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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 13: Bismut connection} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\scriptsize NRU HSE, Moscow} \\[15mm] {\small Misha Verbitsky, March 7, 2018 } \end{center} \newpage {\bf \blue Connections (reminder)} \definition Recall that {\bf \blue a connection} on a bundle $B$ is an operator $\nabla:\; B \arrow B \otimes \Lambda^1 M$ satisfying $\nabla(fb) = b \otimes df + f \nabla(b)$, where $f \arrow df$ is de Rham differential. When $X$ is a vector field, we denote by $\nabla_X(b)\in B$ the term $\langle \nabla(b), X\rangle$. \remark {\bf \purple A connection $\nabla$ on $B$ gives a connection $B^* \stackrel {\nabla^*} \arrow \Lambda^1 M \otimes B^*$ on the dual bundle,} by the formula \[ d(\langle b, \beta\rangle) = \langle \nabla b, \beta\rangle+ \langle b, \nabla^*\beta\rangle \] These connections are usually denoted {\bf \red by the same letter $\nabla$.} \remark For any tensor bundle ${\cal B}_1:= B^*\otimes B^* \otimes ... \otimes B^* \otimes B\otimes B \otimes ... \otimes B$ {\bf \purple a connection on $B$ defines a connection on ${\cal B}_1$} using the Leibniz formula: \[ \nabla(b_1 \otimes b_2) = \nabla(b_1) \otimes b_2 + b_1 \otimes \nabla(b_2). \] \newpage {\bf \blue Parallel transport along the connection (reminder)} \theorem Let $B$ be a vector bundle with connection over $\R$. Then for each $x\in \R$ and each vector $b_x \in B\restrict x$ {\bf \red there exists a unique section $b\in B$ such that $\nabla b=0$, $b\restrict x= b_x$.} \proof This is existence and uniqueness of solutions of an ODE $\frac {db}{dt} + A(b)=0$. \endproof \definition Let $\gamma:\; [0, 1] \arrow M$ be a smooth path in $M$ connecting $x$ and $y$, and $(B, \nabla)$ a vector bundle with connection. Restricting $(B, \nabla)$ to $\gamma([0,1])$, we obtain a bundle with connection on an interval. Solve an equation $\nabla(b)=0$ for $b\in B\restrict{\gamma([0,1])}$ and initial condition $b\restrict x= b_x$. This process is called {\bf\blue parallel transport} along the path via the connection. The vector $b_y:= b\restrict y$ is called {\bf\blue vector obtained by parallel transport of $b_x$ along $\gamma$}. {\bf\blue Holonomy group} of $\gamma$ is the group of endomorphisms of the fiber $B_x$ obtained from parallel transports along all paths starting and ending in $x\in M$ \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 \newpage {\bf \blue Affine space of orthogonal connections (reminder)} \claim Let $B$ be a bundle with a scalar product. Then {\bf \red the space of orthogonal connections on $B$ an affine space over $\Lambda^1 M \otimes \goth{so}(B)$.} \proof Let $s\in B^*\otimes B^*$ be a 2-form on $B$. The action of $A:= \nabla-\nabla_1$ on $B^*\otimes B^*$ is given by $A(s) (x,y)= - s(A(x), y) - s(x, A(y))$. Therefore, a difference $A$ of orthogonal connections satisfies $h(A(x), y) = - h(x, A(y))$ for all $x, y\in B$. This is the same as $A\in \Lambda^1 M \otimes \goth{so}(B)$. \endproof Similarly one proves \claim Let $B$ be a bundle with a Hermitian structure product. Then {\bf \red the space of orthogonal connections on $B$ an affine space over $\Lambda^1 M \otimes \goth u(B)$.} \claim Let $B$ be a bundle with a Hermitian structure and a tensor $\Phi$, and ${\goth g}\subset \End(B)$ the Lie algebra of endomorphisms preserving $\Phi$. Then {\bf \red the space of connections on $B$ preserving $\Phi$ is an affine space over $\Lambda^1 M \otimes \g$.} \newpage {\bf \blue Torsion (reminder)} \definition Let $\nabla$ be a connection on $\Lambda^1 M$, \[ \Lambda^1 \stackrel \nabla \arrow \Lambda^1 M \otimes \Lambda^1M.\] {\bf \blue Torsion of $\nabla$} $T_\nabla:\; \Lambda^1 M \arrow \Lambda^2 M$ is a map $\nabla \circ \Alt - d$, where $\Alt:\; \Lambda^1 M \otimes \Lambda^1M\arrow \Lambda^2 M$ is exterior multiplication. \remark {\bf \red Torsion is often defined as a map $\Lambda^2 TM \arrow TM$ using the formula $\nabla_X(Y)- \nabla_Y(X) - [X,Y]$.} This map coincides with the torsion map $\Lambda^1M\arrow \Lambda^2 M$ defined above. \definition Let $(M,g)$ be a Riemannian manifold. A connection $\nabla$ on $TM$ is called {\bf \blue orthogonal} if $\nabla(g)=0$, and {\bf \blue Levi-Civita connection} if it is orthogonal and has zero torsion. \theorem {\bf \blue (``the fundamental theorem of Riemannian geometry'')} {\bf \red Every Riemannian manifold admits a Levi-Civita connection, and it is unique.} \newpage {\bf \blue Levi-Civita connection, its existence and uniqueness (reminder)} \pstep Chose an orthogonal connection $\nabla_0$ on $\Lambda^1 M$. The space ${\cal A}$ of orthogonal connections is affine and {\bf \purple its linearization is $\Lambda^1 M \otimes {\goth{so}}(TM)$.} We shall identify $\goth{so}(TM)$ and $\Lambda^2 M$. Then {\bf \purple ${\cal A}$ is an affine space over $\Lambda^1 M \otimes \Lambda^2 M$}. {\bf \green Step 2:} Then the linearized torsion map is \[ T_{lin} :\; \Lambda^1 M \otimes {\goth {so}}(TM)= \Lambda^1(M) \otimes \Lambda^2 M \stackrel{\Alt_{12}} \arrow \Lambda^2 M \otimes \Lambda^1M = \Lambda^2 M \otimes T M. \] {\bf \purple It is an isomorphism}. Indeed, on the right and on the left there are bundles of the same rank, hence it would suffice to show that $T_{lin}=\Alt_{12}$ is injective. However, if $\eta \in \ker T_{lin}$, it is a form which is symmetric on first two arguments and antisymmetric on the second two, giving $\eta(x,y,z) = \eta(y,x,z) = - \eta (y,z, x).$ This gives $\sigma(\eta) =-\eta$, where $\sigma$ is a cyclic permutation of the arguments. Since $\sigma^3=1$, this implies $\eta=0$. {\bf \green Step 3:} We have shown that {\bf \purple an orthogonal connection is uniquely determined by its torsion}. Indeed, torsion map is an isomorphism of affine spaces. {\bf \green Step 4:} Let $\nabla:= \nabla_0 -T_{lin}^{-1}(T_{\nabla_0})$. Then $T_\nabla= T_{\nabla_0}-T_{lin}(T_{lin}^{-1}(T_{\nabla_0}))=0$, hence {\bf \red $\nabla$ is torsion-free.} \endproof \newpage {\bf \blue Space of Cartan tensors} \definition Let $C(V) \subset V\otimes V \otimes V$ be $\ker \Sym\cap \ker \Alt$, where $\Sym$ is the symmetrization map $\Sym:\; V\otimes V \otimes V\arrow \Sym^3(V)$ and $\Alt$ the antisymmetrization map $\Alt:\; V\otimes V \otimes V\arrow \Lambda^3(V)$. Then $C(V)$ is called {\bf \blue the space of Cartan tensors on $V$}. \remark Clearly, there is a direct sum decomposition $V\otimes V \otimes V= \Lambda^3(V)\oplus \Sym^3(V) \oplus C(V)$. {\bf \green Lemma 1:} Denote by $\Sym_{ij}$, $\Alt_{ij}$ the operators of symmetrization and antisymmetrization of $\Phi\in V^{\otimes 3}$ using the indices $i, j$. {\bf \red Then \[ C(V) = \Alt_{12}(\Sym_{23}(V^{\otimes 3}))\oplus \Sym_{12}(\Alt_{23}(V^{\otimes 3})). \]} \!\!\:\proof Since $\Sym_{23}(V^{\otimes 3})$ is generated by $x\otimes y \otimes y$, one has $\im(\Alt_{12}\Sym_{23})\supset C(V)$. Similarly, $\im(\Alt_{12}\Sym_{23})\supset C(V)$. For a converse statement, we use the decomposition $V\otimes V=\Sym^2 V \oplus \Lambda^2 V$. This gives \[ V^{\otimes 3}= \im \Alt_{12}\Sym_{23}\oplus \im\Alt_{12}\Alt_{23} \oplus \im\Sym_{12}\Alt_{23} \oplus \im\Sym_{12}\Sym_{23}. \] Then $\ker \Sym \cap \ker\Alt$ is precisely $ \im \Alt_{12}\Sym_{23}\oplus \im\Sym_{12}\Alt_{23}$. \endproof \newpage {\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, which can be used to extend $T_\nabla$ from $\Lambda^1M$ to $\Lambda^* M$: \[ T_\nabla(\lambda \wedge \mu)= T_\nabla(\lambda) \wedge \mu + (-1)^{\tilde \lambda}\lambda \wedge T_\nabla(\mu) \] \newpage {\bf \blue Symplectic connections} \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 2:} 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 $T_\nabla$ of $\nabla$ as a section $\tau\in \Lambda^2 M \otimes \Lambda^1 M$. Let $\rho:=\Alt(T_\nabla)$. {\bf \red Then $d\omega=2\rho$.} \proof $T_\nabla(\omega) =d\omega$, because $\nabla(\omega)=0$. However, $T_\nabla(\omega)= \Alt(A_1(\omega \otimes T_\nabla) - A_2(\omega \otimes T_\nabla))$, where $A_i:\; \Lambda^2 M \otimes TM \otimes \Lambda ^2 M$ is the convolution of $i$-th component of $\omega \otimes T_\nabla$ and the last. Clearly, $A_i(\omega \otimes T_\nabla)=\tau$. This gives $T_\nabla(\omega)= d\omega= 2\rho$. \endproof \newpage {\bf \blue Torsion of almost symplectic structures} {\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 $\Alt_{123}(\tau)= \frac 1 2 d\omega$ 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:} The space of symplectic connections is an affine space with linearization $\Lambda^1 M \otimes \goth{sp}(TM)= \Lambda^1 M \otimes \Sym^2(\Lambda^1 M)$. The image of the linearized torsion map $T_{lin}=\Alt_{12}$ belongs to $C(V)$ (Lemma 1). Therefore, the image of $\Alt_{123}(\tau)$ is independent from the choice of $\nabla$. Any tensor ${\goth T}\in \Lambda^2M \otimes TM$ with $\Alt_{123}(\tau)=\Alt_{123}({\goth T})$ can be obtained as a torsion of an appropriate connection, because the part of $C(V)$ which is antisymmetric in the first two multipliers is precisely $\Alt_{12}(\Lambda^1 M \otimes \Sym^2(\Lambda^1 M))$. {\bf \green Step 3:} $\Alt_{123}(\tau)=\frac 1 2 d \omega$ (Lemma 2). \endproof \newpage {\bf \blue Torsion of Hermitian connection} \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 $\Alt(I_3(T_{lin}(\Lambda^1 M \otimes {\goth u}(TM)))=0$.} \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 \end{document}