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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Symplectic geometry\\[15mm] \small lecture 3: Flow of diffeomorphisms and Moser isotopy lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] September 11, 2021 } \end{center} \newpage {\bf \blue Flow of diffeomorphisms} \definition Let $f:\; M \times [a,b]\arrow M$ be a smooth map such that for all $t\in [a,b]$ the restriction $f_t:= f\restrict{M\times\{t\}}:\; M \arrow M$ is a diffeomorphism. Then $f$ is called {\bf\blue a flow of diffeomorphisms}. \claim Let $V_t$ be a flow of diffeomorphisms, $f\in C^\infty M$, and $V^*_t(f)(x):= f(V_t(x))$. Consider the map $\frac d{dt}V_t\restrict{t=c}:\; C^\infty M \arrow C^\infty M$, with $\frac d{dt}V_t\restrict{t=c}(f)= (V_c^{-1})^*\frac{dV_t}{dt}\restrict{t=c}f$. {\bf \red Then $f\arrow (V_t^{-1})^*\frac d{dt}V_t^*f$ is a derivation} (that is, a vector field). \proof $\frac d{dt}V_t^*(fg)= V_t^*(f)\frac d{dt}V_t^* g+ \frac d{dt}V_t^* f V_t^*(g)$ by the Leignitz rule, giving \[ (V_t^{-1})^*\frac d{dt}V_t^*(fg)= f \cdot (V_t^{-1})^*\frac d{dt}V_t^*g + g\cdot (V_t^{-1})^*\frac d{dt}V_t^*f. \] \endproof\\ \definition The vector field $V_t^{-1}\frac d{dt}V_t\restrict{t=c}$ is called {\bf\blue the vector field tangent to a flow of diffeomorphisms $V_t$ at $t=c$.} \newpage {\bf \blue Flow of diffeomorphisms obtained from vector fields} \exercise Let $M$ be a compact manifold, and $\Psi:\; C^\infty M \arrow C^\infty M$ is a ring automorphism. Prove that {\bf \purple $\Psi$ is induced by an action of a diffeomorphism of $M$.} \theorem Let $M$ be a compact manifold, and $X_t\in TM$ a family of vector fields smoothly depending on $t\in [0,a]$. {\bf \red Then there exists a unique diffeomorphism flow $V_t$, $t\in [0,a]$, such that $V_0=\Id$ and $(V_t^{-1})^*\frac d{dt}V_t^*=X_t$.} \pstep Given $f\in C^\infty M$, we can solve an equation $\frac d{dt}W_t(f)=\Lie_{X_t}(f)$ (here $\Lie_{X_t}(f)$ denotes the derivative along the vector field). The solution $W_t(f)$ exists for all $t\in [0,x]$ and is unique by Peano theorem on existence and uniqueness of solutions of ODE. {\bf \green Step 2:} Since \[ \frac d{dt}W_t(fg)= \Lie_{X_t}(f) g + \Lie_{X_t}(g)f = \frac d{dt}(W_t(f)W_t(g)), \] $W_t$ is multiplicative. Also, it is invertible. Applying the previous exercise, we obtain that $W_t$ is a diffeomorphism. \endproof For another proof see Chapter 5 of {\em Arnold V.I., Ordinary Differential Equations} \newpage {\bf \blue Diffeomorphism flow and vector fields} \definition Let $v_t$ be a vector field on $M$, smoothly depending on the ``time parameter'' $t\in[a,b]$, and $\Psi_t^{-1}\Psi_t:\;M\times[a,b]\arrow M$ a flow of diffeomorphisms which satisfies $\frac d{dt}\Psi_t=v_t$ for each $t\in[a,b]$, and $V_0=\Id$. Then $\Psi_t$ is called {\bf\blue an exponent of $v_t$}. \claim Let $\Psi_t$ be an exponent of a time-dependent vector field $v_t$. {\bf \red Then for any differenial form $\eta$, we have \[ \left(\Psi^{-1}\right)^*\frac{d\Psi_t^*}{dt}\restrict{t=t_0} \eta= \Lie_{v_t}\eta. \]} \proof Both sides of this equation are derivations of de Rham algebra which commute with $d$ and are equal to $\Lie_{vt}$ on functions. \endproof {\bf \green COROLLARY 1:} Let $\Psi_t$ be an exponent of a time-dependent vector field $v_t$, and $\eta_t:= \Psi_t^* \eta$. {\bf \red Then $\Psi_a^*\eta= \eta + \int_0^a \Lie_{v_t}\eta_t dt$.} Conversely, if this equation holds, we have $\eta_t:= \Psi_t^* \eta$. \endproof \newpage {\bf \blue Moser isotopy lemma (variant)} \theorem {\bf \blue (Moser's isotopy lemma)}\\ Let $M$ be a compact symplectic manifold, and $\omega_t$, $t\in [0,1]$ a smooth deformation of a symplectic form. Assume that the cohomology class $[\omega_t]\in H^2(M)$ is constant in $t$. {\bf \red Then there exists a diffeomorphism flow $\Psi_t\in \Diff(M)$ mapping $\omega_t$ to $\omega_0$, for all $t$.} \pstep By Corollary 1, we need to find a family of vector fields $v_t$ such that $\omega_0+ \int_0^a \Lie_{v_t}\omega_t dt= \omega_a$. Then $\omega_0:= \Psi_t^* \omega$. {\bf \purple This would follow if $\frac {d\omega_t}{dt}= \Lie_{v_t} \omega_t$.} Since all $\omega_t$ are cohomologous, the form $\frac {d\omega_t}{dt}$ is exact. Then $\frac {d\omega_t}{dt}=d \eta_t$, where $\eta_t\in\Lambda^1(M)$. {\bf \purple This form can be chosen smoothly in $t$ (Lecture 2).} {\bf \green Step 2:} By Step 1, to prove the theorem, we need to solve the equation \[ \Lie_{v_t}\omega_t=-\frac {d\omega_t}{dt}.\ \ (*) \] By Cartan's formula, $\Lie_{v_t}\omega_t= d(i_{v_t}(\omega_t))$. {\bf \purple Then (*) is equivalent to $d(i_{v_t}\omega_t)= -d\eta_t$.} {\bf \green Step 3:} The map $x \arrow i_x(\omega)$ induces an isomorphism $TM \arrow T^*M$ whenever $\omega$ is a non-degenerate 2-form. Therefore, there exists a unique $v_t\in TM$ such that $i_{v_t}\omega_t=-\eta_t$. This solves (*) and finishes the proof. \endproof \newpage {\bf \blue Darboux' theorem (reminder)} \theorem {\bf \red A symplectic manifold is locally symplectomorphic to a symplectic ball} (in a neighbourhood of each point). \pstep It is sufficient to check that for any symplectic form $\omega_1$ on $\R^n$ there exists a neighbourhood $U\ni 0$ such that $(U, \omega_1)$ is symplectomorphic to a symplectic ball. {\bf \green Step 2:} Choose coordinates $x_i, y_i$ on $\R^{2n}$ in such a way that $\omega_1\restrict{T_0\R^{2n}}=\omega_0\restrict{T_0\R^{2n}}$, where $\omega_0=\sum_i dx_i \wedge dy_i$. The form $\omega_t:=t\omega_1+(1-t)\omega_0$ is non-degenerate in 0, because $\omega_1 \restrict 0=\omega_0\restrict 0$. {\bf \purple Choose a starlike neighborhood $U\ni 0$ such that $\omega_t$ is non-degenerate for all $t\in [0,1]$.} {\bf \green Step 3:} In $U$ the forms $\omega_t$ are all non-degenerate and cohomologous. As in the proof of Moser's lemma, {\bf \purple choose $\eta_t$ such that $\frac {d\omega_t}{dt}=d \eta_t$, and a vector field $v_t:=-\omega_t^{-1}(\eta_t)$, vanishing in 0.} Substracting from $\eta_t$ a constant 1-form, we may assume that $\eta_t \restrict {T_0U}=0$. Then the the coefficients of the form $\eta_t$ grow as $o(r)$, where $r$ is the distance from zero. Therefore, for $U$ sufficiently small, {\bf \purple the vector field $\Psi_t$ integrates in the whole $U$, and defines a diffeomorphism $\Psi$ between $(U, \omega_0)$ and $(\Psi(U), \omega_1)$.} Finally, since $v_t=0$ in 0, the set $\Psi(U)$ contains 0. \endproof \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 \end{document}