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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Geometry of manifolds \\[15mm]
\small Lecture 5: Derivations in rings}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

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

{\tiny\bf Math in Moscow and HSE
\\[2mm]  March 4, 2013
}
\end{center}


\newpage

{\bf \blue Rings and derivations}


\remark
All rings in these handouts are assumed to be 
commutative and with unit. Algebras are associative,
but not necessarily commutative (such as the matrix algebra).
{\bf\blue Rings over a field $k$} are rings containing a field $k$.
We assume that $k$ has characteristic 0.

\definition
Let $R$ be a ring over a field $k$. A $k$-linear map
$D\; R \arrow R$ is called {\bf\blue a derivation}
if it satisfies {\bf\blue the Leibnitz equation}
$D(fg) = D(f) g + gD(f)$. The space of derivations
is denoted as  $\Der_k(R)$.

\example
$\frac{d}{dt}:\; \C[t]\arrow \C[t]$. \ \ 
$\frac{d}{dt}:\; C^\infty \R\arrow C^\infty\R$.

\remark
Any derivation $\delta\in\Der_k(R)$ vanishes on $k\subset R$.
Indeed, $\delta(1) = \delta(1\cdot 1) = 2\delta(1)$.

\claim
Let $K$ be {\bf \blue a finite extension} of a field $k$,
that is, a field containing $k$ and finite-dimensional
as a $K$-linear space. {\bf \purple Then $\Der_k(K)=0$.}

\proof
Indeed, any $x\in K$ satisfies a non-trivial polynomial equation
$P(x)=0$ with coefficients in $k$. Chose $P(t)$ 
of smallest degree possible. {\bf \purple For any $\delta\in\Der_k(R)$, we have
$0=\delta(P(x))= P'(x) \delta(x)$,}
and unless $\delta(x)=0$, one has $P'(x)=0$, giving a contradiction.
\endproof


\newpage

{\bf \blue Modules over a ring}


\definition
Let $R$ be a ring over a field $k$.
{\bf \blue An $R$-module} is a vector space
$V$ over $k$, equipped with an algebra
homomorphism $R\arrow \End(V)$, where $\End(V)$ 
denotes the endomorphism algebra of $V$, that
is, the matrix algebra.

\remark
Let $R$ be a field. Then $R$-modules
are the same as vector spaces over $R$.

\definition
Homomorphisms, isomorphisms,
submodules, quotient modules, direct sums 
of modules are defined in the same way as for the
vector spaces. A ring $R$ is itself an $R$-module.
A direct sum of $n$ copies of
$R$ is denoted $R^n$. Such $R$-module is called 
{\bf\blue a free $R$-module}.

\example
{\bf \purple
$R$-submodules in $R$ are the same as ideals in $R$.}


\definition
{\bf\blue Finitely generated} $R$-module is a quotient
module of $R^n$.


\newpage

{\bf \blue Noetherian rings}

\definition
{\bf\blue A Noetherian ring} is a ring $R$ with all
ideals finitely generated as $R$-modules.

\theorem
Let $R$ be a Noetherian ring. {\bf \red Then
any submodule of a finitely generated
$R$-module is finitely generated.}

{\bf \green Proof. Step 1:}
Consider an exact sequence of $R$-modules\\
$0\arrow M_1 \arrow M \arrow M_2\arrow 0$.
Then $M$ is called 
{\bf\blue an extension} of $M_1$ and $M_2$.
An extension of finitely-generated modules
is finitely generated. Indeed, take a finite set
of generators in $M_2$, and let $\{\xi_i\}$
be preimages of these generators in $M$. Let
$\{\zeta_j\}$ be a finite set of generators
in $M_1\subset M$. {\bf \purple Then $\{\zeta_j +\xi_i\}$
generate $M$.}

{\bf \green Step 2:}
{\bf \blue A filtration}
on a module $M$ as a sequence of submodules \\
$0=M_0\subset M_1 \subset ... \subset M_n =M$.
From Step 1 and induction it follows that
any {\bf \purple $M$ admitting a filtration with finitely-generated
$M_i/M_{i-1}$ is also finitely-generated.}

{\bf \green Step 3:} Let $M\subset R^n=W$.
Consider a filtration $W_0\subset W_1 \subset ... \subset W_{n}=W$,
with $W_i=R^i$, and let $M_i =M\cap W_i$.
{\bf \purple Then $M_i/M_{i-1}$ is a submodule of $W_i/W_{i-1}=R$, hence
finitely-generated.} \endproof

\newpage

{\bf \blue Ring of smooth functions}

\theorem
Let $R$ be a ring of smooth real functions on $\R^n$.
{\bf \red Then $R$ is not Noetherian.} Moreover,
the ideal $I$ of all functions 
with all derivatives vanishing at 0
is not finitely generated.

{\bf \green Proof. Step 1:}
If $I$ is generated by $f_1,..., f_n$,
then for each $g\in I$, one can express $g$ as $g=\sum g_i
f_i$. 
{\bf \purple Then
\[ \lim\limits_{x\rightarrow 0}\sup \frac{g}{\sum f_i^2}\leq
\sum |g_i|\frac{f_1}{\sum f_i^2}< \infty.\]}

{\bf \green Step 2:} Let $x_i$ be coordinate functions.
{\bf \purple The function $F:=\frac{\sum f_i^2}{\sum x_i^2}$ is smooth},
and all its derivatives vanish at 0, however, 
\[ \lim\limits_{x\rightarrow 0}\sup \frac{F}{\sum f_i^2}=
\lim\limits_{x\rightarrow 0}\sup\frac{1}{\sum x_i^2}=
\infty.
\]
\endproof

\newpage

{\bf \blue Derivations as an $R$-module}



\remark
Let $R$ be a ring over $k$. {\bf \blue The space $\Der_k(R)$ 
of derivations is also an $R$-module,} with multiplicative
action of $R$ given by $rD(f) = r D(f)$.


\claim
Let $R= k[t_1, .., t_k]$ be a polynomial ring.
{\bf \red Then $\Der_k(R)$ is a free $R$-module
isomorphic to $R^n$,} with generators $\frac d {dt_1}, 
\frac d{dt_2}, ...,\frac d{dt_n}$.

\proof
Consider a map $\Der_k(R) \arrow R^n$, 
\[ D \arrow (D(t_1), D(t_2), ..., D(t_n))
\]
It is surjective, because it maps each $\frac d {dt_i}$
to $(0, ..., 0, 1, 0, ..., 0)$, and injective, because
each derivation which vanishes on $t_i$, vanishes on the
whole polynomial ring.
\endproof

{\bf \green Now we prove a similar result for $C^\infty \R^n$.}

\newpage

{\bf \blue Hadamard's Lemma}

\lemma {\bf \blue (Hadamard's Lemma)}\\
Let $f$ be a smooth function
$f$ on $\R^n$, and $x_i$ the coordinate functions. {\bf \red Then
$f(x)=f(0)+\sum_{i=1}^n x_i g_i(x)$, for some smooth 
$g_i\in C^\infty \R^n$.}


{\bf \green Proof:} Let $t\in \R^n$.
Consider a function $h(t)\in C^\infty \R^n$, $h(t)=f(tx)$.
Then $\frac{dh}{dt}=\sum \frac{df(tx)}{dx_i}(tx)x_i$, giving
\[
f(x)-f(0)= \int_0^1 \frac{dh}{dt} dt=\sum_i x_i  \int_0^1\frac{df(tx)}{dx_i}(tx)dt.
\]
\endproof

\corollary
Let $\goth m_0$ be an 
ideal of all smooth functions on $\R^n$ vanishing in $0$.
{\bf \purple Then $\goth m_0$ is generated by coordinate functions.}
\endproof

\corollary
Let $f$ be a smooth function on $\R^n$
satisfying $f(x)=0$ and $f'(x)=0$. {\bf \red Then
$f\in {\goth m}_x^2$.}

\proof $f(x)=\sum_{i=1}^n x_i g_i(x)$,
where all $g_i$ vanish in 0. \endproof



\newpage

{\bf \blue Derivations of $C^\infty \R^n$}

\theorem
Let  $x_1, ..., x_n$ be coordinates on $\R^n$, $R=C^\infty\R^n$,
and $\Der(R) \stackrel \Pi \arrow(C^\infty\R^n)^n$
map $D$ to $(D(x_1), D(x_2), ..., D(x_n))$.
{\bf \red Then $D:\; \Der(C^\infty\R^n) \arrow R^n$ is an isomorphism.}

{\bf \green Proof. Step 1:}
Since $\Pi$ maps each $\frac d {dt_i}$
to $(0, ..., 0, 1, 0, ..., 0)$, it is {\bf \purple surjective.}

{\bf \green Step 2:} Let ${\goth m}_0$ be an ideal of 0, 
and $D\subset \ker \Pi$. Then $\Pi(x_i)=0$, where $x_i$ are
coordinate functions.
By Hadamard's Lemma, $f(x)=f(0)+\sum_{i=1}^n x_i g_i(x)$,
hence $D(f)=\sum_{i=1}^n x_i D(g_i)$. {\bf \purple Therefore,
$D(f)$ lies in ${\goth m}_0$.}

{\bf \green Step 3:} {\bf \purple Same argument proves that $D(f)$ vanishes
everywhere,} for all $f\in C^\infty M$. \endproof

\newpage

{\bf \blue Sheaves}

\definition
{\bf\blue An open cover} of a topological space $X$ is a
family of open sets $\{U_i\}$ such that $\bigcup_iU_i=X$.


\remark The definition of a sheaf below is a
more abstract version of the notion of 
``sheaf of functions'' defined previously.

\definition
A {\bf\blue presheaf} on a
topological space $M$ is a collection of vector spaces
${\cal F}(U)$, for each open 
subset $U\subset M$,  together with {\bf \blue restriction maps}
$R_{UW}{\cal F}(U)\arrow {\cal F}(W)$ defined for each $W\subset U$,
such that for any three open sets $W\subset V\subset U$,
$\Psi_{UW}=\Psi_{UV}\circ \Psi_{VW}$. Elements of ${\cal F}(U)$
are called {\bf \blue sections of ${\cal F}$ over $U$},
and restriction map often denoted $f\restrict W$

\definition
A presheaf 
${\cal F}$ is called {\bf\blue a sheaf} 
if for any open set $U$ and any cover $U=\bigcup U_I$
the following two conditions are satisfied.\\
\phantom{xu} 1. Let $f\in {\cal F}(U)$ be a section of ${\cal F}$ on $U$
such that its restriction to each $U_i$ vanishes. {\bf \purple 
Then $f=0$.} \\
\phantom{xu}  2. Let $f_i\in{\cal F}(U_i)$ be a family of
sections 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 $i$.}


\newpage

{\bf \blue Sheaves and exact sequences}



\definition
A sequence $A_1 \arrow A_2 \arrow A_3 \arrow ...$
of homomorphisms of abelian groups or vector spaces
is called {\bf\blue exact} if the image of each map is
the kernel of the next one.


\claim
A presheaf
${\cal F}$ is a sheaf if and only if for every cover $\{U_i\}$ of
an open subset $U\subset M$, {\bf \red the sequence of restriction maps
$$0\to{\cal F}(U)\to\prod\limits_i{\cal F}(U_i)\to\prod\limits_{i\ne
j}{\cal F}(U_i\cap U_j)$$
is exact,} with $\eta\in {\cal F}(U_i)$ mapped to $\eta\restrict{U_i\cap U_j}$
and $-\eta\restrict{U_j\cap U_i}$. 


\newpage

{\bf \blue Ringed spaces (reminder)}

\definition
{\bf\blue A sheaf of rings} is a sheaf ${\cal F}$ such that
all the spaces ${\cal F}(U)$ are rings, and all restriction
maps are ring homomorphisms. 

\definition
{\bf \blue A sheaf of functions} is a subsheaf
in a sheaf of all functions, closed under multiplication.

{\bf \red For simplicity, I assume now that a sheaf of rings
is a subsheaf in a sheaf of all functions}.

\definition
A {\bf\blue ringed space} $(M,{\cal F})$ is a
topological space equipped with a sheaf of
rings. 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.


\newpage

{\bf \blue Smooth manifold (reminder)}

\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 (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) $\sum_if_i=1$


\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 $\{U_\alpha\}$.}


\definition
Let  $U\subset V$ be open subsets in $M$.
We write $U\Subset V$ if the closure of $U$ is contained in $V$.

\definition
Let $f\in {\cal F}(M)$ be a section of a sheaf ${\cal F}$
on $M$. A point $x\in M$ does not lie in the
{\bf\blue support} $\Sup(f)$  of $f$ if $f\restrict U=0$ for
some neighbourhood $U\ni x$.

\remark {\bf \purple Support of a section is obviously closed.}

\newpage

{\bf \blue Smooth functions with prescribed support }

\exercise
Let $X,Y\subset M$ be non-intersecting closed subsets
in a metric space.
{\bf \purple Find non-intersecting open neighbourhoods
$U_1\supset X$ and $U_2\supset U$.}

\claim
Let $U\Subset V$ -- open subsets in a smooth metrizable
manifold. {\bf \red Then there exists a smooth function
$\Phi_{U,V}\in C^\infty M$, supported on $V$,
and equal to 1 on $U$.}

{\bf \green Proof. Step 1:}
Find non-intersecting open neighbourhoods
$U_1$ and $U_2$ of $\bar U$ and $M \backslash V$, and choose
a partition of unity $\{V_i, \phi_i\}$ subordinate to the cover $U_1, U_2,
U_3=V\backslash \bar U$. {\bf \purple Then for each $i$, either 
$\Sup(\phi_i)\cap U_1=\emptyset$, or $\Sup(\phi_i)\cap U_2=\emptyset$.}

{\bf \green Step 2:} Let $\Phi_{U,V}:= \sum_S \phi_i$,
where the sum is taken over the set $S$ all $\phi_i$ satisfying
$\Sup(\phi_i)\cap U_1\neq \emptyset$. {\bf \purple Since support
of all such $\phi_i$ lies in $M \backslash U_2\subset V$, 
one has $\Sup(\Phi_{U,V})\subset V$.} Also, {\bf \purple for each
$x\in U_1$, one has $\sum_{i\in S} \phi_i(x)=1$, hence 
$\Phi_{U,V}=1$ on $U_1\supset U$.} \endproof





\newpage

{\bf \blue Vector fields as derivations }

\definition
Let $M$ be a smooth manifold. A {\bf\blue vector field} on $M$
is an element in $\Der(C^\infty M)$.

\example
For $M=\R^n$, {\bf \red the space $\Der(C^\infty M)$ is a free module
generated by $\frac{d}{dx_i}$, $i=1, ..., n$.}

\remark
We want to prove that vector fields form a sheaf. 
{\bf  \purple However,
it is not immediately clear how to restrict a vector
field from $U$ to $W\subset U$.}

\theorem
Let $U\Subset V$ be open subset of a smooth metrizable 
manifold, and $D\in (C^\infty M)$ a derivation.
Consider a smooth function  $\Phi_{U,V}\in C^\infty M$ 
supported on $V$,  and equal to 1 on $U$.
Given $f\in C^\infty V$, define $D(f)\restrict U:=D(\Phi_{U,V}f)$.
Choosing a cover of such $U_i$, we can glue together
a section $D(f)$ of $C^\infty V$ from such $D(f)\restrict {U_i}$
{\bf \red This operation is independent of all choices we made
and gives an element $D\restrict V\in \Der(V)$.
Moreover, this restriction maps define a structure
of a sheaf on $\Der(M)$.}

{\bf \green Proof: next lecture.} The proof uses germs.



%\newpage

%{\bf \blue A sheaf of vector fields }




\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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