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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small Lecture 22: Kodaira-Nakano vanishing theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, April 15, 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 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 Hodge theory with coefficients in $B$ (reminder)}

\proposition
Let $B$ be a holomorphic Hermitian bundle on a K\"ahler manifold,
and $\nabla= \bar\6 + \6$ its Chern connection.
We use the same letters $\6$, $\bar\6$ for the
Hodge components of $d_\nabla:\; \Lambda^i(M)\otimes B \arrow
\Lambda^{i+1}(M)\otimes B$. Then
{\bf \red on $B$-valued differential forms, the following relations hold.}
\[
 [\Lambda, \6] = \1 \bar\6^*,  \ \ \ 
 [\Lambda, \bar\6] = - \1 \6^*,  \ \ \ 
  [L, \bar\6^*] = - \1 \6, \ \ \ 
 [L, \6^*] = \1 \bar\6.
\]
\!\!\!\!\!\!\!\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.}

\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 $H^{p, q}(M, B)$.

\newpage

{\bf \blue Serre's duality (reminder)}

\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


\newpage

{\bf \blue Laplacians and curvature }

\remark Curvature of Chern connection:
$\Theta_B=\{\nabla^{1,0}, \bar\6\}$,
in the present notation it's
$\Theta_B=\{\6, \bar\6\}$.

\theorem {\bf \blue (Bochner-Kodaira-Nakano identity)}\\
{\bf \red $\Delta_{\bar\6} =  \Delta_{\6} + H_B$,
where $H_B:=-\1 [\Lambda, \Theta_B]$.}

\proof
Super-Jacobi identity gives
\begin{multline*} 
 [\Lambda, \Theta_B]= [\Lambda, \{\6, \bar\6\}] =
 \{[\Lambda, \6], \bar\6\} + \{\6,  [\Lambda, \bar\6]\}\\
 = \1 \{\bar\6^*, \bar\6\} - \1\{\6^*, \6\}= 
 \1 \Delta_{\bar\6} - \1 \Delta_{\6}.
\end{multline*}
\endproof

\remark
The operators $\Delta_{\bar\6}$ and $\Delta_{\6}$
are positive. Therefore, if $(H_Bx, x) >0$,
one has  $(\Delta_{\bar\6}x,x)>0$. {\bf \purple If
 $(H_Bx, x) >0$ for all $x$, the operator
 $\Delta_{\bar\6}$ has no kernel, and the
corresponding cohomology group vanishes.}

\newpage

{\bf \blue Positive line bundles}

\definition
A holomorphic line bundle is called {\bf \blue
positive} if its first Chern class is cohomologous
to a K\"ahler form.

{\bf \green Theorem 1:} {\bf \blue (Kodaira-Nakano)}
Let $L$ be a positive line bundle on a compact K\"ahler
manifold. {\bf \red Then for any bundle $B$ there exists $N>0$
such that $H^i(B\otimes L^N)=0$ for all $i>0$}.

\proof We deduce Theorem 1 from Theorem 2 below.

{\bf \green Theorem 2:} {\bf \blue (Kodaira-Nakano)}
Let $B$ be a holomorphic Hermitian line bundle on $n$-dimensional
K\"ahler manifold, $\Theta_B$ its curvature, and 
$L_{\Theta_B}:\; \Lambda^*(M) \arrow \Lambda^*(M)$ 
the operator of multiplication by $\Theta_B$.
Suppose that the self-adjoint operator 
$H_B:=\1[L_{\Theta_B}, \Lambda]$ satisfies
$(H_B(x), x)>0$ for any non-zero $k$-form $x$, $k>n$.
{\bf \red Then $H^p(B\otimes \Omega^q M)=0$ for all
$p+q >n$.}

\proof
$\Delta_{\bar\6}= \Delta_\6 -H_B$, which gives
 $(\Delta_{\bar\6}x, x)= (\Delta_{\6}x, x) - (H_Bx,x)>0$.
\endproof

\newpage

{\bf \blue Kunihiko Kodaira}


\begin{center}{\epsfig{file=Kodaira_Kunihiko.jpg,width=0.33\linewidth}\\[10mm]
{\bf \small \blue Kunihiko Kodaira (1915-1997)}}
\end{center}


\newpage

{\bf \blue Negative line bundles and cohomology}

\remark
For any vector bundles $E, F$ with connections, their 
curvatures are related as $\Theta_{E\otimes F} = \Theta_E + \Theta_F$.
Also, if $L$ is a line bundle, one has $\Theta_{L^*}=- \Theta_L$.

\definition
Let $L$ be a line bundle such that $L^*$ is positive.
Then $L$ is called {\bf \blue negative}.

\remark
Suppose that $L$ is a negative line bundle on 
$(M,\omega)$, and $\1\omega$  the curvature of $L$.
Then $H_{L}:=\1[L_{\Theta_L}, \Lambda]= H$.
{\bf \red On $k$-forms $H_L$ acts as multiplication by $k-n$.}
{\bf \purple Therefore, $L$ satisfies conditions of Theorem 2, and
$H^p(L\otimes \Omega^q M)=0$ 
for all $p+q >n$.}


\newpage

{\bf \blue Kodaira-Nakano theorem: the proof}

\remark
Suppose $L$ is a negative vector bundle,
with $H_L= -\1[L_{\Theta_L}, \Lambda]= H$.
The operator
$H_{E\otimes F}$ is expressed as 
$H_{E\otimes F}= H_E + H_F$. This gives
$H_{B\otimes L^N} = H_B- N H$. For 
 $N > \alpha$, where $\alpha$ is the
biggest eigenvalue of $H_B$, we have
\[ (H_{B\otimes L^N}x, x)= (H_B x, x)+ N(n-k)|x|^2> 0.
\] 
{\bf \purple Then Theorem 2 gives $H^p(L^N\otimes B \otimes \Omega^q M)=0$ 
for all $p+q >n$.}

\corollary
Let $L$ be a negative line bundle.
Then {\bf \red for all vector bundles $B$ with Chern connection,
there exists $N>0$ such that $H^i(L^{-N} \otimes B)=0$
for all $i>0$.}

\proof
Apply the previous corollary to $B^*\otimes K_M$, and use the Serre's duality
\[ 0 = H^{n-i} (L^N\otimes B^* \otimes \Omega^m M)= H^i(B \otimes
       L^{-N})^*.
\]
\endproof

\remark {\bf \purple We have proved Theorem 1.}
\endproof


\newpage

{\bf \blue Forms realized as a curvature  (reminder)}

{\bf \green Proposition 1:}
Let $\omega$ be an integer (1,1)-form with integer
cohomology class on a compact K\"ahler manifold.
{\bf \red Then $\omega$ is a curvature of a holomorphic
line bundle}.

\pstep Exponential exact sequence 
 $0 \arrow \Z_M \arrow \calo_M \arrow \calo^*_M\arrow 0$
gives
\[
H^1(\calo^*_M) \stackrel c \arrow H^2(M, \Z) \stackrel p \arrow H^2(M, \calo_M),
\]
where $H^1(\calo^*_M)= \Pic(M)$ is the group of 
holomorphic line bundles, $c$ maps a bundle to its first
Chern class, and $p$ projects $H^2(M)$ to its Hodge
component $H^2(M, \calo_M)= H^{0,2}(M)$. Then
{\bf \purple for any integer class $[\omega]\in
  H^{1,1}(M)\cap H^2(M, \Z)$, there exists
a line bundle $L$ such that $[\omega]= c_1(L)$.}


{\bf \green Step 2:}
Take any metric $h$ on $L$. Its curvature
$\omega_h$ is a closed (1,1)-form, cohomologous to
$\omega$. By $dd^c$-lemma, $\omega_h -\omega = -2 \6\bar\6
f$ for some $f\in C^\infty M$. {\bf \purple By Corollary 1,
curvature of $h':=e^{2f}h$ satisfies 
$\omega_h -\omega_{h'}= -2 \6\bar \6 f$,}
giving  $\omega_{h'}= \omega$. \endproof

\newpage

{\bf \blue Base points of  line bundles}

\definition
Let $L$ be a holomorphic line bundle on $M$. 
A point $x\in M$ is called {\bf \blue base point}
for $M$ if any global section of $L$ {\bf \blue vanishes in $x$,}
that is, if it does not generate the stalk $L_x$ over the
ring of germs $\calo_{M, x}$.

A bundle is called {\bf \blue base point free}
if it has no base points.

\example
Let $\C^{n+1}\backslash 0 \stackrel \pi \arrow \C P^n$ be the
natural projection. For each point $x\in \C P^n$, the
fiber $\pi^{-1}(x)$ is $\C^*\subset l_x$, where $l_x$ is the
line associated with $x$. Denote by $\calo(-1)$ the bundle
with the fiber $l_x$ in each $x\in \C P^n$.
Each line functional on $\C^{n+1}$ defines a line functional
on each $l_x$, which holomorphically depends 
on $x$. {\bf \red This means that the dual line bundle,
denoted $\calo(1)$, is base point free.}


\remark
Let $L$ be a base point free line bundle,
and ${\goth m}_x\subset \calo_{M, x}$ the maximal ideal
of a point $x\in M$. Consider {\bf \blue the fiber}
$L\restrict x:= L_x /{\goth m}_x$. It is a vector space
over $\C$ of rank 1. 
A map $H^0(M, L)\arrow L_x$ taking $f$ to $f\restrict x$
defines a map $\phi_x\in \Hom_\C(H^0(M, L), L\restrict x)\cong H^0(M, L)^*$.
{\bf \purple The isomorphism $\Hom_\C(H^0(M, L), L\restrict x)\cong H^0(M, L)^*$
is defined only after we fix an isomorphism $L\restrict x \cong \C$,}
that is, up to a constant multiplier. Therefore, 
{\bf \red the map $x\arrow \psi_x$ defines a holomorphic
map $\phi:\; M \arrow {\Bbb P}H^0(M,L)^*$}. {\bf \purple This map is well-defined
only if $L$ is base point free.}

\newpage

{\bf \blue Very ample line bundles}

\claim
Let $L$ be a base point free holomorhic line bundle on $M$, and
$\phi:\; M \arrow {\Bbb P}H^0(M,L)^*$ the natural map defined above.
{\bf \red Then $L\cong \phi^*(\calo(1))$.}

\pstep This statement is essentially a tautology:
global sections of $\calo(1)$ on ${\Bbb P} V$ are 
identified with $V^*$, hence global sections
of $\phi^*(\calo(1))$ are identified with $H^0(M,L)$.

{\bf\green  Step 2:}
To see that this identification is carried over to 
the corresponding sheaves, we 
consider the tautological line bundle
$\calo(-1)$ with the fiber $l_x$ at each 
$x\in {\Bbb P}H^0(M,L)^*$. By construction, $l_x$ is dual to
$L\restrict x$, and this duality is holomorphic
on $x$, hence $\phi^*(\calo(-1))=L^*$.
\endproof

\definition
A line bundle $L$ is called {\bf \blue very ample}
if it is base point free, and the corresponding
holomorphic map $\phi:\; M \arrow {\Bbb P}H^0(M,L)^*$
is an embedding. It is called {\bf \blue ample}
of $L^{\otimes N}$ is very ample for some $N>0$.

\definition
A complex manifold is called {\bf \blue projective}
if it admits a holomorphic embedding to $\C P^n$.


\newpage

{\bf \blue Kodaira embedding theorem}

\corollary 
Let $(M, \omega)$ be a compact K\"ahler manifold
such that the cohomology class of $\omega$ is rational.
{\bf \red Then $M$ admits a positive line bundle.}

\proof Let $n\omega$ be a K\"ahler form proportional
to $\omega$ with integer K\"ahler class. {\bf \purple Then 
$n\omega$ is realized as a curvature of a holomorphic
line bundle (Proposition 1).} This line bundle
is  positive. \endproof

\theorem {\bf \blue (Kodaira)}
{\bf \red A positive bundle is ample}.\\
\proof {\em (will be proven later in April)} \endproof

\corollary {\bf \blue (Kodaira embedding theorem)}\\
{\bf \red A compact K\"ahler manifold $M$ is projective
if and only if it admits a K\"ahler form with
rational cohomology class.}

\proof
Using the Corollary above, we obtain that
$M$ admits a positive holomorphic line bundle $L$,
and the theorem of Kodaira above implies that
$L$ is ample, hence $L^{\otimes N}$ defines
a projective embedding.

Conversely, if $M\subset \C P^n$ is a projective
manifold, the cohomology class of Fubini-Study form 
is proportional to a rational class, 
hence $M$ admits a K\"ahler form with a rational cohomology class.
\endproof



\end{document}

