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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:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Geometry of manifolds \\[15mm] \small Lecture 12: Poincar\'e lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf Math in Moscow and HSE \\[2mm] May 13, 2013 } \end{center} \newpage {\bf \blue De Rham algebra (reminder)} \definition Let $M$ be a smooth manifold. {\bf \blue A bundle of differential $i$-forms on $M$} is the bundle $\Lambda^i T^* M$ of antisymmetric $i$-forms on $TM$. It is denoted $\Lambda^i M$. \remark $\Lambda^0 M=C^\infty M$. \definition Let $\alpha\in (V^*)^{\otimes i}$ and $\alpha\in (V^*)^{\otimes j}$ be polylinear forms on $V$. Define the {\bf\blue tensor multiplication} $\alpha \otimes \beta$ as\\ \phantom{xxxxxxx} $\alpha \otimes \beta(x_1, ..., x_{i+j}):= \alpha(x_1, ..., x_j) \beta(x_{i+1}, ..., x_{i+j}).$ \definition Let $\bigotimes_k T^*M \stackrel \Pi \arrow \Lambda^k M$ be the antisymmetrization map, \[ \Pi(\alpha)(x_1, ..., x_n):= \frac 1{n!} \sum_{\sigma\in \Sym_n} (-1)^\sigma \alpha(x_{\sigma_1}, x_{\sigma_2}, ..., x_{\sigma_n}). \] Define {\bf \blue the exterior multiplication $\wedge:\; \Lambda^i M\times \Lambda^j M \arrow \Lambda^{i+j} M$} as $\alpha \wedge \beta := \Pi (\alpha \otimes \beta)$, where $\alpha \otimes \beta$ is a section $\Lambda^i M\otimes \Lambda^j M \subset \bigotimes_{i+j} T^*M$ obtained as their tensor multiplication. \remark The fiber of the bundle $\Lambda^* M$ at $x\in M$ {\bf \purple is identified with the Grassmann algebra $\Lambda^* T^*_x M$.} This identification is compatible with the Grassmann product. \newpage {\bf \blue De Rham differential (reminder)} \definition {\bf\blue De Rham differential} $d:\; \Lambda^*M \arrow \Lambda^{*+1}M$ is an $\R$-linear map satisfying the following conditions.\\ \phantom{xxx} * For each $f \in \Lambda^0M=C^\infty M$, $d(f)\in \Lambda^1 M$ is equal to the image of the K\"ahler differential $df\in \Omega^1 M$ in $\Lambda^1 M = \Omega^1 M/K$.\\ \phantom{xxx} * {\bf \blue (Leibnitz rule)} $d(a\wedge b) = da \wedge b + (-1)^j a\wedge db$ for any $a\in \Lambda^i M, b \in \Lambda^j M$.\\ \phantom{xxx} * $d^2=0$. \theorem\\ {\bf \red De Rham differential is uniquely determined by these axioms.} \remark The proof of uniqueness is based on the following lemmas. \lemma Let $A=\bigoplus A^i$ be a graded algebra, $B\subset A$ a set of multiplicative generators, and $D_1, D_2:\; A \arrow A$ two odd derivations which are equal on $B$. {\bf \purple Then $D_1=D_2$.} \endproof \lemma {\bf \purple $\Lambda^* M$ is generated by $C^\infty M$ and $d(C^\infty M)$.} \remark Let $t_1, ..., t_n$ be coordinate functions on $\R^n$, $\alpha_i$ coordinate monomials, and $\alpha:= \sum f_i \alpha_i$. Define $d(\alpha):= \sum_i \sum_j\frac{df_i}{dt_j}dt_j \wedge \alpha_i$. {\bf \red Then $d$ satisfies axioms of de Rham differential.} This proves its existence. \newpage {\bf \blue Lie derivative (reminder)} \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. \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.} \lemma {\bf \purple $\{d, \{d, E\}\}=0$ for each $E\in \End(\Lambda^* M)$.} \proof By the super Jacobi identity, $\{d, \{d, E\}\}=-\{d, \{d, E\}\}+\{\{d, d,\} E\}\}$, however, $\{d, d\} =2d^2 =0$. \endproof \theorem {\bf \blue (Cartan's formula)} Let $i_v$ be a convolution with a vector field. {\bf \red Then $\{d, i_v\}$ is a Lie derivative along $v$.} \proof $\{d, \{d, i_v\}\}=0$ by the lemma above. A supercommutator of two derivations is a derivation. Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$ \endproof \newpage {\bf \blue Pullback of a differential form (reminder)} \definition Let $M \stackrel \phi\arrow N$ be a morphism of smooth manifolds, and $\alpha \in \Lambda^i N$ be a differential form. Consider an $i$-form $\phi^* \alpha$ taking value \[ \alpha\restrict{\phi(m)}(D_\phi(x_1), ... D_\phi(x_i))\] on $x_1, ..., x_i\in T_m M$. It is called {\bf\blue the pullback of $\alpha$}. If $M\stackrel \phi \arrow N$ is a closed embedding, the form $\phi^* \alpha$ is called {\bf \blue the restriction} of $\alpha$ to $M \hookrightarrow N$. \lemma {\bf \blue (*)} Let $\Psi_1, \Psi_2:\; \Lambda^* N \arrow \Lambda^* M$ be two maps which satisfy graded Leibnitz identity, commute with de Rham differential, and satisfy $\Psi_1\restrict{C^\infty M}= \Psi_2\restrict{C^\infty M}$. {\bf \red Then $\Psi_1 = \Psi_2$.} \proof The algebra $\Lambda^* M$ is generated multiplicatively by $C^\infty M$ and $d(C^\infty M)$; restrictions of $\Psi_i$ to these two spaces are equal. \endproof \claim {\bf \red Pullback commutes with the de Rham differential.} {\bf\green Proof:} Follows from Lemma (*). \endproof \newpage {\bf \blue Flow of diffeomorphisms (reminder)} \definition Let $f:\; M \times [a,b]\arrow M$ be a smooth map such that for all $t\in [a,b]$ the restriction $f_t:= f\restrict{M\times\{t\}}:\; M \arrow M$ is a diffeomorphism. Then $f$ is called {\bf\blue a flow of diffeomorphisms}. \claim Let $V_t$ be a flow of diffeomorphisms, $f\in C^\infty M$, and $V^*_t(f)(x):= f(V_t(x))$. Consider the map $\frac d{dt}V_t\restrict{t=c}:\; C^\infty M \arrow C^\infty M$, with $\frac d{dt}V_t\restrict{t=c}(f)= (V_c^{-1})^*\frac{dV_t}{dt}\restrict{t=c}f$. {\bf \red Then $(V_c^{-1})^* \frac d{dt}V_t\restrict{t=c}$ is a derivation} (that is, a vector field). {\bf \green Proof:} \[ \frac d{dt}V_t\restrict{t=c}(fg) = V_c^*f \frac d{dt}V_t\restrict{t=c} (g) + V_c^*g \frac d{dt}V_t\restrict{t=c} (f). \] \endproof \definition The vector field $(V_c^{-1})^*\frac d{dt}V_t\restrict{t=c}$ is called {\bf\blue a vector field tangent to a flow of diffeomorphisms $V_t$ at $t=c$.} \newpage {\bf \blue Lie derivative and a flow of diffeomorphisms (reminder)} \definition Let $v$ be a vector field on $M$, and $V:\;M\times[a,b]\arrow M$ a flow of diffeomorphisms which satisfies $(V_c^{-1})^*\frac d{dt}V_t\restrict{t=c}=v$ for each $c$, and $V_0=\Id$. Then $V_t$ is called {\bf\blue an exponent of $v$}. \claim Exponent of a vector field is unique; it exists when $M$ is compact. This statement is called {\bf \blue ``Picard-Lindel\"of theorem''} or {\bf \blue ``uniqueness and existence of solutions of ordinary differential equations''}. \proposition Let $v$ be a vector field, and $V_t$ its exponent. For any $\alpha \in \Lambda^*M$, consider $V_t^*\alpha$ as a $\Lambda^*M$-valued function of $t$. {\bf \red Then $\Lie_v\alpha = \frac{d}{dt}(V_t^*\alpha)\restrict{t=0}$.} \proof By definition, $\Lie_v=\frac{d}{dt}V_t$ on functions. $\Lie_v$ commutes with de Rham differential, because $\Lie_v=i_v d+di_v$. The map $\frac{d}{dt}V_t$ commutes with de Rham differential, because it is a derivative of a pullback. Now {\bf \purple Lemma (*) is applied to show that $\Lie_v\alpha = \frac{d}{dt}(V_t^*\alpha)$.} \endproof \newpage {\bf \blue Homotopy operators} \definition {\bf\blue A complex} is a sequence of vector spaces and homomorphisms $... \stackrel d\arrow C_{i-1} \stackrel d\arrow C_i\stackrel d\arrow C_{i+1} \stackrel d\arrow...$\\ satisfying $d^2=0$. {\bf \blue Homomorphism} $(C_*, d)\arrow (C_*', d)$ of complexes is a sequence of homomorphism $C_i \arrow C'_i$ commuting with the differentials. \definition An element $c\in C_i$ is called {\bf\blue closed} if $c\in \ker d$ and {\bf \blue exact} if $c\in \im d$. {\bf \blue Cohomology} of a complex is a quotient $\frac{\ker d}{\im d}$. \remark A homomorphism of complexes induces a natural homomorphism of cohomology groups. \definition Let $(C_*, d)$, $(C_*', d)$ be a complex. {\bf \blue Homotopy} is a sequence of maps $h:\; C_* \arrow C'_{*-1}$. Two homomorphisms $f, g:\; (C_*, d)\arrow (C_*', d)$ are called {\bf \blue homotopy equivalent} if $f-g =\{h,d\}$ for some homotopy operator $h$. \claim Let $f, f':\; (C_*, d)\arrow (C_*', d)$ be homotopy equivalent maps of complexes. {\bf \red Then $f$ and $f'$ induce the same maps on cohomology.} {\bf \green Proof. Step 1:} Let $g:= f-f'$. It would suffice to prove that $g$ induces 0 on cohomology. \newpage {\bf \blue Lie derivative and homotopy} \claim Let $f, f':\; (C_*, d)\arrow (C_*', d)$ be homotopy equivalent maps of complexes. {\bf \red Then $f$ and $f'$ induce the same maps on cohomology.} {\bf \green Proof. Step 1:} Let $g:= f-f'$. It would suffice to prove that $g$ induces 0 on cohomology. {\bf \green Step 2:} Let $c\in C_i$ be a closed element. {\bf \purple Then $g(c)= dh(c)+ hd(c)=dh(c)$ exact.} \endproof \definition Let $d$ be de Rham differential. A form in $\ker d$ is called {\bf\blue closed}, a form in $\im d$ is called {\bf\blue exact}. Since $d^2=0$, any exact form is closed. {\bf\blue The group of $i$-th de Rham cohomology of $M$}, denoted $H^i(M)$, is a quotient of a space of closed $i$-forms by the exact: $H^*(M)=\frac{\ker d}{\im d}$. \remark Let $v$ be a vector field, and $\Lie_v:\; \Lambda^* M\arrow \Lambda^* M$ be the corresponding Lie derivative. Then {\bf \red $\Lie_v$ commutes with the de Rham differential, and acts trivially on the de Rham cohomology.} \proof $\Lie_v=i_v d+di_v$ maps closed forms to exact. \endproof \newpage {\bf \blue Poincar\'e lemma} \definition An open subset $U\subset \R^n$ is called {\bf\blue starlike} if for any $x\in U$ the interval $[0,x]$ belongs to $U$. \theorem {\bf \blue (Poicar\'e lemma)} Let $U\subset \R^n$ be a starlike subset. {\bf \red Then $H^i(U)=0$ for $i>0$.} \remark {\bf \purple The proof would follow if we construct a vector field $\vec{r}$ such that $\Lie_{\vec r}$ is invertible on $\Lambda^*(M)$:} $\Lie_{\vec r}R= \Id$. Indeed, for any closed form $\alpha$ we would have $\alpha = \Lie_{\vec r} R\alpha= di_{\vec r} R\alpha+ i_{\vec r} Rd\alpha=di_{\vec r} R\alpha,$ hence any closed form is exact. Then Poincar\'e lemma is implied by the following statement. \proposition Let $U\subset \R^n$ be a starlike subset, $t_1, ..., t_n$ coordinate functions, and $\vec r:= \sum t_i \frac d {dt_i}$ the radial vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on $\Lambda^i(U)$ for $i>0$.} \newpage {\bf \blue Radial vector field on starlike sets} \proposition Let $U\subset \R^n$ be a starlike subset, $t_1, ..., t_n$ coordinate functions, and $\vec r:= \sum t_i \frac d {dt_i}$ the radial vector field. {\bf \red Then $\Lie_{\vec r}$ is invertible on $\Lambda^i(U)$ for $i>0$.} {\bf \green Proof. Step 1:} Let $t$ be the coordinate function on a real line, $f(t)\in C^\infty \R$ a smooth function, and $v:= t \frac d {dt}$ a vector field. Define $R(f)(t):=\int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda$. Then this integral converges whenever $f(0)=0$, and satisfies $\Lie_v R(f)=f$. Indeed, \[ \int^1_0 \frac{f(\lambda t)}{\lambda} d\lambda= \int^t_0 \frac{f(\lambda t)}{t\lambda} d(t\lambda)= \int^t_0 \frac{f(z)}{z} d(z),\] hence $\Lie_v R(f)= t \frac{f(t)}{t}= f(t)$. {\bf \green Step 2:} Consider a function $f\in C^\infty \R^n$ satisfying $f(0)=0$, and $x=(x_1, ..., x_n)\in \R^n$. {\bf \purple Then \[ R(f)(x):=\int_0^1 \frac{f(\lambda x)}{\lambda} d\lambda \] converges, and satisfies $\Lie_{\vec{r}} R(f)=f$.} \newpage {\bf \blue Radial vector field on starlike sets (cont.)} {\bf \green Step 3:} Consider a differential form $\alpha\in\Lambda^i$, and let $h_\lambda x\arrow \lambda x$ be the homothety with coefficient $\lambda\in [0,1]$. Define \[ R(\alpha):= \int^1_0 \lambda^{-1} h_\lambda^*(\alpha) d\lambda. \] Since $h_\lambda^*(\alpha)=0$ for $\lambda=0$, this integral converges. {\bf \red It remains to prove that $\Lie_{\vec r}R=\Id$.} {\bf \green Step 4:} Let $\alpha$ be a coordinate monomial, $\alpha=dt_{i_1} \wedge dt_{i_2} \wedge ... \wedge dt_{i_k}$. Clearly, $\Lie_{\vec r} (T^{-1}\alpha)=0$, where $T=t_{i_1}t_{i_2}... t_{i_k}$. {\bf \purple Since $h_\lambda^* (f\alpha)= h_\lambda^*(Tf) T^{-1}\alpha$, we have $R (f\alpha)= R(Tf) T^{-1}\alpha$ for any function $f\in C^\infty M$.} This gives \[ \Lie_{\vec r} R(f\alpha) = \Lie_{\vec r}R(Tf) T^{-1}\alpha= TfT^{-1}\alpha = f\alpha. \] \endproof \end{document}