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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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 $\dim_H B_i<n$. Using the Claim above,
we obtain that $\mu_n(f(M))=0$.
\endproof

\definition
{\bf \blue Hausdorff dimension of a subset $Z\subset M$ of a manifold}
is supremum of $\dim_h(Z\cup B)$ for all subsets $B\subset M$
equipped with a coordinate system.


\corollary 
Let $f:\; M \arrow N$ be a differentiable map of 
smooth manifolds, $\dim M < \dim N$. Suppose that $M$
is covered by a countable number of open balls
with compact closure. {\bf \red Then $\mu_n (f(M))=0$.}
\endproof

\corollary 
{\bf \blue (a version of Sard's lemma)}
{\bf \red Under these assumptions, $f(M)$ is nowhere dense.}

\proof Indeed, were it dense in an open ball $B$,
one would have $\mu_n(f(M))\geq \mu_n(B)>0$,
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. 




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