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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: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \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:\ }} \newcommand{\exercise}{{\bf \green EXERCISE:\ }} \newcommand{\pstep}{{\bf \green Proof. Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Hodge theory \\[15mm] \small lecture 14: Supersymmetry for K\"ahler manbifolds (1)} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, March 14, 2018 } \end{center} \newpage {\bf \blue Hodge $*$ operator (reminder)} Let $V$ be a vector space. {\bf \blue A metric $g$ on $V$ induces a natural metric on each of its tensor spaces:} $g(x_1\otimes x_2 \otimes ... \otimes x_k, x_1'\otimes x_2' \otimes ... \otimes x_k') = g( x_1, x'_1)g(x_2, x'_2) ... g(x_k, x'_k)$. {\bf \purple This gives a natural positive definite scalar product on differential forms over a Riemannian manifold $(M,g)$:} $g(\alpha, \beta) := \int_M g(\alpha, \beta) \Vol_M$ Another non-degenerate form is provided by the {\bf \blue Poincare pairing:}\\ $\alpha, \beta \arrow \int_M \alpha \wedge \beta$. \definition Let $M$ be a Riemannian $n$-manifold. Define {\bf \blue the Hodge $*$ operator} $*:\; \Lambda^k M \arrow \Lambda^{n-k} M$ by the following relation: {\bf \red $g(\alpha, \beta) = \int_M \alpha \wedge *\beta.$} \remark {\bf \red The Hodge $*$ operator always exists.} It is defined explicitly in an orthonormal basis $\xi_1, ..., \xi_n \in \Lambda^1 M$: \[ * (\xi_{i_1}\wedge\xi_{i_2} \wedge ... \wedge\xi_{i_k}) = (-1)^s \xi_{j_1}\wedge\xi_{j_2} \wedge ... \wedge\xi_{j_{n-k}}, \] where $\xi_{j_1},\xi_{j_2}, ..., \xi_{j_{n-k}}$ is a complementary set of vectors to $\xi_{i_1}, \xi_{i_2}, ..., \xi_{i_k}$, and $s$ the signature of a permutation $(i_1, ..., i_k, j_1, ..., j_{n-k})$. \remark $*^2\restrict{\Lambda^k(M)}=(-1)^{k(n-k)}\Id_{\Lambda^k(M)}$ \newpage {\bf \blue $d^*=(-1)^{nk}*d*$ (reminder)} \claim On a compact Riemannian $n$-manifold, one has $d^*\restrict{\Lambda^k M} = (-1)^{nk} *d*$, where $d^*$ denotes {\bf \blue the adjoint operator}, which is defined by the equation $(d\alpha, \gamma) = (\alpha, d^*\gamma)$. {\bf \green Proof:} Since \[ 0=\int_M d(\alpha\wedge \beta) = \int_M d(\alpha)\wedge \beta + (-1)^{\tilde \alpha} \alpha \wedge d(\beta), \] one has $(d\alpha, *\beta) = (-1)^{\tilde \alpha} (\alpha, *d\beta)$. Setting $\gamma:= *\beta$, we obtain \[ (d\alpha, \gamma) = (-1)^{\tilde \alpha} (\alpha, *d(*)^{-1}\gamma)= (-1)^{\tilde \alpha} (-1)^{\tilde \alpha(\tilde n-\tilde\alpha)} (\alpha, *d*\gamma)= (-1)^{\tilde \alpha\tilde n}(\alpha, *d*\gamma). \] \endproof \remark Since in all applications which we consider, $n$ is even, {\bf \purple I would from now on ignore the sign $(-1)^{nk}$.} \newpage {\bf \blue Graded vector spaces and algebras (reminder)} \definition A {\bf \blue graded vector space} is a space $V^* =\bigoplus_{i\in \Z} V^i$. \remark If $V^*$ is graded, the endomorphisms space $\End(V^*)=\bigoplus_{i\in \Z} \End^i(V^*)$ is also graded, with $\End^i(V^*)= \bigoplus_{j\in \Z} \Hom(V^j, V^{i+j})$ \definition A {\bf \blue graded algebra}(or ``graded associative algebra'') is an associative algebra $A^*=\bigoplus_{i\in \Z} A^i$, with the product compatible with the grading: $A^i \cdot A^j \subset A^{i+j}$. \remark A bilinear map of graded paces which satisfies $A^i \cdot A^j \subset A^{i+j}$ is called {\bf \blue graded}, or {\bf \blue compatible with grading}. \remark The category of graded spaces can be defined as a {\bf\purple category of vector spaces with $U(1)$-action,} with the weight decomposition providing the grading. Then {\bf\purple a graded algebra is an associative algebra in the category of spaces with $U(1)$-action}. \definition An operator on a graded vector space is called {\bf \blue even} ({\bf \blue odd}) if it shifts the grading by even (odd) number. The {\bf\blue parity} $\tilde a$ of an operator $a$ is 0 if it is even, 1 if it is odd. We say that an operator is {\bf \blue pure} if it is even or odd. \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 Supersymmetry in K\"ahler geometry} Let $(M, I, g)$ be a Kaehler manifold, $\omega$ its Kaehler form. {\bf \blue On $\Lambda^*(M)$, the following operators are defined.} 0. $d$, $d^*$, $\Delta$, because it is Riemannian. 1. $L(\alpha):= \omega\wedge \alpha$ 2. $\Lambda(\alpha) := * L * \alpha$. It is easily seen that $\Lambda= L^*$. 3. The Weil operator $W\restrict{\Lambda^{p,q}(M)}=\1(p-q)$ \theorem {\bf \red These operators generate a Lie superalgebra $\goth a$ of dimension $(5|4)$,} acting on $\Lambda^*(M)$. Moreover, the Laplacian $\Delta$ is central in $\goth a$, hence {\bf \purple $\goth a$ also acts on the cohomology of $M$. } \remark This is a convenient way to summarize the K\"ahler relations and the Lefschetz' $\goth{sl}(2)$-action. \newpage {\bf \blue The coordinate operators } Let $V$ be an even-dimensional real vector space equipped with a scalar product, and $v_1, ..., v_{2n}$ an orthonormal basis. Denote by $e_{v_i}:\; \Lambda^k V \arrow \Lambda^{k+1} V$ an operator of multiplication, $e_{v_i}(\eta) = v_i \wedge \eta$. Let $i_{v_i}:\; \Lambda^k V \arrow \Lambda^{k-1} V$ be an adjoint operator, $i_{v_i}= * e_{v_i} *$. \claim The operators $e_{v_i}$, $i_{v_i}$, $\Id$ are a basis of an {\blue \bf odd Heisenberg Lie superalgebra $\goth H$}, with {\bf \purple the only non-trivial supercommutator given by the formula $\{ e_{v_i}, i_{v_j}\} = \delta_{i,j}\Id$.} Now, consider the tensor $\omega= \sum_{i=1}^n v_{2i-1} \wedge v_{2i}$, and let $L(\alpha) = \omega \wedge \alpha$, and $\Lambda:= L^*$ be the corresponding {\bf \blue Hodge operators}. \claim {\bf \blue (Lefschetz triples)} From the commutator relations in $\goth H$, one obtains immediately that \[ H:=[L, \Lambda] = \left [\sum e_{v_{2i-1}} e_{v_{2i}}, \sum i_{v_{2i-1}} i_{v_{2i}}\right] = \sum_{i=1}^{2n} e_{v_i} i_{v_i} - \sum_{i=1}^{2n} i_{v_i} e_{v_i}, \] {\bf \red is a scalar operator acting as $k-n$ on $k$-forms.} \corollary The triple $L, \Lambda, H$ satisfies the relations for the $\goth{sl}(2)$ Lie algebra: $[L,\Lambda]=H$, $[H,L]=2L$, $[H,\Lambda]=2\Lambda$. \newpage {\bf \blue Hodge components of $d$} \claim Let $(M,I)$ be an almost complex manifold, and $d= \oplus d^{i, 1-i}$ be the Hodge components of $d$, with $d^{a,b}:\; \Lambda^{p,q}(M) \arrow \Lambda^{p+a, q+b}(M)$. {\bf \red Then there are only 4 components, $d= d^{2, -1}+ d^{1,0}, + d^{0,1} + d^{-1, 2}$, with $d^{2, -1}$ and $d^{-1, 2}$ $C^\infty$-linear.} \pstep Each of the components $d^{i,j}$ satisfies the Leibniz identity. To see this, take the Leibniz identity for $d$ and consider its Hodge components. {\bf \green Step 2:} $d\restrict {\Lambda^1(M)}$ has only 4 Hodge components, because it is mapped to $\Lambda^2(M)$ which has only 4 Hodge components. {\bf \green Step 3:} Since the de Rham algebra is generated by $\Lambda^1$, Step 2 implies that all components $d^{i, 1-i}$ vanish, except the 4 listed above. {\bf \green Step 4:} $d^{i, 1-i}(f\eta)= d^{i, 1-i}(f)\wedge \eta + f d^{i, 1-i}(\eta)$ for all $f\in C^\infty M$. However, since $\Lambda^1= \Lambda^{1,0}\oplus \Lambda^{0,1}$, one has $d^{i, 1-i}(f)=0$ for all $i\neq 1, 0$. {\bf \purple Therefore $d^{2, -1}$ and $d^{-1, 2}$ are $C^\infty$-linear.} \endproof \remark The map $d^{-1,2}:\; \Lambda^{0,1}(M) \arrow \Lambda^{2,0}(M)$, interpreted as an element in $\Lambda^{2,0}\otimes T^{0,1}M$ {\bf \red is equal to the Nijenhujs tensor.}. Indeed, for any $X, Y \in T^{1,0}(M)$, and any $\eta\in \Lambda^{0,1}(M)$, one has $d\eta (X, Y)=\eta([X,Y]) + \Lie_X\eta(Y) - \Lie_Y\eta(X)= \eta(N(X,Y))$. \newpage {\bf \blue The twisted differential $d^c$} \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 $[{\cal W}, d^{0,1}] = - \1 d^{0,1}$. Then $[{\cal W}, d]= \1 d^{1,0}-\1 d^{0,1}=IdI^{-1}$. \endproof \corollary $\{d, d^c\} = \{d, \{d, {\cal W}\}\}=0$ (Lemma 1). \newpage {\bf \blue Plurilaplacian} \theorem Let $M,I$ be a complex manifold. {\bf \red Then 1. $\6^2=0$.\\ 2. $\bar\6^2=0$.\\ 3. $dd^c =- d^c d$\\ 4. $dd^c= 2 \1 \6\bar\6$.} \proof The first is vanishing of (2,0)-part of $d^2$, and the second is vanishing of its (0,2)-part. Now, $\{d, d^c\}= -\{d, \{d, {\cal W}\}\}=0$ (Lemma 1), this gives $dd^c =- d^c d$. Finally, $2 \1 \6\bar\6= \frac 1 2 (d+ \1 d^c)(d-\1 d^c)= \frac 1 2 (dd^c - d^c d)= dd^c$. \endproof \definition The operator $dd^c$ is called {\bf \blue the pluri-Laplacian}. \exercise Prove that {\bf \purple on a Riemannian surface $(M,I, \omega)$, one has $dd^c(f) =\Delta(f) \omega$.} \end{document}