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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{\pstep}{{\bf \green Proof. Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Hodge theory \\[15mm] \small lecture 17: Poincar\'e-Dolbeault-Grothendieck lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, March 24, 2018 } \end{center} \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 For a different proof of the following proposition, see Lecture 16. \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 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 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 $\eta \in \Lambda^{0,p}(D^n)$ 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.} \remark We have proven PDG-lemma for an $(0,1)$-form $\eta$ with compact support in $\C$. In this case {\bf \purple $\eta= \bar\6\alpha$, where $\alpha$ is a smooth function on $\C$.} This function is not necessarily compactly supported, but {\bf \purple is is bounded by $C/|z|$ for large $z$.} \remark Using the decomposition $\Lambda^{p,q}(D^n) \cong \Lambda^{p,0}(D^n) \otimes \Lambda^{0,q}(D^n)$, any form can be represented by a sum $\sum_i \alpha^{0,q}_i \wedge P^{p,0}_i$, where $P_i$ are monomials on $dz_i$, where $z_i$ are holomorphic coordinate functions. Since $\bar\6(\alpha^{0,q}_i \wedge P^{p,0}_i) = \bar\6(\alpha^{0,q}_i) \wedge P^{p,0}_i$, {\bf \purple it suffices to prove the Poincar\'e-Dolbeault-Grothendieck lemma for $(0,q)$-forms.} \remark To prove vanishing of cohomology of $\bar\6:\; \Lambda^{0,q}(M) \arrow\Lambda^{0,q+1}(M)$, it suffices to construct {\bf \blue the homotopy operator}, that is, a map \\ $\bar\6:\; \Lambda^{0,q}(M) \arrow\Lambda^{0,q-1}(M)$ satisfying $\{\bar\6, \gamma\}=\Id$. {\bf \purple This is how we prove the Poincar\'e lemma in Handout 8.} \newpage {\bf \blue Proof of Poincar\'e-Dolbeault-Grothendieck lemma} \pstep Let $\bar\6_i:\; \Lambda^{0,q}(D^n) \arrow\Lambda^{0,q+1}(D^n)$ 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 $D^n$. {\bf \purple Then $\bar\6= \sum_i \bar\6_i$}. {\bf \green Step 2:} By PDG-lemma in dimension 1, cohomology of $\6_i$ vanish. Denote by $\gamma_i$ the corresponding integral operator $P_\xi$. 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)=0$. This implies that {\bf \purple $\im \big[\{\bar\6_i, \gamma_i\}-\Id\big]$ lies in the space $R_i$ forms without $d\bar z_i$ in monomial decomposition and with all coefficients holomorphic as functions on $z_i$.} {\bf \green Step 3:} Properties $\gamma_i$: \\ {\bf \purple (1). $\im \big[\{\bar\6_i, \gamma_i\}-\Id\big]\subset R_i$. (2). $\{\bar\6_i, \gamma_j\}=0$, if $i\neq j$. (3). $\big[\{\bar\6_i, \gamma_i\}\big]\restrict {R_i}=0$. (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 2, properties (2) and (4) are implied by the exlicit formula for $\gamma_i$. Finally, (3) follows because for all forms $\alpha$ without $d\bar z_i$ in monomial decomposition one has $\{\gamma_i, \bar\6\}(\alpha) = \gamma_i(\bar\6_i(\alpha))$. {\bf \green Step 4:} 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 Proof of Poincar\'e-Dolbeault-Grothendieck lemma (2)} {\bf \green Step 4:} 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 5:} Let $\gamma:= \sum_i\gamma_i$. Since $\{\bar\6_i, \gamma_j\}=0$ for $i\neq j$, Step 4 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 6:} Let $W_0$ be the space of $(0,p)$-forms on $D^n$ which can be smoothly extended in a certain neighbourhood of the closure $\overline{D^n}\subset \C^n$, and $W_k \subset W_{k-1}$ its 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 5 implies $\big[\{\bar\6, \gamma\} -(n-k)\Id\big]\restrict{W_k}\subset W_{k+1}$.} {\bf \green Step 7:} Clearly, $W_n=0$ for any $p>0$: elements of this space are $(0,p)$-forms without any $d\bar z_i$ in its monomial decomposition. Using induction in $d=n-k$, {\bf \purple we can assume that any $\bar\6$-closed form in $W_{k+1}$ is $\bar\6$-closed; to prove PDG-lemma, it would suffice to prove the same for any $\bar\6$-closed form $\alpha\in W_{k}$.} Step 6 gives $(n-k)\alpha - \{\bar\6, \gamma\}(\alpha) = (n-k)\alpha - \bar\6\gamma(\alpha)\in W_{k+1}$, and this for 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 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 (reminder)} \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, f{\bf \purple or 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 \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 \end{document}