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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Locally conformally K\"ahler manifolds \\[15mm]
\small lecture 3: Vaisman manifolds}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE and IUM, Moscow
\\[2mm]  February 24, 2014
}
\end{center}

\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 Conical K\"ahler manifolds}

\definition
Let $(X,g)$ be a Riemannian manifold, and $C(X):= X \times \R^{>0}$,
with the metric $t^2 g+ dt^2$, where $t$ is a coordinate on $\R^{>0}$.
Then $C(X)$ is called {\bf \blue Riemannian cone} of $X$.
{\bf \purple Multiplicative group $\R^{>0}$ acts on $C(X)$ by homotheties,
$(m, t) \arrow (m, \lambda t)$.}

\definition
Let $X$ be a Riemannian manifold,
$(M,g)=C(X):= X \times \R^{>0}$ its Riemannian cone, and $h_\lambda$
the standard homothety action. Assume that $(M,g)$ is equipped with
a complex structure, in such a way that $g$ is K\"ahler,
and $h_\lambda$ acts holomorphically. Then $C(X)$
is called {\bf \blue a conical K\"ahler manifold}.
In this situation, $X$ is called {\bf \blue Sasakian
  manifold}.

\remark A {\bf \blue contact manifold} is defined
as a manifold $X$ with symplectic structure on $C(X)$, and
$h_\lambda$ acting by homotheties. In particular,
{\bf \purple Sasakian manifolds are contact}.
{\bf \blue Sasakian manifold are related to contact in the
same way as K\"ahler manifolds are related to
symplectic.}

\example Any odd-dimensional sphere is Sasakian
{\bf \purple (check this!).}

\example \\ {\bf \blue (we will have a discussion 
of this example in the next lecture)}\\
Let $L$ be a positive holomorphic line bundle
on a projective manifold. {\bf \purple Then the total
space of its unit $S^1$-fibration in $L$ is Sasakian.}


\newpage

{\bf \blue Vaisman manifolds}

\example For any given $\lambda\in \R^{>1}$, {\bf \purple the quotient
$C(X)/h_\lambda$ of a conical K\"ahler manifold is locally
conformally K\"ahler.}

\definition
A compact LCK manifold is called {\bf \blue structurally
Vaisman}, if its is obtained as a quotient of a 
conical K\"ahler manifold $C(X)$ by $\Z$ acting on
$C(X)$ by holomorphic homotheties.

\definition
An LCK manifold $(M, g, \omega, \theta)$ is called
{\bf \blue Vaisman} if $\nabla\theta=0$, where
$\nabla$ is the Levi-Civita connection associated
with $g$.

\theorem
{\bf \blue (Structure Theorem)} \\ {\bf \red A compact LCK manifold
$M$ is Vaisman if and only if it is structurally Vaisman.}

\remark {\bf \purple Locally, this statement is true without
compactness of $M$.} To prove it globally on $M$, one needs
first to show that the monodromy of the weight local
system is $\Z$; this needs compactness.

In this lecture, I will prove the local statement; a
global version of the structure theorem will be proven 
later in the lectures.

\newpage

{\bf \blue Levi-Civita connection (reminder)}


\definition
{\bf \blue The torsion} of a connection 
$\Lambda^1 \stackrel \nabla \arrow \Lambda^1 M \otimes \Lambda^1M$
is a map $\Alt \circ \nabla - d$,
where $\Alt:\;  \Lambda^1 M \otimes \Lambda^1M\arrow \Lambda^2 M$
is exterior multiplication. It is a map 
$T_\nabla:\; \Lambda^1M \arrow \Lambda^2 M$.

\exercise
{\bf \red Prove that torsion is a $C^\infty M$-linear.}

\remark
The dual operator $x, y \arrow \nabla_x Y - \nabla_y X-[X,Y]$
is also called {\bf \blue the torsion of $\nabla$}. It is a map
$\Lambda^2 TM \arrow TM$.

\exercise
{\bf \purple Prove that these two tensors are dual.}

\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}.


\newpage

{\bf \blue Vaisman structure on conical K\"ahler manifolds}

Let us prove  the implication \\
{\bf \blue (Structure Vaisman) $\Rightarrow$ (Vaisman).}

\theorem 
Let $(\tilde M, \tilde g, \tilde \omega) = X\times \R^{>0}$ 
be a conical K\"ahler manifold,
$\Z =\langle \gamma \rangle$ a group acting on
$\tilde M = C(X)$ by K\"ahler homotheties, and $t:\; C(X)\arrow
\R^{>0}$ the projection map. {\bf \red Then 
the form $\omega:= t^{-2}\tilde\omega$ is an
LCK form on $M:= \tilde M \langle \gamma \rangle$,
its Lee form is $t^{-1} dt$, and $\nabla\theta=0$.
Here $\nabla$ is the Levi-Civita connection on 
$(M, g)$, and $g=t^{-2} \tilde g$.}

{\bf \green Proof:} {\bf \purple $t^{-2} \tilde g$ is the product
metric on $C(X)= X\times \R$, where $z=\log t$ is the
coordinate on $\R$ and $dz= t^{-1} dt$ the unit covector. }
To find $\theta$, notice that 
\[ d\omega=d(t^{-2}\tilde\omega)= -t^{-3}dt\wedge \tilde
\omega= -t^{-1}dt \wedge \omega= - dz\wedge \omega.
\]
Then $\nabla(dz)=0$, because it is the unit covector on 
the $\R$ component of $(C(X), g) =X \times \R$.
\endproof

\newpage

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

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

(i) {\bf \red The complex structure $I$ is integrable, and 
the Hermitian form $\omega$ is closed.}

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

\remark {\bf \purple The implication (ii) $\Rightarrow$ (i) is clear.}
Indeed, $[X,Y]=\nabla_X Y - \nabla_Y X$, hence it is
a $(1,0)$-vector field when $X, Y$ are of type (1,0), and 
then {\bf \purple $I$ is integrable}. Also, {\bf \purple $d\omega=0$, because
$\nabla$ is torsion-free,} and $d\omega= \Alt(\nabla\omega)$.

The implication (i) $\Rightarrow$ (ii) is proven by the same argument
as used to construct the Levi-Civita connection.

\newpage


{\bf \blue Holonomy group (reminder)}

\definition (Cartan, 1923)
Let $(B,\nabla)$ be a vector bundle with connection over $M$.
For each loop $\gamma$ based in $x\in M$, let 
$V_{\gamma, \nabla}:\; B\restrict x \arrow B\restrict x$
be the corresponding parallel transport along the connection.
The {\bf \blue holonomy group} of $(B,\nabla)$
is a group generated by $V_{\gamma, \nabla}$,
for all loops $\gamma$. If one takes all contractible
loops instead, $V_{\gamma, \nabla}$ generates
{\bf \blue the local holonomy}, or {\bf \blue
the restricted holonomy} group.

\remark A bundle is {\bf \blue flat} (has vanishing curvature)
{\bf\purple if and only if its restricted holonomy vanishes.}


\remark If $\nabla(\phi)=0$ for some tensor 
$\phi\in B^{\otimes i}\otimes (B^*)^{\otimes j}$,
{\bf \red the holonomy group preserves $\phi$.}

\definition {\bf \blue Holonomy of a Riemannian manifold}
is holonomy of its Levi-Civita connection.

\example Holonomy of a Riemannian manifold lies in
$O(T_x M, g\restrict x)=O(n)$.


\example  Holonomy of a K\"ahler manifold lies in
$U(T_x M, g\restrict x, I \restrict x)=U(n)$.

\remark The holonomy group {\bf \red does not depend
on the choice of a point $x\in M$.}

\newpage


{\bf \blue The de Rham splitting theorem (reminder)}

\corollary 
Let $M$ be a Riemannian manifold, 
and $\Hol_0(M)\stackrel \rho \arrow \End(T_xM)$
a reduced holonomy representation. Suppose that $\rho$ is reducible:
$T_xM = V_1\oplus V_2 \oplus ...\oplus V_k$. {\bf \red Then $G=\Hol_0(M)$ also
splits: $G= G_1\times G_2 \times ...\times G_k$,}
with each $G_i$ acting trivially on all $V_j$ with $j\neq i$.

\theorem (de Rham) A complete, simply connected  
Riemannian manifold with non-irreducible holonomy 
{\bf \red splits as a Riemannian product,} onto factors
corresponding to irreducible components of the holonomy
representation.

{\bf \green Corollary 1:}
Let $X\in TM$ be a vector field saisfying $\nabla X=0$.
{\bf \purple
Then $M$ locally splits as a Riemannian manifold: $M=M_1 \times I$,}
where $I\subset \R$ is interval equipped with a standard metric.
\endproof

\newpage

{\bf \blue The Lee field and conical K\"ahler structures}

The local implication {\bf \blue (Vaisman) $\Rightarrow$ 
(Structure Vaisman)} would follow locally  if we prove

{\bf \green Proposition 1:}
Let $(M,\omega,\theta)$ be a Vaisman manifold,
and $X=\theta^\sharp$ the vector field dual to $\theta$,
called {\bf \blue the Lee field}.
{\bf \red Then $X$ is holomorphic.}

We deduce the local form of {\bf \blue (Vaisman) $\Rightarrow$ 
(Structure Vaisman)} from Proposition 1. Choose a cover
$\tilde M \arrow M$ such that the pullback of $\theta$ 
is exact: $\theta=d\psi$. Since $M$ is locally a product (Corollary 1),
$M=(X \times \R, g_0+dt^2)$, one has $\psi = t$, and the manifold
$(\tilde M, \psi^2 g)$ is a Riemannian cone
of $X=\psi^{-1}(c)$. {\bf \purple To obtain that $\tilde M$ is 
a conical K\"ahler manifold it remains to show that
the standard homotheties of the Riemannian cone 
act on $\tilde M$ holomorphically.} 

However, these
homotheties are obtained by exponentiation of $X$.

{\bf \green Proposition 1 will be proven later in this lecture.}


\newpage

{\bf \blue Lie derivative}

\definition
Let $X\in TM$ be a vector field, and
$e^{tX}$ the corresponding diffeomorphism flow.
For any tensor $A\in TM^{\otimes i}\otimes T^*M^{\otimes j}$,
let $\Lie_X(A):= \frac{d}{dt}\restrict{t=0}(e^{tX})^*(A)$.
This operation is called {\bf \blue the Lie derivative}.

\claim
The Lie derivative satisfies the following properties.

1. {\bf \blue Leibniz identity}.
$\Lie_X(\alpha\otimes \beta)=\Lie_X(\alpha) 
\otimes \beta + \alpha\otimes Lie_X(\beta)$

2. {\bf \blue Contraction}. 
Let $\Pi:\; TM^{\otimes i}\otimes T^*M^{\otimes j} \arrow
TM^{\otimes i-k}\otimes T^*M^{\otimes j-k}$ denote contraction of 
$k$ components of the tensor. Then $\Lie_X(\Pi(\alpha))=\Pi(\Lie_X(\alpha))$.

3. {\bf \blue Differential}.
$\Lie_X(f)=D_X(f)$ for any function $f\in C^\infty M$.

4. {\bf \blue Commutator:} 
$\Lie_x(Y)=[X,Y]$ for any vector field $Y\in TM$.

5. {\bf \blue Cartan formula:}  
$\Lie(\eta)=(d\eta)\cntrct X+ d(\eta\cntrct X)$, for any 
differential form $\eta\in \Lambda^i(M)$.

{\bf \purple 
Moreover, $\Lie_X$ is uniquely determined by the
 properties 1-3.}
\endproof


\newpage

{\bf \blue Killing fields}

\proposition
Let $g\in \Sym^2T^*M$ be a Riemannian form on $TM$, 
$X\in TM$ a vector field, $\nabla$ the Levi-Civita connection, 
and $h:=\Lie_X(g)$. {\bf \red Then $h(Y, Z)=g(\nabla_YX,Z)+ g(\nabla_Z X, Y)$.}

{\bf \blue Proof:} By contraction property, 
$\Lie_X(g(Y,Z))=\Lie_X(g)(Y,Z)+ g([X,Y],Z)+ g([X,Z],Y)$.
Similarly, $\nabla_X(g(Y,Z))=\nabla_X(g)(Y,Z)+ g(\nabla_XY,Z)+ g(\nabla_XZ,Y)$.
However, $\nabla_X(g(Y,Z))=\Lie_X(g(Y,Z))$, giving
\begin{align*}
\Lie_X(g)(Y,Z)&=g(\nabla_XY,Z)+ g(\nabla_XZ,Y)-g([X,Y],Z)- g([X,Z],Y)=\\
& = g(\nabla_YX,Z)+ g(\nabla_Z X, Y)
\end{align*}
using $\nabla_XY-[X,Y]=\nabla_YX$, $\nabla_XZ-[X,Z]=\nabla_ZX$.
\endproof

\remark
A vector field which satisfies $\Lie_X(g)=0$ is called 
{\bf \blue  a Killing vector field.} A vector field 
$X$ is Killing if and only if
the diffeomorphisms $e^{tX}$ are isometries.


\newpage

{\bf \blue Lie derivatives and Levi-Civita connection}


\theorem 
Let $M$ be a Riemannian manifold, $\nabla$ the Levi-Civita connection,
and $X\in TM$ a vector field. Consider an operator 
$A_X(\psi):= \nabla_X(\psi)-\Lie_x(\psi)$ on tensors. Then\\
\phantom{a}\: \ \ \ 1. {\bf \red 
$A_X$ is $C^\infty$-linear, satisfies the Leibnitz rule,
and commutes with contraction.}\\
\phantom{a} 
\ \ \ 2. {\bf \red On 1-forms and vector fields, $A_X=\nabla(X)$,} where
$\nabla(X)\in TM\otimes \Lambda^1 M=\End(TM)=\End(T^*M)$
is understood as an endomorphism of $TM$ and $\Lambda^1 M$.

{\bf \green Proof. Step 1:} Linearity follows from 
$\nabla_X(f\psi)= f \nabla_X(\psi)+\Lie_X f(\psi)$
and $\Lie_X(f\psi)= f \Lie_X(\psi)+\Lie_X f(\psi)$,
and Leibniz and contraction identity from similar identities for $\Lie_X$
and $\nabla$. 

{\bf \green Step 2:} Then $A_X=\nabla(X)$ for $\Lambda^1 M$
would follow from a similar identity for $TM$. 

{\bf \green Step 3:} On vector fields, $\nabla_X Y - [X,Y]=\nabla_Y X$
because $\nabla$ is torsion-free. \endproof


\newpage

{\bf \blue Levi-Civita connection and homothety action}


{\bf \green Theorem 1:} 
Let $\theta$ be a  1-form on a Riemannian manifold, 
and $X$ the dual vector field. {\red \bf Then the following are equivalent.}\\
\phantom{a} 
\ \ \ (i) $\nabla(X)=\lambda \Id$.\\
\phantom{a} 
\ \ \ (ii) $\nabla(\theta)=\lambda g$.\\
\phantom{a} 
\ \ \ (iii) $d\theta=0$ and $\Lie_X g=-2\lambda g$.

{\bf \green Proof. Step 1:} {\bf \purple
$\nabla(X)=\lambda \Id$ is clearly equivalent to
$\nabla\theta= \lambda g$} (one is obtained from another by applying $g^{-1}$,
which is parallel).

{\bf \green Step 2:} (ii) and (i) $\Rightarrow$ (iii):
$d\theta=\Alt(\nabla\theta)=0$, since $g$ is symmetric.
From $\nabla(X)=\lambda \Id$ we obtain 
$\nabla_X g - \Lie_X g= \lambda \Id(g)=2\lambda g$.

{\bf \green Step 3:} (iii) $\Rightarrow$ (i):
since $\Lie_X g=-2\lambda g$ and $\nabla_X g=0$,
we obtain that $\nabla(X)(g)=2\lambda g$. 
{\bf \purple This implies that the
symmetric part of $\nabla(X)$, considered as a section
of $\End(TM)$ is equal to $\lambda \Id$}
(the antisymmetric part acts on $g$ trivially).
To obtain the antisymmetric part, it is more convenient
to replace $\nabla(X)$ by $\nabla(\theta)$. {\bf \purple Then 
the antisymmetric part of $\nabla(\theta)$
is equal to $\Alt(\nabla(\theta))=d\theta=0$.}
\endproof

{\bf \green Remark 1:}
In this situation, for each tensor 
$\Phi\in TM^{\otimes i}\otimes T^*M^{\otimes j}$,
one has $\nabla(X)(\Phi)=(i-j)\lambda\Phi$.


\newpage

{\bf \blue Lee field on Vaisman manifolds}


{\bf \green Proposition 1:}
Let $(M,\omega,\theta)$ be a Vaisman manifold,
and $X=\theta^\sharp$ the vector field dual to $\theta$,
called {\bf \blue the Lee field}.
{\bf \red Then $X$ is holomorphic.}

{\bf \green Proof. Step 1:} Locally, $M$ is a product, $M=S\times \R$,
with the product metric. Let $\tilde M$ be a covering of $M$ such
that the pullback of $\theta$ is exact on $\tilde M$,
$\theta=d\phi$. Then $\tilde\omega:= e^{-\phi}\omega$ is a K\"ahler form.
{\bf \purple The corresponding metric $\tilde g$ on $\tilde M$ a cone metric,
and $X$ acts on $(M, \tilde g)$ by homotheties.}

{\bf \green  Step 2:} Let $\nabla^W$ be the
Levi-Civita connection on the K\"ahler manifold
$(\tilde M, \tilde g, \tilde \omega)$. 
It is called {\bf \blue Weyl connection}. {\bf \purple 
Theorem 1 implies that
$\nabla^W(X)=\lambda \Id$, for some constant $\lambda$.}

{\bf \green  Step 3:}  Since $\tilde M$ is K\"ahler, 
$\nabla_X^W I=0$. By Remark 1, $\nabla^W(X)(I)= 0$. Therefore,
{\bf \purple Theorem 1 implies that $\Lie_X(I)=0$.}
This is equivalent to diffeomorphism flow 
$e^{tX}$ preserving the complex structure, 
and hence to $X$ being holomorphic. \endproof




\end{document}

