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Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{% {\bf \green EXERCISE:\ }} \newcommand{\proof}{% {\bf \green Proof:\ }} \newcommand{\pstep}{% {\bf \green Proof. Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 \end{document}