\documentclass{slides} \usepackage{amssymb, amsmath, amscd, color, epsfig, url} %\usepackage[matrix,arrow]{xy} \newcommand{\green}{\color[rgb]{0,0.4,0}} \newcommand{\purple}{\color[rgb]{0.4,0,0.4}} \newcommand{\red}{\color[rgb]{0.7,0,0}} \newcommand{\blue}{\color{blue}} \def\eqref#1{(\ref{#1})} \newcommand{\goth}{\mathfrak} \newcommand{\g}{{\frak g}} \newcommand{\arrow}{{\:\longrightarrow\:}} \newcommand{\Z}{{\Bbb Z}} \newcommand{\C}{{\Bbb C}} \newcommand{\R}{{\Bbb R}} \newcommand{\Q}{{\Bbb Q}} \renewcommand{\H}{{\Bbb H}} \newcommand{\6}{\partial} \def\1{\sqrt{-1}\:} \newcommand{\restrict}[1]{{\left|_{{#1}}\right.}} \newcommand{\cntrct} % contraction with a vector field {\hspace{2pt}\raisebox{1pt}{\text{$\lrcorner$}}\hspace{2pt}} \def\Bbb#1{\mathbb #1} \newcommand{\calo}{{\cal O}} \newcommand{\cac}{{\cal C}} % Correcting TeX... %\let\oldtilde=\tilde %\renewcommand{\tilde}{\widetilde} \renewcommand{\bar}{\overline} \renewcommand{\phi}{\varphi} \renewcommand{\epsilon}{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant} % Operatornames \newcommand{\even}{{\rm even}} \newcommand{\ev}{{\rm even}} \newcommand{\odd}{{\rm odd}} \newcommand{\const}{{\it const}} \newcommand{\fl}{{\rm fl}} \newcommand{\im}{\operatorname{im}} \newcommand{\End}{\operatorname{End}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Hol}{\operatorname{{\cal H}ol}} \newcommand{\Tot}{\operatorname{Tot}} \newcommand{\Id}{\operatorname{Id}} \newcommand{\id}{\operatorname{\text{\sf id}}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Alt}{\operatorname{Alt}} \newcommand{\Iso}{\operatorname{Iso}} \newcommand{\Sec}{\operatorname{Sec}} \newcommand{\Can}{\operatorname{Can}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Spin}{\operatorname{Spin}} \newcommand{\codim}{\operatorname{codim}} \newcommand{\coim}{\operatorname{coim}} \newcommand{\coker}{\operatorname{coker}} \newcommand{\slope}{\operatorname{slope}} \newcommand{\rk}{\operatorname{rk}} \newcommand{\Def}{\operatorname{Def}} \newcommand{\Lie}{\operatorname{Lie}} \newcommand{\Tw}{\operatorname{Tw}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Diff}{\operatorname{Diff}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\inbfpare}[1]{{% \mbox{\tt (}\hspace{-5pt}\mbox{\tt (} #1 % \mbox{\tt )}\hspace{-5pt}\mbox{\tt )}% }} \newcommand{\comment}[1]{{}} \def\blacksquare{\hbox{\vrule width 10pt height 10pt depth 0pt}} \def\endproof{\blacksquare} \def\shortdash{\mbox{\vrule width 4.5pt height 0.55ex depth -0.5ex}} \makeatletter %\@ifundefined{Bbb} % {\newcommand{\Bbb}[1]{{\mathbb #1}}}% %{}% {\edef\Bbb#1{{\Bbb #1}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Pagestyle % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\ps@verbit}{% \renewcommand{\@oddhead}{% \scriptsize {\it \small Hodge theory, lecture 1 \hfil \tiny M. Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{{\bf \green EXERCISE:\ }} \newcommand{\pstep}{{\bf \green Proof. Step 1:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}