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\begin{document}
\frontmatter
\title[Lectures on Linear Differential Operators]%
{Lectures\\ on Linear Differential Operators\\
over Commutative Algebras\\
The 1st Diffiety School\\
\ \\
\ \\
Forino, Italy\\
\ \\
\ \\
July, 1998
\ \\
\ \\
}
\author{Joseph Krasil'shchik\\
The Independent University of Moscow\\
in collaboration with\\
Barbara Prinari\\
Universit\`{a} di Lecce}
\thanks{This work was partially supported by the \emph{INTAS Grant N 96-0793}.
The author is indebted to \emph{L'istituto Italiano per gli Studi
Filosofici} for the financial support of the \emph{1st Italian Diffiety
School}, where these lectures were given, and to the \emph{Community of
Forino} for the hospitality and stimulating conditions. It is also my
pleasure to express gratitude to the \emph{Department of Applied Mathematics}
of the \emph{University of Seville}, where this text was written in the final
form.}
\maketitle
%\newpage
\tableofcontents
\setcounter{chapter}{-1}
\chapter*{Foreword}
These Lecture Notes contain introductory material to algebraic theory
of linear differential operators over commutative algebras. The basics
of this theory were exposed in a short paper by A.M.~Vinogradov \cite{AlgLog}.
Later, a much more extensive exposition was published in the first chapter
of \cite{KLV}. 
%A very concise version can also be found in \cite{AAM}.

The initial course, held in Forino (Summer 1998) at the \emph{First Italian
Diffiety School}, consisted of ten lectures and a number of practical
lessons, but in preparation of the printed version it became clear that a
more logical structure should contain five parts (which are also called
\emph{lectures} below):
\begin{itemize}
\item general introduction to \emph{commutative algebra} (rings, algebras, and
modules) together with main concepts of the \emph{category theory} --- first two
lectures,
\item main functors of \emph{differential calculus} over commutative algebras ---
lecture 3,
\item their \emph{representative objects} (jets and differential forms) ---
lecture 4,
\item relations to geometry serving as a bridge to the next course in
\emph{geometry of differential equations} --- the last lecture.
\end{itemize}
We conclude these notes with a series of exercises which are an essential
part of the main text. A lot of them were analyzed and solved during
practical lessons held by M.M.~Vinogradov and V.A. and V.N.~Yumaguzhins.

As additional reading, we strongly recommend the books by I.~MacDonald and
M.~Atiyah \cite{AtMac} and S.~Lang \cite{Lang} (commutative algebra),
S.~MacLane \cite{Maclane,Maclane1} (category theory and homology), M.~Atiyah
\cite{Atiyah} and D.~Husemoller \cite{Huse} (vector bundles), and Jet
Nestruev \cite{JetN} (relations between commutative algebra and geometry of
smooth manifolds). An extended discussion of algebraic theory of
differential calculus together with its relation to differential equations
can be found in \cite{Levoca}.
\begin{center}
$\star\ \star\ \star$
\end{center}
This text is based on the notes made by Barbara Prinari during the
lectures. She not only fixed the material with great accuracy, but also
typeset the initial \TeX file. Without her help these lectures would hardly
be published.
\mainmatter
\chapter*{Lecture 1}
We start with the maim concept lying in the base of all our future
constructions. This is a notion of a \emph{ring}.

Let $A$ be an Abelian group. The operation in the group will be denoted by
the symbol ``$+$'':
\[
+\colon A\times A\to A,\qquad (a,b)\mapsto a+b.
\]
Suppose there is another operation in $A$ denoted by ``dot'':
\[
\cdot\colon A\times A\to A,\qquad (a,b\mapsto a\cdot b\equiv ab,
\]
such that one has the distributivity law
\begin{align}
a(b+c) &=ab+ac  \label{ring1} \\
(b+c)a &=ba+ca.  \label{ring2}
\end{align}
for all $a,b,c\in A$. Then one says that $A$ is a \emph{ring}.\index{ring} If
in addition we have associativity
\[
a(bc)=(ab)c,
\]
then $A$ is called an \emph{associative} ring.\index{associative ring} If
there is commutativity
\[
ab=ba,
\]
it is called \emph{commutative}.\index{commutative ring} If there exists an
element $1\in A$ such that
\[
1\cdot a=a,\qquad a\cdot 1 =a
\]
for all $a\in A$, the ring is called \emph{unitary}\index{unitary ring} and
the element $1$ is called the \emph{unity} of the ring. It is easy to check
that if it exists it is unique.
\begin{example}
The groups $\mathbb Z$, $\mathbb R$, $\mathbb C$ are rings with respect to
``ordinary'' operations of summation and multiplication of numbers. They are
all associative, unitary and commutative rings.
\end{example}
\begin{remark}
The rings $\mathbb R$ and $\mathbb C$ possess an additional property: any
element $a\neq0$ is invertible in them, i.e., there exists an element
$b\equiv a^{-1}$ such that $a\cdot a^{-1}= a^{-1}\cdot a=1$. Such rings
are called \emph{fields}.\index{field} Obviously, $\mathbb Z$ is not a
field.
\end{remark}
\begin{example}
Let $V$ be a vector space over $\mathbb R$. The set of all linear
operators acting on $V$,
\[
L\colon V\to V,
\]
forms a ring with composition playing the role of the product and it is
denoted by $\Mat(V)$ (since, in a chosen basis, linear operators correspond
to matrices). Then we know that
\[
L_{1}\circ (L_{2}\circ L_{3})=(L_{1}\circ L_{2})\circ L_{3},
\]
i.e., $\Mat(V)$ is an associative ring, but since in general,
\[
L_{1}\circ L_{2}\neq L_{2}\circ L_{1}
\]
it is not commutative.
\end{example}
\begin{example}
Let $E$ be a set and consider the set of all subsets of $E$ denoted by
$2^E$. It can be shown that $2^E$ forms a commutative ring with respect
to the operations
\[
a+b\defeq a\cap b,\quad
a+b\defeq\bar{a}\cap b)\cup(a\cap\bar{b}),\qquad
a,b\in  2^E,
\]
where $\bar{a}\defeq E\setminus a$ (see Exercise \ref{exc2.1}).
\end{example}
\begin{example}
Let $C({\mathbb R})$ be the set of all continuous functions on the real
line. With respect to the operations
\begin{align*}
(f+g)(x)&\defeq f(x)+g(x) \\
(f\cdot g)(x)&\defeq f(x)g(x)
\end{align*}
$C({\mathbb R})$ is a commutative and unitary ring (the same is valid for
the set of smooth functions $C^\infty(\mathbb{R})$).
\end{example}
\begin{example}\label{exmGrR}
Let $G$ be a finite group and $A$ a commutative ring with unit. Consider the
set $A[G]$ consisting of all formal sums
\[
A[G] =\Big\{\sum_{g_i\in G}a_ig_i\mid a_i\in A\Big\}.
\]
Then, having two such expressions, we can define their sum and product
\begin{align*}
\sum_{g_i\in G}a_ig_i+\sum_{g_i\in G}b_ig_i&\defeq\sum_{g_i\in G}
(a_i+b_i)g_i \\
\sum_{g_i\in G}a_ig_i\cdot\sum_{g_j\in G}b_ig_j&\defeq\sum(a_ib_j)g_ig_j.
\end{align*}
Then this object becomes a ring which is associative but in general not
commutative (if $G$ is not commutative). $A[G]$ is called the \emph{group
ring}\index{group ring} of $G$ (with coefficients in $A$).
\end{example}
\begin{example}
Let $V$ be a vector space and assume that there exists a multiplication in
$V$ denoted by $[~,~]$ such that
\begin{enumerate}
\item $[a,b]=-[b,a]$,
\item $[a,b+c]=[a,b]+[a,c]$,
\item $[a,[b,c]]+[c,[a,b]]+[ b,[ c,a] ]=0$.
\end{enumerate}
Then $V$ is called a \emph{Lie algebra}\index{Lie algebra} (it is neither
commutative, nor associative).

If $A$ is a ring, a Lie algebra structure can naturally be induced by
defining
\[
[a,b]\defeq ab-ba.
\]
for all $a,b\in A$. It is easy to check that it satisfies properties 1-3. If
$A$ is a commutative ring, then $[~,~]$ will be trivial, i.e., $[a,b]=0$
for any $a,b\in A$.
\end{example}
\begin{example}\label{exm1.7}
Let $V$ be a vector space over $\mathbb R$ (or any other field). Then we can
define the \emph{tensor product} $V\otimes V$. This is a vector space. In
a similar way we can consider the spaces $V\otimes V\otimes V=(V\otimes V)
\otimes V$ and so on:
\[
\underbrace{V\otimes V\otimes\dots\otimes V}_{n\ \mathrm{times}}=V^{\otimes
n}
\]
which is called the $n$-th \emph{tensor power} of $V$ and, if we put
$V^{\otimes 0}\defeq\mathbb{R}$, we can define
\[
T(V)=\sum_{n\ge 0}V^{\otimes n}.
\]
Elements of $V^{\otimes n}$ are linear combinations of \emph{homogeneous}
elements $v=v_1\otimes\dots\otimes v_n$, $v_i\in V$. If $w=w_1\otimes\dots
\otimes w_m\in V^{\otimes m}$ is another homogeneous element, we define
their product $v\otimes w$ as a simple concatenation:
\[
v\otimes w\defeq v_1\otimes\dots\otimes v_n\otimes w_1\otimes\dots\otimes
v_m\in V^{\otimes(n+m)}.
\]
This operation is extended to arbitrary elements of $T(V)$ in an obvious way.
Thus $T(V)$ becomes an associative algebra with respect to the tensor
product. It is called the \emph{tensor algebra}\index{tensor algebra of a
vector space} of the space $V$. In a sense, all rings can be obtained from
tensor algebras by a regular procedure.
\end{example}

Let now $A$ and $B$ be two rings and consider a map
\begin{equation}
\label{hom}
f\colon A\to B
\end{equation}
such that for all $a,a^\prime\in A$ one has
\begin{align}
f(a+a^\prime)&=f(a)+f(a^\prime)  \label{homprop1} \\
f(aa^\prime)&=f(a)f(a^\prime)  \label{homprop2}
\end{align}
(if $A$ and $B$ are unitary we should add also
\begin{equation}
f(1_A)=1_B,
\end{equation}
where $1_A$ are $1_B$ units in $A$ and $B$ respectively.). Then $f$ is called
a \emph{ring homomorphism}.\index{ring homomorphism}

The subset of $B$
\[
\im f\defeq\{b\in B\mid b=f(a),\quad a\in A\}
\]
is a ring and is called the \emph{image}\index{image of a homomorphism} of
$f$ and the subset of $A$
\[
\ker f\defeq\{a\in A\mid\ f(a)=0\}
\]
is a ring too, and is called the \emph{kernel}\index{kernel of a
homomorphism} of $f$. If $\ker f=0$, the homomorphism $f$ is called a
\emph{monomorphism}.\index{monomorphism} In the case $\im f=B$, it is
called an \emph{epimorphism}.\index{epimorphism} If a homomorphism is both
epi- and monomorphism, it is called an \emph{isomorphism}.\index{isomorphism}

Elements of $\ker f$ satisfy the following property:
for all $a\in A$ and $a^\prime\in \ker f$ one has $aa^\prime\in \ker f$ and
$a^\prime a\in \ker f$.

Now let $A$ be a ring and $P\sbs A$ be a subgroup of $A$ with respect to
addition. If
\begin{equation}
A\cdot P\sbs P  \label{left}
\end{equation}
(i.e., $ap\in P$ for all $a\in A$ and $p\in P$), $P$ is called a \emph{left
ideal} of $A$.\index{left ideal of a ring} If
\begin{equation}
P\cdot A\sbs P,  \label{right}
\end{equation}
then $P$ is called a \emph{right ideal} of $A$.\index{right ideal of a ring}
If both \eqref{left} and \eqref{right} are valid, then $P$ is called a
\emph{two-sided ideal}.\index{to-sided ideal of a ring} If $A$ is a
commutative ring, all ideals are two-sided ones.

Consider a pair, a ring $A$ and a two-sided ideal $P$. Then the quotient
group $A/P$ is defined
\[
A/P=\{[a]=a\bmod P \mid a\in A\}.
\]
Two elements $a,a^\prime$ are in the same coset in $A/P$ if and only if
$a-a^\prime\in P.$ Then we can define sum and product of cosets
in the natural way
\[
[a]+[b]\defeq[a+b],\qquad
[a]\cdot[b]\defeq[a\cdot b].
\]
If $P$ is a two-sided ideal, these operation are well defined. With respect
to these operations the quotient set is a ring. It is the \emph{quotient
ring}\index{quotient ring} of $A$ with respect to the two-sided ideal $P$.

There is a natural epimorphism $\pi\colon A\to A/P$ taking any element
$a\in A$ to the corresponding coset $[a]\in A/P$. Thus we have the
sequence of homomorphisms
\[
0\to P\xrightarrow{\ i\ }A\xrightarrow{\pi}A/P\to0,
\]
where $i$ is the natural embedding and the kernel of $\pi$ coincides
with the image of $i$. This is the so-called \emph{short exact sequence}
\index{short exact sequence} of rings.
\begin{example}
Consider the real line $\mathbb R$ and the ring of smooth functions $C^\infty
(\mathbb{R})$. Fix a point $x\in\mathbb{R}$ and consider
the set
\begin{equation}
\label{mux}
\mu_x=\{f\in C^\infty(\mathbb{R})\mid f(x)=0\}
\end{equation}
It is easy to check that $\mu_x$ is an ideal. The quotient ring $C^\infty
(\mathbb{R})/\mu_x$ is isomorphic to $\mathbb{R}$ (see Exercise
\ref{exc2.6}). If we choose two points $x,y\in\mathbb{R}$ the set $\mu_{x,y}$
of functions vanishing at both points is an ideal too, and so on.
\end{example}
\begin{example}\label{exm1.9}
Consider an integer $n\in\mathbb{Z}$ and the set
\begin{equation}
n\mathbb{Z}=\{na\mid a\in\mathbb{Z}\}  \label{nZ}
\end{equation}
It is an ideal in the ring of integer numbers so we can consider the quotient
ring $\mathbb{Z}_n\defeq\mathbb{Z}/n\mathbb{Z}$. This ring consists of $n$
elements
\[
\mathbb{Z}_n=\{[0],[1],\dots,[n-1]\}
\]
which are residues modulo $n$. In fact, we can prove (see Exercise
\ref{exc2.8}) that any ideal of $\mathbb{Z}$ is of the form \eqref{nZ}, that
is the following fact is valid:
\begin{proposition}
A subset $P\sbs\mathbb{Z}$ is an ideal of the ring $\mathbb{Z}$ if and
only if there exists $n\in\mathbb{Z}$ such that $P=n\mathbb{Z}$.
\end{proposition}
Note that $\mathbb{Z}_n$ is a field if and only if $n$ is a prime number.
\end{example}
\begin{example}\label{sym-prod}
Consider the tensor algebra $T(V)$ (see Example \ref{exm1.7}) and the
two-sided ideal $P_-$ of this algebra generated by the elements of the
form $a\otimes b-b\otimes a$, $a,b\in V=T^1(V)$. This means that $P_-$
consists of the elements of the form
\[
\sum_iv_i\otimes(a_i\otimes b_i-b_i\otimes a_i)\otimes w_i,\quad
v_i,w_i\in T(V).
\]
Then the quotient ring is commutative and is called the \emph{symmetric
algebra}\index{symmetric algebra} of $V$ and is denoted by $S(V)$. If we
chose a basis $x_1,\dots,x_n$ in $V$, then any element of $S(V)$ can be
uniquely represented in the form of finite sum
\[
\sum_{0\le i_1,\dots,i_n}a_{i_1,\dots,i_n}x_1^{i_1}\dots x_n^{i_n},
\]
where $a_{i_1,\dots,i_n}$ are elements of the basic field, while $S(V)$ is
identified with the \emph{ring of polynomials}\index{ring of polynomials} in
the variables $x_1,\dots,x_n$.
\end{example}
\begin{example}\label{wedge-prod}
Consider another ideal $P_+$ in $T(V)$ generated by the elements
$a\otimes b+b\otimes a$. The corresponding quotient algebra is denoted
by $\bigwedge(V)$ and is called the \emph{external algebra}\index{external
algebra} of $V$. The multiplication in this algebra is denoted by $\wedge
\colon\bigwedge(V)\otimes\bigwedge(V)\to\bigwedge(V)$. Under the natural
projection $\pi\colon T(V)\to\bigwedge(V)$, the images of elements from
$V^{\otimes n}$ constitute elements of \emph{degree} $n$. If $v,w\in
\bigwedge(V)$ are elements of degrees $n$ and $m$ respectively, the one
can show that $v\wedge w=(-1)^{nm}wv$.

If $x_1,\dots,x_n$ is a basis in $V$, then any element of $\bigwedge(V)$ is
uniquely represented in the form
\[
\sum_{1\le i_1<\dots i_k\le n}a_{i_1,\dots,i_n}x_{i_1}\wedge\dots\wedge
x_{i_k}.
\]
\end{example}
\begin{example}
Now let $A$ be the ring of all real polynomials in one variable $A=
\mathbb{R}[x]$. Any ideal $P$ in $A$ is of the form $A\cdot f$, i.e., is
generated by some polynomial. The quotient ring is a field in this
case if and only if the polynomial $f$ is indecomposable. This field is
$\mathbb R$, if $f=x-\lambda$, $\lambda\in\mathbb{R}$, and $\mathbb C$, if
$f$ has complex roots.
\end{example}
The last example, together with Example \ref{exm1.9}, leads to the
following
\begin{definition}
Let $A$ be a ring and $P\sbs A$ be its ideal. $P$ is called \emph{maximal},
\index{maximal ideal} if for any other ideal $P^\prime$ the embedding
$P^\prime\supset P$ implies $P^\prime=P$.
\end{definition}

\begin{theorem}
Let $A$ be a commutative and unitary ring. Let $P\sbs A$ be an ideal. Then
the quotient ring $A/P $ is a field if and only if $P$ is a maximal ideal.
\textup{(}For the proof see Exercise \ref{exc3.6}\textup{)}.
\end{theorem}

Let $A$ be a commutative ring with unity (from now on we shall always
consider such rings only) and let $a,b\in A$. We say that $a\neq0$ is a
\emph{zero divisor},\index{zero divisor} if
\begin{equation}
ab=0  \label{zerodiv}
\end{equation}
for some $b\neq0$.

\begin{definition}
An ideal $P\sbs A$ is called a \emph{prime ideal},\index{prime ideal}
if $ab\in P$ implies that either $a\in P$ or $b\in P$.
\end{definition}

\begin{proposition}
An ideal $P\sbs A$ is prime if and only if the quotient ring $A/P$
possesses no zero divisors \textup{(}see Exercise \ref{exc4.1}\textup{)}.
\end{proposition}

\begin{proposition}
Any maximal ideal $P\sbs A$ is a prime ideal \textup{(}see Exercise
\ref{exc4.2}\textup{)}.
\end{proposition}

\begin{proposition}
Any ideal of $A$ is contained in a maximal ideal \textup{(}for the proof
Exercise \ref{exc4.3}\textup{)}.
\end{proposition}

\begin{example}
Let $A$ be the ring of polynomials over $\mathbb R$ in three variables,
$A=\mathbb{R}[x,y,z]$. Consider the ideals
\[
P_1=\{x-\lambda\},\quad P_2=\{x-\lambda,y-\mu\},\quad
P_3=\{x-\lambda,y-\mu,z-\nu\}.
\]
Then both $P_1$ and $P_2$ are prime ideals, $P_3$ is a maximal and
consequently a prime ideal too.
\end{example}

Let us now consider two rings $A$ and $B$ and a ring homomorphism $\vf
\colon A\to B$. Then we can define an action of $A$ over $B$ in this way:
for any $a\in A$ and $b\in B$ we set
\begin{equation}
a\circ b\defeq\vf(a)b.\label{action}
\end{equation}
Usually we shall omit the ``$\circ$'' symbol and denote the action of $a$
over $b$ simply as $ab$.

It is easy to check, using \eqref{ring1}, \eqref{ring2} and \eqref{homprop1},
\eqref{homprop2}, that this action possesses the following properties
\begin{align}
a\circ(b+b^\prime)&=a\circ b+a\circ b^\prime\label{action1} \\
(a\circ b)b^\prime&=a\circ(bb^\prime)\label{action11}\\
(aa^\prime)\circ b&=a\circ b+a\prime\circ b  \label{action2} \\
(a+a^\prime)\circ b&=a\circ b+a^\prime\circ b  \label{action3}
\end{align}
and, if $A$ is a unitary ring, one has also
\begin{equation}
1_{A}\circ b=b.  \label{action4}
\end{equation}
In this case, $B$ is called an \emph{algebra} over $A$ or an $A$\h algebra.
\index{algebra over a ring}

\begin{example}
The ring of smooth functions $C^\infty(\mathbb{R}^n)$ is an algebra
over the ring $\mathbb R$.
\end{example}
\begin{example}
The polynomial ring $\Bbbk[x_1,\dots,x_n]$ is an algebra over the
field $\Bbbk$.
\end{example}

Take now instead of $B$ just an Abelian group and let us denoted by $P$. Let
us define an action of $A$ over $P$ in a reasonable way. ``Reasonable'' in
this context means that for all $a\in A$ and $p\in P$ one has a correspondence
\[
(a,p)\mapsto ap\in P
\]
and this correspondence possesses the following properties: for all $a,
a^\prime\in A$ and $p,q\in P$ one has
\begin{align}
a(p+q)&=ap+aq  \label{mod1} \\
(a+a^\prime)p&=ap+a^\prime p \label{mod2} \\
(a+a^\prime)p&=ap+a^\prime p.  \label{mod3}
\end{align}
Then we say that $P$ is an \emph{$A$-module}.\index{module over a ring}
We shall always assume the modules under consideration to be \emph{unitary},
\index{unitary module} which means that
\begin{equation}
1_Ap=p.\label{mod4}
\end{equation}
\begin{example}
Any vector space over a field $\Bbbk$ is a module over this field.
\end{example}
\begin{example}
Consider the algebra $C^\infty(\mathbb{R})$ and the set of vector valued
functions $A=\{(f_1,\dots,f_s)\mid f_i\in C^\infty(\mathbb{R})\}$. We can
multiply such functions by any function $g\in C^\infty(\mathbb{R})$ in
a natural way
\[
g(f_1,\dots,f_s)\defeq(gf_1,\dots,gf_s)
\]
and add them to each other as vectors.
So $A$ is a $C^\infty(\mathbb{R})$-module, that is a module over the ring of
smooth functions.
\end{example}
\begin{example}
Let $V$ be a real vector space (i.e., a space over $\mathbb R$) and $\vf
\colon V\to V$
be a linear operator. Consider the ring of all real polynomials in one
variable, $\mathbb{R}[x]$. If $p\in\mathbb{R}[x]$,
\[
p=a_0+a_1x+\dots+a_nx^n,\quad a_0,\dots,a_n\in\mathbb{R},
\]
then we can define the action of this polynomial over $V$ by
\[
p(v)\defeq\left(a_0+a_1\vf+\dots+a_n\vf^n\right)(v)
\]
that is
\[
p(v)=a_0v+a_1\vf(v)+\dots+a_n\vf^n(v).
\]
It is easy to check that $V$ is a module over the ring $\mathbb{R}[x]$ with
respect to this action.
\end{example}
\begin{example}
Let now $G$ be a finite group and $V$ be a vector space over a field $\Bbbk$.
We say that $G$ is \emph{represented} in $V$, if a group homomorphism of $G$
to the group of endomorphisms of $V$ is given, that is for any $g\in G$ one
has $\vf(g)\colon V\to V$ satisfying
\[
\vf(gg^\prime)=\vf(g)\circ\vf(g^\prime).
\]
To say that $G$ is represented in $V$ is the same as to say that we have a
$\Bbbk[G]$-module structure on $V$ (see Example \ref{exmGrR}).
\end{example}

Let $A$ be a ring and $P$ and $Q$ be two $A$\h modules. A mapping $f\colon P
\to Q$ is called an \emph{$A$\h homomorphism} if for any elements $p,p^\prime
\in P$ and $a\in A$ one has
\begin{align}
f(p+p^\prime) &=f(p)+f(p^\prime)  \label{modh1} \\
f(ap) &=af(p).  \label{modh2}
\end{align}
If $A$ is a field and $P$, $Q$ are vector spaces over $A$, an $A$\h
homomorphism is just a linear operator. The set of all $A$\h homomorphisms
is denoted by $\Hom_A(P,Q)$. If $f\in\Hom_A(P,Q)$ and $a\in A$, we can define
an action of $a$ over $f$ as
\begin{equation}
(af)(p)\defeq af(p)  \label{homA}
\end{equation}
and the sum of two homomorphisms by
\[
(f+f^\prime(p)\defeq f(p)+f^\prime(p).
\]
This definition introduces an $A$-module structure in $\Hom_{A}(P,Q)$. The
ring $A$ itself is an $A$\h module, so we can consider the module
$\Hom_A(A,P)$  It is easy to check that
\begin{equation}
\Hom_A(A,P)\simeq P.  \label{homAP}
\end{equation}
Let us also introduce the module $P^*\defeq\Hom_A(P,A)$ which is called the
\emph{dual} or \emph{adjoint} module of $P$.\index{dual module}\index{adjoint
module}

\chapter*{Lecture 2}
At the end of the previous lecture we introduced the notion of a
homomorphism between $A$\h modules and showed that the set $\Hom_A(P,Q)$
is an $A$\h module as well.

Let us now consider three $A$\h modules $P$, $Q$, and $R$ and two $A$\h
module homomorphisms $f$, $g$
\[
P\xrightarrow{f}Q\xrightarrow{g}R.
\]
Then we can consider the composition $h=g\circ f\colon P\to R$. It is easy to
check that $h$ is an $A$\h homomorphism from $P$ to $R$.
A way to express that $h$ is the composition of $f$ and $g$ is to say that
the diagram
\begin{diagram}
P&\rto^{f}&Q\\
\dto^{h}&\ldto_{g}&\\
R&&\\
\end{diagram}
is \emph{commutative}.\index{commutative diagram} We shall often use the
diagram language to make exposition more clear and intuitive.

Let us consider a more complicated situation: namely let $A$ be a ring and
$B$ be an $A$\h algebra. Let $\vf\colon A\to B$ be the corresponding
homomorphism and $P$, $Q$ be $B$\h modules. Then they
are $A$\h modules as well: the $A$\h module structure is defined by
\[
ap\defeq\vf(a)p,\qquad a\in A,\ p\in P.
\]
\begin{example}
Obviously, the field of complex numbers $\mathbb{C}$ is an $\mathbb{R}$\h
algebra. Therefore any complex vector space can be considered as a real
vector space using the previous definition with $\vf\colon\mathbb{R}
\hookrightarrow\mathbb{C}$ being the standard embedding.
\end{example}

So, $P$ and $Q$ are both $A$- and $B$\h modules and, consequently,
$\Hom_A(P,Q)$ is an $A$\h module while $\Hom_B(P,Q)$ is both a $B$\h module
and an $A$\h module as well according to the previous definition. But we can
define also a $B$\h module structure in $\Hom_A(P,Q)$: given $f\in
\Hom_A(P,Q)$ and $b\in B$, we can set
\begin{align}
(bf)(p)&\defeq bf(p),\qquad p\in P \label{firstaction}\\
\intertext{and we can also introduce a different action, which we shall
denote by ${}^+$:}
(b^+f)(p)&\defeq f(bp).  \label{secondact}
\end{align}
It is easy to check that both $bf$ and $b^+f$ are $A$\h homomorphism from $P$
to $Q$ and they are different since $f$ is an $A$\h homomorphism and not
a $B$\h homomorphism.

\begin{example}
Let $B=\mathbb{R}[x]$. Then $B$ an $\mathbb{R}$\h algebra and is a module
over itself. As for any ring, $\Hom_B(B,B)=B$. On the other hand, elements
of $\Hom_\mathbb{R}(B,B)$ are uniquely determined by images of $1,x,x^2,
\dots$ and are arbitrary infinite\h dimensional $\mathbb{R}$\h matrices.
\end{example}

Thus, two module structures exist in $\Hom_{A}(P,Q)$ defined by the actions
\eqref{firstaction} and \eqref{secondact}, and these structures commute
in the following sense:
\[
b_{1}(b_{2}^{+}f)=b_{2}^{+}(b_{1}f).
\]
for any $b_1,b_2\in B$ and $f\in\Hom_A(P,Q)$. In this case we say that a
\emph{bimodule structure}\index{bimodule} is given.

Now let $A$ be a ring and $P,Q$ two $A$-modules. We say that a mapping $f\in
\Hom_{A}(P,Q)$ is an \emph{epimorphism},\index{epimorphism} if it is
surjection, that is for any $q\in Q$ there exists $p\in P$ such that
$f(p)=q$. In this case we also say that the sequence $P\xrightarrow{f}Q\to
0$ is exact. We say that $f$ is an \emph{embedding} (or \emph{monomorphism}),
\index{monomorphism} if $f(p)=0$ implies $p=0$ and represent this fact by
the exact sequence $0\to P\xrightarrow{f}Q$. Epimorphic embeddings are
called \emph{isomorphisms}.\index{isomorphism}

If $P=Q$, elements of module $\Hom_A(P,P)$ are called \emph{endomorphisms}
\index{endomorphism} and the notation
\[
\End_AP\defeq\Hom_A(P,P)
\]
is used. Note that $\End_AP$ is an associative $A$\h algebra with respect
to composition.

Now let $P$ be an $A$\h module and $P^\prime\sbs P$ a subset of $P$. We
say that $P^\prime$ is a \emph{submodule}\index{submodule} in $P$ if
\[
p_1+p_2\in P^\prime,\qquad ap\in P^\prime
\]
for all $p,p_1,p_2\in P^\prime$ and $a\in A$.
\begin{example}
Let $P=A$. Then submodules of $A$ are just its ideals.
\end{example}

Let $P^\prime$ be a submodule of $P$. Then $p\sim q\Leftrightarrow p-q\in
P^\prime$ is an equivalence relation and we can consider the quotient
$P/P^\prime$. Evidently, the operations
\begin{align*}
(p+P^\prime)+(q+P^\prime)&\defeq (p+q)+P^\prime\\
a(p+P^\prime)&\defeq ap+P^\prime
\end{align*}
determine a well\h defined module structure in $P/P^\prime$ and one has a
natural epimorphism $P\to P/P^\prime$, $p\mapsto p+P^\prime$.

Let $P$, $Q$ be $A$-modules and $f\colon P\to Q$ be an $A$-module
homomorphism. Consider the sets
\begin{align*}
\ker f &=\left\{p\in P\mid f(p)=0\right\} \\
\im f &=\left\{ q\in Q\mid q=f(p)\text{ for some }p\in P\right\}.
\end{align*}
Both are submodules (of $P$ and $Q$ respectively) called \emph{kernel}
\index{kernel of a homomorphism} and \emph{image}\index{image of a
homomorphism} of $f$. We can consider the
quotient $P/\ker f$ and  it is easy to check that $P/\ker f$ is isomorphic
to $\im f$.

Now given two modules $P$ and $Q$ we can consider their Cartesian product
\[
P\times Q=\left\{(p,q)\mid p\in P,\ q\in Q\right\}
\]
and introduce an $A$-module structure to it by
\begin{align*}
(p,q)+(p^\prime,q^\prime)&\defeq(p+p^\prime,q+q^\prime) \\
a(p,q)&\defeq (ap,aq).
\end{align*}

The module obtained is denoted by $P\oplus Q$ and is called the \emph{direct
sum} of $P$ and $Q$.\index{direct sum of modules} The simplest $A$\h module is
$A$ itself, so we can construct the module
\[
A^n=\underbrace{A\oplus A\oplus\dots\oplus A}_{n\ \mathrm{times}}.
\]
which is called a \emph{free module} with $n$ generators (or of rank $n$).
\index{free module} In $A^n$ we can consider the \emph{free generators}
\begin{align*}
&e_1=(1_A,0,\dots,0) \\
&e_2=(0,1_A,\dots,0) \\
&\hdotsas{e_2=(0,1_A,\dots,0)} \\
&e_n=(0,0,\dots,1_A)
\end{align*}
and any element $a\in A^n$ can be uniquely written in the form
\[
a=a_1e_1+\dots+a_ne_n.
\]
The following result explains why the modules and generators above are
called free.
\begin{proposition}\label{prop:free}
Let $P$ be an arbitrary $A$-module and $A^n$ be the free module of rank
$n$ with free generators $e_1,\dots,e_n$. Then for any elements $p_1,\dots,
p_n\in P$ there exists a unique homomorphism $f\colon A^n\to P$ such that
$f(e_i)=p_i$, $i=1,\dots,n$.
\end{proposition}

For an arbitrary $A$-module $P$, we say that elements $p_1,\dots,p_n$ are
\emph{generators}\index{generators of a module} of $p$, if any element $p\in
P$ can be represented as a linear combination $p=\sum_{i=1}^na_ip_i$, $a_i\in
A$ (of course, this representation may not be unique). From Proposition
\ref{prop:free} we obtain the following
\begin{corollary}\label{cor:free}
Any $A$-module $P$ with $n$ generators can be represented as a quotient
of the free module $A^n$ of rank $n$.
\end{corollary}
Thus, given a module $P$, we have $P\simeq A^n/R$ for some submodule $R
\sbs A^n$. If $r_1,\dots,r_k$ are generators of $R$, we say that $P$
is described by relations $r_1=0,\dots,r_k=0$.
\begin{example}
The module $\mathbb{Z}_m$ is the quotient of $\mathbb{Z}$ described by
the relation $m=0$.
\end{example}

Now let $A$ be a commutative ring and $P$, $Q$, $R$ be $A$-modules. Consider
the set $P\times Q$ and a mapping $f\colon P\times Q\rightarrow R$. We say
that $f$ is \emph{bilinear}\index{bilinear mapping} if
\begin{enumerate}
\item  $f(ap,q)=af(p,q)$,
\item  $f(p,aq)=af(p,q)$,
\item  $f(p+p^\prime,q)=f(p,q)+f(p^\prime,q)$,
\item  $f(p,q+q^\prime)=f(p,q)+f(p,q^\prime).$
\end{enumerate}
for any $a\in A$, $p,p^\prime\in P$, $q,q^\prime\in Q$. In other words, the
mappings $f_p\colon q\mapsto f(p,q)$ and $f_q\colon p\mapsto f(p,q)$ are
$A$\h homomorphisms from $Q$ to $R$ and from $P$ to $R$ respectively.

Let now $T$ be an $A$-module and $t\colon P\times Q\to T$ be a  bilinear map.
We say that $T$ is the \emph{tensor product}\index{tensor product} of $P$ and
$Q$ if for any bilinear map $f\colon P\times Q\to R$ to an arbitrary $A$\h
module $R$, there exists a unique homomorphism $f^\prime$ such that the
diagram
\begin{diagram}[LaTeXeqno]\label{tensorprod}
P\times Q&\rto^f&R\\
\dto^t&\lddashto_{f^\prime}&\\
T&&\\
\end{diagram}
is commutative.

The tensor product $T$ of $P$ and $Q$ is denoted by $P\otimes_AQ.$ If it
exists, it is unique up to isomorphisms.

%\chapter*{Lecture 3}

Consider the free module generated by the formal expressions $p\otimes q$
with $p\in P$ and $q\in Q$ and denote it by $L(P,Q)$. Let $R\sbs L(P,Q)$
be the submodule generated by the elements
\begin{align*}
&(p+p^{\prime })\otimes q-p\otimes q-p^{\prime }\otimes q \\
&p\otimes (q+q^{\prime })-p\otimes q-p\otimes q^{\prime } \\
&ap\otimes q-p\otimes aq.
\end{align*}
One can show that the quotient module $L(P,Q)/R$ satisfies property
\eqref{tensorprod} and thus we have $L(P,Q)/R\simeq P\otimes_AQ$. This
proves existence of tensor product. Its main properties are formulated
in the following
\begin{proposition}
For any $A$-modules the following isomorphisms take place:
\begin{enumerate}
\item $P\otimes_A(Q\otimes_AR)\simeq(P\otimes_AQ)\otimes_AR$,
\item $A\otimes _{A}Q\simeq Q\otimes _{A}A\simeq Q$,
\item $P\otimes_A(Q\oplus R)\simeq P\otimes_AQ\oplus P\otimes_AR$,
\item $(P\oplus Q)\otimes_AR\simeq P\otimes_AR\oplus Q\otimes_AR$,
\item $\Hom_A(P\otimes_AQ,R)\simeq\Hom_A(P,\Hom_A(Q,R))$.
\end{enumerate}
\end{proposition}

Now we shall introduce an important class of modules.
\begin{definition}
An $A$-module $P$ is called a \emph{projective}\index{projective module} if
for any $A$\h modules $M$ and $M^\prime$, any homomorphism $f\in
\Hom_A(P,M^\prime)$, and any epimorphism $h\colon M\to M^\prime$ there exists
a homomorphism $f^\prime\in\Hom_A(P,M)$ such that the diagram
\begin{diagram}
&&P&\\
&\lddashto^{f^\prime}&\dto_f&\\
M&\rto^h&M^\prime&\rto^{}&0\\
\end{diagram}
is commutative.
\end{definition}

\begin{theorem}\label{proj-free}
An $A$-module $P$ is projective if and only if there exists a free module $F$
and a submodule $P^\prime\sbs F$ such that $F=P\oplus P^\prime$
\end{theorem}
In particular, all free modules are projective, but not vice versa.

Now let us consider a commutative ring $A$ and the set of all $A$\h modules,
which we denote by $\mathcal{M}od(A)$. We can define some nice operations in
this set; for example,
\[
P,Q\in\mathcal{M}od(A)\Longrightarrow\Hom_A(P,Q)\in\mathcal{M}od(A).
\]
What are the properties of this correspondence? First fix $P$ and consider
the correspondence $\Hom_A(P,\cdot)\colon Q\mapsto\Hom_A(P,Q)$. Then you have
a correspondence from $\mathcal{M}od(A)$ to
itself
\[
Q\in\mathcal{M}od(A)\Longrightarrow\Hom_A(P,Q)\in \mathcal{M}od(A).
\]
Now, consider another module $Q^\prime$ and a homomorphism $g\in
\Hom_A(Q,Q^\prime)$. Then to any $f\in\Hom_A(P,Q)$ we can put into
correspondence the composition $g\circ f\in\Hom_A(P,Q^\prime)$. Denoting
$g\circ f$ by $\Hom_A(P,g)(f)$, we obtain the mapping
\[
\Hom_A(P,g)\colon\Hom_A(P,Q)\to\Hom_A(P,Q^\prime).
\]
So, for a fixed $P$, $\Hom_A(P,\cdot)$ is a correspondence taking $A$\h
modules to $A$\h modules and $A$\h homomorphisms to $A$\h homomorphisms.
Moreover, if we have a sequence
\[
Q\xrightarrow{g}Q^\prime\xrightarrow{g^\prime}Q^{\prime\prime},
\]
then it can be easily seen that
\begin{equation}\label{covhom}
\Hom_A(P,g^\prime\circ g)=\Hom_A(P,g^\prime)\circ\Hom_A(P,g)
\end{equation}
and
\begin{equation}\label{idhom}
\Hom_A(P,\id_Q)=\id_{\Hom_A(P,Q)}.
\end{equation}

Let us consider another example of a similar nature. Fix a module $P$
construct the correspondence
\[
T_P\colon\mathcal{M}od(A)\to\mathcal{M}od(A)
\]
by setting $T_P(Q)=P\otimes_AQ$. If we have a homomorphism $g\in
\Hom_A(Q,Q^\prime)$, we can define $T_P(g)\colon P\otimes_AQ\to P\otimes_A
Q^\prime$ by
\[
T_P(g)(p\otimes q)\defeq p\otimes g(q).
\]
Similar to the previous case, we have $T_P(g^\prime\circ g)=T_P(g^\prime)
\circ T_P(g)$, if the composition $g^\prime\circ g$ is defined, and
$T_P(\id_Q)=\id_{P\otimes Q}$.

Let us now fix a module $Q$ and consider the correspondence
\[
\Hom_A(\cdot,Q)\colon\mathcal{M}od(A)\to\mathcal{M}od(A)
\]
taking a module $P$ to the module $\Hom_A(P,Q)$. Then for any homomorphism
$f\in\Hom_A(P^\prime,P)$ we have the homomorphism $\Hom_A(f,Q)\colon
\Hom_A(P,Q)\to\Hom_A(P^\prime,Q)$ defined by $\Hom_A(f,Q)(g)\defeq f\circ g$,
where $g\colon P\to Q$. In this case we have
\begin{equation}
\label{contrvhom}
\Hom_A(f\circ f^\prime,Q)=\Hom_A(f^\prime,Q)\circ\Hom_A(f,Q),
\end{equation}
whenever the composition $f\circ f^\prime$ is defined. Note that the order in
the right-hand side is reversed with respect to \eqref{covhom}.

Let us generalize these constructions. Consider ``something'' (a sort of
\emph{Universum}) denoted by $\mathcal C$, which consists of the collection
$\mathcal O$ of some \emph{objects} such that, given a pair of objects $O_1,
O_2\in\mathcal O$, we can put it into correspondence with a set
$\mathcal{M}or(O_{1},O_{2})$, whose elements are called \emph{morphisms} from
$O_1$ to $O_2$. Suppose that the correspondence between the pair $O_1,O_2$
and this set $\mathcal{M}or(O_1,O_2)$ satisfies the following properties:
\begin{enumerate}
\item  For any object $O$ there exists a morphism $\id_O\in
\mathcal{M}or(O,O)$;
\item  Given a triple $O_1,O_2,O_3$, one has the \emph{composition} mapping
\[
\circ\colon\mathcal{M}or(O_1,O_2)\times\mathcal{M}or(O_2,O_3)\to
\mathcal{M}or(O_1,O_3)
\]
such that:
\begin{enumerate}
\item For any $\vf\in\mathcal{M}or(O_1,O_2)$ one has $\id_{O_2}\circ
\vf =\vf$ and $\vf\circ\id_{O_1}=\vf$;
\item If $\vf\in\mathcal{M}or(O_1,O_2)$, $\psi\in\mathcal{M}or(O_2,O_3)$,
and $\zeta\in\mathcal{M}or(O_3,O_4)$, the associative rule holds: $\zeta
\circ(\psi\circ\vf)=(\zeta\circ\psi)\circ\vf$.
\end{enumerate}
\end{enumerate}
Then $\mathcal{C}$ is called a \emph{category}.\index{category} Let us
consider some examples.
\begin{example}
Consider the category whose objects are sets. If we have two sets $S_1$ and
$S_{2}$, and we define
\[
\mathcal{M}or(S_1,S_2)=\mathcal{M}aps(S_1,S_2),
\]
i.e., objects are mappings of sets. We have a category of sets.
\end{example}
\begin{example}
Consider all finite groups as objects. If we have two finite groups $G_1$ and
$G_2$ and define the set of morphisms as the set of all group homomorphisms
between $G_1$ and $G_2$ we get a category. The same can be done with Abelian
groups or with all groups.
\end{example}
\begin{example}
Consider all modules over a commutative ring $A$ and, as above, denote them
by $\mathcal{M}od(A)$. In this case
\[
\mathcal{M}or(P_1,P_2)=\Hom_A(P_1,P_2).
\]
\end{example}
\begin{example}
In the case of vector spaces over a field $\Bbbk$, morphisms will be
linear maps between these spaces.
\end{example}

How to establish relations between different categories? Let
$\mathcal{C}_1$ and $\mathcal{C}_2$ be two categories. Consider a
correspondence
\begin{equation}
F\colon\mathcal{C}_1\Longrightarrow\mathcal{C}_2  \label{functor}
\end{equation}
taking any object of $\mathcal{C}_1$ to an object of $\mathcal{C}_2$,
\[
O\mapsto F(O)
\]
and if we have a pair of objects $O$ and $O^\prime$ in $\mathcal{C}_1$, then
$F$ takes
\[
\mathcal{M}or(O,O^\prime)\Longrightarrow\mathcal{M}or(F(O),F(O^\prime)).
\]
Assume that three objects of $\mathcal{C}_1$ are given, $O_1,O_2,O_3$,
together with two morphisms $\vf\in\mathcal{M}or(O_1,O_2)$ and $\psi\in
\mathcal{M}or(O_2,O_3)$. Then $F$ is called a \emph{covariant functor}
\index{covariant functor}\index{functor} between $\mathcal{C}_1$ and $\mathcal{C}_2$, if
for all morphisms $\vf,\psi$
\[
F(\vf\circ\psi)=F(\vf )\circ F(\psi) \label{covfunct}.
\]
whenever the composition $\vf\circ\psi$ is defined. If $F$ takes
\[
\mathcal{M}or(O,O^\prime)\Longrightarrow\mathcal{M}or(F(O^\prime),F(O)).
\]
and
\[
F(\vf \circ \psi )=F(\psi )\circ F(\vf), \label{contrvfunct}
\]
then it is called is called a \emph{contravariant functor}.
\index{contravariant functor}\index{functor}
\begin{example}
The correspondence $T_{P}\colon Q\mapsto P\otimes_AQ$
is a covariant functor from the category $\mathcal{M}od(A)$ to itself.
\end{example}
\begin{example}
For a fixed $P$, the correspondence $Q \mapsto\Hom_A(P,Q)$
is also a covariant functor from the category $\mathcal{M}od(A)$ to itself.
\end{example}
\begin{example}
If we fix $Q$, the correspondence $P \mapsto\Hom_A(P,Q)$
is a contravariant functor.
\end{example}
\begin{example}
Let $G$ be a finite group. Consider the normal subgroup generated by
$g_1g_2g_1^{-1}g_2^{-1}$ with $g_1,g_2\in G,$ and denoted by $[G,G]$:
\[
[G,G]=\{g_1g_2g_1^{-1}g_2^{-1}\}\sbs G.
\]
Then $G/[G,G]$ is an Abelian group while the correspondence
\[
G\mapsto G/[G,G]
\]
is a functor from the category of finite groups to the category of Abelian
groups.
\end{example}
\begin{example}
Smooth manifolds make a category with smooth maps playing the role of
morphisms. Then the correspondence $M\mapsto C^\infty(M)$ is a contravariant
functor from this category to the category of $\mathbb{R}$\h algebras.
\end{example}

Let now $\mathcal{C}_1$ and $\mathcal{C}_2$ be two categories and $F,F\colon
\mathcal{C}_1\Longrightarrow\mathcal{C}_2$ be two (say, covariant) functors.
Assume that for any object $O$ of $\mathcal{C}_1$ a morphism $T_O\colon
F(O)\to G(O)$ is given. Then the correspondence $O\mapsto T_O$ is called
a \emph{natural transformation}\index{natural transformation of functors}
of functors $F$ and $G$ if the diagram
\begin{diagram}
F(O)&\rto^{F(\vf)}&F(O^\prime)\\
\dto^{T_O}&&\dto_{T_{O^\prime}}\\
G(O)&\rto^{G(\vf)}&G(O^\prime)\\
\end{diagram}
is commutative for all objects $O,O^\prime$ and all morphisms $\vf\in
\mathcal{M}or(O,O^\prime)$. For example, the isomorphism
$\Hom_A(P\otimes_AQ,R)\simeq\Hom_A(P,\Hom(Q,R))$ gives a natural
transformation of the functors $R\mapsto\Hom_A(P\otimes_AQ,R)$ and
$R\mapsto\Hom_A(P,\Hom(Q,R))$. We shall see other examples of natural
transformations in next lectures.

\chapter*{Lecture 3}
Using a geometrical analogy now, we shall start to construct differential
calculus in the category $\mathcal{M}od(A)$.

Let $U$ be a domain in $\mathbb{R}^n$. Let us fix a point $x\in U$; a
tangent vector $\xi _{x}$ in $x$ is a map
\[
\xi_x\colon C^\infty(U)\to\mathbb{R}
\]
which is linear and such that for all $f,g\in C^\infty(U)$ one has
\begin{equation}
\xi_x(fg)=\xi_x(f)g(x)+f(x)\xi_x(g).  \label{Leibnitz}
\end{equation}
Suppose we have a tangent vector at any point of the domain. Then we can
consider the mapping
\[
\xi\colon C^\infty(U)\to C^\infty(U)
\]
such that $\xi(f)(x)\defeq\xi_x(f)$ smoothly depending on $x$.
Then \eqref{Leibnitz} acquires the from
\begin{equation}
\xi(fg)=\xi(f)g+f\xi(g).  \label{Leib}
\end{equation}
It is easy to check that for all $f,g\in C^\infty(U)$ and $\al\in
\mathbb{R}$ one has also
\begin{enumerate}
\item  $\xi (fg)=\xi (f)g+f\xi (g)$,
\item  $\xi (\al f)=\al \xi (f)$.
\end{enumerate}
Thus, smooth family of tangent vectors is a $\mathbb{R}$\h linear map from
the ring of smooth functions to itself that satisfies the Leibniz rule
\eqref{Leib}.

Now let us try to generalize this concept. Let $K$ be a commutative unitary
ring and $A$ be a $K$\h algebra. A map $X\colon A\to A$ is called a
\emph{derivation},\index{derivation of an algebra} if it is $K$\h linear and
satisfies the Leibniz rule, that is
\begin{enumerate}
\item  $X(\kappa a)=\kappa X(a)$;
\item  $X(a+b)=X(a)+X(b)$;
\item  $X(ab)=X(a)b+bX(a)$.
\end{enumerate}
for all $a,b\in A$ and $W\kappa \in K$. The set of all derivations will be
denoted by $D(A)$. The classical geometric case corresponds to $K=\mathbb{R}$
and $A=C^\infty(\mathbb{R}^n)$.

For any $X,Y\in D(A)$ and $a,b\in A$, we can set
\begin{align*}
(X+Y)(a)&\defeq X(a)+Y(a)\\
(aX)(b)&\defeq aX(b),
\end{align*}
and it is easy to check that $X+Y$ and $aX$ are derivations as well. So the
set of derivations $D(A)$ in an $A$-module. In fact it is more than just a
module. For any $X,Y\in D(A)$ we can consider the compositions $X\circ Y$ and
$Y\circ X$ and define
\begin{equation}
[X,Y]\defeq X\circ Y-Y\circ X.\label{Liebrack}
\end{equation}
It is easy to check that it is a derivation; it is called the \emph{Lie
bracket} of $X $ and $Y$. One can show that for all $X,Y,Z\in D(A)$ and
$\kappa\in K$ the following identities hold
\begin{enumerate}
\item  $[X,Y]+[Y,X]=0$;
\item  $[X+Y,Z]=[X,Z]+[Y,Z]$;
\item  $[\kappa X,Y]=\kappa [X,Y]$;
\item  $[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0$.
\end{enumerate}
Thus, $D(A)$ is a Lie algebra over $K$.

Consider two domains $W\sbs\mathbb{R}^n$ and $V\sbs\mathbb{R}^m$ and
let $F\colon W\to V$ be a smooth mapping. Then this mapping generates a ring
homomorphism $F^*\colon C^\infty(V)\to C^\infty(W)$
which is defined by
\begin{equation}
(F^*(f))(b)=f(F(b)).  \label{pull back}
\end{equation}
So every function $f$ on $V$ is pulled back to a smooth function on $W$ in
this way. Note that $F^*$ is also a homomorphism of $\mathbb{R}$\h algebras.

We can generalize this concept. Let us denote $A=C^\infty(V)$ and $B=C^\infty
(W)$. Having a mapping $F\colon W\to V$, we can construct $F^*$ and so,
having a derivation $\xi$ in $A$ we can define a mapping $\overline{\xi}
\colon A\to B$ such that for all $f\in A$ one has
\[
\overline{\xi}(f)\defeq F^*(\xi(f)).
\]
One can check that $\overline{\xi}$ satisfies the Leibniz rule, that is
\[
\overline{\xi}(fg)=\overline{\xi}(f)F^*(g)+F^*(f)\overline{\xi}(g)
\]
for any $f,g\in A$ (action by $F^*$ on $B$ determines an $A$\h algebra
structure in $B$).

Let $A$ be a $K$-algebra and $P$ be an $A$-module. We can define
\begin{equation}
\label{D(P)}
D(P)=\{f\in\Hom_K(A,P)\mid f(ab)=af(b)+bf(a)\,\forall a,b\in A\}.
\end{equation}
If we have such an $f\in D(P)$ and $a\in A$, then
\[
(af)(b)\defeq a(f(b))
\]
and it is easy to check that $af\in D(P)$, so $D(P)$ is an $A$-module.
However it is not a Lie algebra, since there is no natural way to define a
commutator.

On the other hand, if we have two $A$-modules $P,Q$, an $A$-module
homomorphism $\vf\colon P\to Q$ and a derivation with values in $P$, we
can consider the composition $D(\vf)\defeq\vf\circ f\colon A\to Q$.
It is $K$\h linear and it satisfies the same identities, so $\vf\circ f
\in D(Q)$. Then to any module $P$ we put into correspondence the module
$D(P)$. Moreover, having two homomorphisms $P\xrightarrow{\vf}Q
\xrightarrow{\psi}R$, it results that
\[
D(\psi\circ\vf)=D(\psi)\circ D(\vf),
\]
so $D(\cdot)$ a covariant functor. We say that elements of $D(P)$ are
\emph{$P$-valued derivation} of $A$.\index{derivation}

Fix a commutative ring $A$ which is an algebra over a ring $K$. Let $P$ be an
$A$\h module and $X$ be a derivation, that is $X\in\Hom_K(A,P)$ and
\begin{equation}
X(ab)=aX(b)+bX(a).  \label{genLeib}
\end{equation}
for any $a,b\in A$. Any element $X$ of $D(P)$ can be multiplied by an element
of $A$ in two ways, from the left, that is
\begin{equation}
(aX)(b)\defeq aX(b)  \label{leftD}
\end{equation}
or from the right
\begin{equation}
(a^+X)(b)\defeq X(ab).  \label{rightD}
\end{equation}
Now, $(a^+X)(b)=X(ab)=aX(b)+X(a)b=(aX)(b)+X(a)b$, so we can write
\[
(a^+X-aX)(b)=X(a)b
\]
and this suggests a way to define a commutator
\begin{equation}
[a,X](b)\defeq X(a)b.  \label{comm}
\end{equation}
Now let $a$ be fixed; then if we have another element $\overline{a}$, we can
easily check that for all $b$ one has $[\overline{a},[a,X]](b)=0$, that is
\begin{equation}
[\overline{a},[a,X]]=0.  \label{doublcomm}
\end{equation}

Let now $X\colon A\to P$ be such that $[ \overline{a},[a,X]] =0$ for all $a,
\overline{a}\in A$. Then the difference $X-X(1)$ satisfies \eqref{genLeib}
provided $X$ itself satisfies \eqref{doublcomm}. Then, if we put $X-X(1)=
\overline{X}$, then $X=\overline{X}+f$,
where $\overline{X}$ is a derivation and $f=X(1)$ is a constant. We conclude
that any element satisfying \eqref{doublcomm} is the sum of a derivation and a
constant. Standard first order differential operators (on $\mathbb R$) are of
this form,
\[
\De=g\pd{}{x}+f.
\]

Let $A$ be a $K$-algebra and $P,Q$ be $A$-modules. Then we can consider the
bimodule $\Hom_K(P,Q)$. Let $\De\in\Hom_K(P,Q)$ and consider
\[
[a,\De]=a^+\De -\De a=\De a-a\De.
\]
This difference belongs to $\Hom_K(P,Q)$ so we can take $\overline{a}\in A$
and consider $[\overline{a},[a,\De]]$. We say that $\De $ is a
\emph{first order differential operator} acting from $P$ to $Q$ if for all
$a,\overline{a}\in A$ one has
\[
[\overline{a},[ a,\De ]] =0. \label{diffopk}
\]

Consider two $A$-modules $P,Q$ and a homomorphism $\De\in\Hom_K(P,Q)$. We
say that $\De$ is a \emph{differential operator}\index{differential
operator} of order $\leq k$ if for all $a_0,a_1,\dots,a_k\in A$ it results
\[
[a_0,[a_1,\dots[a_k,\De ]\dots]]=0.
\]
Let us denote the set of all such maps as $\widetilde{\Diff}_k(P,Q)$, $k\in
\mathbb{N}$. What are the structures living in this set? If we have two
differential operators $\De ,\na \in \widetilde{\Diff}_k(P,Q)$, then it
is easy to check that $\De+\na\in\widetilde{\Diff}_k(P,Q)$, so
$\widetilde{\Diff}_k(P,Q)$ is an Abelian group. Moreover, given $a\in A$, it
results that
\[
[a_k,a\De]=a[a_k,\De]
\]
and so $\widetilde{\Diff}_k(P,Q)$ is also an $A$-module, which will be
denoted by $\Diff_k(P,Q)$. If we consider the right multiplication, we get
the second module structure denoted by $\Diff_k^+(P,Q)$, and finally
$\Diff_k^{(+)}(P,Q)$ will denote the bimodule structure.

Consider three modules and two linear maps $P\xrightarrow{\De}Q
\xrightarrow{\na}R$.
If you take $a\in A$ and consider $[a,\na\circ\De]$, you will have
\[
[a,\na\circ\De]=[a,\na]\circ\De+\na\circ[a,\De ].
\]
Let $\De$ be a $K$-linear map. You can prove by induction or directly from
the definition that
\[
\underbrace{[a,[a,\dots[a,\De]\dots]]}_{k+1\ \mathrm{times}}=
\sum_{i=0}^{k+1}(-1)^{k+1-i}\binom{k+1}{i}a^i\na a^{k+1-i}.
\]
Consider $f\in C^\infty(\mathbb{R})$ and $x_0\in\mathbb{R}$. Then that $f$
can be represented in the form
\[
f(x)=f(x_0)+(x-x_0)f^\prime(x_0)+\dots+\frac{(x-x_0)^{k}}{k!}f^{(k)}(x_0)+
(x-x_0)^{k+1}g(x)
\]
with $g\in C^\infty(\mathbb{R})$. Now let $\De\colon C^\infty(\mathbb{R})
\to C^\infty(\mathbb{R})$ satisfy the algebraic definition of a
differential operator of order $\leq k$ and apply $\De $ to the function
$f$. Because of linearity, we have
\begin{multline*}
\De (f)=f(x_0)\De(1)+f^\prime(x_0)\De(x-x_0)+\dotsb\\
\dotsb+\frac{f^{(k)}(x_0)}{k!}\De\left[(x-x_0)^{k}\right]+
\De \left[(x-x_0)^{k+1}g(x)\right].
\end{multline*}
If we compute this expression at $x=x_0$ it suffices to notice that the last
term in the right-hand side vanishes and so the value of $\De(f)$ at the
point $x_0$ is determined by the set of coefficients $\De_0=\De (1)$,
$\De _1=\De(x-x_0),\dots,\De_k=\De(x-x_0)^k$ and
\[
\De =\De_0+\De_1\od{}{x}+\dots+\frac{\De_k}{k!}\od{^k}{x^k}.
\]
A similar proof is valid for arbitrary differential operators $\De
\colon C^\infty(\mathbb{R}^n)\to C^\infty(\mathbb{R}^n)$.
So from the algebraic definition in the case of smooth functions we obtain
the usual expression and, vice versa, if we have
\[
\De=\sum\De_\si\pd{^\si}{x^\si},
\]
where $\De_\si\in C^\infty(\mathbb{R})$, this $\De$ satisfies our
algebraic definition of a differential operator. Thus we can think of
differential operators as of purely algebraic objects.

Consider the bimodule $\Diff_k^{(+)}(P,Q)$. If we have a differential
operator of order $k$, i.e.,
\[
[a_0,[a_1,\dots[a_k,\De]\dots]]=0,
\]
then we can add an arbitrary $a^\prime$ and obtain
\[
[a^\prime,[a_0,[a_1,\dots[a_k,\De]\dots]]] =0,
\]
so $\De$ is a differential operator of order $k+1$ as well and
consequently
\[
\Diff_k^{(+)}(P,Q)\hookrightarrow\Diff_{k+1}^{(+)}(P,Q).
\]
Operators of zero order are those which satisfy $[a_0,\De]=0$ for all $a_0
\in A$, that is $a_0\De =\De a_0$
and so elements of $\Diff_0(P,Q)$ are just $A$\h homomorphisms and the two
module structures coincide in this case.

Having the sequence of embeddings
\[
\Hom_A(P,Q)=\Diff_0(P,Q)\hookrightarrow\dots\hookrightarrow
\Diff_{k}^{(+)}(P,Q)\hookrightarrow\Diff_{k+1}^{(+)}(P,Q)
\hookrightarrow\dotsb,
\]
we can define a new object
\[
\Diff_*^{(+)}(P,Q)\defeq\bigcup_{k\geq 0}\Diff_k^{(+)}(P,Q).
\]
It is a bimodule consisting of all differential operators of any finite
order.

Let us fix a module $P$. Then, having a homomorphism $f\colon Q\to
Q^\prime$ and an operator $\De\in\Diff_k^{(+)}(P,Q),$ we can consider the
composition $P\xrightarrow{\De}Q\xrightarrow{f}Q^\prime$
and associate to any differential operator $\De\in\Diff_k^{(+)}(P,Q)$ the
operator $f\circ\De\in\Diff_k^{(+)}(P,Q^\prime)$. This map, which depends
of $P$ and on $f$, will be denoted by
\[
\Diff_k^{(+)}(P,f)\colon\De\in\Diff_k^{(+)}(P,Q)\mapsto f\circ\De\in
\Diff_{k}^{(+)}(P,Q^\prime),
\]
and it defines a functor from the category of $A$\h modules to the category
of $A$\h bimodules.

Fix $Q$ now. Then, given a homomorphism $f\colon P\to P^\prime$, to any
operator $\De\in\Diff_k^{(+)}(P^\prime,Q)$ we can associate the operator
$P\xrightarrow{f}P^\prime\xrightarrow{\De}Q$.
If $P\xrightarrow{f}P^\prime\xrightarrow{g}P^{\prime\prime}$ are two
homomorphisms, then
\[
\Diff_k^{(+)}(f\circ g,Q)=\Diff_k^{(+)}(g,Q)\circ\Diff_k^{(+)}(f,Q)
\]
and so it is a contravariant functor.

Consider the modules $\Diff_k(P,Q)$ and $\Diff_k^{+}(P,Q).$ They are
isomorphic as Abelian groups, but they are different as far as module
structures are concerned. Then we can consider the identical mappings
\begin{align*}
\Diff_{k}(P,Q)&\xrightarrow{\id^+}\Diff_{k}^{+}(P,Q)\\
\intertext{and}
\Diff_{k}(P,Q)&\xleftarrow{\id_+}\Diff_{k}^{+}(P,Q).
\end{align*}
The mappings $\id_+$ and $\id^+$ are differential operators of order $\leq
k$, but there compositions $\id_+\circ\id^+$ and $\id^+\circ\id_+$ are
operators of zero order (identical homomorphisms).

What happens in general if we consider the compositions of differential
operators $P\xrightarrow{\De}Q\xrightarrow{\na}R$ with $\De\in
\Diff_k^{(+)}(P,Q)$ and $\na\in\Diff_l^{(+)}(Q,R)$? Generally speaking,
$\na\circ\De\in\Hom_K(P,R)$. Let us prove that it is a differential
operator. To this end, take $a\in A$ and consider the identity
\[
[a,\na\circ\De]=[a,\na]\circ\De+\na\circ[a,\De ].
\]
Hence, for $m=k+l$ one has
\begin{multline*}
[a_0,[a_1,\dots[a_m,\na\circ\De]\dots]]\\
=\sum[a_{i_0},[a_{i_1},\dots
[a_{i_s},\na ]\dots]]\circ[a_{j_0},[a_{j_1},\dots[a_{j_t},\De]\dots]].
\end{multline*}
The whole number of $a$'s is $k+l+1$; then either in the first term the
number of commutators is $\geq l$ or in the second one it is $\geq k$. In
any case, one of the two components vanishes and hence $\na\circ\De\in
\Diff_{k+l}^{(+)}(P,R)$, just as in the standard case.

Consider a particular case $P=Q$. Taking the composition of two differential
operators from $P$ to itself, we obtain a new differential operator from $P$
to itself. Composition is obviously associative so $\Diff_*^{(+)}(P,P)$ is an
associative algebra with respect to composition (but not commutative!).

To deal with differential operators from $A$ (as $A$-module over itself) to
an $A$\h module $Q$, we shall use the notation
\[
\Diff_k^{(+)}(Q)\defeq\Diff_k^{(+)}(A,Q).
\]
Derivations with values in $Q$ are first order differential operators so we
have $D(Q)\hookrightarrow\Diff_1^{(+)}(Q)$. Note that the embedding
$D(Q)\hookrightarrow\Diff_1^+(Q)$ ia a first order monomorphic differential
operator, while $D(Q)\hookrightarrow\Diff_1(Q)$ is an $A$\h module
monomorphism.

Let us consider the sequence $0\to D(Q)\to\Diff_1(Q)$.
If we take the quotient of $\Diff_1(Q)$ by the image, we obtain $Q$ (we
are killing the differential part), so we have the exact sequence of
homomorphisms
\begin{equation}
\label{firstexact}
0\to D(Q)\to\Diff_1(Q)\to Q\to 0.
\end{equation}
On the other hand, $\Diff_0^{(+)}(Q)\hookrightarrow\Diff_1^{(+)}(Q)$ and,
since we established that $\Diff_0^{(+)}(Q)\equiv\Diff_0^{(+)}(A,Q)=
\Hom_A(A,Q)=Q$, we have the exact sequence of differential operators
\begin{equation}
\label{secondexact}
0\to Q\to\Diff_1^+(Q)\to D(Q)\to 0.
\end{equation}
From these two sequences, \eqref{firstexact} and \eqref{secondexact}, we get
two important theories: \emph{Spencer cohomology} and \emph{algebraic model
of Hamiltonian mechanics}.


\chapter*{Lecture 4}
In the previous lecture we established that
\begin{align}
Q &\mapsto\Diff_l^{(+)}(P,Q),  \label{QDiff} \\
P &\mapsto\Diff_l^{(+)}(P,Q)  \label{PDiff}
\end{align}
are covariant and contravariant functors from $\mathcal{M}od(A)$ to the
category of $A$\h bimodules.

Let us fix the second module structure and consider the mapping
\[
\rD_l\colon\Diff_l^+Q\to Q
\]
defined by
\begin{equation}
\label{deltal}
\rD_l(\De )\defeq\De(1),\quad\De\in\Diff_l^+Q.
\end{equation}
Then $\rD_l$ is a differential operator of order $l$ (note that in the
first module structure it is a homomorphism). In fact, for any $a\in A$
one has $\rD_l(a^+\De)=\De(a)$.

Consider an arbitrary differential operator of order $\De\in
\Diff_l^+(P,Q)$ and let us establish a correspondence between this module
and the module $\Hom_A(P,\Diff_l^+Q)$. Let us construct a mapping $\vf
\colon A\to Q$ by setting
\[
(\vf(\De)(p))(a)=\De(ap),\quad a\in A,p\in P.
\]
One can easily check that it is a homomorphism and, moreover, it is an
isomorphism. In fact if $f\in\Hom_A(P,\Diff_l^+Q)$ then the inverse mapping
is $f\mapsto\rD_l\circ f$. This means that for any
differential operator $\De$ of order $l$ acting from $P$ to $Q,$ there
exists a homomorphism $\vf(\De )$ such that the diagram
\begin{diagram}
P&&\rto^{\De}&&Q\\
&\rdto_{{\vf}(\De)}&&\ruto_{\rD_l}&\\
&&\Diff_l^+Q&&
\end{diagram}
is commutative.

In fact, we established that the functors
\[
P\mapsto\Diff_l^+(P,Q)\text{ and }P\mapsto\Hom_A(P,\Diff_l^+Q)
\]
act in the same way, they are identical to each other. In category theory
this means that the functor $\Diff_l^+(\cdot,Q)$ is \emph{representable}.
\index{representable functor} The module $\Diff_l^+Q$ is called the
\emph{representative object}\index{representative object} for this functor.

Consider now the operator $\rD_l\colon\Diff_l^+(Q)\to Q$. Then we can also
consider the composition
\[
\Diff_s^+(\Diff_l^+(Q))\xrightarrow{\rD_s}\Diff_l^+(Q)\xrightarrow{\rD_l}Q,
\]
which is a differential operator of order $l+s$. Hence, by the universal
property of the operator $\Diff_{l+s}^+Q\xrightarrow{\rD_{l+s}}Q$ there
exists a unique homomorphism $\vf(\rD_l\circ\rD_s)\defeq c_{l,s}$ such
that the following diagram
\begin{diagram}
\Diff_s^+(\Diff_l^+Q)&\rto^{\rD_s}&\Diff_l^+Q\\
\dto^{c_{l,s}}&&\dto_{\rD_{l}}\\
\Diff_{s+l}^+Q&\rto^{\rD_{s+l}}&Q\\
\end{diagram}
is commutative. The mapping $c_{l,s}$ is called the \emph{universal
composition}.\index{universal composition} Universal composition is
\emph{associative} in the sense that the diagram
\begin{diagram}
\Diff_l^+(\Diff_s^+(\Diff_m^+Q))&\rto^{c_{l,s}}&\Diff_{l+s}^+(\Diff_m^+Q)\\
\dto^{\Diff_l^+(c_{s,m})}&&\dto_{c_{l+s,m}}\\
\Diff_l^+(\Diff_{s+m}^+Q)&\rto^{c_{l,s+m}}&\Diff_{s+l+m}^+Q\\
\end{diagram}
is commutative.

Note also that the universal composition is related to any module $Q$ and,
strictly speaking, we must write $c_{l,s}=c_{l,s}(Q)$. Let $\vf\colon Q
\to Q^\prime$ be a homomorphism. Then the diagram
\begin{diagram}
\Diff_l^+(\Diff_s^+Q)&\rto^{c_{l,s}(Q)}&\Diff_{l+s}^+Q\\
\dto^{\Diff_l^+(c_{s,m})}&&\dto_{\Diff_{l+s}^+(\vf)}\\
\Diff_l^+(\Diff_s^+\vf)&\rto^{c_{l,s}(Q^\prime)}&\Diff_{l+s}^+Q^\prime\\
\end{diagram}
is commutative. It means that $c_{l,s}\colon\Diff_l^+\Diff_s^+\Rightarrow
\Diff_{l+s}^+$ is  a natural transformation of functors.

%\chapter*{Lecture 5}

Now consider the functor
\[
Q\mapsto\Diff_l(P,Q)
\]
for a fixed $P$ and let us try to find the representative object for it. We
need to find something, $?(P)$, which possesses the property
\[
\Diff_l(P,Q)\simeq\Hom_A(?(P),Q).
\]
Consider the tensor product $A\otimes_KP$ and let us introduce two $A$\h
module structures into this object. Namely, we set
\[
b(a\otimes p)=(ba)\otimes p,\qquad b_+(a\otimes p)=a\otimes (bp).
\]
In such a way we obtain bimodule since, given $b,b^\prime$, we have
\[
b( b_+^\prime(a\otimes p))=b_+^\prime(b(a\otimes p)).
\]

Let us measure the difference between these two multiplications. If $\te
\in A\otimes_KP$, we set
\[
[a,\te]\defeq a\te -a_+\te.
\]
Taking arbitrary elements $a_0,a_1,\dots a_l\in A$, we can consider the
iterated commutator
\[
[a_0,[a_1,\dots[a_l,\te]\dots]]\in A\otimes_KP
\]
and generate, for a fixed $l$, the submodule
\[
\mu_l(P)\defeq\{\,[a_0,[a_1,\dots[a_l,\te]]]\,\}
\]
taking all $a_i\in A$ and $\te\in A\otimes_KP$. The quotient module
\[
\CJ^l(P)=A\otimes_KP/\mu_l(P).
\]
is called the \emph{module of $l$-jets}.\index{module of jets}

Let us take an element $a\otimes p\in A\otimes_KP$ and denote its coset in
$\CJ^l(P)$ by $[a\otimes p]$. Having an element $p\in P$, we can consider the
element $1\otimes p$ and the coset $[1\otimes p]$ thus defining the mapping
\[
j_l\colon P\to\CJ^l(P),\qquad p\mapsto j_l(p)\defeq[1\otimes p].
\]
The element $j_l(p)$ is called the \emph{$l$-jet} of $p$.\index{jet of an
element}

The set $\CJ^l(P)$ is an $A$-module with respect to multiplication
$b[a\otimes p]\defeq[ba\otimes p]$. Note that $\CJ^0(P)=P$. It can be easily
proved that $j_l$ is a differential operator of order $l$.
\begin{theorem}\label{jet-uni}
Let $\De\colon P\to Q $ be a differential operator of order $\leq l$. Then
there exists a uniquely defined homomorphism $\psi (\De )$ such that
the diagram
\begin{diagram}
P&&\rto^{\De}&&Q\\
&\rdto_{j_{l}}&&\ruto_{\psi(\De)}&\\
&&J^{l}(P)&&\\
\end{diagram}
is commutative.
\end{theorem}

From this theorem we obtain the isomorphism
\[
\Diff_l(P,Q)\simeq\Hom_A(\CJ^l(P),Q)
\]
which means that the module $\CJ^l(P)$ is the representative object for the
functor $\Diff_l(P,\cdot)$.

Consider the operator $j_l$. Being of order $\leq l$, it is also of order
$\leq l+1$ and consequently we have the commutative diagram
\begin{diagram}
\CJ^{l}(P)&\lto^{\nu_{l+1,l}}&\CJ^{l+1}(P)\\
\uto^{j_l}&\ruto&^{j_{l+1}}\\
P&&\\
\end{diagram}
The mapping $\nu_{l+1,l}$ is dual to the embedding $\Diff_l(P,Q)
\hookrightarrow\Diff_{l+1}(P,Q)$ and, in fact, is an epimorphism. So, we have
the sequence of mappings
\begin{equation}
\label{chain}
P\equiv\CJ^0(P)\leftarrow\CJ^1(P)\leftarrow\dots\leftarrow\CJ^l(P)\leftarrow
\CJ^{l+1}(P)\leftarrow\dotsb
\end{equation}
The kernel of the projection $\nu_{l+1,l}$ coincides with the quotient
$\mu _l(P)/\mu_{l+1}(P)$ and thus we have the exact short sequence of
modules
\[
0\to\mu_l(P)/\mu_{l+1}(P)\to\CJ^{l+1}(P)\xrightarrow{\nu_{l+1,l}}\CJ^l(P)
\to 0.
\]

Now consider \eqref{chain} and a sequence of elements $\te_l$, $l=0,1,
\dotsb$, such that $\te_l\in\CJ^l(P)$ and $\nu_{l+1,l}(\te_{l+1})=
\te_l$. These sequences may be added to each other and multiplied by
elements $a\in A$ component\h wise. So they form an $A$\h module which is
denoted by $\CJ^\infty(P)$ and called the \emph{module of infinite jets}.
\index{module of infinite jets} If we know the sequence as a whole, we,
in particular, know its $l$-th term and thus get the mapping
\[
\nu_{\infty,l}\colon\CJ^\infty(P)\to\CJ^l(P).
\]
From our definition it follows also that we can construct the diagram
\begin{diagram}
J^\infty(P)&&\\
\dto^{\nu_{\infty,l}}&\rdto^{\nu_{\infty,l+1}}&\\
J^l(P)&\lto^{\nu_{l+1,l}}&J^{l+1}(P)\\
\end{diagram}
and it is commutative.

In particular, if $p\in P$, then the sequence $\{j_l(p)\}_{l\geq0}$, is a
element of $\CJ^\infty(P)$. It is denoted by $j_\infty(p)$ and is called th
\emph{infinite jet}\index{infinite jet of an element} of the element $p$.
Obviously, $\nu_{\infty,l}(j_\infty(p))=j_l(p)$. Formally speaking, the
mapping $j_\infty\colon P\to\CJ^\infty(P)$ is not a differential operator,
that is there always exist a sequence $a_0,\dots,a_l$ of an arbitrary high
length such that
\[
[a_0,[a_1,\dots[a_l,j_\infty]\dots]]\neq 0,
\]
but if we restrict ourselves to a finite part, it becomes a differential
operator.

The correspondence $P\mapsto\CJ^l(P)$ is a covariant functor. In fact, if we
consider a homomorphism $f\colon P\to Q$, then the composition $f\circ j_l$ is
a differential operator of order $l$, so there exists a unique homomorphism
from $\CJ^l(P)$ to $\CJ^l(Q),$ denoted by $J^{l}(f),$ such that the diagram
\begin{diagram}
P&\rto^{j_l}&\CJ^l(P)\\
\dto^{f}&&\dto_{\CJ^l(f)}\\
Q&\rto^{j_l}&\CJ^l(Q)\\
\end{diagram}
is commutative.
Due to uniqueness, we have
\[
\CJ^l(f\circ g)=\CJ^l(f)\circ\CJ^l(g)
\]
whenever it makes sense.

Now consider a module $P$ and for any $l$ the operator $j_l\colon P\to
\CJ^l(P)$. Since $\CJ^l(P)$ is an $A$\h module as well, we can take the
composition
\[
P\xrightarrow{j_l}\CJ^{l}(P)\xrightarrow{j_s}\CJ^s(\CJ^l(P)),
\]
which is a differential operator of order $l+s.$ By the universal property,
we can construct the commutative diagram
\begin{diagram}
P&\rto^{j_{l+s}}&\CJ^{l+s}(P)\\
\dto^{j_l}&&\dto_{c^{l,s}}\\
\CJ^l(P)&\rto^{j_s}&\CJ^s(\CJ^l(P))\\
\end{diagram}
and this mapping $c^{l,s}$ is dual to the universal composition map $c_{l,s}$.
The homomorphisms $c^{l,s}$ and $c_{l,s}$ are adjoint in categorical sense
and $c^{l,s}$ is called the \emph{universal co-composition}. \index{universal
co-composition} This operation possesses the \emph{co-associativity}
property expressed by commutativity of the diagram
\begin{diagram}
\CJ^{l+s+m}(P)&\rto^{c^{s,l+m}}&\CJ^s(\CJ^{l+m}(P))\\
\dto^{c^{l+s,m}}&&\dto_{\CJ^s(c^{l,m})}\\
\CJ^{s+l}(\CJ^m(P))&\rto^{c^{s+l,m}}&\CJ^s(\CJ^l(\CJ^m(P)))\\
\end{diagram}

Consider now the case $l=1$ and take the module $\CJ^1(A)$. Then we have
the mapping
\[
i_1\colon A\to\CJ^1(A)
\]
defined by $i_1(a)\defeq[a\otimes 1]$. It is a homomorphism, so its image is
a submodule. Let us set
\begin{equation}\label{1forms}
\La^1(A)\defeq\CJ^1(A)/\im i_1.
\end{equation}
Then we obtain the exact sequence of modules
\[
0\to A\xrightarrow{i_1}\CJ^1(A)\xrightarrow{\pi}\La^1(A)\to 0,
\]
where $\pi$ is the natural projection. The module $\La^1(A)$ is called
the \emph{module of $1$-forms}\index{module of $1$-forms} of algebra $A$.
The composition
\[
d\equiv\pi\circ j_1\colon A\to\La^1(A)
\]
is called the first \emph{de~Rham differential}\index{de~Rham differential}
and is a derivation of $A$ with values in $\La^1(A)$. It is easy to check
that $\La^1(A)$ is generated by the elements $d_a$, $a\in A$, with the
relations
\[
d_{ab}=ad_b+bd_a,\qquad d_{\al a+\be b}=\al d_a+\be d_b,
\]
where $a,b\in A$, $\al,\be\in K$.


%\chapter*{Lecture 5}

We continue to deal with differential forms. An immediate consequence of
Theorem \ref{jet-uni} is
\begin{theorem}\label{thm-lam1}
For any derivation $X\colon A\to P$ there exists a unique homomorphism
$\psi(X)\colon\La^1(A)$ such that the diagram
\begin{diagram}
A&\rto^{d}&\La^1(A)\\
\dto^{X}&\ldto_{\psi(X)}&\\
P&&\\
\end{diagram}
is commutative.
\end{theorem}
Hence, we have the isomorphism
\[
\Hom_A(\La^1(A),P)=D(P),
\]
and the module of $1$-forms is the representative object of the covariant
functor $D(\cdot)$.

Now let us define the module $\La^l(A)$ of differential $l$-forms as the
 $l$-th exterior power of $\La^1(A)$ (see Example \ref{wedge-prod}),
that is
\[
\La^l(A)=\bigwedge^l\La^1(A)
\]
(when $A$ is fixed, we shall skip it as argument and use a simpler notation
$\La^l=\La^l(A)$). Thus, we have a series of modules
\[
\La^0=A,\ \La^1,\ \La^2,\dots,\La^l,\dots
\]

The module $\La^l$ is generated by the elements $a_0,da_1\wedge\,da_2
\wedge\dots\wedge\,da_l$ and we set by definition
\begin{equation}\label{d}
d(a_0\,da_1\wedge\,da_2\wedge\dots\wedge\,da_l)=da_0\wedge\,da_1\wedge\,da_2
\wedge\dots\wedge\,da_l.
\end{equation}
We take \eqref{d} for a definition of $d$ and thus obtain the sequence
of first order differential operators
\begin{equation}
\label{chain2}
0\to A\xrightarrow{d}\La^1\xrightarrow{d}\dots\xrightarrow{d}
\La^{l-1}\xrightarrow{d}\La^l\xrightarrow{d}\dotsb
\end{equation}
We call the mapping $d$ the \emph{de~Rham differentials}\index{de~Rham
differential} and the sequence \eqref{chain2} is called the \emph{de~Rham
complex} of the algebra $A$.\index{de~Rham complex}
\begin{remark}
This is a ``bad" definition of the de~Rham complex. A ``good" one refers
to rather complicated categorial construction, which are beyond the scope
of these lectures.
\end{remark}
The basic properties of the de~Rham differential are as follows:
\begin{enumerate}
\item As it was already mentioned, it is a first order differential
operator.
\item It differentiates the wedge product, i.e.,
\begin{equation}
\label{d-Leib}
d(\om\wedge\te)=d(\om)\wedge\te+(-1)^l\om\wedge\,d(\te)
\end{equation}
where $l$ is the degree of $\om$.
\item $d\circ d=0$
\end{enumerate}
From the last property it follows that
\[
\im d\sbs \ker d
\]
at all terms in the chain \eqref{chain}. Hence, we can introduce the
$K$\h modules
\[
H^l(A)\defeq\ker d/\im d,
\]
which are called the \emph{de~Rham cohomology}\index{de~Rham cohomology} of
the algebra $A$.
\begin{remark}
Note that if you have an arbitrary sequence of homomorphisms
\[
P_0\xrightarrow{\partial_0}P_1\xrightarrow{\partial_1}\dots\to
P_{l-1}\xrightarrow{\partial_l}P_{l}\xrightarrow{\partial_{l+1}}\dotsb
\]
with the property $\partial_{l+1}\circ\partial_l=0$, you can realize the
same construction. Such sequences are called \emph{complexes}\index{complex}
of modules and the modules of the form $\ker\partial_{l+1}/\im\partial_l$ are
called \emph{homologies}\index{homology} (or \emph{cohomologies})
\index{cohomology} associated to these complexes.
\end{remark}

Now consider the composition $j_s\circ d\colon\La^{l-1}\to
\CJ^s(\La^l)$. It is a differential operator of order $s+1$. Due to the
universal property of operators $j_s$, we obtain the following commutative
diagram
\begin{diagram}
\dotsb&\rto^{}&\La^{l-1}&\rto^{d}&\La^l&\rto^{d}&\La^{l+1}&\rto^{}&\dotsb\\
&&\dto^{j_{s+1}}&&\dto^{j_s}&&\dto^{j_{s-1}}&&\\
\dotsb&\rto^{}&\CJ^{s+1}(\La^{l-1})&\rto^{\mathcal{S}}&\CJ^s(\La^l)&\rto^{\mathcal{S}}&\CJ^{s-1}(\La^{l+1})&\rto^{}&\dotsb\\
\end{diagram}
A simple exercise is to prove that the sequence
\[
0\to A\xrightarrow{j_k}\CJ^k(A)\xrightarrow{\mathcal{S}}\CJ^{k-1}(\La^1)
\to\dots\CJ^1(\La^{k-1})\xrightarrow{\mathcal{S}}\La^k\to 0
\]
is a complex. It is called the \emph{Spencer complex} of $A$ and its
cohomologies are called the \emph{Spencer cohomologies}.\index{Spencer
complex}\index{Spencer cohomologies}


%\chapter*{Lecture 5}

We have established already that the module $\La^1$ is the representative
object for the functor $D(\cdot)$, i.e., $D(P)=\Hom_A(\La^1,P)$. Let us
now try to understand what functors are represented by the modules
$\La^l$. Let us set  $\Hom_A(\La^l,P)\defeq D_l(P)$ and start with
the case $l=2$. Thus,
\[
D_2(P)=\Hom_A(\La^2,P).
\]
Note that there exists a natural homomorphism from this module to the module
$\Hom_A(\La^1,\Hom_A(\La^1,P))$; let us denote it by $\eta$. To
construct this $\eta$, we set
\[
[\eta(f)](\om)(\te)\defeq f(\om\wedge\te),
\]
where $f\in\Hom_A(\La^2,P)$, $\om,\te\in\La^1$. Hence, $D_2(P)$
lies in $D_1(D_1(P))$. Consequently, we can take an element $\na\in
D_2(P)$ and consider $\na(a)$, obtaining a new derivation lying in
$D_1(P)$. Obviously,
\[
\na(a)(b)=-\na (b)(a).
\]
and thus we have
\[
D_2(P)=\{\na\colon A\to D_1(P)\mid \na(a,b)=-\na(b,a)\}.
\]
In a similar way, for an arbitrary $l$, we have
\[
\Hom_A(\La ^{l},P)\to\Hom_A\Big(\La^1,
\underbrace{\Hom_A(\La^{l-1},P)}_{D_l(P)}\Big),
\]
which means that we have a mapping from $D_l(P)$ to $D_1(D_{l-1}(P))$. It
means that any element $\na\in D_l(P)$ can be evaluated at $a_1,\dots,
a_l\in A$ and by simple induction on $l$ we obtain that $\na\colon A\times
\dots\times A\to P$ lies in $D_l(P)$ if and only if
\begin{enumerate}
\item It is a derivation with respect to all arguments,
\begin{multline*}
\na (a_1,\dots,a_{i-1},ab,a_{i+1},\dots,a_{l})=
a\na(a_1,\dots,a_{i-1},b,a_{i+1},\dots,a_l)\\
+b\na(a_1,\dots,a_{i-1},a,a_{i+1},\dots ,a_l)
\end{multline*}
\item It is skew\h symmetric,
\[
\na(a_1,\dots,a_l)=(-1)^{|\si|}\na(a_{\si(1)},\dots,
a_{\si (l)}),
\]
where $|\si|$ denotes the parity.
\end{enumerate}

So we have two ways to describe the module $D_{l}(P)$: the first one is to
identify $D_l(P)$ with $\Hom_A(\La^l,P)$ thus obtaining the coupling
$\langle\na,\om\rangle\in P;$ the second one is to describe $D_l(P)$
as skew\h symmetric derivations of the form $A\times\dots\times A\to P$. The
relation between the two interpretation is
\[
\na(a_1,\dots,a_l)=\langle\na,da_1\wedge\dots\wedge da_l\rangle .
\]

Let us now define inductively the \emph{wedge product}\index{wedge product
in polyderivations} operation
\[
\wedge\colon D_l(A)\otimes D_s(P)\to D_{l+s}(P).
\]
When $l=s=0$, we have $D_0(A)=A$, $D_0(P)=P$ and we set $a\wedge p=ap$.
Now, by induction on $l+s$ we set for $\De\in D_l(A)$ and $\na\in
D_s(P)$
\begin{equation}\label{wedge-d}
(\De\wedge\na)(a)=\De\wedge\na(a)+(-1)^{|\na|}\De(a)
\wedge\na.
\end{equation}
We formulate the final result in the following
\begin{proposition}\label{prop-wedge}
Let $A$ be a commutative $K$-algebra and $P$ be an $A$-module. Then\textup{:}
\begin{enumerate}
\item The wedge product $\wedge\colon D_l(A)\otimes D_s(P)\to D_{l+s}(P)$
is well defined by~\eqref{wedge-d}.
\item The module $D_*(A)=\bigoplus_{l\geq0}D_l(A)$ is a graded associative
commutative algebra with respect to the wedge product, i.e.,
\[
\De\wedge\na=(-1)^{|\De|\cdot|\na|}\na\wedge\De.
\]
\item The module $D_*(P)=\bigoplus_{l\geq0}D_l(P)$ is a graded module over
$D_*(A)$.
\end{enumerate}
\end{proposition}
Consequently, the correspondence $P\mapsto D_*(P)$ is a covariant functor from
the category of all $A$\h modules to the category of all graded $D_*(A)$\h
modules.

We shall now define a coupling between the modules $D_1(A)$ and $\La^l$
(the \emph{inner product})\index{inner product} Namely, if $\om\in
\La^l$ and $X\in D_1(A)$, then for any $\na\in D_{l-1}(P)$ we set
\[
\langle\na,X\inner\om\rangle=(-1)^{|\na|}\langle X\wedge\na,
\om \rangle .
\]
Thus, the inner product $\inner$ is the adjoint to the wedge product
$\wedge$ of polyderivations. We shall also use the notation $i_X\om$
instead of $X\inner\om$. The main properties of the inner product
are formulated in the following
\begin{proposition}\label{prop-int}
For any $X,Y\in D(A)$ and $\om,\te\in\La^*$ one has\textup{:}
\begin{enumerate}
\item  The mapping $i_X\colon\La^l\to\La^{l-1}$ is $A$\h linear.
\item  The action of $\inner$ is anticommutative, i.e.,
\[
i_X\circ i_Y=-i_Y\circ i_X
\]
\textup{(}in particular $i_X\circ i_X=0$\textup{)}.
\item  For all $a\in A$ one has $i_X(da)=X(a)$.
\item  Finally, it is a derivation of $\La^*=\bigoplus_{l\geq0}\La^l$,
i.e.,
\[
i_X(\om\wedge\te)=(i_X\om)\wedge\te +(-1)^{|\om|}\om\wedge
i_X\te.
\]
\end{enumerate}
\end{proposition}
Now, we shall write the action of the de~Rham differential in terms of
inner product. Introduce the notation
\[
\om(X_1,\dots,X_l)=i_{X_l}(\dots(i_{X_1}\om)\dots)\in A.
\]
Let $\om $ be a form of degree $l$; then
\begin{multline*}
(d\om)(X_1,\dots,X_{l+1})=\sum_{\al=1}^{l+1}(-1)^{\al-1}
X_\al\om(X_1,\dots,\widehat{X}_\al,\dots,X_{l+1})\\
+\sum_{\al<\be}(-1)^{\al+\be}\om([X_\al,X_\be],X_1,\dots,
\widehat{X}_\al,\dots\widehat{X}_\be,\dots,X_{l+1}),
\end{multline*}
and it is the usual form of the de~Rham differential.

Finally, we define the \emph{Lie derivative}\index{Lie derivative} $L_X$ of a
form $\om\in\La^l$ with respect to a derivation $X\in D_1(A)$. Namely,
we set
\[
L_X(\om)\defeq d(L_X\om)+L_X(d\om)=[d,L_X]\om,
\]
It is easy to check that $L_X\colon\La^l\to\La^l$ satisfies the
following properties:
\begin{proposition}\label{prop-Lied}
For any $X,Y\in D_1(A)$, $\om,\te\in\La^*$, and $a\in A$ one has
\begin{enumerate}
\item  $L_X\colon\La^*\to\La^*$ is a $K$-linear mapping.
\item  $L_X$ is a derivation of $\La^*$, i.e.,
\[
L_X(\om\wedge\te)=(L_X\om)\wedge\te+\om\wedge(L_X\te).
\]
\item  $L_X(d\om)=d(L_X\om)$, i.e., $[L_X,d]=0$.
\item  $L_{aX}(\om)=aL_{X}\om+da\wedge L_X\om$.
\item  $L_{[X,Y]}=[L_X,L_Y]$.
\item  $[L_X,i_Y]=i_{[X,Y]}$.
\end{enumerate}
\end{proposition}

\chapter*{Lecture 5}

Our last step is to construct a bridge between algebraic theory of linear
differential operators and geometry of nonlinear differential equations.
We start with general considerations.

Consider an abstract mechanical system. To observe properties of the system,
you need the notion of observables and these observables are real functions
on the manifold $M$ of positions. But the set of initial positions is not
sufficient to describe system dynamics: we must also know, for example,
momenta or velocities of the points, which adds to the picture cotangent or
tangent bundle of $M$. If particles under consideration possess additional
properties (e.g., spin),we should also add new variables corresponding
to these observables.

Let us begin with simplest situation. Suppose we have a smooth manifold $M$
(which can be understood as a set of states of a physical system). What are
smooth functions on this manifold? Of course, there is an analytical
definition, but how can it be understood in a geometric way? A function is a
smooth map from $M$ to $\mathbb{R}$. But one can consider the Cartesian
product $M\times\mathbb{R}$ and the canonical projection $\pi\colon M\times
\mathbb{R}\to M$. Then to say that $f$ is a smooth function on $M$ is the
same as to say that $f\colon M\to M\times\mathbb{R}$ is a smooth mapping
satisfying
\begin{equation}\label{sec-proj}
\pi \circ f=\id_M.
\end{equation}
This definition can be generalized. Consider an additional manifold $N$ and
the Cartesian product $M\times N$. An $N$\h valued function $f$ is a smooth
mapping from $M$ to $M\times N$ satisfying the same identity \eqref{sec-proj}.
For instance if $N=\mathbb{R}^s$, we obtain the concept of smooth vector\h
valued functions on our manifold $M$, i.e., functions represented by vectors
$(f_1,\dots,f_s)$, whose components are just smooth real\h valued functions.

Consider now a more complicated situation. Let $M$ be a smooth manifold. Then,
by definition, it is locally diffeomorphic to $\mathbb{R}^n$. If a point
moves in the neighborhood $\CU\sbs M$ and $x=(x_1,\dots,x_n)$ are local
coordinates of this point in $\CU$, the velocity is just a tangent vector
$v_1\pd{}{x_1}+\dots+v_n\pd{}{x_n}$. So at one point all possible velocities
are described by points of $\mathbb{R}^n$ with coordinates $(v_1,\dots,v_n)$.
If we consider two intersecting local charts $\CU$ and $\CV$ and a point in
their intersection, we can express the velocities in both local coordinates.
Then we have the sets of all possible velocities $\mathbb{R}^n\times\CU$
and $\mathbb{R}^n\times\CV$ and these two objects, if $\CU\cap\CV$ is not
void, are related to each other by the Jacobi matrix. In this way, spaces of
``local velocities" in different neighborhoods are glued together to a new
manifold $TM$ called the \emph{tangent manifold}. There is a natural
projection $\pi\colon TM\to M$ and locally the projection looks as the
projection $\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n$. Any smooth
mapping $f\colon M\to TM$ satisfying \eqref{sec-proj} may be understood as
a \emph{field of velocities}\index{field of velocities} on~$M$.

This construction is a particular case of the following one. Consider three
smooth manifolds $M$, $O$, and $F$ let $\pi\colon O\to M$ be a surjective
smooth mapping. Suppose any point $x\in M$ possesses a neighborhood $\CU_x=
\CU$ such that there exists a diffeomorphism $\vf=\vf_{\CU}\colon\pi^{-1}(\CU_x)\to
\CU_x\times F$ satisfying $\pi\left|_{\pi^{-1}(\CU_x)}\right.=\pr_{\CU_x}
\circ\vf$, where $\pr_{\CU_x}\colon\CU_x\times F\to\CU_x$ is the
projection to the first component. In this case we say that this $\pi$ is a
\emph{locally trivial fiber bundle}\index{locally trivial fiber bundle} over
$M$, $M$ is called the \emph{base}, $F$ the \emph{fiber}, and $O$ is called
the \emph{total space} of the bundle $\pi$.\index{base of a fiber bundle}
\index{fiber of a fiber bundle}\index{total space of a fiber bundle}
The set $F_x=\pi^{-1}(x)$ is called the \emph{fiber over the point} $x$. From
the definition it follows that for any $x\in M$ there exists a diffeomorphism
\begin{equation}\label{fiber-dif}
\vf_x\colon F_x\to F.
\end{equation}
\begin{example}\label{triv-bund}
For any two manifolds, $M$ and $F$, the projection $M\times F\to M$ is
a fiber bundle.
\end{example}
\begin{example}
The M\"obius band is the total space of the bundle over the circle $S^1$
with the fiber $\mathbb{R}$.
\end{example}
\begin{example}
The natural projection $TM\to M$ is a fiber bundle with the fiber
$\mathbb{R}^n$, $n=\dim M$.
\end{example}
\begin{example}
The natural projection $T^*M\to M$, where $T^*M$ is the cotangent manifold,
is a fiber bundle with the fiber $\mathbb{R}^n$, $n=\dim M$.
\end{example}

Now fix $M$ and consider two fiber bundles $\pi$ and $\pi^\prime$. A
\emph{morphism} of $\pi$ to $\pi^\prime$ is a smooth mapping $\vf\colon
O\to O^\prime$ such that the diagram
\begin{diagram}[LaTeXeqno]\label{bund}
O&&\rto^{\vf}&&O^\prime\\
&\rdto_{\pi}&&\ldto_{\pi^\prime}&\\
&&M&&\\
\end{diagram}
is commutative.\index{morphism of fiber bundles}
A morphism of bundles which is a diffeomorphism is called an
\emph{isomorphism}.\index{isomorphism of fiber bundles} A bundle isomorphic
the Cartesian product $M\times F$ is called \emph{trivial}.\index{trivial
fiber bundle} If the mapping $\vf$is an embedding, we say that $\pi$ is
a \emph{subbundle}\index{subbundle of a fiber bundle} in $\pi^\prime$.

Fiber bundles over a manifold $M$ together with their morphisms form a
category. To any bundle $\pi\colon O\to M$ one can put into correspondence
the set smooth mappings
\[
\Ga(\pi)\defeq\{f\colon M\to O\mid \pi\circ f=\id_M\}.
\]
The elements of $\Ga(\pi)$ are called \emph{sections}.\index{section of
a bundle} The correspondence $\pi\mapsto\Ga(\pi)$ is a functor from the
category of fiber bundles to the category of sets. In fact, if $\vf$ is
a morphism of the form \eqref{bund}, we can set $\Ga(\vf)\colon f\mapsto
\vf\circ f\in\Ga(\pi^\prime)$ for any $f\in\Ga(\pi)$.

We can generalize definition of a morphism in a following way. Let $\pi\colon
O\to M$ and $\pi^\prime\colon O^\prime\to M^\prime$ be two fiber bundles and
$\psi\colon M\to M^\prime$ be a smooth mapping. Then $\vf\colon O\to
O^\prime$ is a morphism,\index{morphism of fiber bundles} if the diagram
\begin{diagram}
O&\rto^{\vf}&O^\prime\\
\dto^{\pi}&&\dto_{\pi^\prime}\\
M&\rto^{\psi}&M^\prime\\
\end{diagram}
is commutative.

Now let us consider a smooth mapping $g\colon M^\prime\to M$ and the subset
in Cartesian product $M^\prime\times O$, which consists of pairs defined by
\[
g^*(O)=\{( x^\prime,o)\mid g(x^\prime)=\pi(o)\}
\]
It is a submanifold in $M^\prime\times O$ and can be mapped both to $O$ and
to $M^\prime$:
\begin{diagram}
(x^\prime,o)\in g^*(O)&\rto^{\pi^*(g)}&O\ni o\\
\dto^{g^*(\pi)}&&\dto_{\pi}\\
x^\prime\in M^\prime&\rto^{g}&M\\
\end{diagram}
By definition of the space $g^*O$, it is a commutative diagram and the
mapping $g^*(\pi)\colon g^*(O)\to M^\prime$ is a fiber bundle. Is called
the \emph{pullback} (or \emph{induced bundle}).\index{pullback}\index{induced
bundle} In particular, if $g$ is an embedding, the bundle $g^*(\pi)$ is
called the \emph{restriction} of $\pi$ to the submanifold $M^\prime$.

When $g$ is a fiber bundle too, the diagonal arrow is a fiber bundle as well:
\begin{diagram}
g^*(O)&\rto^{}&\ O\\
\dto^{g^*(\pi)}&\rdto^{W}&\dto_{\pi}\\
M^\prime&\rto^{g}&\ M\\
\end{diagram}
and is called the \emph{Whitney product} of bundles $g$ and $\pi$.
\index{Whitney product} Note that if $F$ is the fiber of the bundle $\pi$ and
$G$ is that of the bundle $g$, then the fiber of Whitney product will be
$F\times G$.

In what follows, we shall deal with a special type of fiber bundle.
\begin{definition}\label{vect-bund}
A fiber bundle $\pi\colon O\to M$ is called a \emph{vector bundle}, if
\begin{enumerate}
\item The fiber $F$ is a vector space.
\item  For any $\CU,\CV\sbs M$ such that $\vf_{\CU}\colon\pi^{-1}(\CU)
\to\CU\times F$ and $\vf_{\CV}\colon\pi^{-1}(\CV)\to\CV\times F$ the
mappings
\[
\vf_{\CU}\Big|_{\pi^{-1}(\CU\cap\CV)}\circ
\vf_{\CV}^{-1}\Big|_{(\CU\cap\CV)\times F}\colon
\CU\cap\CV)\times F\to(\CU\cap\CV)\times F
\]
are fiber-wise linear.
\end{enumerate}
\end{definition}
We shall consider the case when $F$ is an $\mathbb R$\h vector space.
The dimension of $F$ as a vector space is called the \emph{dimension} of a
fiber bundle.\index{dimension of a fiber bundle}

Let $\pi\colon O\to M$ and $\pi^\prime\colon O^\prime\to M$ be two vector
bundles and $\vf\colon O\to O^\prime$ be a morphism. We say that $\vf$ is
a \emph{morphism of vector bundles}\index{morphism of vector bundles} if it
is fiber-wise linear, i.e., $\vf(a+b)=\vf(a)+\vf(b)$ for any
$a,b\in F_x$ and $x\in M$. Obviously, vector bundles over $M$ together with
vector bundle morphisms form a category which we denote by ${\CV}ect(M)$.

Consider now two sections $f,g\in\Ga(\pi)$ of a vector bundle $\pi$.
Then, due to vector space structure in the fibers, we can set
\[
(f+g)(x)\defeq f(x)+g(x),\qquad x\in M.
\]
Moreover, by the same reason, for any smooth function $r\in C^\infty(M)$ we
set
\[
(rf)(x)\defeq r(x)f(x),\qquad x\in M.
\]
Now, we see that the correspondence $\Ga\colon\pi\mapsto\Ga(\pi)$ is a
functor from the category ${\CV}ect(M)$ to the category
${\CM}od(C^\infty(M))$ of $C^\infty(M)$\h modules. An amazing fact is that
for any finite\h dimensional manifold $M$ and finite\h dimensional vector
bundle $\pi$ the module $\Ga(\pi)$ is projective with finite number of
generators. Moreover, any such a module over the algebra $C^\infty(M)$ can
be realized as $\Ga(\pi)$ for some vector bundle $\pi$! A detailed
discussion of these matters can be found in \cite{JetN}.

%\chapter*{Lecture 8}
Let us fix now a smooth manifold $M$ and establish relations between
calculus in the category ${\CM}od(C^\infty(M))$ and analytical constructions
over $M$ needed for the geometrical theory of differential equations.

Consider two vector bundles, $\pi$ and $\xi$, over $M$. Then a \emph{linear
differential operator} of order $k$ acting from $\pi$ to $\xi$\index{linear
differential operator in fiber bundles} is an element of the module
$\Diff_k(\Ga(\pi),\Ga(\xi))$. Note that this definition is in full
agreement with the classical one. In a similar way, $\pi$\h valued
derivation on $M$ are derivations $C^\infty(M)\to\Ga(\pi)$ and, in
particular, $C^\infty(M)$\h valued derivations coincide with vector fields
on $M$, or with section of the tangent bundle $TM$. We shall meet no problem
in carrying over the theory of functors $\Diff_k^{(+)}$, $D_i$, etc. to the
geometric situation.

The problem will arise when passing to representative objects. Consider
a projective module $P$ with finite number of generators over $C^\infty(M)$.
Then, as we already know, a locally trivial vector bundle $\pi$ over $M$
corresponds to $P$ such that $\Ga(\pi)=P$. Consider a point $x\in M$
and the ideal $\mu_x\defeq\{f\in C^\infty(M)\mid f(x)=0\}$. The value of
an element $p\in P$, as of a section of $\pi$, at $x$ is its coset $p(x)$ in
the quotient $C^\infty(M)/\mu_x$, the latter being a finite\h dimensional
space over the field $C^\infty(M)=\mathbb R$. Projective modules over
$C^\infty(M)$ are characterized by the fact that their elements are
completely determined by the values $p(x)$ at all points $x\in M$. It follows
from Theorem \ref{proj-free} and the fact that the bundles corresponding to
free modules are trivial.
\begin{example}\label{ghost}
Let $M=\mathbb R$ and $A=C^\infty(M)$. The module $\La^1(A)$
contains, for example, the element $d(e^x)-e^x\,dx$. This element vanishes
at all points of $\mathbb R$, but is nontrivial in $\La^1(A)$. It is
easy to find similar examples for the modules of jets. Thus, in
representative objects a sort of ``ghost'' elements arise, which are not
observed geometrically.
\end{example}
This example shows that the category of projective $C^\infty(M)$\h
modules is not closed with respect to construction of representative objects
for the basic functors of differential calculus and a straightforward
approach does not lead to geometrical constructions parallel to these
objects.

To kill the ``ghost'' similar to the one arising in Example \ref{ghost}, we
shall apply the following procedure. For a smooth manifold $M$ and a
$C^\infty(M)$\h module $P$, let us set $\mathfrak{M}(P)\defeq
\bigcap_{x\in M}\mu_x\cdot P$. Let us say that a module is \emph{geometrical},
\index{geometrical module} if $\mathfrak{M}(P)=0$. The we have the
\emph{functor of geometrization}\index{geometrization functor} $\mathfrak{G}
\colon P\mapsto P/\mathfrak{M}(P)$ acting from the category of all
$C^\infty(M)$\h modules to the category of geometrical modules over the
same algebra.
\begin{proposition}\label{geo-rep}
The functors $D_i(\cdot)$ and $\Diff_k(P,\cdot)$ are representable in the
category of geometrical modules over $C^\infty(M)$. Moreover, for any
geometrical module $Q$ one has
\[
D_i(Q)=\Hom_{C^\infty(M)}(\mathfrak{G}(\La^i(C^\infty(M)),Q)
\]
and
\[
\Diff_k(P,Q)=\Hom_{C^\infty(M)}(\mathfrak{G}(\CJ^k(P)),Q).
\]
\end{proposition}
\begin{corollary}\label{geo-forms}
For any smooth manifold $M$ one has
\[
\La^i(M)=\mathfrak{G}(\La^i(C^\infty(M)).
\]
\end{corollary}
To formulate a result similar to Corollary \ref{geo-forms}, we need a new
geometrical construction.

Let $\pi\colon E\to M$ be locally trivial vector bundle. Consider a point
$\te\in E$, $\pi(\te)=x\in M$, and a section $f\in\Ga(\pi)$ whose
graph passes through $\te$: $f(x)=\te$. Denote by $[f]_x^k$ the set
of sections whose graphs are tangent to the graph of $f$ at $\te$ with
order $k$. This class is called the \emph{$k$-jet} of $f$ at $x$.\index{jet
of a section at a point} Obviously, it is completely determined by the Taylor
expansion of $f$ at $x$ of order $k$. Let us choose a coordinate neighborhood
$\CU\ni x$ in such a way that the bundle $\pi$ becomes trivial over $\CU$.
Let $x_1,\dots,x_n$ be local coordinates in $\CU$ and $e_1,\dots,e_m$ be
a basis in the fiber of $\pi$. Then any section is locally represented as
$f=f^1e_1+\dots+f^me_m$, where $f^j=f^j(x_1,\dots,x_n)$ are smooth functions
on $\CU$. Hence, the class $[f]_x^k$ is completely determined by the
values of all partial derivatives $\pd{^{|\si|}f^j}{x_\si}$ at $x$,
$j=1,\dots,m$, $0\leq|\si|\leq k$. Denote the number of these partial
derivatives by $N=N(n,m,k)$.

Now, let us take an atlas $\big\{\CU_\al\big\}$ in $M$ consisting of
coordinate neighborhoods of the above described type. Consider the set
\begin{equation}\label{Jk}
J^k(\pi)\defeq\{[f]_x^k\mid x\in M,\,f\in\Ga(\pi)\}.
\end{equation}
It is covered by the subsets
\begin{equation}\label{atlas}
\tilde{\CU}_\al\defeq\{[f]_x^k\mid x\in \CU_\al,\,f\in\Ga(\pi)\}.
\end{equation}
Define coordinate functions $x_i$ and $p_\si^j$ in $\CU_\al$ by
\begin{equation}\label{coordJk}
x_i([f]_x^k)=x_i,\quad p_\si^j([f]_x^k)=\pd{^{|\si|}f^j}{x_\si}(x).
\end{equation}
Then functions \eqref{coordJk} establish a one-to-one correspondence between
$\tilde{\CU}_\al$ and the space $\CU_\al\times\mathbb{R}^N$.
\begin{proposition}\label{Jet-mf}
The system of sets $\tilde{\CU}_\al$ together with coordinate functions
\eqref{coordJk} constitute an atlas in the set $J^k(\pi)$. Thus $J^k(\pi)$
becomes a smooth manifold. Moreover, the projection $\pi_k\colon J^k(\pi)\to
M$, $\pi_k([f]_x^k)\defeq x$, is a smooth locally trivial vector bundle.
\end{proposition}
\begin{definition}\label{defJk}
The manifold $J^k(\pi)$ is called the \emph{manifold of $k$-jets}
\index\emph{manifold of jets} for the bundle $\pi$. The bundle $\pi_k$ is
called the \emph{bundle of $k$-jets}\index\emph{bundle of jets} for the
bundle $\pi$.
\end{definition}
We can now formulate a statement similar to Corollary \ref{geo-forms}:
\begin{proposition}\label{geo-jets}
Let $M$ be a smooth manifold and $\pi\colon E\to M$ be a locally trivial
vector bundle. Let $P=\Ga(\pi)$. Then
\[
\Ga(\pi_k)=\mathfrak{G}(\CJ^k(P)).
\]
\end{proposition}
Manifolds of jets is a natural environment for geometrical theory of
differential equations. This theory will be discussed in the next lecture
course.

\chapter*{Exercises}
\begin{exercise}
Let $A$ be a ring. Prove that $0\cdot a=a\cdot 0=0$ for all $a\in A$.
\end{exercise}

\begin{exercise}\label{exc2.1}
Let $A$ be a set and $2^E$ the set of all subsets of $A$. Two natural
binary operations exist in $2^E$: the union $\cup$ and the intersection
$\cap$ operations.
\begin{enumerate}
\item Show that $2^E$ is not a ring with respect to these operations.
\item Prove that if we define
\begin{align}
a\cdot b &=a\cap b  \label{prod} \\
a+b &=(\overline{a}\cap b)\cup (a\cap \overline{b})  \label{sum}
\end{align}
where the `` bar'' denotes complement in $E$, then $2^E$ is a commutative
unitary ring.
\end{enumerate}
\end{exercise}

\begin{exercise}\label{exc:charf}
Consider the set $\mathbb{Z}_{2}=\{0,1\}$. Check that it is a commutative
ring with the following rules of summation
\[
0+0=1,\quad 1+0=1,\quad 1+1=0
\]
and the usual ones for multiplication.
\end{exercise}

\small
Let now $E$ be a set. The \emph{characteristic functions}
\index{characteristic functions of a subset} of a subset
$a\in 2^E$ is defined as
\[
\chi_a(x)=\begin{cases}
1& x\in a \\
0& x\notin a
\end{cases}
\]
Since there exists a one-to-one correspondence between $2^E$ and the
set of characteristic functions $X(E)$, we can introduce the structure
of a ring in the latter by setting
\[
\chi_a+\chi_b=\chi_{a+b},\quad \chi_a\chi_b=\chi_{a\cdot b},
\]
where $a,b\in 2^E$ and sum and product of sets are defined by \eqref{prod}
and \eqref{sum}.
\normalsize
\begin{exercise}
Prove that the correspondence $0\mapsto\chi_\emptyset$, $1\mapsto\chi_E$
is a homomorphism of $\mathbb{Z}_2$ to $X(E)$ and thus $X(E)$ is a
$\mathbb{Z}_2$\h algebra.
\end{exercise}
\begin{exercise}
Let $A$ and $B$ be two rings and $f\colon A\to B$ be a ring homomorphism.
Prove that $\ker f$ is a two-sided ideal in $A$.
\end{exercise}

\begin{exercise}
Let $A$ be a ring and $I$ be a two-side ideal of $A$. Consider the quotient
$A/I$ and prove that the operations
\[
[a]+[b]\defeq[a+b],\quad [a]\cdot[b]\defeq[ab]
\]
determine a well\h defined ring structure in $A/I$. What structure will
be carried by $A/I$, if $I$ is a left or right ideal?
\end{exercise}

\begin{exercise}\label{exc2.6}
Let $A=C^\infty(\mathbb{R})$ and $x\in\mathbb{R}$. Consider the set
\[
\mu_x=\left\{ f\in A\mid f(x_)=0\right\}.
\]
Prove that $\mu_x$ is an ideal and that $A/\mu_x$ is isomorphic to
$\mathbb{R}$.
\end{exercise}

\begin{exercise}
Let again $A=C^\infty(\mathbb{R})$ and $x,y\in\mathbb{R}$ be two different
points. Consider the set
\[
\mu_{xy}=\left\{ f\in A\mid f(x)=f(y)=0\right\} .
\]
Prove that $\mu_{xy}$ is an ideal and that $A/\mu_{xy}$, as an $\mathbb{R}$\h
vector space is isomorphic to $\mathbb{R}^2$ and describe the ring structure
in terms of a basis in $\mathbb{R}^2$.
\end{exercise}

\small
This problem has an interesting prolongation. Of course, if we set $x=y$,
the result will the same as in Exercise \ref{exc2.6}. But we can act
in a different way. Namely, let $d=d(x,y)$ be the distance between the
points $x$ and $y$. Then, as $d\to 0$, the ideal $I_{xy}$ becomes
\[
I=\left\{f\in C^\infty(\mathbb{R})\mid f(x)=\od{f}{x}(x)=0\right\}
\]
and you can prove that $A/I$ is still isomorphic to $\mathbb{R}^2$ as a
vector space, but sum and product will be defined as follows
\begin{align*}
(a_1,b_1)+(a_2,b_2)&=(a_1+a_2,b_1+b_2)\\
(a_1,b_1)\cdot(a_2,b_2)&=(a_1a_2,a_1b_2+a_2b_1).
\end{align*}

Now we shall need the following
\begin{definition}
Let $A$ be a ring and $I\sbs A$ an ideal. The ideal $I$ is called
\emph{principal},\index{principal ideal} if there exists an element $a\in A$
such that $I=\{aA\}$.
\end{definition}
If all ideals of $A$ are principal ideals then $A$ is called \emph{principal
ideal domain}.

We shall denote by $I_n$ the principal ideals of $\mathbb{Z}$ (that is
ideals of the form $I_n=n\mathbb{Z}$ with $n\in\mathbb{Z}$).
\normalsize
\begin{exercise}\label{exc2.8}
Prove that $\mathbb{Z}$ is a principal ideal domain.
\end{exercise}

\begin{exercise}
Let $p\in\mathbb{Z}$ Prove that $\mathbb{Z}_p\defeq\mathbb{Z}/I_p$ is a
field if and only if $p$ is prime.
\end{exercise}

\small
Let us, for example, consider the case $p=3$. Then
\[
\mathbb{Z}/3\mathbb{Z}=\{[0],[1],[2]\}
\]
and we can write the tables of sum and multiplication:
\[
\begin{array}{l|lll}
+ & 0 & 1 & 2 \\
\hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1
\end{array}
\qquad
\begin{array}{l|lll}
\cdot & 0 & 1 & 2 \\
\hline
0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 \\
2 & 0 & 2 & 1
\end{array}
\]
\normalsize
\begin{exercise}
Prove that $n\mathbb{Z}\sbs m\mathbb{Z}$ if and only if $m$ divides $n$.
\end{exercise}
\small
For an arbitrary ring $A$ and its ideals $I$ and $I'$ we can define the
ideal $I+I'$ by setting
\[
I+I'\defeq\{a+b\mid a\in I,b\in I'\}.
\]
\normalsize
\begin{exercise}
Consider the ideals $n\mathbb{Z}$ and $m\mathbb{Z}$ with $m,n\in\mathbb{Z}$.
Then $n\mathbb{Z}+m\mathbb{Z}=l\mathbb{Z}$ for some $l\in\mathbb{Z}$. Prove
that $l$ is maximal common divisor (mcd) of $m$ and $n$.
\end{exercise}

\begin{exercise}
Prove also that $n {\mathbb Z}\cap m {\mathbb Z}=s {\mathbb Z}$ with $s$
being the minimal common multiple (mcm) of $m$ and $n$.
\end{exercise}

\begin{exercise}
Let $A$ be a field. Prove that $I\sbs A$ is an ideal if and only if $I=
\{0\}$ or $I=A$. Thus, fields, and they only, have no nontrivial ideals.
\end{exercise}

\begin{exercise}\label{exc3.6}
Let $A$ be a (commutative unitary) ring. Prove $I\sbs A$ is a maximal
ideal if and only if $A/I$ is a field.
\end{exercise}

\begin{exercise}\label{exc4.1}
Prove that an ideal $I\sbs A$ is prime if and only if the quotient
$A/I$ possesses no zero divisors.
\end{exercise}

\begin{exercise}\label{exc4.2}
Prove that any maximal ideal is prime.
\end{exercise}

\begin{exercise}\label{exc4.3}
Prove that any ideal is contained in a maximal ideal.
(\emph{Hint}: Use the Zorn Lemma.)
\end{exercise}

\begin{exercise}
Let $A$ be a ring and $I\sbs A$ be its ideal. Consider the natural
projection $\pi\colon A\to A/I$, $\pi(a)=[a]$. Prove that:
\begin{enumerate}
\item if $J\sbs A/I$ is an ideal then $\pi^{-1}(J)\sbs A$ is an ideal
in $A$ containing $I$;
\item if $I'\sbs A$ is an ideal, then $\pi(J)\sbs A/I$ is an ideal
too.
\end{enumerate}
\end{exercise}

\begin{exercise}
Construct an example of projective module which is not free.
\end{exercise}
\begin{exercise}
Let $x\in\mathbb{R}$ and $\mu_x\sbs C^\infty(\mathbb{R})$ be the ideal
of functions vanishing at $x$. Prove that $\mu_x$ is a projective
$C^\infty(\mathbb{R})$\h module. Is it true for the case $\mathbb{R}^n$?
\end{exercise}

\begin{exercise}
Let $x\in S^1$ and $\mu_x\sbs C^\infty(S^1)$ be the ideal
of functions vanishing at $x$. Prove that $\mu_x$ is a projective
$C^\infty(S^1)$\h module.
\end{exercise}

\begin{exercise}
Prove that $P$ is projective if and only if the exists $P^\prime$ such that
$P^\prime\oplus P$ is a free module.
\end{exercise}

\begin{exercise}
Prove that $P$ is a projective module if and only if for any $A$\h modules
$Q$ and $R$ an exact sequence
\[
0\to Q\xrightarrow{\ i\ }R\xrightarrow{\pi}P\to 0
\]
splits, i.e., there exists a homomorphism $j\colon P\to R$ such that $\pi\circ
j=\id_P$.
\end{exercise}

\begin{exercise}
Let
\[
0\to R\to S\to T\to 0
\]
Be an exact sequence of modules. Prove that $P$ is projective if and
only if the sequence
\[
0\to\Hom_A(P,R)\to\Hom_A(P,S)\to\Hom_A(P,T)
\]
is an exact as well.
\end{exercise}

\begin{exercise}
Consider $C^\infty(\mathbb{R})$ and vector valued functions (i.e., rows
consisting of smooth functions)
\[
C^\infty(\mathbb{R})\oplus\dots\oplus C^\infty(\mathbb{R}).
\]
Compute derivations of $C^\infty(\mathbb{R})$ with values in this module.
\end{exercise}

\begin{exercise}
Let $A=C^\infty(\mathbb{R})$, $P=A\oplus A$ be the set of vector valued
functions with two components. Describe first order differential operators
acting from $P$ to itself.
\end{exercise}

\begin{exercise}\label{cross}
Consider the cross $M=\{(x,y)\mid xy=0\}\sbs\mathbb{R}^2\}$ and let us
define the algebra $A=C^\infty(M)$ by
\[
C^\infty(M)\defeq\{f\colon M\to\mathbb{R}\mid f=f^\prime|_M,\,f^\prime\in
C^\infty(\mathbb{R}^2)\}
\]
\begin{enumerate}
\item Describe all tangent vectors at the point $0$.
\item Describe all derivations $A\to A$.
\item Describe $\Diff_*(A)$. Show that this algebra is not generated by
functions $\Diff_0(A)=A$ and derivations $D(A)$ (contrary to the case of
smooth manifolds).
\end{enumerate}
\end{exercise}

\begin{exercise}
Let $A=C(\mathbb{R})$ the algebra of all continuous functions on $\mathbb{R}$.
Prove that $D(A)=0$.
\end{exercise}

\begin{exercise}
Let $A=\mathbb{R}[x]/(x^n)$. Describe $D(A)$, $\Diff_*(A)$.
\end{exercise}

\begin{exercise}
Let $A=\mathbb{Z}_m[x]/(x^n)$. Describe $D(A)$, $\Diff_*(A)$.
\end{exercise}
\small
For any commutative $K$-algebra $A$, let us introduce the quotient modules
\[
\Smbl_kA\defeq\Diff_k^{(+)}A/\Diff_{k-1}^{(+)}A
\]
and denote by $\smbl_k\De$ the coset of $\De\in\Diff_k^{(+)}A$ in
$\Smbl_kA$. This coset is called the \emph{symbol} of the operator $\De$.
\index{symbol of a differential operator}
\normalsize
\begin{exercise}
Prove that the module structures inherited by $\Smbl_kA$ from $\Diff_k^{+}A$
and from $\Diff_kA$ coincide.
\end{exercise}
\small
Let $f=\smbl_k\De$ and $g=\smbl_l\na$, $\De\in\Diff_kA$, $\na\in
\Diff_lA$. Let set $f\cdot g\defeq\smbl_{k+l}(\De\circ\na)$. Then
$\Smbl_*A=\bigoplus_{i\geq0}\Smbl_iA$ becomes an algebra, which is call
the \emph{algebra of symbols}\index{algebra of symbols} for $A$. Let us
also set $\{f,g\}\defeq\smbl_{k+l-1}(\De\circ\na-\na\circ\De)$.
\normalsize
\begin{exercise}
Prove that $\Smbl_*A$ is a commutative algebra with respect to the
multiplication $(f,g)\mapsto f\cdot g$.
\end{exercise}
\begin{exercise}
Prove that $\Smbl_*A$ is a Lie $K$-algebra with respect to the
multiplication $(f,g)\mapsto\{f,g\}$ and
\[
\{f,g\cdot h\}=\{f,g\}\cdot h+g\cdot\{f,h\}
\]
for any $f,g,h\in\Smbl_*A$.
\end{exercise}
\begin{exercise}
Write down the action of $c_{l,s}$ in terms of elements and prove
commutativity of all the diagrams above.
\end{exercise}

\begin{exercise}
Prove the associativity law for the universal composition $c_{l,s}$.
\end{exercise}

\begin{exercise}
Prove that the universal composition $c_{l,s}$ is a natural transformation
of functors.
\end{exercise}

\begin{exercise}
Let $\De\in\Diff_l^{(+)}(A)$ with $A=C^\infty\mathbb{R})$. Prove in
coordinate form that $\na=[a,\De]$ lies in $\Diff_{l-1}^+A$.
\end{exercise}

\begin{exercise}
Prove that $\id_+$ and $\id^+$ are differential operators.
\end{exercise}

\begin{exercise}
Consider the operator $\rD_l\colon\Diff_l^+A\to A$ defined in \eqref{deltal}.
Check in coordinate form that it is a differential operator of order $l$ (in
algebraic form it has already been proved) for $A=C^\infty(\mathbb{R}^n)$.
\end{exercise}

\begin{exercise}
Prove the isomorphism between the modules $\Diff_l^+(P,Q)$ and
$\Hom_A(P,\Diff_l^+Q)$.
\end{exercise}

\begin{exercise}
Let $P=Q=\left\{\binom{f_{1}}{f_{2}}\mid f_{2}\in C^\infty(\mathbb{R})\equiv
A\right\}$. Describe the homomorphism $\vf(\De)\colon P\to
\Diff^+P$ for an arbitrary differential operator $\De\colon P\to P$.
\end{exercise}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}
Prove that $\CJ^k(P)$ is the representative object for the functor
$\Diff_k(P,\cdot)$ (Theorem \ref{jet-uni}).
\end{exercise}

\begin{exercise}\label{muinmu}
Prove the embeddings $\mu_{l+1}\sbs\mu_l$ for the submodules $\mu_l\sbs
A\otimes_KP$ defining $\CJ^l(P)$.
\end{exercise}

\begin{exercise}
Prove that $\mu_l/\mu_{l+1}=\ker\nu_{l+1,l}$, where the projections
$\nu_{l+1,l}\colon\CJ^{l+1}(P)\to\CJ^l(P)$ is defined by the universal
property of $\CJ^l(P)$.
\end{exercise}

\begin{exercise}
Prove co-associativity of the co-composition $c^{k,l}\colon\CJ^{k+l}\to
\CJ^k(\CJ^l(P))$.
\end{exercise}

\begin{exercise}
Prove that the co-composition $c^{k,l}$ is a natural transformation of
functors $\CJ^{k+l}$ and $\CJ^k(\CJ^l(P))$.
\end{exercise}

\begin{exercise}
Prove that $\La^1$ is the representative object for the functor
$D_1(\cdot)$ (Theorem \ref{thm-lam1}).
\end{exercise}
\small
Let $P$ be an $A$-module and $N\sbs P$ be a \emph{subset} in $P$.
Introduce the notation
\[
D(N)\defeq\{\De\in D(P)\mid\im\De\sbs N\}.
\]
Of course, in general, $D(N)$ does not possess an $A$\h module structure.
\normalsize
\begin{exercise}
Prove that the modules $D_i(P)$ may be defined in the following inductive
way: $D_0(P)=P$, $D_1(P)=D(P)$ and
\[
D_{i+1}(P)=D\big(D_i(P)\sbs(\Diff_1^+)^i(P)\big)
\]
\end{exercise}

\begin{exercise}
Prove the basic properties inner product (Proposition \ref{prop-int}):
\begin{enumerate}
\item  The mapping $i_X\colon\La^l\to\La^{l-1}$ is $A$\h linear.
\item  The action of $\inner$ is anticommutative, i.e.,
\[
i_X\circ i_Y=-i_Y\circ i_X
\]
\textup{(}in particular $i_X\circ i_X=0$\textup{)}.
\item  For all $a\in A$ one has $i_X(da)=X(a)$.
\item  Finally, it is a derivation of $\La^*=\bigoplus_{l\geq0}\La^l$,
i.e.,
\[
i_X(\om\wedge\te)=(i_X\om)\wedge\te +(-1)^{|\om|}\om\wedge
i_X\te.
\]
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove the basic properties of the de~Rham differential $d$:
\begin{enumerate}
\item It is a first order differential operator.
\item It differentiates the wedge product, i.e.,
\[
d(\om\wedge\te)=d(\om)\wedge\te+(-1)^l\om\wedge\,d(\te)
\]
\item $d\circ d=0$
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove the formula
\begin{multline*}
d\om(X_1,\dots,X_n)=\sum_iX_i(\om(X_1,\dots,\widehat{X}_i,\dots,X_n))\\
+\sum_{i<j}(-1)^{i+j}\om([X_i,X_j],X_1,\dots,\widehat{X}_i,\dots
\widehat{X}_j,\dots,X_n).
\end{multline*}
\end{exercise}

\begin{exercise}
Prove the basic properties of Lie derivatives (Proposition~\ref{prop-Lied}):
\begin{enumerate}
\item  $L_X\colon\La^*\to\La^*$ is a $K$-linear mapping.
\item  $L_X$ is a derivation of $\La^*$, i.e.,
\[
L_X(\om\wedge\te)=(L_X\om)\wedge\te+\om\wedge(L_X\te).
\]
\item  $L_X(d\om)=d(L_X\om)$, i.e., $[L_X,d]=0$.
\item  $L_{aX}(\om)=aL_{X}\om+da\wedge L_X\om$.
\item  $L_{[X,Y]}=[L_X,L_Y]$.
\item  $[L_X,i_Y]=i_{[X,Y]}$.
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove the formula
\[
(L_X\om)(X_1,\dots,X_n)=X\om(X_1,\dots,X_n)+
\sum_{i=1}^n\om(X_1,\dots,[X,X_i],\dots,X_n).
\]
\end{exercise}
\small
A $K$-algebra $A$ is called \emph{smooth}\index{smooth algebra} if (a) $K$
is an algebra over the field $\mathbb{Q}$ of rational numbers and (b)
$\La^1(A)$ is a projective module with finite number of generators.
\normalsize

\begin{exercise}
Prove that when $A$ is a smooth algebra, then one has the isomorphism
\[
\ker\nu_{l,l-1}= S^{\otimes k}\La^1(A),
\]
where $S^{\otimes k}$ denotes the $k$-th symmetric power of a module.
\end{exercise}

\begin{exercise}
Prove that when $A$ is a smooth algebra and $P$ is a projective module,
then $\CJ^k(P)$ is projective as well.
\end{exercise}

\begin{exercise}
Prove that for smooth algebras the algebra $\Diff_*A$ is multiplicatively
generated by elements of $\Diff_1A$ (cf.~Exercise \ref{cross}).
\end{exercise}
\small
Let $A$ be a commutative ring. The set $\Spec A$ consisting of all its
prime ideals is called the \emph{spectrum}\index{spectrum of a ring} of $A$.
Let $a\in A$. Let us define
\[
U_a\defeq\{I\in\Spec A\mid a\notin I\}
\]
and take the system $\left\{U_a\right\}_{a\in A}$ for a base of open sets
in $\Spec A$.  Thus, $\Spec A$ becomes a topological space and the
corresponding topology in $\Spec A$ is called the \emph{Zarissky topology}.
\index{Zarissky topology} If $f\colon A\to B$ is an algebra homomorphism,
then for any $I\in\Spec B$ the set $f^{-1}I\sbs A$ is a prime ideal in
$A$. Thus we obtain the mapping $f^*\colon\Spec B\to\Spec A$ which is
continuous in Zarissky topology. It is easy to see that the correspondence
$A\mapsto\Spec A$ determines a contravariant functor from the category
of commutative rings to that of topological spaces.

The set $\Spec_mA$ of maximal ideals contains in $\Spec A$, but the
correspondence $A\mapsto\Spec_mA$ is not functorial. Nevertheless, we can
consider the following construction. Fix a field $\Bbbk$ and call a
homomorphism $\vf\colon A\to\Bbbk$ a \emph{$\Bbbk$-point}. Denote the
set of $\Bbbk$-points by $\Spec_\Bbbk A$. Let $a\in A$ and $U\defeq\{\vf
\in\Spec_\Bbbk A\mid\vf(a)\neq 0$. Taking $\left\{U_a\right\}_{a\in A}$
for a base of open sets, we make a topological space of $\Spec_\Bbbk A$. The
correspondence $A\mapsto\Spec_\Bbbk A$ determines a functor and there exists
natural continuous mappings $\Spec_\Bbbk A\to\Spec_m A\to\Spec A$ (the
composition being a natural transformation of functors $\Spec_\Bbbk A$ and
$\Spec A$!).
\normalsize
\begin{exercise}
Consider the polynomial algebra $A=\mathbb{R}[x]$. Describe $\Spec A$,
$\Spec_{\mathbb{R}}A$ and $\Spec_{\mathbb{C}}A$.
\end{exercise}

\begin{exercise}
Consider the ring of periodic functions
\[
A=\{f\in C^\infty(\mathbb{R})\mid f(x)=f(x+1)\,\forall x\in\mathbb{R}\}.
\]
Describe $\Spec_\mathbb{R}A$.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}^2)\mid f(x,y)=f(x,y+1)\, \forall(x,y)\in
\mathbb{R}^2\}.
\]
Prove that $\Spec_\mathbb{R}A\simeq S^1\times\mathbb{R}$.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}^2)\mid f(x+1,y)=f(x,y)=f(x,y+1)\,\forall (x,y)
\in \mathbb{R}^2\}.
\]
Prove that $\Spec_\mathbb{R}A\simeq S^1\times S^1$.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}^2)\mid f(x+1,y)=f(x,y)=f(x,-y)\, \forall (x,y)
\in\mathbb{R}^2\}.
\]
Prove that $\Spec_\mathbb{R}A$ is the Klein bottle.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}^2)\mid f(x,y)=f(x+1,-y)\, \forall (x,y)\in
\mathbb{R}^2\}.
\]
Prove that $\Spec_\mathbb{R}A$ is the M\"{o}bius band.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}3\setminus\{0\})\mid f(x,y,z)=
f(\lambda x,\lambda y,\lambda z)\,\forall\lambda >0\}.
\]
Prove that $\Spec_\mathbb{R}A\simeq S^2$.
\end{exercise}

\begin{exercise}
Let
\[
A=\{f\in C^\infty(\mathbb{R}^3\setminus\{0\})\mid f(x,y,z)=f(\lambda x,
\lambda y,\lambda z)\, \forall\lambda\neq 0\}.
\]
Prove that $\Spec_\mathbb{R}A\simeq\mathbb{RP}^2$.
\end{exercise}

\begin{exercise}
Let
\[
A=C^\infty(\mathbb{R}^3)/(x^2+y^2+z^2).
\]
Prove that $\Spec_\mathbb{R}A\simeq S^2$.
\end{exercise}

\begin{exercise}
Consider the ring of characteristic functions (see Exercise~\ref{exc:charf})
and describe its spectrum and Zarissky topology.
\end{exercise}

\begin{exercise}
Consider the ring of smooth bounded functions on $\mathbb{R}^n$
and describe its spectrum and Zarissky topology.
\end{exercise}

\begin{exercise}
Prove that functions \eqref{coordJk} on p.~\pageref{coordJk} determine a
smooth manifold structure on $J^k(\pi)$ (Proposition~\ref{Jet-mf}).
\end{exercise}

\begin{exercise}
Prove that $\pi_k\colon J^k(\pi)\to M$ is a smooth locally trivial vector
bundle.
\end{exercise}

\begin{exercise}
Let $M$ be a smooth manifold and $\tau^*\colon T^*M\to M$ be its cotangent
bundle. Prove that $\Ga(\tau^*)=\mathfrak{G}(\La^i(C^\infty(M))$,
where $\mathfrak{G}$ is the geometrization functor.
\end{exercise}

\begin{exercise}
Let $\pi\colon E\to M$ be a smooth finite-dimensional vector bundle and
$\pi_k\colon J^k(\pi)\to M$ be the bundle of its $k$-jets. Prove that
$\Ga(\pi_k)=\mathfrak{G}(\CJ^k(P))$, where $P=\Ga(\pi)$.
\end{exercise}
\backmatter
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\input{forino.ind}
\end{document}
