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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 4: Hausdorff measure and Whitney's theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf Math in Moscow and HSE \\[2mm] February 25, 2013 } \end{center} \newpage {\bf \blue Sheaves of functions (reminder)} \definition {\bf\blue An open cover} of a topological space $X$ is a family of open sets $\{U_i\}$ such that $\bigcup_iU_i=X$. \definition A {\bf\blue presheaf of functions} on a topological space $M$ is a collection of subrings ${\cal F}(U)\subset C(U)$ in the ring $C(U)$ of all functions on $U$, for each open subset $U\subset M$, such that the restriction of every $\gamma\in{\cal F}(U)$ to an open subset $U_1\subset U$ belongs to ${\cal F}(U_1)$. \definition A presheaf of functions ${\cal F}$ is called {\bf\blue a sheaf of functions} if these subrings satisfy the following condition. Let $\{U_i\}$ be a cover of an open subset $U\subset M$ (possibly infinite) and $f_i\in{\cal F}(U_i)$ a family of functions defined on the open sets of the cover and compatible on the pairwise intersections: $$f_i|_{U_i\cap U_j}=f_j|_{U_i\cap U_j}$$ for every pair of members of the cover. {\bf \purple Then there exists $f\in{\cal F}(U)$ such that $f_i$ is the restriction of $f$ to $U_i$ for all $i$.} \remark {\bf \purple A presheaf of functions} is a collection of subrings of functions on open subsets, compatible with restrictions. {\bf\purple A sheaf of fuctions is a presheaf allowing ``gluing''} a function on a bigger open set if its restrictions to smaller open sets are compatible. \newpage {\bf \blue Ringed spaces (reminder)} A {\bf\blue ringed space} $(M,{\cal F})$ is a topological space equipped with a sheaf of functions. A~{\bf \blue morphism} $(M,{\cal F})\stackrel\Psi\longrightarrow(N,{\cal F}')$ of ringed spaces is a continuous map $M\stackrel\Psi\longrightarrow N$ such that, for every open subset $U\subset N$ and every function $f\in{\cal F}'(U)$, the function $\psi^* f:=f\circ\Psi$ belongs to the ring ${\cal F}\big(\Psi^{-1}(U)\big)$. An {\bf\blue isomorphism} of ringed spaces is a homeomorphism $\Psi$ such that $\Psi$ and $\Psi^{-1}$ are morphisms of ringed spaces. \definition Let $(M,{\cal F})$ be a topological manifold equipped with a sheaf of functions. It is said to be a {\bf\blue smooth manifold of class} $C^\infty$ or $C^i$ if every point in $(M,{\cal F})$ has an open neighborhood isomorphic to the ringed space $({\Bbb B}^n,{\cal F}')$, where ${\Bbb B}^n\subset \R^n$ is an open ball and ${\cal F}'$ is a ring of functions on an open ball ${\Bbb B}^n$ of this class. \definition {\bf\blue Diffeomorphism} of smooth manifolds is a homeomorphism $\phi$ which induces an isomorphims of ringed spaces, that is, $\phi$ and $\phi^{-1}$ map (locally defined) smooth functions to smooth functions. {\bf \red Assume from now on that all manifolds are Hausdorff and of class $C^\infty$}. \newpage {\bf \blue Partition of unity: a formal definition (reminder)} \definition Let $M$ be a smooth manifold and let $\{U_\alpha\}$ a locally finite cover of $M$. A {\bf \blue partition of unity} subordinate to the cover $\{U_\alpha\}$ is a family of smooth functions $f_i:M\to[0,1]$ with compact support indexed by the same indices as the $U_i$'s and satisfying the following conditions.\\ (a) Every function $f_i$ vanishes outside $U_i$\\ (b) $\sum_if_i=1$ The argument of previous page proves the following theorem. \theorem Let $\{U_\alpha\}$ be a countable, locally finite cover of a manifold $M$, with all $U_\alpha$ diffeomorphic to $\R^n$. {\bf \red Then there exists a partition of unity subordinate to $\{U_\alpha\}$.} \endproof \newpage {\bf \blue Embedding to $\R^\infty$ (reminder)} \definition Define $\R^I_f$ as a direct sum of several copies of $\R$ indexed by a set $I$, that is, the set of points in a product where only finitely meny of coordinates can be non-zero. {\bf \blue The set $\R^I_f$ has metric \[ d((x_1, ..., x_n, ...), (y_1, ..., y_n, ...)):= \sqrt{|x_1-y_1|^2+|x_2-y_2|^2+...+|x_n-y_n|+...}. \]} {\bf \purple It is well-defined, because only finitely many of $x_i, y_i$ are non-zero.} \theorem Let $M$ be a compact smooth manifold, $\{V_i, \phi_i:\; V_i \arrow \R^n, i\in I\}$ be a locally finite atlas, and $\mu_i:\; M \arrow [0,1]$ a subordinate partition of unity. Define $\nu_i:=\alpha(\mu_i)$ and $\Phi_i$ as above, and let \[ \Psi:=\prod_I:\;\Phi_i:\; M \arrow \underbrace{S^n\times S^n \times ...\times S^n}_{\text{$I$ times}} \subset (\R^{n+1})^I \] be the corresponding product map. Then {\bf \red $\Psi$ is a homeomorphism to its image.} \newpage {\bf \blue Borel measure} \definition Let $C$ be a cube in $\R^n$ with edges parallel to coordinate axes of length $r$. Such a cube is called {\bf \blue normal}. Its {\bf \blue volume} is $r^n$. \definition Let $S\subset \R^n$ be a closed subset. The {\bf \blue volume}, or {\bf \blue Borel measure} of $S$ is an infimum of $\sum_i \Vol(S_i)$ for all (possibly, infinite) covers of $S$ by normal cubes. \claim A subset $Z\subset\Bbb R^n$ has {\bf \blue measure zero} if for every $\varepsilon>0$ there exists a countable cover of $Z$ by cubes $C_i$ such that $\sum_i\Vol C_i<\varepsilon$. \remark Borel measure is a weaker form of Lebesgue measure, defined on closed subsets of $\R^n$, and equal to Lebesgue measure on those subsets. \newpage {\bf \blue Borel measure: axiomatic definition} \theorem {\bf \blue (Properties of the volume)} \\ Let $\mu(S)$ denote the measure of $S$. Then\\ \phantom{hu} (a) $\mu(\bigcup S_i)\leq \sum_i \mu(S_i)$. \\ \phantom{hu} (b) Measure is {\bf \blue monotonous:} for any $A\subset B$, $\mu(A)\leq \mu(B)$.\\ \phantom{hu} (c) Let $A=\bigcup A_i$ be an intersection of closed sets $A_0\supset A_1 \supset ...$. \\Then $\mu(A)=\lim \mu(A_i)$. \\ \phantom{hu} (d) Measure is {\bf\blue additive}. Let $S=\bigcup_i S_i$ and $\mu(S_i\cap S_j)=0$ for all $i\neq j$. Then $\mu(S)=\sum_i \mu(S_i)$.\\ \phantom{hu} (e) Measure of a normal cube is $l^n$, where $l$ is a length of its side. Moreover, {\bf \red for any closed set, its measure is determined uniquely by these properties.} \exercise {\bf \purple Prove this theorem, using the following lemma.} \lemma Let $S\subset \R^n$ be a closed subset. Then $S=\bigcap S_i$, where $S_0\supset S_1\supset S_2 \supset ...$, and each $S_i$ is a countable union of normal cubes, intersecting only in their faces. \newpage {\bf \blue Hausdorff measure and Hausdorff dimension} \definition Let $M$ be a metric space. The {\bf \blue diameter} $\diam M\in[0,\infty]$ is the number $\sup\limits_{x,y\in M}d(x,y)$. \definition In a metric space, {\bf\blue a ball} $B_\epsilon(x)$ of radius $\varepsilon$ centered at $x$ is defined as the set of all points $y$ satisfying $d(x,y)<\varepsilon$. \definition Let $\{S_i\}$ be a cover of a metric space $M$ by balls of radius $r$ with $r<\varepsilon$. Define $\mu_{d,\varepsilon}\in[0,\infty]$ as $\mu_{d,\varepsilon}(M):=\inf_{\{S_i\}}\sum_i(\diam S_i)^d,$ where the infimum is taken over all such covers. The limit $\mu_dM:=\sup\lim_{\varepsilon\to0}\mu_{d,\varepsilon}(M)$ is called {\bf\blue $d$-dimensional Hausdorff measure} of $M$. \example Let $M=\Bbb R^n$ with a metric $d_\infty$ given by the norm $|(x_1,\dots,x_n)|:=\max|x_i|.$ {\bf \purple The balls in this metric are cubical}, and the (usual) volume of such a ball $B$ is equal to $(\diam B)^n$. This gives $\mu_n (S)= \Vol S$ for each cube with sides parallel to coordinate planes. \corollary For $M=\Bbb R^n$ with the metric described above, {\bf \purple Hausdorff measure is equal to the Borel measure.} \newpage {\bf \blue Lipschitz maps} \definition A map $f:M\to N$ of metric spaces is called {\bf \blue Lipschitz with constant $C$} if $d(x,y)\ge C\cdot d\big(f(x),f(y)\big)$ for all $x,y\in M$. A map is called {\bf \blue bi-Lipschitz} if it is bijective and the inverse map is also Lipschitz. \example {\bf \purple A differentiable map $f:\; \R^n \arrow \R^m$ is Lipschitz on each compact set $B$.} Indeed, $d(x,y) \geq C d(f(x), f(y))$, where $C=\sup_B |Df|$. \example Let $\nu_1, \nu_2$ be norms on a vector space $V$, and $d_1, d_2$ the corresponding metrics. {\bf \purple The identity map $(V, d_1) \arrow (V, d_2)$ is $C$-Lipschitz if and only if the unit ball $B_1(x, d_1)$ belongs to $B_C(x, d_2)$.} \claim Let $f:M\to N$ be a $C$-Lipschitz map. {\bf \red Then the corresponding Hausdorff measures are related as $\mu_n(S) \geq C^n \mu_n(f(S))$.} \proof Let $\{S_i=B_{\epsilon_i}(x_i)\}$ be a cover of $S$. Then $\{B_{C\epsilon_i}(f(x_i))\}$ is a cover of $f(S)$. \endproof \corollary \\ {\bf \purple Let $f:M\to N$ be a $C$-Lipschitz map. Then $\dim_H(M)\geq \dim_H(f(M))$.} \endproof \newpage {\bf \blue Equivalent norms and Hausdorff measure} \definition Two norms on a vector space $V$ are called {\bf \blue equivalent} if the identity map $(V, d_1) \arrow (V, d_2)$ is bi-Lipschitz. \example Since a unit cube in $\R^n$ contains a ball of radius 1, and is contained in a ball of radius $\sqrt n$, one has $|x|_{L^2}\geq |x|_{L^\infty} \geq \sqrt{n^{-1}} |x|_{L^2}$, where $L^2$ is the usual norm, and $L^\infty$ the norm $|(x_1,\dots,x_n)|:=\max|x_i|.$ Therefore, {\bf \purple the norms $L^2$ and $L^\infty$ are equivalent}. \corollary Let $\mu^{L^2}_n$ denote the Hausdorff measure associated with the Euclidean metric on $\R^n$, and $\mu$ the usual (Borel) measure. {\bf \red Then $\mu^{L^2}_n(S)\geq \mu(S)\geq \sqrt{n^{-n}}\mu^{L^2}_n(S)$.} \proof See the Claim above. \endproof \newpage {\bf \blue Hausdorff dimension} \theorem Let $M$ be a metric space. Consider $\mu_d(M)$ as a function of $d$. {\bf \red Then there exists a number $d_0\in [0,\infty]$ such that $\mu_d(M)=\infty$ for $d< d_0$, and $\mu_d(M)=0$ for $d>d_0$.} \proof Whenever $d'>d$, one has \begin{multline*} \mu_{d',\varepsilon}(M)=\inf\limits_{\{S_i\}}\sum\limits_i (\diam S_i)^{d'}=\inf\limits_{\{S_i\}}\sum\limits_i (\diam S_i)^{d}(\diam S_i)^{d'-d} <\\ < \varepsilon^{d'-d}\inf\limits_{\{S_i\}}\sum\limits_i (\diam S_i)^d=\varepsilon^{d'-d}\mu_{d,\varepsilon}(M) \end{multline*} Passing to the limit $\epsilon\rightarrow 0$, we obtain that $\mu_{d'}(M)\leq 0 \mu_d(M)$. Therefore, {\bf \purple $\mu_{d'}(M)=0$ whenever $\mu_{d}(M)$ is finite, and $\mu_{d}(M)=\infty$ whenever $\mu_{d'}(M)>0$.} \endproof \definition {\bf\blue Hausdorff dimension} $\dim_H(M)$ of a metric space is the number $\sup_{d\geq 0} \{ \mu_d(M)=\infty\}$. \exercise {\bf \purple Prove that set $M$ has Hausdorff dimension 0 iff it is finite.} \newpage {\bf \blue Lipschitz maps and Hausdorff dimension} \claim {\bf \purple Let $f$ be a Lipschitz map. Then $\dim_H(f(M))\leq \dim_H(M)$. } \endproof \corollary {\bf \red A cube $C$ in $\R^n$ has Hausdorff dimension $n$. } \proof Indeed, $C$ is bi-Lipschitz equivalent to the cube in $L^\infty$-metric, but the Hausdorff measure associated with the $L^\infty$-metric is equal to the usual volume. \endproof \claim Let $C=\bigcup C_i$ be a union of a countably many sets with $\mu_d C_i=0$. {\bf \red Then $\mu_d C=0$.} \proof Take a cover $S_j(i)$ of $C_i$ with $\sum_j(\diam S_j(i))^d< \delta^{i+1}$. Then $\{S_j(i)\}$ is a cover of $C$ with $\sum_{j,i}(\diam S_j(i))^d<\delta$. \endproof \corollary {\bf \purple $\R^n$ with the usual metric has Hausdorff dimension $n$.} \proof Take a cover of $\R^n$ by coutably many unit cubes $C_i$. For $d>n$, $\mu_d(C_i)=0$, hence $\mu_d(\R^n)=0$. Since $\mu_n(C_i)>0$, $\mu_n(\R^n)$ is also positive. \endproof \newpage {\bf \blue Hausdorff dimension of a manifold} \theorem Let $f:\; M \arrow \R^n$ be a smooth map from a manifold, $\dim M < n$. {\bf \red Suppose that $M$ admits a countable cover by open balls with compact closure. Then $\mu_n f(M)=0$. } \proof Let $B\subset \R^m$ be a closed ball, and $\phi:\; B \arrow \R^n$ a differentiable map. Then $\phi$ is Lipschitz, with the Lipschitz constant $C\leq \sup |D\phi|$. The set $M$ is covered by closed balls $B_i$, and $\mu_n (f(B_i))=0$, because $f\restrict B_i$ is Lipschitz, and $\dim_H B_i0$, giving $\dim_H(f(M))\geq n$, in contradiction to the corollary above. \endproof \newpage {\bf \blue Whitney's theorem (with a bound on dimension): strategy of the proof } \theorem Let $M$ be a smooth $n$-manifold. {\bf \red Then $M$ admits a closed embedding to $\R^{2n+2}$.} {\bf \green Strategy of the proof:}\\ \phantom{huy} 1. $M$ is embedded to $\R^\infty$.\\ \phantom{huy} 2. We find a linear projection $\R^\infty\stackrel \pi \arrow \R^{2n+2}$ such that $\pi\restrict M$ is a closed embedding of manifolds. \lemma Let $M\subset \R^I$ be a subset, and $\pi:\; \R^I \arrow \R^J$ a linear projection. Consider the set $W$ of all vectors $\R(x-y)$, where $x,y \in M$ are distinct points. {\bf\purple Then $\pi\restrict M$ is injective if and only if $\ker \pi\cap W=0$.} \proof $\pi\restrict M$ is not injective if and only if $\pi(x)=\pi(y)$, which is equivalent to $\pi(x-y)=0$. \endproof \newpage {\bf \blue Whitney's theorem: injectivity of projections } \remark Let $M\subset \R^I$ be a submanifold, and $W\subset \R^I$ the set of all vectors $\R(x-y)$, where $x,y \in M$ are distinct points. {\bf \red Then $W$ is an image of a $2m+1$-dimensional manifold}, hence (by Sard's Lemma) {\bf \purple for any projection of $\R^I$ to a $(2m+2)$-dimensional space, image of $W$ has measure 0.} \corollary Let $M\subset \R^I$ be an $m$-dimensional submanifold, and $S\subset \R^I$ a maximal linear subspace not intersecting $W$. {\bf \purple Then the projection of $W$ to $\R^I/S$ is surjective.} \proof Suppose it's not surjective: $v\notin S$. Then $S \oplus \R v$ satisfies assumptions of lemma, hence $M \arrow \R^I/(S+ \R v)$ is also injective. \endproof \theorem Let $M$ be a smooth $n$-manifold, $M\hookrightarrow \R^I$ an embedding constructed earlier. {\bf \red Then there exists a projection $\pi:\; \R^I\arrow \R^{2n+2}$ which is injective on $M$.} \proof Let $S$ be the maximal linear subspace such that the restriction of $\pi:\; \R^I\arrow \R^I/S$ to $M$ is injective. Then the $2m+1$-dimensional manifold $W$ is mapped surjectively to $\R^I/S$, hence $\dim \R^i/S\leq 2m+1$ by Sard's lemma. \endproof \newpage {\bf \blue Tangent space to an embedded manifold } \definition Let $M\hookrightarrow\Bbb R^n$ be a smooth $m$-submanifold. The {\bf\blue tangent plane} at $p\in M$ is the plane in $\Bbb R^n$ tangent to $M$ (i.e, the plane lying in the image of the differential given in local coordinates). A {\bf\blue tangent vector} is an arbitrary vector in this plane with the origin at $p$. The space of all tangent vectors at $p$ is denoted by $T_pM$. Given a metric on $\Bbb R^n$, we can define the space of {\bf\blue unit tangent vectors} $\Bbb S^{m-1}M$ as the set of all pairs $(p,v)$, where $p\in M$, $v\in T_pM$, and $|v|=1$. \remark $\Bbb S^{m-1}M$ is a smooth manifold, projected to $M$ with fibers isomorphic to $m-1$-spheres, hence {\bf \purple $\Bbb S^{m-1}M$ is $(2m-1)$-dimensional.} \lemma Let $M\subset \R^I$ be a subset, and $\pi:\; \R^I \arrow \R^J$ a linear projection. Consider the set $W'$ of all vectors $\R t$, where $t\in T_xM$ {\bf\purple Then the differential $D\pi\restrict M$ is injective if and only if $\ker \pi\cap W'=0$.} \endproof Now the above argument is repeated: we take a maximal space $S\supset \R^I$ such that the restriction of $\pi:\; \R^I\arrow \R^I/S$ to $M$ is injective and has injective differential, and the projection of $W\cup W'$ to $\R^I/S$ has to be surjective. However, $W'$ is an image of an $2m$-dimensional manifold ${\Bbb S}^{m-1}M\times \R$, hence {\bf \red the projection of $W\cup W'$ to $\R^I/S$ can be surjective only if $\dim \R^I/S\leq 2m +2$.} This proves Whitney's theorem. \end{document}