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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 Complex geometry\\[15mm] \small lecture 15: Poincar\'e-Dolbeault-Grothendieck lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] November 11, 2020 } \end{center} \newpage {\bf \blue Weight decomposition for $T^n$-action (reminder)} \exercise Consider the $n$-dimensional torus $T^n$ as a Lie group, $T^n = U(1)^n$. {\bf \purple Prove that any finite-dimensional Hermitian representation of $T^n$ is a direct sum of 1-dimensional representations,} with action of $T^n$ given by $\rho(t_1, ..., t_n)(x)= \exp(2\pi\1\sum_{i=1}^np_it_i) x$, for some $p_1, ..., p_n\in \Z^n$, called {\bf \blue the weights} of the 1-dimensional representation. \definition Let $V$ be a Hermitian space (possibly infinitely-dimensional) equipped with an action of $T^n$, and $V_\alpha\subset V$ weight $\alpha$ representations, $\alpha\in \Z^n$. The direct sum $\bigoplus_{\alpha\in \Z^n} V_\alpha$ is called {\bf \blue the weight decomposition} for $V$ if it is dense in $V$. \theorem Let $W$ be a Hermitian vector space. Then {\bf \red the Fourier series provide the weight decomposition on $L^2(T^n, W)$.} \endproof \theorem Let $W$ be a Hermitian representation of $T^n$. {\bf \red Then $W$ admits a weight decomposition $V = \widehat{\bigoplus_{\alpha\in \Z^n} W_\alpha}$.} \proof We realize $W$ as a subrepresentation in $L^2(T^n, W)$, and use the Fourier series to obtain the weight decomposition of $L^2(T^n, W)$. \endproof \newpage {\bf \blue Weight decomposition for $T^n$-action on differential forms (reminder)} \remark Let $M$ be a manifold with the $T^n$-action, and \[ \Lambda^*(M)= \hat\bigoplus_{\alpha \in \Z^n} \Lambda^*(M)_{p_1, ..., p_k}\] be the weight decomposition on the differential forms. Then the {\bf \red de Rham differential preserves each term $\Lambda^*(M)_{p_1, ..., p_k}$.} Indeed, {\bf \purple $d$ commutes with the action of the Lie algebra of $T^n$, and $\Lambda^*(M)_{p_1, ..., p_k}$ are its eigenspaces.} \remark Let $\alpha= \sum \alpha_{p_1, ..., p_k}$ be the weight decomposition. {\bf \purple The forms $\alpha_{p_1, ..., p_k}$ are obtained by averaging \[ e^{2\pi\1\sum_{i=1}^np_it_i}\alpha = \Av_{T^n} e^{2\pi\1\sum_{i=1}^n-p_it_i}\alpha \] hence they are smooth.} \theorem Let $M$ be a smooth manifold, and $T^n$ a torus acting on $M$ by diffeomorphisms. Denote by $\Lambda^*(M)^{T^n}$ the complex of $T^n$-invariant differential forms. {\bf \red Then the natural embedding $\Lambda^*(M)^{T^n}\hookrightarrow \Lambda^*(M)$ induces an isomorphism on de Rham cohomology.} \newpage {\bf \blue Constant forms on a torus (reminder)} \definition Let $T^n=(S^1)^n$ be a compact torus equipped with a action on itself by shifts, and $\Lambda^*_\const(M)$. the space of $T^n$-invariant forms on $T^n$. These forms are called {\bf \blue constant differential forms}. Clearly, {\bf \purple constant forms have constant coefficients in the usual (flat) coordinates on the torus.} \theorem The natural embedding $\Lambda^*_\const(T^n)\hookrightarrow \Lambda^*(T^n)$ {\bf \red induces an isomorphism $\Lambda^*_\const(T^n)=H^*(T^n)$.} \proof The embedding $\Lambda^*_\const(T^n)= \Lambda^*(T^n)^{T_n} \hookrightarrow \Lambda^*(T^n)$ induces an isomorphism on cohomology, however, all constant forms are closed, hence $H^*(\Lambda^*_\const(T^n), d)=\Lambda^*_\const(T^n)$. \endproof \newpage {\bf \blue Holomorphic vector fields (reminder)} \definition Let $(M,I)$ be a complex manifold, and $X\in TM$ a real vector field. It is called {\bf \blue holomorphic} if $\Lie_X(I)=0$, that is, if the corresponding flow of diffeomorphisms is holomorphic. \claim Let $(M,I)$ be a complex manifold, and $X\in TM$ a holomorphic vector field. {\bf \red Then $X^c:= I(X)$ is also holomorphic, and commutes with $X$.} \lemma Let $X$ be a holomorphic vector field, and $X^c=I(X)$. {\bf \red Then $\{d^c, i_X\}= - \Lie_{X^c}$.} \proof Using $\{IdI^{-1}, i_X\}= I\{d, I^{-1}i_XI\}I^{-1}$, we obtain $\{d^c, i_X\}= - I \{d, i_{X^c}\}I^{-1}= I\Lie_{X^c}I^{-1}$. However, $X^c$ is holomorphic, hence $I\Lie_{X^c}I^{-1}= \Lie_{X^c}$. \endproof \proposition Let $X$ be a holomorphic vector field, and $X^c=I(X)$. {\bf \red Then $\{\bar\6, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})$.} \proof $\bar\6= \frac 1 2 (d +\1 d^c)$, hence \[ \{\bar\6, i_X\}= \frac 1 2 \Lie_X + \1 \{d^c, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c}). \] \endproof \newpage {\bf \blue Dolbeault cohomology of an elliptic curve (reminder)} \proposition Let $X=\C/\Z^2$ be an elliptic curve, and $\Lambda^*(X)= \bigoplus_{\alpha \in \Z^2} \Lambda^*(X)_{p_1,p_2}$ its weight decomposition under the $T^2$-action. Consider the space $T^2$-invariant forms $\Lambda^*(X)^{T^2}=\Lambda^*(X)_{0,0}$. {\bf \red Then the natural embedding $\Lambda^*(X)^{T^2} \hookrightarrow \Lambda^*(X)$ induces an isomorphism of Dolbeault cohomology.} \proof Let $\alpha\in \Lambda^*(X)_{p_1,p_2}$ be a $\bar\6$-closed form, with $(p, q)\neq (0,0)$. Suppose, for example, that $p\neq 0$, and $X$ is the generator of the corresponding component of the Lie algebra such that $\Lie_X\alpha=p\1\alpha$. Since $X^c$ belongs to the same Lie algebra, we have $\Lie_{X^c}(\alpha)= v\alpha$, where $v\in \1\R$. Then \[ \frac{\1p+ v}{2} \alpha= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})\alpha=\{ \bar\6, i_X\}\alpha=\bar\6 i_X\alpha, \ \ \ (***) \] hence $\alpha$ is $\bar\6$-exact. This implies that {\bf \purple $\bar\6$ has no cohomology on \[ \bigoplus_{p_1, p_2\neq (0,0)}\Lambda^*(X)_{p_1,p_2}.\]} \endproof \newpage {\bf \blue Dolbeault cohomology of a disk} \corollary Let $K\subset \C$ be a compact subset, $K^0$ its interior, and $\eta\in \Lambda^{0,1}(K^0)$ a form smoothly extending to a neighbourhood of $K$. {\bf \red Then $\eta$ is $\bar\6$-exact.} \proof Choosing an appropriate lattice $\Z^2\subset \C$, we may assume that $K$ is a subset of an elliptic curve $X$. Since $\eta$ extends to a neighbourhood of $K$, we can use partition of unity to extend it to a smooth form $\tilde \eta$ on $X$. Applying the weight decomposition $\tilde \eta =\sum_{\alpha\in \Z^2} \eta_\alpha$, we obtain that the form $\eta- \eta_{0,0}$ is $\bar\6$-exact. However, the constant part $\eta_{0,0}=\const\cdot dz \wedge d\bar z= \const \cdot\bar\6(\bar z dz)$ (for (1,1)-form) or $\eta_{0,0}=\const\cdot d\bar z= \const \cdot\bar\6(\bar z)$ for (0,1)-form is also $\bar\6$-exact. \endproof \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma} \definition {\bf \blue Polydisc} $D^n$ is a product of $n$ discs $D\subset \C$. \theorem {\bf \blue (Poincar\'e-Dolbeault-Grothendieck lemma)}\\ Let $\eta \in \Lambda^{p,q}(D^n)$, $q>0$, be a $\bar\6$-closed form on a polydisc, smoothly extended to a neighbourhood of its closure $\overline{D^n}\subset \C^n$. {\bf \red Then $\eta$ is $\bar\6$-exact.} {\bf \green We proved it for $n=1$.} Now we prove it for all $n$. %\newpage % %{\bf \blue Inverting $\bar\6$ using the Hodge theory} % % %\claim %Let $\beta$ be a $\bar\6$-exact form, and $\gamma:= \Delta^{-1}\bar\6^* \beta$. %{\bf \red Then $\bar\6(\gamma)=\beta$. } % %\proof %Indeed, %\[ % \bar\6^*\beta=\{\bar\6,\bar\6^*\}(\Delta^{-1}\bar\6^* \beta)= % \bar\6^*\bar\6\gamma %\] %because $(\bar\6^*)^2=0$ and $\Delta^{-1}$ commutes %with $\bar\6^*$. However, $\ker \bar\6^*$ is orthogonal %to $\im \bar\6$, hence $\bar\6^*\restrict{\im \bar\6}$ is injective. %Then $\bar\6^*\beta=\bar\6^*\bar\6\gamma$ implies %$\beta=\bar\6\gamma$. %\endproof % % %\remark A different proof (independent from the Hodge theory) of the following %proposition is given using the weight decomposition on a torus. %\newpage % %{\bf \blue Poincar\'e-Dolbeault-Grothendieck (dimension 1)} % %\proposition %Let $\alpha$ be a (p,1)-form on a disk $D_{r+\epsilon}\subset \C$ %of radius $r+\epsilon$. %{\bf \red Then $\alpha\restrict{D_r} = P_\xi(\alpha)$, %where $P_\xi:\; \Lambda^{p,1}(D_{r+\epsilon})\arrow %\Lambda^{p,0}(D_{r})$ is an operator %which depends only on $r$ and $\epsilon$. }\\[2mm] %\pstep %Hodge theory implies that {\bf \purple cohomology of $\bar\6$ on %any K\"ahler manifold are identified with cohomology %of $d$.} Indeed, the corresponding Laplacians coinside. \\[2mm] %{\bf \green Step 2:} Then the cohomology of $\bar\6$ on %a 1-dimensional complex torus $T$ are 1-dimensional in each %bidegree $(p,q)$. However, averaging of a $(p,q)$-form %on a torus gives a map from $\Lambda^{p,q}(T)$ to $T$-invariant %forms, which are clearly parallel, hence harmonic. Since %torus acts on itself by isometries, this action commutes %with the harmonic projection. Therefore, {\bf \purple a form is cohomologous %to 0 if and only if is average on $T$ vanishes.} This is %true for the de Rham differential and for the Dolbeault differential.\\[2mm] %{\bf \green Step 3:} Fix an embedding from $D_{r+\epsilon}$ %to a torus, and let $\psi$ be a cutoff function which is 1 on %$D_r$ and 0 outside of $D_{r+\epsilon}$. Then $\psi\alpha$ %is can be extended to a smooth $(p,1)$-form on $T$. %Chose a $(p,1)$-form which is supported on $T\backslash D_{r+\epsilon}$ %and has non-zero $\bar\6$-cohomology class $v\in H^{p,1}(T)$ in $T$, and let %$A(\psi\alpha)\in H^{p,1}(T)$ be the $\bar\6$-cohomology class of $\psi\alpha$. %Denote by $\lambda(\psi\alpha)\in \C$ the number such that %$\psi\alpha-\lambda(\psi\alpha)v$ is cohomologous to 0. %Then $P_\xi(\alpha):= \Delta^{-1} \bar\6^*(\psi\alpha-\lambda(\psi\alpha)v)$ %satisfies $\bar\6(P_\xi(\alpha))=\alpha + \lambda(\psi\alpha)v$, %and this form is equal to $\alpha$ on $D_r\subset T$. %\endproof \newpage {\bf \blue $\bar\6$-homotopy operator on $T^2$} From now on, {\bf \purple 1-dimensional complex torus is always $\C/\Z[\1]$ and the $n$-dimensional complex torus $T^{2n}$ is a product of $n$ copies of $T^2=\C/\Z[\1]$.} \claim Let $\mu \in \Lambda^{p,q}(M)_{a,b}$ be a form of weight $(a,b)$ on a torus $T^2=\C/\Z[\1]$, and $X$ the coordinate vector field along the real axis. {\bf \red Then $\{\bar\6, i_X\}(\mu)= \frac 1 2 (b+\1a)$.} \proof $\{\bar\6, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})$, and $X^c$ is the coordinate vector field along the imaginary axis, acting on $\mu$ by multiplication by $\1 b$. \endproof \definition Given $\mu= \sum_{a,b\in \Z^2}\mu_{a,b}$ define \[ P(\mu) := \sum_{(a,b)\neq (0,0)} 2(b+\1a)^{-1}\mu_{a,b}. \] The operator $P$ {\bf \purple commutes with all operators which commute with the $T^2$-action on itself:} with $d$, $d^c$, $i_X$, $i_{X^c}$, etc. \corollary {\bf \red Then $\{\bar\6, P i_X\})= \mu-\mu_{0,0}$.} In particular, {\bf \red if $\mu$ is $\bar\6$-closed, we also have $\bar\6P(i_X(\mu))=\mu-\mu_0$.} \endproof \newpage {\bf \blue Homotopy operator $\gamma_k$ on $T^{2n}$} Let $U\subset T^{2n}$ be a polydisk. Since $U$ is contractible, all constant $(p, q)$-forms on a torus with $q>0$ are $\bar\6$-exact on $U$: $\bar\6\bar z_i = \bar\6(\bar z_i)$, which can be well defined on $U$ because it is contractible. For any disk $U\subset T^2$, fix a cutoff function $\rho_\epsilon$ which is 1 on $U$ and 0 outside of a contractible $\epsilon$-neighbourhood of $\bar U$. Consider the map $Q:\; \Lambda^{p,1}(T^2)\arrow \Lambda^{p,0}(T^2)$ taking $\mu$ to $\mu_{0,0}$ and replacing any constant summand of form $\alpha \wedge \bar\6\bar z_i$ by $\rho_\epsilon \bar z_i \alpha$. \claim In these assumptions, {\bf \red we have $\{\bar\6, \gamma\}(\mu)=\mu$ on $U$ for any form $\mu\in \Lambda^{p,1}(T^2)$,} where $\gamma(\alpha)=P(i_X(\alpha))+ Q(\mu)$. \proof If $\mu_{0,0}=0$, we have $Q(\mu)=0$, and this expression becomes $\{\bar\6, P(i_X)\})= \mu-\mu_{0,0}$ proven above. If $\mu=\mu_{0,0}$, it becomes $\bar\6(Q(\mu))\restrict U = \mu$. \endproof {\bf \green Corollary 1:} Let $U\subset T^{2n}$ be a polydisk, and $\rho_\epsilon$ a cutoff function which is 1 on $U$ and 0 outside of a contractible $\epsilon$-neighbourhood of $\bar U$. We chose $\rho_\epsilon$ in such a way that $\Lie_{d/dx_i}(\rho_\epsilon)=0$ at any point $(x_1, ..., x_n)$ such that $|x_i|<1$. Let $\gamma_k$ denote the operator $\gamma$ along the $k$-th component in $T^{2n}=(T^2)^n$, and $\bar\6_k$ the $\bar\6$ along this component. {\bf \red Then $\{\bar\6_k, \gamma_k\}(\mu)=\mu$ on $U$ for any form $\mu$ divisible by $d\bar z_k$, and $\{\bar\6_k, \gamma_l\}\restrict U=0$ for $l\neq k$.} %\proof $\{\bar\6_k, \gamma_k\}(\mu)=\mu$ is proven above, %and $\{\bar\6_k, \gamma_l\}\restrict U=0$ follows from %$\Lie_{d/dx_i}(\rho_\epsilon)=0$. \endproof \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma} \theorem {\bf \blue (Poincar\'e-Dolbeault-Grothendieck lemma)}\\ Let $\eta \in \Lambda^{0,p}(D^n)$ be a $\bar\6$-closed form on a polydisc, smoothly extended to a neighbourhood of its closure $\overline{D^n}\subset \C^n$. {\bf \red Then $\eta$ is $\bar\6$-exact.} We prove the following version of Poincar\'e-Dolbeault-Grothendieck. \theorem Let $U\subset T^{2n}$ be a sufficiently small polydisk, and $\mu \in \Lambda^{p,q}(T^{2n})$ a form with $q>0$ which is $\bar\6$-closed on $U$. {\bf \red Then there exists $\alpha \in \Lambda^{p,q-1}(T^{2n})$ such that $\bar\6\alpha=\mu$ on $U$.} \pstep Let $\bar\6_i:\; \Lambda^{p,q}(T^n) \arrow\Lambda^{p,q+1}(T^n)$ be the operator $\alpha \arrow d\bar z_i \wedge \frac {d}{d\bar z_i}\alpha$, where $z_i$ is $i$-th coordinate on $T^n$. {\bf \purple Then $\bar\6= \sum_i \bar\6_i$}. Denote by $\gamma_i$ the homotopy operator defined above. If $\alpha= d\bar z_i\wedge \beta$, one has $\{\bar\6_i, \gamma_i\}(\alpha)=\alpha$. If $\alpha$ contains no monomials divisible by $d\bar z_i$, one has \[ \bar \6_i \{\bar\6_i, \gamma_i\}(\alpha)=\bar\6_i \gamma_i \bar \6_i (\alpha)= \{\bar\6_i, \gamma_i\}\bar \6_i\alpha= \bar \6_i\alpha, \] hence $\bar\6_i(\alpha-\{\bar\6_i, \gamma_i\})\restrict U=0$. This implies that {\bf \purple $\im \big[\{\bar\6_i, \gamma_i\}-\Id\big]\big|_ U$ lies in the space $R_i(U)$ of forms without $d\bar z_i$ in monomial decomposition and with all coefficients holomorphic as functions on $z_i$.} \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma (2)} \theorem Let $U\subset T^{2n}$ be a sufficiently small polydisk, and $\mu \in \Lambda^{p,q}(T^{2n})$ a form with $q>0$ which is $\bar\6$-closed on $U$. {\bf \red Then there exists $\alpha \in \Lambda^{p,q-1}(T^{2n})$ such that $\bar\6\alpha=\mu$ on $U$.} \pstep Let $\bar\6_i:\; \Lambda^{p,q}(D^n) \arrow\Lambda^{p,q+1}(D^n)$ be the operator $\alpha \arrow d\bar z_i \wedge \frac {d}{d\bar z_i}\alpha$, and $\gamma_i$ the homotopy defined above. Then $\im \big[\{\bar\6_i, \gamma_i\}-\Id\big]\restrict U$ lies in the space $R_i(U)$ of forms without $d\bar z_i$ in monomial decomposition and with all coefficients holomorphic as functions on $z_i$. {\bf \green Step 2:} Let $R_i$ denote the space of forms $\alpha$ on $T^{2n}$ such that $\alpha \restrict U$ belongs to the space $R_i(U)$ defined above. Properties of $\gamma_i$: \\ {\bf \purple (1). $\im \big[\{\bar\6_i, \gamma_i\}-\Id\big]\subset R_i$. (2). $\{\bar\6_i, \gamma_j\}\restrict U=0$, if $i\neq j$. (3). the restriction $\big[\{\bar\6_i, \gamma_i\}\big]\restrict {R_i}$ vanishes on $U$. (4). $\gamma_i(R_j)\subset R_j$, $\bar\6_i(R_j) \subset R_j$ for all $i\neq j$.}\\ Property (1) is proven in Step 1, property (2) and (4) follow because $\gamma_i$ is independent from the $z_j$-coordinate for all $j\neq i$. Finally, (3) follows because for all forms $\alpha$ without $d\bar z_i$ in its monomial decomposition one has $\{\gamma_i, \bar\6\}(\alpha) = \gamma_i(\bar\6_i(\alpha))$. {\bf \green Step 3:} Properties (1), (3) and (4) give $\left[\{\bar\6_i, \gamma_i\}-\Id\right]( {R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}}) \subset R_i\cap R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}$ for $i\notin \{i_1, i_2, ..., i_k\}$, and $\{\bar\6_i, \gamma_i\}\restrict {R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}}=0$ otherwise. \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma (3)} {\bf \green Step 3:} Properties (1), (3) and (4) give $\left[\{\bar\6_i, \gamma_i\}-\Id\right]( {R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}}) \subset R_i\cap R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}$ for $i\notin \{i_1, i_2, ..., i_k\}$, and $\{\bar\6_i, \gamma_i\}\restrict {R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}}=0$ otherwise. {\bf \green Step 4:} Let $\gamma:= \sum_i\gamma_i$. Since $\{\bar\6_i, \gamma_j\}=0$ for $i\neq j$, Step 3 gives \[ \big[\{\bar\6, \gamma\}-(n-k)\Id\big] (R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}) \subset \sum_{i\neq i_1, i_2, ..., i_k} R_i\cap R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k} \] {\bf \green Step 5:} Let $W_0=\Lambda^*(T^{2n})$, and $W_k \subset W_{k-1}$ the subspace generated by all $R_{i_1}\cap R_{i_2} \cap ...\cap R_{i_k}$ for $i_1 < i_2 < ... < i_k$. {\bf \purple Step 4 implies $\big[\{\bar\6, \gamma\} -(n-k)\Id\big]\restrict{W_k}\subset W_{k+1}$.} {\bf \green Step 6:} $W_n$ is the space of $(p,0)$-forms holomorphic on $U$, and it does not contain any $(p,q)$-forms for $q>0$. Using induction in $d=n-k$, {\bf \purple we can assume that any $\bar\6$-closed $(p,q)$-form in $W_{k+1}$ is $\bar\6$-exact when $q>0$. To prove PDG-lemma, it would suffice to prove the same for any $\bar\6$-closed form $\alpha\in W_{k}$.} Step 5 gives $(n-k)\alpha - \{\bar\6, \gamma\}(\alpha) = (n-k)\alpha - \bar\6\gamma(\alpha)\in W_{k+1}$, and this form is $\bar\6$-exact by the induction assumption. {\bf \red This gives $(n-k)\alpha - \bar\6\gamma(\alpha) = \bar\6\eta$}, hence $\alpha$ is $\bar\6$-exact. \endproof \newpage {\bf \blue Hartogs theorem} \theorem Let $f$ be a holomorphic function on $\C^n \backslash K$, where $K\subset \C^n$ is a compact, and $n>1$. {\bf \red Then $f$ can be extended to a holomorphic function on $\C^n$.}\\[2mm] % \pstep Replacing $K$ by a bigger compact, we can assume that $f$ is smoothly extended to a small neighbourhood of the closure $\overline{M\backslash K}$. Then $f$ can be extended to a smooth function on $\C^n$, holomorphic outside of $K$. {\bf \purple Then $\alpha:=\bar\6\tilde f$ is a $\bar\6$-closed $(0,1)$-form with compact support}.\\[2mm] % {\bf \green Step 2:} Using the standard open embedding of $\C^n$ to $\C P^n$, we may consider $\alpha$ as a $\bar\6$-closed $(0,1)$-form on $\C P^n$. Since $H^1(\C P^n)=0$, this gives $\alpha =\bar\6\phi$, where $\phi$ is a continuous function on $\C P^n$. In particular, {\bf \purple $\phi$ is bounded on $\C^n\subset \C P^n$}.\\[2mm] % {\bf \green Step 3:} Since $\bar\6\phi$ vanishes outside of $K$, the function $\phi$ is holomorphic outside of $K$. Since bounded holomorphic functions on $\C$ are constant, {\bf \purple $\phi$ is constant on any affine line not intersecting $K$}. \\[2mm] {\bf \green Step 4:} This implies that $\phi=\const$ on the union of all affine lines not intersecting $K$. Since $n>1$, the complement of this set is compact. Substracting constant if necessary, we obtain that {\bf \purple $\phi$ is a function with compact support}.\\[2mm] % {\bf \green Step 5:} $\bar\6(\tilde f -\phi)= \alpha-\alpha=0$, {\bf \red hence $\tilde f -\phi$ is holomorphic}. However, $\phi$ has compact support, and therefore $f=\tilde f -\phi$ outside of a compact. \endproof \newpage {\bf \blue Algebra of supersymmetry of a K\"ahler manifold: reminder} Let $(M, I, g)$ be a Kaehler manifold, $\omega$ its Kaehler form. {\bf \blue On $\Lambda^*(M)$, the following operators are defined.} 0. $d$, $d^*$, $\Delta$, because it is Riemannian. 1. $L(\alpha):= \omega\wedge \alpha$ 2. $\Lambda(\alpha) := * L * \alpha$. It is easily seen that $\Lambda= L^*$. 3. The Weil operator $W\restrict{\Lambda^{p,q}(M)}=\1(p-q)$ \theorem {\bf \red These operators generate a Lie superalgebra $\goth a$ of dimension $(5|4)$,} acting on $\Lambda^*(M)$. Moreover, the Laplacian $\Delta$ is central in $\goth a$, hence {\bf \purple $\goth a$ also acts on the cohomology of $M$. } The odd part of this algebra generates ``odd Heisenberg algebra'' $\langle d, d^c, d^*, (d^c)^*, \Delta\rangle$, with the only non-zero anticommutator $\{d, d^*\}= \{d^c, (d^c)^*\}=\Delta$. The even part of this algebra contains an $\goth{sl}(2)$-triple $\langle L, \Lambda, H\rangle$ acting on $\goth a^{\odd}$ as on a direct sum of two weight 1 representations (``Kodaira relations''). The Weil element commutes with $\langle L, \Lambda, H, \Delta\rangle$ and acts on $\goth a^{\odd}$ via $[W, d]= d^c$, $[W, d^*]= (d^c)^*$. \newpage {\bf \blue Inverting $\bar\6$ using the Hodge theory } \claim Let $\beta$ be a $\bar\6$-exact form, and $\gamma:= \Delta^{-1}\bar\6^* \beta$. {\bf \red Then $\bar\6(\gamma)=\beta$. } \proof Indeed, \[ \bar\6^*\beta=\{\bar\6,\bar\6^*\}(\Delta^{-1}\bar\6^* \beta)= \bar\6^*\bar\6\gamma \] because $(\bar\6^*)^2=0$ and $\Delta^{-1}$ commutes with $\bar\6^*$. However, $\ker \bar\6^*$ is orthogonal to $\im \bar\6$, hence $\bar\6^*\restrict{\im \bar\6}$ is injective. Then $\bar\6^*\beta=\bar\6^*\bar\6\gamma$ implies $\beta=\bar\6\gamma$. \endproof \remark Similarly, {\bf \purple for any $d$-exact form $\beta$, one has $\beta=\Delta^{-1}d^* \beta$.} \newpage {\bf \blue $dd^c$-lemma} \theorem Let $\eta$ be a form on a compact K\"ahler manifold, satisfying one of the following conditions.\\ (1). $\eta$ is an exact $(p,q)$-form. (2). $\eta$ is $d$-exact, $d^c$-closed. \\ (3). $\eta$ is $\6$-exact, $\bar\6$-closed. \\ {\bf \red Then $\eta\in \im dd^c=\im \6\bar\6$.} \proof Notice immediately that in all three cases {\bf \purple $\eta$ is closed and orthogonal to the kernel of $\Delta$}, hence its cohomology class vanishes. Since $\eta$ is exact, it lies in the image of $\Delta$. Operator $G_\Delta:=\Delta^{-1}$ is defined on $\im \Delta= \ker \Delta^\bot$ and commutes with $d, d^c$. In case (1), $\eta$ is $d$-exact, and $I(\eta)= \bar\eta$ is $d$-closed, hence $\eta$ is $d$-exact, $d^c$-closed like in (2). Then $\eta= d \alpha$, where $\alpha:=G_\Delta d^*\eta$. Since $G_\Delta$ and $d^*$ commute with $d^c$, the form $\alpha$ is $d^c$-closed; since it belongs to $\im \Delta=\im G_\Delta$, it is $d^c$-exact, $\alpha=d^c \beta$ which gives $\eta = dd^c \beta$. In case (3), we have $\eta= \6 \alpha$, where $\alpha:=G_\Delta \6^*\eta$. Since $G_\Delta$ and $\6^*$ commute with $\bar\6$, the form $\alpha$ is $\bar\6$-closed; since it belongs to $\im \Delta$, it is $\bar\6$-exact, $\alpha=\bar\6 \beta$ which gives $\eta = \6\bar\6 \beta$. \endproof \newpage {\bf \blue Massey products} Let $a, b, c\in \Lambda^*(M)$ be closed forms on a manifold $M$ with cohomology classes $[a], [b], [c]$ satisfying $[a][b]=[b][c]=0$, and $\alpha, \gamma\in \Lambda^*(M)$ forms which satisfy $d(\alpha)= a\wedge b$, $d(\gamma) = b \wedge c$. Denote by $ L_{[a]}, L_{[c]}:\; H^*(M) \arrow H^*(M)$ the operation of multiplication by the cohomology classes $[a], [c]$. {\bf \purple Then $\alpha \wedge c - a\wedge \gamma$ is a closed form, and its cohomology class is well-defined modulo $\im L_{[a]}+\im L_{[c]}$}. \definition Cohomology class $\alpha \wedge c - a\wedge \gamma$ is called {\bf \blue Massey product of $a, b, c$.} \proposition {\bf \red On a K\"ahler manifold, Massey products vanish.} \proof Let $a, b, c$ be harmonic forms of pure Hodge type, that is, of type $(p, q)$ for some $p, q$. Then $ab$ and $bc$ are exact pure forms, hence $ab, bc\in \im dd^c$ by $dd^c$-lemma. This implies that $\alpha:=d^* G_\Delta(ab)$ and $\gamma:=d^* G_\Delta(bc)$ are $d^c$-exact. Therefore $\mu:=\alpha \wedge c - a\wedge \gamma$ is a $d^c$-exact, $d$-closed form. {\bf \purple Applying $dd^c$-lemma again, we obtain that $\mu$ is $dd^c$-exact, hence its cohomology class vanish.} \endproof \end{document}