\documentclass{slides} \usepackage{amssymb, amsmath, amscd, color, epsfig, dutchcal} %\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}} \renewcommand{\a}{{\frak a}} \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{\Sets}{\operatorname{{\cal S}ets}} \newcommand{\Tot}{\operatorname{Tot}} \newcommand{\Id}{\operatorname{Id}} \newcommand{\id}{\operatorname{\text{\sf id}}} \newcommand{\lin}{{\operatorname{\text{\sf lin}}}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\St}{\operatorname{St}} \newcommand{\Alt}{\operatorname{Alt}} \newcommand{\Iso}{\operatorname{Iso}} \newcommand{\Sec}{\operatorname{Sec}} \newcommand{\Can}{\operatorname{Can}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Spin}{\operatorname{Spin}} \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 Complex geometry, lecture 10 \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: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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{\proof}{% {\bf \green Proof:\ }} \newcommand{\pstep}{% {\bf \green Proof. Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 10: Bismut connection} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 24, 2020 } \end{center} \newpage {\bf \blue Lie algebra and tensors (reminder)} \definition Let $V$ be a representation of a Lie algebra $\g$. {\bf \purple Then $V^*$ is also a representation;} the action of $\g$ on $V^*$ is given by the formula $\langle g(x), \lambda \rangle= - \langle x, g(\lambda) \rangle$, for all $x\in V, \lambda \in V^*$. A tensor product of two $\g$-representations $V_1, V_2$ is also a $\g$-representation, with the action of $\g$ defined by $g(x\otimes y)= g(x) \otimes y + x\otimes g(y)$. This defines the action of $\g$ on all tensor powers $V^{\otimes i} \otimes (V^*)^{\otimes j}$, which are called {\bf \blue the tensor representations} of $\g$. We say that $\g$ {\bf \blue preserves a tensor $\Phi$} if $g(\Psi)=0$ for all $g\in \g$. \example The algebra of all $g\in \End(V)$ preserving a non-degenerate bilinear symmetric form $h\in \Sym^2(V^*)$ is called {\bf \blue orthogonal algebra}, denoted $\goth{so}(V,h)$ or $\goth{so}(V)$. Since $g\in \goth{so}(V)$ if and only if $h(g(x), y)= - h(x, g(y))$, {\bf \red $\goth{so}(V)$ is represented by antisymmetric matrices.} \claim Let $h\in \Sym^2(V^*)$ be a non-degenerate bilinear symmetric form. Using $h$, we identify $V$ and $V^*$. This gives an isomorphism $V^*\otimes V^* \stackrel \tau \arrow V^*\otimes V= \End(V)$. {\bf \red Then $\tau(\Lambda^2V^*)= \goth{so}(V)$.} \proof For any $f\in \End(V)$, the 2-form $\tau^{-1}(f)$ is written as $x, y \arrow h(f(x), y)$. By definition, $f\in \goth{so}(V)$ means that $h(f(x), y) = - h(x, f(y))$ and this happens if and only if $\tau^{-1}(f)$ is antisymmetric. \endproof \newpage {\bf \blue The Lie algebra $\goth u(V)$ (reminder)} \example Let $(V,I)$ be a real vector space with a complex structure map $I:\; V \arrow V$, $I^2=-\Id$, and a Hermitian (that is, $I$-invariant) scalar product. Define {\bf \blue the unitary Lie algebra} $\goth u(V)=\{ f\in \End(V)\ \ |\ \ f(I)=f(h)=0\}$. This is the same as the space of $I$-invariant orthogonal matrices. \claim Consider the natural map $V^*\otimes V^* \stackrel \tau \arrow V^*\otimes V= \End(V)$ associated with $h$. {\bf \red Then $\tau(\Lambda^{1,1}(V^*))= \goth u(V)$.} \proof The isomorphism $\tau$ is $I$-invariant, because $h$ is $I$-invariant. {\bf \purple Then $\tau^{-1}(\goth u(V))$ is the space of $I$-invariant 2-forms,} which is precisely $\Lambda^{1,1}(V^*)$. \endproof \corollary Let $B$ be a bundle with a Hermitian structure product. Then {\bf \red the space of orthogonal connections on $B$ an affine space over $\Lambda^1 M \otimes \goth u(B)$.} \newpage {\bf \blue The space of intrinsic torsion (reminder)} \remark Let $\Phi$ be a tensor on a manifold, and $\nabla$ a connection preserving $\Phi$. Denote by $\a(M)\subset \End(TM)$ the bundle of Lie algebras consisting of all $A\in \End(TM)$ such that $A(\Phi)=0$. Clearly, a connection $\nabla_1$ preserves $\Phi$ if and only if $\nabla-\nabla_1 \in \Lambda^1(M)\otimes \a(M)$. In other words, {\bf \purple connections preserving $\Phi$ are an affine space over $\Lambda^1(M)\otimes \a(M)$.} \definition Consider the linearized torsion operator $\Alt_{12}:\; \Lambda^1(M)\otimes \a(M)\arrow \Lambda^2(M)\otimes TM$. The quotient bundle \[ {\cal T}_{\a}:= \frac{\Lambda^2(M)\otimes TM}{\Alt_{12}(\Lambda^1(M)\otimes \a(M))} \] is called {\bf\blue the space of intrinsic torsion for $\a(M)$-valued connections}. \definition Let $\Phi$ be a tensor on a manifold, and $\nabla$ a connection preserving $\Phi$. {\bf \blue Intrinsic torsion} of $\Phi$ is the image of the torsion of $\nabla$ in ${\cal T}_{\a}$. \newpage {\bf \blue Intrinsic torsion (reminder)} \theorem Let $\Phi$ be a tensor on a manifold, $\nabla$ a connection preserving $\Phi$, and $\tau(\Phi)$ the intrinsic torsion. {\bf \red Then $\tau(\Phi)$ is independent from the choice of $\nabla$.} Moreover, $M$ admits a torsion-free connection preserving $\Phi$ {\bf \red if and only if $\tau(\Phi)=0$.} \pstep For any $\nabla$ and $\nabla'$ preserving $\Phi$, and $A:= \nabla - \nabla'$, one has $A\in \Lambda^1(M)\otimes \a(M)$, hence $T_\nabla - T_{\nabla'}\in \Alt_{12}(\Lambda^1(M)\otimes \a(M))$. Therefore, $T_\nabla$ represents the same vector in ${\cal T}_{\a}$ as $T_{\nabla'}$ {\bf \green Step 2:} The map $\nabla \mapsto T_\nabla$ takes an affine space of all connections preserving $\Phi$ and puts it to an affine subspace $W\subset \Lambda^2(M)\otimes TM$. The linearization of $W$ is the image of $T_\lin$, hence {\bf \purple $W$ is an affine space $\im(T_\lin)+ T_\nabla$.} It contains zero if and only if $T_\nabla\in \im(T_\lin)$. \endproof \example The space of intrinsic torsion for $\goth{so}(TM)$ is zero {\bf \purple (prove it).} \example The space of intrinsic torsion for the symplectic Lie algebra $\goth{sp}(TM)$ is {\bf \purple naturally identified with the space $\Lambda^3(M)$} {\bf \red (this is proven later today).} \newpage {\bf \blue Symplectic connections (reminder)} %{\bf \blue Torsion and the differential forms} \definition When $B=\Lambda^1M$, consider the exterior multiplication map $\Alt:\; \Lambda^i M \otimes \Lambda^1 M\arrow \Lambda^{i+1} M$. Define {\bf \blue the torsion map} $T_\nabla(\eta) := \Alt(\nabla(\eta)) -d\eta$. Then $T_\nabla$ is equal to torsion on $\Lambda^1 M$ and satisfies the Leibnitz identity: \[ T_\nabla(\lambda \wedge \mu)= T_\nabla(\lambda) \wedge \mu + (-1)^{\tilde \lambda}\lambda \wedge T_\nabla(\mu)\ \ \ \ (**) \] \!\!\!\!\!\!\!\definition {\bf \blue An almost symplectic structure} on a manifold is a non-degenerate 2-form. \exercise Let $(M, \omega)$ be an almost symplectic manifold. Prove that there exists a connection $\nabla$ on $TM$ such that $\nabla(\omega)=0$. We call such connection {\bf \blue a symplectic connection}. {\bf \green Lemma 1:} Let $\omega\in \Lambda^ 2 M$ be an almost symplectic structure, and $\nabla$ a symplectic connection. Using $\omega$, we will identify $TM$ and $\Lambda^1 M$, and then we can consider the torsion tensor ${\goth T}\in \Lambda^2M \otimes TM$ of $\nabla$ as $\tau\in \Lambda^2 M \otimes \Lambda^1 M$. Let $\rho:=\Alt(\tau)$. {\bf \red Then $d\omega=-2\rho$.} \proof Clearly, $T_\nabla(\omega) =-d\omega$, because $\nabla(\omega)=0$ and $T_\nabla(\omega)=\Alt(\nabla(\omega)) -d\omega$. By (**), we have $T_\nabla(\omega)= \Alt(A_1(\omega \otimes {\goth T}) - A_2(\omega \otimes {\goth T}))$, where $A_i$ is the convolution of $i$-th component of $\omega \otimes T_\nabla$ and the last, taking $\Lambda^2 M \otimes \Lambda ^2M\otimes TM$ to $\Lambda^2M\otimes \Lambda^1 M$ and $\Lambda^1M\otimes \Lambda^2 M$. Clearly, $\Alt(A_1(\omega \otimes T_\nabla))=-\Alt(A_2(\omega \otimes T_\nabla))=\rho$. This gives $T_\nabla(\omega)= d\omega= -2\rho$. \endproof \newpage {\bf \blue Torsion of almost symplectic structures (reminder)} {\bf \green Theorem 1:} Let $(M, \omega)$ be an almost symplectic manifold, and $\nabla$ a symplectic connection. Denote its torsion by $T_\nabla\in \Lambda^2M \otimes TM$. Using the form $\omega$, we identify $TM$ and $\Lambda^1 M$ and consider $T_\nabla$ as a section $\tau\in\Lambda^2M \otimes \Lambda^1M$. Denote by $\Alt_{123}$ the multiplication map $\Lambda^2M \otimes \Lambda^1M\arrow \Lambda^3 M$. {\bf \red Then $\Alt_{123}(\tau)= -\frac 1 2 d\omega$.} Moreover, {\bf \red any tensor ${\goth T} \in \Lambda^2M \otimes \Lambda^1M$ such that $\Alt_{123}({\goth T})= -\frac 1 2 d\omega$ can be realized as a torsion of a symplectic connection.} \pstep Let $\goth{sp}(TM)$ be the Lie algebra of all tensors $a\in \End(TM)$ such that $\omega(a(x), y)=- \omega(x, a(y))$. The same argument as the one used to show $\goth{so}(TM)= \Lambda^2 M$ shows that $\goth{sp}(TM)=\Sym^2(\Lambda^1 M)$. {\bf \green Step 2:} Under this identification, the linearized torsion map $T_{\lin}:\; \Lambda^1M \otimes \goth{sp}(TM)\arrow \Lambda^2M \otimes TM$ becomes $\Alt_{12}:\; \Lambda^1M \otimes \Sym^2(\Lambda^1M)\arrow \Lambda^2M \otimes \Lambda^1M$. Kernel of this map is clearly $\Sym^3(\Lambda^1 M)$. This gives an exact sequence {\bf \purple (check it).} \[ 0 \arrow \Sym^3(\Lambda^1 M)\hookrightarrow \Lambda^1M \otimes \Sym^2(\Lambda^1M)\stackrel {\Alt_{12}} \arrow \Lambda^2M \otimes \Lambda^1M\stackrel {\Alt_{123}}\arrow \Lambda^3M\arrow 0. \] We identified $\Lambda^3M$ with the space of intrinsic torsion for $\goth{sp}(TM)$. {\bf \green Step 3:} $\Alt_{123}(\tau)=-\frac 1 2 d \omega$ (Lemma 1). This is precisely the intrinsic torsion of $\nabla$. \endproof \newpage {\bf \blue Torsion of unitary connection on a complex manifold} \proposition Let $(M,I,\omega)$ be an Hermitian complex manifold, $\nabla$ a connection on $TM$ preserving $I$ and $\omega$, and $T_\nabla\in \Lambda^2 M \otimes TM =\Lambda^2 M\otimes \Lambda^1 M$ (we identify $TM$ and $\Lambda^1 M$ using the Riemannian structure). {\bf \red Then \[ T_\nabla\in\bigg(\Lambda^{2,0}(M)\otimes \Lambda^{0,1}(M)\bigg)\oplus \bigg(\Lambda^{0,2}\otimes \Lambda^{1,0}(M)\bigg)\oplus \bigg(\Lambda^{1,1}(M)\otimes \Lambda^1 M\bigg). \ \ \ \ \ (**) \]} \!\!\pstep Integrability of $I$ implies that $[T^{1,0}M, T^{1,0}M] \subset T^{1,0}M$. Since $\nabla(I)=0$, one also has $\nabla_X(T^{1,0}M) \subset T^{1,0}M$ for any vector field $X\in TM$. This gives $\nabla_X(Y)- \nabla_Y(X) - [X,Y]\in T^{1,0}M$ for any $X, Y \in T^{1,0}M$. We have shown that \[ T_\nabla\in \bigg(\Lambda^{2,0}(M)\otimes T^{1,0}(M)\bigg)\oplus \bigg(\Lambda^{0,2}\otimes T^{0,1}(M)\bigg)\oplus \bigg(\Lambda^{1,1}(M)\otimes \Lambda^1 M\bigg). \] {\bf \green Step 2:} Since the Riemannian form $g$ is of type (1,1), it pairs (0,1)-vectors and (1,0)-vectors. Therefore, it identifies $T^{1,0}M$ with $\Lambda^{0,1}(M)$. This proves (**). \endproof \newpage \begin{center}{\epsfig{file=bismut_jean_michel.jpg,width=0.69\linewidth}\\ {\bf \blue Jean-Michel Bismut (born 26 February 1948) }} \end{center} \newpage {\bf \blue Bismut connection} \theorem {\bf \blue (Bismut)} Let $(M,I,\omega)$ be an Hermitian complex manifold. Then there exists a unique connection $\nabla$ preserving $I$ and $\omega$, such that its torsion $T_\nabla\in \Lambda^2 M \otimes TM =\Lambda^2 M\otimes \Lambda^1 M$ (we identify $TM$ and $\Lambda^1 M$ using the Riemannian metric) {\bf \red is antisymmetric:} $T_\nabla \in \Lambda^3 M \subset \Lambda^2 M\otimes \Lambda^1 M$. Moreover, {\bf \red in this case $T_\nabla=- \frac 1 2 I(d\omega)$.} \remark This connection is called {\bf \blue the Bismut connection}. When $(M, I, \omega)$ is K\"ahler, it is torsion-free and orthogonal, hence {\bf \purple $\nabla$ is the Levi-Civita connection.} We obtain that {\bf \red on a K\"ahler manifold, Levi-Civita connection satisfies $\nabla(I)=0$.} \pstep There are two different ways to identify $\Lambda^2 M \otimes TM$ and $\Lambda^2 M\otimes \Lambda^1 M$: using $g:\; TM \tilde \arrow \Lambda^1 M$ and using $\omega:\; TM \tilde \arrow \Lambda^1 M$. Denote the first tensor by $\tau_g$ and the second by $\tau_\omega$. It is clear that $I_3(\tau_g)=\tau_\omega$, where $I_3(x\otimes y\otimes z)= x\otimes y\otimes I(z)$. Torsion of symplectic connections was described earlier today (Theorem 1): we have shown that $\Alt(\tau_\omega)=-\frac 1 2 d\omega$. {\bf \purple This implies that the image of the linearized torsion $T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM))$ satisfies $\Alt(I_3(T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM)))=0$.} Indeed, $\Alt(I_3(T_\nabla))$ is independent from $\nabla$ for any Hermitian connection $\nabla$, {\bf \purple hence the linearization of the affine map $\nabla \mapsto \Alt(I_3(T_\nabla))$ vanishes.} \newpage {\bf \blue Bismut connection (2)} \pstep {\bf \purple The image of the linearized torsion $T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM))$ satisfies $\Alt(I_3(T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM)))=0$.} {\bf \green Step 2:} The torsion of $\nabla$ belongs to the space \[ {\goth W} :=\bigg(\Lambda^{2,0}(M)\otimes \Lambda^{0,1}(M)\bigg)\oplus \bigg(\Lambda^{0,2}\otimes \Lambda^{1,0}(M)\bigg)\oplus \bigg(\Lambda^{1,1}(M)\otimes \Lambda^1 M\bigg), \] as shown above. The linearized torsion map is $T_{\lin}:\; \Lambda^1 M \otimes {\goth u}(TM)\arrow {\goth W}$. By the same argument as in the proof of existence of Levi-Civita connection, this map is injective. {\bf \purple This gives an exact sequence \[ 0 \arrow \Lambda^1 M \otimes {\goth u}(TM) \stackrel {T_{\lin}} \arrow {\goth W} \stackrel{I_3 \circ \Alt}\arrow \Lambda^{2,1}(M)\oplus \Lambda^{1,2}(M) \arrow 0, \ \ \ (***) \]} The last arrow of (***) is surjective because any (2,1)+(1,2)-form can be obtained as anti-symmetrization of $\alpha\in I_3({\goth W})$. The sequence (***) is exact in the middle term because dimension of the middle term is equal to sum of dimensions of the left and right terms. \newpage {\bf \blue Bismut connection (3)} {\bf \green Step 2:} Let ${\goth W} :=(\Lambda^{2,0}(M)\otimes \Lambda^{0,1}(M))\oplus (\Lambda^{0,2}\otimes \Lambda^{1,0}(M))\oplus (\Lambda^{1,1}(M)\otimes \Lambda^1 M).$ Then {\bf \purple the sequence \[ 0 \arrow \Lambda^1 M \otimes {\goth u}(TM) \stackrel {T_{\lin}} \arrow {\goth W} \stackrel{I_3 \circ \Alt}\arrow \Lambda^{2,1}(M)\oplus \Lambda^{1,2}(M) \arrow 0 \ \ \ (***) \] is exact.} {\bf \green Step 3:} Let ${\goth U}\subset {\goth W}$ be a subspace consisting of all antisymmetric 3-forms, ${\goth U}=\Lambda^{2,1}(M)\oplus \Lambda^{1,2}(M)$. Clearly, for any differential form $\eta$, one has $\Alt(I_3(\eta))=W(\eta)$, where $W$ is {\bf \blue the Weil operator} acting as $W(\eta)(x, y, z) = \eta(Ix, y, z)+ \eta(x, Iy, z) + \eta(x, y, Iz)$. Then ${\goth U} \stackrel{I_3 \circ \Alt}\arrow \Lambda^{2,1}(M)\oplus \Lambda^{1,2}(M)$ is bijective. Therefore, {\bf \purple there exists a unique form $\sigma \in {\goth U}$ such that $\Alt(I_3(\sigma))=-\frac 1 2 d\omega$. } {\bf \green Step 4:} Let $\nabla_0$ be a connection on $TM$ which satisfies $\nabla_0(g)=\nabla_0(I)=0$ {\bf \red (prove that it exists)}, and $\tau_g\in {\goth W}$ its torsion. Then $\Alt(I_3(\tau_g))= \Alt(I_3(\sigma))=-\frac 1 2 d\omega$ by Theorem 1. Therefore, {\bf \purple there exists a unique $A\in \Lambda^1 M \otimes {\goth u}(TM)$ such that $T_{\lin}(A)+\tau_g=\sigma$, and the torsion of connection $\nabla:=\nabla_0+A$ is equal to $\sigma$.} {\bf \green Step 5:} Step 3 gives $\sigma= \frac 1 2 W^{-1}(d\omega)$. However, $d\omega$ is (2,1)+(1,2)-form, and for such forms $W=I$, hence $\sigma= - \frac 1 2 I (d\omega)$. \endproof \newpage {\bf \blue Levi-Civita connection on K\"ahler manifolds} When $d\omega=0$, this immediately implies \theorem On a K\"ahler manifold $(M,I, g, \omega)$, {\bf \red the Levi-Civita connection $\nabla$ satisfies $\nabla(I)=\nabla(\omega)=0$.} Indeed, the Bismut connection is torsion-free in this case, hence coincides with Levi-Civita. Let us prove this theorem directly. \pstep Let $\nabla$ be a unitary connection on $M$, that is, one which satisfies $\nabla(I)=\nabla(\omega)=0$ {\bf \purple (prove that it exists).} There are two different ways to identify $\Lambda^2 M \otimes TM$ and $\Lambda^2 M\otimes \Lambda^1 M$: using $g:\; TM \tilde \arrow \Lambda^1 M$ and using $\omega:\; TM \tilde \arrow \Lambda^1 M$. Denote the first tensor by $\tau_g$ and the second by $\tau_\omega$. It is clear that $I_3(\tau_g)=\tau_\omega$, where $I_3(x\otimes y\otimes z)= x\otimes y\otimes I(z)$. Torsion of symplectic connections was described earlier today (Theorem 1): we have shown that $\Alt(\tau_\omega)=-\frac 1 2 d\omega$. {\bf \purple This implies that $\Alt(I_3(T_\nabla))=0$.} Since this is true for any unitary connection, {\bf \purple one also has $\Alt(I_3(T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM)))=0$.} \newpage {\bf \blue Levi-Civita connection on K\"ahler manifolds (2)} \pstep {\bf \purple We proved that $\Alt(I_3(T_\nabla))=0$,} where $I_3$ is $I$ acting on the third tensor component. {\bf \purple Moreover, $\Alt(I_3(T_{\lin}(\Lambda^1 M \otimes {\goth u}(TM)))=0$.} {\bf \green Step 2:} The torsion of $\nabla$ belongs to the space \[ {\goth W} :=\bigg(\Lambda^{2,0}(M)\otimes \Lambda^{0,1}(M)\bigg)\oplus \bigg(\Lambda^{0,2}\otimes \Lambda^{1,0}(M)\bigg)\oplus \bigg(\Lambda^{1,1}(M)\otimes \Lambda^1 M\bigg), \] as shown above. The linearized torsion map is $T_{\lin}:\; \Lambda^1 M \otimes {\goth u}(TM)\arrow {\goth W}$. By the same argument as in the proof of existence of Levi-Civita connection, this map is injective. {\bf \purple This gives an exact sequence \[ 0 \arrow \Lambda^1 M \otimes {\goth u}(TM) \stackrel {T_{\lin}} \arrow {\goth W} \stackrel{I_3 \circ \Alt}\arrow \Lambda^{2,1}(M)\oplus \Lambda^{1,2}(M) \arrow 0, \ \ \ (***) \]} The last arrow of (***) is surjective because any (2,1)+(1,2)-form can be obtained as anti-symmetrization of $\alpha\in I_3({\goth W})$. The sequence (***) is exact in the middle term because dimension of the middle term is equal to sum of dimensions of the left and right terms. {\bf \green Step 3:} Now, $T_\nabla$ satisfies $\Alt(I_3(T_\nabla))=0$, hence belongs to the image of $T_\lin$. {\bf \purple Therefore, the connection $\nabla - T_\lin^{-1}(T_\nabla)$ is a torsion-free unitary connection.} \endproof \newpage {\bf \blue Laplacian on differential forms} \definition 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 $(M,g)$:} $g(\alpha, \beta) := \int_M g(\alpha, \beta) \Vol_M$ \definition Let $M$ be a Riemannian manifold. {\bf \blue Laplacian on differential forms} is $\Delta:= dd^* + d^* d$. \remark {\bf \purple Laplacian is self-adjoint and positive definite:} $(\Delta x, x)= (dx, dx) + (d^*x, d^*x).$ Also, $\Delta$ commutes with $d$ and $d^*$. \theorem {\bf \blue (The main theorem of Hodge theory)}\\ {\bf \red There is an orthonormal basis in the Hilbert space $L^2(\Lambda^*(M))$ consisting of eigenvectors of $\Delta$.} Moreover, {\bf \red each eigenspace is finitely-dimensional, and the eigenvalues convergeto zero.} \theorem {\bf \blue (``Elliptic regularity for $\Delta$'')}\\ Let $\alpha\in L^2(\Lambda^k(M))$ be an eigenvector of $\Delta$. {\bf \red Then $\alpha$ is a smooth $k$-form.} \newpage {\small \bf \green Fritz Alexander Ernst Noether \\ (October 7, 1884 - September 10, 1941)} \begin{center} \epsfig{file=noetherfritzundemmy.jpg,width=0.35\linewidth}\\ {Emmy Noether und Fritz Noether, 1933} \end{center} \newpage {\bf \blue De Rham cohomology} \definition The space $H^i(M):= \frac {\ker d\restrict{\Lambda^iM}}{d\left(\Lambda^{i-1}M\right)}$ is called {\bf \blue the de Rham cohomology of $M$}. \definition A form $\alpha$ is called {\bf\blue harmonic} if $\Delta(\alpha)=0$. \remark Let $\alpha$ be a harmonic form. {\bf \red Then $(\Delta x, x)= (dx, dx) + (d^*x, d^*x),$} hence $\alpha \in \ker d \cap \ker d^*$. \remark {\bf \purple The projection ${\cal H}^i(M) \arrow H^i(M)$ from harmonic forms to cohomology is injective.} Indeed, a form $\alpha$ lies in the kernel of such projection if $\alpha=d\beta$, but then $(\alpha,\alpha)=(\alpha, d\beta) = (d^* \alpha, \beta) =0$. \theorem {\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism}\\ (see the next page). \remark {\bf \purple Poincare duality immediately follows from this theorem.} \newpage {\bf \blue Hodge theory and the cohomology} \theorem {\bf \red The natural map ${\cal H}^i(M) \arrow H^i(M)$ is an isomorphism.} {\bf \green Proof. Step 1:} Since $d^2=0$ and $(d^*)^2=0$, one has $[d, \Delta]= dd^*d - d d^* d=0$. This means that {\bf \red $\Delta$ commutes with the de Rham differential.} {\bf \green Step 2:} Consider the eigenspace decomposition $\Lambda^*(M) \tilde = \bigoplus_\alpha \Lambda^*_\alpha(M)$, where $\alpha$ runs through all eigenvalues of $\Delta$, and $\Lambda^*_\alpha(M)$ is the corresponding eigenspace. {\bf \purple For each $\alpha$, de Rham differential defines a complex \[ \Lambda^0_\alpha(M) \stackrel d \arrow \Lambda^1_\alpha(M) \stackrel d \arrow \Lambda^2_\alpha(M) \stackrel d \arrow ... \] } {\bf \green Step 3:} On $\Lambda^*_\alpha(M)$, one has $dd^* + d^* d= \alpha$. When $\alpha \neq 0$, and $\eta$ closed, this implies $dd^*(\eta) + d^* d(\eta)= dd^* \eta = \alpha\eta$, hence $\eta= d\xi$, with $\xi:= \alpha^{-1}d^* \eta$. This implies that {\bf \purple the complexes $(\Lambda^*_\alpha(M), d)$ don't contribute to cohomology.} {\bf \green Step 4:} We have proven that \[ H^*(\Lambda^* M, d) = \bigoplus_\alpha H^* (\Lambda^*_\alpha(M),d)= H^* (\Lambda^*_0(M),d)={\cal H}^*(M). \] \endproof \end{document}