer by an exact form, that is to say,

α − β

is exact. This

is all good and well, but this is just another cohomological theory, what does it

have to do with singular cohomology? The key theorem is that of Georges de

Rham.

Theorem 1.36

(De Rham’s Theorem)

.

Let

M

be a smooth manifold. Then the

map

Ψ

∗

: H

k

dR

(M) → H

k

(M; R)

[ω] 7−→

[c] 7→

Z

c

ω

is an isomorphic of vector spaces.

Note here that the cycle

[c]

is in

H

k

, not in

H

k

dR

. This theorem provides

us with analytic tools to understand the topology of manifolds. Interestingly

enough, this also shows that the diﬀerential forms depend on the manifold

M

,

but not on its smooth structure. This is one clear advantage of working with

cohomology.

11

2 Symplectic Structures

2.1 Symplectic Geometry

We ﬁnally start our study of symplectic geometry in earnest. Now that we

are familiar with diﬀerential forms and how to work with them, we can ﬁnally

deﬁne what symplectic geometry is about. Following the astonishing work of

Felix Klein, modern geometry is commonly understood as studying a space and

a group of transformations associated with it. In Riemannian Geometry for

instance, we study smooth manifolds equipped with a Riemannian metric. It is

not so much the space that is interesting but the structure on that space, as is

often the case in mathematics as a whole. The question then becomes, what is

the inherent structure on a symplectic manifold? As it turns out, it’s a special

kind of 2-form.

Definition 2.1 (Symplectic form on

M

). A symplectic form on

M

is a closed

non-degenerate 2-form ω.

What do we mean by non-denegerate exactly? From the 2-form, we can

induce a bilinear pairing on the tangent spaces

T

p

M

. The form being non-

degenerate means that if

ω

(

v, w

) = 0 for all

w ∈ T

p

M

, then

v

= 0. Notice that

we need our manifold to be even-dimensional here for this to make sense. Let’s

look at the simplest example there is, R

2n

.

Example 2.2. Let

M

=

R

2n

. This is a symplectic manifold (usually called a

symplectic vector space). The canonical symplectic form is

ω

=

P

i=1

dx

i

∧ dy

i

where

R

2n

= (

x

1

, . . . , x

n

, y

1

, . . . , y

n

). Remember that

dx

i

=

e

∗

i

, so applying our

standard form is akin to taking multiple determinants.

Having deﬁned what symplectic spaces look like, it is only natural to ask

what the morphisms look like in this category. The structure is the symplectic

form, so a morphism should preserve this.

Definition 2.3. Suppose (

M, ω

) and (

N, ω

0

) are two symplectic manifolds. A

symplectomorphism is a smooth map ϕ : M → N such that

ϕ

∗

ω = ω

0

If

M

and

N

are nothing but

R

2n

, we usually say the symplectomorphisms

are linear. Those linear symplectomorphisms form a group that we denote

Sp

(2

n

). In dimension 1, this is nothing but the volume-preserving maps, but in

higher dimensions, symplectomorphisms have more structure. Indeed, any linear

symplectomorphism is volume-preserving.

Exercise 2.4. Show that if

ϕ ∈ Sp

(2

n

), then

det ϕ

= 1. Conclude that any

linear symplectomorphism preserves volume.

Historically, these notions came out of classical mechanics. More speciﬁcally,

the key equation in classical mechanics is Newton’s Law.

12

Principle 2.5 (Newton’s Law).

F = m¨x.

It is not immediate what this equation has to do with our symplectic manifolds.

And indeed, it doesn’t really look like geometry at this point. It turns out that

Lagrange (and especially Hamilton) managed to give another equation (equivalent

to Newton’s) that also governs classical mechanics. Instead of considering the

object’s position function, we instead focus our attention on the energy of the

system. More formally, let us deﬁne a new set of coordinates that we will

call canonical coordinates. We let

r

= (

p, q

) where

q

is the vector of so-called

generalized coordinates, and

p

is the vector of conjugate momenta. What

matters is that, as you can see, this is even-dimensional. This is because if

p

= (

p

1

, p

2

, p

3

) (say), then

q

= (

q

1

, q

2

, q

3

). The Hamiltonian function is then

deﬁned as

H

=

T

+

V

where

T

is potential energy and

V

is kinetic energy.

It turns out that the phase space of our coordinates is a symplectic manifold.

Locally, the form is given as

ω

=

P

n

i=1

dp

i

∧ dq

i

. We can now reformulate

Newton’s equations in a set of two equations.

Proposition 2.6. Newton’s Law is equivalent to the following:

(

˙x(t) =

∂H

t

∂y

(x(t), y(t))

˙y(t) = −

∂H

t

∂x

(x(t), y(t))

where H

t

(x(t), y(t)) =

1

2

kyk

2

− V

t

(x).

The ﬁrst term is kinetic energy, the second is potential energy. And now we

can kind of see why symplectic geometry comes in. Notice how the second term

has a negative sign. This is key. We can now deﬁne a vector ﬁeld on

R

2n

in the

following way.

Proposition 2.7.

Let

ω

0

be the standard symplectic form on

R

2n

. We deﬁne

the vector ﬁeld X

H

t

to be the one that satisﬁes the following equation:

ω

0

(X

H

t

(z), ·) = dH

t

(z)(·).

More concretely, we have

X

H

t

(z) =

∂H

∂y

(z), −

∂H

∂x

(z)

.

We have

˙z

(

t

) =

X

H

t

(

z

(

t

)) which gives us a ﬂow

ϕ

t

H

. The important fact is

now to notice that this ﬂow is symplectic.

Proposition 2.8. The ﬂow ϕ

t

H

is symplectic.

13

Proof.

We want to show that (

ϕ

t

H

)

∗

ω

0

=

ω

0

. Since

ϕ

0

H

is the identity, it is

enough to show that

d

dt

(ϕ

t

H

)

∗

ω

0

= 0. Using Cartan’s formula, we get

d

dt

(ϕ

t

H

)

∗

ω

0

= (ϕ

t

H

)

∗

L

X

H

ω

0

= (ϕ

t

H

)

∗

(dι

X

H

+ ι

X

H

d)ω

0

= (ϕ

t

H

)

∗

(d(ω(X

H

, ·)) + ι

X

H

0)

= (ϕ

t

H

)

∗

(d(dH(·)))

= 0.

where dω

0

= 0 since the form is closed.

Exercise 2.9. Show Liouville’s Theorem:

Vol

(

ϕ

t

H

(

U

)) =

Vol

(

U

) with

U

in

phase space.

2.2 Symplectic Embeddings

So far, we have only discussed the geometry of symplectic manifolds. It turns

out topological properties of symplectic manifolds are also quite interesting.

The problem of symplectic embeddings give us a way to understand how rigid

symplectic structures actually are. Interestingly, the ﬁrst fundamental result

will require us to step into the world of complex geometry.

Definition 2.10 (Almost complex structures). Let

M

be a smooth manifold.

An almost complex structure on

M

is a smooth vector bundle endomorphism

J

:

T M → T M

such that

J

2

=

−Id

. Eﬀectively, this means a family of linear

maps J

p

: T

p

M → T

p

M such that J

p

◦ J

p

= −Id

T

p

M

.

The idea will then be to equip our symplectic manifold (

M, ω

) with an almost

complex structure

J

to show interesting things about

M

. We ﬁrst need to settle

the question of existence. We assume our symplectic manifold also possesses a

Riemannian metric g.

Proposition 2.11.

Let (

M, ω, g

) be a symplectic manifold. We set

J

(

M

) :=

{J almost complex structure on M }

. Then,

J

(

M

) is non-empty and contractible.

Proof.

Let

g

(

M

) denote all the metrics we can put on

M

. By convexity,

g

(

M

)

is contractible. We can always construct our almost-complex structure from our

symplectic form and our metric. Indeed, we have

J

(

u

) = (

ι

u

(

g

))

−1

(

ι

u

(

ω

(

u

)).

Since

ω → ω

is the identity, and

g 7→ g

0

is a homeomorphism, we indeed have

that J (M) is contractible.

Definition 2.12 (

ω

-tame almost complex structures). Let (

M, ω, J

) be a sym-

plectic manifold with an almost complex structure

J

. We say that

J

is

ω

-tame if

ω

(

u, Ju

)

>

0 for all

u 6

= 0. The family of all

ω

-tame almost complex structures

is denoted by J

ω

(M).

14

This lets us deﬁne the key notion that will help us prove our principal

theorem.

Definition 2.13 (

J

-holomorphic curves). Let (

M, ω, J

) be a symplectic manifold

with almost complex structure

J

. A curve

u

is said to be

J

-holomorphic if we

have

J ◦ d

u

= d

u

◦ i

where i is the canonical complex structure on C.

We are now ready to outline the most fundamental theorem in the theory.

Theorem 2.14

(Gromov’s Nonsqueezing Theorem)

.

If

B

2n

(

a

) embeds symplec-

tically in

Z

2n

(

A

), then

a ≤ A

. Here

Z

2n

(

A

) means the 2

n

-dimensional cylinder

whose disk section has area A.

The proof involves moduli spaces and cohomology. We sketch it here in case

the reader is suﬃciently proﬁcient with said material to understand it and ﬁll in

the gaps.

Proof.

Suppose ﬁrst that our symplectic embedding

ϕ

also preserves the standard

complex structure on

C

n

, that is

J

0

=

i ⊕ i ⊕ i ⊕ · · · ⊕ i

. First, notice that

Z

2n

=

D

(

A

)

× C

n−1

. Let

D

z

0

(

A

) =

D

(

A

)

× {z

0

}

be the disc that contains

ϕ

(0).

Set

S

:=

ϕ

−1

ϕ(B

2n

(a)) ∩ D

z

0

(A)

is a 2-dimensional complex submanifold

passing through the origin. This means that the area of

S

is greater or equal to

a by Lelong’s inequality. Since J is complex, we have that

a ≤ area

J

0

(S)

= area

ω

0

(S)

=

Z

S

ω

0

=

Z

S

ϕ

∗

ω

0

=

Z

ϕ(S)

ω

0

≤

Z

D

z

0

(A)

ω

0

= A.

The third equality holds because

ϕ

∗

ω

0

=

ω

0

(the embedding is symplectic), the

fourth one is just the change of variable formula, and the ﬁfth one comes from the

monotonicity of the integral. This proof is slightly absurd. If

ϕ

preserves both

the symplectic form and the complex structure, it preserves the metric. Here,

the metric is Euclidean which means the embedding is a translation followed by

a rotation, which makes the entire proof trivial. Our goal will now be to mimick

this idea using the contractibility of J (M).

Suppose then that

ϕ

is only symplectic. Let

J

ϕ

be an almost complex structure

15

on

Z

2n

(

A

) such that

J

ϕ

=

ϕ

∗

J

0

. Here,

ϕ

∗

J

0

means

dϕ ◦ J

0

◦ dϕ

−1

. We can ﬁnd

such a

J

ϕ

because both

J

0

and

ϕ

∗

J

0

are

ω

-tame. Indeed, a simple calculation

shows that

ω

0

(

ϕ

∗

J

0

u, u

) =

ω

0

J

0

◦ dϕ

−1

(·)u, dϕ

−1

(·)u

>

0. We’re now looking

for a

J

ϕ

-holomorphic disk in

Z

2n

(

A

) passing through

ϕ

(0) with boundary on

the boundary of

Z

2n

(

A

) with

ω

0

non-negative everywhere along the disc. If such

a disk exists, we can repeat the argument above. To simplify matters, we start

by compactifying the disk. Take

A

0

> A

and compactify

D

(

A

) into

S

2

(

A

0

). We

can endow

M

:=

S

2

(

A

0

)

× C

n−1

with the product symplectic form

ω

=

ω

S

2

⊕ ω

0

because there always exists a symplectic form on

S

2

. We thus need to show the

following lemma:

Lemma 2.15. The moduli space

M(J

0

) := {u :

S

2

, i

→ C

n−1

× S

2

(A

0

) | ∂

J

0

u = 0, u(N ) = ϕ(0) and

u(S

2

)

=

{∗} × S

2

(A

0

)

/P SL(2, C)}

for J

0

∈ J

ω

(M) is not empty. Here N is the North pole on S

2

(A

0

).

Proof.

First, notice that

M

(

J

0

) =

{u

(

x

) = (

∗, x

)

}

. There’s only one solution

because for all

u J

0

-holomorphic, by the maximum principle we can we can

always squeeze restrict

u

to be constant. Choose a path from

J

0

to

J

, with

J

t

,

t ∈ [0, 1]

,

J

t

∈ J

ω

(

C

n−1

× S

2

(

A

0

). Let

M

:=

`

0≤t≤1

{t} × M

(

J

t

)

⊂

[0, 1]×C

∞

(

S

2

, C

n−1

×S

2

)

/reparametrization

. If we can show that

M

is compact,

we are done because then

M

(

J

) is non-empty. Suppose instead that

M

t

becomes empty at

t

∗

. We have a sequence

{t

∗

k

} → t

∗

<

1. For all

k

, we

have

u

k

(

S

2

) :=

S

k

⊂ U

(

J

t

k

). Gromov’s compactness theorem tells us that,

after passing to a subsequence, the spheres

S

k

converge in a suitable sense to

a ﬁnite union of

J

t

∗

-holomorphic spheres

S

1

, . . . , S

m

whose homology classes

[C

i

]

add up to

[C]

=

S

2

(A

0

) × {∗}

. But now we’re done, because each sphere

is

J

t

∗

-holomorphic and thus 0

<

R

S

i

ω

=

[ω] C

i

. This means that

m

= 1 and

n

1

= 1, because

C

i

=

n

i

C

i

in

H

2

(

M

;

Z

) =

Z

with

n

1

≥

1, and

P

i

= 1

m

n

i

= 1.

This in turn shows that M

t

∗

is not empty.

The rest of the proof is easy. Let

u

be such a

J

ϕ

-holomorphic disk in

M

(

J

ϕ

).

We then have that

u

is

J

ϕ

-holomorphic and thus setting

S

:=

ϕ

−1

ϕ(B

2n

(a) ∩ u(S

2

)

,

we get a ≤ A using the Lelong inequality.

16