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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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



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