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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Symplectic geometry\\[15mm] \small lecture 4: Hamiltonian symplectomorphisms} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] September 15, 2021 } \end{center} \newpage {\bf \blue Symplectic structure on the total space of cotangent bundle} From now on, {\bf \blue the total space of the cotangent bundle} to $M$ is denoted as $T^*M$. The {\bf \blue diffeomorphism group} of $M$ is denoted $\Diff(M)$. \theorem Let $M$ be a smooth manifold. {\bf \red Then $T^*M$ is equipped with a natural, $\Diff(M)$-invariant symplectic form $\omega$.} \proof Let $\pi:\; TM \arrow M$ denote the projection. Consider a point $(x, \xi)\in T^*M$, where $x\in M$ and $\xi \in T^*_x M$. Let $\theta\in \Lambda^1 (TM)$ be a 1-form which takes a tangent vector $v \in T_{(x, \xi)}T^*M$ and maps it to $\langle D\pi(v), \xi\rangle$. Here, $D\pi:\; T(T^*M) \arrow TM$ is the differential, and $\langle \cdot, \cdot\rangle$ the pairing between $T_xM$ and $T^*_xM$. If we introduce the coordinates $p_1, ..., p_n$ on $M$ and $q_1, ..., q_n$ on the fibers of $T^*M\arrow M$ dual to $p_1, ..., p_n$, the form $\theta$ at a point $p_1,..., p_n, q_1, ... q_n$ takes a vector $\sum_i f_i \frac d{dp_i} + \sum_i g_i \frac d{dpq_i}$ to $\sum_i f_i q_i$. Therefore, $\theta=\sum_i q_i dp_i$. {\bf \purple In the same coordinates $d\theta = \sum_i dq_i \wedge dp_i$; this form is symplectic and by construction $\Diff(M)$-invariant.} \endproof \newpage {\bf \blue Lagrangian subspaces} \remark Let $(V, \omega)$ be a symplectic vector space, $\dim_\R V=2n$, and $W\subset V$ be a subspace of dimension $m$. Denote by $W^\bot$ the symplectic orthogonal to $W$. Then $\frac{W}{W^\bot \cap W}$ is a symplectic vector space, hence $\rk (\omega\restrict W)= \dim W - \dim (W^\bot \cap W)$. Since $\dim W^{\bot}= 2n - \dim W$, the intersection $W^\bot \cap W$ has dimension at most $n$. \definition A subspace $W\subset V$ is called\\ (i) {\bf \blue isotropic} if $\omega\restrict W=0$; in this case $W\subset W^\bot$, hence $\dim W \leq n$.\\ (ii) {\bf \blue coisotropic} if $\omega\restrict W^\bot=0$, or, equivalently, $W^\bot \subset W$. In this case $W\supset W^\bot$, hence $\dim W \geq n$.\\ (iii) {\bf \blue Lagrangian} if it is isotropic and coisotropic, that is, $W^\bot=W$. \remark Isotropic and coisotropic subspaces {\bf \purple have minimal possible rank of $\omega$ for a given dimension.} \remark {\bf \purple An $\omega$-orthogonal complement to an isotropic space is coisotropic,} and vice versa. \newpage {\bf \blue Lagrangian submanifolds} \definition Let $X\subset M$ be a submanifold in a symplectic manifold. It is called {\bf \blue Lagrangian} if $T_x X \subset T_xM$ is Lagrangian for all $x\in X$. \theorem Let $\xi\in \Lambda^1 M$ be a 1-form, and $\Gamma_\xi\subset T^* M$ its graph, considered as a submanifold in the total space of the cotangent bundle. {\bf \red Then $\Gamma_\xi$ is Lagrangian if and only if $d\xi=0$.} \proof Let $\sigma:\; x \mapsto (x, \xi(x))$ be the standard diffeomorphism from $M$ to $\Gamma_\xi$. Consider the restriction of $\theta$ to $\Gamma_\xi$. For each $u\in T_{(x, \xi(x))}\Gamma_\xi$, the form $\theta$ takes $u$ to $\xi(d\pi(u))$. This implies that $\sigma^* \theta\restrict{\Gamma_\xi} = \xi$, hence $\sigma^* \omega\restrict{\Gamma_\xi} = d\xi$. \endproof \newpage {\bf \blue Hamiltonian vector fields} \definition Let $v\in TM$ be a vector field on a symplectic manifold $(M, \omega)$. We say that $v$ is {\bf \blue symplectomorphic} if $\Lie_v\omega=0$, that is, if $\omega$ is invariant under the corresponding diffeomorphism flow. \remark From Cartan's formula, we have $\Lie_v\omega=d (i_v\omega)$, hence {\bf \purple $v$ is symplectomorphic if and only if the $\omega$-dual 1-form is closed.} \definition Let $v\in TM$ be a symplectomorphic vector field on a symplectic manifold $(M, \omega)$, and $\eta:= i_v\omega$ the corresponding 1-form. We say that $v$ is {\bf \blue a Hamiltonian vector field} if $i_v\omega$ is exact. Its {\bf\blue Hamiltonian} is a function $f$ such that $df= i_v\omega$. The group of {\bf \blue Hamiltonian symplectomorphisms} is generated by diffeomorphisms obtained by exponents of a time-dependent vector field $v_t$, which is Hamiltonian for all $t\in [0,1]$. \remark We have an exact sequence \[ 0 \arrow \R \arrow C^\infty(M) \stackrel \delta \arrow \Ham(M) \arrow 0 \] If we identify $\Ham(M)$ with exact 1-forms, the differential $\delta:\; C^\infty(M)\arrow \Ham(M)$ {\bf \purple is identified with the de Rham differential.} \newpage {\bf \blue The flux} \definition Let $\Psi_t$ be a flow of symplectomorphisms of $(M,\omega)$, with $t\in [0, a]$ and $\Psi_0=\Id_M$. Given a homology class $[u]\in H_1(M, \Z)$, consider its representative as a smooth circle $u:\; S^1\arrow M$. Then $\Psi_t(u)$ takes $(s\in S^1, t\in [0,1])$ to $\Psi_t(u(s))$, giving a smooth map $\Psi_t(u):\; S^1 \times [0, 1]\arrow M$. Denote its image by $\Psi_{[0,1]}(u)$. Define {\bf \blue the flux} of $\Psi_t$ on $u$ as $\int_{\Psi_{[0,1]}(u)}\omega$. \claim The flux of $\Psi_t$ on $u$ {\bf \red is independent from the choice of $u$ in $[u]$.} \proof Suppose that $u_\tau$, $\tau\in [0,1]$ is a homotopy of $u_0$ to $u_1$. Consider the map $\tilde\Psi:\; S^1\times [0,1] \times [0,1]$ mapping $(s, t, \tau)$ to $\Psi_t(u_\tau)$. The surfaces $\Psi_{[0,1]}(u_0)$, $\Psi_{[0,1]}(u_1)$, $\cup_{\tau\in [0,1]} u_\tau$ and $\cup_{\tau\in [0,1]} \Psi_1(u_\tau)$ bound a solid torus $\im \tilde \Psi$, hence {\bf \purple to prove that $\int_{\Psi_{[0,1]}(u_1)}\omega = \int_{\Psi_{[0,1]}(u_2)}\omega$, it would suffice to show that $\int_{\cup_{\tau\in [0,1]} u_\tau}\omega= \int_{\cup_{\tau\in [0,1]} \Psi_a(u_\tau)}\omega$,} which is clear because $\Psi_1$ is a symplectomorphism. To pass from homotopy to homology, we notice that flux is additive on unions of curves, hence {\bf \purple vanishes on a commutator of $\pi_1(M)$, } and defines a map from $\frac{\pi_1(M)}{[\pi_1(M), \pi_1(M)]}=H_1(M)$. \endproof \newpage {\bf \blue The flux conjecture} \claim {\bf \red Flux of a Hamiltonian diffeomorphism vanishes.} \proof Let $\Psi_t$ be a Hamiltonian symplectomorphism flow on $(M,\omega)$, with $t\in [0, 1]$ and $\Psi_t^{-1}\frac {d\Psi_t}{dt}=v_t$, where $v_t$ is a Hamiltonian vector field with Hamiltonian $H_t$, and $u:\; S^1 \arrow M$ a circle. To prove that the flux vanishes it would suffice to show that $\frac{d}{dt} \int_{\Psi_{[0,t]}(u)}\omega=0$. This derivative can be computed as $\int_{\Psi_t(u))} \omega(\frac {d\Psi_t}{dt}, \cdot)$. However, the 1-form $\omega(\frac {d\Psi_t}{dt}, \cdot)= \Psi_t^*(dH_t)$ is exact, because $v_t$ is Hamiltonian, hence $\frac{d}{dt} \int_{\Psi_{[0,t]}(u)}\omega=0$. \endproof \remark The famous ``flux conjecture'' states that vanishing of the flux is a necessary and sufficient condition: {\bf \blue if $\Psi_t$, $t\in [0,1]$ is a flow of symplectomorphisms, and the flux of $\Psi_1$ vanishes, then $\Psi_1$ is a Hamiltonian symplectomorphism. } Flux conjecture was proven by Kaoru Ono using Floer theory. \remark Flux conjecture implies that {\bf \purple Hamiltonian symplectomorphisms is closed in the group of symplectomorphisms.} \newpage {\bf \blue Commutator of Hamiltonian vector fields} \example Let $u_\epsilon$ be a rotation of a 2-dimensional torus $S^1\times S^1$ with the product metric which comes from unit circles and the symplectic structure given by the volume form. Consider a rotation along the first circle with angle $\epsilon$. Then its flux through the second circle is the area of the segment bounded by the first circle and its image, that, is $2\pi\epsilon$. In particular, {\bf \red rotation of a torus is not a Hamiltonian symplectomorphism}. \theorem Let $X$, $Y$ be a Hamiltonian vector fields on $(M, \omega)$. {\bf \red Then the commutator $[X,Y]$ is Hamiltonian.} \proof Let $f, g\in C^\infty M$, and let $X_f, X_g$ be the corresponding Hamiltonian vector fields. Define {\bf \blue the Poisson bracket} as $\{f, g\}:= \omega(X_f, X_g)$. By definition, $\omega(X_f, X_g)= \langle dg, X_f\rangle = \Lie_{X_f}(g)$, which gives $[X_f, X_g] = \Lie_{X_f}(X_g)= X_{\Lie_{X_f}(g)}= X_h$, where $h= \omega(X_f, X_g)$. Therefore, {\bf \purple commutator of Hamiltonian vector fields corresponds to the Poisson bracket on their Hamiltonian functions.} \endproof \newpage {\bf \blue Hamiltonian vector fields on Lagrangian fibrations} \definition Let $(M, \omega)$ be a symplectic manifold, and $\pi:\; M \arrow X$ a smooth submersion. It is called {\bf \blue a Lagrangian fibration} if all its fibers are Lagrangian. \claim Let $\pi:\; M \arrow X$ be a Lagrangian fibration, and $H$ a function on $X$. Then {\bf \red the corresponding Hamiltonian vector field $v$ is tangent to the fibers of $\pi$.} Moreover, {\bf \red $v$ is non-degenerate everywhere on a fiber $\pi^{-1}(x)$ if and only if $dH\neq 0$ in $x$;} otherwise $v\restrict \pi^{-1}(x)=0$. \proof Let $L:= \pi^{-1}(x)$. Consider $\omega$ as a map from $T_m M$ to $T_m^* M$. Then $\omega^{-1}$ takes 1-forms vanishing in $T_m L$ to vectors $v\in T_m M$ such that $\omega(v, \cdot)$ vanishes on $T_mL$. However, $T_mL^\bot= T_mL$ because $L$ is Lagrangian. {\bf \purple Therefore, $\omega^{-1}$ takes 1-forms vanishing on $T L$ to the vector fields tangent to $L$.} The forms vanishing on $TL$ are generated by $\pi^* \Lambda^1 X$, hence the corresponding Hamiltonian vector fields are tangent to the fibers of $\pi$. \endproof \remark Let $L= \pi^{-1}(x)$ be a fiber of a Lagrangian fibration. Then $\omega$ defines a non-degenerate pairing between $T_mL$ and $T_x X$. This implies that {\bf\red the bundle $TL$ is trivial.} \newpage {\bf \blue Collections of commuting Hamiltonians} \remark Let $\pi:\; M \arrow X$ be a Lagrangian fibration, and $v_1, v_2$ two Hamiltonian vector fields obtained from $H_1, H_2\in C^\infty X$ as above. {\bf \purple Then $H_1$, $H_2$ commute.} Indeed, the Poisson commutator $\{H_1, H_2\}$ is expressed as $\omega(v_t, u_t)$, vanishing for the vector fields tangent to fibers of $\pi$. \proposition Let $\pi:\; M \arrow X$ be a Lagrangian fibration, and $L$ its fiber. Then {\bf \red the bundle $TL$ has a basis of globally defined commuting vector fields.} Moreover, {\bf \red if $L$ is compact, it is a torus.} \proof Let $L=\pi^{-1}(x)$ be a fiber, and $H_1, ..., H_n\in C^\infty X$ be a collection of functions such that $dH_1, ..., dH_n\restrict T_x X$ is a basis in $T^*_x X$. Then {\bf \purple the corresponding vector fields commute and define a basis in $TL$}. If these vector fields are {\bf \blue complete} (that is, can be integrated to a flow for all $t\in \R$), the manifold $L$ is equipped with a transitive $\R^n$-action, which gives $L= \R^n/\Lambda$, where $\Lambda$ is a discrete subgroup. If $L$ is compact, $\Lambda$ is a cocompact lattice, hence $L$ is a torus. \endproof \end{document}