\documentclass{slides} \usepackage{amssymb, amsmath, amscd, color, epsfig} %\usepackage[matrix,arrow]{xy} \newcommand{\green}{\color[rgb]{0,0.4,0}} \newcommand{\purple}{\color[rgb]{0.4,0,0.4}} \newcommand{\red}{\color[rgb]{0.7,0,0}} \newcommand{\blue}{\color{blue}} \def\eqref#1{(\ref{#1})} \newcommand{\goth}{\mathfrak} \newcommand{\g}{{\frak g}} \newcommand{\arrow}{{\:\longrightarrow\:}} \newcommand{\Z}{{\Bbb Z}} \newcommand{\C}{{\Bbb C}} \newcommand{\R}{{\Bbb R}} \newcommand{\Q}{{\Bbb Q}} \renewcommand{\H}{{\Bbb H}} \newcommand{\6}{\partial} \def\1{\sqrt{-1}\:} \newcommand{\restrict}[1]{{\left|_{{#1}}\right.}} \newcommand{\cntrct} % contraction with a vector field {\hspace{2pt}\raisebox{1pt}{\text{$\lrcorner$}}\hspace{2pt}} \def\Bbb#1{\mathbb #1} \newcommand{\calo}{{\cal O}} \newcommand{\cac}{{\cal C}} % Correcting TeX... %\let\oldtilde=\tilde %\renewcommand{\tilde}{\widetilde} \renewcommand{\bar}{\overline} \renewcommand{\phi}{\varphi} \renewcommand{\epsilon}{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant} % Operatornames \newcommand{\even}{{\rm even}} \newcommand{\ev}{{\rm even}} \newcommand{\odd}{{\rm odd}} \newcommand{\const}{{\it const}} \newcommand{\fl}{{\rm fl}} \newcommand{\im}{\operatorname{im}} \newcommand{\End}{\operatorname{End}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Hol}{\operatorname{{\cal H}ol}} \newcommand{\Tot}{\operatorname{Tot}} \newcommand{\Id}{\operatorname{Id}} \newcommand{\id}{\operatorname{\text{\sf id}}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Alt}{\operatorname{Alt}} \newcommand{\Iso}{\operatorname{Iso}} \newcommand{\Sec}{\operatorname{Sec}} \newcommand{\Can}{\operatorname{Can}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Mass}{\operatorname{Mass}} \newcommand{\codim}{\operatorname{codim}} \newcommand{\coim}{\operatorname{coim}} \newcommand{\coker}{\operatorname{coker}} \newcommand{\slope}{\operatorname{slope}} \newcommand{\rk}{\operatorname{rk}} \newcommand{\Def}{\operatorname{Def}} \newcommand{\Lie}{\operatorname{Lie}} \newcommand{\Tw}{\operatorname{Tw}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Diff}{\operatorname{Diff}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\inbfpare}[1]{{% \mbox{\tt (}\hspace{-5pt}\mbox{\tt (} #1 % \mbox{\tt )}\hspace{-5pt}\mbox{\tt )}% }} \newcommand{\comment}[1]{{}} \def\blacksquare{\hbox{\vrule width 10pt height 10pt depth 0pt}} \def\endproof{\blacksquare} \def\shortdash{\mbox{\vrule width 4.5pt height 0.55ex depth -0.5ex}} \makeatletter %\@ifundefined{Bbb} % {\newcommand{\Bbb}[1]{{\mathbb #1}}}% %{}% {\edef\Bbb#1{{\Bbb #1}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Pagestyle % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\ps@verbit}{% \renewcommand{\@oddhead}{% \scriptsize {\it \small K\"ahler manifolds, lecture 2 \hfil \tiny M. 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 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 \end{document}