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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf K\"ahler manifolds and holonomy \\[15mm]
\small lecture 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf Tel-Aviv University
\\[2mm]  December 16, 2010, 
}
\end{center}

\newpage

{\bf \blue K\"ahler manifolds}

{\bf\green DEFINITION:} An Riemannian metric $g$ on
an almost complex manifiold $M$ is called 
{\bf \blue Hermitian} if $g(Ix, Iy)= g(x,y)$.
In this case, $g(x, Iy)= g(Ix, I^2y) = - g(y, Ix)$,
hence $\omega(x,y):= g(x, Iy)$ is skew-symmetric.

{\bf\green DEFINITION:} The differential 
form $\omega\in \Lambda^{1,1}(M)$ is called
{\bf \blue the Hermitian form} of $(M,I,g)$.

\remark It is $U(1)$-invariant, hence {\bf \purple of Hodge type (1,1)}.

{\bf\green DEFINITION:} A complex Hermitian manifold $(M,I,\omega)$
is called {\bf \blue K\"ahler} if $d\omega=0$. 
The cohomology class $[\omega]\in H^2(M)$ of a form $\omega$ 
is called {\bf \blue the K\"ahler class} of $M$, and
$\omega$ {\bf \blue the K\"ahler form}. 

\newpage

{\bf \blue Levi-Civita connection and K\"ahler geometry}


\definition
Let $(M, g)$ be a Riemannian manifold. A connection $\nabla$ 
is called {\bf \blue orthogonal} if $\nabla(g) =0$.
It is called {\bf \blue Levi-Civita} if it is torsion-free.

\theorem (``the main theorem of differential geometry'')\\
{\bf \red For any Riemannian manifold, the
Levi-Civita connection exists,\\ and it is unique}.

{\bf \green THEOREM:} Let $(M,I,g)$ be an almost complex Hermitian
manifold. {\bf \purple Then the following conditions are equivalent.}

(i) {\bf \red  $(M,I,g)$ is K\"ahler}

(ii) One has {\red $\nabla(I)=0$,} where $\nabla$ is the Levi-Civita connection.


\newpage

{\bf \blue Holomorphic vector bundles}

\definition A (smooth) {\bf \blue vector bundle} on a smooth manifold
is a locally trivial sheaf of $C^\infty M$-modules.

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

\remark {\bf \red A section $b$ of a bundle $B$ is often 
denoted as $b\in B$}.

\claim
Let $B$ be a holomorphic vector bundle.
Consider the sheaf $B_{C^\infty}:=B \otimes_{\calo_M} C^\infty M$.
It is clearly locally trivial, hence {\bf \purple $B_{C^\infty}$ is 
a smooth vector bundle.}

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


\newpage

{\bf \blue A holomorphic structure operator}

\definition
Let $d= d^{0,1}+d^{1,0}$ be the Hodge decomposition
of the de Rham differential on a complex manifold, 
$d^{0,1}:\; \Lambda^{p,q}(M) \arrow \Lambda^{p,q+1}(M)$
and $d^{1,0}:\; \Lambda^{p,q}(M) \arrow \Lambda^{p+1,q}(M)$.
The operators $d^{0,1}$, $d^{1,0}$ are denoted $\bar\6$
and $\6$ and called {\bf \blue the Dolbeault differentials}.

\remark
From $d^2=0$,
one obtains {\bf \red $\bar\6^2=0$ and $\6^2=0$.}

\remark
{\bf \purple The operator $\bar\6$ is $\calo_M$-linear.}

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

\remark The kernel of {\bf \purple $\bar\6$ coincides with the set 
of holomorphic sections} of $B$.

\newpage

{\bf \blue The $\bar\6$-operator on vector bundles}

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

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

\remark If $\bar\6$ is a holomorphic structure operator,
then $\bar\6^2=0$.

\theorem (Atiyah-Bott) 
Let $\bar\6:\; V \arrow \Lambda^{0,1}(M)\otimes V$
be a $\bar\6$-operator, satisfying $\bar\6^2=0$. {\bf \red Then
$B:=\ker \bar\6\subset V$ is a holomorphic vector
bundle of the same rank.}

\remark This statement is a vector bundle analogue
of Newlander-Nirenberg theorem.

\definition
$\bar\6$-operator $\bar\6:\; V \arrow \Lambda^{0,1}(M)\otimes V$
on a smooth manifold is called {\bf \blue
a holomorphic structure operator}, if $\bar\6^2=0$.

\newpage 

{\bf Connections and holomorphic structure operators}

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

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

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

\theorem
On any holomorphic Hermitian vector bundle, {\bf \red the
Chern connection exists, and is unique.}

\newpage

{\bf \blue Curvature of a connection}

\definition
Let  $\nabla:\; B \arrow B \otimes \Lambda^1 M$
be a connection on a smooth budnle. Extend it to an
operator on $B$-valued forms
\[
B \stackrel{\nabla}\arrow \Lambda^{1}(M)\otimes B
\stackrel{\nabla}\arrow \Lambda^{2}(M)\otimes B 
\stackrel{\nabla}\arrow \Lambda^{3}(M)\otimes B \stackrel{\nabla}\arrow ...
\]
using $\nabla(\eta \otimes b) = d\eta + (-1)^{\tilde \eta} \eta \wedge \nabla b$.
The operator $\nabla^2:\; B \arrow B\otimes \Lambda^{2}(M)$
is called {\bf \blue the curvature} of $\nabla$.

\remark The algebra of $\End(B)$-valued forms naturally acts
on $\Lambda^* M \otimes B$. The curvature
satisfies $\nabla^2(fb) = d^2 f b + df \wedge \nabla b - df \wedge
\nabla b + f \nabla^2 b= f \nabla^2 b$, hence it is
$C^\infty M$-linear. {\bf \purple We consider it as an $\End(B)$-valued
2-form on $M$. }

\proposition (Bianchi identity)
Using the {\bf \purple graded Jacobi identity}, we obtain 
$[\nabla, \nabla^2]=[\nabla^2,\nabla]+
[\nabla, \nabla^2]=0$, hence $[\nabla, \nabla^2]=0$.
This gives {\bf \blue Bianchi identity:}
$\nabla(\Theta_B)=0$.

\remark If $B$ is a line bundle, $\End B$ is trivial,
and {\bf \red the curvature $\Theta_B$ of $B$ is a closed 2-form.}

\definition
The cohomology class
$c_1(B):=\frac{\1}{2\pi}[\Theta_B]\in H^2(M)$
is called {\bf \blue the real first Chern class of a line bunlde $B$}.

{\bf \green An exercise:} Check that $c_1(B)$
is independent from a choice of $\nabla$.

\newpage

{\bf \blue Curvature of a holomorphic line bundle}

\remark When speaking of a {\bf \blue ``curvature of a holomorphic
bundle'',} one usually means the curvature of a Chern connection.

\remark 
Let $B$ be a holomorphic Hermitian line bundle, and $b$ 
its non-degenerate section. Denote by $\eta$ a (1,0)-form
which satisfies $\nabla^{1,0} b=\eta\otimes b$.
Then $d|b|^2= \Re g(\nabla^{1,0} b, b) = \Re\eta|b|^2$.
{\bf \purple This gives $\nabla^{1,0} b= \frac{\6 |b|^2}{|b|^2}b=
   2\6\log|b| b.$}

\remark
Then $\Theta_B(b)= 2\bar\6\6\log|b| b$, {\bf \red  that is,
$\Theta_B = -2 \6\bar\6\log|b|$.}

\corollary
If $g' = e^{2f} g$ -- two metrics on a holomorphic line bundle,
$\Theta, \Theta'$ their curvatures, {\bf \purple one has
 $\Theta' - \Theta = -2 \6\bar\6 f$}


\claim Let $\eta$ be a closed (1,1)-form in the
same cohomology class as $\Theta_{B,h}$. {\bf \red Then $\eta$
is a curvature of a Chern connection} on $B$, for 
some metric $h'$. 

{\bf \green Proof:}
The difference $\Theta_{B,h}-\eta$ is an exact
(1,1)-form, hence {\bf \red belongs to an image of
$\6\bar\6$ (``$\6\bar\6$-lemma''):
$\Theta_{B,h}-\eta=-2 \6\bar \6 f$.} Then
the curvature of a metric $h':=e^{2f}h$
satisfies $\Theta_{B,h} - \Theta_{B,h'} = -2 \6\bar \6 f,$
hence $\eta=\Theta_{B,h'}$.
\endproof

\remark {\bf \purple
Such metric is unique, up to a constant.}


\newpage

{\bf \blue Calabi-Yau manifolds}

\remark
Let $B$ be a line bundle on a manifold. Using 
the long exact sequence of cohomology associated with the
exponential sequence
\[ 
0 \arrow \Z_M \arrow C^\infty M \arrow (C^\infty M)^* \arrow 0,
\] 
{\bf \red we obtain $0 \arrow H^1(M, (C^\infty M)^*) \arrow H^2(M, \Z) \arrow 0$.}

\definition
Let $B$ be a complex line bundle, and $\xi_B$ its defining element
in $H^1(M, (C^\infty M)^*)$. Its image in $H^2(M, \Z)$ is called
{\bf\blue the integer first Chern class} of $B$.

\remark
{\bf \purple A complex line bundle $B$ is (topologically)
trivial if  and only if
$c_1(B)=0$.}

\theorem (Gauss-Bonnet) 
A real Chern class of a vector bundle {\bf \purple is an image
of the integer Chern class $c_1(B,\Z)$} under the natural
homomorphism $H^2(M, \Z)\arrow H^2(M, \R)$. 

\definition 
{\bf\blue A first Chern class} of a complex $n$-manifold
is $c_1(\Lambda^{n,0}(M))$.

\definition\\
{\bf\blue A Calabi-Yau manifold} is a compact 
Kaehler manifold with $c_1(M,\Z)=0$.

\newpage

{\bf \blue Calabi-Yau theorem}

\newcommand{\Ric}{\operatorname{Ric}}
\definition Let $(M,I, \omega)$ be a Kaehler $n$-manifold, and
$K(M):= \Lambda^{n,0}(M)$ its {\bf \blue canonical bundle.} We consider
$K(M)$ as a colomorphic line bundle, $K(M)= \Omega^n M$.
The natural Hermitian metric on $K(M)$ is written as
\[ (\alpha, \alpha') \arrow \frac{\alpha\wedge \bar \alpha'}{\omega^n}.\]
Denote by $\Theta_K$ the curvature
of the Chern connection on $K(M)$.
The {\bf\blue Ricci curvature} $\Ric$ of $M$ is symmetric
2-form $\Ric(x,y)= \Theta_K(x, Iy)$.

\definition
A K\"ahler manifold is called {\bf \blue Ricci-flat}
if its Ricci curvature vanishes. 

\theorem
(Calabi-Yau) \\
Let $(M, I, g)$ be Calabi-Yau manifold. {\bf \red Then there exists
a unique Ricci-flat Kaehler metric in any given
Kaehler class.}

\newpage

{\bf \blue Calabi-Yau theorem and Monge-Amp\`ere equation}

\remark
Let $(M, \omega)$ be a K\"ahler $n$-fold, and
$\Omega$ a non-degenerate section of $K(M)$,
Then $|\Omega|^2 = \frac{\Omega\wedge \bar \Omega}{\omega^n}$
If $\omega_1$ is a new Kaehler metric on $(M,I)$, $h, h_1$ the
associated metrics on $K(M)$, then
$\frac {h} {h_1}= \frac {\omega_1^n}{\omega^n}$

\corollary
A metric $\omega_1= \omega+\6\bar\6\phi$ {\bf \red is Ricci-flat if and only if
$(\omega+\6\bar\6\phi)^n = \omega^n e^f$,} where $-2\6\bar\6 f= \Theta_{K,\omega}$.

{\bf\green Proof:} For such $f$, $\phi$, one has
$\log\frac {h} {h_1}= - f$. This gives
\[
 \Theta_{K,\omega_1}= \Theta_{K,\omega}+ \6\bar\6\frac {h} {h_1}
 = \Theta_{K,\omega}-2\6\bar\6 f = 0.
\]
\endproof


\theorem
(Calabi-Yau) Let $(M, \omega)$ be a compact Kaehler $n$-manifold, and
$f$ any smooth function. {\bf \red Then there exists
a unique up to a constant function $\phi$} such that
$(\omega+ dd^c \phi)^n = A e^f \omega^n,$
where $A$ is a positive constant obtained from the
formula $\int_M A e^f \omega^n= \int_M \omega^n$.

\remark
\[
(\omega+ dd^c \phi)^n = A e^f \omega^n,
\]
is called {\bf\blue the Monge-Ampere equation.}


\newpage

{\bf \blue Uniqueness of solutions of
complex Monge-Ampere equation}

\proposition (Calabi)
{\bf \red A complex Monge-Ampere equation has at most one solution,}
up to a constant.

{\bf \green Proof. Step 1:}
Let $\omega_1, \omega_2$ be solutions of Monge-Ampere equation.
Then $\omega_1^n = \omega_2^n$. By $dd^c$-lemma, one has
$\omega_2= \omega_1 + dd^c \psi$. {\bf \purple We need to show $\psi=const$.}

{\bf \green  Step 2:}
This gives
\[
0 = (\omega_1 + dd^c \psi)^n - \omega_1^n= 
dd^c \psi\wedge \sum_{i=0}^{n-1}\omega_1^i \wedge\omega_2^{n-1-i}.
\]

{\bf \green  Step 3:} Let 
$P:=\sum_{i=0}^{n-1}\omega_1^i \wedge\omega_2^{n-1-i}$.
This is a positive $(n-1, n-1)$-form. {\bf \purple There exists
a Hermitian form $\omega_3$ on $M$ such that $\omega_3^{n-1}=P$.}

{\bf \green  Step 4:} Since $dd^c \psi\wedge P =0$,
this gives $\psi dd^c \psi\wedge P=0$. Stokes' formula implies
\[
0 = \int_M \psi \wedge\6 \bar\6\psi\wedge P=
- \int_M \6\psi \wedge\bar\6 \psi\wedge P = - \int_M  |\6\psi|_3^2\omega_3^n.
\]
where $|\cdot|_3$ is the metric associated to $\omega_3$.
{\bf \red Therefore $\bar\6 \psi=0$.}
\endproof

\end{document}

