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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}
