Basic course in symplectic geometry:

A symplectic manifold $(M,\omega)$ is a manifold $M$ along with a closed
2-form $\omega$ which is everywhere nondegenerate (i.e., 
$\omega_x:T_x M \times T_x M \to \R$ is an everywhere nondegenerate
skew-symmetric pairing); thus $\omega$ induces fiberwise isomorphism
between $T_x M$ and $T_x^* M$.

Ex: If $M$ is an oriented surface, $\omega \in \Omega^2(M)$ everywhere
positive.

If $M$ is Kahler, i.e., we have $(M,g,J)$, $J:TM \to TM$, $J^2=-1$,
$J$ is {\em integrable} (i.e., local complex coordinates exist in which $J$
is the standard complex structure on $\C^n$) and $g$ is a Riemannian
metric s.t. $g(JX,JY)=g(X,Y)$ and $\nabla J=0$ (where $\nabla$ is the
Levi-Civita connection).

Then $\omega(X,Y)=g(JX,Y)$ is a symplectic form on $M$.

Why is this nondegenerate? $g$ is nondegenerate, $J$ is an isomorphism...

Why is it skew-symmetric? Follows from $J$ being an isometry...

Why is it closed? It's covariantly constant because $g$ and $J$ are (by
definition and assumption, respectively); all covariantly constant things
are closed by basic differential geometry.

If $M \subset \C^n$ is a smooth algebraic variety, then $(M, J_M=i, 
g=const|_M)$ is Kahler.  In fact, $\omega$ as given above is the restriction
of the standard symplectic form on $\C^n$, which is
$\frac{i}{2} \sum_k dx_k \wedge d\overline{x}_k$.

How do we prove that this is nondegenerate? (Clearly closed, as it's the
restriction of a closed form).  Note that $\omega(X,JX)=g(X,X)>0$, and
so $\omega$ is nondegenerate.  (Notice that, since $M$ is a variety, its
tangent spaces are complex, and so the above makes sense).

In fact, if $M$ is Kahler and $N \subset M$ is a complex submanifold, then $N$
is Kahler.

So we can also look at projective varieties, with the restriction of the
Fubini-Study metric.

There are many non-Kahler symplectic manifolds...

How far is an arbitrary symplectic manifold from being Kahler?

Let $(M, \omega)$ be a symplectic manifold.  Then the tangent bundle is
a symplectic vector bundle (i.e., a vector bundle with a smoothly varying
symplectic structure on each fiber).  That is, this is an $Sp(2n,\R)$-bundle.

Recall that we can put a metric on any bundle.  Why? Because arbitrary bundles
are $GL(n,\R)$-bundles, bundles with metrics are $O(n)$-bundles, and
$GL(n,\R)$ has max'l compact subgroup $O(n)$ and hence retracts to $O(n)$.

The maximal compact subgroup of $Sp(2n,\R)$ is $U(n)$; hence any symplectic
manifold admits an almost-complex structure $J$ such that $\omega(X,Y)=
\omega(JX,JY)$, $\omega(X, JX)>0$ for all $X \neq 0$.

Hence $g(X,Y)=\omega(X,JY)$ is a metric.  We call $J$ a {\em compatible
almost-complex structure}.

More concretely:

Given a point, it's easy to see that we can find a $J$ {\em at that point}
which satisfies the above condition.  Note that we have different choices;
these are parametrized by {\em Siegel upper half-space}, which is contractible.
So we can make different choices at different points, and patch things together
with partitions of unity.

Or:

Pick a Riemannian metric $g$, set $\omega(X,Y)=g(X,AY)$.  Consider
$A_x:TM_x \to TM_x$.  Then $A_x$ is skew-symmetric, invertible.  Thus
$A_x$ has eigenvalues which are purely imaginary.  If we project
all the eigenvalues to $\pm i$, we get a transformation $J$ with
$J^2=-1$.

So we can always find an almost complex structure $J$.  Moreover, the space
of all $J$ which are compatible with some fixed $\omega$ is contractible
(in particular, it's path-connected).

For instance, the Chern classes $c_k(TM,\omega) \in H^{2k}(M)$ are 
well-defined (note that these depend on the almost-complex structure).

Can also think of this as saying that, as $U(n)$ and $Sp(2n,\R)$ are homotopy-
equivalent, they have the same characteristic classes.

In particular, let $K_M=\Lambda^n_{\C}(TM,J)^*, n=\frac{\dim_\R M}{2}$.
Note that $K_M$ isn't strictly well-defined, but it's well-defined up
to isomorphism.  Then
$c_1(TM,\omega)=c_1(\Lambda^n_\C(TM,J))=-c_1(K_M) \in H^2(M)$.

So how far are we from being Kahler?

"The key word is 'almost'-complex structure".

Note that, if $J$ is $\omega$-compatible {\em and integrable}, then
$M$ is Kahler.

Philosophically, we go through algebraic geometry and figure out where we
don't need integrability.  (Of course, the answer is "almost nowhere", but
there is some stuff we can recover...)

Let $(M,\omega)$ be a symplectic manifold.  What are its automorphisms?

Let $X \in C^\infty(TM)$ be a symplectic vector field; i.e., the flow of
$X$ locally preserves $\omega$, so $L_X \omega=0$.  Since $d \omega=0$, this
is equivalent by Cartan's formula to $d(i_X \omega)=0$.

We have a correspondence between vector fields and 1-forms:
$X \mapsto -i_X \omega$; by the above, the symplectic vector fields
correspond to closed 1-forms.

"There's always tons of closed 1-forms!"

In particular, given a point and a cotangent vector at that point, we
can always find a closed 1-form with that value at that
point.  So we can always use symplectic automorphisms to move a point in
any given direction.

So: The group $\Symp(M,\omega)=Aut(M,\omega)=\{\phi(M \to M):\phi^* \omega=
\omega\}$ acts transitively on $M$ (in fact, it's $n$-transitive for any
finite $n$).

Hence $(M,\omega)$ "looks the same" everywhere, locally.

$\Symp(M,\omega)$ is an infinite-dimensional Lie group with Lie algebra
$\LSymp(M,\omega) \cong {\alpha \in \Omega^1(M):d\alpha=0\}$ (in a very loose
sense: if we have a path of symplectic diffeomorphisms starting with
the identity, we can differentiate it to get a symplectic vector field).

Note: $\LSymp(M,\omega)$ is not simple: we have the exact sequence

$0 \to \R \to  C^\infty(M,\R) \to \LSymp(M,\omega) \to H^1(M,\R) \to 0$,
by basic de Rham cohomology.

(note that map from $C^\infty(M,\R)$ to $\LSymp(M,\omega)$ is given
by $H \mapsto X_H$, $\omega(.,X_H)=dH$.

The last map is a map of Lie algebras, where $H^1(M,\R)$ carries the
trivial Lie bracket.  So $C^\infty(M,\R)$ is a Lie algebra.

For $f,g \in C^\infty(M,\R)$, we define the {\em Poisson bracket}
$\{f,g\}=\omega(X_f,X_g)=dg(X_f)=-df(X_g)$.

Claim: $X_{\{f,g\}}=[X_f,X_g]$.

Pf: $X_{\{f,g\}}=X_{X_f \cdot g}$.  What does this mean?

We take the function $g$, differentiate it in the direction of the symplectic
vector field $X_f$, and then use the correspondence.

The correspondence depends only on $\omega$, and $X_f$ preserves $\omega$
as it is symplectic.  So we can instead do the above two operations in
the opposite order: use the correspondence (yielding $X_g$) and then
differentiate in the direction of the symplectic vector field $X_f$
(yielding $[X_f,X_g]$, as desired).  For more detail, write things out
in terms of flow...

"This fact that you have vector fields given by functions is kind of neat,
because to give a vector field is to give $n$ functions, but to give one
function is much easier."

Moser's theorem: Let $(\omega_t)$ be a smooth family of symplectic forms on
the closed manifold $M$.  If $[\omega_t] \in H^2(M;\R)$ is constant, then
there exists a family of diffeomorphisms $\phi_t:M \to M$
$\phi_0=id$, $\phi_t^*(\omega_t)=\omega_0$.

Pf: Since $[\omega_t]$ is constant, $\frac{d\omega_t}{dt}=d\rho_t$.
Now we have a $t$-dependent vector field $Z_t$: the {\em Moser vector field},
defined by $\omega_t(-,Z_t)=\rho_t$.  Then integrate $Z_t$ to get
the family $\phi_t$.  Claim: $\phi_t$ satisfies the above relation.

(Note that we need $M$ to be closed in order to guarantee that $Z_t$ is
integrable).

To see this, it suffices to show that $\frac{d}{dt} \phi_t^*\omega_t=0$
(since the condition is clearly satisfied at $t=0$.

But this is equal to
$\phi_t^*(L_{Z_t} \omega_t+\frac{d\omega_t}{dt})=
\phi_t^*(d(i_{Z_t} \omega_t+\frac{d \omega_t}{dt})
=\phi_t^*(-d\rho_t+\frac{d\omega_t}{dt})=0$, by the
definition of $\rho_t$.

Cor: Let $M_t subset \C P^n$ be a smooth family of smooth
projective varieties.  Equip $M_t$ with $\omega_t$, the restriction
of the Fubini-Study symplectic form on $\C P^n$.  Then all $(M_t,\omega_t)$
are symplectically isomorphic; indeed, there is a smooth family of symplectic
isomorphisms.

First, note that they're diffeomorphic (locally, we can find the 
diffeomorphism). Moreover, the cohomology classes of the symplectic forms
stay the same; thus we can apply Moser's theorem.

For instance, we can speak of "the" smooth hypersurface of degree $d$ in
$\C P^n$ (i.e., we talk talk about this symplectically).

Submanifolds of symplectic manifolds:

If $(M,\omega)$ is symplectic, what kinds of submanifolds should we look
at?

Given a submanifold $N$, look at $TN^{\perp}=
\{x \in TM|_N:\omega(x,y)=0, y \in TN\}$.

-Symplectic submanifolds: $N \subset M$ with $\omega|_N$ symplectic;
equivalently, $TN^\perp+TN=TM|_N$.
-isotropic submanifolds: $N \subset M$ with $TN \subset TN^{\perp}$ (i.e.,
$\omega|N$ vanishes). 
-coisotropic submanifolds: $TN^{\perp} \subset TN$
-Lagrangian submanifolds:  $TN^{\perp}=TN$

More on these next time...

Thm (Darboux): Suppose $(M,\omega)$ is symplectic.  Then for any $x \in $M$
there exists a
n'hood $U$ containing $x$ and local coordinates $(p_1,\dots,p_n,q_1,\dots,q_n)$
on $U$ such that $\omega|_U=\sum_i dp_i \wedge dq_i$.

Pf: "Local version of Moser's theorem"; look at coordinates in which it
agrees with the standard form at $x$...