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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 6: Laplace operator is Fredholm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, February 10, 2018 } 
\end{center}

\newpage

{\bf \blue Fredholm operators (reminder)}

\definition
A continuous operator $F:\; H_1\arrow H_2$
of Hilbert spaces is called {\bf\blue Fredholm}
if its image is closed and kernel and 
cokernel are finite-dimensional.

\remark
{\bf \red ``Cokernel'' of a 
morphism $F:\; H_1\arrow H_2$ of topological
vector spaces is often defined as 
$\frac{H_2}{\overline {\im F}}$.}

\definition
An operator $F:\; H_1\arrow H_2$
{\bf \blue has finite rank} if its image has finite rank.

\claim
An operator 
 $F:\; H_1\arrow H_2$ {\bf \red is Fredholm} if and only if
there exists  $F_1:\; H_2\arrow H_1$
such that {\bf \red the operators $\Id-FF_1$ and $\Id - F_1 F$ have 
finite rank.}

\proof This is because $F$ defines an isomorphism
$F:\; H_1/\ker F \arrow \im F$ as shown above.
\endproof


\newpage

{\bf \blue Fredholm operators and compact operators (reminder)}

\theorem 
{\bf \red
The set of Fredholm operators is open in the operator norm
topology.}\\ 
\pstep
Let $F:\; U \arrow V$ be a Fredholm operator,
and $U_1:= (\ker F)^\bot$. Since $F$ is invertible
on $U_1$, it satisfies $\inf_{x\in U_1} \frac{|F(x)|}{|x|} > 2\epsilon$.
Then, for any operator $A$ with $\|A\| < \epsilon$, one has
$\inf_{x\in U_1} \frac{|F+A(x)|}{|x|} > \epsilon$.
{\bf \purple  This implies that $F\restrict{U_1}$ is 
an invertible map to its image,
which is closed.} In particular, $\ker(F+A)$ is finite-dimensional.

{\bf \green Step 2:} To obtain that $\coker (F+A)$ is finite-dimensional
for $\|A\|$ sufficiently small, we observe that $\coker (F+A)=
\ker (F^* +A^*)$, and $F^*$ is also Fredholm. Then Step 1 implies that
{\bf \purple 
$\ker(F^*+A^*)$ is finite-dimensional for $\|A\|$ sufficiently small.}
\endproof

\corollary
Let $A$ be compact and $F$ Fredholm. 
{\bf \red Then $A+F$ is Fredholm.}

\proof
Let $A_i$ be a sequence of operators with
finite rank converging to $A$. Then 
$F+(A-A_i)$ is Fredholm for $i$ sufficiently
big, because the set of Fredholm operators is open.
However, a sum of Fredholm operator and
operator of finite rank is Fredholm, hence
$F+A= F+(A-A_i)+A_i$ is also Fredholm.
\endproof

\newpage

{\bf \blue Equivalent scalar products on vector spaces}

\theorem
Let $V$ be a vector space, and $g_1$, $g_2$
two scalar products. We say that
{\bf \blue $g_1$ is equivalent to $g_2$}
if these two scalar product induce the same topology.

\theorem
The topology induced by $g_1$ {\bf \red is equivalent to topology
induced by $g_2$ if and only if
 $C^{-1} g_2 \leq g_1 \leq C g_2$}
for some $C>0$.

\proof
Consider the identity operator $A:\; (V, g_1) \arrow (V, g_2)$.
Its operator norm is $\sup_{x\neq 0} \frac{g_2(x,x)}{g_1(x,x)}$.
Operator norm is bounded if and only if $\Id$ is continuous,
and this is equivalent to existence of a constant $C>0$ such 
that $C^{-1} g_2 \leq g_1$. Existence of a constant $C$
such that $g_1 \leq C g_2$ is equivalent to continuity of 
$A^{-1}$.
\endproof

\newpage

{\bf \blue Equivalent scalar products and symmetric operators}


\lemma
Let $V$ be a vector space, and $g$, $g_1$
scalar products. Consider the symmetric operator
$B_1$ such that $g_1(x,y)= g(B_1(x), y)$.
{\bf \red Then 
\[ 
  \sup_x \frac{g(B_1(x), B_1(x))}{g(x, x)}
= \left(\sup_x \frac{g_1(x,x)}{g(x,x)}\right)^2.
\]
}
\proof
By Cauchy-Schwarz, $g(x,x)g(B_1(x), B_1(x))\geq g(B_1(x), x)^2 = g_1(x,x)^2$.
This gives $\sup_x \frac{g(B_1(x), B_1(x))}{g(x,x)^2}\geq 
\left(\sup_x \frac{g_1(x,x)}{g(x,x)}\right)^2$.
On the other hand, $\sup_x \frac{{g(B_1(x), B_1(x))}}{g(x,x)}$
is norm of $B_1^2$, which gives 
\[ 
   \sup_{x} \frac{g(B_1(x), B_1(x))}{g(x,x)} =\| B_1^2\|\leq \| B_1\|^2= 
   \sup_{x} \left(\frac{g_1(x,x)}{g(x,x)}\right)^2
\]
hence $\sup \frac{g(B_1(x), B_1(x))}{g(x,x)^2}\leq 
\left(\sup_x \frac{g_1(x,x)}{g(x,x)}\right)^2$.
\endproof

\newpage

{\bf \blue Equivalent scalar products and Fredholm operators}

\remark
A continuous operator $F:\; H_1\arrow H_2$
in vector spaces with scalar product is
called {\bf \blue Fredholm} if it is Fredholm on their completions
(which are Hilbert spaces).


{\bf \green Corollary 1:}
Let $g, g_1, g_2$ be metrics on $V$, and 
consider the symmetric operators
$B_i$ such that $g_i(x,y)= g(B_i(x), y)$.
Denote by $\tilde g_2$ the metric $\tilde g_2(x,y):= g_2(B_2(x), B_2(y)$.
{\bf \red Then $g_1$ is equivalent to $g_2$
if and only if $B_1:\; (V,  \tilde g_2) \arrow (V,g)$ is Fredholm.}

\proof $B_1:\; (V,  \tilde g_2) \arrow (V,g)$ is Fredholm if and only if
it for some constant $C>0$, one has
$ C^{-1}  g(B_2(x), B_2(x)) \leq  
g(B_1(x), B_1(x)) \leq C  g(B_2(x), B_2(x))$.
This is the same as $ C^{-1}  g_2(x,x) \leq  
g_1(x,x) \leq C  g_2(x,x)$ by the previous lemma.
\endproof

\newpage

{\bf \blue Sobolev's $L^2$-norm on $C^\infty_c(\R^n)$ (reminder)}

\definition
Denote by $C^\infty_c(\R^n)$ the space of smooth functions
with compact support. For each differential monomial
\[
   P_\alpha = \frac
   {\partial ^{k_1}}{\partial x_1^{k_1}}\frac
   {\partial ^{k_2}}{\partial x_2^{k_2}}...
   \frac{\partial ^{k_n}}{\partial x_1^{k_n}}
\]
consider the corresponding partial derivative
\[
P_\alpha(f) = \frac
   {\partial ^{k_1}}{\partial x_1^{k_1}}
   \frac{\partial ^{k_2}}{\partial x_2^{k_2}}... 
   \frac{\partial ^{k_n}}{\partial x_1^{k_n}}f.
\]
Given $f\in C^\infty_c(\R^n)$, one defines   
{\bf \blue the $L^2_p$ Sobolev's norm $|f|_{p}$}
as follows:
\[
|f|_s^2= \sum_{\deg P_\alpha\leq p} 
 \int \left|P_\alpha(f) \right |^2\Vol
\]
where the sum is taken over all differential
monomials $P_\alpha$ of degree $\leq p$,
and  $\Vol= dx_1\wedge dx_2\wedge ... dx_n$ -
the standard volume form. 

\remark Same formula defines {\bf \blue Sobolev's $L^2$-norm
$L^2_p$ on the space of smooth functions on a torus $T^n$.}

\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 
In local coordinates, connection on $B$ is a sum of differential
and a form $A\in \End B \otimes \Lambda^1 M$. Therefore,
$\nabla_X$ is a derivation along $X$ plus linear endomorphism.
This implies that {\bf\red any first order differential operator on $B$ 
is expressed as a linear combination
of the compositions of covariant 
derivatives $\nabla_X$ and linear maps.}

This follows from the definition of the first order differential
operator: {\bf \purple by definition, it is a linear combination of 
partial derivatives combined with linear maps.}


\newpage

{\bf \blue Connection and a tensor product (reminder)}

\remark 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 $L^2_p$-metrics and connections}


\definition
Let $F$ be a vector bundle on a compact manifold.
The {\bf \blue $L^2_p$-topology} on the space of sections
of $F$ is a topology defined by the norm $|f|_p$ with 
$|f|^2_p=\sum_{i=0}^p \int_M |\nabla^if|^2\Vol_M$,
for some connection and scalar product on $F$ and $\Lambda^1M$.

\remark
{\bf \red The metric $|f|^2_p$ is equivalent to the Sobolev's $L^2_p$-metric
on $C^\infty(M)$.} Indeed, all partial derivatives
of a function $f$ are expressed through $\nabla^if$,
hence an $L^2$-bound on partial derivatives gives 
$L^2$-bound on $\nabla^if$, and is given by such a bound.

From now on, {\bf \blue we write $(x, y)$ instead of $\int_M (x, y) \Vol_M$.}
This metric is also denoted $L^2$; the space of sections of $B$
with this metric $(B, L^2)$.

\definition
We define the {\bf \blue Sobolev's $L^2_p$-metric on vector bundles} by
$L^2_p(x, y)= \sum_{i=0}^p (\nabla^i(x), \nabla^i(y))$.

\newpage

{\bf \blue $L^2_p$-metrics and Fredholm maps}

First, let's show that we can drop all terms in this sum,
except two.

{\bf \green Theorem 1:}
{\bf \red The Sobolev's $L^2_p$-metric is equivalent
to \\ $g(x, y):=  (\nabla^p(x), \nabla^p(y)) +  (x,y)$.}

\pstep
Let $D_1=  \sum_{i=0}^p\nabla^i$
mapping $B$ to $\left(\bigoplus_{i=0}^p(\Lambda^1)^{\otimes p} 
\right)\otimes B$ 
and $D_2(x)= \nabla^p + x$
mapping $B$ to $(\Lambda^1 M)^{\otimes p} \otimes B \oplus B$.
Then $L^2_p(x,y) = (D_1(x), y)$ and  $g(x,y) = (D_2(x), y)$.
Notice that $L^2_p(x,y)= (D_1^* D_1 x, y)$
and $g(x,y)= (D_2^* D_2 x, y)$.

{\bf \green Step 2:}
To prove that these two metrics are equivalent, we need to show
that $D^*_2D_2:\; (B, h) \arrow (B, L^2)$ is Fredholm, where
$h(x, y) = (D_1^* D_1 x, D_1^* D_1 y)$ (Corollary 1).

{\bf \green Step 3:}
On a flat torus, the metric $h$
is equivalent to $L^{2}_{2p}$. Using the same argument
as proves the Rellich lemma, we obtain that any
differential operator $\Phi$ of order $<2p$ defines a compact
operator $\Phi:\; (B, h) \arrow (B, L^2)$.

{\bf \green Step 4:}
The map $D_1^*D_1 :\; (B, h) \arrow (B, L^2)$ 
is by definition an isometry, and $D_1^*D_1-D^*_2D_2$
is a differential operator of lower order, which is compact
as a map $(B, h) \arrow (B, L^2)$ by the Rellich lemma.
Then $D^*_2D_2-D_1^*D_1$ is a compact operator, and
$D^*_2D_2$ is Fredholm whenever $D^*_1D_1$ is Fredholm.
\endproof

\newpage

{\bf \blue $L^2_p$-metrics and symbols of elliptic operators}

The same argument proves the following result.

\theorem
Let $B$ be a vector bundle, and
$D:\; B \arrow B$ a differential operator
which has the same symbol as $(\nabla^p)^* \nabla^p$.
{\bf \red Then $D:\; (B, L^2_{2p}) \arrow (B, L^2)$ is Fredholm.}

\pstep
Denote by $U$ the differential operator
$(\nabla^{2p})^* \nabla^{2p}$.
To show that $D:\; (B, L^2_{2p}) \arrow (B, L^2)$
is Fredholm, it would suffice to prove that 
the metric $ (x,y) + (D(x), D(y))$
is equivalent to $L^2_{2p}(x, y)$.
The $L^2_{2p}$-metric is 
equivalent to $(x,y) + (U(x), y)$, as shown in
Theorem 1.


{\bf \green Step 2:}
For any two differential operators $A, B$,
symbol of $AB$ is equal to the symbol of $BA$.
Therefore, the symbol of 
$U=(\nabla^{2p})^* \nabla^{2p}$ is equal to the symbol
of $(\nabla^p)^* \nabla^p(\nabla^p)^* \nabla^p$
and this is equal to the symbol of $D^*D$.
This implies that $U-D^*D$
is an operator of order less than $2p$, hence
defines a compact map $(B, L^2_{2p}) \arrow (B, L^2)$. 
Therefore, the metric $(x,y) + (D^*Dx, y)$ is equivalent to 
$(x,y) + (Ux, y)$ which is equivalent to $L^2_{2p}$-metric, 
as shown in Theorem 1.
\endproof

\newpage

{\bf \blue Laplace operators}

\definition
Let $M$ be a Riemannian manifold, and
 $d:\; \Lambda^*(M) \arrow \Lambda^{*+1}(M)$
de Rham differential. Then $dd^*+d^*d$ is called
{\bf \blue the Laplacian}.

\definition
Let $M$ be a Riemannian manifold, and $B$
a bundle with orthogonal metric and a connection
$\nabla:\; B \arrow B \otimes \Lambda^1 M$.
Using the formula $\nabla(b\otimes \eta)= 
\nabla(b) \wedge \eta + b \otimes d\eta$, we extend
$\nabla$ to an operator $\nabla:\; B \otimes \Lambda^i M
\arrow B \otimes \Lambda^{i+1} M$ satisfying the
Leibnitz equation. This operator is denoted $d_\nabla$
to distinguish it from the connection.
{\bf \blue The Laplacian with coefficients in $B$}
is $d_\nabla d^*_\nabla +d^*_\nabla d_\nabla$.

\theorem
{\bf \red The Laplacian has the same symbol 
$\sigma\in \Sym^2(TM) \otimes \End(\Lambda^*M \otimes B)$
as $\nabla^*\nabla$, and it is equal to 
$g^{-1} \otimes \Id_{B\otimes \Lambda^*M}$,}
where $g^{-1}\in \Sym^2TM$ is the bivector which corresponds
to the Riemannian metric.

We shall prove it next week.
The following corollary is immediate.

\corollary
{\bf \red The Laplacian is a Fredholm map} from 
$(\Lambda^*(M) \otimes B, L^2_{p})$ to $(\Lambda^*(M) \otimes B, L^2_{p-2})$.

\proof Indeed, {\bf \purple Laplacian is a sum of a Fredholm
map $(\nabla^*)\nabla$ and a compact operator}
(all lower order differential operators are compact
by Rellich lemma).
\endproof

\end{document}



