% -------------------------------------------------------------------------- % University of Cambridge, Centre for Mathematical Sciences (CMS), % Department of Pure Mathematics & Mathematical Statistics (DPMMS) % Part III Mathematics Essay 2002-2003 % Title: Enveloping algebras of finite dimensional nilpotent Lie algebras % Author: Dierk Philipp Fahr, Trinity Hall % Supervisor: Dr. C.J.B. Brookes % ------------------------------------------------------------------------- \documentclass[12pt,a4paper,titlepage]{article} \pagestyle{plain} \usepackage{eufrak} % -------------------------- Theorems & Definitions -------------------------- \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newcommand{\bq}{\begin{equation}} \newcommand{\eq}{\end{equation}} \newenvironment{proof}{\noindent \textnormal{\textbf{Proof:}}}{\begin{flushright}$\mathbf{Q.E.D.}$\end{flushright}} % Proofs %\newtheorem{defn}{Definition}[section] % numbering afresh %- OR ? \newtheorem{defn}[thm]{Definition} % numbering depending on the theorem \newenvironment{rem}{\noindent \textnormal{\textbf{Remark:}}}{} % Remark %\newcounter{remark} % Remarks should be counted from 1 for each section. %\newtheorem{rem}[remark]{Remark} % ------------------ Commands, Environments & Abbreviations ------------------ \newcommand{\st}{such that\ } \newcommand{\IFF}{if and only if\ } \newcommand{\vs}{vector space\ } \newcommand{\fd}{finite dimensional\ } \newcommand{\la}{Lie algebra} \newcommand{\ID}{integral domain} \newcommand{\uea}{universal enveloping algebra} \newcommand{\ea}{enveloping algebra} \newcommand{\lief}{\EuFrak{f}} \newcommand{\lieg}{\EuFrak{g}} \newcommand{\lieh}{\EuFrak{h}} \newcommand{\mbf}[1]{$\mathbf{#1}$} % bold math \newcommand{\ca}[1]{\mathcal{#1}} % written italic math \newcommand{\Ug}{\ca{U}(\lieg)} \newcommand{\ex}{\noindent \textbf{Example: }} \newcommand{\exs}{\noindent \textbf{Examples: }} \newcommand{\rems}{\noindent \textnormal{\textbf{Remarks:}}} %\newcommand{\remafresh}{\setcounter{remark}{0}} % Reset Remark counter to 1 %\newcommand{\qedproof}{\begin{flushright}$\mathbf{Q.E.D.}$\end{flushright}} %\newcommand{\qed}{\hfill \mbox{$\Box$}} % Beweisende \newcommand{\MAT}{\mathcal{M}(k)} % matrix set M(k) \newcommand{\MATn}{\mathcal{M}_n(k)} % matrix set Mn(k) \newcommand{\MATRIX}[4]{\left(\begin{array}{cc} #1 & #2 \\#3 & #4 \end{array}\right)} % A 2x2 matrix \newenvironment{numlist}[1]{\newcounter{#1}\begin{list}{\textnormal{\textbf{(\roman{#1})}}}{\usecounter{#1}}}{\end{list}} % Create list of definitions with number i, ii, etc. \begin{document} % -------------------------- TITLE & CONTENTS -------------------------- % \begin{titlepage} % \end{titlepage} \title{Enveloping algebras of finite dimensional nilpotent Lie algebras} \author{Dierk Philipp Fahr} \date{\today} \thanks{\\University of Cambridge, DPMMS\\Part III Essay 2002-03\\Supervisor: Dr.~C.J.B.~Brookes} \maketitle \pagenumbering{roman} % Start page numbering at roman i here for the table of contents. \tableofcontents \newpage \pagenumbering{arabic} % Start page numbering at 1 here. % -------------------------- INTRODUCTION -------------------------- \section{Introduction} The aim of this essay is to give an introduction to the theory of \uea s of \fd \la s. We will assume a \la\ $\lieg$ to be over a field $k$ of characteristic zero, and we will concentrate on \fd nilpotent \la s. Furthermore the approach will be one of ring-theoretic aspects rather than from a categorist's point of view. The theory of \la s was initiated by the Norwegian mathematician Marius Sophus Lie (1842-1899), who called them ``infinitesimal transformations''. H.~Weyl introduced the current terminology in 1934, when the general structure theory of complex \la s was already well developed by E.~Cartan, F.~Engel, and W.~Killing in 1888 - 1894. The \uea\ $\Ug$ of a \la\ $\lieg$ is the analogue of the usual group algebra of a group. Nowadays \ea s and their deformations are studied in a number of interrelated fields such as quantum groups or Lie superalgebras. We will consider \la s over arbitrary fields of characteristic zero. For positive characteristic, new phenomena arise and the classical methods no longer work, leading to the theory of modular \la s, which was first studied by E.~Witt in 1937.\\ In more detail, this essay will lead to the construction of the \uea\ $\Ug$ and show some of its properties. The most fundamental result is the Poincare-Birkhoff-Witt Theorem, which basically says that every \la\ $\lieg$ can be embedded in a \uea\ $\Ug$. The latter is an associative algebra, and passing from $\lieg$ to $\Ug$ has the disadvantage of introducing an infinite-dimensional algebra, but allows us to use associative methods (maximal left ideals, etc.). Further it turns out, that the ring-theoretic properties of $\Ug$ are closely related to the properties of its \emph{associated graded algebra}, which will be looked at in detail. It will be clear that in the case of an \fd abelian \la\ $\lieg$, $\Ug$ is commutative and isomorphic to a polynomial ring. We will also see that every $\lieg$-module is also a module for the \ea\ $\Ug$. Finally we will take a closer look at the ideal structure one can find in $\Ug$.\\ The essay is structured as follows: Some basic definitions and terminology of ring theory used will be briefly outlined in the first part of chapter~\ref{Ringtheorybasics}. Then the notions of \emph{filtered} and \emph{graded rings} (chapter~\ref{Filteredrings}) are introduced, as these concepts will show to be important and useful. At the end of the first chapter some examples of rings, which are closely related to \ea s, are given. The basic results needed about \emph{\la s} and their representations are dealt with in chapter~\ref{LieAlgebras}, and the terms \emph{solvable} and \emph{nilpotent} \la s explained. The next chapter~\ref{UniversalEnvelopingAlgebras} forms the core of the essay, since \emph{\uea s} are defined, examples of \ea s constructed, and the \emph{Poincare-Birkhoff-Witt Theorem}~(\ref{PBWThm}) is proved in full. As a consequence of this, some properties of the \ea s are shown in chapter~\ref{CaseNoetherian}. Finally, in chapter~\ref{IdealsinUg}, we will concentrate on ideals in $\Ug$ and, in particular, look at prime ideals.\\ The main sources of reference used for the completion of this essay were the book ``Enveloping Algebras'' by Jacques Dixmier~\cite{Dixmier} and the volume ``Noncommutative Noetherian Rings'' by J.~McConnell and J.~C.~Robson~\cite{Robson}. %- %\newpage % -------------------------- Ring theory basics -------------------------- \subsection{Ring theory basics}\label{Ringtheorybasics} %\remafresh We begin by introducing some standard notation and definitions from ring theory which are used in this essay. First of all by a ``ring'' we mean a nonzero associative ring $R$ with an identity element $1_R$ which is not necessarily commutative. A subring of a ring $R$ always contains the identity element $1_R$. We shall also differentiate between \emph{left ideals} and \emph{right ideals}. An \emph{ideal} always means a two-sided ideal in $R$. \begin{defn}\label{completeprime} $ $ % INTEGRAL DOMAIN \begin{numlist}{id} \item A nonzero ring $R$ is called an \emph{(integral) domain}, if the product of nonzero elements is always nonzero, i.e.~for all $a,b\in R,\ ab=0 \Rightarrow\, a=0$ or $b=0$. \item An ideal $I$ of a ring $R$ is \emph{completely prime} if $R/I$ is an \ID. \end{numlist} \end{defn} \ex % INTEGRAL DOMAIN The standard example of rings which fail to be \ID s arise as matrix rings. Take $R$ to be any (nontrivial) ring, and let $\ca{M}_{2}(R)$ be the set of all $2\times2$ matrices over $R$. Consider the following two matrices: \[ r=\MATRIX{0}{0}{1}{0},\ s=\MATRIX{0}{0}{0}{1},\] then clearly $rs=0$, but both $r,s\neq0$. We note, that this ring is not commutative either because $sr\neq0$.\\ % M(R) Throughout we will use the notation $\MATn$ to denote the set of all $n\times n$ matrices with entries in $k$, and $\MAT$ to be the set of all square matrices (of any size). \begin{defn}\label{primering} $ $ % FIELD \begin{numlist}{field} \item A \emph{(skew\footnote{`Skew fields' are also called `division rings'.}) field} is a nonzero (non-)commutative ring $k$, in which every nonzero element is an invertible (also called unit) of $k$. \item A ring $R$ is \emph{simple} if $R$ has precisely the two ideals $0$ and $R$. Equivalently $0$ is the unique maximal ideal of $R$. \item $R$ is called a \emph{prime ring} if for $a,b$ nonzero elements in $R$,\/ $aRb\neq0$. Equivalently, if $I,J$ are nonzero ideals in $R$, then $IJ\neq 0$. \end{numlist} \end{defn} For the description of \ea s as will be done in chapter~\ref{UniversalEnvelopingAlgebras}, we need two further definitions: \begin{itemize} \item Let $A$ be any ring and $\{x_i\, |\, i\in I\}$ be a system of independant, non-commuting indeterminantes over $A$. Then we can form the \emph{free $A$-ring} generated by $\{x_i\, |\, i\in I\}$, which we denote by $R=A\langle x_i\, |\, i\in I\rangle.$ The elements of $R$ are polynomials of the non-commuting variables $\{x_i\}$ with coefficients from $A$, which commute with each $x_i$. Note that in comparison to the usual polynomial ring $A[x_i\, |\, i\in I]$, the $x_i$ do not commute with each other. \item Suppose $S$ is a ring with a subring $R$ and that $S$ is generated, as a ring, by $R$ together with elements $\{x_i\ |\ i\in I\}$. Then $S$ is called a \emph{ring extension} of $R$ and \emph{finitely generated ring extension} if $I$ is finite. \end{itemize} Some final definitions: \begin{defn}\label{primeid} $ $ % Nilpotent ideal \begin{numlist}{nilpoten} \item An element $a$ of a ring $R$ is \emph{nilpotent} if $a^n=0$ for some $n\geq1$. \item An ideal $I$ of $R$ is \emph{nilpotent} if $I^n=0$ for some $n\geq1$. \item An ideal $I$ is called \emph{prime} if $R/I$ is a prime ring. \item The set of prime ideals of $R$ is denoted by \emph{$Spec(R)$}, the \emph{prime spectrum} of $R$ \end{numlist} \end{defn} \noindent Note that in a commutative ring every prime ideal is completely prime. It is also immediate to check that a nilpotent ideal is contained in any prime ideal.\\ % -------------------------- Filtered rings -------------------------- \subsection{Filtered \& graded rings}\label{Filteredrings} %\remafresh Here the important definitions of filtered, graded and associated graded rings are introduced. These concepts will become very useful, since often information can be obtained by passing from a ring with a ``natural'' filtration to the associated graded ring, and then translating the results back to the original ring. In particular, this is useful if the ring is filtered by finite (additive) subgroups, \st the associated graded ring is commutative. We will consider finite gradings of type $\mathbf{Z}$, where the components are indexed by integers. \begin{defn} % Filtration A \emph{$\mathbf{Z}$-filtration} (or simply a \emph{filtration}) of a ring $R$ is a sequence of additive subgroups $R_i$ in $R$, \st $$1\in R_0,\quad \ldots\subseteq R_{i-1}\subseteq R_i\subseteq R_{i+1}\subseteq \ldots\, i\in \mathbf{Z}, \mbox{ with}$$ $$ R_iR_j\subseteq R_{i+j}, \mbox{ for all } i,j\in \mathbf{Z},\ \mbox{ and }\ R=\bigcup_{i\in \mathbf{Z}} R_i.$$ The filtration is called \emph{finite} if each $R_i$ is finite. \end{defn} A similar definitions holds for a right $R$-module $M$, where $R$ is filtered by $\{R_i\}_{i \in \mathbf{Z}}$, then getting $M_iR_j\subseteq M_{i+j}$ for all $i,j\in \mathbf{Z}$, and $M=\bigcup M_i$ with $\{M_i\}_{i\in \mathbf{Z}}$ a sequence of additive subgroups forming the compatible filtration. \begin{defn} % Grading A \emph{grading} $\{R_i\}_{i\in \mathbf{Z}}$ of the ring $R$ is a sequence of additive subgroups $R_i$ of $R$ \st $$ R=\bigoplus_{i\in \mathbf{Z}}R_i\, \mbox{ and }\, R_iR_j\subseteq R_{i+j} \mbox{ for all } i,j\in \mathbf{Z}.$$ A ring $R$ with a grading $\{R_i\}$ is called \emph{graded}, and \emph{finitely graded} if each of the components $R_i$ is finite. The elements of $R_i$ are called \emph{homogeneous of degree i}. \end{defn} \begin{rem} % Grading The component $R_0$ is a subring of $R$ containing $1_R$. \end{rem} \begin{defn} % Graded Modules Let $R$ be a graded ring with grading $\{R_i\}_{i\in \mathbf{Z}}$. A right $R$-module $M$ is \emph{graded}, if there exist subspaces $M_i$ \st $$ M=\bigoplus_{i\in \mathbf{Z}}M_i \mbox{ and } M_iR_j\subseteq M_{i+j} \mbox{ for all } i,j\in \mathbf{Z}.$$ Again, if each $M_i$ is \fd then $M$ is said to be \emph{finitely graded}. The elements of $M_i$ are called \emph{homogeneous of degree i}. \end{defn} \begin{defn} % Associated graded rings Let $R$ be a filtered ring with filtration $\{R_i\}_{i\in \mathbf{Z}}$. The \emph{associated graded ring} is $$gr(R)=\bigoplus_{i\in \mathbf{Z}}R_i/R_{i-1}\ \mbox{ (as additive groups)},$$ equipped with obvious addition and multiplication given by $$ (r+R_{i-1})(s+R_{j-1})=(rs+R_{i+j-1}),\ \mbox{ for } r\in R_i,\, s\in R_j.$$ \end{defn} Note that if the associated graded ring of a filtered ring $R$ is finitely generated, then $R$ is itself finitely generated. Again a similar definition holds for a filtered right $R$-module $M$: We get $gr(M)=\bigoplus M_i/M_{i-1}$ being the graded right $gr(R)$-module via $(m+M_{j-1})(r+R_{i-1})=(mr+M_{i+j-1})$ for $r\in R_i,\, m\in M_j.$ Finally to conclude the list of definitions we have: \begin{defn}\label{standardfiltration} % standard filtration Let $R$ be a ring with a filtration $\{R_i\}_{i\in \mathbf{Z}}$, and let $M$ be a finitely generated right $R$-module by a set $A$. The \emph{standard filtration} of $M$ is the filtration $\{AR_i\}_{i\in \mathbf{Z}}$. \end{defn} % -------------------------- Weyl, tensor & symmetric Algebra -------------------------- \subsection{Weyl, tensor \& symmetric algebras}\label{Tensorsymmetric} %remafresh This section gives some interesting examples of rings, which will help us later to describe properties of the \uea s in chapter~\ref{UniversalEnvelopingAlgebras}.\\ \exs \begin{enumerate} % Weyl algebra \item If $R=k\langle x,y\rangle$, $k$ a field of characteristic zero, and $J=\{xy-yx-1\}$, write $(J)$ for the ideal generated by $J$ in $R$. Then we can form the quotient ring $R/(J)$, which is the \emph{(first) Weyl algebra} over $k$. This is denoted by $A_1(k)$, and can also be viewed as a ring of differential operators on the polynomial ring $k[y]$, or as being generated by the subring $k$ together with $x,y$ and the above relation. Higher Weyl algebras can be defined inductively: $A_n(k)=A_1(A_{n-1}(k))$. They are not commutative. In terms of generators and relations, the $n$th Weyl algebra $A_n(k)$ over $k$, is the $k$-algebra with $2n$ generators $x_1,\ldots,x_n,y_1,\ldots,y_n$ and relations $x_iy_j-y_jx_i=\delta_{ij}$, and $x_ix_j-x_jx_i=0$, $y_iy_j-y_jy_i=0$. For the $n$th Weyl algebra (working in n variables), the associated graded ring is in fact a polynomial algebras in $2n$ variables. Weyl algebras are simple domains (but not division rings) and belong to the family of rings of differential operators. \item Let $R$ be a ring, and $M$ an $R$-module. The \emph{tensor algebra} $T(M)$ is a graded $R$-algebra with $n$th graded component $T_n(M)=\underbrace{M\otimes\ldots\otimes M}_n$, and multiplication given by $ab=a\otimes b\in T_{n+m}(M)$ for $a\in T_n(M),\, b\in T_m(M)$. Thus $$T(M)=\bigoplus_{n=0}^\infty T_n(M)=R\oplus M\oplus (M\otimes M)\oplus (M\otimes M\otimes M)\ldots$$ If $V$ is a \fd \vs over a field $k$ with dimension $n$, another (easier) description of the \emph{tensor algebra} $T(V)$ over $k$ is as follows: If $\{e_1,\ldots,e_n\}$ is a $k$-basis on $V$, $T(V)$ is essentially the free $k$-algebra $k\langle e_1,\ldots,e_n\rangle$. A special quotient algebra is of interest, called the \emph{symmetric algebra} $S(V)$, defined next. % Symmetric algebra \item The \emph{symmetric algebra} $S(V)$ on $V$ is obtained by quotiening out the ideal $I$ generated by all $u\otimes v-v\otimes u\ (u,v\in V):\quad S(V)=T(V)/I$. $S(V)$ inherits the grading of $T(V)$ and the effect of factoring out $I$ is just to make the elements of $V$ commute. Thus, with the basis $\{e_1,\ldots,e_n\}$ of $V$, $S(V)$ is isomorphic to the ordinary polynomial algebra $k[e_1,\ldots,e_n]$ (with commuting $e_i$'s). \end{enumerate} The tensor algebra will play an important role in the construction of the \uea, but all of the above rings are closely related to \ea s. We will describe some aspects of how these relations look like in chapter~\ref{UniversalEnvelopingAlgebras}. For this we need the following useful definition: \begin{defn}\label{deriv} %derivation A \emph{derivation} of a $k$-algebra $R$ is a linear operator $d$ satisfying the \emph{Leibniz rule}: $D(a,b)=D(a)b+aD(b)$ for all $a,b\in R$. Given any element $a$ of a ring $R$, there is a derivation of $R$, namely $r\mapsto ra-ar=[r,a]$, which is a so-called \emph{inner derivation}, denoted by $ad(a)$. \end{defn} We finally note that it can be shown that every derivation of $A_n$ is an inner derivation. % -------------------------- Lie Algebras -------------------------- \section{Lie Algebras}\label{LieAlgebras} %remafresh This chapter presents basic material about \fd \la s and their representations needed in order to study the \uea s. We will start defining \la s and show some useful properties, also giving the definitions of \emph{solvable} and \emph{nilpotent} \la s. Throughout this chapter $k$ will denote a field of characteristic $0$. \subsection{Basic definitions}\label{Basicdefinitions} %\remafresh \begin{defn}\label{Lieg} % Lie ALgebra A \emph{\la} $\lieg$ over a field $k$ is a $k$-\vs equipped with a \emph{Lie product}, i.e.~a $k$-bilinear map $\lieg\times\lieg\rightarrow\lieg,\ (x,y)\mapsto [x,y]$, \st the following hold: \begin{numlist}{LA} \item $[x,x]=0$\, for all $x\in\lieg$. \item $[x,y]=-[y,x]$\, for all $x,y\in\lieg$. \item $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$\, for all $x,y,z\in\lieg$, called the \emph{Jacobi identity}. \end{numlist} We say that a \la\ $\lieg$ is \emph{abelian} if $[x,y]=0$ for all $x,y\in\lieg$. \end{defn} \begin{rem} % Lie bracket If $A$ is an associative $k$-algebra one can define a \la\ structure on $A$ by setting $[a,b]=ab-ba$, the so called \emph{bracket product}. A \emph{Lie algebra homomorphism} of \la s $\lieg$ into $\lieg'$ is a linear mapping $\varphi:\lieg\rightarrow\lieg'$ \st $\varphi([x,y])=[\varphi(x),\varphi(y)]$, for all $x,y\in\lieg$. \end{rem} \begin{defn} % Representation A \emph{representation} of a \la\ $\lieg$ is a \la\ homomorphism $\rho: \lieg\rightarrow A$, where $A$ is an associative algebra with the bracket product. \end{defn} % \rho: g->MATn always exists In fact, each \fd \la\ can be represented as a subalgebra of $\MATn$ for some $n$. \begin{defn} % adjoint representation The \emph{adjoint representation} of $\lieg$ is a linear transformation \st given $x\in\lieg,\ ad(x): \lieg\rightarrow\lieg$ maps $y\mapsto [x,y]$. \end{defn} For two subspaces $I,J$ of a \la\ $\lieg$, let $[I,J]\subseteq\lieg$ denote the subspace spanned by all $[x,y]$ with $x\in I, y\in J.$ Then an \emph{ideal} can be defined as a vector subspace $\ca{I}$ of $\lieg$ \st $[\lieg,\ca{I}]\subseteq\ca{I}$.\\ \rems % adjoint kernel, ideal \begin{enumerate} \item As usual the centre of $\lieg$ is the set of commuting elements in $\lieg$, i.e.~the set $\{x \in \lieg\ |\ [x,y]=0, \mbox{ for all } y\in\lieg\}$, and using the bracket product we have $xy=yx$. The centre is an ideal of $\lieg$ and this ideal is precisely the kernel of the adjoint representation of $\lieg$. \item The \emph{ideals} of $\lieg$ are the $k$-subspaces of $\lieg$ closed under all $ad(x),\ x\in\lieg$. \item If $I,J$ are ideals in $\lieg$, then $[I,J]$ is an ideal of $\lieg$, which can easily be seen by computing using the Jacobi identity. In particular, for any \la\ $\lieg$, $[\lieg,\lieg]\subseteq\lieg$ is an ideal. \end{enumerate} \begin{defn}\label{gmodule} % Module A \emph{right $\lieg$-module} is a \vs $M$ together with a $k$-bilinear map $M\times\lieg\rightarrow M$, sending $(m,x)\mapsto [m,x]$, \st $$[m,[x,y]]=[[m,x],y]-[[m,y],x]$$ for each $m\in M,\, x,y\in\lieg$. \end{defn} \ex $\lieg$ is itself a right $\lieg$-module under its own product, and its ideals are the submodules.\\ Let $M$ be a nonzero $\lieg$-module. The nonzero elements $m\in M$ \st $km\subseteq M$ are called the \emph{eigenvectors} (or \emph{semi-invariants}) of $\lieg$ in $M$. The set of eigenvectors is denoted by $E(M)$. Those eigenvectors which also have $m\lieg=0$ are the \emph{invariants} of $\lieg$. Let $M^\lieg$ be the set of invariants. If $m\in M$ is an eigenvector then, for each $x\in\lieg$, there is an element $\lambda(x)\in k$ \st$[m,x]=\lambda(x)m$. This gives a linear form $\lambda\in\lieg^*$, i.e.~a $k$-linear map from $\lieg$ to $k$. This is called an \emph{eigenvalue} of $\lieg$ on $M$, and is zero if $m$ is an ivariant. Given a linear form $\lambda$ on $\lieg$ we let $$ M_\lambda=M_\lambda(\lieg)=\{m\in M\, |\, [m,x]=\lambda(x)m\ \mbox{ for all }\, x\in\lieg\}.$$ This is a subset of $E(M)\cup\{0\}$. In particular $M_0=M^\lieg\cup\{0\}$. %- needed ? \begin{lem} % Let $M$ be a nonzero $\lieg$-module, $m\in M_\lambda,\, m\neq0$ and $\lieg'=ker(\lambda)$. Then $[\lieg,\lieg]\subseteq\lieg'$ and $\lieg'\triangleleft\lieg$. \end{lem} \begin{proof} If $x,y\in\lieg$ then $[m,[x,y]]=[[m,y],x]-[[m,x],y]$ and so $$ \lambda([y,x])m=\lambda(y)\lambda(x)m-\lambda(x)\lambda(y)m=0.$$ Thus $[\lieg,\lieg]\subseteq\lieg'$ and so $\lieg'\triangleleft\lieg$. \end{proof} %- more here after ? \begin{rem} $m\in M_\lambda(\lieg)$ \IFF\ $m(ad(x)-\lambda(x))=0$ for all $x\in \lieg$. The \emph{weight space} of $M$ corresponding to $\lambda$, $M^\lambda=M^\lambda(\lieg)$, is: $$\{m\in M\, | \mbox{ for each } x\in\lieg,\, m(ad(x)-\lambda(x))^n=0,\, n \mbox{ a natural number\}}$$ If $\lieg$ is abelian and $M$ triangular (see definition~(\ref{triangular})) then we have: $$M=\bigoplus_{\lambda\in\lieg^*}M^\lambda.$$ \end{rem} Finally, assume $k$ is a commutative ring and $\lieg$ a $k$-\la, by which we mean a free $k$-module (with basis $\{x_i\, |\, i\in I\}$) with a Lie product. Then: \begin{defn}\label{derivation} %derivation If $R$ is any non-commutative $k$-algebra containing $k$, we have an action of $\lieg$ on the coefficient $k$-algebra $R$, where elements of $\lieg$ act as derivations: A \emph{$k$-derivation} of $R$ is a $k$-module homomorphism $D:R\rightarrow R$ which is also a derivation. \end{defn} % -------------------------- Nilpotent & solvable Lie Algebras -------------------------- \subsection{Nilpotent \& solvable \la s}\label{Nilpotent} %\remafresh Here we define nilpotent and solvable \la s. These play an important role in the general theory of \la s, as do semisimple \la s, however, which will neither be defined nor used here. We will assume $\lieg$ to be a \fd $k$-\la. \begin{defn} % Nilpotent If $\lieg$ has a chain of ideals $0=\lieg_0\subset\lieg_1\subset\ldots\subset\lieg_m=\lieg$ with $[\lieg_{i+1},\lieg]\subseteq\lieg_i$ for each $i$, then $\lieg$ is said to be \emph{nilpotent}. \end{defn} \rems % nilpotent \begin{enumerate} \item This condition asserts that $\lieg$ acts trivially on $\lieg_{i+1}/\lieg_i$. It can also be arranged that each factor $\lieg_{i+1}/\lieg_i$ is one-dimensional. \item Note that the adjoint representation, with respect to the appropriate basis, is then by strictly lower triangular matrices. Also, the set of all strictly triangular $n\times n$ matrices over $k$ forms a nilpotent \la. We will come back to these points later. \end{enumerate} \begin{prop} % g nilpotent <-> g/I nilpotent Let $\lieg$ be a \la\ and let $\lieh$ be an ideal contained in the centre of $\lieg$. Then: $$\lieg \mbox{ is nilpotent } \iff \lieg / \lieh \mbox{ is nilpotent.}$$ \end{prop} %- proof einbauen ? \noindent F.~Engel was the first to prove the following two theorems: \begin{thm}% ENGEL1 A \la\ $\lieg$ is nilpotent \IFF $ad(x)$ is nilpotent for each $x\in\lieg$. \end{thm} %- Einbauen ? p. 14 Dixmier ! % \begin{proof} % (\leftarrow) Assume that $ad(x)$ is nilpotent for all $x\in\lieg$. Let $\lieg'$ be a solvable Lie % subalgebra of maximal dimension of $\lieg$ and suppose that $\lieg'\neq\lieg$. Consider the adjoint representation of $\lieg$ and let $\rho$ be its restriction to $\lieg'$. Then $\lieg'$ is stable under $\rho(\lieg')$. Let $\sigma$ be the quotient representation of $\rho$ in $\lieg/ \lieg'$. % Now $\sigma(x)$ is nilpotent for all $x\in \lieg'$, and consider the canonical mapping, $\varphi$ say of $\lieg$ onto $\lieg 7 \lieg#$. Then there exists $y\in\lieg$ \st $\varphi(y)\neq0$ and $\sigma(\lieg')\varphi(y)=0$. % \end{proof} \begin{thm}% ENGEL2 Let $\varphi: \lieg \rightarrow End(V)$ be a representation of a \la\ $\lieg$ on a nonzero \fd \vs $V$. Suppose that $\varphi(x)$ is nilpotent for all $x\in\lieg$. Then there exists a nonzero element $v\in V$, which is invariant under $\lieg$. \end{thm} %- proof einbauen ? \noindent A weaker condition on a \la\ than nilpotency is the property to be solvable: \begin{defn} % Solvable A \la\ $\lieg$ is \emph{solvable} if $\lieg$ has a chain of Lie subalgebras $0=\lieg_0\triangleleft\lieg_1\triangleleft\ldots\triangleleft\lieg_m=\lieg$ with each an ideal in the next and each factor being abelian. \noindent Equivalently $\lieg$ over $k$ is called \emph{solvable} if it has a $k$-basis $\{x_1,\ldots,x_n\}$ \st $kx_1+\ldots +kx_i\triangleleft kx_1+\ldots +kx_{i+1}$ for each $i$. \end{defn} \noindent There is an additional condition on $\lieg$ to make it \emph{completely solvable}: \begin{defn} % Completely solvable $\lieg$ is \emph{completely solvable} if it has a $k$-basis $\{x_1,\ldots,x_n\}$ \st for each $iU->A UNIVERSAL $$\begin{array}{ccc} \lieg & \stackrel{\varphi}{\longrightarrow} &\Ug\\ &\searrow &\quad \downarrow \psi\\ &\theta&A \end{array}$$ \noindent The universal object\, $\Ug$ is determined up to isomorphism. \end{defn} If $\{x_i\ |\ i\in I\}$ is a $k$-basis for $\lieg$, then $\Ug$ may be described as the associative $k$-algebra generated by elements $\{x_i\ |\ i\in I\}$ with the relations $x_ix_j-x_jx_i=[x_i,x_j]$. So $\Ug$ is the factor of the free $k$-algebra on the set $\{x_i\}$ by the ideal generated by the elements $x_ix_j-x_jx_i-[x_i,x_j]$.\\ % k[x,y,z] \ex If we take $\lieg$ to be the three-dimensional \la\ with basis $\{x,y,z\}$ and products as follows: $$ [x,y]=xy-yx=z,\quad [y,z]=yz-zy=x,\quad [z,x]=zx-xz=y$$ Then the $k$-algebra $k[x,y,z]$ with these relations is the \uea\ $\Ug$ of $\lieg$.\\ \begin{rem} If $M$ is a $\lieg$-module (see definition~(\ref{gmodule})) then the action of $\lieg$ on $M$ gives a representation of $\lieg$ which, by the universal property of\/ $\Ug$, extends to an action of $\Ug$ on $M$. This makes $M$ into a\/ $\Ug$-module. For example if $m\in M,\, x,y\in \lieg$, and $xy$ is the monomial in\/ $\Ug$ then $m(xy)=[[m,x],y]$.\\ Conversely, if $M$ is any $\Ug$-module then $M$ is also a $\lieg$-module by restriction, so $[m,x]=mx$. We can therefore interchange the two terms.\\ \end{rem} Recall from chapter~\ref{Tensorsymmetric} the definition of the tensor algebra $T(V)$, where $V$ is a \fd \vs over a field $k$ with a basis $\{e_1,\ldots,e_n\}$. We can define the \uea\ as follows: If V has a given structure as a \la\ $\lieg$ over $k$ with a Lie product, then we can form the \emph{\uea} $\Ug$ of $V=\lieg$ by quotiening out the ideal of T(V) generated by $u\otimes v-v\otimes u - [u,v]$ for all $u,v\in V$. In the special case when $\lieg$ is an abelian \la\ ($[x,y]=0$ for all $x,y\in\lieg$), we get back the symmetric algebra: $\Ug\simeq S(\lieg).$\\ % -------------------------- Poincare-Birkhoff-Witt Theorem -------------------------- \subsection{Poincare-Birkhoff-Witt Theorem}\label{PBW} %\remafresh This chapter is devoted to the important Poincare-Birkhoff-Witt Theorem (PBW). After an introduction, which recalls some previous definitions and details of the construction, we will be led to the definition of the \emph{standard monomials}. We will prove the PBW theorem on the structure of \ea s of \la s in full generality. The argument here follows a similar one given in~\cite{Varadarajan}, originally coming from Jacobson~\cite{Jacobson}, and applies to, not necessarily \fd \la s over arbitrary fields.\\ Let $\lieg$ be a \la\ over a field $k$ of characteristic zero. Let\/ $T=T(\lieg)$ be the tensor algebra of $\lieg$: $$T=k\oplus\lieg\oplus(\lieg\otimes\lieg)\oplus(\lieg\otimes\lieg\otimes\lieg)\ldots$$ with an associative product defined by the obvious rule $$(x_1\otimes\ldots\otimes x_k)(y_1\otimes\ldots\otimes y_m)=x_1\otimes\ldots x_k\otimes y_1\otimes\ldots\otimes y_m\, \in T_{k+m},$$ where $T_n=\underbrace{\lieg\otimes\ldots\otimes\lieg}_{n}$. This makes $T$ an associative graded algebra with $1$, generated by $1$ along with any basis of $\lieg$. \noindent For $x,y\in\lieg$ let $$u_{x,y}=(x\otimes y-y\otimes x)-[x,y]\, \in T.$$ Let $\ca{I}$ be the two-sided ideal in $T$ generated by the set of all elements $u_{x,y}$, so $\ca{I}$ is the $k$-subspace of\/ $T$ spanned by all elements of the form $a\otimes u_{x,y}\otimes b$ with $a,b\in T$. Since $u_{x,y}\in T_1+T_2$, the ideal $\ca{I}$ contains no elements of $T_0\simeq k$, so $\ca{I}$ is a proper ideal in $T$. Then, as before, the \uea\ $\Ug$ of $\lieg$ is the quotient $\Ug=T/ \ca{I}$. Let $q: T\rightarrow \Ug$ be the quotient map. It is clear that for any basis $\{x_i\, |\, i\in I\}$ of $\lieg$ the images $q(x_{i_1}\otimes\ldots\otimes x_{i_n})$ in\/ $\Ug$ of tensor monomials $x_{i_1}\otimes\ldots\otimes x_{i_n}$ span the \ea\ over $k$, since they span the tensor algebra. Suppose now, that the indexing set $I$ is ordered. Using the Lie product $[\ ,\ ]$ we can do a certain amount of rearranging of the $x_{i_j}$ in a monomial. We \emph{anticipate} that everything in $\Ug$ can be rewritten to be as a sum of monomials $x_{i_1}\ldots x_{i_n}$, where $i_1\leq i_2\leq \ldots\leq i_n$. A monomial in which the indices possess this ordering is called a \emph{standard} (or \emph{ordered}) \emph{monomial}. \begin{thm}[Poincare-Birkhoff-Witt]\label{PBWThm} Let $\lieg$ be a \la\ over a field $k$ of characteristic zero. If\/ $\{x_i\, |\, i\in I\}$ is an ordered basis of $\lieg$, then the set $\{x_{i_1}\cdots x_{i_n}\, |\, i_1\leq \ldots\leq i_n\}$ is a basis for the \uea\ $\Ug$ of $\lieg$. That is, the standard monomials are a $k$-basis for $\Ug$. In particular, the canonical map $\lieg\rightarrow\Ug$ is an embedding. \end{thm} \begin{proof} % PBW. We need to show that the set $\{x_{i_1}\cdots x_{i_n}\, |\, i_1\leq \ldots\leq i_n\}$ spans\/ $\Ug$ and is linearly independant. Let us first proceed by induction to show that the (images of) standard monomials span\/ $\Ug$. Indeed, given a monomial $x_{i_1}\cdots x_{i_n}$ with some indices not correctly ordered, it must be that there is at least one index, say $j$ \st $i_j>i_{j+1}$. Since $$ x_{i_j}x_{i_{j+1}}-x_{i_{j+1}}x_{i_j}-[x_{i_j},x_{i_{j+1}}]\, \in \ca{I},\ \mbox{ we have}$$ $$x_{i_1}\cdots x_{i_n}=x_{i_1}\cdots x_{i_{j-1}}\cdot (x_{i_j}x_{i_{j+1}}-x_{i_{j+1}}x_{i_j}-[x_{i_j},x_{i_{j+1}}])\cdot x_{i_{j+2}}\cdots x_{i_n} +$$ $$ x_{i_1}\cdots x_{i_{j-1}}x_{i_{j+1}}x_{i_j}x_{i_{j+2}}\cdots x_{i_n}+x_{i_1}\cdots x_{i_{j-1}}[x_{i_j},x_{i_{j+1}}]x_{i_{j+2}}\cdots x_{i_n}$$ The first summand lies inside the ideal $\ca{I}$, while the third is a tensor of smaller degree. Thus we do an induction on the degree of tensors, and for each fixed degree we do an induction on the number of pairs of indices out of order.\\ The far more difficult part to prove is linear independance. Note, although it will not be clear until the end, that the Jacobi identity~(see definition~(\ref{Lieg})) plays an essential role to complete the proof. The argument is mostly very natural, apart from the use of the endomorphism `L', which we will define, and which was originally a device introduced by Jacobson. Let us briefly define the following: given a tensor monomial $x_{i_1}\otimes \ldots\otimes x_{i_n}$, the \emph{defect} of this monomial is the number of pairs of indices $i_j,i_{j'}$ \st $ji_{j'}$. Suppose that we can define a linear map $L: T\rightarrow T$ with the properties that, first, $L$ is the identity map on a standard monomial, and second, that whenever $i_j>i_{j'}$ we have \begin{eqnarray*} L(x_{i_1}\otimes \ldots\otimes x_{i_n})&=&L(x_{i_1}\otimes \ldots\otimes x_{i_{j+1}}\otimes x_{i_j}\otimes \ldots\otimes x_{i_n}) +\\ &\ &L(x_{i_1}\otimes \ldots\otimes [x_{i_j},x_{i_{j+1}}]\otimes \ldots\otimes x_{i_n}) \end{eqnarray*} If so, then $L(\ca{I})=0$, while $L$ acts as the identity map on any linear combination of standard monomials. This would prove that the subspace $T$ consisting of linear combinations of standard monomials meets the ideal $\ca{I}$ just at 0, so maps injectively to the enveloping algebra. Note that, incidentally, $L$ would have the property: \begin{eqnarray*} L(y_{i_1}\otimes \ldots\otimes y_{i_n})&=&L(y_{i_1}\otimes \ldots\otimes y_{i_{j+1}}\otimes y_{i_j}\otimes \ldots\otimes y_{i_n}) +\\ &\ &L(y_{i_1}\otimes \ldots\otimes [y_{i_j},y_{i_{j+1}}]\otimes \ldots\otimes y_{i_n}) \end{eqnarray*} for any $y_{i_j}$ in $\lieg$. Thus, the problem reduces to defining such a map $L$. This is done by the help of induction. Firstly define $L$ to be the identity on $T_0+T_1$. Note that there is no second condition on $L$ here, and the first condition is met since every monomial tensor of degree $1$ or $0$ is standard. Now fix $n\geq 2$, and attempt to define $L$ on monomials in $T_{\leq n}$ inductively by using the second required property: \emph{define} $L(x_{i_1}\otimes \ldots\otimes x_{i_n})$ by \begin{eqnarray*} L(x_{i_1}\otimes \ldots\otimes x_{i_n})&=&L(x_{i_1}\otimes \ldots\otimes x_{i_{j+1}}\otimes x_{i_j}\otimes \ldots\otimes x_{i_n}) +\\ &\ &L(x_{i_1}\otimes \ldots\otimes [x_{i_j},x_{i_{j+1}}]\otimes \ldots\otimes x_{i_n}), \end{eqnarray*} where $i_j>i_{j+1}$. One term on the right-hand side is Žof lower degree, and the other is of smaller defect. Thus, we do an induction on the degree of the tensor monomials, and for each fixed degree we do an induction on the defect. Now we have to check if this definition is well-defined, which could be a potential problem. Monomials of degree $n$ and of defect $0$ are already standard. For monomials of degree $n$ and of defect $1$ the definition is unambiguous, since there is just one pair of indices that are out of order. So suppose that the defect is at least two. Let $ji_{j+1}$ and $i_{j'}>i_{j'+1}$. To prove well-definedness it suffices to show that the two right-hand sides of the defining relation for $L(x_{i_1}\otimes \ldots\otimes x_{i_n})$ are actually the same element of $T$. Consider the case that $j+1i_{j'+1}$. Without loss of generality, we can simplify notation by letting $j=1,j'=3$ and $n=4$, since all the other factors in the monomials are inert. To simplify notation further, write $x$ for $x_{i_1}$, $y$ for $x_{i_2}$, $z$ for $x_{i_3}$ and $w$ for $x_{i_4}$. We use the inductive definition to obtain: \begin{eqnarray*} L(x\otimes y\otimes z\otimes w) & = &L(y\otimes x\otimes z\otimes w) + L([x,y]\otimes z\otimes w)\\ &=& L(y\otimes x\otimes w\otimes z) + L(y\otimes x\otimes [z,w]) + \\ &\ & L([x,y]\otimes w\otimes z) + L([x,y]\otimes [z,w]) \end{eqnarray*} But then it is clear, by an analogous computation, that the same expression is obtained when the roles of $j$ and $j'$ are reversed. Thus, the induction step is completed in the case $j+1i_2>i_3.$ Then on one hand, applying the inductive definition by first interchanging $x$ and $y$, and then doing further reshufflings, we have: \begin{eqnarray*} L(x\otimes y\otimes z) & = &L(y\otimes x\otimes z) + L([x,y]\otimes z)\\ &=& L(y\otimes z\otimes x) + L(y\otimes [z,x]) + L([x,y]\otimes z)\\ &=& L(z\otimes y\otimes x) + L([y,z]\otimes x) + L(y\otimes [z,x]) + L([x,y]\otimes z) \end{eqnarray*} On the other hand, starting by doing the interchange of $y$ and $z$ gives: \begin{eqnarray*} L(x\otimes y\otimes z) & = &L(x\otimes z\otimes y) + L(x\otimes [y,z])\\ &=& L(z\otimes x\otimes y) + L([x,z]\otimes y) + L(x\otimes [y,z])\\ &=& L(z\otimes y\otimes x) + L(z\otimes [x,y]) + L([x,z]\otimes y) + L(x\otimes [y,z]) \end{eqnarray*} It remains to see that the right-hand sides are the same. Since $L$ is already well-defined, by induction, for tensors of degree $n-1$ (here in effect $n-1=2$), we can invoke the property $L(v\otimes w)=L(w\otimes v)+L([v,w])$, for all $v,w\in\lieg$. Therefore, the first of the two previous computations can be extended to give: $$L(x\otimes y\otimes z)= L(z\otimes y\otimes x) + L(x\otimes [y,z]) + L([[y,z],x]) +$$ $$+ L([z,x]\otimes y) + L([y,[z,x]]) + L(z\otimes [x,y]) + L([[x,y],z])$$ The latter differs from the right-hand side of the second computation by the expression: $$ L([[y,z],x]) + L([y,[z,x]]) + L([[x,y],z]).$$ Thus, we wish to prove that the latter is $0$. It suffices to show that $$ [[y,z],x] +[y,[z,x]] +[[x,y],z]=0.$$ Or equivalently, since $[[x,y],z]=-[z,[x,y]]$: $$ [[y,z],x] +[y,[z,x]]=[z,[x,y]].$$ That is, writing $(ad\, y)(x)=[y,x]$, as defined in chapter~\ref{Basicdefinitions}, we want to show that $(ad\, y)\circ (ad\, z)=ad[y,z].$ So we want to show that $ad$ is a \la\ homomorphism, which it is, since the above comes out to be exactly the Jacobi identity which holds. This proves the theorem. \end{proof} %PBW The \ea\ $\Ug$ is an associative algebra. Although passing from $\lieg$ to $\Ug$ has the disadvatage of introducing an infinite-dimensional algebra, it allows us to use associative methods, to do calculations in $\Ug$ and study its properties, as in the next chapter. %- comments p.32 7.8 %-checken: SOERGEL ? (Varadnadjan ? Rentschler ? % -------------------------- Properties of U(g) ----------------------- \subsection{Properties of\/ $\Ug$}\label{CaseNoetherian} %\remafresh Here we show some ring-theoretic properties of the \ea\ $\,\Ug$ as constructed in the last chapter. Again let $\{x_i\ |\ i\in I\}$ be a $k$-basis for a \la\ $\lieg$ and $\Ug$ the associative $k$-algebra generated by elements $\{x_i\ |\ i\in I\}$ with the relation $x_ix_j-x_jx_i=[x_i,x_j]$. This makes\/ $\Ug$ an extension (as defined in~\ref{Ringtheorybasics}) of the subring $k$ with generators $\{x_i\ |\ i\in I\}$. We can define a standard filtration $\{F_n\}_{n\in\mathbf{Z}}$ (see definition~(\ref{standardfiltration}), chapter~\ref{Filteredrings}) and we say that an element $a\in F_n\setminus F_{n-1}$ has \emph{degree $n$} and $\overline{a}=a + F_{n-1}$ is the \emph{symbol} of a. This leads naturally to an associated graded ring $gr(\Ug)$, with multiplication defined in the obvious way. \begin{lem}\label{use} % gr U(g) is generated by {x_i bar} $gr(\Ug)$ is a ring extension of $k$ with generators $\{\overline{x}_i\ |\ i\in I\}$ and each $\overline{x}_i$ is homogeneous of degree $1$. \end{lem} \begin{proof} % gr U(g) is generated by {x_i bar} Let $z$ be a nonzero element of $gr(\Ug)_j$. So $z=\overline{y}$ for some $y\in F_j\setminus F_{j-1}$ with $y$ being a finite sum of $r_0x_{i_1}r_1\ldots x_{i_j}r_j$, each belonging to $F_j\setminus F_{j-1}$. The symbol is $\overline{r_0x_{i_1}r_1\ldots x_{i_j}r_j}$, i.e.~$\overline{r}_0\overline{x}_{i_1}\overline{r}_1\ldots\overline{x}_{i_j}\overline{r}_j$. Thus $z=\overline{y}$ is a finite sum of this type and so $gr(\Ug)$ is generated over $k$ by the $\overline{x}_i$. \end{proof} Now let us look at properties of $gr(\Ug)$, which we will later use to show some properties of\/ $\Ug$. As usual, $\lieg$ will be a \la\ over a field $k$ of characteristic zero. \begin{prop}\label{comm}\label{us} % gr U(g) is commutative $gr(\Ug)$ is a commutative $k$-algebra. \end{prop} \begin{proof} % gr U(g) is commutative The relation $x_ix_j-x_jx_i=[x_i,x_j]$ shows that $$\overline{x}_i\overline{x}_j-\overline{x}_j\overline{x}_i=[x_i,x_j]+F_1$$ in $F_2/F_1$. But $[x_i,x_j]\in\lieg\subset F_1$ and so $\overline{x}_i\overline{x}_j=\overline{x}_j\overline{x}_i$. \end{proof} \begin{prop}\label{ugid} $ $ % if gr(Ug) ID --> U(g) ID \begin{numlist}{doof} \item If $gr({\Ug})$ is an \ID, then\/ $\Ug$ is an \ID. \item If $gr({\Ug})$ is a prime ring, then\/ $\Ug$ is prime. \end{numlist} \end{prop} \begin{proof} \begin{numlist}{doo} \item Let $a,c\in\Ug$ be nonzero elements of degree $m,n$ respectively. Then $\overline{a},\overline{c}$ are nonzero elements of $gr(\Ug)$ and so $\overline{a}\,\overline{c}\neq 0$. Since $\overline{a}\,\overline{c}$ is defined to be $ac+F_{m+n-1}$ and so if $ac\in F_{m+n-1}$ then $\overline{a}\,\overline{c}=0$. Otherwise $\overline{a}\,\overline{c}=\overline{a}\overline{c}$. This means that $ac\not\in F_{m+n-1}$ and so $ac\neq0$. \item If $a,c\in S$ are nonzero then $\overline{a}gr(\Ug)\overline{c}\neq0$. Thus there is a homogeneous element $\overline{u}$ of $gr(\Ug)$ with $\overline{a}\,\overline{u}\,\overline{c}\neq0$. If $u$ is a representative of $\overline{u}$ in $\Ug$, then $auc\neq0$ and so $a\Ug c\neq0$, i.e.~$\Ug$ is prime. \end{numlist} \end{proof} We note that the converses are not true. Concerning the Noetherian property we have: \begin{prop} $ $ % gr U(g) noetherian --> Ug noetherian \begin{numlist}{noeth} \item If\/ $\Ug$ is filtered and $gr({\Ug})$ is right Noetherian, then $\Ug$ is right Noetherian. \item If $\lieg$ is finite dimensional, then $\Ug$ is right and left Noetherian. \end{numlist} \end{prop} %- PROOF ! \begin{proof} \begin{numlist}{noet} %- PROOF ! \item \ldots (Will most likely include proof of lectures in~\cite{Brookes} (Theorem 5.9), which was given there for general graded rings. Alternatively there is Theorem 1.6.9 in~\cite{Robson}, both very long.) %- p.28 + SEE NOETHERIEN NOTES LECTURE NOTES p.53,54,55 + p. 63 \item Since here $\lieg$ is finite dimensional, $gr(\Ug)$ is finitely generated using lemma~(\ref{use}) and thus we have that $gr(\Ug)$ is Noetherian and commutative by~(\ref{us}). So by \textbf{(i)}, $\Ug$ is right and left Noetherian. \end{numlist} \end{proof} Interestingly, there are finitely generated Noetherian algebras with a natural filtration for which the associated graded algebra is not Noetherian. This was found by J.~C.~McConnell.\footnote{see~\cite{Krause}, p.~71.} By the PBW-Theorem~(\ref{PBWThm}) we know that the canonical mapping $\lieg\rightarrow\Ug$ is an embedding. Another corollary is: \begin{cor}\label{corthree} % gr(U(g))=S(g) $gr(\Ug)$ a polynomial ring over $k$ in the commuting indeterminantes $\{\overline{x}_i\ |\ i\in I\}$. Also if $gr(\Ug)$ is the graded algebra associated with the filtration $U_o=k,\, U_n=(k+\lieg)^n$ for $n>0$ of\/ $\Ug$, then $gr(\Ug)\simeq S(\lieg)$, the symmetric algebra on $\lieg$. \end{cor} %- Proof ? This gives $gr(\Ug)$ to be an \ID, which is very useful, since then: \begin{cor}\label{PBWCor} $ $ % U(g) is an ID \begin{numlist}{pbbd} \item $\Ug$ is an \ID. \item If $\lieg\neq0$, then $\Ug$ is not a simple ring. \end{numlist} \end{cor} \begin{proof} \begin{numlist}{pbd} \item Is now clear, since $gr(\Ug)$ is an \ID\ and so we can apply proposition~(\ref{ugid}). \item If $\lieg\neq0$, $\lieg$ has a trivial representation in $k$ via the zero map. This gives a $k$-algebra homomorphism $\Ug\rightarrow k$ with kernel the ideal generated by the $\{x_i\}$ and so $\Ug$ is not simple. \end{numlist} \end{proof} So, for a \fd \la\ $\lieg$ and using the above propositions, we see that $\Ug$ is a Noetherian domain and thus can be embedded in a skew field. This means that $\Ug$ has a field of fractions, which is called the \emph{enveloping field} of $\lieg$ and is denoted by $K(\lieg)$. All the known machinery and process of passing through the localization works as usual\footnote{For the details see~\cite{Dixmier}, chapter 3.6.}.\\ \begin{rem} % U(g) commutative Suppose $\lieg$ is abelian, i.e. $[x,y]=0$ for all $x,y\in\lieg$. This is the same as asserting that the adjoint representation of $\lieg$ is zero, or that $\Ug$ is commutative. So in the case of an abelian \la, the relations in $\Ug$ simply mean that its generators commute. Thus\/ $\Ug$ is a commutative polynomial ring with $\{x_i\}$ as generators and $\Ug\simeq gr(\Ug) \, (\simeq S(\lieg))$. Note that for every \uea\ of a \fd \la\ the associated graded algebra is commutative (see proposition~(\ref{comm})).\\ \end{rem} \ex Let $\lieg$ be the two-dimensional \la\ with basis $\{x,y\}$ and product $[x,y]=y$. Then $\Ug=k[x,y]$ subject to $xy-yx=y$, which can also be written as $xy=y(x+1)$. Then $\Ug$ becomes $k[x][y;\sigma]$, where $\sigma$ is the $k$-automorphism of $k[x]$ defined by $x\mapsto x+1$.\\ We can say that for any $\lieg,\ \Ug$ is a form of non-commutative polynomial ring over $k$. But note that it cannot always be built up by adjoining generators one at a time. %- Drinlassen ? Since we will look at prime ideals in chapter~\ref{PrimeIdeals}, we will need the definition of a \emph{crossed product}. Suppose $R$ is a $k$-algebra, with $k\subseteq R$, and that $\lieg$ is a $k$-\la\ with basis $\{x_i\, |\, i\in I\}$. \begin{defn} % crossed product A $k$-algebra $S$ containing $R$ is called a \emph{crossed product} of $R$ by $\Ug$, written $R*\Ug$, provided there is a $k$-module embedding of $\lieg$ into $S$, $x\mapsto\overline{x}$, \st \begin{numlist}{crossed} \item $\overline{x}r-r\overline{x}\in R$ and $r\mapsto \overline{x}r-r\overline{x}$ is a $k$-derivation of $R$, \item $\overline{x}\,\overline{y}-\overline{y}\,\overline{x}\in\overline{[x,y]}+R$ for all $x,y\in\lieg$, and \item $S$ is a free right (and left) $R$-module with the standard monomials in $\{\overline{x}_i\}$ as a basis. \end{numlist} \end{defn} \ex If $\lieg$ is the abelian \la\ of dimension $2n$ over a field $k$, then $$A_n(K)\simeq k*\Ug$$ under the obvious embedding of $\lieg$.\\ We also have the following, which can checked in a similarly way to the above: \begin{prop}\label{propthree} $ $ % Similar R*Ug properties \begin{numlist}{propthre} \item $gr(R*\Ug)$ is a polynomial ring over $R$ in the commuting indeterminantes $\{\overline{x}_i\ |\ i\in I\}$. \item If $R$ is an \ID, then $R*\Ug$ is an \ID. \item If $R$ is prime, then $R*\Ug$ is prime. \end{numlist} \end{prop} Finally, it can be shown that the Weyl algebras occur as simple homomorphic images of \ea s of nilpotent \la s, of which the next chapter gives an example. % -------------------------- Prime Ideals -------------------------- \section{Prime Ideals}\label{PrimeIdeals}\label{IdealsinUg} %\remafresh Let $\lieg$ denote a \fd \la\ and\/ $\Ug$ its \uea. Recall the following from corollary~(\ref{corthree}): If $\lieg$ is a \fd \la, its \uea\ $\Ug$ has a filtration by the \fd subspaces $U_0=k$ and $U_n=(k+\lieg)^n$ for $n>0$. The associated graded algebra $gr(\Ug)$ is finitely generated and by using the PBW-Theorem~(\ref{PBWThm}) it is isomorphic to the algebra $S(\lieg)$, which is a polynomial ring. Thus the following corollary: %- Wirklich by PWF ? --> CoUTHINO \begin{cor} % If $I$ is an ideal of $\Ug$, then the above filtration induces a filtration by \fd subspaces on $\Ug/I$, and the resulting associated graded algebra is isomorphic to $S(\lieg)/gr(I)$. Since the latter is finitely generated, it is a commutative Noetherian algebra. \end{cor} \ex We also note from last chapter~\ref{CaseNoetherian}, that the Weyl algebra $A_n(k)$ is not an \ea. But it is a homomorphic image of the \uea\ of the $(2n+1)$-dimensional \emph{Heisenberg \la} $\lieg$: This has generators $x_1,\ldots,x_n,y_1,\ldots,y_n,z$ \st $[x_i,y_i]=z$ for $i=1,\ldots,n$ and all other products being zero: $[x_i,x_j]=0,\, [y_i,y_j]=0$ for all $i,j$, and $[x_i,y_j]=0$ for all $i\neq j$, and $z$ is central. These are basically the easiest examples of \la s. We have: $$A_n(k)\simeq\Ug/(z-1).$$ Recall from definitions~(\ref{completeprime}) and~(\ref{primeid}) in chapter~\ref{Ringtheorybasics} that an ideal $I$ of a ring $R$ is called \emph{prime} if $I\neq R$ and, in $R/I$, the product of two nonzero ideals is nonzero. $I$ was said to be \emph{completely prime} if $R/I$ is an \ID. There is two further useful definitions: \begin{defn} $ $ %semi-prime \begin{numlist}{primeprime} \item An ideal $I$ of a ring $R$ is \emph{semiprime}, if $I\neq R$ and if in $R/I$ every nilpotent ideal is zero. \item An ideal of $I$ is called \emph{$\lieg$-stable} if\/ $[\lieg,I]\subseteq I$. \end{numlist} \end{defn} %semi-prime \begin{prop}\label{root} % Noetheiren ? Assume that a ring $R$ is Noetherian. Let $I$ be an ideal of $R$, and $\ca{I}$ the set of ideals of $R$, a power of which is contained in $I$. Then: \begin{numlist}{noetherr} \item $\ca{I}$ has a largest element $J$. \item If $I\neq R,\ J$ is the smallest semiprime ideal of $R$ containing $I$. We then call $J$ the \emph{root} of $I$. \end{numlist} \end{prop} \begin{proof} Let $J$ be the sum of the elements of $\ca{I}$. Since $R$ is Noetherian, there exists $I_1,\ldots,I_r\in\ca{I}$ \st $J=I_1+\ldots+I_r$. A sufficiently large power of $I_1,\ldots,I_r$ is contained in $I$, an hence $J\in \ca{I}$. Then $J$ is the largest element of $\ca{I}$. If $I\neq R$, then $J$ is clearly semiprime. If $J'$ is a semiprime ideal containing $I$, the image of $J$ in $R/J'$ is nilpotent, and hence $J\subseteq J'$. \end{proof} Let $\lieh$ be a subalgebra of $\lieg$, and $I$ an ideal of\/ $\Ug$. If $I$ is completely prime, then $I\cap\ca{U}(\lieh)$ is a completely prime ideal of $\ca{U}(\lieh)$. Apart from this observation, the properties of $I\cap \ca{U}(\lieh)$ do not appear closely linked to the properties of $I$. Let us look at some properties when $\lieh$ is an ideal of $\lieg$. \begin{prop} Let $\lieh$ be an ideal of $\lieg$, $I$ an ideal of\/ $\Ug$, and $J$ be the ideal $I\cap\mathcal{U}(\lieh)$ of\/ $\mathcal{U}(\lieh)$. Then we have the following: \begin{numlist}{dontknow} \item If $I$ is semiprime, then $J$ is semiprime. \item If $I$ is prime, then $J$ is prime. \end{numlist} \end{prop} \begin{proof} For the proof of the proposition we need a lemma, for which a more general proof can be found in Dixmier~\cite{Dixmier}\footnote{see~\cite{Dixmier}, Lemma 3.3.3 (p.108).}: \begin{lem} Let $R$ be a Noetherian algebra. Let $I$ be an ideal of $R$, which is $\lieg$-stable, $J$ its root, and $P_1,\ldots,P_s$ the minimal prime ideals of $R$ containing $I$. Then $J,P_1,\ldots,P_s$ are $\lieg$-stable. \end{lem} \noindent To prove \textbf{(i)}, let us assume that $J$ is not semiprime. Let $J'$ be the root of $J$ as in proposition~(\ref{root}). There exists a power $J''$ of $J'$ \st $J''\not\subseteq J$ and $J''^2\subseteq J$. We have $[\lieg,J]\subseteq J$, hence $[\lieg,J']\subseteq J'$ (by the lemma) and $[\lieg,J'']\subseteq J''$, whence $\Ug J''=J''\Ug$. Let $K$ be the ideal $\Ug J''\Ug$ of $\Ug$. Then $J''\subseteq K\cap\mathcal{U}(\lieh)$, hence $K\cap \mathcal{U}(\lieh) \not\subseteq J$ and so $K\not\subseteq I$. But $$K^2=\Ug J''\Ug J''\Ug=\Ug K''^2\Ug \subseteq \Ug J \Ug \subseteq I,$$ so that $I$ is not semiprime, which proves \textbf{(i)}. Let us now assume that $I$ is prime. Then, from \textbf{(i)}, $J$ is semiprime. Let $J_1,\ldots,J_s$ be minimal elements, pairwise distinct, in the set of prime ideals of $\mathcal{U}(\lieh)$ containing $J$. We assume that $J$ is not prime. Then $s>1$ and we set $J_0=J_2\cap\ldots \cap J_s$. We have $J_0J_1\subseteq J,\, J_0\neq J$ and $J_1\neq J$. By the lemma, $[\lieg,J_i]\subseteq J_i$ for all $i$, whence $$\Ug J_i= J_i \Ug = \Ug J_i \Ug.$$ Then $\Ug J_o \Ug \not\subseteq I$, $\Ug J_1 \Ug \not\subseteq I$ and $(\Ug J_0)(J_1 \Ug)\subseteq I,$ which is a contradiction, since we assumed that $I$ is prime. This proves \textbf{(ii)}. \end{proof} \begin{lem}\label{lemmaone} If elements $x,y$ of an algebra satisfy $[x,y]=y$, then $yx^n=(x-1)^ny$ for $n\geq 0$. \end{lem} \begin{proof} For $n=0$ this clearly holds, since then $y=y$. Now assume it is true for $n$, then: $$yx^{n+1}=(x-1)^nyx=(x-1)^n(xy-y)=(x-1)^{n+1}y,$$ and we can conclude by induction. \end{proof} \begin{prop} Let $I$ be a prime ideal of\/ $\Ug$ and consider the adjoint representation $\epsilon:\lieg\rightarrow \Ug/I$. Let $\lambda \in \lieg^*$, and $u\in\Ug/I,\, u\neq0$ \st $\epsilon(x)u=\lambda(x)u$ for all $x\in\lieg$. Let $\lieg'=ker(\lambda)$, which is an ideal of $\lieg$, and $I'=I\cap\mathcal{U}(\lieg')$. Then $I$ is the ideal of\/ $\Ug$ generated by $I'$. \end{prop} \begin{proof} For the rather technical proof we need a further lemma: \begin{lem}\label{lemmatwo} Let $I,\epsilon,\lambda,u$ be as in the statement of the proposition. Then $u$ is not a zero divisor in $\Ug/I$. \end{lem} \begin{proof} % of lemma Let $v\in\Ug/I$. Let us assume that, for example, $uv=0$ and prove that $v=0$. let $u',v'$ be representatives of $u,v$ in $\Ug$. Then $u',v'\in I$. if $u'\lieg^nv'\subseteq I$, we deduce that \begin{eqnarray*} u'\lieg^{n+1}v'&\subseteq&\lieg u'\lieg^nv'+ [\lieg,u']\lieg^nv'\\ &\subseteq&\lieg(u'\lieg^nv')+(ku'+I)\lieg^nv'\subseteq I. \end{eqnarray*} Hence $u'\Ug v'\subseteq I$, and consequently $u'\in I$ or $v' \in I$ since $I$ is prime. As $u\neq0$, we have indeed $v=0$. \end{proof} \noindent For the proof of the proposition we may assume that $\lambda\neq0$. Let $x\in\lieg$ \st $\lambda(x)=1$. Each $a\in\Ug-\{0\}$ can be uniquely written as $x^na_n+x^{n-1}a_{n-1}+\ldots+a_0$ with $a-n,\ldots,a_0\in\ca{U}(\lieg'),\, a_0\neq0$. We assume that $a\in I$, and show that $a_0,\ldots,a_n\in I'$. This is obvious if $n=0$. Assume $n>0$ and that the assertion has been proved for all integers less than $n$. It is then sufficient to prove that $a_n\in I'$. Let $z_0$ be a representative of $u$ in $\Ug$. Then \begin{numlist}{asdf} \item $[x,z_0]\in z_0+I$, \item $[y,z_0]\in I$ for $y\in\lieg'$, hence for $y\in\ca{U}(\lieg').$ \item It is clear that $[\ldots[[a,z_0],z_0]\ldots,z_0]\in I.$ \end{numlist} We show that the first term of \textbf{(iii)}, with $p$ factors $z_0$, is congruent modulo $I$ to $$n(n-1)\cdots(n-p+1)x^{n-p}z^{p}_{0}a_n+$$ $$+x^{n-p-1}b_{n-p-1}+x^{n-p-2}b_{n-p-2}+\ldots+b_0,$$ where $b_{n-p-1},\ldots,b_0$ are elements of\/ $\Ug$ which are permutable to $z_0$ modulo $I$. This is clear, taking \textbf{(ii)} into account, if $p=0$. Let us assume it for $p$. From \textbf{(i)},\textbf{(ii)} and the lemma~(\ref{lemmaone}), we have (modulo $I$): $$[n(n-1)\ldots(n-p+1)x^{n-p}z^{p}_0 a_n+x^{n-p-1}b_{n-p-1}+\ldots+b_0,\, z_0]\equiv n(n-1)\ldots(n-p+1)$$ $$\left( \left( \begin{array}{c} n-1\\p \end{array} \right)x^{n-p-1}z_0-\left( \begin{array}{c} n-p\\2 \end{array}\right)x^{n-p-2}z_0+\ldots+(-1)^{n-p-1}z_0\right)z^{p}_0 a_n$$ $$+\left(\begin{array}{c} n-p-1\\1\end{array} \right)x^{n-p-2}z_0-\left(\begin{array}{c} n-p-1\\2\end{array} \right)x^{n-p-3}z_0+\ldots$$ $$+(-1)^{n-p}z_0 b_{n-p-1}+\ldots+z_0b_1$$ $$=n(n-1)\ldots(n-p)x^{n-p-1}z^{p+1}_0 a_n+x^{n-p-2}b'_{n-p-2}+\ldots+b'_0,$$ where $b'_{n-p-1},\ldots,b_0$ are permutable with $z_0$ modulo $I$. The assertion has thus been established for $p+1$, and hence for all integers. In particular, for $p=n$, relation \textbf{(iii)} becomes $n!z^{n}_0 a_n\in I$. We conclude that $a_n\in I$ (hence that $a_n\in I'$) by the lemma~(\ref{lemmatwo}) above. \end{proof} Let us now sum up some facts concerning the relationship between ideals of\/ $\Ug$ and ideals of certain subalgebras: \begin{itemize} \item If $\lieh$ is an ideal of $\lieg$, then $\ca{U}(\lieh)$ is a subalgebra of $\Ug$ and $$[\lieg,\ca{U}(\lieh)]\subseteq \ca{U}(\lieh).$$ \item Let $I$ be a $\lieg$-stable ideal. Then in this case, $I\Ug$ is an ideal of $\Ug$ and $$I\Ug\cap \ca{U}(\lieh)=I.$$ \item If $I$ is prime in $\ca{U}(\lieh)$, then $I\Ug$ is prime in $\Ug$. Conversely, if $J$ is an ideal of $\Ug$, then $J\cap \ca{U}(\lieh)$ is a $\lieg$-stable ideal of $\ca{U}(\lieh)$ which is prime whenever $J$ is prime. \end{itemize} %- ???? \noindent (Include the things below ?) \\ Let us now take a closer look at the prime and simple factor rings of $\Ug$ when $\lieg$ is nilpotent. We have already mentioned in chapter~\ref{Basicdefinitions}, that viewed as a $\Ug$-module, $\Ug$ is also, by restriction, a $\lieg$-module. However it has another $\lieg$-module structure given by the adjoint representation of $\lieg$: $$[u,x]=ux-xu,\ \mbox{ for } u\in\Ug,\, x\in \lieg.$$ In particular, each ideal of $\Ug$ is $\lieg$-stable and so is a $\lieg$-submodule under this action. %- NEED 2.2 p.533 \begin{lem} If $R$ is a factor ring of\/ $\Ug$ and $S$ is a right denominator set in $R$, then both $R$ and $R_S$ inherit the adjoint $\lieg$-module structure from $\Ug$, and each of their ideals is a $\lieg$-submodule. \end{lem} \ldots \\ \noindent We will always view a ring $R$ or $R_S$ as a $\lieg$-module via the adjoint representation. Recall that if $\{x_i\ |\ i\in I\}$ is a $k$-basis for a \la\ $\lieg$ and $I$ totally ordered by $\leq$, then if $i,j\in I$ with $i>j$ one can write: $$x_ix_j=x_jx_i+[x_i,x_j],\ \mbox{and here $j