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  \tiny M. Verbitsky }}
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\begin{document}
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\begin{center}
{\Large\bf Symplectic geometry\\[15mm]
\small lecture 11: Proof of Gromov's Non-Squeezing Theorem (again)}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
October 9, 2021
}
\end{center}



\newpage 

{\bf \blue Almost complex structures}

{\bf\green DEFINITION:} Let $M$ be a smooth manifold. 
An {\bf \blue almost complex structure} is an operator
$I:\; TM \arrow TM$ which satisfies $I^2 = - \Id_{TM}$.

\definition
Let $(M, \omega)$ be a symplectic manifold, and
$I$ an almost complex structure. We say that
{\bf \blue $I$ is compatible with the symplectic
structure} if $g(x, y):= \omega(Ix, y)$
for some Riemannian form $g$. In that case,
$g$ is called {\bf \blue compatible with $I$} as well.

{\bf \green LEMMA 1:}
Let $(M,\omega)$ be a symplectic manifold,
and $(B_r, \sum dx_i \wedge dy_i) \hookrightarrow (M, \omega)$
a symplectic embedding. Then for any $\epsilon >0$ 
{\bf \red there exists an almost complex structure $I$ on $(M, \omega)$
compatible with $\omega$ and equal to the
standard complex structure on $B_{r-\epsilon} \subset B_r$.}

\proof
Let $g_1$ be any Riemannian metric on $M$, compatible with $\omega$ and
$g_0$ the standard Riemannian metric on $B_r$.
Denote by $\lambda\in C^\infty M$ the cut-off function vanishing
outside of $B_r$ and equal to 1 on $B_{r-\epsilon}$.
Then $\tilde g:= \lambda g_0 + (1-\lambda) g_1$ is equal to
$g_0$ on $B_{r-\epsilon}$ and $g_1$ outside of $B_r$.
This form is compatible with $\omega$ in $B_{r-\epsilon}$
and $M \backslash B_r$. To make it compatible with
$\Omega$ everywhere, we use the argument
from Theorem 1 in Lecture 9: produce
a symmetric matrix $B_1 = e^{-\frac{1}2\log (-A^2)}$,
where $A:= g^{-1}\omega$, and $g(\cdot, \cdot):=\tilde g(B_1\cdot, \cdot)$
is a metric compatible with $\omega$.
This gives an almost complex structure
$I:= \omega^{-1} g$, equal to the standard one on 
$B_{r-\epsilon}\subset M$
because $B_1=\Id$ on $B_{r-\epsilon}$. \endproof

\newpage

{\bf \blue Symplectic capacity and the pseudoholomorphic curves}

{\bf \green THEOREM 2:}
Let  $M=\C P^1 \times T^{2n}$ be the product of 
$\C P^1$ and a torus, equipped with the standard
symplectic structure, and $J$ a compatible
almost complex structure. {\bf \red Then
for any $x\in M$ there exists a pseudo-holomorphic
curve $S$ homologous to $\C P^1 \times \{m \}$
and passing through $x$.}

This theorem implies Gromov's non-squeezing theorem.

\theorem {\bf \blue (Gromov)}
{\bf \red Symplectic capacity of a symplectic cylinder
$\Cyl_1$ is equal to $\pi$.}

{\bf \green Scheme of the proof:}
We map $\Cyl_1= B_1^2 \times \R^{2n-2}$ to  
$M=\C P^1 \times T^{2n-2}$
in such a way that the disk $B_1^2$ 
is bijectively mapped to $\C P^1 \backslash \infty$,
and $\R^{2n-2}$ is mapped to $\R^{2n-2}/\Z^{2n-2} =
T^{2n-2}$. Then we show that the volume of
an intersection of a symplectic ball 
$B_r \subset \Cyl_1$ and a pseudoholomorphic
curve passing through $0\in B_r$ is 
$\geq \pi r^2$, which is impossible
when $r>1$ and the curve is obtained from the 
Gromov's family obtained in Theorem 2,
because all these curves have volume $\pi$.


\newpage

{\bf\blue Proof of Gromov's theorem}

\theorem {\bf \blue (Gromov)}\\
{\bf \red Symplectic capacity of a symplectic cylinder
$\Cyl_1$ is equal to $\pi$.}

\pstep
Let $f_1:\; B_r \arrow \Cyl_1$
be a symplectic embedding, $r>1$, and $I$ the usual
(flat) almost complex structure on
$B_r\subset \C^{n+1}$. Consider the manifold
 $M=\C P^1 \times T^{2n}$, equipped with the
standard symplectic structure, and let
$f_2:\; \Cyl_1 \arrow \C P^1 \times
T^{2n}$ be a symplectic map
taking $\Cyl_1=\Delta \times \R^{2n}$  to $\C P^1 \times T^{2n}$
applying the $\Z^{2n}$ quotient on the second
argument and the natural symplectomorphism
$\Delta\tilde \arrow \C P^1 \backslash \infty$
on the first argument.

{\bf \green Step 2:} Choose the lattice
$\Z^{2n}\subset \R^{2n}$ in such a way that
its fundamental domain contains 
$f_1(B_r)$. Then the composition
$f_1 \circ f_2$ gives a symplectic
embedding $B_r \arrow M=\C P^1 \times T^{2n}$.

We obtained that {\bf \purple Gromov's non-squeezing
theorem is deduced from the following result.}


\newpage

{\bf \blue Proof of Gromov's theorem (2)}

\theorem
Let $M= \C P^1 \times T^{2n}$
be equipped with the standard
symplectic form, with the symplectic
volume of $\C P^1$ equal to $\pi$,
and $\phi:\;  B_r \arrow M$ a symplectic
embedding. {\bf \red Then $r\leq 1$.}

\pstep
Choose a flat complex structure and the
flat Hermitian metric on $B_r$. Denote by
$g_0$ the corresponding Hermitian metric
on $\phi(B_r)$. Then $g_0$
can be extended to a Riemannian metric
$g_1$ on $M$ such that
$g_0=g_1$ in a ball
$\phi(B_{r-\epsilon})$, for some
$\epsilon$ such that $r-\epsilon >1$.
Lemma 1  gives a metric $g$ compatible with the
symplectic structure on $M$ and coinciding
with $g_0$ in a ball $\phi(B_{r-\epsilon})$.
{\bf \purple Replacing $B_r$ by $B_{r-\epsilon}$,
we can add to the assumptions of the
theorem the following assumption.}

{\bf \green There exists a compatible
almost complex structure such that uts
restriction to $\phi(B_r)$ is equal
to the standard complex structure
on  $B_r\subset \C^{n+1}$.}

\newpage

{\bf\blue Proof of Gromov's theorem (3)}

\theorem
Let $M= \C P^1 \times T^{2n}$
equipped with the standard
symplectic form, with the symplectic
volume of $\C P^1$ equal to $\pi$,
and $\phi:\;  B_r \arrow M$ a symplectic
embedding. {\bf \purple Assume that  there exists a compatible
almost complex structure such that its
restriction to $\phi(B_r)$ is equal
to the standard complex structure
on  $B_r\subset \C^{n+1}$.} 
{\bf \red Then $r\leq 1$.}

{\bf \green Step 2:} 
Let $x\in M$ be the image of the center of $B_r$,
and $S\subset M$ the pseudo-holomorphic curve
which passes through $x$ by Theorem 2. 
Then $\pi=\int_S \omega_M \geq
\int_{\phi^{-1}(S)}\omega$,
where $\omega_M$ is the symplectic form on $M$, 
and $\omega$ the symplectic form on $B_r$.
Since $S$ is pseudo-holomorphic, 
{\bf \purple $\int_{\phi^{-1}(S)}\omega_{B_r}$
is the Riemannian volume of its 
intersection with $\phi(B_r)$.}

{\bf \green Step 3:} 
We obtained a complex curve
$D:= \phi^{-1}(S)$ passing through 0 
in a ball $B_r$ with flat Riemannian metric and
the standard complex structure,
with the Riemannian volume $\Vol(D)\leq \pi$.
Applying the homothety, we obtain a properly
embedded complex disk in the ball $B$ of radius 1,
passing through 0 and with area $r^{-1} \pi$.
{\bf \purple For any $r>1$, this is impossible,
as follows from the following statement,}
proven later today. 

{\bf \green PROPOSITION 1:}
Let $D\subset B_1$
be a closed complex disk in a unit ball
$B_1 \subset \C^n$, with $0\in D$. 
{\bf \red Then $\Vol(D) \geq \pi$.}

\newpage

{\bf \blue Monotonicity formula (1)}


\proposition
Let $D\subset B_1$
be a complex curve (that is, a closed 1-dimensional
complex subvariety) in a unit ball
$B_1 \subset \C^n$, with $0\in D$. 
{\bf \red Then $\Vol(D) \geq \pi$.} 

We deduce it from the {\bf \blue monotonicity formula.}

\lemma {\bf \blue (monotonicity formula)}\\
Let $(B, \omega)$ be the unit ball $B_1(0)\subset \R^{2n}$
equipped with the standard symplectic form $\omega$
and the almost complex structure $J$, and
$\Delta\subset B$ a $J$-holomorphic Riemann surface
submanifold passing through 0.
We assume that $\Delta$ is immersed to $B$
with its boundary in such a way that
$\6\Delta$ is mapped to $\6 B$.
{\bf \red Let $\Delta_t$ be the intersection
of $\Delta$ with a ball $B_t(0)$ of radius
$t$. Then the function $t\arrow t^{-2} \Vol(\Delta_t)$
is non-decreasing.}

Clearly,  $\lim\limits_{t\rightarrow 0} \Vol(\Delta_t)= \pi t^2$
(on a very small scale, any smooth disk is flat).
{\bf \purple Then the monotonicity lemma gives
$\Vol\Delta \geq \pi$.} This proves Proposition 1.

\newpage

{\bf \blue Monotonicity formula (2)}

\lemma {\bf \blue (monotonicity formula)}\\
Let $(B, \omega)$ be the unit ball $B_1(0)\subset \R^{2n}$
equipped with the standard symplectic form $\omega$
and the almost complex structure $J$, and
$\Delta\subset B$ a $J$-holomorphic Riemann surface
submanifold passing through 0.
We assume that $\Delta$ is immersed to $B$
with its boundary in such a way that
$\6\Delta$ is mapped to $\6 B$.
{\bf \red Let $\Delta_t$ be the intersection
of $\Delta$ with a ball $B_t(0)$ of radius
$t$. Then the function $t\arrow t^{-2} \Vol(\Delta_t)$
is non-decreasing.}


\pstep
Let $C(\6\Delta)\subset B$ be the 3-dimensional cone
obtained as a union of all intervals connecting
$0$ and points of $\6\Delta\subset \6 B$.
Then
\[ 
\Vol C(\6\Delta)\geq \int_{C(\6\Delta)}\omega=\int_\Delta\omega=\Vol\Delta.
\]
The first equality follows from the Stokes' theorem,
because $\6\Delta= \6 C(\6\Delta)$, and the first inequality
holds because $\omega$ is a calibration.


{\bf \green Step 2:} The formula for the 
cone volume gives
$\Vol C(\6\Delta)=\frac 1 2 l(\6\Delta)$,
where $l(\6\Delta)$ is the area of the boundary
$\6\Delta$. {\bf \purple 
This implies $\Vol\Delta\leq \frac 1 2 l(\6\Delta)$.}


\newpage

{\bf \blue Monotonicity formula (3)}

\lemma {\bf \blue (monotonicity formula)}\\
Let $(B, \omega)$ be the unit ball $B_1(0)\subset \R^{2n}$
equipped with the standard symplectic form $\omega$
and the almost complex structure $J$,
$\Delta\subset B$ a $J$-holomorphic Riemann surface
submanifold passing through 0.
We assume that $\Delta$ is immersed to $B$
with its boundary in such a way that
$\6\Delta$ is mapped to $\6 B$.
{\bf \red Let $\Delta_t$ be the intersection
of $\Delta$ with a ball $B_t(0)$ of radius
$t$. Then the function $t\arrow t^{-2} \Vol(\Delta_t)$
is non-decreasing.}


{\bf \green Step 2:} We have obtained
 $\Vol\Delta\leq \frac 1 2 l(\6\Delta)$.

{\bf \green Step 3:} The same argument applied to $\Delta_t$
gives 
$t^{-2}\Vol\Delta_t\leq \frac 1 {2t} l(\6\Delta_t)$.
(homothety maps the $2$-dimensional Riemann volume $\mu$
to $t^2 \mu$). On the other hand,
$\frac d {dt} \Vol\Delta_t\geq l(\6\Delta_t)$,
because the volume of the strip of $\Delta$
between $\6\Delta_t$ and $\6\Delta_{t+\epsilon}$
is bounded by $\epsilon l(\6\Delta_{t+\epsilon})$.
This gives
\[
\Vol\Delta_t \leq \frac t {2} l(\6\Delta_t) \leq \frac t {2}\frac d {dt} \Vol\Delta_t.
\]
{\bf \green Step 4:}
Let $f(t):= \Vol\Delta_t$.
The last formula of Step 3 gives
$f(t) \leq \frac t 2 f'(t)$, hence
\[
\frac d{dt} t^{-2}f(t) = 
t^{-2}f'(t)- 2t^{-3}f(t)
\geq  t^{-2}f'(t)-t^{-2}f'(t)=0.
\]
\endproof



\end{document}
