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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Locally conformally K\"ahler manifolds \\[15mm]
\small lecture 4: Sasakian manifolds}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf HSE and IUM, Moscow
\\[2mm]  March 3, 2014
}
\end{center}


\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  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 $\{d, d^c\} =0$.

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

\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 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 
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\blue 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 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 holomorphic 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,
$\Theta_B = -2 \6\bar\6\log|b|$.}

\remark
{\bf \purple
The 2-form $2\6\bar\6\log|b|$ is independent from the choice of 
a holomorphic section $b$.}
Indeed, let $b_1=b f$, where $f$ is a non-vanishing holomorphic function.
Then 
\[ 
  \6\bar\6\log(|b_1|^2)= 2\6\bar\6\log(|b|)+ \6\bar\6\log (f\bar f)=
  2\6\bar\6\log(|b|)+ \6\bar\6\log f+ \6\bar\6\log \bar f.
\]
The last two terms vanish, because a logarithm of a holomorphic
function is also holomorphic, and logarithm of an antiholomorphic
function is antiholomorphic. 


\newpage

{\bf \blue K\"ahler potentials and plurisubharmonic functions}

\definition
A real-valued smooth function on a complex manifold
is called {\bf\blue plurisubharmonic (psh)} if the (1,1)-form $dd^c f$
is positive, and {\bf\blue strictly plurisubharmonic} if $dd^c f$
is an Hermitian form.

\remark
Since $dd^c f$ is always closed, {\bf \purple it is also K\"ahler when
it is strictly positive.}

\definition
Let $(M,I,\omega)$ be a K\"ahler manifold.
{\bf \blue K\"ahler potential} is a function $f$
such that $dd^c f=\omega$.

\remark {\bf \red Locally, K\"ahler potentials always exist}.
This is a non-trivial theorem which follows from
Poincare-Dolbeault-Grothendieck lemma.

\example $z\arrow |z|^2$ is a K\"ahler potential for the
usual (flat) Hermitian metric on $\C^n$.

\newpage

{\bf \blue Positive line bundles}

\definition
Let $L$ be a holomorphic Hermitian line bundle,
and $\Theta\in \Lambda^{1,1}(M)$ the curvature of its
Chern connection. $L$ is called {\bf \blue positive}
if $\1 \Theta$ is a strictly positive form.

\remark {\bf \purple In this case it is also K\"ahler;} 
indeed, $dd^c\log|h|$ is closed.

\exercise Prove that {\bf \purple the bundle $\calo(1)$ on
$\C P^n$ equipped with the natural $U(n+1)$-equivariant
metric is positive,} and {\bf \purple 
its curvature is the Fubini-Study form.}

\remark 
Let $L$ be a positive line bundle on $M$, $\Tot(L)\stackrel \pi\arrow M$ 
its total space, $\Tot^*(L)$ be the space of non-zero vectors in $L$,
and  $\psi$ a function on $\Tot^*(L)$ defined 
by $\psi(v)= |v|^2$. {\bf \red Then $dd^c\log\psi= \1 \pi^* \Theta_L$,
where $\Theta_L$ is the curvature of $L$.} Indeed, on fibers
of $L$, $\psi= |z|^2$, and $dd^c\log\psi$ vanishes.

\claim The {\bf \purple semipositive 
form $dd^c\log\psi$ on $\Tot^*(L)$ has one 
zero eigenvalue} (along the fibers) and the rest is positive. \endproof

\corollary 
$dd^c \psi = \1 \psi \pi^* \Theta_L + d\psi\wedge I(d\psi)$,
hence {\bf \red this form is strictly positive.}
\endproof

\newpage

{\bf \blue Regular Vaisman manifolds (reminder)}

\theorem {\bf \blue (Kodaira theorem)}\\
Let $X$ be a compact complex manifold. Then {\bf \red $X$ is
projective if and only if it admits a positive line bundle.}

\definition
Let $X$ be a complex manifold, and $L$ a positive
line bundle on $X$. Consider the $\C^*$-bundle $\Tot^*(L)$
with the K\"ahler metric $\tilde \omega= dd^c \psi$ defined above.
Fox $\lambda\in \C$, $|\lambda|>1$, and let 
$M:= \Tot^*(L)/x\sim \lambda x$ be the corresponding quotient.
Clearly, the map $x\arrow \lambda x$ is a K\"ahler homothety on
$(\Tot^*(L), dd^c \psi)$, hence $M$ is an LCK manifold, Such a manifold
is called {\bf \blue regular Vaisman manifold}.

\remark A regular Vaisman manifold is smoothly fibered on $X$;
{\bf \purple the fibers are elliptic curves $\C^*/x\sim \lambda x$.}

\remark The {\bf \blue classical Hopf manifold} 
$\C^n\backslash 0 /x\sim \lambda x$
is an example of a regular Vaisman manifold, with $X=\C P^{n-1}$.

\remark By Kodaira theorem, {\bf \red all regular Vaisman manifolds
admit a holomorphic embedding to classical Hopf manifolds.}

\newpage

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

\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 $(X,g)$ be a Riemannian manifold,
$C(X):= X \times \R^{>0}$ its Riemannian cone, and $h_\lambda$
the homothety action. Assume that $(X,g)$ 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 Let $L$ be a positive holomorphic line bundle
on a projective manifold. {\bf \purple Then the total
space of its unit $S^1$-fibration is Sasakian.}

\newpage

{\bf \blue Contact manifolds: three equivalent definitions}

{\bf \purple All manifolds are assumed to be oriented here.}

{\bf \green Definition 1:} Let 
$C(S)=(S\times \R^>0)$ be a
cone, equipped with the standard action
$h_\lambda(x,t)=(x, \lambda t)$. Assume that
$C(S)$ is equipped with a symplectic form $\omega$
such that $h_\lambda^*\omega=\lambda^2\omega$.
Then $S$ is called {\bf \blue contact manifold}.

{\bf \green Definition 2:} Let $S$ be an odd-dimensional
manifold, and $B\subset TS$ an oriented sub-bundle of codimension 1,
with Frobenius form $\Lambda^2 B \stackrel \Phi \arrow TS/B$
non-degenerate. Then $S$ is called {\bf \blue contact 
manifold}, $B\subset TS$ {\bf\blue the contact bundle}.


{\bf \green Definition 3:} Let $S$ be manifold of
dimension $2k+1$,
$B\subset TS$ an oriented sub-bundle of codimension 1.
Assume that for any nowhere vanishing 1-form  $\theta\in
\Lambda^1 S$, the form $\theta \wedge (d\theta)^k$
is a non-degenerate volume form. Then $(S,B)$
is called {\bf \blue a contact manifold}, and
$\theta$ {\bf \blue a contact form}.

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

See the proof further on in this lecture.

\newpage

{\bf \blue Basic forms (reminder)}

\definition
Let $M$ be a manifold, $B\subset TM$ a sub-bundle,
$\theta\in \Lambda^i M$ a differential form.
It is called {\bf \blue basic} with respect to 
$B$ if for each $b\in B$, one has $\theta\cntrct b=0$
and $\Lie_b \theta=0$.

\definition 
A sub-bundle $B\subset TM$ is called 
{\bf \blue involutive} if $[B,B]\subset B$.

\theorem {\bf \blue ``Frobenius theorem''}\\
Let $B\subset TM$ be an involutive sub-bundle.
{\bf \red Then for each point $x\in M$ there exists a
neighbourhood $U\ni x$ and a smooth projection
$\pi:\; U\arrow N$ such that $B=\ker \pi$.}
\endproof

\theorem
Let $M$ be a manifold, $B\subset TM$ an involutive sub-bundle,
$\theta\in \Lambda^i M$ a differential form. {\bf \red Then the following
are equivalent. \\
\phantom{a}\ \ \ (i) $\eta$ is basic.\\
\phantom{a}\ \ \ (ii) for any open subset $U\subset M$
and a projection $\pi:\; U\arrow N$ such that $B=\ker \pi$,
one has $\eta=\pi^*\eta'$ for some $\eta'\in \Lambda^i N$.}



\newpage

{\bf \blue Contact manifolds: three equivalent definitions (proofs)}


{\bf \green Definition 2:} Let $S$ be an odd-dimensional
manifold, and $B\subset TS$ an oriented sub-bundle of codimension 1,
with Frobenius form $\Lambda^2 B \stackrel \Phi \arrow TS/B$
non-degenerate. Then $S$ is called {\bf \blue contact 
manifold}, $B\subset TS$ {\bf\blue the contact bundle}.


{\bf \green Definition 3:} Let $S$ be manifold of
dimension $2k+1$,
$B\subset TS$ an oriented sub-bundle of codimension 1.
Assume that for any nowhere vanishing 1-form  $\theta\in
\Lambda^1 S$, the form $\theta \wedge (d\theta)^k$
is a non-degenerate volume form. Then $(S,B)$
is called {\bf \blue a contact manifold}, and
$\theta$ {\bf \blue a contact form}.


{\bf \green Proof. Step 1:} {\bf \blue (2) $\Leftrightarrow$ (3): }\\
for each $x, y \in B$, $d\theta(x,y)= \theta([x,y])= \Phi(x,y)$.
Therefore, the Frobenius form $\Lambda^2 B \stackrel \Phi \arrow TS/B$
can be expressed as $\langle \Phi(x,y), \theta\rangle= d\theta(x,y)$.
{\bf \purple Non-degeneracy of $\theta \wedge (d\theta)^k$ on $TM$
is equivalent to non-degeneracy of $d\theta=\Phi$ on $B=\ker \theta$.}
Therefore, $\langle \Phi(x,y), \theta\rangle= d\theta(x,y)$
is of maximal rank if and only if $\theta \wedge (d\theta)^k$
is non-degenerate.

\newpage

{\bf \blue Contact manifolds: three equivalent definitions (proofs, part two)}


{\bf \green Definition 1:} Let 
$C(S)=(S\times \R^>0)$ be a
cone, equipped with the standard action
$h_\lambda(x,t)=(x, \lambda t)$. Assume that
$C(S)$ is equipped with a symplectic form $\omega$
such that $h_\lambda^*\omega=\lambda^2\omega$.
Then $S$ is called {\bf \blue contact manifold}.



{\bf \green Step 2:} {\bf \blue (3) $\Rightarrow$ (1):}\\
Let $M\stackrel \pi\arrow S$ be the space of positive vectors in the oriented
1-dimensional bundle $L:=TS/B$, which is trivialized by
the form $\theta$, $V\in T M$ a 
unit vertical vector field, and $t:\; M \arrow \R$
a map which associates $\theta(v)$ to a point $(s,v)\in M$,
$s\in S, v \in L\restrict x$. Let
$T:=t\pi^*\theta\in \Lambda^1 M$, and let $\omega:=dT$.
Consider the vector field $r=tV\in TM$. Clearly,
$\Lie_r T=2T$, giving $\Lie_r dT=2dT$.
{\bf \purple To prove that $M$ is a symplectic cone of $S$,
it remains to show that $dT$ is symplectic.}

{\bf \green Step 3:} {\bf \blue (3) $\Rightarrow$ (1), second part:}\\
Since $\ker dt=\pi^*S$, 
any vector field $X\in TS$ can be naturally lifted to a vector
field $\pi^{-1}(X)\in \ker dt\subset TM$. For each 
$Y:= \pi^{-1}(y), x,y \in B$, one has $dT(X,Y)= T([X,Y])=T(\pi^{-1}([x,y]))$,
hence {\bf \purple $dT$ is non-degenerate on $\pi^{-1}(B)$.} Also, 
$dT\cntrct V= T$, and $\ker T= \langle \pi^{-1}B, V\rangle$,
hence {\bf \purple $dT$ is non-degenerate on the symplectic
orthogonal complement to $\pi^{-1}B$.}



\newpage

{\bf \blue Contact manifolds: three equivalent definitions (proofs, part three)}




{\bf \green Definition 1:} Let 
$C(S)=(S\times \R^>0)$ be a
cone, equipped with the standard action
$h_\lambda(x,t)=(x, \lambda t)$. Assume that
$C(S)$ is equipped with a symplectic form $\omega$
such that $h_\lambda^*\omega=\lambda^2\omega$.
Then $S$ is called {\bf \blue contact manifold}.

{\bf \green Definition 3:} Let $S$ be manifold of
dimension $2k+1$,
$B\subset TS$ an oriented sub-bundle of codimension 1.
Assume that for any nowhere vanishing 1-form  $\theta\in
\Lambda^1 S$, the form $\theta \wedge (d\theta)^k$
is a non-degenerate volume form. Then $(S,B)$
is called {\bf \blue a contact manifold}, and
$\theta$ {\bf \blue a contact form}.


{\bf \green Step 4:} {\bf \blue (1) $\Rightarrow$ (3):}\\
Let $M=C(S)=S\times \R^{>0}$, and $t\in C^\infty M$ the 
standard coordinate along $\R^{>0}$. Consider the vector field
$r:= t\frac d {dt}$, and the form $\theta:= \omega \cntrct r$.
Since $\theta\cntrct r=0$ and 
\[ \Lie_{r}t^{-1}\theta=d(t^{-1}\theta)\cntrct r  + d(\theta\cntrct r) 
= t^{-1}\theta-t^{-1} \theta + d(1)=0,
\]
{\bf \purple the form $t^{-1}\theta$ is basic with respect to the projection
$C(S)\arrow S$.} This gives a form $\theta$ on $S$.
Finally, $(d\theta)^{k+1}$
is non-degenerate because $d\theta$ is symplectic. {\bf \purple Therefore,
$(d\theta)^{k+1}\cntrct r= (k+1)(d\theta)^{k}\wedge \theta$
is non-degenerate on $S$.}
\endproof

\newpage

{\bf \blue Reeb field}
\newcommand{\Reeb}{\operatorname{\sf Reeb}}

\definition {\bf \blue A Sasakian manifold} is a contact manifold $S$ with
a Riemannian structure, such that the symplectic cone
$C(S)$ with its Riemannian metric is K\"ahler.

\definition
Let $S$ be a Sasakian manifold, $\omega$ the K\"ahler form
on $C(S)$, and $r=t\frac{d}{dt}$ the homothety vector field.
Then $\Lie_{Ir}t= \langle dt, Ir\rangle=0$, hence $iR$
is tangent to $S\subset C(S)$. This vector field 
(denoted by $\Reeb$) is called 
{\bf \blue the Reeb field} of a Sasakian manifold.

\remark {\bf \purple The Reeb field is dual to the contact 
form} $\theta=\omega\cntrct r$.

\theorem {\bf \red The Reeb field acts on a Sasakian manifold
by contact isometries.} 

(see the next slide)

\definition
A Sasakian manifold is called {\bf \blue regular} if the
Reeb field generates a free action of $S^1$, {\bf \blue
  quasiregular} if all orbits of $\Reeb$ are closed, and
{\bf \blue irregular} otherwise.

\newpage

{\bf \blue Reeb field acts by contact isometries}

\theorem {\bf \red The Reeb field acts on a Sasakian manifold
by contact isometries.} 

{\bf \green Proof. Step 1:} Let $(C(S),\omega)$ be 
the cone of a Sasakian manifold with its K\"ahler form, 
and $t$ the standard coordinate function.
A {\bf \blue holomorphic vector field} is a vector field
$v$ such that its diffeomorphism flow $e^{tv}$ is holomorphic.
The homothety vector field $r=d \frac d{dt}$ is holomorphic,
because $\Lie_r \tilde \omega=2\tilde\omega$, $\Lie_r g =2g$,
giving $\Lie_r I=\Lie_rg\omega^{-1} =0$.

{\bf \green Step 2:} If $X$ is a holomorphic vector field,
then $IX$ is also holomorphic. To see this, chose 
(locally) a K\"ahler metric; then
$\Lie_X(I)=A(I)$, where $A=\nabla(X)$ acts by the formula
$A(I)(v)=A(Iv)-IA(v)$. Therefore, {\bf \purple $X$ is holomorphic
if and only if $\nabla(X)$ is complex linear.} Since
$\nabla(I)=0$, one has $\nabla(IX)=I(\nabla(X))$,
hence $\nabla(X)$ is complex linear $\Leftrightarrow$
$\nabla(IX)$  is complex linear. Then {\bf \red
$\Reeb$ acts on $C(X)$ holomorphically}.

{\bf \green Step 3:}
$\Lie_{\Reeb}\omega=d(\tilde\omega\cntrct Ir)= d (tdt)=0$.
Therefore, $\Lie_{\Reeb}\omega=0$. Since $\Lie_{\Reeb}I=0$
as well, this implies that {\bf \red $\Reeb$ is Killing.}

{\bf \green Step 4:} Contact sub-bundle $B\subset TS$
is defined as $\ker \omega\cntrct \frac d{dt}$; since
{\bf \purple the Reeb field preserves $t$ and $\omega$, it preserves
the contact sub-bundle.}
\endproof

\newpage

{\bf \blue Regular Sasakian manifolds}

\definition
A Sasakian manifold is called {\bf \blue regular} if the
Reeb field generates a free action of $S^1$, {\bf \blue
  quasiregular} if all orbits of $\Reeb$ are closed, and
{\bf \blue irregular} otherwise.

\theorem
Let $S$ be a regular Sasakian manifold. {\bf \red Then there
exists a K\"ahler manifold $X$ and a positive holomorphic
Hermitian  line bundle
$L$ such that $S$ is the space of unit vectors in $L$.}

{\bf \green Proof. Step 1:} Let $X=S/\Reeb$. This quotient is well
defined and smooth, because $\Reeb$ is regular.
Then $X=\C(S)/\C^*$, where the $C^*$-action is generated
by $ r=t\frac{d}{dt}, I(r)$, hence holomorphic. {\bf
  \purple Therefore,
$X$ is a complex manifold} (it's a quotient of a complex
manifold by holomorphic action of a Lie group)

{\bf \green Proof. Step 2:}
Since $2\omega=d\theta=d(tIdt)=dd^c(t^2)$,
the function $t^2$ gives a K\"ahler potential on the 
cone of $S$. The form
$dd^c \log t^2= \frac{\omega}{t}-\frac {dt\wedge
  Idt}{t^2}$
vanishes on $\langle r, I(r)\rangle$ and the rest of its
eigenvalues are positive. Therefore, {\bf \purple $dd^c \log t^2$ is
basic with respect to $\langle r, I(r)\rangle$, and is
equal to a pullback of a K\"ahler form $\omega_X$ on $X$.}

{\bf \green Proof. Step 3:} Consider a holomorphic
Hermitian line bundle
obtained from a $\C^*$-bundle $C(S)\arrow X$. Clearly,
$S$ is its space of unit vectors. Its curvature is
expressed by $dd^c\log |v|=\omega_X$, {\bf \red hence this line 
bundle is positive.} \endproof


\newpage

{\bf \blue Quasiregular Sasakian manifold: an example}

\example
Let $M= \C^2\backslash 0=C(S^3)$ considered as a conical K\"ahler
manifold with the standard structure, 
and $M_1=M/G$, where $G=\Z/4$ is generated by 
$\tau(x,y)=(y,-x)$. Since $\tau^2=-1$, the action of $G$
on $M$ is free. {\bf \purple Therefore, $M_1$ is also a conical K\"ahler
manifold.}

\claim {\bf \red $M_1$ is quasiregular, but not regular.}

{\bf \green Proof:} Free orbits are those which satisfy
$(tx,ty)\neq (y,-x)$ for each $t\in U(1)=\{\lambda\in
\C\ \ |\ \ |\lambda|=1\}$, $t\neq 1$. A non-free orbit
gives $tx=y, ty=-x =t^2x$, hence $t=\pm\1$ and 
$x=\pm y$. \endproof

\remark
{\bf \red For each quasiregular Sasakian manifold $S$, the quotient
$S/\Reeb$ is a K\"ahler orbifold.} Then $S$ is a space of 
unit vectors in a positive line bundle, considered in
the orbifold category. 



\end{document}

