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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small Lecture 21: Hodge theory with coefficients in a bundle and Serre's duality}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, April 11, 2018 } 
\end{center}

\newpage

{\bf \blue Curvature (reminder)}

\definition
Let $\nabla:\; B \arrow B \otimes \Lambda^1 M$ be a connection
on a vector bundle $B$. We extend $\nabla$ to an operator
\[
V \stackrel{\nabla}\arrow \Lambda^{1}(M)\otimes V
\stackrel{\nabla}\arrow \Lambda^{2}(M)\otimes V 
\stackrel{\nabla}\arrow \Lambda^{3}(M)\otimes V \stackrel{\nabla}\arrow ...
\]
using the Leibnitz identity
$\nabla(\eta \otimes b) = d\eta + (-1)^{\tilde \eta} \eta \wedge \nabla b$.
Then the operator $\nabla^2:\; B \arrow B\otimes \Lambda^{2}(M)$
is called {\bf \blue the curvature} of $\nabla$.

\remark
{\bf \purple The algebra of differential forms
with coefficients in $\End B$ acts on
$\Lambda^* M \otimes B$} via
$\eta \otimes a (\eta' \otimes b) = \eta \wedge \eta'
\otimes a(b)$, where
$a\in \End(B)$, $\eta, \eta'\in \Lambda^* M$, and $b\in B$.



\remark
$\nabla^2(fb) = d^2 f b + df \wedge \nabla b - df \wedge
\nabla b + f \nabla^2 b$, hence {\bf \purple
the curvature is a $C^\infty M$-linear operator.}
{\bf \red We shall consider the curvature $B$ as 
a 2-form with values in $\End B$}.
Then $\nabla^2 := \Theta_B \in \Lambda^2 M \otimes \End B$,
where  an $\End(B)$-valued form acts on $\Lambda^* M \otimes B$
as above.

\newpage

{\bf \blue Holomorphic bundles (reminder)}

\definition
{\bf \blue Holomorphic vector bundle} on a complex
manifold $M$ is a locally trivial sheaf of $\calo_M$-modules.

\definition
{\bf \blue The total space} $\Tot(B)$ of a holomorphic bundle $B$ over $M$ is
the space of all pairs $\{x\in M, b \in B_x/{\goth m}_x B\}$,
where $B_x$ is the stalk of $B$ in $x\in M$ and ${\goth m}_x$ the
maximal ideal of $x$. We equip $\Tot(B)$  with the natural topology
and holomorphic structure, in such a way that 
$\Tot(B)$ becomes a locally trivial holomorphic
fibration with fiber $\C^r$, $r=\rk B$.

\remark {\bf \purple The set of holomorphic sections of a map
$\Tot(B)\arrow M$ is naturally identified with the
set of sections of the sheaf $B$.}

\claim 
Ler $B$ be a holomorphic bundle. Consider the sheaf
$B_{C^\infty}:=B \otimes_{\calo_M} C^\infty M$.
{\bf \red Then $B_{C^\infty}$ is a locally trivial sheaf of
$C^\infty M$-modules}.

\definition
$B_{C^\infty}$ is called {\bf \blue smooth vector bundle
underlying the holomorphic vector bundle $B$}.

\remark The natural map
$\Tot(B)\arrow \Tot(B_{C^\infty})$ is a diffeomorphism.

\newpage

{\bf \blue $\bar\6$-operator on vector bundles (reminder)}

\definition
Let $B$ be a holomorphic vector bundle on $M$.
Consider an operator
$\bar\6:\; B_{C^\infty}\arrow B_{C^\infty}\otimes \Lambda^{0,1}(M)$
mapping $b\otimes f$ to $b\otimes \bar\6 f$,
where $b$ is a holomorphic section of $B$, and $f$ smooth.
This operator is called {\bf\blue a holomorphic structure operator}
on $B$. {\bf \red It is well-defined because $\bar\6$
is $\calo_M$-linear}, and $B_{C^\infty}=B \otimes_{\calo_M} C^\infty M$.

\definition
A {\bf \blue $\bar\6$-operator}
on a smooth complex vector bundle $V$ over a complex manifold is
a differential operator $V \stackrel {\bar\6}\arrow \Lambda^{0,1}(M)\otimes V$
satisfying $\bar\6(fb) = \bar\6(f)\otimes b + f\bar\6(b)$
for any $f\in C^\infty M, b\in V$.

\remark
A $\bar\6$-operator {\bf \purple can be extended to 
$\bar\6:\; \Lambda^{0,i}(M)\otimes V \arrow \Lambda^{0,i+1}(M)\otimes V,$}
using the Leibnitz identity 
 $\bar\6 (\eta \otimes b) = \bar\6(\eta)\otimes b + 
(-1)^{\tilde \eta}\eta\wedge\bar\6(b)$,
for all $b\in V$ and $\eta \in \Lambda^{0,i}(M)$.


\theorem {\bf \blue (Malgrange)}
Let  $\bar\6:\; V \arrow \Lambda^{0,1}(M)\otimes V$
be a $\bar\6$-operator on a complex vector bundle,
satisfying  $\bar\6^2=0$, where $\bar\6$ is extended 
to \[
V \stackrel{\bar\6}\arrow \Lambda^{0,1}(M)\otimes V
\stackrel{\bar\6}\arrow \Lambda^{0,2}(M)\otimes V 
\stackrel{\bar\6}\arrow \Lambda^{0,3}(M)\otimes V \stackrel{\bar\6}\arrow ...
\]
as above.
{\bf \red Then
 $B:=\ker \bar\6\subset V$ is a holomorphic bundle of the
same rank, and  $V=B_{\C^\infty}$.}

\newpage

{\bf \blue Chern connection (reminder)}

\definition
Let $V$ be a smooth complex vector bundle with connection
 $\nabla:\; V \arrow \Lambda^1(M)\otimes V$
and holomorphic structure $\bar\6:\; V \arrow \Lambda^{0,1}(M)\otimes V$.
Consider the Hodge type decomposition of $\nabla$,
$\nabla= \nabla^{0,1} + \nabla^{1,0}$, where
\[
\nabla^{0,1}:\; V \arrow \Lambda^{0,1}(M)\otimes V, \ \ \ 
\nabla^{1,0}:\; V \arrow \Lambda^{1,0}(M)\otimes V.
\]
We say that {\bf \blue the connection $\nabla$ is compatible
with the holomorphic structure} if  $\nabla^{0,1}=\bar\6$.

\definition
{\bf \blue A holomorphic Hermitian vector buncle}
is a smooth complex vector bundle equipped with
a Hermitian metric and a holomorphic structure.

\definition
{\bf \blue Chern connection} on a holomorphic Hermitian 
vector bundle is a unitary connection compatible with the
holomorphic structure.

\theorem
Every holomorphic Hermitian vector bundle
{\bf \red admits a Chern connection, which is unique.}

\remark
When people say about ``curvature of a holomorphic
Hermitian line bundle'', {\bf \red they speak about curvature
of a Hermitian connection.}


\newpage

{\bf \blue Curvature of Chern connection (reminder)}


\remark $[d_\nabla, \{d_\nabla, d_\nabla\}]=[\{d_\nabla, d_\nabla\},d_\nabla]+
[d_\nabla, \{d_\nabla, d_\nabla\}]=0$ by the super Jacobi identity.
This gives {\bf \blue the Bianchi identity:}
$d_\nabla(\Theta_B\wedge \eta) = \Theta_B \wedge d_\nabla(\eta)$,
giving $d_\nabla(\Theta_B)=0$.

\remark
When $B$ is a line bundle, $\End(B)$ is trivial,
and $\Theta_B$ is a 2-form. {\bf \purple In this case $d_\nabla=\nabla$
and $\Theta_B$ is closed.}

\definition
When $B$ is a line bundle, the cohomology class
of $-\frac{\1}{\pi} \Theta_B$ is called
{\bf \blue first Chern class of $B$}, denoted
$c_1(B)$.

\theorem
The exponential exact sequence
$0 \arrow \Z_M \arrow C^\infty M \arrow (C^\infty M)^* \arrow 0$
gives an isomorphism $H^1(M, (C^\infty M)^*) \stackrel{c_1^\Z} \arrow H^2(M, \Z)$.
Let $L$ be a line bundle associated with
a cocycle $\eta$. {\bf \red Then the image of $c_1^\Z(\eta)$
in $H^2(M, \R)$ is $c_1(L)$.}

\newpage

{\bf \blue Supersymmetry in K\"ahler geometry (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. $L(\alpha):= \omega\wedge \alpha$, $\Lambda(\alpha) := * L * \alpha$. 
It is easily seen that $\Lambda= L^*$.\\
3. The Weil operator $W\restrict{\Lambda^{p,q}(M)}=\1(p-q)$

\theorem {\bf \blue (K\"ahler package)}\\
{\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 {\bf \blue an odd Heisenberg 
superalgebra}
$\langle d, d^c, d^*, (d^c)^*, \Delta\rangle$, 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 Coordinate operators (reminder) }

Let $V$ be an even-dimensional real vector space equipped with
a scalar product, and $v_1, ..., v_{2n}$ an orthonormal basis.
Denote by $e_{v_i}:\; \Lambda^k V  \arrow \Lambda^{k+1} V$
an operator of multiplication, $e_{v_i}(\eta) = v_i \wedge \eta$.
Let $i_{v_i}:\; \Lambda^k V  \arrow \Lambda^{k-1} V$
be an adjoint operator, $i_{v_i}= * e_{v_i} *$.

\claim
The operators $e_{v_i}$, $i_{v_i}$, $\Id$ are a basis of an {\blue \bf 
odd Heisenberg Lie superalgebra $\goth H$}, 
with {\bf \purple the only non-trivial 
supercommutator given by the formula $\{ e_{v_i}, i_{v_j}\} = \delta_{i,j}\Id$.}

Now, consider the tensor $\omega= \sum_{i=1}^n v_{2i-1} \wedge v_{2i}$,
and let $L(\alpha) = \omega \wedge \alpha$, and $\Lambda:= L^*$
be the corresponding {\bf \blue Hodge operators}.

\remark Relations in $\goth H$ imply that
$H:=[L, \Lambda] = 
\left [\sum e_{v_{2i-1}} e_{v_{2i}}, \sum i_{v_{2i-1}} i_{v_{2i}}\right]
= \sum_{i=1}^{2n} e_{v_i} i_{v_i} - \sum_{i=1}^{2n} i_{v_i} e_{v_i}$
{\bf \red is the scalar operator acting on $k$-forms 
as multiplication by $n-k$.}

\corollary 
{\bf \red The triple $L, \Lambda, H$ satisfies the relations for the
standard generators of $\goth{sl}(2)$}:
$[L,\Lambda]=H$, $[H,L]=2L$, $[H,\Lambda]=-2\Lambda$.

\newpage

{\bf \blue Kodaira identities (reminder)}


\theorem 
Let $M$ be a K\"ahler manifold. One has the following
identities {\bf \red (``K\"ahler idenitities'', ``Kodaira idenities'').}
\[
 [\Lambda, \6] = \1 \bar\6^*,  \ \ \ 
 [L, \bar\6] = - \1 \6^*,  \ \ \ 
  [\Lambda, \bar\6^*] = - \1 \6, \ \ \ 
 [L, \6^*] = \1 \bar\6.
\]
Equivalently,
\[ 
  [\Lambda, d] = (d^c)^*,\ \ \  
\ \ [ L, d^*] = - d^c,\ \ \ \ \ [\Lambda, d^c] = - d^*,
\ \ \ \ \ [ L, (d^c)^*] = d.
\]

\pstep The first set of identities implies 
the second set. Indeed, by adding up appropriate
identities in the top set of their complex conjugate, 
we obtain ones in the bottom set; 
for example, adding $[\Lambda, \6] = \1 \bar\6^*$
and $[\Lambda, \bar\6] = - \1 \6^*,$, we obtain $[\Lambda, d] = (d^c)^*$.
Each of top identities is related to the other three 
by complex conjugation or by Hermitian conjugation, hence
they are all equivalent. Each of the bottom identities
implies the rest by Hermitian conjugation and conjugating
with $I$. Finally, $[\Lambda, \6] = \1 \bar\6^*$
can be obtained as a sum of $[\Lambda, d] = (d^c)^*$
and $[\Lambda, d^c] = - d^*$ with appropriate coefficients.
{\bf \red We obtained that all Kodaira identities are
implied by just one, say, $[L, d^*] =- d^c$}.

\newpage

{\bf \blue Kodaira identities 2 (reminder)}

\pstep {\bf \red We reduced the Kodaira identities to just one, 
$[L, d^*] =- d^c$.}

{\bf \green Step 2:} Let ${\goth E}:\; \Lambda^iM 
\otimes \Lambda^1 M\arrow \Lambda^{i+1}(M)$ be the multiplication, 
and ${\goth I}:\; \Lambda^iM \otimes 
\Lambda^1 M\arrow \Lambda^{i-1}(M)$ the map that takes
$\alpha \wedge \theta$ and puts it to $*(*\alpha \wedge
\theta)$. In other words, ${\goth I}$ takes a tensor 
$\alpha \otimes \theta$, with $\alpha \in \Lambda^iM$
and $\theta \in \Lambda^1 M$, uses the metric $g$
to produce a vector field $X$ from $\theta$,
and maps $\alpha$ to $\alpha\cntrct X$ (convution
of $\alpha$ and $X$). 


{\bf \green Step 3:}
Let $\nabla$ be the Levi-Civita connection.
Then $d\alpha = {\goth E}(\nabla(\alpha))$, because $\nabla$ is
torsion-free. Since $d^*=*d*$, one has 
$d^*(\alpha)= {\goth I}(\nabla(\alpha))$.
Let $x_1, y_1, ..., x_n, y_n\in \Lambda^1_m M$
be an orthonormal basis such that 
$\omega= \sum x_i \wedge y_i$. 
Then ${\goth I}(\nabla(\alpha))= \sum_{i} i_{x_i}(\nabla_{x_i}
\alpha)+ i_{y_i}(\nabla_{y_i} \alpha)$.
Taking a commutator with $L= \sum e_{x_i}e_{y_i}$
and using the commutator relations between $e_{v}$ and
$i_w$ found earlier, we obtain
\[
[L, d^*]= \sum_i \nabla_{x_i}[e_{x_i}e_{y_i}, i_{x_i}]+
\nabla_{y_i}[e_{x_i}e_{y_i}, i_{y_i}]= \sum_i \nabla_{y_i}
e_{x_i}- \nabla_{x_i}e_{y_i}.
\]
(the operator $\nabla_{w}$ commutes with $L$, because
$\omega$ is parallel).
However, 
\[ \sum_i \nabla_{y_i}
e_{x_i}- \nabla_{x_i}e_{y_i}= 
-I\left(\sum_i\nabla_{x_i}e_{x_i} +
\nabla_{y_i}e_{y_i}\right) = - d^c
\]
which gives $[L, d^*] =- d^c$. \endproof

\newpage

{\bf \blue Conventions}

Let $B$ be a holomorphic Hermitian bundle,
and $\nabla= \bar\6 + \nabla^{1,0}$ its Chern connection.
In this lecture, {\bf \red I would use $\6$ instead of $\nabla^{1,0}$.}

As usual, we define the sequence 
\[
V \stackrel{\nabla}\arrow \Lambda^{1}(M)\otimes V
\stackrel{\nabla}\arrow \Lambda^{2}(M)\otimes V 
\stackrel{\nabla}\arrow \Lambda^{3}(M)\otimes V \stackrel{\nabla}\arrow ...
\]
using the Leibnitz identity 
$\nabla(\eta \otimes b) = d\eta + (-1)^{\tilde \eta} \eta \wedge \nabla b$.

Here the operators $\nabla$ are denoted by $d_\nabla$, their
(1,0) and (0,1)-parts $\6_\nabla$ and $\bar\6_\nabla$.
{\bf \red To simplify notations, I shall often omit the subscript 
${}_\nabla$.} 

These conventions are sloppy but more or less standard.

\newpage

{\bf \blue Kodaira relations}

\proposition
Let $B$ be a holomorphic Hermitian bundle on a K\"ahler manifold,
and $\nabla= \bar\6 + \6$ its Chern connection.
{\bf \red On $B$-valued differential forms, the following relations hold.}
\[
 [\Lambda, \6] = \1 \bar\6^*,  \ \ \ 
 [\Lambda, \bar\6] = - \1 \6^*,  \ \ \ 
  [L, \bar\6^*] = - \1 \6, \ \ \ 
 [L, \6^*] = \1 \bar\6.
\]
\proof It suffices
to show, for example, that $[L, \6^*] = -\1\bar\6$.
Let ${\goth E}:\; \Lambda^iM \otimes B
\otimes \Lambda^1 M\arrow \Lambda^{i+1}(M)\otimes B$
denote the multiplication map, and  ${\goth I}:\; \Lambda^iM\otimes B \otimes 
\Lambda^1 M\arrow \Lambda^{i-1}(M)\otimes B$ 
denote the convolution with the dual vector field.
{\bf \purple Then $\bar\6(\eta)= 
{\goth E}^{0,1}(\bar\6(\eta))$, and 
$\6^*(\eta)= {\goth I}^{-1,0}(\bar\6(\eta))$.}
Here $\bar\6$, $\6$ are $(0,1)$ and
$(1,0)$-parts of the connection in
$\Lambda^*(M) \otimes B$ induced from
the Levi-Civita connection on $\Lambda^*(M)$
and the Chern connection on $B$. %, and 
%${(\cdot)}^{p,q}$ means the Hodge component
%of an operator.

{\bf \green Step 2:}
Since $\nabla$ commutes with $L$, we have
 $[L, \6^*] = [L, {\goth I}^{-1,0}]\circ \nabla$,
where 
$L:\;\Lambda^iM \otimes \Lambda^1 M\otimes B \arrow 
\Lambda^{i+2}M \otimes \Lambda^1 M\otimes B$
acts on the first component.
Taking an orthonormal basis $z_i$ in 
$\Lambda^{1,0}(M)$, we obtain 
$L=-\1 \sum e_{z_i} e_{\bar z_i}$. Using the relations in
the odd Heisenberg algebra, we obtain
$[e_x, {\goth I}]= 1$. This gives
$[L, {\goth I}^{-1,0}](z_{i}\otimes \eta)= -\1 e_{\bar z_i}$ É 
$[L, {\goth I}^{-1,0}](\bar z_i\otimes \eta)= 0$, hence
{\bf \purple  $[L, {\goth I}^{-1,0}](a\otimes \eta)=
-\1 {\goth E}^{0,1}(a\otimes \eta)$.}

{\bf \green Step 3:}
Comparing the results of step 1 and 2,
$\bar\6(\eta)= 
{\goth E}^{0,1}(\bar\6(\eta))$, and 
$\6^*(\eta)= {\goth I}^{-1,0}(\bar\6(\eta))$ with 
$[L, {\goth I}^{-1,0}](a\otimes \eta)=
-\1 {\goth E}^{0,1}(a\otimes \eta)$, we obtain 
$[L, \6^*] = -\1\bar\6$.
\endproof

\newpage


{\bf \blue Hodge theory with coefficients in a bundle }

\definition
Let  $\bar\6:\; \Lambda^{p,q}(M)\otimes B\arrow 
\Lambda^{p,q+1}(M)\otimes B$ be the holomorphic structure
operator, extended to differential forms using the
Leibniz identity. The anticommutator
$\Delta_{\bar\6}:=\{\bar\6, \bar\6^*\}= \bar\6\bar\6^* + \bar\6^* \bar\6$
is called {\bf \blue Dolbeault Laplacian with coefficients in $B$}.
It is self-adjoint, positive elliptic operator:
 $(\Delta_{\bar\6} x, x)= (\bar\6x, \bar\6x) + (\bar\6^*x, \bar\6^*x).$

{\bf \green Hodge theory with coefficients in a bundle}
{\bf \red There is a basis in the Hilbert space
$L^2(\Lambda^*(M)\otimes B)$, consisting of eigenvalues
of $\Delta_{\bar\6}$, and each eigenspace is finite-dimensional.}

{\bf \green Elliptic regularity} {\bf \red Each eigenvector of
$\Delta_{\bar\6}$ on $L^2(\Lambda^*(M)\otimes B)$ is 
a smooth form.}

\definition 
A $B$-valued form $\eta$ is called 
{\bf \blue $\Delta_{\bar\6}$-harmonic}
if $\eta\in \Delta_{\bar\6}$.

\remark If $A, B$ holomorphic vector bundles,
$A \otimes B$ means $A\otimes_{\calo M} B$.
If $A$ is smooth, $B$ is holomorphic,
$A \otimes B$ means $A\otimes_{\calo M} B$.
If both $A$ and $B$ are smooth bundles,
$A \otimes B$ means $A\otimes_{C^\infty M} B$.
{\bf \purple The ranks of the bundles are multiplied in all
three cases.}

\newpage


{\bf \blue Dolbeault cohomology with coefficients in a bundle}

\theorem
{\bf \red The space of $\Delta_{\bar\6}$-harmonic forms 
 is identified with} the {\bf \blue Dolbeault cohomology
with coefficients in $B$}, that is, with $\frac{\ker \bar \6}{\im\bar\6}$.
We denote the $(p,q)$-part of Dolbeault cohomology by $H^{p, q}(M, B)$.

\proof Since $\Delta_{\bar\6}$ is elliptic,  
$\Lambda^*(M)=\im \Delta_{\bar\6}\oplus \ker  \Delta_{\bar\6}$. This
gives an orthogonal decomposition
$\Lambda^*(M)=\im\bar\6\oplus \ker\Delta_{\bar\6} \oplus \im\bar\6^*$.
Since  $\im\bar\6^*=  (\ker\bar\6)^\bot$,
this decomposition identifies $\ker\Delta_{\bar\6}$
with cohomology of $\bar\6$.
\endproof

\claim 
{\bf \red The cohomology $H^{p, q}(M, B)$ are identified with the
$H^q(B \otimes \Omega^pM)$,} where $\Omega^pM$ is  the
sheaf of holomorphic $p$-forms.

\proof
The sequence of sheaves
\[
0 \arrow \Omega^pM\otimes B\hookrightarrow \Lambda^{p,0}M\otimes B
\stackrel {\bar\6}\arrow \Lambda^{p,1}M\otimes B 
{\bar\6}\arrow \Lambda^{p,2}M\otimes B
\]
is exact by Poincar\'e-Dolbeault-Grothendieck lemma,
hence it gives an acyclic resolution of $\Omega^p(M)\otimes B$,
and cohomology of its global sections are identified with
the cohomology of the sheaf $B \otimes \Omega^pM$.
\endproof

\newpage

{\bf \blue Serre's duality }

\remark
The operator $*:\;\Lambda^{p,i}(M)\otimes_{\calo_M} B\arrow
\Lambda^{n-p,n-i}(M)\otimes_{\calo_M} B^*$
exchanges $\bar\6$  and $\pm\bar\6^*$.
{\bf \purple Therefore, it preserves the space 
$\ker \Delta_{\bar\6}$} of $\bar\6$-harmonic forms.

\remark
By definition, $H^n(\Omega^n M)= \frac{\Lambda^{n,n}(M)}{\im \bar\6}$.
Stokes formula implies that $\int_M \bar\6(\alpha)=0$ for all
$\alpha$. This gives a natural map
$H^n(\Omega^n M)\stackrel{\int_M}\arrow \C$.

\theorem {\bf \blue (Serre's duality)}\\
Let $M$ be an $n$-dimensional, compact complex
manifold, and $B$ an Hermitian holomorphic bundle. Then
{\bf \red the multiplication
\[ H^i(\Omega^p M\otimes B) \times H^{n-i}(\Omega^{n-p} M\otimes B^*)\arrow
   H^n(\Omega^n M) \stackrel{\int_M}\arrow \C
\]
defines a non-degenerate pairing.}

\proof
Indeed, for each $\eta \in \ker \Delta_{\bar\6}$,
the form $*\eta$ also belongs to $\ker \Delta_{\bar\6}$,
but $\int_M \eta\wedge *\eta>0$, hence this pairing is non-degenerate.
\endproof

\corollary {\bf \blue (Serre's duality for $p=n$)}
Let $M$ be a compact complex manifold,  $\dim_\C M =n$.
{\bf \red Then  $H^i(B)\cong H^{n-i}(B^*\otimes K_M)^*$,
where $K_M= \Omega^n M$.}

\definition $K_M= \Omega^n M$ is called {\bf \blue canonical bundle}.

\newpage

{\bf \blue Schedule for April and May }


\vfil

{\bf \purple Lectures: Misha Verbitsky (April 14,
  May 16) \\
Ekaterina Amerik (April 18, 21, 25)\\[10mm]
\red \Large Exam: Saturday, May 19.}

\vfil


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