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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Symplectic geometry\\[15mm]
\small lecture 2: Moser isotopy Lemma}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
September 08, 2021
}
\end{center}

\newpage

{\bf \blue Superalgebras}

\definition
Let  $A^* = \oplus_{i\in \Z}A^i$ be a graded algebra over a field.
It is called
{\bf\blue graded commutative}, or {\bf\blue supercommutative},
if $ab = (-1)^{ij} ba$
for all $a\in A^i, b \in A^j$. 

\example
{\bf \purple Grassmann algebra $\Lambda^* V$
is clearly supercommutative.}



\definition
Let $A^*$ be a graded commutative algebra, 
and $D:\; A^* \arrow A^{*+i}$ be a map which
shifts grading by $i$. 
It is called a
{\bf\blue  graded derivation}, if
$D(ab) = D(a) b + (-1)^{ij} a D(b)$,
for each $a \in A^j$. 

\remark
If $i$ is even, graded derivation is a usual derivation.
If it is even, it an odd derivation.

\definition
Let $M$ be a smooth manifold, and  $X\in TM$ 
a vector field. Consider an operation of {\bf \blue 
contraction with a vector field} 
$i_X:\; \Lambda^i M \arrow \Lambda^{i-1}M$, 
mapping an $i$-form $\alpha$ to an $(i-1)$-form
$v_1, ..., v_{i-1} \arrow \alpha(X, v_1, ..., v_{i-1})$

\exercise
{\bf \purple Prove that $i_X$ is an odd derivation.}

\newpage

{\bf \blue Supercommutator}


\definition
Let $A^*$ be a graded vector space, and 
$E:\; A^*\arrow A^{*+i}$, $F:\; A^*\arrow A^{*+j}$
operators shifting the grading by $i, j$.
Define  {\bf\blue the supercommutator} $\{E, F\}:= EF - (-1)^{ij} FE$.

\definition
An endomorphism of a graded vector space
which shifts grading by $i$ is called {\bf \blue even}
if $i$ is even, and {\bf \blue odd} otherwise.

\exercise
Prove that the supercommutator satisfies 
{\bf\blue graded Jacobi identity,}
\[
\{ E, \{F, G\}\} = \{\{ E, F\}, G\} + (-1)^{\tilde E
\tilde F}  \{ F, \{E, G\}\}
\]
where  $\tilde E$ and $\tilde F$ are 0 if
$E, F$ are even, and 1 otherwise.

\remark
There is a simple mnemonic rule
which allows one to remember a superidentity,
if you know the commutative analogue.
{\bf\red Each time when in commutative case
two letters $E$, $F$ are exchanged, in supercommutative
case one needs to multiply by $(-1)^{\tilde E\tilde F}$.}


\exercise
{\bf \purple Prove that a supercommutator of superderivations
is again a superderivation.}




\newpage

{\bf \blue Lie derivative}

\definition
Let $B$ be a smooth manifold, and $v\in TM$ a vector field.
An endomorphism $\Lie_v:\; \Lambda^* M \arrow \Lambda^* M$,
preserving the grading is called
{\bf\blue a Lie derivative along $v$} if it satisfies the 
following conditions.\\
\phantom{XX} (1) On functions $\Lie_v$ is 
equal to a derivative along $v$.  (2) $[\Lie_v, d]=0$.\\
 \phantom{XX} (3) $\Lie_v$ is a derivation of the de Rham algebra.

\remark
The algebra $\Lambda^*(M)$ is generated 
by $C^\infty M=\Lambda^0(M)$ 
and $d(C^\infty M)$. The restriction
$\Lie_v\restrict{C^\infty M}$ is determined by the first axiom.
On $d(C^\infty M)$ is also determined
because $\Lie_v(df) =d(\Lie_v f)$. {\bf \red Therefore,
$\Lie_v$ is uniquely defined by these axioms.}


\newpage

{\bf \blue Cartan's formula}

\exercise Prove that
{\bf \purple $\{d, \{d, E\}\}=0$
for each $E\in \End(\Lambda^* M)$.}

\theorem
{\bf \blue (Cartan's formula)}
Let $i_v$ be a convolution with a vector field, 
$i_v(\eta) = \eta(v, \cdot, \cdot, ..., \cdot)$
{\bf \red Then $\{d, i_v\}$ is equal to the Lie derivative along $v$.}

\proof
$\{d, \{d, i_v\}\}=0$ by the lemma above.
A supercommutator of two graded derivations is a graded derivation.
Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$
\endproof

\theorem
{\bf \blue (Cartan's formula)}
Let $i_v$ be a convolution with a vector 
field, $i_v(\eta) = \eta(v, \cdot, \cdot, ..., \cdot)$
{\bf \red Then $\{d, i_v\}$ is equal to the Lie derivative along $v$.}

\proof
$\{d, \{d, i_v\}\}=0$ by the lemma above.
A supercommutator of two graded derivations is a graded derivation.
Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$
\endproof

\newpage

\ \ \\[2mm]

{\bf \blue Cartan's magic formula:} $d\circ i_x + i_x
\circ d =\Lie_x$. 

\centerline{\red \bf Which Cartan?} 

{\small
\centerline{\begin{tabular}{cc}
\'Elie Cartan (1869-1951) & Henri Cartan (1904-2008)\\ 
\epsfig{file=Cartan_big.jpg,height=0.4\linewidth} &
\epsfig{file=Henri-Cartan.jpg,height=0.4\linewidth}\\
 {\bf \green \'Elie Cartan?} & {\bf \green Henri Cartan?}\\
(Robert Bryant and Dick Palais,
& (S.S. Chern: Lectures \\
Mathoverflow) & on differential geometry)
\end{tabular}}
}

\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
 $f \arrow (V_c^{-1})^*\frac{dV_t}{dt}\restrict{t=c}f$. 
{\bf \red Then
$f\arrow (V_t^{-1})^*\frac d{dt}V_t^*f$ is a derivation of
$C^\infty M$} 
(that is, a vector field).

\newpage

{\bf \blue Poincar\'e lemma}


\definition
An open subset $U\subset \R^n$
is called {\bf\blue starlike} if for any
$x\in U$ the interval $[0,x]$ belongs to $U$.


\theorem {\bf \blue (Poicar\'e lemma)}
Let $U\subset \R^n$ be a starlike subset.
{\bf \red Then $H^i(U)=0$ for $i>0$.}

\remark
{\bf \purple The proof would follow if we construct a vector field
$\vec{r}$ such that $\Lie_{\vec r}$ is invertible on
$\Lambda^*(M)$:} $\Lie_{\vec r}R= \Id$. Indeed, for any
closed form $\alpha$ we would have 
$\alpha = \Lie_{\vec r} R\alpha= di_{\vec r} R\alpha+ 
i_{\vec r} Rd\alpha=di_{\vec r} R\alpha,$
hence any closed form is exact.

Then Poincar\'e lemma is implied by the following 
statement.

\proposition
Let $U\subset \R^n$ be a starlike subset, $t_1, ..., t_n$ 
coordinate functions,
and $\vec r:= \sum t_i \frac d {dt_i}$ the radial
vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on
$\Lambda^i(U)$ for $i>0$.}

\newpage


{\bf \blue Radial vector field on starlike sets}

\proposition
Let $U\subset \R^n$ be a starlike subset, 
 $t_1, ..., t_n$ coordinate functions,
and $\vec r:= \sum t_i \frac d {dt_i}$ the radial
vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on
$\Lambda^i(U)$ for $i>0$.}

{\bf \green Proof. Step 1:}
Let $t$ be the coordinate function on a real line,
$f(t)\in C^\infty \R$ a smooth function,
and $v:= t \frac d {dt}$ a vector field. Define
$R(f)(t):=\int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda$.
Then this integral converges whenever $f(0)=0$, 
and satisfies $\Lie_v R(f)=f$. Indeed, 
\[ \int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda=
\int^t_0 \frac{f(\lambda t)}{t\lambda} d(t\lambda)=
\int^t_0 \frac{f(z)}{z} dz,\] hence
$\Lie_v R(f)= t \frac{f(t)}{t}= f(t)$.


{\bf \green Step 2:}
Consider a function $f\in C^\infty \R^n$ satisfying
$f(0)=0$, and  $x=(x_1, ..., x_n)\in \R^n$. {\bf \purple Then
\[
R(f)(x):=\int_0^1 \frac{f(\lambda x)}{\lambda} d\lambda
\]
converges, and satisfies $\Lie_{\vec{r}} R(f)=f$.}


\newpage

{\bf \blue Radial vector field on starlike sets (2)}

{\bf \green Step 3:}
Consider a differential form  $\alpha\in\Lambda^i$,
and let $h_\lambda x\arrow \lambda x$ be the homothety with coefficient
$\lambda\in [0,1]$. Define
\[ 
R(\alpha):= \int^1_0 \lambda^{-1} h_\lambda^*(\alpha) d\lambda.
\]
Since $h_\lambda^*(\alpha)=0$ for $\lambda=0$, this integral
converges. {\bf \red It remains to prove that $\Lie_{\vec r}R=\Id$.}

{\bf \green Step 4:} Let $\alpha$ be a coordinate monomial, $\alpha=dt_{i_1}
\wedge dt_{i_2} \wedge ... \wedge dt_{i_k}$.
Clearly, $\Lie_{\vec r} (T^{-1}\alpha)=0$, where
$T=t_{i_1}t_{i_2}... t_{i_k}$. 
{\bf \purple Since $h_\lambda^* (f\alpha)= h_\lambda^*(Tf) T^{-1}\alpha$, we have
$R (f\alpha)= R(Tf) T^{-1}\alpha$ for any function $f\in C^\infty M$.}
This gives 
\[ \Lie_{\vec r} R(f\alpha) = \Lie_{\vec r}R(Tf) T^{-1}\alpha=
TfT^{-1}\alpha = f\alpha.
\]
\endproof

\newpage

{\bf \blue A smooth choice of anti-differential}

{\bf \green THEOREM 1:}
Let $\alpha_t\in\Lambda^k(M)$, $t\in [0,1]$ be
a smooth family of exact forms. Then
{\bf \red 
there exists a smooth family of forms $\eta_t\in\Lambda^{k-1}(M)$, $t\in [0,1]$,
such that $d\eta_t = \alpha_t$.}

\pstep
{\bf \purple Let $\{U_i\}$ be a covering of $M$ by open balls,
with all successive intersection of balls diffeomorphic to balls or empty.}
Such a covering can be obtained, for example, using Voronoy
partitions, or a triangulation.

We use induction by the number of open balls.
When there is only one ball, the statement follows from
the explicit proof of Poincar\'e lemma.
Suppose that we proved the theorem for a union of $n$ open balls.

Let $M = U \cup V$, where $U$ is a ball,
and $V$ a union of $n$ balls for which the theorem is already proven.
Then $\alpha_t = du_t$ on $U$ and $\alpha_t = dv_t$ on $V$.
The form $u_t- v_t$ is closed. {\bf \red Suppose it is exact.}
Since $U\cap V$ is a union of $n$ balls, the theorem
is true for it, and $u_t- v_t= d w_t$.
Extending $w_t$ to $U$ and replacing
$u_t$ by $u'_t:=u_t-d w_t$, we obtain 
forms $u_t', v_t$ which agree on $U\cap V$,
and can be glued into $\eta_t$ such that $d\eta_t = \alpha_t$.

\newpage

{\bf \blue A smooth choice of anti-differential (2)}

{\bf \green Step 1, remainder:}
Let $M = U \cup V$, where $U$ is a ball,
and $V$ a union of $n$ balls for which the theorem is already proven.
Then $\alpha_t = du_t$ on $U$ and $\alpha_t = dv_t$ on $V$.
{\bf \purple The form $u_t- v_t$ is closed. If it is exact, we are done.}


{\bf \green Step 2:} It remains to prove that $u_t- v_t$
can be chosen exact. Consider the Mayer-Vietoris exact sequence
\[
H^{i-1}(U)\oplus H^{i-1}(V)\arrow
H^{i-1}(U\cap V)\stackrel \delta \arrow H^i(U\cup V)\arrow H^i(U)\oplus H^i(V).
\]
The cohomology class of $u_t- v_t$ is by construction
mapped to the cohomology class of $\alpha_t$ under
the coboundary map $\delta$. {\bf \purple Since $\alpha_t$ is exact,
this class comes from $H^{i-1}(U)\oplus H^{i-1}(V)$.}

Denote the corresponding family of cohomology
classes by $[s_t]:= [u_t-v_t]$. 
Choosing a basis $x_1, ..., x_n$ in cohomology
of $U\cap V$ and representing each $x_i$ by a smooth form,
we may represent $[s_t]$ by a closed form $s_t$ which smoothly depends on
$t$. Since $[s_t]$ comes from $H^{i-1}(U)\oplus H^{i-1}(V)$, 
there are families of closed forms $u''_t$ on $U$ and 
$v''_t$ on $V$ such that
$s_t$ is cohomologous to $u''_t-v''_t$.
{\bf \purple Replacing $u_t$ by $u_t-u''_t$ and
$v_t$ by $v_t - v''_t$, we obtain a new family $u_t''', v_t'''$ such that
$u_t'''- v_t'''= u_t - v _t - s_t $ is exact.}
\endproof




\newpage

{\bf \blue Moser isotopy lemma}

\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
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 \red By Theorem 1, this form can be chosen smoothly in  $t$.}


{\bf \green Step 2:} Let $v_t$ be the tangent vector field
to $\Psi_t$, with $v_t:= \Psi_t^{-1} \frac{d\Psi_t}{dt}$.
The equation $\Psi_t^*\omega_t=\omega_0$ (for all $t\in [0,1]$)
is equivalent to $\Psi_0=\Id,$ $\frac{d\Psi_t}{dt} \omega_t=-\Psi_t\frac
   {d\omega_t}{dt},$
which is the same as 
\[ \Lie_{v_t}\omega_t=-\frac {d\omega_t}{dt}.\ \ (*)
\]
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:} Since $\omega_t$ is non-degenerate,
there exists a unique $v_t\in TM$ such that 
$i_{v_t}\omega_t=-\eta_t$. Integrating the time-dependent
vector field $v_t$ to a flow of diffeomorphisms,
{\bf \purple we obtain $\Psi_t$ satisfying $\Psi_t^*\omega_t=\omega_0$.} \endproof





\newpage

{\bf \blue Hodge theory on finite-dimensional spaces}


Let 
\[ ... \stackrel d \arrow C_{-1} \stackrel d \arrow C_0 \stackrel d \arrow C_1 \stackrel d \arrow C_2 \stackrel d \arrow ...
\]
be a complex of finite-dimensional
vector spaces. Put a scalar product $g$
on each $C_i$, and let $d^*:\; C_i\arrow C_{i-1}$ be adjoint operators.
Since $g(dx, y) = d(x, d^*y)$,
the orthogonal complement to $\im d$ is $\ker d^*$, and
orthogonal complement to $\ker d$ is $\im d^*$.

Let $\Delta:= dd^*+d^* d$ be {\bf\blue the Laplacian operator}.
Then $(\Delta x, y) = (dx, dy) + (d^*x, d^*y)$,
hence $x\in \ker \Delta \Leftrightarrow dx = d^*x=0
\Leftrightarrow x\in (\im d)^\bot \cap (\im d^*)^\bot$.
{\bf \purple This gives a direct sum decomposition
$C_i = \ker \Delta \oplus \im d \oplus \im d^*$.}
Since $\im d^*= (\ker d)^\bot$, this also gives
$\ker \Delta \oplus \im d = \ker d$, 
and {\bf \red identifies $\ker \Delta$ with cohomology of $d$.}

Since $\im \Delta \bot \ker \Delta$, {\bf \purple the operator
$\Delta$ is invertible on $\im \Delta$.}
Denote by $G_\Delta$ the operator which acts
as zero on $\ker \Delta$ and as $\Delta^{-1}$ on $\im \Delta$.
This operator is called {\bf \blue the Green operator}.

Write $G_d:= d^* G_\Delta$.
For any vector $\alpha\in \im d$, one has
$\alpha = G_\Delta \Delta(\alpha)$ becase
$\alpha \bot \ker \Delta$, {\bf \red which gives
$\alpha = G_\Delta dd^* \alpha = d G_\Delta\alpha$.}

\newpage

{\bf \blue A smooth choice of anti-differential (2)}

\theorem
Let $\alpha_t\in\Lambda^k(M)$, $t\in [0,1]$ be
a smooth family of exact forms. Then
{\bf \red 
there exists a smooth family of forms $\eta_t\in\Lambda^{k-1}(M)$, $t\in [0,1]$,
such that $d\eta_t = \alpha_t$.}

{\bf \green A proof using Hodge theory} 
{\bf \red (works only when $M$ is compact)}.\\
{\bf \green Step 1:} The decomposition
$C_*= \ker \Delta \oplus \im d \oplus \im d^*$
is always valid
in finite-dimensional situation. For infinite-dimensional
vector spaces, it works if and only if the spaces $\im d$ 
and $\im d^*$ are closed. {\bf \blue Hodge theory} claims that this
is true, and also defines {\bf \blue the Green operator}
$G_\Delta$ which inverts $\Delta$ on $\im \Delta = \ker \Delta^\bot$.
Since $\Delta$ commutes with $d, d^*$, the same is true for 
$G_\Delta$.

{\bf \green Step 2:} 
Write $G_d:= d^* G_\Delta$.
For any exact form $\alpha$, one has
$\alpha = G_\Delta \Delta(\alpha)$ because
$\alpha \bot \ker \Delta$, which gives
$\alpha = G_\Delta dd^* \alpha = d G_\Delta\alpha$.
Writing $\eta_t=G_\Delta\alpha_t$, {\bf \purple we obtain
a smooth family $\eta_t$ such that $d\eta_t=\alpha_t$.}
\endproof

\newpage

{\bf \blue Fine resolutions}

\definition
Recall that a sheaf ${\cal F}$ over a manifold $M$
is called {\bf\blue fine} if for every covering $\{U_i\}$ of $M$
admitting a partition of unity, and any section
$f\in {\cal F}(M)$, there exists compactly supported
sections $f_i\in {\cal F}(U_i)$ such that
$\sum_i f_i =f$. 

\example 
Any sheaf of $C^\infty M$-modules is clearly fine.

\remark Fine sheaves are clearly {\bf \blue acyclic},
in other words, for any fine resolution 
$0 \arrow {\cal F} \arrow {\cal F}_1 \arrow {\cal F}_2 \arrow ...$
of a sheaf  ${\cal F}$, one has $H^i({\cal F}) = H^i(H^0({\cal F}_*))$
where $H^i(H^0({\cal F}_*))$ denotes the cohomology of the complex
of global sections
$0 \arrow H^0({\cal F}) \arrow H^0({\cal F}_1) \arrow H^0({\cal F}_2) \arrow ...$

\newpage

{\bf \blue Fine resolutions and fiberwise closed differential forms}

\remark
Let  $\R^\sigma_{M \times [0,1]}$
the sheaf of smooth functions which are constant
along the fibers of the projection
$\sigma:\; M\times [0,1] \arrow [0,1]$. The sheaf 
$\R^\sigma _{M \times [0,1]}$
{\bf \purple admits a fine resolution}
\[
0 \arrow \R^\sigma _{M \times [0,1]}\arrow C^\infty(M \times [0,1]) 
\stackrel {d_\sigma} \arrow \Lambda^1_\sigma(M \times [0,1]) \stackrel 
{d_\sigma} \arrow...
\]
where $d_\sigma$ is de Rham differential taken along
the fibers of $\sigma$,
and $\Lambda^*_\sigma(M \times [0,1])$ denotes the fiberwise differential
forms.

\claim
Any smooth family of closed forms 
$\alpha_t\in \Lambda^*(M)$, $t\in [0,1]$,
represents an element $[\alpha_t]\in H^i(\R^\sigma _{M \times [0,1]})$.
{\bf \red There exists a smooth   family $\eta_t\in \Lambda^*(M)$
such that $d\eta_t = \alpha_t$ if and only if
$[\alpha_t]=0$.}
\endproof


\newpage

{\bf \blue A smooth choice of anti-differential (3)}

\theorem
Let $\alpha_t\in\Lambda^k(M)$, $t\in [0,1]$ be
a smooth family of exact forms. Then
{\bf \red 
there exists a smooth family of forms $\eta_t\in\Lambda^{k-1}(M)$, $t\in [0,1]$,
such that $d\eta_t = \alpha_t$.}


{\bf \green Proof using the sheaf theory. Step 1}
The proof will follow if we prove that
{\bf \purple the family $\alpha_t$ represents 0 in the cohomology of 
$\R^\sigma _{M \times [0,1]}$.}


{\bf \green Step 2:}
Let $R^*\sigma_*$ denote the functor of
derived direct image of the sheaf. This is the
same as the functor mapping a sheaf to its 
fiberwise cohomology sheaf.

Let $\R_{M \times [0,1]}$ denote the constant sheaf on
$M \times [0,1]$.
Since $\R^\sigma _{M \times [0,1]}= 
\R_{M \times [0,1]}\otimes_\R \R^\sigma _{M \times [0,1]}$,
and the functor $\otimes_\R \R^\sigma _{M \times [0,1]}$ is exact,
one has 
\[ R^i\sigma_*(\R^\sigma_{M \times [0,1]})= 
   R^i\sigma_*(\R_{M \times [0,1]})\otimes_\R \R^\sigma _{M \times [0,1]}
\]
(``the base change theorem'').
Here $R^i\sigma_*(\R^\sigma_{M \times [0,1]})$ is the constant sheaf
on $[0,1]$ with fiber $H^i(M, \R)$. 
Therefore, a fiberwise exact form $\alpha_t$ represents
zero in $R^i\sigma_*(\R^\sigma_{M \times [0,1]})$,
and in 
\[ 
H^0(R^i\sigma_*(\R^\sigma_{M \times [0,1]})) = 
H^i(\R^\sigma _{M \times [0,1]}).
\]
\endproof





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