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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Symplectic geometry\\[15mm]
\small lecture 10: Proof of Gromov's Non-Squeezing Theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
October 6, 2021
}
\end{center}

\newpage 

{\bf \blue Almost complex structures (reminder)}

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

The eigenvalues of this operator are $\pm \1$.
The corresponding eigenvalue 
decomposition is denoted $TM=T^{0,1}M\oplus T^{1,0}(M)$.


{\bf\green DEFINITION:}
An almost complex structure is {\bf \blue integrable}
if $\forall X,Y \in T^{1,0}M$, one has $[X,Y]\in T^{1,0}M$.
In this case $I$ is called {\bf \blue a complex structure operator}.
A manifold with an integrable almost complex structure
is called {\bf \blue a complex manifold}. 

{\bf\green THEOREM:} (Newlander-Nirenberg)\\
{\bf \purple This definition is equivalent to the usual one.}


\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 $g$.

{\bf \green THEOREM 1:}
Let $(M, \omega)$ be a manifold equipped
with a non-degenerate skew-symmetric 2-form.
Then {\bf \red the space $C$ of almost complex structures
compatible with $\omega$ is contractible.}

\newpage

{\bf \blue Calibrations (reminder)}

\newcommand{\comass}{\operatorname{\sf comass}}

\definition
(Harvey-Lawson, 1982)\\
 Let $W\subset V$ be a $p$-dimensional subspace in a Euclidean space,
and $\Vol(W)$ denote the Riemannian volume form of 
$W \subset V$, defined up to a sign. For any $p$-form $\eta
\in \Lambda^p V$, let {\bf\blue comass} 
$\comass(\eta)$ be  the maximum of $\frac{\eta(v_1, v_2, ...,
v_p)}{|v_1||v_2|...|v_p|}$, for all $p$-tuples $(v_1, ...,
v_p)$ of vectors in $V$ and {\bf\blue face} be the set of  
planes $W\subset V$ where $\frac\eta {\Vol(W)}=\comass(\eta)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\definition\label{_precalibra_Definition_}
A {\bf\blue precalibration} on a Riemannian 
manifold is a differential form with comass 
$\leq 1$ everywhere.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\definition
A {\bf\blue calibration} is a precalibration which is closed.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\definition
Let $\eta$ be a $k$-dimensional precalibration on a
Riemannian manifold, and $Z\subset M$  a $k$-dimensional
subvariety (we always assume that the Hausdorff dimension
of the set of singular points of $Z$ is $\leq k-2$,
because in this case a compactly supported differential
form can be integrated over $Z$).  We say that $Z$ is
{\bf\blue calibrated by $\eta$} if at any 
smooth point $z\in Z$, the space $T_zZ$ is a face of 
the precalibration $\eta$.

\newpage 

{\bf \blue Calibrations and minimal submanifolds (reminder)}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\remark Clearly, for any precalibration $\eta$,
one has \[
\Vol(Z) \geq \int_Z\eta,\ \ \ \ \ (*)
\] where
$\Vol(Z)$ denotes the Riemannian volume of a compact $Z$, 
and the equality happens iff $Z$
 is calibrated by $\eta$. If, in
addition, $\eta$ is closed, the number
$\int_Z\eta$ is a cohomological invariant.
Then, {\bf \purple the inequality (*) implies that $Z$ 
minimizes the Riemannian volume in its 
homology class.}

\definition
A subvariety $Z$ is called {\bf \blue minimal}
if for any sufficiently small deformation 
$Z'$ of $Z$ in class $C^1$, one has 
$\Vol(Z')\geq \Vol(Z)$.

\remark
{\bf \red Calibrated subvarieties are obviously minimal.}


\newpage 

{\bf \blue Pseudoholomorphic curves (reminder)}

\definition
Let $(M,J)$ be an almost complex manifold,
$(\Sigma, I)$ a Riemann surface,
and $\phi:\; \Sigma \arrow M$ an $I$-holomorphic map, 
that is, a smooth map with $D\phi(Ix)= J(D\phi(x))$.
Then $\phi(\Sigma)$ is called {\bf \blue
a pseudo-holomorphic curve}, or
{\bf \blue a $J$-holomorphic curve}.

\theorem
{\bf \blue (Wirtenger's inequality):} \\
Let $(M, I, \omega)$ be an almost K\"ahler manifold.
{\bf \purple Then $\frac{1}{2}\omega$ is a calibration
which calibrates pseudo-holomorphic curves.}

\proof
Let $g_S$ be the Riemannian volume form on $S$,
and $x, y\in T_sS$ be orthogonal vectors of length 1.
Then $g_S(x, y)=1$ and $\omega(x, y)= g(x, Iy)\leq 1$,
and the equality is realized if and only if $x=Iy$,
by Cauchy-Bunyakovsky-Schwarz inequality.
\endproof

\corollary {\bf \red Pseudoholomorphic curves are
minimal.}

\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 $\R^{2n-2}/\Z^{2n-2} =
T^{2n-2}$.


\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 $r-\epsilon >1$.
The operation $g_1(\cdot, \cdot) \arrow g_1(B_1\cdot, \cdot)$
constructed in the proof of Theorem 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. 

\proposition
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 $(B, \eta)$ be a unit ball $B_1(0)\subset \R^n$
equipped with a calibration $\eta\in \Lambda^k B_1$ 
with constant coefficients, and 
the standard flat metric, and 
$\Delta\subset B$ a calibrated $k$-dimensional
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 $t\arrow t^{-k} \Vol(\Delta_t)$
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
$\Vol\Delta \geq \pi$.}

\newpage

{\bf \blue Monotonicity formula (2)}

\lemma {\bf \blue (monotonicity formula)}\\
Let $(B, \eta)$ be a unit ball equipped
equipped with a calibration $\eta\in \Lambda^k B_1$ 
with constant coefficients, and 
$\Delta\subset B$ a calibrated $k$-dimensional
submanifold passing through 0. {\bf \red Let $\Delta_t$ be the intersection
of $\Delta$ with a ball $B_t(0)$ of radius
$t$. Then the function $t\arrow t^{-k} \Vol(\Delta_t)$
is non-decreasing.}


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


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


\newpage

{\bf \blue Monotonicity formula (3)}

\lemma {\bf \blue (monotonicity formula)}\\
Let $(B, \eta)$ be a unit ball equipped
equipped with a calibration $\eta\in \Lambda^k B_1$ 
with constant coefficients, and 
$\Delta\subset B$ a calibrated $k$-dimensional
submanifold passing through 0. {\bf \red Let $\Delta_t$ be the intersection
of $\Delta$ with a ball $B_t(0)$ of radius
$t$. Then the function $t\arrow t^{-k} \Vol(\Delta_t)$
is non-decreasing.}

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



{\bf \green Step 3:} The same argument applied to $\Delta_t$
gives 
$t^{-k}\Vol\Delta_t\leq \frac 1 {kt^{k-1}} l(\6\Delta_t)$.
(homothety maps the $k$-dimensional Riemann volume $\mu$
to $t^k \mu$). 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
\[
\Vol\Delta_t \leq \frac t {k} l(\6\Delta_t) \leq \frac t {k}\frac d {dt} \Vol\Delta_t.
\]
{\bf \green Step 4:}
Let $f(t):= \Vol\Delta_t$.
The last formula of Step 3 gives
$f(t) \leq \frac t k f'(t)$, hence
\[
\frac d{dt} t^{-k}f(t) = 
t^{-k}f'(t)- kt^{-k-1}f(t)
\geq  t^{-k}f'(t)-t^{-k}f'(t)=0.
\]
\endproof



\end{document}
