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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Locally conformally K\"ahler manifolds \\[15mm]
\small lecture 2: Vaisman theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE and IUM, Moscow
\\[2mm]  February 17, 2014
}
\end{center}

\newpage

{\bf \blue Complex manifolds (reminder)}


{\bf\green DEFINITION:} Let $M$ be a smooth manifold. 
An {\bf \blue almost complex structure} is an operator
$I:\; TM \arrow TM$ which satisfies $I^2 = - \Id_{TM}$.

{\bf \purple The eigenvalues of this operator are $\pm \1$.}
The corresponding eigenvalue 
decomposition is denoted $TM=T^{0,1}M\oplus T^{1,0}(M)$.


{\bf\green DEFINITION:}
An almost complex structure is {\bf \blue integrable}
if $\forall X,Y \in T^{1,0}M$, one has $[X,Y]\in T^{1,0}M$.
In this case $I$ is called {\bf \blue a complex structure operator}.
A manifold with an integrable almost complex structure
is called {\bf \blue a complex manifold}. 

{\bf\green THEOREM:} (Newlander-Nirenberg)\\
{\bf \red This definition is equivalent to the usual one.}

\remark The commutator defines a $\C^\infty M$-linear map\\
$N:=\Lambda^2(T^{1,0})\arrow T^{0,1}M$, called {\bf \blue 
the Nijenhuis tensor} of $I$. {\bf \purple One can represent $N$ as a section 
of $\Lambda^{2,0}(M) \otimes T^{0,1}M$.}

\newpage


{\bf \blue K\"ahler manifolds (reminder)}

{\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 $\omega\in \Lambda^2(M)$ 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}. 


{\bf \green Definition:} Let $M=\C P^n$ be a complex projective
space, and $g$ a $U(n+1)$-invariant Riemannian form. It is called
{\bf \blue Fubini-Study form on $\C P^n$}. The Fubini-Study
form is obtained by taking arbitrary Riemannian form
and averaging with $U(n+1)$ using the Haar measure on $U(n+1)$.

\exercise
Prove that {\bf \red the Fubini-Study form is unique} (up to a constant
multiplier).



\newpage

{\bf \blue REMINDER: The Hodge decomposition in linear algebra}

\definition
Let $(V,I)$ be a space equipped with a complex structure.
{\bf \blue The Hodge decomposition} 
$V\otimes_\R \C:= V^{1,0}\oplus V^{0,1}$ is defined in such a way that
$V^{1,0}$ is a $\1$-eigenspace of $I$, and $V^{0,1}$ a $-\1$-eigenspace.

\claim 
$\Lambda^*(V\oplus W)= \Lambda^*(V)\otimes \Lambda^*(W)$

\remark 
Let $V_\C:= V \otimes_\R \C$.
The decomposition $V_\C=V^{1,0}\oplus V^{0,1}$ induces
 $\Lambda^*_\C(V) =
\Lambda^*_\C(V^{0,1})\otimes \Lambda^*_\C(V^{1,0})$, giving
\[ 
  \Lambda^d V_\C= \bigoplus_{p+q=d} \Lambda^p V^{1,0} 
  \otimes \Lambda^q V^{0,1}.
\]
We denote $\Lambda^p V^{1,0} \otimes \Lambda^q V^{0,1}$ by $\Lambda^{p,q}V$.
The resulting decomposition $\Lambda^n V_\C= \bigoplus_{p+q=n}\Lambda^{p,q}V$
is called {\bf \blue the Hodge decomposition of the Grassmann algebra}.



\newpage

{\bf \blue REMINDER: Cauchy-Riemann equation and Hodge decomposition}

The $(p,q)$-decomposition is defined on differential forms
on complex manifold, in a similar way.

\definition
Let $(M,I)$ be a complex manifold
A differential form $\eta\in \Lambda^1(M)$
is {\bf \blue of type (1,0)} if $I(\eta)=\1\eta$, and
{\bf \blue of type (0,1)} if $I(\eta)=-\1\eta$. The corresponding
vector bundles are denoted by $\Lambda^{1,0}(M)$ and
$\Lambda^{0,1}(M)$.

\remark 
Cauchy-Riemann equations can be written as $df\in \Lambda^{1,0}(M)$.
That is, {\bf \red a function $f\in C^{\infty}_\C(M)$ is holomorphic
if and only if $df \in \Lambda^{1,0}(M)$.}

\remark 
Let $(M,I)$ be a complex manifold, and $z_1, ..., z_n$
holomorphic coordinate system in $U\subset M$, with $z_i$ being 
holomorphic functions on $U$. {\bf \purple Then $dz_1, ..., dz_n$ generate
the bundle $\Lambda^{1,0}(M)$, and $d\bar z_1, ..., d\bar z_n$
generate $\Lambda^{0,1}(M)$.}

\exercise {\bf \red Prove this.}


\newpage

{\bf \blue REMINDER: The Hodge decomposition on a complex manifold}


\definition
Let $(M,I)$ be a complex manifold, $\{U_i\}$ its covering,
and and $z_1, ..., z_n$ holomorphic coordinate system
on each covering patch.
{\bf \blue The bundle $\Lambda^{p,q}(M,I)$ 
of $(p,q)$-forms on $(M,I)$} is generated
locally on each coordinate patch by monomials 
$dz_{i_1}\wedge dz_{i_2}\wedge ...\wedge dz_{i_p}
\wedge d\bar z_{i_{p+1}}\wedge ...\wedge dz_{i_{p+q}}$.
{\bf \blue The Hodge decomposition} is a decomposition
of vector bundles: 
\[ \Lambda^d_\C(M)=\bigoplus_{p+q=d} \Lambda^{p,q}(M).
\]

%\remark 
%One has $\Lambda^{p,q}(M)= \Lambda^{p,0}(M)\otimes \Lambda^{0,q}(M)$.
%This gives \\ $\rk \Lambda^{p,q}(M)= \binom{n}{p}\cdot\binom{n}{q},$
%where $n=\dim_\C M$.

\exercise
Prove that the {\bf \purple de Rham differential on a complex manifold
has only two Hodge components:} 
\[
 d\left(\Lambda^{p,q}(M)\right)\subset
 \Lambda^{p+1,q}(M)\oplus  \Lambda^{p,q+1}(M).
\]

\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}.

\exercise
Show that {\bf \purple $\6^2=0$ is equivalent to integrability of the
complex structure.}


\newpage


{\bf \blue Supercommutator (reminder)}

\definition
A {\bf \blue supercommutator}  of pure operators
on a graded vector space is defined by a formula
$\{a,b\}= ab - (-1)^{\tilde a \tilde b}ba$.


\definition
A graded associative algebra is called {\bf \blue
graded commutative} (or ``supercommutative'')
if its supercommutator vanishes.

\example {\bf \purple The Grassmann algebra is supercommutative.}

\definition
{\bf \blue A graded Lie algebra} (Lie superalgebra)
is a graded vector space $\g^*$
equipped with a bilinear graded map 
$\{\cdot,\cdot\}:\; \g^*\times \g^* \arrow \g^*$
which is graded anticommutative:
$\{a,b\} = - (-1)^{\tilde a \tilde b}\{b,a\}$
and satisfies {\bf \blue the super Jacobi identity}
$\{c, \{a,b\}\} = \{\{c, a\},b\}+ (-1)^{\tilde a \tilde c}\{a,\{c, b\}\}$


\example
Consider the algebra $\End(A^*)$ of operators on a 
graded vector space, with supercommutator as above.
{\bf \purple Then $\End(A^*), \{\cdot,\cdot\}$ is a graded Lie algebra.}

{\bf \green Lemma 1:}
Let $d$ be an odd element of a Lie superalgebra, satisfying
$\{d,d\}=0$, and $L$ an even or odd element. {\bf \red Then $\{\{L, d\}, d\}=0$.}


{\bf \green Proof:} 
$0=\{L,\{d,d\}\}= \{\{L, d\}, d\}+
(-1)^{\tilde L}\{d,\{L, d\}\}=2\{\{L, d\}, d\}.$ 
\endproof

\newpage

{\bf \blue  The twisted differential $d^c$ (reminder)}

\definition The {\bf \blue twisted differential}
is defined as $d^c:=I d I^{-1}$.

\claim Let $(M,I)$ be a complex
manifold. {\bf \blue Then 
$\6:= \frac{d + \1 d^c}2$, $\bar \6:= \frac{d - \1 d^c}2$
are the Hodge components of $d$}, $\6= d^{1,0}$, 
$\bar\6= d^{0,1}$. 

{\bf \green Proof:} The Hodge components of $d$ are expressed as
$d^{1,0}=\frac{d + \1 d^c}2$, $d^{0,1}=\frac{d - \1
  d^c}2$. Indeed,
$I(\frac{d + \1 d^c}2)I^{-1}=\1\frac{d + \1 d^c}2$, hence 
{\bf \purple $\frac{d + \1 d^c}2$ has Hodge type (1,0);} the same
argument works for $\bar\6$.
\endproof

\claim
On a complex manifold, one has 
{\bf \red $d^c = [{\cal W}, d]$. }

{\bf \green Proof:} Clearly, $[{\cal W}, d^{1,0}]= \1 d^{1,0}$ and
$[{\cal W}, d^{0,1}] = - \1 d^{0,1}$. Adding these equations,
obtain $d^c = [{\cal W}, d]$.

\corollary $\{d, d^c\} = \{d, \{d, {\cal W}\}\}=0$ (Lemma 1).

\remark
Clearly, $d= \6 + \bar\6$, $d^c=-\1 (\6-\bar\6)$,
$dd^c=-d^cd = 2\1\6\bar\6$.

%\newpage
%
%{\bf \blue  Pluri-Laplacian}
%
%\definition
%The operator $dd^c:\; \Lambda^{p,q}(M)\arrow
%\Lambda^{p+1,q+1}(M)$ is called {\bf\blue
%  pluri-Laplacian}. A function $f\in \ker dd^c$
%is called {\bf \blue pluriharmonic}.
%
%\remark
%Let $D$ be a disk, $z$ holomorphic coordinate, $x, y$ its real and
%imaginary parts. Then $\6\bar\6(f)= \frac{d^2f}{dzd\bar z}= \frac{d^2}{dx^2}+
%\frac{d^2}{dy^2}$, that is, {\bf 
%\red in dimension 1, pluri-Laplacian is Laplacian.}
%
%\corollary
%Let $f$ be a pluriharmonic function on a complex manifold $M$,
%and $C\subset M$ a curve. 
%{\bf \purple Then $f\restrict C$ is a sum of a holomorphic
%and antiholomorphic function.}
%
%\corollary
%{\bf \blue (maximum principle for pluriharmonic functions)}\\
%{\bf \purple A pluriharmonic function which reaches its
%  maximum is constant.}
%
%\proof Indeed, it is constant on any curve (harmonic
%functions on complex curves cannot have a maximum; {\bf
%  \red prove it}).
%\endproof




\newpage

{\bf \blue  Holomorphic forms}

\definition
A $(p,0)$-form $\eta$ on a complex manifold $M$ is called
{\bf \blue holomorphic}, if $\bar\6\eta=0$. 

\definition Let $\Omega^1 M\subset \Lambda^1 M$ be a sheaf over
$M$ generated by $f dg$, where $f, g$ are holomorphic.
This sheaf is called {\bf \blue the sheaf of holomorphic
differentials} on $M$.

\claim
{\bf \purple The sheaf of holomorphic $p$-forms coincides with 
$\Lambda^p_{\calo_M}\Omega^1M$,} where $\calo_M$
is the sheaf of holomorphic functions.

\proof
Clearly, all sections of $\Lambda^p_{\calo_M}\Omega^1M$
are holomorphic. Conversely, any $(p,0)$-form
can be written locally as $\eta=\sum\limits_{I=\{i_1,...,i_p\}} 
\alpha_I dz_{i_1}\wedge dz_{i_2}\wedge...\wedge dz_{i_p}$,
where $z_i$ are holomorphic coordinates.
Then $\bar\6\eta=\sum 
\bar\6\alpha_I dz_{i_1}\wedge dz_{i_2}\wedge...\wedge  dz_{i_p}=0$
implies that $\bar\6\alpha_I$, because
$\Lambda^{p,1}(M)=\Lambda^{p,0}(M)\otimes
\Lambda^{0,1}(M)$, hence for any basis $e_I$ in
$\Lambda^{p,0}(M)$ and any $\{g_I\}\in\Lambda^{0,1}(M)$,
\[ \sum_I g_I\wedge e_I =0 \Leftrightarrow
\text{\ \ all\ \ }g_I=0.
\]
\endproof

\exercise
Prove that {\bf \red on a compact K\"ahler manifold,
any holomorphic form is closed.}


\newpage

{\bf \blue  Holomorphic 1-forms and first cohomology}

\lemma
{\bf \purple Let $\theta$ be an exact holomorphic 1-form
on a compact manifold. Then $\theta=0$.}

\proof $\theta=df$, where $f$ is a function satisfying
$\bar\6\eta=0$, hence holomorphic. Then $f=\const$
by maximum principle.
\endproof

\definition
A $(0,p)$-form $\eta$ is called {\bf \blue antiholomorphic}
if $\bar\eta$ is holomorphic.


The following result is implied by the Hodge theory.

\theorem
Let $(M,I)$ be a compact K\"ahler manifold,
and $[\theta]\in H^2(M,\C)$ is a cohomology class.
{\bf \red
Then $[\theta]$ can be represented by a form $\theta=\theta^{1,0}+\theta^{0,1}$,
where $\theta^{1,0}$ is holomorphic and $\theta^{0,1}$ antiholomorphic.}

\exercise
{\bf \purple Prove this statement for compact complex curves.}

\newpage

{\bf\blue Positive $(1,1)$-forms}

\definition
A {\bf\blue positive (1,1)-form} is a real (1,1)-form
on a complex manifold which can be written as
$\eta=\sum_i \alpha_i \theta_i \wedge I(\theta_i)$,
where $\theta_i$ are real 1-forms, and $\alpha_i$ positive functions.

\remark Hermitian forms are clearly positive. 
Moreover, the cone of positive forms is a closure
of the cone of Hermitian forms. {\bf \purple One may think of
positive forms as of positive semi-definite Hermitian forms.}

\definition
Hermitian forms are called {\bf \blue strictly positive}.

\claim
Let $(M,I)$ be a complex manifold, and $\eta$ a real (1,1)-form.
Then for each 2-dimensional real subspace $W\subset T_x M$
such that $I(W)=W$, {\bf \purple the restriction of $\eta$ to $W$
is proportional to its volume form with non-negative
coefficient.} Conversely, {\bf \purple if $\eta\restrict W$
is non-negative for all such $W$, the $\eta$ is positive.}

\proof A (1,1)-form is Hermitian if and only if $\eta(x,I(x))>0$
for each $x$; it is positive if and only if $\eta(x,I(x))\geq 0$.
\endproof


\newpage

{\bf\blue Mass of a positive $(1,1)$-form}


\definition
Let $(M,I, \omega)$ be a Hermitian manifold, $\dim_\C M=n$.
{\bf\blue Mass} of positive (1,1)-form $\eta$ is a volume form
$\eta \wedge \omega^{n-1}$.

\theorem 
{\bf \blue (normal form for a pair of Hermitian forms)}\\
Let $g$ be a Hermitian metric on $V$, $h$ a pseudo-Hermitian form.
Then there exists an orthonormal (with respect to $g$) basis 
$x_1, I(x_1), x_2, I(x_2), ..., x_n, I(x_n)$ in $V^*$ 
such that $h=\sum a_i x_i \wedge I(x_i)$.

\claim
Let $(M,I, \omega)$ be a Hermitian manifold, 
 $x_1, I(x_1), x_2, I(x_2), ..., x_n, I(x_n)$
an orthonormal basis in $\Lambda^1(M,\R)$, 
and $\eta=\sum_i \alpha_i x_i \wedge I(x_i)$
a positive (1,1)-form (such a basis always exists
because of a normal form theorem).
{\bf \red Then $\eta \wedge \omega^{n-1}= \sum \alpha_i \omega^n$.}

\corollary 
Mass of a positive form is always a (not strictly) positive volume form.
{\bf \red A positive form vanishes if and only if its mass vanishes.}


%\newpage
%
%{\bf \blue  Holomorphic 1-forms and first cohomology: \\
%uniqueness of decomposition}
%
%
%
%\claim On any compact complex manifold, {\bf \red a sum of non-zero
%holomorphic and antiholomorphic forms cannot be exact.}
%
%\proof
%Let $df= \theta^{1,0}+\theta^{0,1}$ be a sum of holomorphic and antiholomorphic
%forms. Then $dd^c f= \1 d (\theta^{1,0}-\theta^{0,1})=0$.
%Then $f=\const$, because {\bf \purple 
%pluriharmonic functions are constant.}
%\endproof
%
%\corollary Let $M$ be a compact complex manifold, and
%$[\theta]\in H^2(M,\C)$ represented by a
%sum $\theta=\theta^{1,0}+\theta^{0,1}$ of holomorphic
%and antiholomorphic forms. {\bf \red Then (a) such a representation
%is unique. Moreover, (b) if $[\theta]\in H^2(M,\R)$
%is real, then $\theta=\Re \theta^{1,0}$.}
%
%\proof (a) is clear, because a difference between
%two ways of representing $\theta$ is an exact
%sum of holomorphic and antiholomorphic forms. 
%
%To prove (b), note
%that $\Im\theta=\Im\theta^{1,0}+\Im\theta^{0,1}$
%vanishes, because it is exact. Since the maps
%$\Im:\; \Lambda^{1,0}(M)\arrow \Lambda^1(M,\R)$
%and $\Im:\; \Lambda^{0,1}(M)\arrow \Lambda^1(M,\R)$ are isomorphisms,
%$\Im\theta^{1,0}=-\Im\theta^{0,1}$ implies 
%$\theta^{1,0}=\overline{\theta^{1,0}}$.
%\endproof

\newpage

{\bf\blue LCK manifolds (reminder)}


\definition
Let $(M,I, \omega)$ be a Hermitian manifold, $\dim_\C M >1$.
Then $M$ is called {\bf \blue locally conformally K\"ahler}
(LCK) if $d\omega=\omega\wedge\theta$, where $\theta$ is a closed
1-form, called {\bf \blue the Lee form}.

\definition
{\bf \blue A manifold is locally conformally K\"ahler}
iff it admits a K\"ahler form taking values in a positive,
flat vector bundle $L$, called {\bf \blue the weight bundle}.


\definition {\bf \blue Deck transform}, or {\bf \blue monodromy maps}
of a covering $\tilde M \arrow M$ are elements of the group $\Aut_M(\tilde M)$.
{\bf \purple
When $\tilde M$ is a universal cover, one has $\Aut_M(\tilde M)=\pi_1(M)$.}

\definition
{\bf \blue An LCK manifold} is a complex manifold such that
its universal cover $\tilde M$ is equipped with a K\"ahler
form $\tilde \omega$, and the deck transform acts on $\tilde M$ by
K\"ahler homotheties.

\theorem {\bf \red These three definitions are equivalent}.

\newpage

{\bf\blue Vaisman's theorem}

\theorem
Let $(M,\omega, \theta)$ be a compact LCK manifold, such that
$\theta$ is not cohomologous to 0. 
{\bf \red Then $M$ does not admit a K\"ahler structure.}

{\bf \green Proof. Step 1:} Let $d\omega=\omega\wedge\theta$,
$\theta'=\theta+d\phi$. Then $d(e^\phi\omega)= e^\phi\omega\wedge \theta+
e^\phi\omega\wedge d\phi =e^\phi\omega\wedge \theta'$. This means that
{\bf \purple we can replace the triple $(M,\omega, \theta)$ by
$(M,e^\phi \omega, \theta')$ for any 1-form $\theta'$
cohomologous to $\theta$. }

{\bf \green Step 2:} Assume that $M$ admits a K\"ahler structure.
Then $\theta$ is cohomologous to a sum of a holomorphic and antiholomorphic form.
Replacing $\omega$ in its conformal class as in Step 1, {\bf \purple we may assume
that $\theta$ is a sum of a holomorphic and antiholomorphic form.}

{\bf \green Step 3:} Then $dd^c\theta= \1 d \bar\6\theta=0$,
giving $dd^c(\omega^{n-1})= \omega^{n-1}\wedge \theta\wedge I(\theta)$.
{\bf \purple 
Then $0=\int_M dd^c(\omega^{n-1})=\int \Mass(\theta\wedge I(\theta))$,
hence $\theta\wedge I(\theta)=0$.}
\endproof



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