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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 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}