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





