 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
k
(
R
n
). This is a real
inﬁnite-dimensional vector space. We also set
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,
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
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
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
h0
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
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 :
k
(R
n
)
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
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)
=
xdyydx
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
:
k
(R
n
)
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
(
αβ
) = (
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
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
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
(k1)
(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
k1
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
k1
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
:
k
(
M
)
k+1
(
M
) denote
the exterior derivative. The cochain complex
· · ·
k1
k
k+1
· · ·
d
k1
d
k
induces cohomology groups H
k
dR
(M) := ker(d
i
)/im(d
i1
).
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
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
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
n1
. 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
n1
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
n1
× 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
n1
× S
2
(
A
0
). Let
M
:=
`
0t1
{t} × M
(
J
t
)
[0, 1]×C
(
S
2
, C
n1
×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
]
[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 