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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Hodge theory \\[15mm]
\small lecture 1: differential operators}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]
{\scriptsize NRU HSE, Moscow} \\[15mm]
{\small Misha Verbitsky, January 24, 2018 } 
\end{center}




\newpage

\begin{center}
{\bf \green William Vallance Douglas Hodge\\
17 June 1903 - 7 July 1975 }\\
\epsfig{file=Hodge_3.jpeg,width=0.35\linewidth}\\
{Photograph by P. Halmos}
\end{center}


\newpage

{\bf \blue Differential operators}


{\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}:
\[ D = f_0 + \sum_{i=1}^n f_i \frac d {dx_i} + 
\sum_{i,j=1}^n f_{ij} \frac {d^2} {dx_idx_j} + 
\sum_{i,j,k=1}^n f_{ijk} \frac {d^3} {dx_idx_jdx_k} + ...
\]

\newpage

{\bf \blue Differential operators with coefficients in a
  trivial vector bundle}



\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 $C^\infty M$.}

\newpage

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

% \hspace{-12mm}
\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 section $e=\sum_{i=1}^n \alpha_i e_i$ 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 $F$.}

\example
{\bf \blue Differential} is a map $d:\; C^\infty M \arrow \Lambda^1 M$
mapping a function to its differential. Prove that it is
a first order differential
operator.

\example
{\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.
Prove that connection is a first order differential
operator.


\newpage

{\bf \blue Local operators}

\definition
Let $E, F$ be vector bundles on $M$.
An operator $D$ mapping sections of $E$ to sections
of $F$ is called {\bf\blue local} if it maps
any section with support in $K \subset M$
to a section with support in $K$.

\remark {\bf \purple Differential operators are clearly local.}

\exercise {\bf \red (difficult)} \\
Let $M$ be a compact manifold, $F, G$ -- vector bundles. 
Prove that {\bf \red any local operator from $F$ to $G$
is a differential operator.} Find a  counterexample
when $M$ is non-compact.

\newpage

{\bf \blue Differential operators: algebraic definition}

\definition 
{\bf \blue (Grothendieck)}\\
Let $R$ be a commutative ring over a field $k$,
and $A, B$ $R$-modules.
{\bf \blue Differential operator of order 0}
from $A$ to $B$ is an $R$-linear map $\phi\in \Hom_R(A,B)$.
Differential operator of order $i>0$ is defined inductively:
$\alpha \in \Diff^i(A,B)$ if for any $r\in R$,
the commutator $\alpha L_r-L_r\alpha$ belongs to $\Diff^{i-1}(A,B)$,
where $L_r(x)=rx$.


\definition
Given a vector bundle on a smooth manifold $M$, we
may consider its space of sections as an $C^\infty M$-module.
{\bf \blue Differential operators} $\Diff^i(F,G)$
on vector bundles $F$, $G$ are defined as differential
operators on the corresponding spaces of sections
in the sense of the Grothendieck's definition.
{\bf \blue Differential operator on $M$}
is an element of 
$\Diff^i(M):= \Diff^i(C^\infty M,C^\infty M)$.

\exercise {\bf 
\red Prove that this definition is equivalent to the usual one.}


\newpage

{\bf \blue Discuss}

0. Explain the format.

Discuss:

1. Language.

2. Time of the lectures and seminars.

The course's page:\\
\url{http://bogomolov-lab.ru/KURSY/Hodge-2018/}

\newpage

{\bf \blue REMINDER: de Rham algebra}

\definition
Let $\Lambda^* M$ denote the vector bundle
with the fiber $\Lambda^*T^*_xM$ at $x\in M$
($\Lambda^*T^*M$ is the Grassman algebra of the cotangent
space $T^*_x M$). The sections of $\Lambda^i M$
are called {\bf \blue differential $i$-forms}.
 The algebraic operation ``wedge product'' defined
on differential forms is $C^\infty M$-linear; 
the space $\Lambda^* M$ of all differential forms
is called {\bf \blue the de Rham algebra}.

\remark $\Lambda^0 M = C^\infty M$.


\theorem
{\bf \red There exists a unique operator 
$C^\infty M\stackrel d \arrow \Lambda^1 M\stackrel d
\arrow \Lambda^2 M \stackrel d \arrow \Lambda^3 M \stackrel d \arrow ...$
satisfying the following properties}

1. On functions, $d$ is equal to the differential.\\
2. $d^2=0$ \\
3. $d(\eta \wedge \xi) = 
d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge
d(\xi)$,
where $\tilde \eta=0$ where $\eta\in \lambda^{2i}M$ is
{\bf \blue an
even form,} and $\eta\in \lambda^{2i+1}M$ is {\bf \blue odd.}


\definition
The operator $d$ is called {\bf \blue de Rham differential}.

\exercise
{\bf \purple Prove it}.


\definition
A form $\eta$ is called {\bf \blue closed} if $d\eta=0$,
{\bf \blue exact} if $\eta \in \im d$. The group $\frac{\ker d}{\im d}$
is called {\bf \blue de Rham cohomology} of $M$.


\newpage

{\bf \blue Graded algebras }

\definition
An algebra $A$ is called {\bf\blue graded} if $A$ is 
represented as $A= \bigoplus A^i$, where $i\in \Z$,
and the product satisfies $A^i \cdot A^j
\subset A^{i+j}$. Instead of $\bigoplus A^i$
one often writes  $A^*$, 
where $*$ denotes all indices together.
Some of the spaces $A^i$ can be zero, but the 
ground field is always in $A^0$, so that it is non-empty. 

\example
{\bf \purple The tensor algebra 
$T(V)$ and the polynomial algebra $\Sym^*(V)$
are obviously graded.}

\definition
Let  $A^* = \oplus_{i\in \Z}A^i$ be a graded algebra over a field.
It is called
{\bf\blue graded commutative}, or {\bf\blue supercommutative},
if $ab = (-1)^{ij} ba$
for all $a\in A^i, b \in A^j$. 

\example
{\bf \purple Grassmann algebra $\Lambda^* V$
is clearly supercommutative.}

\newpage

{\bf \blue Graded derivations }


\definition
Let $A^*$ be a graded commutative algebra, 
and $D:\; A^* \arrow A^{*+i}$ be a map which
shifts grading by $i$. 
It is called a
{\bf\blue  graded derivation}, if it satisfies the Leibniz rule:
$D(ab) = D(a) b + (-1)^{ij} a D(b)$,
for each $a \in A^j$. 


\definition
Let $M$ be a smooth manifold, and  $X\in TM$ 
a vector field. Consider an operation of {\bf \blue 
convolution with a vector field} 
\[ i_X:\; \Lambda^i M \arrow \Lambda^{i-1}M,\] 
mapping an $i$-form $\alpha$ to an $(i-1)$-form
$v_1, ..., v_{i-1} \arrow \alpha(X, v_1, ..., v_{i-1})$


\exercise
{\bf \purple Prove that $i_X$ is an odd derivation.}


\newpage

{\bf \blue Supercommutator}


\definition
Let $A^*$ be a graded vector space, and 
$E:\; A^*\arrow A^{*+i}$, $F:\; A^*\arrow A^{*+j}$
operators shifting the grading by $i, j$.
Define  {\bf\blue the supercommutator} $\{E, F\}:= EF - (-1)^{ij} FE$.

\definition
An endomorphism of a graded vector space
which shifts grading by $i$ is called {\bf \blue even}
if $i$ is even, and {\bf \blue odd} otherwise.

\exercise
Prove that the supercommutator satisfies 
{\bf\blue graded Jacobi identity,}
\[
\{ E, \{F, G\}\} = \{\{ E, F\}, G\} + (-1)^{\tilde E
\tilde F}  \{ F, \{E, G\}\}
\]
where  $\tilde E$ and $\tilde F$ are 0 if
$E, F$ are even, and 1 otherwise.



\remark
There is a simple mnemonic rule
which allows one to remember a superidentity,
if you know the commutative analogue.
{\bf\red Each time when in commutative case
two letters $E$, $F$ are exchanged, in supercommutative
case one needs to multiply by $(-1)^{\tilde E\tilde F}$.}


\exercise
{\bf \purple Prove that a supercommutator of superderivations
is again a superderivation.}



\newpage

{\bf \blue Lie derivative}

\definition
Let $B$ be a smooth manifold, and $v\in TM$ a vector field.
An endomorphism $\Lie_v:\; \Lambda^* M \arrow \Lambda^* M$,
preserving the grading is called
{\bf\blue a Lie derivative along $v$} if it satisfies the 
following conditions.\\
\phantom{XX} (1) On functions $\Lie_v$ is 
equal to a derivative along $v$.  (2) $[\Lie_v, d]=0$.\\
 \phantom{XX} (3) $\Lie_v$ is a derivation of the de Rham
 algebra (that is, satisfies the Leibniz rule).

\remark
The algebra $\Lambda^*(M)$ is generated 
by $C^\infty M=\Lambda^0(M)$ 
and $d(C^\infty M)$. The restriction
$\Lie_v\restrict{C^\infty M}$ is determined by the first axiom.
On $d(C^\infty M)$ is also determined
because $\Lie_v(df) =d(\Lie_v f)$. {\bf \red Therefore,
$\Lie_v$ is uniquely defined by these axioms.}

\exercise Prove the
{\bf \purple anticommutator identity: $[d, \{d, E\}]=0$
for each $E\in \End(\Lambda^* M)$.}


\theorem
{\bf \blue (Cartan's formula)}
Let $i_v$ be a convolution with a 
vector field, $i_v(\eta) = \eta(v, \cdot, \cdot, ..., \cdot)$
{\bf \red Then the anticommutator
$\{d, i_v\}$ is equal to the Lie derivative along $v$.}

\proof
$\{d, \{d, i_v\}\}=0$ by the lemma above.
A supercommutator of two graded derivations is a graded derivation.
Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$
\endproof

\newpage

{\bf \blue Connections}


\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 each 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 a linear maps.}

\newpage

{\bf \blue Connection and a tensor product}

\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 Adjoint connection}

\definition Given a connection $\nabla$ on
a vector bundle $B$ equipped with a scalar product
$(\cdot, \cdot)$, define $\nabla^*$ by the formula
\[
d(b, b') = (\nabla(b), b') + (b, \nabla^*(b')). \ \ \  (**)
\]
Here, $b, b'$ are sections of $B$, 
$d(b, b')$ is a differential of a function, and
$(\nabla(b), b')$ is the 1-form obtained from the
bilinear pairing $B\otimes (B \otimes \Lambda^1 M) \arrow
\Lambda^1 M$.

\claim {\bf \purple The map $\nabla^*: 
 B \arrow B \otimes \Lambda^1 M$ is well defined by (**).
Moreover, it is also a connection.}

\proof
The first statement is clear, because any linear map
$B \arrow \Lambda^1 M$ can be  represented by
$b \arrow (b, A)$ for some $A\in B \otimes \Lambda^1 M$.
To check the second statement, we take $f\in C^\infty M$, 
and write 
\[
(b, b') df + 
f d(b,b')=  d(b, f b') = f (\nabla(b), b') + (b, \nabla^*(f b')).  (**)
\]
which gives $\big(b , \nabla^*(f b')- f
   \nabla^*(b')\big) = (b, b') df$,
hence $\nabla^*(f b')- f
   \nabla^*(b') = b' \otimes df$.
\endproof

\definition
The connection $\nabla^*$ is called {\bf\blue adjoint
  connection} to $\nabla$. Relation $\nabla=\nabla^*$
happens precisely when $\nabla$ preserves the metric
tensor, considered as a section of $B^*\otimes B^*$, 
and in this case $\nabla$ is called
{\bf \blue an orthogonal connection}.



\newpage

{\bf \blue Adjoint connection and $L^2$-product}

\definition
Fix a volume form $\Vol$ on a manifold $M$
Consider a $C^\infty M$-linear scalar product
on a vector bundle $B$. Then {\bf \purple the space of sections
of $B$ is also equipped with a scalar product:}
$(b, b')_{L^2}= \int_M (b,b') \Vol$. It is called
{\bf \blue the standard $L^2$-scalar product on the
space of sections}.

\lemma
{\bf \blue (integration by parts)}\\
Let $B$ be a bundle on $M$ with scalar product and 
connection $\nabla$, and $b, b'\in B$ its sections.
{\bf \red Then, for any vector fields $X\in TM$, one has
\[
\int_M (\nabla_X b, b') + \int_M (b, \nabla_X^* b')
= \int_M (b, b') \Lie_X\Vol \ \ \ (***)
\]}
\proof
By definition, one has
\[
\int_M ((\nabla_X)^*(b), b')\Vol =  
\int_M (b, \nabla_X b')\Vol= - \int_M (\nabla_X^* b, b')\Vol -
\int_M \Lie_X(b,b') \Vol,
\]
where $\Lie_X(b,b')$ is differential of the function
$(b,b')$ along $X\in TM$. However, for any top form $\eta$, one has
$ \Lie_X (\eta)=  d(i_x \eta)$ by Cartan's formula, giving
$\int_M \Lie_X (\eta)=0$, hence
\[ 0 = \int_M \Lie_X((b,b') \Vol) = 
   \int_M \Lie_X(b,b') \Vol + \int_M (b,b') \Lie_X\Vol,
\]
giving the last term in (***).
\endproof


\newpage

{\bf \blue Adjoint operators}

\remark
Operators $A:\; F \arrow G$ and $A^*:\; G \arrow F$
on spaces with a scalar product
are called {\bf\blue orthogonal adjoint}, 
or {\bf \blue adjoint}, if $(A(f), g)= (f, A^*(g))$
for each $f\in F$, $g\in G$.

\claim
An orthogonal adjoint $D^*$ to a differential operator $D$
{\bf \red  is a differential operator again.}

\pstep
This is clear for $C^\infty M$-linear operators (just take the
pointwise adjoint map). {\bf \purple If we prove it
for first order operators, we are done,} because
$(XY)^*= Y^* X^*$.

{\bf \green Step 2:} First order operators 
are expressed as linear combination of linear
maps and derivatives $\nabla_X:\; F \arrow F$ 
combined with linear maps. Therefore, {\bf \purple it would
suffice to show that $(\nabla_X)^*$ is a differential
operator.}

{\bf \green Step 3:}
The map $(\nabla_X)^*$ is a differential operator:
$(\nabla_X)^*(b)= -\nabla^*_X- \frac{\Lie_X(\Vol)}{\Vol}b$, because 
\[ \int_M ((\nabla_X)^*(b), b')\Vol =
   - \int_M (\nabla^*_Xb, b')\Vol - \int(b, b') \Lie_X
   (\Vol)
\]
by ``integration by parts'', as shown above.
\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 $\Delta$.} 

\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.}

These two theorems will be proven later.


\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 
$\{d, \Delta\}=0$.
This means that {\bf \red $\Delta$ commutes with the de Rham differential.}

{\bf \green Step 2:} Consider the eigenspace decomposition
$\Lambda^*(M) \tilde = \bigoplus_\alpha {\cal H}^*_\alpha(M)$,
where $\alpha$ runs through all eigenvalues of $\Delta$,
and ${\cal H}^*_\alpha(M)$ is the corresponding eigenspace.
{\bf \purple For each $\alpha$, de Rham differential defines a complex
\[
{\cal H}^0_\alpha(M) \stackrel d \arrow 
{\cal H}^1_\alpha(M) \stackrel d \arrow 
{\cal H}^2_\alpha(M) \stackrel d \arrow ...
\]
}
{\bf \green Step 3:} On ${\cal H}^*_\alpha(M)$, 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 $({\cal H}^*_\alpha(M), d)$
don't contribute to cohomology.}

{\bf \green Step 4:} We have proven that
\[
H^*(\Lambda^* M, d) = \bigoplus_\alpha H^* ({\cal H}^*_\alpha(M),d)=
H^* ({\cal H}^*_0(M),d)={\cal H}^*(M).
\]
\endproof




\end{document}

