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