\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 9 \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 9: Symplectic connections} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 21, 2020 } \end{center} \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 \begin{align*} T_\nabla(f\eta) = & \Alt(f\nabla\eta + df\otimes \eta) - d(f\eta)\\ = &f\bigg [\Alt(\nabla\eta) - d\eta\bigg] + df\wedge \eta - df\wedge \eta= f T_\nabla(\eta). \end{align*} {\bf \purple Therefore $T_\nabla$ is linear}. \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.} {\bf \green Proof: Lecture 8.} \newpage {\bf \blue Gregorio Ricci-Curbastro, Tullio Levi-Civita} {\setlength\tabcolsep{20mm} \begin{tabular}{cc} \epsfig{file=Ricci-Curbastro.jpeg,width=0.32\linewidth} & \epsfig{file=Levi-Civita_3.jpeg,width=0.35\linewidth}\\ Gregorio Ricci-Curbastro, & Tullio Levi-Civita, \\ 1853-1925 & 1873-1941 \end{tabular} } {\green \em ...With his former student Tullio Levi-Civita, he wrote his most famous single publication, a pioneering work on the calculus of tensors, signing it as Gregorio Ricci. This appears to be the only time that Ricci-Curbastro used the shortened form of his name in a publication, and continues to cause confusion.} \newpage {\bf \blue Torsion and commutator of vector fields (reminder)} \remark Cartan formula gives \begin{align*} T_\nabla(\eta)(X,Y) = &\nabla_X(\eta)(Y) - \nabla_Y(\eta)(X)- d\eta(X,Y) \\ =& \nabla_X(\eta)(Y) - \nabla_Y(\eta)(X) -\eta([X,Y])- \Lie_X(\eta(Y))+ \Lie_Y(\eta(X)). \end{align*} On the other hand, $\nabla_X(\eta)(Y)= \Lie_X(\eta(Y)) - \eta(\nabla_X(Y))$. Comparing the equations, we obtain \[ T_\nabla(\eta)(X,Y)=\eta\bigg(\nabla_X(Y)- \nabla_Y(X) - [X,Y]\bigg). \] {\bf \blue Torsion is often defined as a map $\Lambda^2 TM \arrow TM$ using the formula $\nabla_X(Y)- \nabla_Y(X) - [X,Y]$.} We have just proved \claim {\bf \red The torsion tensor $\nabla_X(Y)- \nabla_Y(X) - [X,Y]$ is dual to the torsion $\nabla \circ \Alt - d:\; \Lambda^1 M \arrow \Lambda^2 M$ defined above}. \endproof \newpage {\bf \blue Linearization of the torsion (reminder)} \remark Consider the space ${\cal A}(\Lambda^1 M)$ of connections on $\Lambda^1 M$. The torsion defines an affine map \[ {\cal A}(\Lambda^1 M) \arrow \Hom(\Lambda^1M, \Lambda^2 M)= TM \otimes\Lambda^2 M . \] because $T(\nabla + \alpha) = T(\nabla) + \Alt_{12}(\alpha)$, where $\Alt_{12}:\; \Lambda^1M \otimes \End(\Lambda^1M) \arrow \Lambda^2M \otimes TM$ is antisymmetrization in the first two indices. \definition {\bf \blue Linearized torsion} is a map \[ T_{\lin}:\; \Lambda^1(M) \otimes \Lambda^1(M) \otimes TM \arrow \Lambda^2 M \otimes TM \] obtained as a linearization of the torsion map. {\bf \purple It is equal to $\Alt_{12}$.} \newpage {\bf \blue Existence of orthogonal connections (reminder)} \claim Let $B$ be a vector bundle equipped with a scalar product. {\bf \red Then $B$ admits an orthogonal connection}. \proof Chose a covering $\{U_i\}$, such that $B$ is trivial on each $U_i$ and admits an orthonormal basis in each $U_i$. On each $U_i$ we chose a connection $\nabla_i$ preserving this basis. Let $\psi_i$ be a partition of unit subjugated to $\{U_i\}$. Then {\bf \purple the formula $\nabla(b):= \sum \nabla_i(\psi_i b)$ defines an orthogonal connection}. \exercise Let $\omega\in \Lambda^2 B^*$ be a non-degenerate skew-symmetric 2-form on $B$. Use the same argument to prove that there exists a connection $\nabla:\; B \arrow B\otimes \Lambda^1 M$ such that $\nabla(\omega)=0$. \theorem {\bf \blue (``the fundamental theorem of Riemannian geometry'')} {\bf \red Every Riemannian manifold admits a Levi-Civita connection, and it is unique.} \proof See the next slide. \newpage {\bf \blue Levi-Civita connection (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 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 $\goth u(V)$} \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} \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} \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} %{\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} {\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 \end{document}