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Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\proof}{{\bf \green Proof:\ }} \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Observation \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{exercise}[section] \setcounter{exercise}{0} \renewcommand{\theexercise}{\noindent{Exercise \thesection.\arabic{exercise}}} \newcommand{\exercise}{% \setcounter{exercise}{\value{Mycounter}} \refstepcounter{exercise} \stepcounter{Mycounter} {\bf \green EXERCISE:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}