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