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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{\proof}{% {\bf \green Proof:\ }} \newcommand{\pstep}{% {\bf \green Proof. Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 13: Cohomology and a circle action} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] November 4, 2020 } \end{center} \newpage {\bf \blue Stone-Weierstrass approximation theorem} \definition Let $M$ be a topological space, and $\|f\|:= \sup_M |f|$ {\bf\blue the sup-norm on functions}. {\bf \blue $C^0$-topology} on the space $C^0(M)$ of continuous, bounded real-valued functions is the topology defined by the sup-norm. \exercise Prove that {\bf \purple $C^0(M)$ with sup-norm is a complete metric space.} \definition Let $A\subset C^0 M$ be a subspace in the space of continuous functions. We say that $A$ {\bf\blue separates the points} of $M$ if for all distinct points $x, y\in M$, there exists $f\in A$ such that $f(x) \neq f(y)$. \theorem {\bf\blue (Stone-Weierstrass theorem)}\\ Let $A\subset C^0 M$ be a subring separating points, and $\bar A$ its closure. {\bf \blue Then $\bar A= C^0M$.} \newpage {\bf \blue Hilbert spaces} \definition {\bf \blue Hilbert space} over $\C$ is a complete, infinite-dimensional Hermitian space which is second countable (that is, has a countable dense set). \definition {\bf \blue Orthonormal basis} in a Hilbert space $H$ is a set of pairwise orthogonal vectors $\{x_\alpha\}$ which satisfy $|x_\alpha|=1$, and such that $H$ is the closure of the subspace generated by the set $\{x_\alpha\}$. \theorem {\bf \red Any Hilbert space has a basis, and all such bases are countable.} \proof A basis is found using Zorn lemma. If it's not countable, open balls with centers in $x_\alpha$ and radius $\epsilon < 2^{-1/2}$ don't intersect, which means that the second countability axiom is not satisfied. \endproof \theorem {\bf \red All Hilbert spaces are isometric}. \proof Each Hilbert space has a countable orthonormal basis. \endproof \newpage {\bf \blue Fourier series} \example Let $(M,\mu)$ be a space with measure. Consider the space $V$ of measurable functions $f:\; M \arrow \C$ such that $\int_M |f|^2 \mu< \infty$. For each $f, g \in V$, the integral $\int f\bar g\mu$ is well defined, by Cauchy inequality: $\int |fg|\mu < \sqrt{\int_M |f|^2 \mu \int_M |g|^2 \mu}$. This gives a Hermitian form on $V$. Let $L^2(M)$ denote the completion of $V$ with respect to this metric. It is called {\bf \blue the space of square-integrable functions on $M$}. Its elements are called {\bf \red $L^2$-functions.} \claim {\bf\blue ("Fourier series")} Functions $e_k(t)= e^{ 2\pi\1 k t}$, $k\in \Z$ on $S^1=\R/\Z$ {\bf \red form an orthonormal basis in the Hilbert space $L^2(S^1)$}. \pstep Orthogonality is clear from $\int_{S^1} e^{ 2\pi\1 k t} dt =0$ for all $k\neq 0$ (prove it). {\bf \green Step 2:} The space of Fourier polynomials $\sum_{i=-n}^n a_k e_k(t)$ is dense in the space of continuous functions on the circle by the Stone-Weierstrass approximation theorem. Therefore, the closure of the space of functions which admit Fourier series is $L^2(S^1)$. \endproof \newpage {\bf \blue Fourier series on a torus} \remark Let $t_1,..., t_n$ be coordinates on $\R^n$. We can think of $t_i$ as of angle coordinates on the torus $T^n=\R^n/\Z^n$, considered as a product of $n$ copies of $S^1$. Consider the {\bf \blue Fourier monomials} $F_{l_1, ..., l_n}:= \exp(2\pi\1\sum_{i=1}^n l_it_i)$, where $l_1, ..., l_n$ are integers. Clearly, \[ L^2(T^n)\cong \underbrace{L^2(S^1) \hat\otimes L^2(S^1)\hat \otimes ...\hat \otimes L^2(S^1)}_{\text{$n$ times}}. \] where $\hat\otimes$ denotes the completed tensor product. This implies that the {\bf \purple Fourier monomials form a Hilbert basis in $L^2(T^n)$. } \remark This also follows directly from the Stone-Weierstrass theorem. \theorem Let $V$ be a Hilbert space, $\Map(T^n, V)$ continuous maps, and $L^2(T^n, V)$ a completion of $\Map(T^n, V)$ with respect to the $L^2$-norm $|v|^2= \int_{T^n} |v(x)|^2 dx$. Consider an orthonormal basis $u_1, ..., u_n, ...$ in $V$. Then {\bf \red an orthonormal basis in $\Map(T^n, V)$ is given by monomial maps $F_{l_1, ..., l_n} u_j$} taking $s\in T^n$ to $F_{l_1, ..., l_n}(s)u_j$. \proof Orthonormality of the collection $\{F_{l_1, ..., l_n} u_j\}$ is clear. To prove its completeness (that is, the density of the subspace generated by $\{F_{l_1, ..., l_n} u_j\}$), notice that $\Map(T^n, V)$ is a completion of $\oplus_i \Map(T^n,V_i)$, where $V_i=\langle v_i\rangle$. Now, $\{F_{l_1, ..., l_n} u_i\}$ is an orthonormal basis in $V_i=\Map(T^n, \C)$. \endproof \newpage {\bf \blue Weight decomposition for $U(1)$-representations} \exercise Let $\rho:\; U(1) \arrow GL(V)$ be a finite-dimensional irreducible complex representation of the Lie group $U(1)$. {\bf \purple Prove that $\dim \C=1$ and there exists $n\in \Z$ such that $t\in U(1)=\R/\Z$ acts on $V$ as $\rho(t)(v)= e^{2\pi\1 nt}v$.} \definition A representation of $U(1)$ with $\rho(t)(v)= e^{2\pi\1 nt}v$ is called {\bf \blue weight $n$} representation. \definition Let $V$ be a Hermitian space (possibly infinitely-dimensional) equipped with an action of $U(1)$, and $V_k\subset V$ weight $k$ representations, $k\in \Z$. The direct sum $\bigoplus V_k$ is called {\bf \blue the weight decomposition} for $V$ if it is dense in $V$. \example Let $L^2(S^1, W)$ the space of maps from $S^1$ to a Hermitian space $W$. We define $U(1)$-action on $L^2(S^1, W)$ by $\rho(t)(f)= R_t(f)$ where $R_t(f(x))= f(x+t)$ shifts $S^1$ by $t$. Clearly, this is a Hermitian representation, and {\bf \purple its weight decomposition is its Fourier decomposition.} \newpage {\bf \blue Weight decomposition for $U(1)$-representations (2)} \claim Let $\bigoplus V_k\subset V$ be the weight decomposition of a Hermitian representation $\rho$ of $U(1)$. Then {\bf \red any vector $v\in V$ can be decomposed onto a converging serie $v=\sum_{i\in \Z} v_i$, with $v_i\in V_i$}. This decomposition is called {\bf \blue the weight decomposition} for $v$. \pstep Clearly, all $V_i$ are pairwise orthogonal; indeed, for any $t\in U(1)$ and $x_p\in V_p$, $x_q\in V_q$, $i\neq j$, we have \begin{multline*} e^{2\pi\1 pt} (x_p, x_q)= (\rho(t)(x_p), x_q)= (x_p, \rho(-t)x_q)= \\= (x_p, e^{-2\pi\1 qt}x_q)=e^{2\pi\1 qt}(x_p, x_q) \end{multline*} giving $p=q$ whenever $(x_p, x_q)\neq 0$. {\bf \green Step 2:} Let $\pi_i:\; V \arrow V_i$ be the orthogonal projection. Then $|x|^2 \geq \sum_{i=-p}^{p} |\pi_i(x)|^2$ because orthogonal projection is always distance-decreasing. Therefore, the serie $\sum_{i\in \Z} \pi_i(x)$ converges. Its limit is a vector $x'$ which satisfies $(x, u)=(x', u)$ for any $u\in \bigoplus_{k\in \Z} V_k$. Since $\bigoplus_{k\in \Z} V_k$ is dense in $V$, this implies $x=x'$. \endproof \newpage {\bf \blue Weight decomposition and Fourier series} \lemma Let $W$ be a Hermitian representation of $U(1)$ admitting a weight decomposition. Then {\bf \red any subquotient of $W$ also admits a weight decomposition.} \proof This is clear for quotients. Any closed subspace $V\subset W$ gives a direct sum decomposition $W=V\oplus V^\bot$, hence it also can be realized as a quotient. \endproof \lemma Let $\rho:\; U(1)\arrow U(W)$ be a Hermitian representation of $U(1)$, and $L^2(S^1, W)$ the space of maps from $S^1$ to $W$ with the $U(1)$-action by translation as defined earlier. {\bf \red Then $W$ can be realized as a sub-representation of $L^2(S^1, W)$}. \proof For any $x\in W$ consider $\alpha_x\in L^2(S^1, W)$ taking $t\in U(1)=\R/\Z$ to $\rho(t)(x)$. Clearly, {\bf \purple $x\mapsto \alpha_x$ defines a homomorphism of representations.} \endproof \theorem Let $W$ be a Hermitian representation of $U(1)$. {\bf \red Then $W$ admits a weight decomposition $W = \widehat{\bigoplus_{i\in \Z} W_i}$.} \proof We realize $W$ as a subrepresentation in $L^2(S^1, W)$, and use the Fourier series to obtain the weight decomposition of $L^2(S^1, W)$. \endproof \newpage {\bf \blue Weight decomposition for $T^n$-action} \exercise Consider the $n$-dimensional torus $T^n$ as a Lie group, $T^n = U(1)^n$. {\bf \purple Prove that any finite-dimensional Hermitian representation of $T^n$ is a direct sum of 1-dimensional representations,} with action of $T^n$ given by $\rho(t_1, ..., t_n)(x)= \exp(2\pi\1\sum_{i=1}^np_it_i) x$, for some $p_1, ..., p_n\in \Z^n$, called {\bf \blue the weights} of the 1-dimensional representation. \definition Let $V$ be a Hermitian space (possibly infinitely-dimensional) equipped with an action of $T^n$, and $V_\alpha\subset V$ weight $\alpha$ representations, $\alpha\in \Z^n$. The direct sum $\bigoplus_{\alpha\in \Z^n} V_\alpha$ is called {\bf \blue the weight decomposition} for $V$ if it is dense in $V$. \theorem Let $W$ be a Hermitian vector space. Then {\bf \red the Fourier series provide the weight decomposition on $L^2(T^n, W)$.} \endproof \theorem Let $W$ be a Hermitian representation of $T^n$. {\bf \red Then $W$ admits a weight decomposition $V = \widehat{\bigoplus_{\alpha\in \Z^n} W_\alpha}$.} \proof We realize $W$ as a subrepresentation in $L^2(T^n, W)$, and use the Fourier series to obtain the weight decomposition of $L^2(T^n, W)$. \endproof \newpage {\bf \blue Weight decomposition for $T^n$-action on differential forms} \remark Let $M$ be a manifold with the $T^n$-action, and \[ \Lambda^*(M)= \hat\bigoplus_{\alpha \in \Z^n} \Lambda^*(M)_{p_1, ..., p_k}\] be the weight decomposition on the differential forms. Then the {\bf \red de Rham differential preserves each term $\Lambda^*(M)_{p_1, ..., p_k}$.} Indeed, {\bf \purple $d$ commutes with the action of the Lie algebra of $T^n$, and $\Lambda^*(M)_{p_1, ..., p_k}$ are its eigenspaces.} \remark The weight decomposition $\alpha= \sum \alpha_{p_1, ..., p_k}$ converges, generally speaking, only in $L^2$, however, if action of $T^n$ is smooth, it converges uniformly in $t\in T^n$ because {\bf \purple the Fourier serie of a smooth function converge uniformly}. \remark Let $\alpha= \sum \alpha_{p_1, ..., p_k}$ be the weight decomposition. {\bf \purple The forms $\alpha_{p_1, ..., p_k}$ are obtained by averaging \[ e^{2\pi\1\sum_{i=1}^np_it_i}\alpha = \Av_{T^n} e^{2\pi\1\sum_{i=1}^n-p_it_i}\alpha \] hence they are smooth.} \newpage {\bf \blue De Rham cohomology and $T^n$-action} \theorem Let $M$ be a smooth manifold, and $T^n$ a torus acting on $M$ by diffeomorphisms. Denote by $\Lambda^*(M)^{T^n}$ the complex of $T^n$-invariant differential forms. {\bf \red Then the natural embedding $\Lambda^*(M)^{T^n}\hookrightarrow \Lambda^*(M)$ induces an isomorphism on de Rham cohomology.} \pstep Let $\alpha\in \Lambda^*(M)$ be a form and $\alpha= \sum \alpha_{p_1, ..., p_n}$ its weight decomposition, with $\alpha_{p_1, ..., p_n}\in \Lambda^*_{p_1, ..., p_n}(M)$ a form of weight $p_1, ... , p_n$. Since $T^n$-action commutes with de Rham differential, these forms are closed when $\alpha$ is closed. {\bf \green Step 2:} Let $r_1, ..., r_n$ be the standard generators of the Lie algebra of $T^n$ rescaled in such a way that $\Lie_{r_k}(\exp(2\pi\1\sum_{i=1}^np_it_i))=\1 p_k$, and $i_{r_k}:\; \Lambda^i(M)\arrow \Lambda^{i-1}(M)$ the convolution operator. Since $\Lie_{r_k}= \{d, i_{r_k}\}$, we have $p_k\alpha_{p_1, ..., p_n}= d (i_{r_k}\alpha_{p_1, ..., p_n})$ whenever $\alpha_{p_1, ..., p_n}$ is closed. Therefore, {\bf \purple all terms in the weight decomposition $\alpha= \sum \alpha_{p_1, ..., p_n}$ are exact except $\alpha_{0,0, ...,0}$.} {\bf \green Step 3:} In the direct sum decomposition of the de Rham complex \[ \Lambda^*(M)= \Lambda^*(M)^{T^n}\oplus \hat\bigoplus_{p_1, ..., p_k\neq (0,0,..., 0)}\Lambda^*_{p_1, ..., p_k}(M) \] the second component has trivial cohomology, because $\Lie_{r_k}$ is invertible on $\bigoplus_{p_k\neq 0}\Lambda^*_{p_1, ..., p_n}(M)$ {\bf \purple (deduce it from $p_k\alpha_{p_1, ..., p_k}= d (i_{r_k}\alpha_{p_1, ..., p_k})$),} and\\ $\Lie_{r_k}(\text{closed form})$ is exact. \endproof \end{document}