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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 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}