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