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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 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 $l0$ 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)