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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Geometry of manifolds \\[15mm]
\small lecture 3: partition of unity}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf Math in Moscow and HSE
\\[2mm]  February 18, 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 Charts and coordinates (reminder)}

\definition
{\bf \blue Coordinate system} on a manifold $M$
is an open subset $V\subset M$ equipped with 
an isomorphism of ringed spaces
$(V, C^\infty V)\cong({\Bbb B}^n, C^\infty{\Bbb B}^n)$ 
per definition of a manifold.

\definition
{\bf \blue A chart} on a smooth manifold 
$(M, C^\infty M)$ is an open subset $U\subset M$
together with an embedding $\psi:\; U \arrow \R^n$
given by smooth functions 
$\phi_1, ..., \phi_n\in C^\infty M$
inducing a diffeomorphism on any open 
subset $V\subset U$ equipped with a coordinate
system $(V, C^\infty V)\cong ({\Bbb B}^n, C^\infty{\Bbb B}^n)$.

\definition
{\bf \blue Transition map}
between two charts $\psi_1:\; U_1 \arrow \R^n$
and $\psi_2:\; U_2 \arrow \R^n$ is a map 
$\Psi_{ij}:\;\psi_1(U_1\cap U_2) \arrow \psi_2(U_1\cap U_2)$
defined as $\Psi_{ij}(x)= \psi_2(\psi_1^{-1}(x))$.

\claim
{\bf \red Transition maps are smooth.}

\proof In local coordinates all functions 
$\phi_1, ..., \phi_n$ used in the definition of the
transition map are smooth. \endproof

\newpage

{\bf \blue Atlases on manifolds (reminder)}

\definition
{\bf \blue An atlas} on a smooth manifold
is a collection of charts $\{U_i, \psi_i:\; U_i \arrow
\R^n\}$ satisfying $\bigcup U_i =M$ together with their
transition maps.

\remark In such a situation, the {\bf \purple charts $U_i$ are
usually identified with their images $\psi(U_i)\subset \R^n$.}

\remark
{\bf \purple
The sheaf $C^\infty M$ can be reconstructed from an atlas}
as follows: a function $f$ on $U\subset M$ is smooth
if and only if its restrictions to $U\cap U_i$ are
smooth on all charts.

\newpage

{\bf \blue Locally finite covers (reminder)}

\definition
An open cover $\{U_\alpha\}$ of a topological space $M$
is called {\bf\blue locally finite} if every point in $M$ has a
neighborhood intersecting only a finite number of $U_\alpha$.

\claim
Let $\{U_\alpha\}$ be a locally finite atlas on a manifold $M$. 
{\bf \purple Then there exists 
a refinement $\{V_\beta\}$ of $\{U_\alpha\}$ such that
a closure of each $V_\beta$ is compact in $M$.}

\theorem
Let $\{U_\alpha\}$ be a countable locally finite
cover of a Hausdorff topological space,
such that a closure of each $U_\alpha$ is compact.
{\bf \red Then there exists another cover
$\{V_\alpha\}$ indexed by the same set,
such that $V_\alpha\Subset U_\alpha$.}

\remark
If all $U_\alpha$ are diffeomorphic to $\R^n$,
all $V_\alpha$ can be chosen diffeomorphic to an open ball.
Indeed, any compact set is contained in an open ball.


\corollary
Let $M$ be a manifold admitting a locally finite countable
cover $\{V_\alpha\}$, with $\phi_\alpha:\; V_\alpha
\arrow \R^n$ diffeomorphisms. {\bf \purple Then there exists 
another atlas $\{U_\alpha, \phi_\alpha':\; U_\alpha
\arrow \R^n\}$, such that $\phi'_\alpha({\Bbb B})$
is also a cover of $M$, and ${\Bbb B}\subset \R^n$ a
unit ball.}
\endproof


\newpage

{\bf \blue Construction of a partition of unity}


\claim
{\bf \red There exists a smooth function $\nu:\; \R^n \arrow [0,1]$ which
vanishes outside of a unit ball ${\Bbb B}\subset \R^n$ and is positive
on ${\Bbb B}$.}

{\bf \green Proof. Step 1:} There exists a {\bf \purple smooth function 
$a:\; \R \arrow \R$ which is positive on $\R^{>0}$ and 0 on 
$\R^{\leq 0}$.} Take $a=e^{-x^{-2}}$ on $\R^{>0}$ and $a=0$
on $\R^{\leq 0}$.

{\bf \green Step 2:} There exists a {\bf \purple smooth function 
$b:\; \R \arrow \R$ vanishing outside of $[0,1]$
and positive on the open interval $]0,1[$.}
Take $b(x)=a(x) a(1-x)$.

{\bf \green Step 3:} There exists a {\bf \purple smooth function 
$c:\; \R \arrow \R$ equal to 0 on $]1,\infty[$
and equal to 1 on $]-\infty,0[$.}
Take $c(x)= 1- \lambda^{-1}\int_{-\infty}^x b(x) dx$,
where $\lambda:=\int_{-\infty}^\infty b(x) dx$.

{\bf \green Step 4:} Now, let 
$\nu(z):= c(|z|^2)$. This function is smooth,
vanishes on $|z|\geq1$, and positive on $|z|<1$.
\endproof


\remark In assumptions of Corollary,
let $\nu_\alpha(z):= \nu(\phi_\alpha')$, and
$\mu_i:=\frac{\nu_i}{\sum_\alpha\nu_\alpha}$.
Then $\mu_\alpha:\; M \arrow [0,1]$ are smooth
functions with support in $U_\alpha$ satisfying $\sum_\alpha \mu_\alpha=1$.
Such a set of functions is called {\bf \blue a partition
  of unity}.


\newpage

{\bf \blue Partition of unity: a formal definition}

\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 Whitney's theorem for compact manifolds}

\theorem
Let $M$ be a compact smooth manifold. {\bf \red Then $M$ admits a closed
smooth embedding to $\R^N$.}

{\bf\green  Proof. Step 1:}
Choose a finite atlas 
$\{V_i, \phi_i:\; V_i \arrow \R^n, i=1, 2, ..., m\}$,
and subordinate partution of unity $\mu_i:\; M \arrow
[0,1]$. Let $\alpha:\; [0,1]\arrow [0,1]$
be a smooth, monotonous function mapping 0 to 0 and
$[1/2m,1]$ to 1, and $\nu_i:=\alpha(\mu_i)$.

{\bf\green Step 2:}
Denote by $W_i$ the set of interior points of 
$\bar W_i:=\{z \ \ |\ \ \nu_i(z)=1\}= 
\{z \ \ |\ \ \mu_i(z)\geq \frac{1}{2m}\}.$  {\bf \purple Since
$\sum_{i=1}^m \mu_i =1$, the set
$\{W_i\}$ is a cover of $M$.}

{\bf\green Step 3:}
For each $i$, the map  
$\Phi_i(z):= 
\frac{(\nu_i\phi_i(z),1-\nu_i(z))}{|(\nu_i\phi_i(z),1-\nu_i(z))|}$
{\bf \purple is smooth and induces a diffeomorphism of
$W_i$ and an open subset of $S^{n}\subset \R^{n+1}$.}

{\bf\green Step 4:}
The product map
\[ \Psi:=\prod_{i=1}^m:\;\Phi_i:\; M \arrow \underbrace{S^n\times
S^n \times ...\times S^n}_{\text{$m$ times}}
\]
is an injective, continuous map from a compact,
hence {\bf \purple it is a homeomorphism to its image.}
It is a smooth embedding, because its differential
is injective.
\endproof

\newpage

{\bf \blue Embedding to $\R^\infty$}

\question
{\bf \purple What if $M$ is non-compact?}

\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  Embedding to $\R^\infty$ (cont.)}

\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.}


{\bf\green Proof. Step 1:} 
$\Psi$ is injective by construction.
To prove that it is a homeomorphism, it suffices to check
that an image of an open set $U$ is open in $\Psi(M)$, for 
each $U\subset W_i$, for some open cover $\{W_i\}$ 

{\bf\green Step 2:} 
However, the set
$\Psi(W_i)$ is determined by $\nu_i(z)=1$, that is, by
$\Phi_i(z)_{n+1}=1$,
where $\Phi_i(z)_{n+1}$ is the last coordinate of
$\Phi_i(z)$. Therefore, {\bf \purple $\Psi$ maps $W_i$ to an open
subset of $\Psi(M)$.}

{\bf\green Step 3:} Since $\Phi_i \restrict{\bar W_i}$
(restriction to a closure) is a continuous, bijective
map from a compact, it's a homeomorphism. Therefore,
{\bf \purple an image of any open subset $U\subset W_i$
is open in $\Psi(W_i)$, which is open in $\Psi(M)$
as follows from Step 2.}
\endproof

\newpage

{\bf \blue  Paracompactness}

\definition
{\bf\blue An open cover} of a topological space $X$ is a
family of open sets $\{U_i\}$ such that $\bigcup_iU_i=X$. A cover
$\{V_i\}$ is a {\bf \blue refinement} of a cover $\{U_i\}$ if every $V_i$ is
contained in some $U_i$.

\definition
A cover of $M$ is called {\bf \blue locally finite}
if any point $x\in M$ has a neighbourhood intersecting
only finitely many of the elements of a cover.

\definition
A topological space is called {\bf \blue paracompact}
if any cover admits a locally finite refinement.

\exercise
Let $M$ be a paracompact topological space,
and $Z\subset M$ a closed supset. {\bf \purple Prove that
$Z$ is paracompact.}

\newpage

{\bf \blue  Paracompactness and partitions of unity}


\theorem
Let $M$ be a manifold. {\bf \red Then the following
conditions are equivalent: \\
\phantom{huyak} (i) $M$ is metrizable \\
\phantom{huyak} (ii) $M$ admits a partition of unity \\
\phantom{huyak} (iii) $M$ is paracompact.}

\proof
Implication (iii) $\Rightarrow$ (ii) is proven above.
Metrizability of $M$ follows from existence of partition
of unity, because $M$ admits a homeomorphism to 
a subset of a metric space $\R^I$. This proves
(ii) $\Rightarrow$ (i). It remains to prove that
metrizability implies paracompactness.

We don't need it, but I will give a short 
sketch of a proof.

\newpage

{\bf \blue  Paracompactness and partitions of unity (cont.)}


{\bf \green Step 1:}
Consider the function $\rho:\; M \arrow \R^{\geq 0}$
mapping $x\in M$ to a supremum of all $r$ such that
the open ball $B_r(x)$ is contained in one of $U_i$,
and its closure is compact.
It is easy to check that {\bf \purple $\rho$ is continuous,}
and, indeed, 1-Lipschitz {\bf \red (prove it).}

{\bf \green Step 2:}
Now we can replace $\{U_i\}$ by a cover
$\{B_{\rho(x)}(x)\ \ |\ \ x\in M\}$, which is
its refinement. 

{\bf \green Step 3:}
Take a maximal subset
$Z\subset M$ such that for each distinct $x,y\in Z$,
one has $d(x,y) \geq 1/8\rho(x)$. 
Such a subset exists by Zorn's Lemma.
Since $\rho$ is 1-Lipschitz,
$\{W_i\} := \{B_{1/2\rho(x)}(x)\ \ |\ \ x\in Z\}$
is also a cover of $M$. It is a refinement of $\{U_i\}$,
as follows from Step 2.

{\bf \green Step 4:} Now, each $W_i=B_{1/2\rho(z_i)}{(z_i)}$ 
intersects only those $W_j=B_{1/2\rho(z_j)}{(z_j)}$ for which
$d(z_i, z_j)\geq 1/8\rho(z_i)$, and there are only
finitely many of them, by compactness
of $\bar W_i$.
\endproof


\newpage

{\bf \blue Measure 0 subsets and Sard's theorem }

\definition
A subset  $Z\subset\R^n$ has {\bf \blue measure
zero} if, for every $\varepsilon>0$, there exists a countable cover of
$Z$ by open balls $U_i$ such that $\sum_i\Vol U_i<\varepsilon$.

\definition
A subset $Z\subset M$ of a manifold $M$ 
has {\bf \blue measure 0} if intersection of $M$
with each chart $U_i\hookrightarrow \R^n$
has measure 0.

{\bf \green Properties of measure 0 subsets}.\\
\phantom{huy} {\bf \red A countable union of measure 0 subsets
has measure 0.}
\phantom{huy} A measure 0 subset $Z\subset M$ is {\bf \blue nowhere
dense}, that is, $(M \backslash Z)\cap U \neq \emptyset$
for any non-empty open subset $U\subset M$.

\theorem
{\bf \blue (a special case of Sard's Lemma)}
Let $f:\; M\arrow N$ be a smooth map of manifolds,
$\dim M < \dim N$. {\bf \red Then $f(M)$ has measure 
zero in $N$.}

{\bf \green Its proof will be given in the next lecture}
(if needed).


\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 $W$ surjects 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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