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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 2: elliptic operators}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, January 27, 2018 } 
\end{center}


\newpage

{\bf \blue Differential operators  (reminder)}


{\bf \green Notation:}
Let $M$ be a smooth manifold, $TM$ its tangent bundle,
$\Lambda^iM$ the bundle of differential $i$-forms,
$C^\infty M$ the smooth functions. {\bf \purple The space of sections 
of a bundle $B$ is denoted by $B$.}


\definition
Let $M$ be a manifold. The ring of {\bf\blue differential operators
on the ring of functions on $M$} is a subalgebra of
$\End_\R(C^\infty M, C^\infty M)$  is defined  as
follows. {\bf \blue Operator of order 0} is a $C^\infty M$-linear map, 
that is, a map $L_\alpha:\; f \mapsto \alpha f$,
where $\alpha\in C^\infty M$ is a smooth function.
{\bf \blue Operator of order 1} is a sum of a differentiation
along a vector field and a $C^\infty M$-linear map.
{\bf \blue Differential operator of order $k$} is a linear combination
of products of $k$ first order differential operators.


\remark
In coordinates $x_1, ..., x_n$ , differential
operators can be expressed as sums of {\bf \blue differential monomials}:
\[ D = f_0 + \sum_{i=1}^n f_i \frac \6 {\6x_i} + 
\sum_{i,j=1}^n f_{ij} \frac {\6^2} {\6x_i\6x_j} + 
\sum_{i,j,k=1}^n f_{ijk} \frac {\6^3} {\6x_i\6x_j\6x_k} + ...
\]


\newpage

{\bf \blue Differential operators with coefficients in a
  vector bundle (reminder)}



\definition
Let $E, F$ be trivial vector bundles on $M$,
with basis $e_1, ..., e_n$ in $E$, $f_1, ..., f_m$ in $F$.
{\bf\blue  A differential operator from $E$ to $F$}
is a function mapping $\sum_{i=1}^n \alpha_i e_i$, where $\alpha_i\in
C^\infty M$, to 
\[ D\left(\sum_{i=1}^n \alpha_i e_i\right) =
\sum_{j=1}^m \sum_{i=1}^n  D_{ij}(\alpha_i) f_j, \  \ \ \ (*)
\]
where $D_{ij}$ are differential operators on $C^\infty M$.
One can think of $D$ as a {\bf \red $n \times m$-matrix
with coefficients in differential operators on $C^\infty M$.}

\definition
We say that a section $b$ of a vector
bundle $B$ on $M$ {\bf \blue has support in a set
  $K\subset M$} if $b$ vanishes in an open set which
contains $M \backslash K$. 
The smallest of all such $K$ is called {\bf \blue
  support} of $b$. 

\definition
Let $E, F$ be vector bundles on $M$.
Let $D$ be an operator mapping sections of $E$ to sections of $F$.
Suppose that for any open set $U\subset M$ such that
$E$ and $F$ are trivial on $U$ with bases $\{e_i\}$, $\{f_j\}$,
and for any $e=\sum_{i=1}^n \alpha_i e_i$ with
support in $U$,  the section $D(e)$ is expressed as in (*):
\[ D\left(\sum_{i=1}^n \alpha_i e_i\right) =
\sum_{j=1}^m \sum_{i=1}^n  D_{ij}(\alpha_i) f_j.
\] Then $D$ is called {\bf \blue a differential operator
from $E$ to $F$.}

\newpage

{\bf \blue Associated graded rings }

\remark
{\bf \blue Algebra} is an associative ring over a field.
Rings in this lecture {\bf \purple are not necessarily commutative,
but always associative.}

\definition
Let $R$ be an associative ring.
{\bf \blue Filtration} on $R$ is a collection of subspaces
$R_0 \subset R_1\subset R_2 \subset ...$ such that
$R_i R_j \subset R_{i+j}$. 

\remark
Let $x\in R_k, y \in R_l$. Then the product $xy$ modulo
$R_{k+l-1}$ depends only on the class of $x$ modulo
$R_{k-1}$. Indeed, $R_{k-1} R_l\subset R_{k+l-1}$.
{\bf \purple This defines the product map
$(R_k/R_{k-1}) \otimes (R_l/R_{l-1}) \arrow
  R_{k+l}/R_{k+l-1}$.} {\bf \red We obtained the
associative product structure on the space 
$\bigoplus_{i=0}^\infty R_i/R_{i-1}$.}

\definition 
Let $R_0 \subset R_1\subset R_2 \subset ...$ be a filtered
ring. The ring $\bigoplus_{i=0}^\infty R_i/R_{i-1}$
is called {\bf \blue the associated graded ring}
of this filtration.

\example 
Consider the filtration $\Diff^0(M)\subset
\Diff^1(M)\subset \Diff^2(M)\subset ...$ on the ring
of differential operators. The associated graded ring
is called {\bf \blue the ring of symbols of differential
operators}. 

\newpage

{\bf \blue Order of zeroes of a function}

\definition
Let $m\in M$, and $x_1, ..., x_n$ a coordinate
system around $m$, with $x_1(m)=x_2(m) = ... = x_n(m)=0$.
We say that {\bf \blue a function $f$ has zero of order
  $\geq k$ at $m$}
if $\frac{\6^{l}f}{\6x_{i_1}\6x_{i_2} ... \6x_{i_l}}(m)=0$
for any $l<k$. 

\claim
Let ${\goth m}\subset C^\infty M$ be the ideal of all functions vanishing
in $m\in M$. {\bf \red Then $f$ has zero of order
  $\geq k$ at $m$ if and only if $f \in {\goth m}^k$.}

\pstep
Let $f \in {\goth m}^k$. Then $\frac{\6 f}{\6x_i}\in
{\goth m}^{k-1}$ by Leibniz rule. {\bf \purple Hence $f$ has zero of order
  $\geq k$ at $m$.}

{\bf \green Step 2:} The function $f$ has zero of order
  $\geq k$ at $m$ if and only if $\frac{\6 f}{\6x_i}$
has zero of order $\geq k-1$ at $m$. 

{\bf \green Step 3:}
If $f$ has zero of order
  $\geq 1$ at $m$, this means that $f \in {\goth m}$
by definition. Using induction in $k$ and 
step 2, we obtain that {\bf \purple $f$ has zero of order
  $\geq k$ at $m$ $\Leftrightarrow$  
$\frac{\6 f}{\6x_i}\in {\goth m}^{k-1}$ $\Leftrightarrow$
$f\in {\goth m}^k$.}
\endproof


\corollary
Consider a function $f$ with the Taylor series decomposition
$f= \sum P_i$ in $m$, where $P_i\in \R[x_1, ..., x_n]$ a
homogeneous polynomial of degree $i$. {\bf \red Then $f$ has
zero of order $\geq k$ in $m$ if and only if $P_0=P_1=...=P_{k-1}=0$.}
\endproof

\newpage

{\bf \blue Differential operators: algebraic definition}

\definition {\bf \blue (Grothendieck)}
Let $R$ be a commutative ring over a field $k$. Given $a\in R$, consider
the map $L_a:\; R \arrow R$ mapping $x$ to $ax$.
Define $\Diff^k(R)\subset \Hom_k(R, R)$ inductively as
follows. The $\Diff^0(R)$ is the space of all
$R$-linear maps from $R$ to $R$, that is, the space of
all $L_a$, $a\in R$. The space  $\Diff^k(R)$, $k>0$
is 
\[ 
\Diff^k(R):=
\{ D\in \Hom_k(R, R)\ \ |\ \ [L_a, D]\in \Diff^{k-1}(R)
\forall a\in R.\}
\]

\exercise
Prove that {\bf \red $\Diff^k(C^\infty M)$ in this sense 
is the same as the space of differential operators defined
in the standard way.}

\claim
For any $D\in\Diff^{k-1}(C^\infty M)$, and any $a\in
C^\infty M$, {\bf \red one has $[L_a, D]\in \Diff^{k-1}(C^\infty M)$.}

\proof
Indeed, for a vector field $X\in TM$, one has
$[L_a, \Lie_X]= L_{\Lie_X(a)}$, which means that
$[L_a, \Diff^1(C^\infty M)]\subset\Diff^0(C^\infty M)$.
Now, if we take a commutator of $L_a$ and a product of
$k$ elements from $\Diff^1(C^\infty M)$, we obtain
a linear combination of products of $k-1$ elements
of $\Diff^1(C^\infty M)$, by Leibniz formula:
\[
[L_a, D_1D_2...D_n]=
[[L_a, D_1] D_2...D_n+ D_1[L_a, D_2] D_3... D_n+ ... + 
D_1D_2...D_{n-1} [L_a, D_n].
\]
\endproof



\newpage

{\bf \blue Differential operators and homogeneous polynomials}

\lemma
Let $x_1, ..., x_n$ be coordinates on $\R^n$,
and $D\in \Diff^k(\R^n)$ a differential operator
vanishing on all homogeneous polynomials
$P\in \R[x_1, ..., x_n]$ of degree $k$.
{\bf \red Then $D=0$.}

\pstep
We prove lemma by induction. For $k=0$ it is clear.
Let $L_{x_i}:\; C^\infty M \arrow C^\infty M$ 
be the multiplication by $x_i$. Then $[L_{x_i}, D]$
is a differential operator of order $k-1$
vanishing on all homogeneous polynomials of degree $\leq k-1$.
Using induction on $k$, we obtain that $[L_{x_i}, D]=0$.
Then $D$ is $\R[x_1,..., x_n]$-linear. Since $D$ vanishes on
polynomials of degree $k$ and is $\R[x_1,..., x_n]$-linear,
it vanishes on all polynomials.

{\bf\green Step 2:} Let $f\in C^\infty \R^n$ and $x\in \R^n$
be a point. Using Hadamar's
lemma, we obtain that $f= P+ f_0$, where $P\in \R[x_1,...,  x_n]$
is a polynomial of degree $k$ and $f_0$ has zero of order 
$\geq k+1$ in $x$. Then $D(f)(x)= D(P)+D(f_0)=0$
(the first summand vanishes because $D$ is $\R[x_1,...,
  x_n]$-linear, and the second because $D$ is of order $k$
and  $f_0$ has zero of order 
$\leq k+1$ in $x$).
\endproof


\newpage

{\bf \blue The ring of symbols }

\theorem
Consider the filtration $\Diff^0(M)\subset
\Diff^1(M)\subset \Diff^2(M)\subset ...$ on the ring
of differential operators. Then
{\bf \red its associated graded ring is isomorphic to 
the ring $\bigoplus_i \Sym^i(TM)$.} \vspace{-2mm}

\pstep Let $f$ be a function with zero of order $\geq k$
in z, and ${\goth m}$ its maximal ideal. 
Then $\Diff^{k-1}(f)=0$. This gives a bilinear pairing
\[
 (\Diff^{k}(M)/ \Diff^{k-1}(M))\times (\goth m^k)/(\goth m^{k-1})
\arrow \R
\]
mapping $D\otimes f$ to $Df(0)$.
Since $(\goth m^k)/(\goth m^{k-1})= \Sym^i(T_zM)^*$,
this pairing, applied to all $z\in M$ gives a natural map 
$\Diff^{k}(M)/ \Diff^{k-1}(M)\stackrel \sigma \arrow \Sym^i(TM)$.
{\bf \purple It remains to prove that $\sigma$ is an isomorphism.}
Since this pairing is local, it suffices to prove that
it is an isomorphism for $M=\R^n$.\vspace{-2mm}

{\bf \green Step 2:} Let $x_1, ..., x_n$ be coordinates in
$M=\R^n$, and $D\in \Diff^{k}(M)$. Then 
$\Sym^*(TM)=C^\infty M[\frac d {dx_1},\frac d {dx_2}, ...,  \frac d {dx_n}]$.
Consider a homogeneous differential monomial
$D= f \frac {\6^k} {\6x_{i_1}...\6x_{i_k}}$. Then
$\sigma(D)= f   \frac d {dx_{i_1}}\frac d
{dx_{i_2}}...\frac d {dx_{i_k}}$.
{\bf \purple Therefore, $\sigma$ is surjective.}

{\bf \green Step 3:}
Let $D\in \Diff^{k}(\R^n)$ be a differential operator
and $\underline D$ its class in $\Diff^{k}(\R^n)/ \Diff^{k-1}(\R^n)$
such that $\sigma(\underline D)=0$. Since $\sigma$ is
evaluation on polynomials, $D$ vanishes on
all homogeneous polynomials of degree $k$.
By Lemma 1 above, $D=0$.
\endproof

\newpage

{\bf \blue Symbols}

\theorem
Consider the filtration $\Diff^0(M)\subset
\Diff^1(M)\subset \Diff^2(M)\subset ...$. Then
{\bf \red its associated graded ring is isomorphic to 
$\bigoplus_i \Sym^i(TM)$,} identified with the
ring if fiberwise polynomial functions on $T^*M$.


\corollary
Let $F, G$ be vector bundles, 
and $\Diff^0(F,G) \subset \Diff^1(F,G) \subset \Diff^2(F,G)$
the corresponding spaces of differential operators.
{\bf \red 
Then \[
\Diff^i(F,G)/\Diff^{i-1}(F,G)=\Sym^i(TM)\otimes\Hom(F,G),
\]}
where $\Sym^i$ denotes the symmetric power (symmetric part
of the tensor power).

\definition
Let $F, G$ be vector bundles, 
and $D\in \Diff^i(F,G)$ a differential operator.
Consider its class in $\Diff^i(F,G)/\Diff^{i-1}(F,G)$ 
as a $\Hom(F,G)$-valued function on $T^*(M)$ (polynomial
of order $i$ on each cotangent space).
This function is called {\bf \blue the symbol}
of $D$.

\exercise
Let $D:\; B \arrow B \otimes \Lambda^1 M$ be a first
order differential operator. {\bf \purple Prove that $D$ is
a connection if and only if its symbol
is equal to the identity operator} $\Id\in \Hom(\Lambda^1 M
\otimes (\Hom(B, B \otimes \Lambda^1 M))$

\exercise
Prove that the symbol of the Laplacian
operator $\Delta:\; \Lambda^*M \arrow \Lambda^* M$
on a Riemannian manifold $M$ at $\xi\in T^*M$ {\bf \purple is equal
to $|\xi|^2 \Id_{\Lambda^* M}$.}

\newpage

{\bf \blue Elliptic operators}

\definition
Let $F$, $G$ be vector bundles of the same rank.
A differential
operator $D:\; F \arrow G$ is called {\bf \blue elliptic}
if its symbol $\sigma(D)\in \Hom(F,G)\otimes \Sym^i(TM)$
is invertible at each non-zero $\xi \in T^*M$.

\example
Consider an operator of second order on $C^\infty(\R^n)$,
\[ D = f_0 + \sum_{i=1}^n f_i \frac \6 {\6x_i} + 
\sum_{i,j=1}^n f_{ij} \frac {\6^2} {\6x_i\6x_j}.
\]
Then the symbol of $D$ is $f_{ij} \frac {\6^2} {\6x_i\6x_j}$;
{\bf \purple it is elliptic if and only if the symmetric form
$f_{ij} dx_i dx_j$ is positive or negative definite
everywhere in $\R^n$.}

\exercise 
Prove that symbol of $D^*$ is a $\Hom(F,G)$-valued
function on $T^*(M)$ which is Hermitian adjoint to
$\symb(D)$.


\newpage

{\bf \blue Elliptic operators: main properties}


The rest of the slides today are introduction to the
main results about elliptic operators; complete proofs 
for some of them will be given  later. 

\theorem {\bf \blue (Elliptic regularity)}\\
Let $D$ be an elliptic operator with smooth coefficients
on a manifold (not necessarily compact), and $f\in \ker D$.
{\bf \red Then $D$ is smooth, and real analytic if coefficients of
$D$ were real analytic.}

\theorem 
Let $D$ be an elliptic operator with smooth coefficients
on a compact manifold. {\bf \red Then its kernel is
finite-dimensional.} If $D:\; F \arrow F$ is self-adjoint,
then {\bf \red $D$ can be diagonalized in an appropriate 
orthonormal basis in the space of sections of $F$, and 
its eigenvalues are discrete.}

\newpage

{\bf \blue Elliptic operators of second order}

Let $D=f_0 + \sum_{i=1}^n f_i \frac \6 {\6x_i} + 
\sum_{i,j=1}^n f_{ij} \frac {\6^2} {\6x_i\6x_j}$ be an
elliptic operator of second order. For second order
operators, {\bf \purple we always assume that the
symbol $f_{ij} dx_i dx_j$ is positive definite.}

\theorem {\bf \blue (E. Hopf's maximum principle)}\\
Let $D$ be a second order elliptic operator
on a manifold (not necessarily compact)
such that $D(\const)=0$, and
$f$ a solution of an equation $D(f)=0$.
Assume that $f$ has a local maximum. {\bf \red Then $f=\const$.}

\theorem {\bf \blue (Harnack inequality)}\\
Let $D$ be a second order elliptic operator
on a manifold $M$ (not necessarily compact), and
$\Omega\subset M$ an open subset with compact closure.
For any $f\in \C^\infty (\Omega)$, denote by 
$\Var_\Omega(f)$ the number $\sup_\Omega(f)-\inf_\Omega(f)$.
{\bf \red Then there exists a constant $C$, depending only
on $D$, $M$ and $\Omega$, such that for any
$f\in \ker D$, one has $\Var_\Omega(f)<C$.}

\corollary
{\bf \red Any pointwise converging sequence of functions $f\in \ker D$
converges uniformly.}

\proof
Indeed, by Harnack's inequality, the solutions of $D(f)=0$
are uniformly continuous, hence the pointwise convergence
implies uniform convergence.
\endproof


\newpage

\begin{center}
{\bf \green Eberhard Hopf (1902-1983) }\\[4mm]
\epsfig{file=Hopf_Eberhard.jpeg,width=0.30\linewidth}
\end{center}


\newpage

{\bf \blue Fredholm operators}


\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 a quadratic form
$|f|^2=\sum_{i=0}^p \int_M |\nabla^if|^2$,
for some connection and scalar product on $F$ and $\Lambda^1M$.

\exercise
{\bf \purple Prove that this topology is independent from the
choice of a connection and a metric.}

\definition
A continuous operator $\psi:\; A \arrow B$ on topological
vector spaces is called {\bf \blue Fredholm}
if its kernel is finite-dimensional, and
its image is closed, and has finite codimension.

\theorem
Let $D:\; F \arrow G$ be an elliptic operator of order
$d$. Clearly, $D$ defines a continuous map
$L^2_{p}(F) \arrow L^2_{p-d}(G)$. {\bf \red Then this map is
Fredholm.}

\remark
{\bf \purple This difficult theorem is a foundation of Hodge theory}
(and many other things).

%\newpage\
%\
%{\bf \blue Elliptic complexes}\
%\
%\definition\
%Let $F$, $G$, $H$ be vector bundles,\
%and $F\stackrel D \arrow G\stackrel D \arrow H$ a complex\
%of differential operators (that is, $D^2=0$).\
%It is called {\bf\blue  elliptic complex} if its symbols \
%$F\stackrel {\sigma(D)} \arrow G\stackrel  {\sigma(D)} \arrow H$\
%give an exact sequence at each non-zero $\xi \in T^*M$.\
%\
%\definition\
%Let $A$, $B$, $C$ be topological vector spaces\
%and $A\stackrel D \arrow B\stackrel D \arrow C$ a complex\
%of continuous maps. It is called {\bf \blue\
%Fredholm complex} if $\im D$ is closed, \
%and $\frac{\ker D}{\im D}$ is finite-dimensional.\
%\
%\
%\exercise Prove that {\bf \purple this is equivalent to the operator \
%$DD^*+D^*D$ being Fredholm.}\
%\
%\theorem\
%Let $F\stackrel {D_1} \arrow G\stackrel {D_2} \arrow H$ be an \
%elliptic complex of differential operators, with\
%$D_1$ of order $d_1$ and $D_2$ of order $d_2$.\
%{\bf \red Then the complex\
%$L^2_{p}(F) \stackrel{D_1}\arrow \
%L^2_{p-d_1}(G)\stackrel{D_2}\arrow L^2_{p-d_1-d_2}(H)$\
%is Fredholm.} \
%\
%\corollary\
%{\bf \purple \
%Cohomology of any elliptic complex are finite-dimensional.}

\newpage

{\bf \blue Index of an elliptic operator}

\remark Let $D:\; F \arrow G$ be an elliptic operator of order
$d$, and \\ $L^2_{p}(F) \arrow L^2_{p-d}(G)$ the
corresponding maps on $L^p$-spaces.

\definition
{\bf \blue Index} of a Fredholm operator $D$
is the number $\dim \ker D - \dim \coker D=
\dim \ker D - \dim\ker D^*$

\remark
Index of an elliptic operator $D:\; L^2_{p}(F) \arrow L^2_{p-d}(G)$ 
a priori depends on $p$, however,
by elliptic regularity, {\bf \purple all elements
of $\ker D$ and $\ker D^*$ are smooth,} hence
$\dim \ker D\restrict {L^2_{p}(F)}$ is independent
from $p$.

\exercise
Let $F_t$ be a continuous family of Fredholm operators.
{\bf \purple Prove that the index of $F_t$ is constant in
  $t$.}

\corollary
Let $D_t$ be a continuous family of elliptic operators of order 
$k$. {\bf \red Then $\ind D_0= \ind D_1$.}

\corollary
Index of an elliptic operator {\bf \red is determined by
  its symbol.}

\proof
For any elliptic operators $D_0, D_1$ with the same symbol,
the operator $D_t:= t D_1 + (1-t) D_0$ has the same
symbol, hence $D_0$ can be deformed to $D_0$ continuously
and {\bf \purple gives a continuous family of Fredholm operators.}
\endproof


\newpage

{\bf \blue Atiyah-Singer index theorem}

\remark
The index of an elliptic operator is clearly
constant under continuous change of its symbol. Therefore,
{\bf \red index depends only on the homotopy class of its
symbol}, which can be considered as a non-degenerate
section of $\Hom(F,G)$ over $T^*M \backslash 0$.
Homotopy classes of such sections are described explicitly
in terms of characteristic classes of $F$, $G$ and 
the topological K-theory of $M$. {\bf \blue
Atiyah-Singer index formula} expresses the index of
an elliptic operator as a polynomial function
of these topological invariants. 

\example
Let $M$ be a compact manifold, and
$D:\; C^\infty M \arrow C^\infty M$ elliptic operator
of second order. {\bf \red Then $\ind D=0$.}

\proof Locally, we can write
$D=f_0 + \sum_{i=1}^n f_i \frac \6 {\6x_i} + 
\sum_{i,j=1}^n f_{ij} \frac {\6^2} {\6x_i\6x_j}$.
The Laplacian $\Delta$ associated with the metric tensor $f_{ij}$
has the same symbol, hence it suffices to prove
$\ind \Delta=0$. However, $\Delta$ is self-adjoint,
hence $\dim \ker \Delta=\dim\coker \Delta$.
\endproof

\corollary
For any second order elliptic operator
$D$ on $C^\infty M$ with $D(\const)=0$,
{\bf \red one has $\dim \coker D=1$. }

\proof
By the strong maximum principle, 
all functions $\ker D$ are constant,
hence $\dim \ker D=1$. {\bf \purple Then $\dim \coker D=1$
by the index theorem.}
\endproof

\end{document}
