"We all have to pass the time until the electoral results come in, so we might
as well do some geometry..."

Recall def'ns:

Stein: $(M,J,\phi)$
Weinstein: $(M,\omega,\theta,\phi)$
Def'ns of complete and finite type in each case

Stein: $\omega=-d(d\phi \circ J)$ is a Kahler form.

Weinstein: $\omega$ symplectic, $d\theta=\omega$, $\phi$ an
exhausting function; $\theta$ is associated to a "compressing" form.

Weinstein, complete, of finite type => has a contact-type cone at $\infty$.

Lots of Stein manifolds, all of which are Weinstein.  When will they
be of finite type?

If $X$ is a smooth projective variety, $\L$ an ample line bundle, and
$\sigma$ a nonzero holomorphic section, then $X-\sigma^{-1}(0)=X-D$ is Stein
(e.g., $X \subset \C P^n$, $M=X-\C P^{n-1}$.

Lemma: Suppose $D$ has normal crossings (locally, $D=\{z_1 \dots z_k=0\}$
in holomorphic coordinates, $z=(z_1,\dots,z_n)$ (i.e., if $D$ is smooth).
Then $M=X-D$ is of finite type.

How do we get a Stein structure to be complete?

Certainly, the given Stein structure on $X-D$ isn't complete (since
the symplectic form is restricted from something closed, and thus has
finite volume).  But we can fix this...

Lemma: Suppose $(M,J,\phi)$ Stein.   Let $h:\R \to \R$ be a smooth function, 
$h'>0$, $h'' >0$, and $h' \in O(h'')$.  Then $(M,J,\tilde \phi=h(\phi))$ is
Stein and complete.

Clearly, $\tilde \phi$ has the same critical points as $\phi$; thus, with the
new function, $M$ is of finite type iff it was originally.

Pf: Let $X$ be the Liouville vector field associated to $(M,J,\phi)$,
so $X=\nabla \phi$.  Similarly, let $\tilde X=\nabla \tilde \phi$.

Then $\tilde X=X h'(\phi)/(h'(\phi)+h''(\phi)|\nabla \phi|^2)$, where
the denominator is evaluated using the original metric.

Why is $\tilde \phi$ plurisubharmonic? "Obvious."

We'll prove completeness.

What is $d\phi(\tilde X)$?  It's $|\nabla \phi|^2/(1+h''(\phi)/h'(\phi)
|\nabla \phi|^2 \leq 1/\delta$, where $\delta$ is the $O$-constant.

So the flow is of bounded speed; hence it's defined for all time.

P.S. Suppose $(M,J,\phi)$ is already complete and of finite type.  
Then $(M,J,\tilde \phi)$ is
isomorphic to $(M,J,\phi)$ as a Weinstein manifold: i.e., there is a
diffeomorphism $f:M \to M$ such that $f^*(\tilde \theta)=\theta+dR$,
$R \in C^{\infty}_c(M)$. (Thus, the isomorphism is compatible with the
Liouville flow outside some compact subset.)

Why is this true? By the previous proof given, we can drop the last assumption
above in this case, and still get the conclusion.  Do it with a collection
of smooth functions interpolating between the original one and the new one,
and apply Moser.

Thm: Take $(M,J)$; let $\phi,\tilde \phi$ be two exhausting strictly
plurisubharmonic functions on $M$.  Suppose that $(M,J,\tilde \phi)$ is
complete and of finite type.  Then there is an embedding $i:M \to M$ such
that $i^*(\tilde \theta-\theta)$ is exact.

Pf: First, assume that $\tilde \ohi$ grows faster than $\phi$, in
the sense that for $\delta=\tilde \phi-\phi$, $\delta \geq 1$ everywhere,
$\delta$ is exhausting.

Consider $U_k=\{x \in M|\delta(x)<k\}$.  By definition, 
$U_1=\emptyset\subset \overline{U_1} \subset U_2 \subset \overline{U_2}
\subset \dots$.

Fix "some kind of stupid function" $l:\R \to \R$,
$l(r)=0$, $r \leq -1$, $l''(r)=0$ everywhere, $l(r)=r$ for $r \geq 1.

For $k=0,1,2,\dots,$ consider $\tilde \phi_k=\phi+l(\delta-k)$.  This
satisfies $\tilde \phi_k=\phi$ on $\overline{U_{k-1}}$,
$\tilde \phi_k=\phi-k$ on $M-U_{k+1}$, and is plurisubharmonic (exercise).

Naive idea: the space of plurisubharmonic functions is convex, so we want to
interpolate between them and then symplectically follow via Moser. 
Unfortunately, our manifold isn't convex, so the vector fields we get might not
be integrable.  So we'll use our $\phi_k$ to make the change "one piece at a
time".

Define $\tilde \theta_k=-d\tilde \phi_k \circ J$, $\tilde \omega_k=
d\theta_k$.  Apply Moser to the linear deformation from $\tilde \phi_k$
and $\tilde \phi_{k+1}$.  Then everything is compactly supported, so
the Moser argument applies, and gives $f_k:M \to M$ a diffeomorphism, $f$
is a diffeomorphism away from $\overline{U_{k+2}}-U_{k-1}$, such that
$f_k^* \tilde \omega_{k+1}=\tilde \omega_k$.  
Trick: Set $i=f_0 \circ f_1 \circ f_2 \circ \dots:M \to M$.  Why does this
make sense?

For any fixed point $x$, for large $k$, $f_k$ preserves $x$; thus we need
only consider any finite number of compositions for any fixed point.

Then $i^* \tilde \omega=\omega$ (since it's true locally).

Why is this not a diffeomorphism? If you do it backwards, the above argument
wouldn't make sense (maybe each $f_k$ takes $x$ further out...)

It's an injective local diffeomorphism, though, so it's an embedding.

We made the assumption that $\tilde \phi$ grows faster than $\phi$.  Suppose
not.  But $h(\tilde \phi)$ does, for some $h$.  Use the previous lemma and
the fact that we're complete and of finite type.

Note that this is the only place where we use the fact that is complete
and of finite type.

Contact surgery:

Weinstein, "Contact surgery and symplectic handlebodies" (Hokkaido J. Math)
Cieliebak, "Handle attaching in symplectic topology and the chord conjecture"
 (J. Euro. Math. Society, full text on MathSciNet)

Suppose $(N^{2n-1},\alpha)$ is a contact manifold; 
let $K^k \subset N$ be a submanifold
which is contact isotropic (i.e., $\alpha|K=0 \in \Omega^1(K)$, and hence
$d\omega|K=0$).

As usual, we have a tubular n'hood thm.  Along $K$, we have the following
inclusions:

$TK \subset TK^{\perp} \subset \xi=\ker(\alpha) \subset TN$
(where the $\perp$ is taken w.r.t. the symplectic structure $d\alpha$ on
$\xi$).

We have the normal bundle $TN/TK=TN/\xi \oplus \xi/TK^{\perp} \oplus
TK^{\perp}/TK$.

Now, $TK^{\perp}/TK$ has a symplectic structure induced from $d\alpha$.

$TN/\xi=\R$, via $\alpha$.

What is $\xi/TK^\perp$? It's $TK^*$, by basic linear algebra.

Now, suppose $K=i(S^k)$, where $i:S^k \to N$ is an embedding.  Then 
$TK^* \oplus \R \cong T^* S^k \oplus \R$ has a canonical trivialization.

Def'n: A {\em framed contact isotropy sphere} is a $K^r \subset N^{2n-1}$,
contact isotropic, $K=i(S^r)$, with a symplectic trivialization of 
$TK^{\perp}/TK$.

Then we have the following restricted version of the tubular n'hood thm:
If $K$ is a framed contact isotropy sphere, then the n'hood of $K \subset N$, 
up to contact diffeomorphism, is $\R^{>0} \times T^*(S^k \times \R^{n-k-1})$.

Why is this contact? Let $\theta$ be the Liouville form; pull it back to
$\bar \theta$, and let $\alpha=t \bar \theta+dt$ ($t$ is the extra variable).

Then $\alpha \wedge (d\alpha)^{n-1}=t^{n-1} dt \wedge 
d\overline{\theta}^{n-1}$, after some algebra; this is nonzero by inspection.

The contact isotropy sphere is $\{1 \times (S^k \times \{0\}\}$.

Legendrian handle attachment:

Assume $M$ is Weinstein, complete, and of finite type.  If we take
$N=\phi^{-1}(c_0)$, $\alpha=\theta|N$ is contact.

Suppose $i:S^r \to N$, $K=i(S^r)$ is a framed contact isotropic sphere.

Do the following 3 things:

1) Remove the cone to get a compact manifold-with-bdry
2) Attach a handle $D^{r+1} \times D^{2n-r-1}$
3) Reattach a (different) cone.

This gives another finite-type complete Weinstein manifold.

Topologically, this is easy...

Locally, what does this handle look like?

Take $C^n$, $\omega$ standard, take $\phi=\frac{1}{4}|z|^2-\frac{1}{2}
(\im(z_1))^2-\frac{1}{2}(\im(z_2))^2-\dots$.

Look at the associated gradient flow; make an argument involving Morse
theory and pictures I can't reproduce to make a handle.

In fact, symplectically, every Stein manifold can be obtained by
repeated such attachments.