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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}