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\title{Lie Groups and Lie Algebras}
\author{this document by Theo Johnson-Freyd \\
based on: Mark Haiman, {\em Math 261A: Lie Groups}\\
Fall 2008, UC-Berkeley}
\date{Last updated \today}

\begin{document}
\maketitle

\tableofcontents
\addcontentsline{toc}{chapter}{Contents}

\Section{List of Theorems}
\theoremlisttype{allname}
\listtheorems{thm}

\Chapter{Introduction}

In the Fall Semester, 2008, Prof.\ Mark Haiman taught Math 261A: Lie Groups, at the University of California Berkeley.  The course covered the structure of Lie groups, Lie algebras, and their (complex) representations.  The textbooks were \cite{Borel} and \cite{Knapp}.  I was one of many students in that class, and typed detailed notes \cite{notes}, including all the motivation, discussion, questions, errors, and personal confusions and commentary.  
What you're reading right now is a first attempt to make those notes more presentable.  It is also a study aid for my qualifying exam.  As such, we present only the definitions, theorems, and proofs, with little  motivation.  I have made limited rearrangements of the material.  Each subsection corresponds to one or two one-hour lectures.
Needless to say, the pedagogy (and, since I was taking dictation, many of the words), are due to M.\ Haiman.  In particular, I have quoted almost verbatim the problem sets M.\ Haiman assigned in the class (the reader may find my answers to some of the exercises in the appendices of \cite{notes}).  Of course, any and all errors are mine.

In addition to \cite{Borel,Knapp}, the reader might be interested in getting a sense of previous renditions of UC Berkeley's Math 261.  In 2006 a three-professor tag-team taught a one-semester Lie Groups and Lie Algebras course; detailed notes are available \cite{anton}, and occasionally I have referenced those notes, especially when I was absent or lost, or when my notes are otherwise lacking.  They go quickly through the material --- about twice as fast as we did --- eschewing most proofs.  For a very different version of the course, the reader may be interested in the year-long Lie Groups and Lie Algebras course taught in 2001 \cite{Knutson}.

These notes are typeset using \TeX Shop Pro on a MacBook running OS 10.5; the backend is pdf\LaTeX.  Pictures and diagrams are drawn using \textsc{pgf}/\TikZ.
 For a full list of packages used, you may peruse the \LaTeX\ source code for this document, available at \url{http://math.berkeley.edu/~theojf/LieGroupsBook.tex}.
 
In addition to Mark Haiman, the people without whom I would not have been able to put together this text were Alex Fink, Dustin Cartwright, Manuel Reyes, and the other participants in the class.

\section{Notation}
We will generally use $\Aa,\Bb,\Cc,\dots$ for categories; named categories are written in small-caps, so that for example \cat{$A$-mod} is the category of (finite-dimensional) $A$-modules.  Objects in a category are generally denoted $A,B,C,\dots$, with the exception that for Lie algebras we use fraktur letters $\aa,\bb,\cc$.  The classical Lie groups we refer to with roman letters ($\GL(n,\CC)$, etc.),  and we write ${\rm M}(n)$ for the algebra of $n\times n$ matrices.   Famous fields and rings are in black-board-bold: $\RR,\CC,\QQ,\ZZ$, and we use $\KK$ rather than $k$ for a generic field.  The \emindexing{natural numbers} $\NN$ is always the set of {\em non-negative} integers, so that it is a \emindexing{rig}, or ``ring with identities but without negations''. 

In a category with products\index{product}, we use $\pt$ for the terminal object, and $\times$ for the monoidal structure; a general monoidal category is written with $\otimes$ for the product.  We do not include associators and other higher-categorical things.

For morphisms in a category we use lower-case Greek and Roman letters $\alpha,\beta,a,b,c,\dots$\index{map}\index{homomorphism}.  An {\em element}\index{element\addcontentsline{toc}{chapter}{Index}} of an object $A$ in monoidal category is a morphism $a: \pt \to A$, or with $\pt$ replaced by the monoidal unit; we will write this as ``$a\in A$'' following the usual convention.  When the category is concrete with products, this agrees with the set-theoretic meaning.  The identity map on any object $A \in\Aa$ we write as $1_A$.  We always write the identity matrix as $1$, or $1_n$ for the $n\times n$ identity matrix when we need to specify its size.  Similarly, $0$ and $0_n$ refer to the zero matrix.

If $M$ is a manifold (we will never need more general geometric spaces), we will write $\Ccc(M)$ for the continuous, smooth, analytic, or holomorphic functions on it, depending on what is natural for the given space.  Thus if $M$ is a real manifold, we will always use the symbol $\Ccc$ for the sheaf of infinitely-differentiable or analytic functions on it, depending on whether the ambient category is that of infinitely-differentiable manifolds or analytic manifolds.  When working over the complex numbers, $\Ccc$ may refer to the sheaf of complex analytic functions or of holomorphic functions.  Moreover,  the word ``\indexing{smooth}'' may mean any of ``infinitely-differentiable'', ``analytic'', or ``holomorphic'', depending on the choice of ambient category.  When a statement does not hold in this generality, we will specify.  We write ${\rm T}M$ for the \indexing{tangent bundle} of $M$, and ${\rm T}_pM$ for the fiber over the point $p$.

%\part{Lie Groups}

\chapter{Motivation: Closed Linear Groups}
\setcounter{page}{1}
\renewcommand{\thepage}{\arabic{page}}

\section{Definition of a Lie Group}

\cite[Lecture 1]{notes}

\subsection{Group objects}

\begin{defn}
  Let $\Cc$ be a category with finite products; denote the terminal object by $\pt$.  A \emindexing{group object}\index{group} in $\Cc$ is an object $G$ along with maps $\mu: G\times G \to G$, $i: G\to G$, and $e: \pt \to G$, such that the following diagrams commute:
  \eqn[GroupAssoc]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (GGG) {$G\times G \times G$} +(2,0) node (GG1) {$G\times G$} +(0,-1) node (GG2) {$G\times G$} +(2,-1) node (G) {$G$} +(0,-.5) node (X) {};
    \draw[->] (GGG) -- node[auto] {$\scriptstyle 1_G\times \mu$} (GG1);
    \draw[->] (GGG) -- node[auto] {$\scriptstyle \mu \times 1_G$} (GG2);
    \draw[->] (GG1) -- node[auto] {$\scriptstyle \mu$} (G);
    \draw[->] (GG2) -- node[auto] {$\scriptstyle \mu$} (G);
  \end{tikzpicture} }
  \eqn[GroupId]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (G) {$G$} +(-1.5,0) node (G1) {$\pt \times G$} +(0,1) node (GG) {$G\times G$} +(1.5,0)    node (G2) {$G\times \pt$} +(0,.5) node (X) {};
    \draw[->] (G1) -- node[auto] {$\scriptstyle e\times 1_G$} (GG);
    \draw[->] (GG) -- node[auto] {$\scriptstyle \mu$} (G);
    \draw[->] (G1) -- node[auto,swap] {$\sim$} (G);
    \draw[->] (G2) -- node[auto,swap] {$\scriptstyle  1_G \times e$} (GG);
    \draw[->] (G2) -- node[auto] {$\sim$} (G);
  \end{tikzpicture}  }
  \eqn[GroupInv]{ \begin{tikzpicture}[baseline=(X.base),scale=3]
    \path node (X) {$\pt$} +(-1,0) node (G1) {$G$} +(-.5,.5) node (GG1a) {$G\times G$} +(.5,.5) node (GG1b) {$G\times G$} +(1,0) node (G2) {$G$} +(-.5,-.5) node (GG2a) {$G\times G$} +(.5,-.5) node (GG2b) {$G\times G$};
  \draw[->] (G1) -- (X); \draw[->] (X) -- node[auto] {$\scriptstyle e$} (G2);
  \draw[->] (G1) -- node[auto] {$\scriptstyle \Delta$} (GG1a);
  \draw[->] (G1) -- node[auto,swap] {$\scriptstyle \Delta$} (GG2a);
  \draw[->] (GG1b) -- node[auto] {$\scriptstyle \mu$} (G2);
  \draw[->] (GG2b) -- node[auto,swap] {$\scriptstyle \mu$} (G2);
  \draw[->] (GG1a) -- node[auto] {$\scriptstyle 1_G\times i$} (GG1b);
  \draw[->] (GG2a) -- node[auto,swap] {$\scriptstyle i\times 1_G$} (GG2b);
  \end{tikzpicture} }
  In \cref{GroupId}, the isomorphisms are the canonical ones.  In \cref{GroupInv}, the map $G\to \pt$ is the unique map to the terminal object, and $\Delta: G\to G\times G$ is the canonical diagonal map.
  
  If $(G,\mu_G,e_G,i_G)$ and $(H,\mu_H,e_H,i_H)$ are two group objects, a map $f:G\to H$ is a {\em group object homomorphism}\index{homomorphism!of group objects}\index{group object!homomorphism} if the following commute:
  \eqn{ \begin{tikzpicture}[baseline=(A.base),scale=2]
    \path node (GG) {$G \times G$} +(1.5,0) node (G) {$G$} +(0,-1) node (HH) {$H\times H$} +(1.5,-1) node (H) {$H$} +(0,-.5) node (A) {};
    \draw[->] (GG) -- node[auto] {$\scriptstyle \mu_G$} (G);
    \draw[->] (HH) -- node[auto] {$\scriptstyle \mu_H$} (H);
    \draw[->] (GG) -- node[auto] {$\scriptstyle f\times f$} (HH);
    \draw[->] (G) -- node[auto] {$\scriptstyle  f$} (H);
  \end{tikzpicture} \;\;\;\;\;\;\;\;\;
  \begin{tikzpicture}[baseline=(pt.base),scale=2]
   \path node (pt) {$\pt$} +(1,.5) node (G) {$G$} +(1,-.5) node (H) {$H$};
   \draw[->] (pt) -- node[auto] {$\scriptstyle e_G$} (G);
   \draw[->] (pt) -- node[auto,swap] {$\scriptstyle e_H$} (H);
    \draw[->] (G) -- node[auto] {$\scriptstyle  f$} (H);
  \end{tikzpicture}
  }
  (That $f$ intertwines $i_G$ with $i_H$ is then a corollary.)
\end{defn}

\begin{defn}
  A {\em (left)} \emindexing{group action}\index{action} of a group object $G$ in a category $\Cc$ with finite products is a map $\rho: G\times X \to X$ such that the following diagrams commute:
  \eqn[ActionAssoc]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (GGG) {$G\times G \times X$} +(2,0) node (GG1) {$G\times X$} +(0,-1) node (GG2) {$G\times X$} +(2,-1) node (G) {$X$} +(0,-.5) node (X) {};
    \draw[->] (GGG) -- node[auto] {$\scriptstyle 1_G\times \rho$} (GG1);
    \draw[->] (GGG) -- node[auto] {$\scriptstyle \mu \times 1_X$} (GG2);
    \draw[->] (GG1) -- node[auto] {$\scriptstyle \rho$} (G);
    \draw[->] (GG2) -- node[auto] {$\scriptstyle \rho$} (G);
  \end{tikzpicture} }
  \eqn[ActionId]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (G) {$X$} +(0,1) node (GG) {$G\times X$} +(-1.5,0)    node (G2) {$\pt \times X$} +(0,.5) node (X) {};
    \draw[->] (GG) -- node[auto] {$\scriptstyle \rho$} (G);
    \draw[->] (G2) -- node[auto] {$\scriptstyle e\times 1_X$} (GG);
    \draw[->] (G2) -- node[auto,swap] {$\sim$} (G);
  \end{tikzpicture}  }
  (The diagram corresponding to \cref{GroupInv} is then a corollary.)  A {\em right action} is a map $X\times G\to X$ with similar diagrams.  We denote a left group action $\rho: G\times X \to X$ by $\rho: G\acts X$.
  
  Let $\rho_X : G\times X \to X$ and $\rho_Y: G\times Y \to Y$ be two group actions.  A map $f: X\to Y$ is {\em $G$-equivariant}\index{equivariant map}\index{map!equivariant} if the following square commutes:
  \eqn[equiv]{ \begin{tikzpicture}[baseline=(A.base),scale=2]
    \path node (GX) {$G \times X$} +(2,0) node (X) {$X$} +(0,-1) node (GY) {$G\times Y$} +(2,-1) node (Y) {$Y$} +(0,-.5) node (A) {};
    \draw[->] (GX) -- node[auto] {$\scriptstyle \rho_X$} (X);
    \draw[->] (GX) -- node[auto] {$\scriptstyle 1_G\times f$} (GY);
    \draw[->] (X) -- node[auto] {$\scriptstyle f$} (Y);
    \draw[->] (GY) -- node[auto] {$\scriptstyle \rho_Y$} (Y);
  \end{tikzpicture} }  
\end{defn}

\subsection{Analytic and Algebraic Groups}

\begin{defn} \label{AGroupDefn}
  A \emindexing{Lie group}\index{group!Lie} is a group object in a category of manifolds.  In particular, a Lie group can be infinitely differentiable\index{Lie group!infinitely differentiable} (in the category \cat{$\Ccc^\infty$-Man}) or {\em analytic}\index{Lie group!analytic}\index{group!analytic} (in the category \cat{$\Ccc^\omega$-Man}) when over $\RR$, or {\em complex analytic}\index{Lie group!complex analytic} or {\em almost  complex}\index{Lie group!almost complex} when over $\CC$.  We will take ``Lie group'' to mean \emindexing{analytic Lie group} over either $\CC$ or $\RR$.  In fact, the different notions of real Lie group coincide, a fact that we will not directly prove, as do the different notions of complex Lie group.  As always, we will use the word ``smooth'' for any of ``infinitely differentiable'', ``analytic'', or ``holomorphic''.
  
  A \emindexing{Lie action} is a group action in the category of manifolds.
  
  A {\em (linear)} \emindexing{algebraic group}\index{group!algebraic} over $\KK$ (algebraically closed) is a group object in the category of (affine) algebraic varieties over $\KK$.
\end{defn}

\begin{eg}
  The \emindexing{general linear group} $\GL(n,\KK)$\index{GLnK@$\GL(n,\KK)$} of $n\times n$ invertible matrices is a Lie group over $\KK$ for $\KK = \RR$ or $\CC$.  When $\KK$ is algebraically closed, $\GL(n,\KK)$ is an algebraic group.  It acts algebraically on $\KK^n$ and on projective space $\PP(\KK^n) = \PP^{n-1}(\KK)$.
\end{eg}

\section{Definition of a Closed Linear Group}

\cite[Lectures 2 and 3]{notes}

We write $\GL(n,\KK)$ for the group of $n\times n$ invertible matrices over $\KK$, and ${\rm M}(n,\KK)$\index{MnK@${\rm M}(n,\KK)$} for the algebra of all $n\times n$ matrices.  We regularly leave off the $\KK$.

\begin{defn}
  A \emindexing{closed linear group}\index{group!closed linear} is a subgroup of $\GL(n)$ (over $\CC$ or $\RR$) that is closed as a topological subspace.
\end{defn}

\subsection{Lie algebra of a closed linear group}

\begin{lemmadef}
  The following describe the same function $\exp: {\rm M}(n) \to \GL(n)$, called the \emindexing{matrix exponential}.
  \begin{enumerate}
    \item $\displaystyle \exp(a) \defined \sum_{n\geq 0} \frac{a^n}{n!}$.
    \item $\displaystyle  \exp(a)  \defined \left. e^{ta} \right|_{t=1}$, where for fixed $a\in {\rm M}(n)$ we define $e^{ta}$ as the solution to the initial value problem $e^{0a} = 1$, $\frac{d}{dt} e^{ta} = ae^ta$.
    \item $\displaystyle \exp(a) \defined \lim_{n\to\infty} \left( 1 + \frac{a}{n} \right)^n$.
  \end{enumerate}
  If $ab - ba = 0$, then $\exp(a+b) = \exp(a) + \exp(b)$.
  
  The function $\exp: {\rm M}(n) \to \GL(n)$ is a local isomorphism of analytic manifolds.  In a neighborhood of $1\in \GL(n)$, the function $\displaystyle \log a \defined -\sum_{n > 0} \frac{(1-a)^n}{n} $ is an inverse to $\exp$.
\end{lemmadef}
\begin{lemmadef} \label{LieAlgGLn}
  Let $H$ be a closed linear group.  The {\em Lie algebra of $H$}\index{Lie algebra!of a closed linear group} is the set
  \eqn{\Lie(H) = \{x\in {\rm M}(n) : \exp(\RR x) \subseteq H\} }
  \begin{enumerate}
    \item $\Lie(H)$ is a $\RR$-subspace of ${\rm M}(n)$.
    \item $\Lie(H)$ is closed under the \emindexing{bracket} $[,] : (a,b) \mapsto ab - ba$.
  \end{enumerate}
\end{lemmadef}
\begin{defn}
  A \emindexing{Lie algebra} over $\KK$ is a vector space $\gg$ along with an antisymmetric map $[,]: \gg \otimes \gg \to \gg$ satisfying the \emindexing{Jacobi identity}:
  \eqn{ [[a,b],c] + [[b,c],a] + [[c,a],b] = 0 }
  A {\em homomorphism of Lie algebras}\index{homomorphism!of Lie algebras}\index{Lie algebra!homomorphism of} is a linear map preserving the bracket.  A {\em Lie subalgebra} is a vector subspace closed under the bracket.
\end{defn}
\begin{eg}
  The algebra $\gl(n) = {\rm M}(n)$\index{gln@$\gl(n)$} of $n\times n$\index{gln@$\gl(n)$} matrices is a Lie algebra with $[a,b] = ab - ba$.  It is $\Lie(\GL(n))$.  \Cref{LieAlgGLn} says that $\Lie(H)$ is a Lie subalgebra of ${\rm M}(n)$.
\end{eg}

\subsection{Some analysis}

\begin{lemma}
  Let ${\rm M}(n) = V\oplus W$ as a real vector space.  Then there exists an open neighborhood $U \ni 0$ in ${\rm M}(n)$ and an open neighborhood $U' \ni 1$ in $\GL(n)$ such that $(v,w) \mapsto \exp(v)\exp(w): V\oplus W \to \GL(n)$ is a homeomorphism $U \to U'$.
\end{lemma}
\begin{lemma} \label{lemmaclosure}
  Let $H$ be a closed subgroup of $\GL(n)$, and $W \subseteq {\rm M}(n)$ be a linear subspace such that $0$ is a limit point of the set $\{w\in W \st \exp(w) \in H\}$.  Then $W\cap \Lie(H) \neq 0$.
\end{lemma}
\begin{proof}
  Fix a Euclidian norm on $W$.  Let $w_1,w_2,\dots \to 0$ be a sequence in $\{w\in W \st \exp(w) \in H\}$, with $w_i \neq 0$.  Then $w_i/|w_i|$ are on the unit sphere, which is compact, so passing to a subsequence, we can assume that $w_i/|w_i| \to x$ where $x$ is a unit vector.  The norms $|w_i|$ are tending to $0$, so $w_i/|w_i|$ is a large multiple of $w_i$.  We approximate this: let $n_i = \lceil 1/|w_i| \rceil$, whence $n_i w_i \approx w_i/|w_i|$, and $n_iw_i \to x$.  But $\exp w_i \in H$, so $\exp(n_i w_i) \in H$, and $H$ is a closed subgroup, so $\exp x \in H$.

Repeating the argument with a ball of radius $r$ to conclude that $\exp (rx)$ is in $H$, we conclude that $x \in \Lie(H)$.
\end{proof}
\begin{prop}
  Let $H$ be a closed subgroup of $\GL(n)$.  There exist neighborhoods $0\in U \subseteq {\rm M}(n)$ and $1 \in U' \subseteq \GL(n)$ such that $\exp: U\isom U'$ takes $\Lie(H) \cap U \isom H \cap U'$.
\end{prop}
\begin{proof}
  We fix a complement $W \subseteq {\rm M}(n)$ such that ${\rm M}(n) = \Lie(H) \oplus W$.  By \cref{lemmaclosure}, we can find a neighborhood $V\subseteq W$ of $0$ such that $\{v \in V \st \exp(v)\in H\} = \{0\}$.  Then on $\Lie(H) \times V$, the map $(x,w) \mapsto \exp(x)\exp(w)$ lands in $H$ if and only if $w = 0$.  By restricting the first component to lie in an open neighborhood, we can approximate $\exp(x+w) \approx \exp(x)\exp(w)$ as well as we need to --- there's a change of coordinates that completes the proof.
\end{proof}
\begin{cor}
  $H$ is a submanifold of $\GL(n)$ of dimension equal to the dimension of $\Lie(H)$.
\end{cor}
\begin{cor}
  $\exp(\Lie(H))$ generates the identity component $H_0$ of $H$.
\end{cor}
\begin{remark}
  In any topological group, the connected component of the identity is normal.
\end{remark}
\begin{cor}
  $\Lie(H)$ is the tangent space ${\rm T}_1H \defined \{ \gamma'(0) \st \gamma:\RR \to H, \gamma(0) = 1\} \subseteq {\rm M}(n)$.
\end{cor}

\section{Classical Lie groups}

\cite[Lectures 4 and 5]{notes}

We mention only the classical compact semisimple Lie groups and the classical complex semisimple Lie groups.  There are other very interesting classical Lie groups, c.f.\ \cite{Lang}.

\subsection{Classical Compact Lie groups}

\begin{lemmadef}
  The \emindexing{quaternians} $\HH$ is the unital $\RR$-algebra generated by $i,j,k$ with the multiplication $i^2 = j^2 = k^2 = ijk = -1$; it is a non-commutative division ring.  Then $\RR \mono \CC \mono \HH$, and $\HH$ is a subalgebra of ${\rm M}(4,\RR)$.  We defined the \emindexing{complex conjugate}\index{conjugate!complex} linearly by $\bar{i} = -i$, $\bar{j} = -j$, and $\bar{k} = -k$; complex conjugation is an anti-automorphism, and the fixed-point set is $\RR$.    The \emindexing{Euclidean norm} of $\zeta \in \HH$ is given by $\|\zeta\| = \bar\zeta \zeta$.

  The \emindexing{Euclidean norm} of a column vector $x\in \RR^n,\CC^n,\HH^n$ is given by $\|x\|^2 = \bar{x}^Tx$, where $\bar{x}$ is the component-wise complex conjugation of $x$.
  
  If $x \in {\rm M}(n,\RR),{\rm M}(n,\CC),{\rm M}(n,\HH)$ is a matrix, we define its \emindexing{Hermetian conjugate}\index{conjugate!Hermetian} to be the matrix $x^* = \bar{x}^T$; Hermetian conjugation is an antiautomorphism of algebras ${\rm M}(n) \to {\rm M}(n)$.  ${\rm M}(n,\HH) \mono {\rm M}(2n,\CC)$ is a $*$-embedding.
  
  Let $j = \left( \begin{array}{c:c} 0 & -1 \\ \hdashline 1 & 0 \end{array} \right) \in {\rm M}(2,{\rm M}(n,\CC)) = {\rm M}({2n},\CC)$ be a block matrix.  We define $\GL(n,H) \defined \{ x \in \GL({2n},\CC) \st jx = \bar{x}j\}$.  It is a closed linear group.
\end{lemmadef}
\begin{lemmadef}  The following are closed linear groups, and are compact:
\begin{itemize}
  \item The {\em (real)} \emindexing{special orthogonal group} $\SO(n,\RR) \defined \{x \in {\rm M}(n,\RR) \st x^*x = 1 \textrm{ and } \det x = 1\}$\index{SOnR@$\SO(n,\RR)$}.  
  \item The {\em (real)} \emindexing{orthogonal group} ${\rm O}(n,\RR) \defined \{x \in {\rm M}(n,\RR) \st x^*x = 1\}$\index{OnR@${\rm O}(n,\RR)$}.  
  \item The \emindexing{special unitary group} $\SU(n) \defined \{x \in {\rm M}(n,\CC) \st x^*x = 1 \textrm{ and } \det x = 1\}$.  \index{SUnR@$\SU(n,\RR)$}
  \item The \emindexing{unitary group} ${\rm U}(n) \defined \{x \in {\rm M}(n,\CC) \st x^*x = 1\}$.  \index{Un@${\rm U}(n)$}
  \item The {\em (real)} \emindexing{symplectic group} $\SP(n,\RR) \defined \{x \in {\rm M}(n,\CC) \st x^*x = 1 \}$.  \index{SpnR@$\SP(n,\RR)$}
  \end{itemize}
\end{lemmadef}
There is no natural quaternionic determinant.

\subsection{Classical Complex Lie groups}

The following groups make sense over any field, but it's best to work over an algebraically closed field.  We work over $\CC$.
\begin{lemmadef}
  The following are closed linear groups over $\CC$, and are algebraic:
  \begin{itemize}
  \item The {\em (complex)} \emindexing{special linear group} $\SL(n,\CC) \defined \{ x \in \GL(n,\CC) \st \det x = 1\}$.\index{SLnC@$\SL(n,\CC)$}
  \item The {\em (complex)} \emindexing{special orthogonal group} $  \SO(n,\CC)  \defined \{ x \in \SL(n,\CC) \st x^Tx = 1\}$.\index{SOnC@$\SO(n,\CC)$}
  \item The {\em (complex)} \emindexing{symplectic group} $  \SP(n,\CC)  \defined \{ x \in \GL(2n,\CC) \st x^Tjx = j \} $.\index{SpnC@$\SP(n,\CC)$}
  \end{itemize}
\end{lemmadef}

\subsection{The Classical groups}

In full, we have defined the following ``classical\index{classical groups}\index{group!classical}'' \indexing{closed linear group}s\index{group!closed linear}:
\[ \hspace{-1 in} \begin{array}{clllll} 
                & \textbf{Group} &\textbf{Group}  & \textbf{Algebra} & \textbf{Algebra} & \\
                & \textbf{Name} &\textbf{Description} & \textbf{Name} & \textbf{Description} & \dim_\RR\\ \\
     \multirow{3}{*}{\textrm{\begin{sideways} Compact \end{sideways}}}
                & \SO(n,\RR) & \{x \in {\rm M}(n,  \RR  ) \st x^*x = 1, \det x = 1\} & \so(n,\RR)\index{sonR@$\so(n,\RR)$} & \{ x \in {\rm M}(n,\RR) \st x^* + x = 0 \} & {n\choose 2} \\ 
                & \SU(n) & \{x \in {\rm M}(n,  \CC  ) \st x^*x = 1, \det x = 1\} & \su(n)\index{sun@$\su(n)$} & \{ x \in {\rm M}(n,\CC) \st x^* + x = 0, \tr x = 0 \} & n^2 - 1 \\
                & \SP(n,\RR) & \{x \in {\rm M}(n,  \HH  ) \st x^*x = 1\}  &  \sp(n,\RR)\index{spnR@$\sp(n,\RR)$} &  \{ x \in {\rm M}(n,\HH) \st x^* + x = 0 \} & 2n^2 + n \\ \\
                & \GL(n,\HH) & \{ x \in \GL({2n},\CC) \st jx = \bar{x}j \} & \gl(n,\HH)\index{glnH@$\gl(n,\HH)$} & \{ x \in M_{2n}(\CC) \st jx = \bar{x}j \} & 4n^2  \\ \\
     \multirow{3}{*}{\textrm{\begin{sideways} Complex \end{sideways}}}
                & \SL(n,\CC) &  \{x\in {\rm M}(n,\CC) \st  \det x = 1\} & \sl(n,\CC)\index{sln@$\sl(n,\CC)$} & \{ x \in {\rm M}(n,\CC) \st \tr x = 0 \} & 2(n^2 - 1)  \\
                & \SO(n,\CC) & \{x\in {\rm M}(n,\CC) \st   x^T x = 1, \det x = 1 \} & \so(n,\CC)\index{sonC@$\so(n,\CC)$} & \{ x \in {\rm M}(n,\CC) \st x^T + x  = 0 \} & n(n-1) \\
                & \SP(n,\CC) &  \{x\in {\rm M}(n,\CC) \st   x^Tjx = j \} & \sp(n,\CC)\index{spnC@$\sp(n,\CC)$} & \{x \in {\rm M}(n,\CC) \st x^Tj + jx = 0 \} & 2{2n+1 \choose 2}
\end{array} \hspace{-1 in} \]
\begin{prop}Via the natural embedding ${\rm M}(n,\HH) \mono {\rm M}({2n},\CC)$, we have:
\begin{align}
  \SP(n) & = \GL(n,\HH) \cap U(2n) \\
   & = \GL(n,\HH) \cap \SP(n,\CC) \\
   & = U(2n) \cap \SP(n,\CC)
\end{align}
\end{prop}

\section{Homomorphisms of closed linear groups}

\begin{defn}
  Let $H$ be a closed linear group.  The \emindexing{adjoint action} $H\acts H$ is given by by $^g h \defined ghg^{-1}$, and this action fixes $1\in H$.  This induces the {\em adjoint action} $\Ad: H \acts {\rm T}_1H = \Lie(H)$.  It is given by $g\cdot y = gyg^{-1}$, where now $y \in \Lie(H)$.
\end{defn}
\begin{lemma}
  Let $H$ and $G$ be closed linear groups and $\phi: H \to G$ a smooth homomorphism.  Then $\phi(1) = 1$, so $d\phi: {\rm T}_1H \to {\rm T}_1G$ by $X \mapsto (\phi(1 + tX))'(0)$.  The diagram of actions commutes:
\[ \begin{tikzpicture}
  \path node (H) {$H$} (H.east)  node[anchor=west] (a1) {$\acts$} (a1.east) node[anchor=west] (TH) {${\rm T}_1H$} (H) ++(0,-1.5) node (G) {$G$} (a1) ++(0,-1.5)  node(a2) {$\acts$} (TH) ++(0,-1.5) node (TG) {${\rm T}_1G$};
  \draw[->] (H) -- node[auto] {$\scriptstyle \phi$} (G);
  \draw[->] (TH) -- node[auto] {$\scriptstyle d\phi$} (TG);
\end{tikzpicture}\]
This is to say:
\eqn{d\phi(\Ad(h)Y) = \Ad(\phi(h)) \,d\phi(Y) }
Thus $d\phi\,[X,Y] = [d\phi\, X, d\phi \,Y]$, so $d\phi$ is a {\em Lie algebra homomorphism}\index{homomorphism!of Lie algebras}\index{Lie algebra!homomorphism of}.

If $H$ is connected, the map $d\phi$ determines the map $\phi$.
\end{lemma}

\Section{Exercises}

\begin{enumerate}
  \item \begin{enumerate}
    \item Show that the orthogonal groups ${\rm O}(n,\RR)$\index{OnR@${\rm O}(n,\RR)$} and ${\rm O}(n,\CC)$\index{OnC@${\rm O}(n,\CC)$} have two connected components, the identity component being the special orthogonal group $SO_n$, and the other consisting of orthogonal matrices of determinant $-1$. 
    \item Show that the center of ${\rm O}(n)$ is $\{\pm I_n\}$. 
     \item Show that if $n$ is odd, then $\SO(n)$ has trivial center and ${\rm O}(n) \cong \SO(n) \times (\ZZ/2\ZZ)$ as a Lie group. 
    \item Show that if $n$ is even, then the center of $\SO(n)$ has two elements, and ${\rm O}(n)$ is a semidirect 
product $(\ZZ/2\ZZ) \ltimes \SO(n)$, where $\ZZ/2\ZZ$ acts on $\SO(n)$ by a non-trivial outer automorphism of 
order $2$. 
  \end{enumerate}
  \item Construct a smooth 
group homomorphism $\Phi : \SU (2) \to \SO(3)$ which induces an isomorphism of Lie algebras 
and identifies $\SO(3)$ with the quotient of $\SU (2)$ by its center $\{ \pm I\}$.
  \item Construct an isomorphism of $\GL(n, \CC)$ (as a Lie group and an algebraic group) with 
a closed subgroup of $\SL(n + 1, \CC)$. 
  \item Show that the map $\CC^\ast \times \SL(n, \CC) \to \GL(n, \CC)$ given by $(z , g) \mapsto z g$ is a surjective homomorphism of Lie and algebraic groups, find its kernel, and describe the corresponding 
homomorphism of Lie algebras.
  \item Find the Lie algebra of the group $U \subseteq \GL(n, \CC)$ of upper-triangular matrices with $1$ 
on the diagonal. Show that for this group, the exponential map is a diffeomorphism of the 
Lie algebra onto the group. 
  \item A \emindexing{real form} of a complex Lie algebra $\gg$ is a real Lie subalgebra $\gg_\RR$ such that that 
$\gg = \gg_\RR \oplus i\gg_\RR$, or equivalently, such that the canonical map $\gg_\RR \otimes_\RR \CC \to \gg$ given by scalar 
multiplication is an isomorphism. A real form of a (connected) complex closed linear group 
$G$ is a (connected) closed real subgroup $G_\RR$ such that $\Lie(G_\RR)$ is a real form of $\Lie(G)$. 
  \begin{enumerate}
    \item Show that ${\rm U}(n)$ is a compact real form of $\GL(n, \CC)$ and $\SU (n)$ is a compact real 
form of $\SL(n, \CC)$. 
    \item Show that $\SO(n,\RR)$\index{SOnR@$\SO(n,\RR)$} is a compact real form of $\SO(n, \CC)$\index{SOnC@$\SO(n,\CC)$}. 
     \item Show that $\SP(n,\RR)$\index{SpnR@$\SP(n,\RR)$} is a compact real form of $\SP(n, \CC)$\index{SpnC@$\SP(n,\CC)$}.
   \end{enumerate}
\end{enumerate}





\chapter{Mini-course in Differential Geometry}

\section{Manifolds}

\cite[Lectures 6, 7, and 8]{notes}

\subsection{Classical definition}

\begin{defn}
  Let $X$ be a (Hausdorff) topological space.  A \emindexing{chart} consists of the data $U \open X$ and a homeomorphism $\phi: U \isom V \open \RR^n$.  $\RR^n$ has \emindexing{coordinates} $x_i$, and $\xi_i \defined x_i\circ \phi$ are \emindexing{local coordinates}\index{coordinates!local} on the chart.  Charts $(U,\phi)$ and $(U',\phi')$ are {\em compatible}\index{compatible charts}\index{chart!compatible} if on $U\cap U'$ the $\xi_i'$ are smooth functions of the $\xi_i$ and conversely.  I.e.:
  \eqn[CompatChart]{\begin{tikzpicture}[baseline=(X.base),scale=2]
  \path node (UU) {$U\cap U'$} +(-3,0) node (U1) {$U$} +(3,0) node (U2) {$U'$} +(0,-.5) node (X) {};
  \draw[->] (U1) -- node[auto] {$\scriptstyle \phi$} ++(0,-1) node[anchor=north] (V1) {$V$};
  \draw[->] (U2) -- node[auto] {$\scriptstyle \phi'$} ++(0,-1) node[anchor=north] (V2) {$V'$};
  \draw[->] (UU) -- node[auto,swap] {$\scriptstyle \bar\phi$} ++(-1.5,-1) node[anchor=north] (W1) {$W$};
  \draw[->] (UU) -- node[auto] {$\scriptstyle \bar\phi'$} ++(1.5,-1) node[anchor=north] (W2) {$W'$};
  \path (W1.west) node[anchor=east] (s1) {$\supseteq$} (s1.west) node[anchor=east] {V};
  \path (W2.east) node[anchor=west] (s2) {$\subseteq$} (s2.east) node[anchor=west] {V'};
  \draw[->] (W1) -- node[auto] {$\scriptstyle \bar\phi' \circ \bar\phi^{-1}$} node[auto,swap] {\scriptsize \rm smooth with smooth inverse}(W2);
  \end{tikzpicture}}
  An \emindexing{atlas} on $X$ is a covering by pairwise compatible charts.
\end{defn}
\begin{lemma}
  If $U$ and $U'$ are compatible with all charts of $\Aa$, then they are compatible with each other.
\end{lemma}
\begin{cor}
  Every atlas has a unique maximal extension.
\end{cor}
\begin{defn}
  A \emindexing{manifold} is a Hausdorff topological space with a maximal atlas.  It can be {\em real}, {\em infinitely-differentiable}, {\em complex}, {\em analytic}, etc., by varying the word ``smooth'' in the compatibility condition \cref{CompatChart}.
\end{defn}

\begin{defn}
  Let $U$ be an open subset of a manifold $X$.  A function $f: U \to \RR$ is {\em smooth}\index{smooth!function} if it is smooth on local coordinates in all charts.
\end{defn}

\subsection{Sheafs}

\begin{defn}
  A \emindexing{sheaf of functions} $\Sss$ on a topological space $X$ assigns a ring $\Sss(U)$ to each open set $U\open X$ such that:
  \begin{enumerate}
    \item if $V \subseteq U$ and $f\in \Sss(U)$, then $f|_V \in \Sss(V)$, and 
  \item if $U = \bigcup_\alpha U_\alpha$ and $f : U\to \RR$ such that $f|_{U_\alpha} \in \Sss(U_\alpha)$ for each $\alpha$, then $f\in \Sss(U)$.
  \end{enumerate}
  The \emindexing{stalk} of a sheaf at $x\in X$ is the space $\Ss_x \defined \lim_{U\ni x} \Sss(U)$.
\end{defn}
\begin{lemma}
  Let $X$ be a manifold, and assign to each $U\open X$ the ring $\Ccc(U)$ of smooth functions on $U$.  Then $\Ccc$ is a sheaf.  Conversely, a topological space $X$ with a sheaf of functions $\Sss$ is a manifold if and only if there exists a covering of $X$   by open sets $U$ such that $(U,\Sss|_U)$ is isomorphic as a space with a sheaf of functions to $(V, \Sss^{\RR^n}|_V)$ for some $V \subseteq \RR^n$ open.
\end{lemma}

\subsection{Manifold constructions}

\begin{defn}
  If $X$ and $Y$ are smooth manifolds, then a {\em smooth map}\index{smooth!map}\index{map!smooth} $f:X \to Y$ is a continuous map such that for all $U \subseteq Y$ and $g\in \Ccc(U)$, then $g\circ f \in \Ccc(f^{-1}(U))$.  Manifolds form a category \cat{Man} with products: a {\em product of manifolds}\index{product!of manifolds} $X \times Y$ is a manifold with charts $U\times V$.  
\end{defn}
\begin{defn}
  Let $M$ be a manifold, $p\in M$ a point, and $\gamma_1,\gamma_2: \RR \to M$ two paths with $\gamma_1(0) = \gamma_2(0) = p$.  We say that $\gamma_1$ and $\gamma_2$ are \emindexing{tangent} {\em at $p$} if $(f\circ \gamma_1)'(0) = (f\circ \gamma_2)'(0)$ for all smooth $f$ on a nbhd of $p$, i.e.\ for all $f \in \Ccc_p$.  Each equivalence class of tangent curves is called a \emindexing{tangent vector}.
\end{defn}
\begin{defn} Let $M$ be a manifold and $\Ccc$ its sheaf of smooth functions.
  A \emindexing{point derivation} is a linear map $\delta: \Ccc_p \to \RR$ satisfying the \emindexing{Leibniz rule}:
  \eqn{ \delta(fg) = \delta f\,g(p) + f(p)\,\delta g }
\end{defn}
\begin{lemma}
  Any tangent vector $\gamma$ gives a point derivation $\delta_\gamma: f\mapsto (f\circ \gamma)'(0)$.  Conversely, every point derivation is of this form.
\end{lemma}
\begin{lemmadef}
  Let $M$ and $N$ be manifolds, and $f:M\to N$ a smooth map sending $p\mapsto q$.  The following are equivalent, and define $(df)_p:{\rm T}_pM \to {\rm T}_qN$, the \emindexing{differential} of $f$ at $p$:
  \begin{enumerate}
  \item If $[\gamma] \in {\rm T}_pM$ is represented by the curve $\gamma$, then $(df)_p(X) \defined [f\circ \gamma]$.
  \item If $X \in {\rm T}_pM$ is a point-derivation on $\Ss_{M,p}$, then $(df)_p(X) : \Ss_{N,q} \to \RR$ (or $\CC$) is defined by $\psi \mapsto X[\psi\circ f]$.
  \item In coordinates, $p\in U \underset{\textrm{open}}{\subseteq} \RR^m$ and $q\in W \underset{\textrm{open}}{\subseteq} \RR^n$, then locally $f$ is given by $f_1,\dots,f_n$ smooth functions of $x_1,\dots,x_m$.  The tangent spaces to $\RR^n$ are in canonical bijection with $\RR^n$, and a linear map $\RR^m \to \RR^n$ should be presented as a matrix:
  \eqn{ \textrm{\upshape Jacobian}(f,x) \defined \frac{\partial f_i}{\partial x_j} }
\end{enumerate}
\end{lemmadef}

\begin{lemma}
We have the \emindexing{chain rule}: if $M \overset{f}\to N \overset{g}\to K$, then $d(g\circ f)_p = (dg)_{f(p)} \circ (df)_p$.
\end{lemma}

\begin{thm}[Inverse Mapping Theorem] \begin{enumerate} \item
Given smooth $f_1,\dots, f_n : U \to \RR$ where $p\in U \open \RR^n$, then $f : U \to \RR^n$ maps some neighborhood $V \ni p$ bijectively to $W \open \RR^n$ with s/a/h inverse iff $\det \textrm{\upshape Jacobian}(f,x) \neq 0$.
  \item 
  A smooth map $f: M\to N$  of manifold restricts to an isomorphism $p\in U \to W$ for some neighborhood $U$ if and only if $(df)_p$ is a linear isomorphism.\end{enumerate}
\end{thm}


\subsection{Submanifolds}

\begin{prop} \label{immersionprop}
  Let $M$ be a manifold and $N$ a topological subspace with the induced topology such that for each $p\in N$, there is a chart $U\ni p$ in $M$ with coordinates $\{\xi_i\}_{i=1}^m:U\to \RR^m$ such that $U\cap N = \{q\in U \st \xi_{n+1}(q) = \dots = \xi_m(q) = 0\}$.  Then $U\cap N$ is a chart on $N$ with coordinates $\xi_1,\dots,\xi_n$, and $N$ is a manifold with an atlas given by $\{U\cap N\}$ as $U$ ranges over an atlas of $M$.  The sheaf of smooth functions $\Ccc_N$ is the sheaf of continuous functions on $N$ that are locally restrictions of smooth functions on $M$.  The embedding $N\mono M$ is smooth, and satisfies the universal property that any smooth map $f:Z\to M$ such that $f(Z) \subseteq N$ defines a smooth map $Z\to N$.
\end{prop}
\begin{defn}
  The map $N\mono M$ in \cref{immersionprop} is an \emindexing{immersed submanifold}.  A map $Z\to M$ is an \emindexing{immersion} if it factors as $Z\isom N \mono M$ for some immersed submanifold $N\mono M$.
\end{defn}
\begin{prop}
  If $N\mono M$ is an immersed submanifold, then $N$ is locally closed.
\end{prop}
\begin{prop}
  Any closed linear group $H \subseteq \GL(n)$ is an immersed analytic submanifold.  If $\Lie(H)$ is a $\CC$-subspace of ${\rm M}(n,\CC)$, then $H$ is a holomorphic submanifold.
\end{prop}
\begin{proof}
  The following diagram defines a chart near $1\in H$, which can be moved by left-multiplication wherever it is needed:
  \eqn{\begin{tikzpicture}[baseline=(X.base)]
    \path node (U) {$U$} ++(3,0) node (V) {$V$} +(0,-.5) node (X) {};
    \path (U.east) +(0,2pt) node (U1) {} +(0,-2pt) node (U2) {} (V.west) +(0,2pt) node (V1) {} +(0,-2pt) node (V2) {};
    \draw[->] (U1) -- node[auto] {$\scriptstyle \exp$} (V1);
    \draw[->] (V2) -- node[auto] {$\scriptstyle \log$} (U2);
    \path (U.west) +(1ex,0) node[anchor=east] {${\rm M}(n) \supseteq$} (U.north) ++(0,2pt) node[rotate=270] (i1) {$\in$} ++(0,2pt) node[anchor=south] {$0$};
    \path (V.east) +(-1ex,0) node[anchor=west] {$\subseteq \GL(n)$} (V.north) ++(0,2pt) node[rotate=270] (i2) {$\in$} ++(0,2pt) node[anchor=south] {$1$};
    \path (U) ++(0,-1.5) node (u) {$\Lie(H) \cap U$};
    \path (V) ++(0,-1.5) node (v) {$H \cap V$};
    \path (u.east) +(0,2pt) node (u1) {} +(0,-2pt) node (u2) {} (v.west) +(0,2pt) node (v1) {} +(0,-2pt) node (v2) {};
    \draw[->] (u1) --  (v1);
    \draw[->] (v2) -- (u2);
    \draw[right hook-to] (u) -- (U);
    \draw[right hook-to] (v) -- (V);
  \end{tikzpicture}}
\end{proof}


\begin{lemma}
Given ${\rm T}_pM = V_1 \oplus V_2$, there is an open neighborhood $U_1 \times U_2$ of $p$ such that $V_i = {\rm T}_pU_i$.
\end{lemma}

\begin{lemma} If $s: N \to M\times N$ is a s/a/h section, then $s$ is a (closed) immersion.
\end{lemma}

\begin{prop} A smooth map $f: N \to M$ is an immersion on a neighborhood of $p\in N$ if and only if $(df)_p$ is injective.\end{prop}


\section{Vector Fields}

\cite[Lectures 8, 9, and 10]{notes}

\subsection{Definition}

\begin{defn}
  Let $M$ be a manifold.  A \emindexing{vector field} assigns to each $p\in M$ a vector $x_p$, i.e.\ a point derivation:
  \eqn{ x_p(fg) = f(p)\,x_p(g) + x_p(f)\,g(p) }
  We define $(xf)(p) \defined x_p(f)$.  Then $x(fg) = f\,x(g) + x(f)\,g$, so $x$ is a \emindexing{derivation}.  But it might be discontinuous.  A vector field $x$ is {\em smooth}\index{smooth!vector field}\index{vector field!smooth} if $x : \Ccc_M \to \Ccc_M$ is a map of sheaves.  Equivalently, in local coordinates the components of $x_p$ must depend smoothly on $p$.  By changing (the conditions on) the sheaf $\Ccc$, we may define analytic or holomorphic vector fields.
  
  Henceforth, the word ``vector field'' will always mean ``smooth (or analytic or holomorphic) vector field''.  Similarly, we will use the word ``smooth'' to mean smooth or analytic or holomorphic, depending on our category.
\end{defn}

\begin{lemma} The commutator $[x,y] \defined xy - yx$ of derivations is a derivation. \end{lemma}
\begin{proof}An easy calculation:
 \begin{align}
   xy(fg) & = xy(f)\,g + x(f)\,y(g) + y(f)\,x(g) + f\,xy(g) \\
   \textrm{Switch $X$ and $Y$, and subtract: } \nonumber \\
    [x,y](fg) & = [ x,y](f)\,g + f\,[x,y](g)
 \end{align}
\end{proof}

\begin{defn}
  A \emindexing{Lie algebra} is a vector space $\ll$ with a bilinear map $[,] : \ll \times \ll \to \ll$ (i.e.\ a linear map $[,]: \ll \otimes \ll \to \ll$), satisfying
\begin{enumerate}
  \item Antisymmetry\index{antisymmetry}: $[x,y] + [y,x] = 0$
  \item Jacobi\index{Jacobi identity}: $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$
\end{enumerate}
\end{defn}

\begin{prop}
  Let $V$ be a vector space.  The bracket $[x,y] \defined xy-yx$ makes $\End(V)$ into a Lie algebra.
\end{prop}

\begin{lemmadef}
  Let $\ll$ be a Lie algebra.
  The \emindexing{adjoint action} $\ad: \ll \to \End(\ll)$ given by $\ad x: y \mapsto [x,y]$ is a \emindexing{derivation}\index{Lie derivation}:
  \eqn{ (\ad x)[y,z] = [(\ad x)y,z] + [y,(\ad x)z] }
  Moreover, $\ad : \ll \to \End(\ll)$ is a {\em Lie algebra homomorphism}\index{homomorphism!of Lie algebras}\index{Lie algebra!homomorphism of}:
  \eqn{ \ad([x,y]) = (\ad x)(\ad y) - (\ad y)(\ad x) }
\end{lemmadef}

\subsection{Integral Curves}

Let $\partial_t$ be the vector field $f \mapsto \frac{d}{dt}f$ on $\RR$.

\begin{prop}
  Given a smooth vector field $x$ on $M$ and a point $p \in M$, there exists an open interval $I \open \RR$ such that $0\in I$ and a smooth curve $\gamma: I \to M$ satisfying:
  \begin{align}
  \label{integral1} \gamma(0) & = p \\
  \label{integral2} (d\gamma)_t(\partial_t) & = x_{\gamma(t)} \, \forall t\in I
\end{align}
  When $M$ is a complex manifold and $x$ a holomorphic vector field, we can demand that $I\open \CC$ is an open domain containing $0$, and that $\gamma : I \to M$ be holomorphic.
\end{prop}

\begin{proof}
In local coordinates, $\gamma: \RR \to \RR^n$, and we can use existence and uniqueness theorems for solutions to differential equations; then we need that a smooth (analytic, holomorphic) differential equation has a smooth (analytic, holomorphic) solution.

But there's a subtlety.  What if there are two charts, and solutions on each chart, that diverge right where the charts stop overlapping?  Well, since $M$ is \indexing{Hausdorff}, if we have two maps $I \to M$, then the locus where they agree is closed, so if they don't agree on all of $I$, then we can go to the maximal point where they agree and look locally there.
\end{proof}

\begin{defn}
  The \emindexing{integral curve} $\int_{x,p}(t)$ of $x$ at $p$ is the maximal curve satisfying \cref{integral1,integral2}.
\end{defn}
\begin{prop}
  The integral curve $\int_{x,p}$ depends smoothly on $p\in M$.
\end{prop}

\begin{prop}
  Let $x$ and $y$ be two vector fields on a manifold $M$.  For $p\in M$ and $s,t\in \RR$, define $q$ by the following picture:
  \[ \begin{tikzpicture}
    \path node[circle,fill,inner sep=1pt] (p) {}
    ++($ (3,1) + .5*(rand,rand) $) node[circle,fill,inner sep=1pt] (p1) {} +(90:2ex) node {$p_1$}
    ++($ (1,-2) + .5*(rand,rand) $) node[circle,fill,inner sep=1pt] (p2) {} +(0:2ex) node {$p_2$}    
    ++($ (-3,-1) - .5*(rand,rand) $) node[circle,fill,inner sep=1pt] (p3) {} +(270:2ex) node {$p_3$}    
    ++($ (-1,2) - .5*(rand,rand) $) node[circle,fill,inner sep=1pt] (q) {} 
    ;
    \node at ($ (p) ! -2ex ! (q) $) {$p$};
    \node at ($ (q) ! -2ex ! (p) $) {$q$};
    \draw[->,bend left] (p) to node[auto] {$\int_x$} node[auto,swap] {$\scriptstyle t$} (p1);
    \draw[->,bend right] (p1) to node[auto] {$\int_y$} node[auto,swap] {$\scriptstyle s$} (p2);
    \draw[->,bend right] (p2) to node[auto,swap] {$\int_x$} node[auto] {$\scriptstyle -t$}(p3);
    \draw[->,bend left] (p3) to node[auto,swap] {$\int_y$} node[auto] {$\scriptstyle - s$} (q);
  \end{tikzpicture} \]
  Then for any smooth function $f$, we have $f(q) - f(p) = st[x,y]_pf + O(s,t)^3$.
\end{prop}

\begin{proof}
Let $\alpha(t) = \int_{x,p}(t)$, so that $f(\alpha(t))' = xf(\alpha(t))$.  Iterating, we see that $\left( \frac{d}{dt} \right)^n f(\alpha(t)) = x^nf(\alpha(t))$, and by Taylor series expansion, 
\eqn{f(\alpha(t)) = \sum \frac1{n!} \left( \frac{d}{dt} \right)^n f(\alpha(0))t^n = \sum \frac1{n!} x^nf(p)t^n = e^{tx}f(p).}  By varying $p$, we have:
\begin{align}
  f(q) & = \left( e^{-sy}f\right)(p_3) \\
   &  = \left(e^{-tx}e^{-sy} f\right)(p_2) \\
   &  = \left( e^{sy}e^{-tx}e^{-sy}f\right)(p_1) \\
   &  = \left( e^{tx}e^{sy}e^{-tx}e^{-sy}f\right)(p)
 \end{align}
  We already know that $e^{tx}e^{sy}e^{-tx}e^{-sy} = 1 + st[x,y] + \textrm{higher terms}$.  Therefore $f(q) - f(p) = st[x,y]_pf + O(s,t)^3$.
\end{proof}

\subsection{Group Actions}

\begin{prop} \label{leftinvvfieldprop}
  Let $M$ be a manifold, $G$ a Lie group, and $G\acts M$ a Lie group action, i.e.\ a smooth map $\rho: G\times M \to M$ satisfying \cref{ActionAssoc,ActionId}.  Let $x\in {\rm T}_eG$, where $e$ is the identity element of the group $G$.  The following descriptions of a vector field $\ell x \in \Vect(M)$ are equivalent:
  \begin{enumerate}
    \item Let $x = [\gamma]$ be the equivalence class of tangent paths, and let $\gamma: I \to G$ be a representative path.  Define $(\ell x)_m = [\tilde\gamma]$ where $\tilde\gamma(t) \defined \rho\bigl(\gamma(t)^{-1},m\bigr)$.  On functions, $\ell x$ acts as:
    \eqn{ (\ell x)_mf \defined \dzero f\bigl(\gamma(t)^{-1}m\bigr) }
    \item Arbitrarily extend $x$ to a vector field $\tilde x$ on a neighborhood $U\subseteq G$ of $e$, and lift this to $\tilde\tilde x$ on $U\times M$ to point only in the $U$-direction: $\tilde{\tilde x}_{(u,m)} \defined (\tilde x_u,0) \in {\rm T}_uU \times {\rm T}_mM$.  Let $\ell x$ act on functions by:
    \eqn{ (\ell x)f \defined -\tilde{\tilde x} (f\circ p) \bigr|_{\{e\} \times M = M} }
    \item $\displaystyle (\ell x)_m \defined -(d\rho)_{(e,m)}(x,0) $
  \end{enumerate}
\end{prop}
\begin{prop}
  Let $G$ be a Lie group, $M$ and $N$ manifolds, and $G\acts M$, $G\acts N$ Lie actions, and let $f: M\to N$ be $G$-equivariant.  Given $x \in {\rm T}_eG$, define $\ell ^Mx$ and $\ell ^Nx$ vector fields on $M$ and $N$ as in \cref{leftinvvfieldprop}.  Then for each $m\in M$, we have:
  \eqn{ (df)_m(\ell ^Mx) = (\ell ^Nx)_{f(m)} }
\end{prop}

\begin{defn}
  Let $G\acts M$ be a Lie action.  We define the \emindexing{adjoint action} of $G$ on $\Vect(M)$ by $^g y \defined dg(y)_{gm} = (dg)_m(y_m)$.  Equivalently, $G\acts \Ccc_M$ by $g: f\mapsto f\circ g^{-1}$, and given a vector field thought of as a derivation $y: \Ccc_M \to \Ccc_M$, we define $^g y \defined g y g^{-1}$.
\end{defn}
\begin{eg}
  Let $G\acts G$ by right multiplication: $\rho(g,h) \defined hg^{-1}$.  Then $G\acts {\rm T}_eG$ by the \emindexing{adjoint action} $\Ad(g) = d(g-g^{-1})_e$, i.e.\ if $x = [\gamma]$, then $\Ad(g)x = [g\,\gamma(t)\,g^{-1}]$.
\end{eg}
\begin{defn}
  Let $\rho: G\acts M$ be a Lie action.  For each $g\in G$, we define $^g M$ to be the manifold $M$ with the action ${^g\rho}: (h,m) \mapsto \rho(ghg^{-1},m)$.
\end{defn}
\begin{cor}
  For each $g\in G$, the map $g: M \to {^g M}$ is $G$-equivariant.  We have:
  \eqn{ ^g \ell x = dg(\ell x) = \ell^{^g M}x = \ell (\Ad(g)x) }
\end{cor}
\begin{prop} \label{leftinvprop}
  Let $\rho: G\acts G$ by $\rho_g: h \mapsto hg^{-1}$.  Then $\ell : {\rm T}_eG \to \Vect(G)$ is an isomorphism from ${\rm T}_eG$ to {\em left-invariant} vector fields\index{left-invariant vector field}, such that $(\ell x)_e = x$.
\end{prop}
\begin{proof}
  Let $\lambda: G\acts G$ be the action by left-multiplication: $\lambda_g(h) = gh$.  Then for each $g$, $\lambda_g$ is $\rho$-equivariant.  Thus $d\lambda_g(\ell x) = {^{\lambda_g}(\ell x)} = \ell x$, so $\ell x$ is left-invariant, and $(\ell x)_e = x$ since $\rho(g,e) = g^{-1}$.  Conversely, a left-invariant field is determined by its value at a point:
  \eqn{ (\ell x)_g = (d\lambda_g)_e(\ell x_e) = (d\lambda_g)_e(x) }
\end{proof}

\subsection{Lie algebra of a Lie group}

\begin{lemmadef} \label{abstractlie}
  Let $G\acts M$ be a Lie action.  The subspace of $\Vect(M)$ of $G$-invariant derivations is a Lie subalgebra of $\Vect(M)$.
  
  Let $G$ be a Lie group.  The {\em Lie algebra of $G$}\index{Lie algebra!of a Lie group} is the Lie subalgebra $\Lie(G)$ of $\Vect(G)$ consisting of left-invariant vector fields, i.e.\ vector fields invariant under the action $\lambda: G\acts G$ given by $\lambda_g:h \mapsto gh$.
  
  We identify $\Lie(G) \defined {\rm T}_eG$ as in \cref{leftinvprop}.
\end{lemmadef}

\begin{lemma}
  Given $G\acts M$ a Lie action, $x\in \Lie(G)$ represented by $x = [\gamma]$, and $y\in \Vect(M)$, we have:
  \eqn{ \dzero {^{\gamma(t)}y}f = [\ell x,y]f }
\end{lemma}
\begin{proof}
    \begin{align}
    \left. \frac{d}{dt}\right|_{t=0} { ^{\gamma(t)}y} f \, (p) & = \left. \frac{d}{dt}\right|_{t=0} { \gamma(t)\, y \, \gamma(t)^{-1}} f\, (p) \\
     & = \left. \frac{d}{dt} \right|_{t=0} \gamma(t)\,y\,f(\gamma(t)\,p) \\
     & = \left. \frac{d}{dt} \right|_{t=0} \gamma(t)\,y\,f(\gamma(0)\,p) + \gamma(0)\, \left.\frac{d}{dt} \right|_{t=0} y f(\gamma(t)p) \\
     & = \ell x(yf)(p) + y\, \left. \frac{d}{dt}\right|_{t=0} f(\gamma(t)p) \\
     & = \ell x(yf)(p) + y(-\ell x \,f)(p) \\
     & = [\ell x,y]f\,(p)
  \end{align}
\end{proof}
\begin{cor}
  Let $G\acts M$ be a Lie action.  If $x,y\in \Lie(G)$, where $x = [\gamma]$, then
  \eqn{\ell ^M\bigl( \ell ^{\Ad}(-x)\,y \bigr) =  \left. \frac{d}{dt} \right|_{t=0} \ell \bigl( \Ad( \gamma(t)) y \bigr) f = \left[ \ell x,\ell y \right] f}
\end{cor}

\begin{lemma}
  The Lie bracket defined on $\Lie(\GL(n)) = \gl(n) = {\rm T}_e\GL(n) = {\rm M}(n)$ defined in \cref{abstractlie} is the matrix bracket $[x,y] = xy - yx$.
\end{lemma}
\begin{proof}
  We represent $x\in \gl(n)$ by $[e^{tx}]$.  The adjoint action on $\GL(n)$ is given by $\Ad_G(g)\,h = ghg^{-1}$, which is linear in $h$ and fixes $e$, and so passes immediately to the action $\Ad: \GL(n)\acts {\rm T}_e\GL(n)$ given by $\Ad_\gg(g)\,y = gyg^{-1}$.  Then
  \eqn{ [x,y] = \dzero e^{tx} y e^{-tx} = xy - yx. }
\end{proof}
\begin{cor}
  If $H$ is a closed linear group, then \cref{abstractlie,LieAlgGLn} agree.
\end{cor}

\Section{Exercises}

\begin{enumerate}
  \item \begin{enumerate}
    \item Show that the composition of two immersions is an immersion. 
    \item Show that an immersed submanifold $N \subseteq M$ is always a closed submanifold of an open submanifold, but not necessarily an open submanifold of a closed submanifold.
  \end{enumerate}
  \item Prove that if $f : N \to M$ is a smooth map, then $(df)_p$ is surjective if and only if there are open neighborhoods $U$ of $p$ and $V$ of $f(p)$, and an isomorphism $\psi : V \times W \to U$, such that $f \circ \psi$ is the projection on $V$.
  
  In particular, deduce that the fibers of $f$ meet a neighborhood of $p$ in immersed closed submanifolds of that neighborhood. 
    
  \item Prove the implicit function theorem: a map (of sets) $f : M \to N$ between manifolds is smooth if and only if its graph is an immersed closed submanifold of $M \times N$.
  \item Prove that the curve $y^2 = x^3$ in $\RR^2$ is not an immersed submanifold. 
  \item Let M be a complex holomorphic manifold, $p$ a point of $M$, $X$ a holomorphic vector field. Show that $X$ has a complex integral curve $\gamma$ defined on an open neighborhood $U$ of $0$ in $\CC$, and unique on $U$ if $U$ is connected, which satisfies the usual defining equation but in a complex instead of a real variable $t$.
  
  Show that the restriction of $\gamma$ to $U \cap \RR$ is a real integral curve of $X$, when $M$ is regarded as a real analytic manifold. 
  \item Let $\SL(2,\CC)$ act on the Riemann sphere $\PP^1(\CC)$ by fractional linear transformations $\left[ \begin{array}{cc}a & b \\ c & d \end{array}\right]  z = (az + b)/(cz + d)$. Determine explicitly the vector fields $f(z)\dd_z$ corresponding to 
the infinitesimal action of the basis elements
  \[ E = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], \;\; H = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right], \;\; F = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right] \]
  of $\sl(2,\CC)$, and check that you have constructed a Lie algebra homomorphism by computing 
the commutators of these vector fields.
  \item \begin{enumerate}
    \item Describe the map $\gl(n,\RR) = \Lie(\GL(n,\RR)) = {\rm M}(n,\RR) \to \Vect(\RR^n)$ given by the infinitesimal action of $\GL(n,\RR)$. 
    \item Show that $\so(n,\RR)$ is equal to the subalgebra of $\gl(n,\RR)$ consisting of elements whose infinitesimal action is a vector field tangential to the unit sphere in $\RR^n$.
  \end{enumerate}
  \item \begin{enumerate}
    \item Let $X$ be an analytic vector field on $M$ all of whose integral curves are unbounded (i.e., they are defined on all of $\RR$). Show that there exists an analytic action of $R = (\RR, +)$ on $M$ such that $X$ is the infinitesimal action of the generator $\partial_t$ of $\Lie(\RR)$.
    \item More generally, prove the corresponding result for a family of n commuting vector fields $X_i$ and action of $\RR^n$. 
  \end{enumerate}
  \item \begin{enumerate}
    \item Show that the matrix $\left[ \begin{array}{cc} -a & 0 \\ 0 & -b \end{array}\right]$ belongs to the identity component of $\GL(2,\RR)$ for all positive real numbers $a$, $b$. 
    \item Prove that if $a\neq b$, the above matrix is not in the image $\exp(\gl(2,\RR))$ of the exponential map.
  \end{enumerate}
\end{enumerate}








\chapter{General theory of Lie groups}

\section{From Lie algebra to Lie group}

\subsection{The exponential map}

\cite[Lecture 10]{notes}

We state the following results for Lie groups over $\RR$.  When working with complex manifolds, we can replace $\RR$ by $\CC$ throughout, whence the interval $I\open \RR$ is replaced by a connected open domain $I\open \CC$.  As always, the word ``\indexing{smooth}'' may mean ``infinitely differentiable'' or ``analytic'' or \dots.

\begin{lemma} \label{oneparamsubgplemma}
  Let $G$ be a Lie group and $x\in \Lie(G)$.  Then there exists a unique Lie group homomorphism $\gamma_x: \RR \to G$ such that $(d\gamma_x)_0(\partial t) = x$.  It is given by $\gamma_x(t) = (\int_e\ell x)(t)$.
\end{lemma}
\begin{proof}
  Let $\gamma: I \to G$ be the maximal integral curve of $\ell x$ passing through $e$.  Since $\ell x$ is left-invariant, $g\gamma(t)$ is an integral curve through $q$.  Let $g = \gamma(s)$ for $s\in I$; then $\gamma(t)$ and $\gamma(s)\gamma(t)$ are integral curves through $\gamma(s)$, to they must coincide: $\gamma(s+t) = \gamma(s)\gamma(t)$, and $\gamma(-s) = \gamma(s)^{-1}$ for $s\in I\cap (-I)$.  So $\gamma$ is a groupoid homomorphism, and by defining $\gamma(s+t) \defined \gamma(s)\gamma(t)$ for $s,t\in I,\, s+t\not\in I$, we extend $\gamma$ to $I+I$.  Since $\RR$ is archimedean, this allows us to extend $\gamma$ to all of $\RR$; it will continue to be an integral curve, so really $I$ must have been $\RR$ all along.
\end{proof}
\begin{cor}
  There is a bijection between one-parameter subgroups of $G$ (homomorphisms $\RR \to G$) and elements of the Lie algebra of $G$.
\end{cor}
\begin{defn}
  The \emindexing{exponential map}\index{map!exponential} $\exp: \Lie(G) \to G$ is given by $\exp x \defined \gamma_x(1)$, where $\gamma_x$ is as in \cref{oneparamsubgplemma}.
\end{defn}

\begin{prop}
  Let $x^{(b)}$ be a {\em smooth family} of vector fields\index{vector field!smooth family}\index{smooth!family of vector fields} on $M$ parameterized by $b\in B$ a manifold, i.e.\ the vector field $\tilde x$ on $B\times M$ given by $\tilde x_{(b,m)} = (0,x^{(b)}_m)$ is smooth.  Then $(b,p,t) \mapsto \left(\int_p x^{(b)}\right)(t)$ is a smooth map from an open neighborhood of $B\times M \times \{0\}$ in $B\times M \times \RR$ to $M$.  When each $x^{(b)}$ has infinite-time solutions, we can take the open neighborhood to be all of $B\times M \times \RR$.
\end{prop}
\begin{proof}
  Note that 
\eqn{ \left( \int_{(b,p)}\tilde{x}\right)(t) = \left(b, \left( \int_p x^{(b)}\right)(t) \right) }
So $B \times M \times \RR \to B \times M \overset{\pi}{\to} M$ by $(b,p,t) \mapsto \left( \int_{(b,p)}\tilde{x}\right)(t) \mapsto \left( \int_p x^{(b)}\right)(t)$ is a composition of smooth functions, hence is smooth.
\end{proof}

\begin{thm}[Exponential Map] \index{exp}
  For each Lie group $G$, there is a unique smooth map $\exp: \Lie(G) \to G$ such that for $x\in \Lie(G)$, the map $t\mapsto \exp(tx)$ is the integral curve of $\ell x$ through $e$; $t\mapsto \exp(tx)$ is a Lie group homomorphism $\RR \to G$.
\end{thm}

\begin{eg}
  When $G = \GL(n)$, the map $\exp: \gl(n)\to \GL(n)$ is the matrix exponential.
\end{eg}

\begin{prop}
  The differential at the origin $(d\exp)_0$ is the identity map $1_{\Lie(G)}$.
\end{prop}
\begin{proof}
  $d(\exp tx)_0(\partial_t) = x$.
\end{proof}
\begin{cor}
  $\exp$ is a local homeomorphism.
\end{cor}
\begin{defn}
  The local inverse of $\exp: \Lie(G) \to G$ is called ``$\log$''.\index{log}
\end{defn}
\begin{prop}
  If $G$ is connected, then $\exp(\Lie(G))$ generates $G$.
\end{prop}
\begin{prop} \label{expsquareprop}
  If $\phi: H\to G$ is a group homomorphism, then the following diagram commutes:
  \eqn{\begin{tikzpicture}[baseline=(X.base),scale=1.5]
    \path node (X) {} ++(0,-.5) node (LH) {$\Lie(H)$} +(0,1) node (H) {$H$} ++(2,0) node (LG) {$\Lie(G)$} +(0,1) node (G) {$G$};
    \draw[->] (LH) -- node[auto] {$\scriptstyle \exp$} (H);
    \draw[->] (LG) -- node[auto] {$\scriptstyle \exp$} (G);
    \draw[->] (LH) -- node[auto] {$\scriptstyle (d\phi)_e$} (LG);
    \draw[->] (H) -- node[auto] {$\scriptstyle \phi$} (G);
  \end{tikzpicture}}
   If $H$ is connected, then $d\phi$ determines $\phi$.
\end{prop}

\subsection{The Fundamental Theorem}

\cite[Lecture 11]{notes}

Like all good algebraists, we assume that Axiom of Choice.

\begin{thm}[Fundamental Theorem of Lie Groups and Algebras] \label{fundthm} \mbox{}
  \begin{enumerate}
    \item The functor $G\mapsto \Lie(G)$ gives an equivalence of categories between the category \cat{scLieGp} of simply-connected Lie groups (over $\RR$ or $\CC$) and the category \cat{LieAlg} of finite-dimensional Lie algebras (over $\RR$ or $\CC$).
    \item ``The'' inverse functor $\hh \mapsto \Grp(\hh)$ is left-adjoint to $\Lie: \cat{LieAlg} \to \cat{LieGp}$.
  \end{enumerate}
\end{thm}

We outline the proof.  Consider open neighborhood $U$ and $V$ so that the horizontal maps are a homeomorphism:
\eqn{ \begin{tikzpicture}[baseline=(U.base)]
  \path node (U) {$U$} ++(2,0) node (V) {$V$};
  \path (U.east) +(0,2pt) node (U1) {} +(0,-2pt) node (U2) {};
  \path (V.west) +(0,2pt) node (V1) {} +(0,-2pt) node (V2) {};
  \path (U.north) node[rotate=90,anchor=west] (s1) {$\subseteq$} (s1.east) +(0,4pt) node[anchor=base] {$\Lie(G)$};
  \path (V.north) node[rotate=90,anchor=west] (s2) {$\subseteq$} (s2.east) +(0,4pt) node[anchor=base] {$G$};
  \path (U.south) node[rotate=90,anchor=east] (i1) {$\in$} (i1.west) +(0,-10pt) node[anchor=base] {$0$};
  \path (V.south) node[rotate=90,anchor=east] (i2) {$\in$} (i2.west) +(0,-10pt) node[anchor=base] {$e$};
  \draw[->] (U1) -- node[auto] {$\scriptstyle \exp$} (V1);
  \draw[->] (V2) -- node[auto] {$\scriptstyle \log$} (U2);
\end{tikzpicture} }
Consider the restriction $\mu: G\times G\to G$ to $(V\times V)\cap \mu^{-1}(V) \to V$, and use this to define a ``partial group law'' $b : \textrm{open} \to U$, where $\textrm{open} \subseteq U\times U$, via
\eqn{ b(x,y) \defined \log(\exp x \exp y) }
We will show that the Lie algebra structure of $\Lie(G)$ determines $b$.

Moreover,  given $\hh$ a finite-dimensional Lie algebra, we will need to define $b$ and build $\tilde{H}$ as the group freely generated by $U$ modulo the relations $xy = b(x,y)$ if $x,y,b(x,y) \in U$.  We will need to prove that $\tilde H$ is a Lie group, with $U$ as an open submanifold.

\begin{cor} \label{fundcor}
  Every Lie subalgebra $\hh$ of $\Lie(G)$ is $\Lie(H)$ for a unique connected subgroup $H \mono G$, up to equivalence.
\end{cor}

The standard proof of \cref{fundthm} is to first prove \cref{fundcor} and then use \cref{AdoThm}.  We will use \cref{LieThm3} rather than \cref{AdoThm}.

\begin{thm}[Baker-Campbell-Hausdorff Formula (second part only)] \label{BCHthm2}
  \mbox{} \begin{enumerate}
    \item Let $\Tt(x,y)$ be the free tensor algebra generated by $x$ and $y$, and $\Tt(x,y)[[s,t]]$ the (non-commutative) ring of formal power series in two commuting variables $s$ and $t$.  Define $b(tx,sy) \defined \log(\exp(tx)\exp(sy)) \in \Tt(x,y)[[s,t]]$, where $\exp$ and $\log$ are the usual formal power series.  Then
    \eqn[BCHeqn]{ b(tx,sy) = tx + sy + st\frac12[x,y] + st^2\frac1{12}[x,[x,y]] + s^2t\frac1{12}[y,[y,x]] + \dots }
    has coefficients all coefficients given by Lie bracket polynomials in $x$ and $y$.
    \item Given a Lie group $G$, there exists a neighborhood $U' \ni 0$ in $\Lie(G)$ such that $U' \subseteq U \underset{\log}{\overset{\exp}{\rightleftarrows}} V\subseteq G$ and $b(x,y)$ converges on $U' \times U'$ to $\log(\exp x\exp y)$.
  \end{enumerate}
\end{thm}
  We need more machinery than we have developed so far to prove part 1.  We work with analytic manifolds; on $\Ccc$ manifolds, we can make an analogous argument using the language of differential equations.
\begin{proof}[of part 2.]
  For a clearer exposition, we distinguish the maps $\exp: \Lie(G) \to G$ from $e^x \in \RR[[x]]$.
  
  We begin with a basic identity.  $\exp(tx)$ is an integral curve to $\ell x$ through $e$, so by left-invariance, $t\mapsto g\exp(tx)$ is the integral curve of $\ell x$ through $g$.  Thus, for $f$ analytic on $G$,
\eqn{ \frac{d}{dt} \bigl[ f(g \exp tx) \bigr] = \bigl((\ell x) f\bigr)(g\exp tx) }
We iterate:
\eqn{ \left( \frac{d}{dt}\right)^n \bigl[ f(g\exp tx)\bigr] = \bigl( (\ell x)^n f\bigr)(g\exp tx) }
If $f$ is analytic, then for small $t$ the Taylor series converges:
\begin{align}
  f(g\exp tx) & = \sum_{n=0}^\infty\left. \left( \frac{d}{dt}\right)^n \bigl[ f(g\exp tx)\bigr] \right|_{t=0} \frac{t^n}{n!} \\
   &  = \sum_{n=0}^\infty \left. \bigl( (\ell x)^n f\bigr)(g\exp tx) \right|_{t=0} \frac{t^n}{n!} \\
   & = \sum_{n=0}^\infty \bigl( (\ell x)^n f\bigr)(g)  \frac{t^n}{n!} \\
   & = \sum_{n=0}^\infty \left(\frac{ (t\,\ell x)^n }{n!}f\right)(g)  \\
   & = \bigl( e^{t\,\ell x}f\bigr) (g)
\end{align}
We repeat the trick:
\eqn{ f(\exp tx \exp sy) = \bigl( e^{s\,Ly}f\bigr) (\exp tx) = \bigl( e^{t\,\ell x} e^{s\,Ly}f\bigr) (e) = \bigl( e^{tx} e^{sy}f\bigr) (e) }
The last equality is because we are evaluating the derivations at $e$, where $\ell x = x$.

We now let $f = \log: V\to U$, or rather a coordinate of $\log$.  Then the left-hand-side is just $\log(\exp tx \exp sy)$, and the right hand side is $\bigl( e^{tx} e^{sy}\log\bigr) (e) = \bigl( e^{b(tx,sy)}\log\bigr) (e)$, where $b$ is the formal power series from part 1.\ --- we have shown that the right hand side converges.  But by interpreting the calculations above as formal power series, and expanding $\log$ in Taylor series, we see that the formal power series $\bigl( e^{b(tx,sy)}\log\bigr) (e)$ agrees with the formal power series $\log\bigl(e^{b(tx,sy)}\bigr) = b(tx,sy)$.  This completes the proof of part 2.
\end{proof}

\section{Universal Enveloping Algebras}

\subsection{The Definition}

\cite[Lecture 12]{notes}

\begin{defn}
  A \emindexing{representation} of a Lie group\index{Lie group!representation of} is a homomorphism $G \to \GL(n,\RR)$ (or $\CC$).  A {\em representation} of a Lie algebra\index{Lie algebra!representation of} is a homomorphism $\Lie(G) \to \gl(n)= \End(V)$; the space $\End(V)$ is a Lie algebra with the bracket given by $[x,y] = xy - yx$.
\end{defn}
\begin{defn}
  Let $V$ be a vector space.  The \emindexing{tensor algebra} over $V$ is the free unital non-commuting algebra $\Tt V$ generated by a basis of $V$.  Equivalently:
  \eqn{ \Tt V \defined \bigoplus_{n\geq 0} V^{\otimes n}}
  The multiplication is given by $\otimes: V^{\otimes k} \times V^{\otimes l} \to V^{\otimes(k+l)}$.  $\Tt$ is a functor, and is left-adjoint to $\Forget: \cat{Alg} \to \cat{Vect}$.
\end{defn}
\begin{defn}
  Let $\gg$ be a Lie algebra.  The \emindexing{universal enveloping algebra} is
  \eqn{ \Uu\gg \defined \Tt\gg / \bigl< [x,y] - (xy - yx) \bigr> }
  $\Uu: \cat{LieAlg} \to \cat{Alg}$ is a functor, and is left-adjoint to $\Forget: \cat{Alg} \to \cat{LieAlg}$.
\end{defn}
\begin{cor}
  The category of $\gg$-modules is equal to the category of $\Uu\gg$-modules.
\end{cor}
\begin{eg}
  A Lie algebra $\gg$ is {\em abelian}\index{abelian Lie algebra}\index{Lie algebra!abelian} if the bracket is identically $0$.  If $\gg$ is abelian, then $\Uu\gg = \Ss\gg$, where $\Ss V$ is the symmetric algebra generated by the vector space $V$ (so that $\Ss$ is left-adjoint to $\Forget: \cat{ComAlg} \to \cat{Vect}$).
\end{eg}
\begin{eg}
  If $\ff$ is the free Lie algebra on generators $x_1,\dots,x_d$, defined in terms of a universal property, then $\Uu\ff = \Tt(x_1,\dots,x_d)$.
\end{eg}

\begin{defn}
  A vector space $V$ is {\em graded}\index{graded vector space}\index{vector space!graded} if it comes with a direct-sum decomposition $V = \bigoplus_{n \geq 0}V_n$.  A morphism of graded vector spaces\index{homomorphism!of graded vector spaces} preserves the grading.  A \emindexing{graded algebra}\index{algebra!graded} is an algebra object in the category of graded vector spaces.  I.e.\ it is a vector space $V = \bigoplus_{n\geq 0}$ along with a unital associative multiplication $V\otimes V\to V$ such that if $v_n\in V_n$ and $v_m\in V_m$, then $v_nv_m \in V_{n+m}$.
  
  A vector space $V$ is {\em filtered}\index{filtered vector space}\index{vector space!filtered} if it comes with an increasing sequence of subspaces 
 \eqn{  \{0\} \subseteq V_{\leq 0} \subseteq V_{\leq 1} \subseteq \dots \subseteq V}
 such that $V = \bigcup_{n\geq 0} V_n$.
  A morphism of graded vector spaces\index{homomorphism!of filtered vector spaces} preserves the filtration.  A \emindexing{filtered algebra}\index{algebra!filtered} is an algebra object in the category of filtered vector spaces.  I.e.\ it is a filtered vector space along with a unital associative multiplication $V\otimes V\to V$ such that if $v_n\in V_{\leq n}$ and $v_m\in V_{\leq m}$, then $v_nv_m \in V_{\leq (n+m)}$.
  
  Given a filtered vector space $V$, we define $\gr V \defined \bigoplus_{n\geq 0} \gr_n V$, where $\gr_n V \defined V_{\leq n} / V_{\leq (n-1)}$.
\end{defn}
\begin{lemma}
  $\gr$ is a functor.  If $V$ is a filtered algebra, then $\gr V$ is a graded algebra.
\end{lemma}
\begin{eg}
  Let $\gg$ be a Lie algebra over $\KK$.  Then $\Uu\gg$ has a natural filtration inherited from the filtration of $\Tt\gg$, since the ideal $\<xy - yx = [x,y]\>$ preserves the filtration.  Since $\Uu\gg$ is generated by $\gg$, so is $\gr\Uu\gg$; since $xy - yx = [x,y] \in \Uu_{\leq 1}$, $\gr\Uu\gg$ is commutative, and so there is a natural projection $\Ss\gg \onto \gr\Uu\gg$.
\end{eg}

\subsection{Poincar\'e-Birkhoff-Witt Theorem}

\cite[Lectures 12 and 13]{notes}

\begin{thm}[Poincar\'e-Birkhoff-Witt] \label{PBWthm}
  The map $\Ss\gg \to \gr\Uu\gg$ is an isomorphism of algebras.
\end{thm}
\begin{cor}
  $\gg\mono \Uu\gg$.  Thus every Lie algebra is isomorphic to a Lie subalgebra of some $\End(V)$, namely $V = \Uu\gg$.
\end{cor}

\begin{proof}[of \cref{PBWthm}]
  Pick an ordered basis $\{x_\alpha\}$ of $\gg$; then the monomials $x_{\alpha_1}\dots x_{\alpha_n}$ for $\alpha_1\leq \dots \leq \alpha_n$ are an ordered basis of $\Ss\gg$, where we take the ``deg-lex'' ordering: a monomial of lower degree is immediately smaller than a monomial of high degree, and for monomials of the same degree we alphabetize.  Since $\Ss\gg \onto \gr\Uu\gg$ is an algebra homomorphism, the set $\{ x_{\alpha_1}\dots x_{\alpha_n} \st \alpha_1\leq \dots \leq \alpha_n\}$ spans $\gr\Uu\gg$.  It suffices to show that they are independent in $\gr\Uu\gg$.  For this it suffices to show that the set $S \defined \{ x_{\alpha_1}\dots x_{\alpha_n} \st \alpha_1\leq \dots \leq \alpha_n\}$ is independent in $\Uu\gg$.
  
  Let $I = \<xy - yx - [x,y]\>$ be the ideal of $\Tt\gg$ such that $\Uu\gg = \Tt\gg / I$.  Define $J \subseteq \Tt\gg$ to be the span of expressions of the form
  \eqn[PBWpf1]{ \xi = x_{\alpha_1}\cdots x_{\alpha_k} \left( x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma] \right) x_{\nu_1}\cdots x_{\nu_l} }
  where $\alpha_1 \leq \dots \leq \alpha_k \leq \beta > \gamma$, and there are no conditions on $\nu_i$, so that $J$ is a right ideal.  We take the deg-lex ordering in $\Tt\gg$.  The leading monomial in \cref{PBWpf1} is $x_{\vec\alpha} x_\beta x_\gamma x_{\vec\nu}$.  Thus $S$ is an independent set in $\Tt\gg/J$.  We need only show that $J = I$.
  
  The ideal $I$ is generated by expressions of the form $x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma]$ as a two-sided ideal. If $\beta > \gamma$ then $\bigl(x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma]\bigr) \in J$; by antisymmetry, if $\beta < \gamma$ we switch them and stay in $J$.  If $\beta = \gamma$, then $\bigl(x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma]\bigr) = 0$.  Thus $J$ is a right ideal contained in $I$, and the two-sided ideal generated by $J$ contains $I$.  Thus the two-sided ideal generated by $J$ is $I$, and it suffices to show that $J$ is a two-sided ideal.
  
  We multiply $x_\delta \xi$.  If $k>0$ and $\delta \leq \alpha_1$, then $x_\delta \xi \in J$.  If $\delta > \alpha_1$, then $x_\delta \xi \equiv x_{\alpha_1} x_\delta x_{\alpha_2} \dots + [x_\delta,x_{\alpha_1}]x_{\alpha_2} \dots$ mod $J$.  And both $x_\delta x_{\alpha_2}\dots$ and $[x_\delta,x_{\alpha_1}]x_{\alpha_2} \dots$ are in $J$ by induction on degree.  Then since $\alpha_1 < \delta$, $x_{\alpha_1}x_\delta x_{\alpha_2}\dots \in J$ by (transfinite) induction on $\delta$.

So suffice to show that if $k = 0$, then we're still in $J$.  I.e.\ if $\alpha > \beta > \gamma$, then we want to show that $x_\alpha\left(x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma] \right) \in J$.  Well, since $\alpha > \beta$, we see that $x_\alpha x_\beta - x_\beta - [x_\alpha,x_\beta] \in J$, and same with $\beta \tofrom \gamma$.  So, working modulo $J$, we have
\begin{align*}
  x_\alpha\left(x_\beta x_\gamma - x_\gamma x_\beta - [x_\beta,x_\gamma] \right) & \equiv \left( x_\beta x_\alpha + [x_\alpha,x_\beta] \right) x_\gamma - \left( x_\gamma x_\alpha + [x_\alpha,x_\gamma] \right) x_\beta - x_\alpha[x_\beta,x_\gamma] \\
    & \equiv  x_\beta \left( x_\gamma x_\alpha + [x_\alpha,x_\gamma] \right) + [x_\alpha, x_\beta] x_\gamma - x_\gamma \left( x_\beta x_\alpha + [x_\alpha,x_\beta] \right)  \\
    & \;\;\;\;\; - [x_\alpha,x_\gamma] x_\beta - x_\alpha [x_\beta,x_\gamma] \\
    & \equiv x_\gamma x_\beta x_\alpha + [x_\beta,x_\gamma]x_\alpha + x_\beta [x_\alpha,x_\gamma]  + [x_\alpha, x_\beta] x_\gamma - x_\gamma \left( x_\beta x_\alpha + [x_\alpha,x_\beta] \right) \\
    & \;\;\;\;\; - [x_\alpha,x_\gamma] x_\beta - x_\alpha [x_\beta,x_\gamma] \\
    & =  [x_\beta,x_\gamma]x_\alpha + x_\beta [x_\alpha,x_\gamma]  + [x_\alpha, x_\beta] x_\gamma - x_\gamma  [x_\alpha,x_\beta]  - [x_\alpha,x_\gamma] x_\beta - x_\alpha [x_\beta,x_\gamma] \\
    & \equiv  -[x_\alpha,[x_\beta,x_\gamma]] + [x_\beta,[x_\alpha,x_\gamma]] - [x_\gamma,[x_\alpha,x_\beta]] \\
    & =  0 \textrm{ by Jacobi.}
\end{align*}
\end{proof}

\subsection{$\Uu\gg$ is a bialgebra}

\cite[Lecture 13]{notes}

\begin{defn}
  An \emindexing{algebra} over $\KK$ is a vector space $U$ along with a  $\KK$-linear ``\indexing{multiplication}'' map $\mu: U\underset{\KK}{\otimes} U \to U$ which is \emindexing{associative}, i.e.\ the following diagram commutes:
    \eqn[AlgAssoc]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (GGG) {$U \otimes U \otimes U$} +(2,0) node (GG1) {$U \otimes U$} +(0,-1) node (GG2) {$U\otimes U$} +(2,-1) node (G) {$U$} +(0,-.5) node (X) {};
    \draw[->] (GGG) -- node[auto] {$\scriptstyle 1_U\otimes \mu$} (GG1);
    \draw[->] (GGG) -- node[auto] {$\scriptstyle \mu \otimes 1_U$} (GG2);
    \draw[->] (GG1) -- node[auto] {$\scriptstyle \mu$} (G);
    \draw[->] (GG2) -- node[auto] {$\scriptstyle \mu$} (G);
  \end{tikzpicture} }
  We demand that all our algebras be \emindexing{unital}, meaning that there is a linear map $e: \KK \to U$ such that the maps $U = \KK \otimes U \overset{e\otimes 1_U}\to U\otimes U \overset{\mu}\to U$ and $U = U\otimes K \overset{1_U\otimes e}\to U\otimes U \overset{\mu}\to U$ are the identity maps.  We will call the image of $1\in \KK$ under $e$ simply $1\in U$.
  
  A \emindexing{coalgebra} is an algebra in the opposite category.  I.e.\ it is a vector space $U$ along with a ``\indexing{comultiplication}'' map $\Delta:U \to U\otimes U$ so that the following commutes:
      \eqn[CoalgAssoc]{ \begin{tikzpicture}[baseline=(X.base),scale=2]
    \path node (GGG) {$U $} +(2,0) node (GG1) {$U \otimes U$} +(0,-1) node (GG2) {$U\otimes U$} +(2,-1) node (G) {$U\otimes U \otimes U$} +(0,-.5) node (X) {};
    \draw[->] (GGG) -- node[auto] {$\scriptstyle \Delta$} (GG1);
    \draw[->] (GGG) -- node[auto] {$\scriptstyle \Delta$} (GG2);
    \draw[->] (GG1) -- node[auto] {$\scriptstyle 1_U \otimes \Delta$} (G);
    \draw[->] (GG2) -- node[auto] {$\scriptstyle \Delta \otimes 1_U$} (G);
  \end{tikzpicture} }
  In elements, if $\Delta x = \sum x_{(1)} \otimes x_{(2)}$, then we demand that $\sum x_{(1)} \otimes \Delta(x_{(2)}) = \sum \Delta(x_{(1)}) \otimes x_{(1)} $.  We demand that our coalgebras be \emindexing{counital}, meaning that there is a linear map $\epsilon: U\to \KK$ such that the maps $U \overset{\Delta} \to U\otimes U \overset{\epsilon \otimes 1_U} \KK \otimes U = U$ and $U \overset{\Delta} \to U\otimes U \overset{ 1_U \otimes \epsilon} U\otimes \KK = U$ are the identity maps.
  
  A \emindexing{bialgebra} is an algebra in the category of coalgebras, or equivalently a coalgebra in the category of algebras.  I.e.\ it is a vector space $U$ with maps $\mu: U\otimes U \to U$ and $\Delta: U\to U\otimes U$ satisfying \cref{AlgAssoc,CoalgAssoc} such that $\Delta$ and $\epsilon$ are (unital) algebra homomorphisms or equivalently such that $\mu$ and $e$ are (counital) coalgebra homomorphism.  We have defined the multiplication on $U \otimes U$ by $(x\otimes y)(z\otimes w) = (xz) \otimes (yw)$, and the comultiplication by $\Delta(x\otimes y) = \sum \sum x_{(1)} \otimes y_{(1)} \otimes x_{(2)} \otimes y_{(2)}$, where $\Delta x = \sum x_{(1)} \otimes x_{(2)}$ and $\Delta y = \sum y_{(1)} \otimes y_{(2)}$; the unit and counit are $e\otimes e$ and $\epsilon\otimes \epsilon$.
\end{defn}
\begin{defn}
  Let $U$ be a bialgebra, and $x\in U$.  We say that $x$ is \emindexing{primitive} if $\Delta x = x\otimes 1 + 1\otimes x$, and that $x$ is \emindexing{grouplike} if $\Delta x = x\otimes x$.  The set of primitive elements of $U$ we denote by $\prim U$.
\end{defn}

\begin{prop} \label{UgBiAlgProp}
  $\Uu\gg$ is a bialgebra with $\prim \Uu\gg = \gg$.
\end{prop}
\begin{proof}
  To define the comultiplication, it suffices to show that $\Delta:\gg \to \Uu\gg \otimes \Uu\gg$ given by $x\mapsto x\otimes 1 + 1\otimes x$ is a Lie algebra homomorphism, whence it uniquely extends to an algebra homomorphism by the universal property.  We compute:
  \begin{align}
  [x \otimes 1 + 1 \otimes x , y \otimes 1 + 1 \otimes y ]_{\Uu \otimes \Uu} & = [ x\otimes 1,y\otimes 1] + [1\otimes x,1\otimes y] \\
   & = [x,y]\otimes 1 + 1\otimes [x,y]
\end{align}
  To show that $\Delta$ thus defined is coassiciative, it suffices to check on the generating set $\gg$, where we see that $\Delta^2(x) = x\otimes 1 \otimes 1 + 1\otimes x \otimes 1 + 1\otimes 1 \otimes x$.
  
  By definition, $\gg \subseteq \Uu\gg$.  To show equality, we use \cref{PBWthm}.  We filter $\Uu\gg \otimes \Uu\gg$ in the obvious way, and since $\Delta$ is an algebra homomorphism, we see that $\Delta\bigl(\Uu\gg_{\leq 1}\bigr) \subseteq \bigl(\Uu\gg \otimes \Uu\gg\bigr)_{\leq 1}$, whence $\Delta\bigl(\Uu\gg_{\leq n}\bigr) \subseteq \bigl(\Uu\gg \otimes \Uu\gg\bigr)_{\leq n}$.  Thus $\Delta$ induces a map $\bar\Delta$ on $\gr\Uu\gg = \Ss\gg$, and $\bar\Delta$ makes $\Ss\gg$ into a bialgebra.
  
  Let $\xi \in \Uu\gg_{\leq n}$ be primitive, and define its image to be $\bar\xi \in \gr_n\Uu\gg$; then $\bar\xi$ must also be primitive.  But $\Ss\gg \otimes \Ss\gg = \KK[y_\alpha,z_\alpha]$, where $\{x_\alpha\}$ is a basis of $\gg$ (whence $\Ss\gg = \KK[x_\alpha]$), and we set $y_\alpha = x_\alpha\otimes 1$ and $z_\alpha = 1\otimes x_\alpha]$.  We check that $\bar\Delta(x_\alpha) = y_\alpha + z_\alpha$, and so if $f(x)\in \Ss\gg$, we see that $\Delta f(x) = f(y+z)$.  So $f\in \Ss\gg$ is primitive if and only if $f(y+z) = f(y) + f(z)$, i.e.\ iff $f$ is homogenoues of degree $1$.  Therefore $\prim \gr\Uu\gg = \gr_1\Uu\gg$, and so if $\xi \in \Uu\gg$ is primitive, then $\bar\xi \in \gr_1\Uu\gg$ so $\xi = x + c$ for some $x\in \gg$ and some $c\in \KK$.  Since $x$ is primitive, $c$ must be also, and the only primitive constant is $0$.
\end{proof}

\subsection{Geometry of the Universal Enveloping Algebra}

\cite[Lecture 14]{notes}

\begin{defn}\label{GDiffOpDef}
  Let $X$ be a space and $\Sss$ a sheaf of functions on $X$.  We define the sheaf $\Ddd$ of {\em grothendieck differential operators}\index{differential operator} inductively.  Given $U \open X$, we define $\Ddd_{\leq 0}(U) = \Sss(U)$, and $\Ddd_{\leq n}(U) = \{x: \Sss(U) \to \Sss(U) \st [x,f] \in \Ddd_{\leq (n-1)}(U)\,\forall f\in \Sss(U)\}$, where $\Sss(U) \acts \Sss(U)$ by left-multiplication.  Then $\Ddd(U) = \bigcup_{n\geq 0} \Ddd_{\leq n}(U)$ is a filtered sheaf; we say that $x\in \Ddd_{\leq n}(U)$ is an ``$n$th-order differential operator on $U$''.
\end{defn}

\begin{lemma}
  $\Ddd$ is a sheaf of filtered algebras, with the multiplication on $\Ddd(U)$ inherited from $\End(\Sss(U))$.  For each $n$, $\Ddd_{\leq n}$ is a sheaf of Lie subalgebras of $\Ddd$.
\end{lemma}

\begin{thm}[Grothendieck Differential Operators]
  Let $X$ be a manifold, $\Ccc$ the sheaf of smooth functions on $X$, and $\Ddd$ the sheaf of differential operators on $\Ccc$ as in \cref{GDiffOpDef}.  Then $\Ddd(U)$ is generated as a noncommutative algebra by $\Ccc(U)$ and $\Vect(U)$, and $\Ddd_{\leq 1} = \Ccc(U) \oplus\Vect(U)$.
\end{thm}

\begin{prop} \label{ugleftinvprop}
  Let $G$ be a Lie group, and $\Ddd(G)^G$ the subalgebra of left-invariant differential operators on $G$.  The natural map $\Uu\gg \to \Ddd(G)^G$ generated by the identification of $\gg$ with left-invariant vector fields is an isomorphism of algebras.
\end{prop}

\section{The Baker-Campbell-Hausdorff Formula}

\cite[Lecture 14]{notes}

\begin{lemma} \label{BCHlemma1}
  Let $U$ be a bialgebra with comultiplication $\Delta$.  Define $\hat\Delta: U[[s]] \to (U\otimes U)[[s]]$ by linearity; then $\hat\Delta$ is an $s$-adic-continuous algebra homomorphism, and so commutes with formal power series.
  
   Let $\psi \in U[[s]]$ with $\psi(0) = 0$.  Then $\psi$ is primitive term-by-term --- $\hat\Delta(\psi) = \psi \otimes 1 + 1\otimes \psi$, if and only if $e^\psi$ is ``group-like'' in the sense that $\hat\Delta(e^\psi) = e^\psi \otimes e^\psi$, where we have defined $\otimes: U[[s]] \otimes U[[s]] \to (U\otimes U)[[s]]$ by $s^n \otimes s^m \mapsto s^{n+m}$.
\end{lemma}
\begin{proof}
  $e^\psi \otimes e^\psi = (1\otimes e^\psi)(e^\psi \otimes 1) = e^{1\otimes \psi}e^{\psi\otimes 1} = e^{1\otimes \psi + \psi \otimes 1}$
\end{proof}

\begin{lemma} \label{BCHlemma2}
  Let $G$ be a Lie group, $\gg = \Lie(G)$, and identify $\Uu\gg$ with the left-invariant differential operators on $G$, as in \cref{ugleftinvprop}.  Let $\Ccc(G)_e$ be the stalk of smooth functions defined in some open set around $e$ (we write $\Ccc$ for the sheaf of functions on $G$; when $G$ is analytic, we really mean the sheaf of analytic functions on $G$).  Then if $u \in \Uu\gg$ satisfied $uf(e) = 0$ for each $f\in \Ccc(G)_e$, then $u = 0$.
\end{lemma}
\begin{proof}  For $g\in G$, we have $uf(g) = u (\lambda_{g^{-1}}f)(e) = \lambda_{g^{-1}}(uf)(e) = 0$. \end{proof}

\begin{thm}[Baker-Campbell-Hausdorff Formula]  \label{BCHthm}
  \begin{enumerate}
    \item Let $\ff$ be the free Lie algebra on two generators $x,y$; recall that $\Uu\ff = \Tt(x,y)$.  Define the formal power series $b(tx,sy) \in \Tt(x,y)[[s,t]]$, where $s$ and $t$ are commuting variables, by
    \eqn{ e^{b(tx,sy)} \defined e^{tx}e^{sy}}
    Then $b(tx,sy) \in \ff[[s,t]]$, i.e.\ $b$ is a series all of whose coefficients are Lie algebra polynomials in the generators $x$ and $y$.
    \item If $G$ is a Lie group (in the analytic category), then there are open neighborhoods $0\in U'\open U \open \Lie(G) = \gg$ and $0\in V' \open V \open G$ such that $U \overset{\exp}{\underset{\log}{\rightleftarrows}} V$ and $U' \rightleftarrows V'$ and such that $b(x,y)$ converges on $U' \times U'$ to $\log(\exp x \exp y)$.
  \end{enumerate}
\end{thm}

\begin{proof}
  \begin{enumerate}
    \item Let $\hat\Delta : \Tt(x,y)[[s,t]] = \Uu\ff[[s,t]] \to (\Uu\ff\otimes \Uu\ff)[[s,t]]$ as in \cref{BCHlemma1}.  Since $e^{tx}e^{sy}$ is grouplike --- 
    \eqn{ \hat\Delta(e^{tx}e^{sy}) = \hat\Delta(e^{tx})\,\hat\Delta(e^{sy}) = \left(e^{tx}\otimes e^{tx}\right) \left(e^{sy}\otimes e^{sy}\right) = e^{tx}e^{sy}\otimes e^{tx}e^{ty}} --- we see that $b(tx,sy)$ is primitive term-by-term.
    \item Let $U,V$ be open neighborhoods of $\Lie(G)$ and $G$ respectively, and pick $V'$ so that $\mu:G\times G\to G$ restricts to a map $V'\times V' \to V$; let $U' = \log(V')$.  Define $\beta(x,y) = \log(\exp x \exp y)$; then $\beta$ is an analytic function $U' \times U' \to U'$.
    
    Let $x,y\in \Lie(G)$ and $f\in \Ccc(G)_e$.  Then $\bigl(e^{tx}e^{sy}f\bigr)(e)$ is the Taylor series expansion of $f(\exp tx \exp sy)$, as in the proof of \cref{BCHthm2}.  Let $\tilde\beta$ be the formal power series that is the Taylor expansion of $\beta$; then $e^{\tilde\beta(tx,sy)}f(e)$ is also the Taylor series expansion of $f(\exp tx \exp sy)$.  This implies that for every $f \in  \Ccc(G)_e$, $e^{\tilde\beta(tx,sy)}f(e)$ and $e^{tx}e^{sy}f(e)$ have the same coefficients.  But the coefficients are left-invariant differential operators applied to $f$, so by \cref{BCHlemma2} the series $e^{\tilde\beta(tx,sy)}$ and $e^{tx}e^{sy}$ must agree.  Upon applying the formal logarithms, we see that $b(tx,sy) = \tilde\beta(tx,sy)$.
    
    But $\tilde\beta$ is the Taylor series of the analytic function $\beta$, so by shrinking $U'$ (and hence $V'$) we can assure that it converges.
  \end{enumerate}
\end{proof}

\section{Lie Subgroups}

\subsection{Relationship between Lie subgroups and Lie subalgebras}

\cite[Lecture 15]{notes}

\begin{defn}
  Let $G$ be a Lie group.  A \emindexing{Lie subgroup} of $G$ is a subgroup $H$ of $G$ with its own Lie group structure, so that the inclusion $H\mono G$ is a local immersion.  We will write ``$H\leq G$'' when $H$ is a Lie subgroup of $G$.
\end{defn}
\begin{thm}[Identification of Lie subalgebras and Lie subgroups] \label{subthm}
  Every Lie subalgebra of $\Lie(G)$ is $\Lie(H)$ for a unique connected Lie subgroup $H \leq G$.
\end{thm}
\begin{proof}
  We first prove uniqueness.  If $H$ is a Lie subgroup of $G$, with $\hh = \Lie(H)$ and $\gg = \Lie(G)$, then the following diagram commutes:
  \eqn{ \begin{tikzpicture}[baseline=(X.base),scale=1.1]
    \path node (X) {} ++(0,-.5) node (h) {$\hh$} +(1,0) node (g) {$\gg$} ++(0,1) node (H) {$H$} +(1,0) node (G) {$G$};
    \draw[->] (h) -- node[auto] {$\scriptstyle \exp$} (H);
    \draw[->] (g) -- node[auto,swap] {$\scriptstyle \exp$} (G);
    \draw[right hook->] (h) -- (g);
    \draw[right hook->] (H) -- (G);
  \end{tikzpicture} }
  This shows that $\exp_G(\hh) \subseteq H$, and so $\exp_G(\hh) = \exp_H(\hh)$, and if $H$ is connected, this generates $H$.  So $H$ is uniquely determined by $\hh$ as a group.  Its manifold structure is also uniquely determined: we pick $U,V$ so that the vertical arrows are an isomorphism:
  \eqn[subgpthmeqn]{\begin{tikzpicture}[baseline=(X.base)]
    \path node (X) {} +(0,-.5) node (U) {$0 \in U \subseteq \gg$} +(0,.5) node (V) {$\; e\in V \subseteq G$};
    \path (U.north) +(-2pt,0) coordinate (u1) +(2pt,0) coordinate (u2);
    \path (V.south) +(-2pt,0) coordinate (v1) +(2pt,0) coordinate (v2);
    \draw[->] (u1) -- node[auto] {$\scriptstyle \exp$} (v1);
    \draw[->] (v2) -- node[auto] {$\scriptstyle \log$} (u2);
  \end{tikzpicture}}
  Then $\exp(U\cap \hh) \underset{\log}{\isom} U\cap \hh$ is an immersion into $\gg$, and this defines a chart around $e\in H$, which we can push to any other point $h\in H$ by multiplication by $h$.  This determines the topology and manifold structure of $H$.
  
  We turn now to the question of existence.  We pick $U$ and $V$ as in \cref{subgpthmeqn}, and then choose $V' \open V$ and $U' \defined \log V'$ such that:
  \begin{enumerate}
    \item $(V')^2 \subseteq V$ and $(V')^{-1} = V'$
    \item $b(x,y)$ converges on $U' \times U'$ to $\log(\exp x \exp y)$
    \item $h V' h^{-1} \subseteq V$ for $h\in V'$
    \item $e^{\ad x}y$ converges on $U' \times U'$ to $\log\bigl((\exp x)(\exp y)(\exp x)^{-1}\bigr)$
    \item $b(x,y)$ and $e^{\ad x}y$ are elements of $h\cap U$ for $x,y \in \hh \cap U'$
  \end{enumerate}
  Each condition can be independently achieved on a small enough open set.  In condition 4., we consider extend the formal power series $e^t$ to operators, and remark that in a neighborhood of $0\in \gg$, if $h = \exp x$, then $\Ad h = e^{\ad x}$.  Moreover, the following square always commutes:
  \eqn{ \begin{tikzpicture}[baseline=(X.base),scale=1.1]
    \path node (X) {} ++(0,-.5) node (g1) {$\gg$} +(2,0) node (g2) {$\gg$} ++(0,1) node (G1) {$G$} +(2,0) node (G2) {$G$};
    \draw[->] (g1) -- node[auto] {$\scriptstyle \exp$} (G1);
    \draw[->] (g2) -- node[auto,swap] {$\scriptstyle \exp$} (G2);
    \draw[->] (g1) -- node[auto] {$\scriptstyle \Ad(h)$} (g2);
    \draw[->] (G1) -- node[auto] {$\scriptstyle g \mapsto hgh^{-1}$} (G2);
  \end{tikzpicture} }
  Thus, we define $W = \exp(\hh \cap U')$, which is certainly an immersed submanifold of $G$, as $\hh \cap U'$ is an open subset of the immersed submanifold $\hh \mono \gg$.  We define $H$ to be the subgroup generated by $W$.  Then $H$ and $W$ satisfy the hypotheses of \cref{subprop}.
\end{proof}

\begin{prop}  \label{subprop}
  We use the word ``manifold'' to mean ``object in a particular chosen category of sheaves of functions''.  We use the word ``smooth'' to mean ``morphism in this category''.
  
  Let $H$ be a group and $U\subseteq H$ such that $e\in U$ and $U$ has the structure of a manifold.  Assume further that the maps $U\times U \to H$, ${^{-1}} : U \to H$, and (for each $h$ in a generating set of $H$) $\Ad(h): U \to H$ mapping $u\mapsto huh^{-1}$ have the following properties:
  \begin{enumerate}
    \item The preimage of $U \subseteq H$ under each map is open in the domain.
    \item The restriction of the map to this preimage is smooth.
  \end{enumerate}
  Then $H$ has a unique structure as a group manifold such that $U$ is an open submanifold.
\end{prop}
\begin{proof}
  The conditions 1.\ and 2.\ are preserved under compositions, so $\Ad(x)$ satisfies both conditions for any $x\in H$.  Let $e\in U'\open U$ so that $(U')^3 \subseteq U$ and $(U')^{-1} = U'$.
  
  For $x\in H$, view each coset $xU'$ as a manifold via $U'\overset{x\cdot}\to xU'$.  For any $U'' \open U'$ and $x,y\in G$, consider $yU'' \cap xU'$; as a subset of $xU'$, it is isomorphic to $x^{-1}yU'' \cap U'$.  If this set is empty, then it is open.  Otherwise, $x^{-1}yu_2 = u_1$ for some $u_2\in U''$ and $u_1\in U'$, so $y^{-1}x = u_2u_1^{-1} \in (U')^2$ and so $y^{-1}x U' \subseteq U$.  In particular, the $\{y^{-1}x\} \times U' \subseteq \mu^{-1}(U) \cap (U\times U)$.  By the assumptions, $U' \to y^{-1}xU$ is smooth, and so $x^{-1}yU'' \cap U'$, the preimage of $U''$, is open in $U'$.  Thus the topologies and smooth structures on $xU'$ and $yU'$ agree on their overlap.
  
  In this way, we can put a manifold structure on $H$ by declaring that $S\open H$ if $S\cap xU' \open xU'$ for all $x\in H$ --- the topology is locally the topology of $U \ni e$, and so is Hausdorff --- , and that a function $f$ on $S\open H$ is smooth if its restriction to each $S\cap xU'$ is smooth.
  
  If we were to repeat this story with right cosets rather than left cosets we would get the a similar structure: all the left cosets $xU'$ are compatible, an all the right cosets $U'x$ are compatible.  To show that a right coset is compatible with a left coset, it suffices to show that for each $x\in H$, $xU'$ and $U'x$ have compatible smooth structures.  We consider $xU' \cap U'x \subseteq xU'$, which we transport to $U' \cap x^{-1}U'x \subseteq U'$.  Since we assumed that conjugation by $x$ was a smooth map, we see that right and left cosets are compatible.
  
  We now need only check that the group structure is by smooth maps.  We see that $(xU')^{-1} = (U')^{-1}x^{-1} = U'x^{-1}$, and multiplication is given by $\mu: xU' \times U'y \to xUy$.  Left- and right-multiplication maps are smooth with respect to the left- and right-coset structures, which are compatible, and we assumed that $\mu: U' \times U' \to U$ was smooth.
\end{proof}

\subsection{Review of Algebraic Topology}

\cite[Lecture 16]{notes}

\begin{defn}
  A \emindexing{groupoid} is a category all of whose morphisms are invertible.
\end{defn}
\begin{defn}
  A space $X$ is \emindexing{connected} if the only subsets of $X$ that are both open and closed are $\emptyset$ and $X$.
\end{defn}

\begin{defn}
  Let $X$ be a space and $x,y\in X$.  A \emindexing{path} from $x$ to $y$, which we write as $x\leadsto y$, is a continuous function $[0,1] \to X$ such that $0\mapsto x$ and $1\mapsto y$.  Given $p: x\leadsto y$ and $q: y\leadsto z$, we define the \emindexing{concatenation} $p\cdot q$ by
  \eqn{ p\cdot q(t) \defined \left\{ \begin{array}{ll} p(2t), & 0 \leq t \leq \frac12 \\ q(2t - 1), & \frac12 \leq t \leq 1 \end{array}\right. }
  We write $x\sim y$ if there is a path connecting $x$ to $y$; $\sim$ is an equivalence relation, and the equivalence classes are {\em path components}\index{path component} of $X$.  If $X$ has only one path component, then it is \emindexing{path connected}\index{connected!path}.
  
  Let $A$ be a distinguished subset of $Y$ and $f,g:Y\to X$ two functions that agree on $A$.  A \emindexing{homotopy} $f\underset{A}\sim g$ relative to $A$ is a continuous map $h: Y\times[0,1] \to X$ such that $h(0,y) = f(y)$, $h(1,y) = g(y)$, and $h(t,a) = f(a) = g(a)$ for $a\in A$.  If $f\sim g$ and $g\sim h$, then $f\sim h$ by concatenation.  The \emindexing{fundamental groupoid}\index{groupoid!fundamental} $\pi_1(X)$ of $X$ has objects the points of $X$ and arrows $x\to y$ the homotopy classes of paths $x\leadsto y$.  We write $\pi_1(X,x)$ for the set of morphisms $x\to x$ in $\pi_1(X)$.  The space $X$ is \emindexing{simply connected}\index{connected!simply} if $\pi_1(X,x)$ is trivial for each $x\in X$.
\end{defn}

\begin{eg}
  A path connected space is connected, but a connected space is not necessarily path connected.  A path is a homotopy of constant maps $\pt \to X$, where $A$ is empty.
\end{eg}

\begin{defn}
  Let $X$ be a space.  A \emindexing{covering space} of $X$ is a space $E$ along with a ``projection'' $\pi: E\to X$ such that there is a non-empty discrete space $S$ and a covering of $X$ by open sets such that for each $U$ in the covering, there exists an isomorphism $\pi^{-1}(U) \isom S\times U$ such that the following diagram commutes:
  \eqn{\begin{tikzpicture}[baseline=(X.base),scale=1.5]
    \path node (X) {} +(0,-.5) node (U) {$U$} +(0,.5) node (c) {$\cong$};
    \path (c.east) node[anchor=west] (S) {$S\times U$};
    \path (c.west) node[anchor=east] (P) {$\pi^{-1}(U)$};
    \draw[->] (P) -- node[auto,swap,pos=.3] {$\scriptstyle \pi$} (U);
    \draw[->] (S) -- node[auto,pos=.3] {\scriptsize \upshape project} (U);
  \end{tikzpicture} }
\end{defn}
\begin{prop}
  Let $\pi_E: E \to X$ be a covering space.
  \begin{enumerate}
    \item Given any path $x\leadsto y$ and a lift $e\in \pi^{-1}(x)$, there is a unique path in $E$ starting at $e$ that projects to $x\leadsto y$.
    \item Given a homotopy $\underset{A}\sim: Y \tto X$ and a choice of a lift of the first arrow, there is a unique lift of the homotopy, provided $Y$ is locally compact.
  \end{enumerate}
  Thus $E$ induces a functor $E: \pi_1(X) \to \cat{Set}$, sending $x\mapsto \pi_E^{-1}(X)$.
\end{prop}

\begin{defn}
  A space $X$ is \emindexing{locally path connected}\index{connected!locally path}\index{path connected!locally} if each $x\in X$ has arbitrarily small path connected neighborhoods.  A space $X$ is \emindexing{locally simply connected}\index{connected!simply path}\index{simply connected!locally} if it has a covering by simply connected open sets.
\end{defn}

\begin{prop}
  Assume that $X$ is path connected, locally path connected, and locally simply connected.  Then:
  \begin{enumerate}
    \item $X$ has a simply connected covering space $\tilde\pi: \tilde X \to X$.
    \item $\tilde X$ satisfies the following universal property: Given $f: X\to Y$ and a covering $\pi: E \to Y$, and given a choice of $x\in X$, an element of $\tilde{x} \in \tilde\pi^{-1}(x)$, and an element $e\in \pi^{-1}(f(x))$, then there exists a unique $\tilde f: \tilde X \to E$ sending $\tilde x \mapsto e$ such that the following diagram commutes:
    \eqn{ \begin{tikzpicture}[baseline=(B.base),scale=1.5]
      \path node (B) {} +(0,.5) node (tX) {$\tilde X$} +(1,.5) node (E) {$E$} +(0,-.5) node (X) {$X$} +(1,-.5) node (Y) {$Y$};
      \draw[->] (tX) -- node[auto] {$\scriptstyle \tilde\pi$} (X);
      \draw[->] (E) -- node[auto] {$\scriptstyle \pi$} (Y);
      \draw[->,dotted] (tX) -- node[auto] {$\scriptstyle \tilde f$} (E);
      \draw[->] (X) -- node[auto] {$\scriptstyle f$} (Y);
    \end{tikzpicture}}
    \item If $X$ is a manifold, so is $\tilde X$.  If $f$ is smooth, so is $\tilde f$.
  \end{enumerate}
\end{prop}

\begin{prop}
  \begin{enumerate}
  \item Let $G$ be a connected Lie group, and $\tilde{G}$ its simply-connected cover.  Pick a point $\tilde{e} \in \tilde{G}$ over the identity $e\in G$.  Then $\tilde{G}$ in its given manifold structure is uniquely a Lie group with identity $\tilde{e}$ such that $\tilde{G} \to G$ is a homomorphism.  This induces an isomorphism of Lie algebras $\Lie(\tilde{G}) \isom \Lie(G)$.
  \item $\tilde{G}$ satisfies the following universal property:  Given any Lie algebra homomorphism $\alpha: \Lie(G) \to \Lie(H)$, there is a unique homomorphism $\phi: \tilde{G} \to H$ inducing $\alpha$.
\end{enumerate}
\end{prop} 

\begin{proof}
  \begin{enumerate}
  \item If $X$ and $Y$ are simply-connected, then so is $X\times Y$, and so by the universal property $\tilde{G} \times \tilde{G}$ is the universal cover of $G\times G$.  We lift the functions $\mu: G\times G\to G$ and $i: G\to G$ to $\tilde{G}$ by declaring that $\tilde\mu(\tilde e,\tilde e) = \tilde e$ and that $i(\tilde e) = \tilde e$; the group axioms (\cref{GroupAssoc,GroupId,GroupInv}) are automatic.
  \item Write $\gg = \Lie(G)$ and $\hh = \Lie(H)$, and let $\alpha: \gg\to \hh$ be a Lie algebra homomorphism.  Then the graph $\ff \subseteq \gg\times \hh$ is a Lie subalgebra.  By \cref{subthm}, $\ff$ corresponds to a subgroup $F \leq \tilde{G} \times H$.  We check that the map $F \mono \tilde G \times H \to G$ induces the map $\ff \to \gg$ on Lie algebras.
  \eqn{ \begin{tikzpicture}[baseline=(X.base),scale=1.5]
    \path node (G) {$G$} ++(0,.5) node (X) {} ++(0,.5) node (H) {$\tilde G \times H$} (H.west) node[anchor=east] (s) {$\subseteq$} (s.west) node[anchor=east] (F) {$F$};
    \draw[->] (H) -- (G);
    \draw[->,dotted] (F) -- node[auto,swap] {$\scriptstyle \ff \isom \gg$} (G);
  \end{tikzpicture} }
  $F$ is connected and simply connected, and so by the universal property, $F\cong\tilde G$.  Thus $F$ is the graph of a homomorphism $\phi: \tilde{G} \to H$.
  \end{enumerate}
\end{proof}

\section{A dictionary between algebras and groups}

\cite[Lecture 17]{notes}

We have completed the proof of \cref{fundthm}, the equivalence between the category of finite-dimensional Lie algebras and the category of simply-connected Lie groups, subject only to \cref{LieThm3}.  Thus a Lie algebra includes most of the information of a Lie group.  We foreshadow a dictionary, most of which we will define and develop later: \\
\\
\begin{tabular}{ll}
  \underline{Lie Algebra $\gg$} & \underline{Lie Group $G$ (with $\gg = \Lie(G)$)} \\ \\
  Subalgebra $\hh \leq \gg$ & Connected Lie subgroup $H \leq G$ \\ \\
  Homomorphism $\hh \to \gg$ & $\tilde{H} \to G$ provided $\tilde{H}$ simply connected \\ \\
  Module/representation $\gg \to \gl(V)$ & Representation $\tilde{G} \to \GL(V)$ ($\tilde{G}$ simply connected) \\ \\
  Submodule $W \leq V$ with $\gg: W \to W$ & Invariant subspace $G: W \to W$ \\ \\
  $V^\gg \defined \{ v \in V \st \gg v = 0\}$ & $V^{\tilde{G}} = \{ v\in V \st Gv = v\}$ \\ \\
  $\ad: \gg \acts \gg$ via $\ad(x)y = [x,y]$ & $\Ad: G \acts G$ via $\Ad(x)y = xyx^{-1}$ \\  \\
  An \emindexing{ideal} $\aa$, i.e.\ $[\gg,\aa] \leq \aa$, i.e.\ sub-$\gg$-module & $A$ is a normal Lie subgroup, provided $G$ is connected \\ \\
  $\gg / \aa$ is a Lie algebra & $G/A$ is a Lie group only if $A$ is closed in $G$ \\ \\
  Center $Z(\gg) = \gg^\gg$ & $Z_0(G)$ the identity component of center; this is closed \\ \\
  Derived subalgebra $\gg' \defined [\gg,\gg]$, an ideal & Should be commutator subgroup, but that's not closed:\\
   & the closure also doesn't work, although if $G$ is compact, \\
   & then the commutator subgroup is closed. \\ \\
  {\em Semidirect product}\index{semidirect product}\index{product!semidirect} $\gg = \hh \oplus \aa$ with  & If $A$ and $H$ are closed, then $A \cap H$ is discrete, and \\
  $\hh \acts \aa$ and $\aa$ an ideal & $\tilde{G} = \tilde{H} \ltimes \tilde{A}$
\end{tabular}

\subsection{Basic Examples: one- and two-dimensional Lie algebras}

We classify the one- and two-dimensional Lie algebras and describe their corresponding Lie groups.  We begin by working over $\RR$.

The only one-dimensional Lie algebra is abelian.  Its connected Lie groups are the line $\RR$ and the circle $S^1$.

There is a unique abelian two-dimensional Lie algebra, given by a basis $\{x,y\}$ with relation $[x,y]=0$.  This integrates to three possible groups: $\RR^2$, $\RR \times (\RR/\ZZ)$, and $(\RR/\ZZ)^2$.

There is a unique nonabelian Lie algebra up to isomorphism, which we call $\bb$.  It has a basis $\{x,y\}$ and defining relation $[x,y] = y$:
\eqn{ \begin{tikzpicture}[scale=1,auto,baseline=(b.base)]
  \node (X) at (0,0) {$-x$};
  \node (Y) at (2,0) {$y$};
  \node (b) at (0,-.5) {};
  \draw[->] (X) to node {$\scriptstyle \ad y$} (Y);
  \draw[->] (Y) .. controls (1,-1) and (3,-1) .. node [swap] {$\scriptstyle \ad X$} (Y);
\end{tikzpicture}}

We can represent $\bb$ as a subalgebra of $\gl(2$ by $x = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$ and $y = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$.  Then $\bb$ exponentiates under $\exp: \gl(2 \to \GL(2$ to the group
\eqn{ B = \left\{ \left[ \begin{array}{cc} a & b \\ 0 & 1 \end{array} \right] \st a\in \RR_+, b\in \RR \right\} }
We check that $B = \RR_+ \ltimes \RR$, and $B$ is connected and simply connected.

\begin{lemma}
  A discrete normal subgroup $A$ of a connected Lie group $G$ is in the center.  In particular, any discrete normal subgroup is abelian.
\end{lemma}

\begin{cor}
  The group $B$ defined above is the only connected group with Lie algebra $\bb$.
\end{cor}
\begin{proof}
  Any other must be a quotient of $B$ by a discrete normal subgroup, but the center of $B$ is trivial.
\end{proof}

We turn now to the classification of one- and two-dimensional Lie algebras and Lie groups over $\CC$.  Again, there is only the abelian one-dimensional algebra, and there are two Lie algebras: the abelian one and the nonabelian one.

The simply connected abelian one-(complex-)dimensional Lie group is $\CC$ under $+$.  Any quotient factors (up to isomorphism) through the cylinder $\CC \to \CC^\times : z\mapsto e^z$.  For any $q\in \CC^\times$ with $|q|\neq 1$, we have a discrete subgroup $q^\ZZ$ of $\CC^\times$, by which we can quotient out; we get a torus $E(q) = \CC^\times / q^\ZZ$.  For each $q$, $E(q)$ is isomorphic to $(\RR/\ZZ)^2$ as a real Lie algebra, but the holomorphic structure depends on $q$.  This exhausts the one-dimensional complex Lie groups.

The groups that integrate the abelian two-dimensional complex Lie algebra are combinations of one-dimensional Lie groups: $\CC^2, \CC\times E, \CC^\times\times\CC^\times,$ etc.

In the non-abelian case, the Lie algebra $\bb_+ \leq \gl(2$ integrates to $B_\CC \leq \GL(2$ given by:
\eqn{ B_\CC = \left\{ \left[ \begin{array}{cc} a & b \\ 0 & 1 \end{array} \right] \st a\in \CC^\times, b\in \CC \right\} = \CC^\times \ltimes \CC }
This is no longer simply connected.  $\CC \acts \CC$ by $z\cdot w = e^zw$, and the simply-connected cover of $B$ is 
\eqn{ \tilde{B}_\CC = \CC \ltimes \CC \;\;\; (w,z)(w',z') \defined (w+e^zw',z+z') }
This is an extension:
\eqn{ 0 \to \ZZ \to \tilde{B}_\CC \to B_\CC \to 0}
with the generator of $\ZZ$ being $2\pi i$.  Other quotients are $\tilde{B}_\CC / n\ZZ$.

\Section{Exercises}

\begin{enumerate}
  \item \begin{enumerate} \item Let $S$ be a commutative $\KK$-algebra. Show that a linear operator $d : S \to S$ is a derivation if and only if it annihilates $1$ and its commutator with the operator of multiplication by every function is the operator of multiplication by another function. 

    \item Grothendieck's inductive definition of differential operators on $S$ goes as follows: the differential operators of order zero are the operators of multiplication by functions; the space $D_{\leq n}$ of operators of order at most $n$ is then defined inductively for $n > 0$ by $D_{\leq n} = \{d \st [d, f ] \in D_{\leq n-1} \textrm{ for all } f \in S \}$. Show that the differential operators of all orders form a filtered algebra $D$, and that when $S$ is the algebra of smooth functions on an open set in $\RR^n$ [or $\CC^n$], $D$ is a free left $S$-module with basis consisting of all monomials in the coordinate derivations $\partial/\partial x^i$.

  \end{enumerate}
  \item Calculate all terms of degree $\leq 4$ in the Baker-Campbell-Hausdorff formula (\cref{BCHeqn}). 

  \item Let $F(d)$ be the free Lie algebra on generators $x_1 , \dots , x_d$. It has a natural $\NN^d$ grading 
in which $F(d)_{(k_1 ,\dots,k_d)}$ is spanned by bracket monomials containing $k_i$ occurences of each generator $X_i$. Use the PBW theorem to prove the generating function identity 
  \[ \prod_{\bf k} \frac{1}{(1 - t_1^{k_1}\dots t_d^{k_d})^{\dim F(d)_{(k_1,\dots,k_d)}}} = \frac{1}{1 - (t_1 + \dots + t_d)} \]
  \item Words in the symbols $x_1 , \dots , x_d$ form a monoid under concatentation, with identity the empty word. Define a \emindexing{primitive word}\index{word!primitive} to be a non-empty word that is not a power of a shorter word. A \emindexing{primitive necklace}\index{necklace!primitive} is an equivalence class of primitive words under rotation.  Use the generating function identity in Problem 3 to prove that the dimension of $F(d)_{k_1,\dots,k_d}$ is equal to the number of primitive necklaces in which each symbol $x_i$ appears $k_i$ times. 
  \item A \emindexing{Lyndon word}\index{word!Lyndon} is a primitive word that is the lexicographically least representative of its primitive necklace. 
  \begin{enumerate} \item Prove that $w$ is a Lyndon word if and only if $w$ is lexicographically less than $v$ for every factorization $w = uv$ such that $u$ and $v$ are non-empty. 

    \item Prove that if $w = uv$ is a Lyndon word of length $> 1$ and $v$ is the longest proper right factor of $w$ which is itself a Lyndon word, then $u$ is also a Lyndon word. This factorization of $w$ is called its \emindexing{right standard factorization}. 
  
    \item To each Lyndon word $w$ in symbols $x_1 , \dots , x_d$ associate the \indexing{bracket polynomial} $p_w = x_i$ if $w = x_i$ has length $1$, or, inductively, $p_w = [p_u , p_v ]$, where $w = uv$ is the right standard factorization, if $w$ has length $> 1$. Prove that the elements $p_w$ form a basis of $F(d)$.

  \end{enumerate} 
  \item Prove that if $q$ is a power of a prime, then the dimension of the subspace of total degree $k_1 + \dots + k_q = n$ in $F(q)$ is equal to the number of monic irreducible polynomials of degree $n$ over the field with $q$ elements. 
  \item This problem outlines an alternative proof of the PBW theorem (\cref{PBWthm}).

  \begin{enumerate} \item Let $L(d)$ denote the Lie subalgebra of $\Tt(x_1 , \dots , x_d )$ generated by $x_1 , \dots , x_d$. Without using the PBW theorem---in particular, without using $F(d) = L(d)$---show that the value given for $\dim F(d)_{(k_1 ,\dots,k_d)}$ by the generating function in Problem 3 is a lower bound for $\dim L(d)_{(k_1,\dots,k_d)}$. 
    \item Show directly that the Lyndon monomials in Problem 5(b) span $F(d)$.    
    \item Deduce from (a) and (b) that $F(d) = L(d)$ and that the PBW theorem holds for $F(d)$.
    \item Show that the PBW theorem for a Lie algebra $\gg$ implies the PBW theorem for $\gg/\aa$, where $\aa$ is a Lie ideal, and so deduce PBW for all finitely generate Lie algebras from (c).
    \item Show that the PBW theorem for arbitrarty Lie algebras reduces to the finitely generated case.
  \end{enumerate}
  \item Let $b(x, y )$ be the Baker-Campbell-Hausdorff series, i.e., $e^{b(x,y )} = e^x e^y$ in noncommuting variables $x$, $y$.  Let $F (x, y )$ be its linear term in $y$, that is, $b(x, sy ) = x + sF (x, y ) + O(s^2 )$. 
  \begin{enumerate} \item Show that $F (x, y )$ is characterized by the identity 
    \eqn[HW38a]{ \sum_{k,l\geq 0} \frac{x^k\,F(x,y)\,x^l}{(k+l+1)!} = e^xy. }

    \item Let $\lambda, \rho$ denote the operators of left and right multiplication by $x$, and let $f$ be the series in two commuting variables such that $F (x, y ) = f (\lambda, \rho)(y )$. Show that 
    \[ f(\lambda,\rho) = \frac{\lambda - \rho}{1 - e^{\rho-\lambda}} \]

    \item Deduce that 
    \[ F(x,y) = \frac{\ad x}{1 - e^{-\ad x}}(y).\]

  \end{enumerate}
  \item Let $G$ be a Lie group, $\gg = \Lie(G)$, $0 \in U' \subseteq U \subseteq \gg$ and $e \in V' \subseteq V \subseteq G$ open 
neighborhoods such that $\exp$ is an isomorphism of $U$ onto $V$, $\exp(U') = V'$, and $V'V' \subseteq V$.  Define $\beta: U' \times U' \to U$ by $\beta (x, y ) = \log(\exp(x ) \exp(y ))$, where $\log : V \to U$ is the inverse of $\exp$. 
  \begin{enumerate} \item Show that $\beta (x, (s + t)y ) = \beta (\beta (x, ty ), sy )$ whenever all arguments are in $U'$. 
     
    \item Show that the series $(\ad x )/(1 - e^{- \ad x})$, regarded as a formal power series in the coordinates of $x$ with coefficients in the space of linear endomorphisms of $\gg$, converges for all $x$ in a neighborhood of $0$ in $\gg$. 

    \item Show that on some neighborhood of $0$ in $\gg$, $\beta (x, ty )$ is the solution of the initial value problem 
    \begin{align}
      \label{HW39c1} \beta(x,0) & = x \\
      \label{HW39c2} \frac{d}{dt} \beta(x,ty) & = F(\beta(x,ty),y),
    \end{align}
    where $F(x,y) = \left( (\ad x)/(1 - e^{-\ad x}) \right)(y)$.

    \item Show that the Baker-Campbell-Hausdorff series $b(x, y )$ also satis�es the identity in part (a), as an identity of formal power series, and deduce that it is the formal power series solution to the IVP in part (c), when $F (x, y )$ is regarded as a formal series. 

    \item Deduce from the above an alternative proof that $b(x, y )$ is given as the sum of a series of Lie bracket polynomials in $x$ and $y$, and that it converges to $\beta (x, y )$ when evaluated on a suitable neighborhood of $0$ in $\gg$. 

    \item Use part (c) to calculate explicitly the terms of $b(x, y )$ of degree $2$ in $y$.

  \end{enumerate}
  \item \begin{enumerate} \item Show that the Lie algebra $\so(3, \CC)$ is isomorphic to $\sl(2, \CC)$. 
    \item Construct a Lie group homomorphism $\SL(2, \CC) \to \SO(3, \CC)$ which realizes the isomorphism of Lie algebras in part (a), and calculate its kernel.
  \end{enumerate}

  \item \begin{enumerate} \item Show that the Lie algebra $\so(4, \CC)$ is isomorphic to $\sl(2, \CC) \times \sl(2, \CC)$.
    \item Construct a Lie group homomorphism $\SL(2, \CC) \times \SL(2, \CC) \to \SO(4, \CC)$ which realizes the isomorphism of Lie algebras in part (a), and calculate its kernel.
  \end{enumerate}

  \item Show that every closed subgroup $H$ of a Lie group $G$ is a Lie subgroup, so that the inclusion $H \mono G$ is a closed immersion.

  \item Let $G$ be a Lie group and $H$ a closed subgroup. Show that $G/H$ has a unique manifold structure such that the action of $G$ on it is smooth.

  \item Show that the intersection of two Lie subgroups $H_1$, $H_2$ of a Lie group $G$ can be given a canonical structure of Lie subgroup so that its Lie algebra is $\Lie(H_1) \cap \Lie(H_2 ) \subseteq \Lie(G)$.

  \item Find the dimension of the closed linear group $\SO(p, q, \RR) \subseteq \SL(p + q, \RR)$ consisting of elements which preserve a non-degenerate symmetric bilinear form on $\RR^{p+q}$ of signature $(p, q)$. When is this group connected?

  \item Show that the kernel of a Lie group homomorphism $G\to H$ is a closed subgroup of $G$ whose Lie algebra is equal to the kernel of the induced map $\Lie(G)\to \Lie(H)$.

  \item Show that if $H$ is a normal Lie subgroup of $G$, then $\Lie(H)$ is a Lie ideal in $\Lie(G)$.

\end{enumerate}















\chapter{General theory of Lie algebras}

\section{$\Uu\lowercase{\gg}$ is a Hopf algebra}

\cite[Lecture 18]{notes}

\begin{defn} \label{HopfDefn}
  A \emindexing{Hopf algebra}\index{algebra!Hopf} over $\KK$ is a (unital, counital) bialgebra $(U,\mu,e,\Delta,\epsilon)$ along with a bialgebra map $\anti: U \to U^\op$ called the \emindexing{antipode},   where $U^\op$ is $U$ as a vector space, with the opposite multiplication and the opposite comultiplication.  I.e.\ we define $\mu^\op :U \otimes U \to U$ by $\mu^\op(x\otimes y) = \mu(y\otimes x)$, and $\Delta^\op : U \to U\otimes U$ by $\Delta^\op(x) = \sum x_{(2)} \otimes x_{(1)}$, where $\Delta(x) = \sum x_{(1)} \otimes x_{(2)}$.   The antipode $\anti$ is required to make the following pentagons commute:
    \eqn[Antipode]{ \begin{tikzpicture}[baseline=(X.base),scale=3]
    \path node (X) {$\KK$} +(-1,0) node (G1) {$U$} +(-.5,.5) node (GG1a) {$U\otimes U$} +(.5,.5) node (GG1b) {$U\otimes U$} +(1,0) node (G2) {$U$} +(-.5,-.5) node (GG2a) {$U\otimes U$} +(.5,-.5) node (GG2b) {$U\otimes U$};
  \draw[->] (G1) -- node[auto] {$\scriptstyle \epsilon$} (X); \draw[->] (X) -- node[auto] {$\scriptstyle e$} (G2);
  \draw[->] (G1) -- node[auto] {$\scriptstyle \Delta$} (GG1a);
  \draw[->] (G1) -- node[auto,swap] {$\scriptstyle \Delta$} (GG2a);
  \draw[->] (GG1b) -- node[auto] {$\scriptstyle \mu$} (G2);
  \draw[->] (GG2b) -- node[auto,swap] {$\scriptstyle \mu$} (G2);
  \draw[->] (GG1a) -- node[auto] {$\scriptstyle 1_G\otimes \anti$} (GG1b);
  \draw[->] (GG2a) -- node[auto,swap] {$\scriptstyle \anti\otimes 1_G$} (GG2b);
  \end{tikzpicture} }
\end{defn} 
\begin{defn}
  An algebra $(U,\mu,e)$ is \emindexing{commutative} if $\mu^\op = \mu$.  A coalgebra $(U,\Delta,\epsilon)$ is \emindexing{cocommutative} if $\Delta^\op = \Delta$.
\end{defn}
\begin{eg}
  Let $G$ be a finite group and $\Ccc(G)$ the algebra of functions on it.  Then $\Ccc(G)$ is a commutative Hopf algebra with $\Delta(f)(x,y) = f(xy)$, where we have identified $\Ccc(G) \otimes \Ccc(G)$ with $\Ccc(G\times G)$, and $\anti(f)(x) = f(x^{-1})$.
  
  Let $G$ be an algebraic group, and $\Ccc(G)$ the algebra of polynomial functions on it.  Then Then $\Ccc(G)$ is a commutative Hopf algebra with $\Delta(f)(x,y) = f(xy)$, where we have identified $\Ccc(G) \otimes \Ccc(G)$ with $\Ccc(G\times G)$, and $\anti(f)(x) = f(x^{-1})$.
  
  Let $G$ be a group and $\KK[G]$ the group algebra of $G$, with mutliplication defined by $\mu(x\otimes y) = xy$ for $x,y\in G$.  Then $G$ is a cocommutative Hopf algebra with $\Delta(x) = x\otimes x$ and $\anti(x) = x^{-1}$ for $x\in G$.
  
  Let $\gg$ be a Lie algebra and $\Uu\gg$ its universal enveloping algebra.  We have seen already (\cref{UgBiAlgProp}) that $\Uu\gg$ is naturally a bialgebra with $\Delta(x) = x\otimes 1 + 1\otimes x$ for $x\in \gg$; we make $\Uu\gg$ into a Hopf algebra by defining $\anti(x) = -x$ for $x\in \gg$.
\end{eg}
\begin{lemmadef}
  Let $U$ be a cocommutative Hopf algebra.  Then the antipode is an involution.  Moreover, the category of (algebra-) representations of $U$ has naturally the structure of a symmetric monoidal category with duals.  In particular, to each pair of representations $V,W$ of $U$, there are natural ways to make $V\otimes_\KK W$ and $\Hom_\KK(V,W)$ into $U$-modules.  Then any (di)natural functorial contruction of vector spaces --- for example $V\otimes W \cong W\otimes V$, $\Hom(U\otimes V,W)\cong \Hom(U,\Hom(V,W)$, and $W:V\mapsto \Hom(\Hom(V,W),W)$ --- in fact corresponds to a homomorphism of $U$-modules.
\end{lemmadef}
\begin{proof}
  Proving the last part would require we go further into category theory than we would like.  We describe the $U$-action on $V\otimes_\KK W$ and on $\Hom_\KK(V,W)$ when $V$ and $W$ are $U$-modules.  For $u\in U$, let $\Delta(u) = \sum u_{(1)}\otimes u_{(2)} = \sum u_{(2)} \otimes u_{(1)}$, and write the actions of $u$ on $v\in V$ and on $w\in W$ as $u\cdot v\in V$ and $u\cdot w \in W$.  Let $\phi \in \Hom_\KK(V,W)$.  Then we define:
  \begin{align}
   \label{UgHopfeqn1} u\cdot (v\otimes w) &\defined \sum (u_{(1)}\cdot v)\otimes (u_{(2)}\cdot w) \\
   \label{UgHopfeqn2} u\cdot \phi &\defined \sum u_{(1)} \circ \phi \circ \anti(u_{(2)})
  \end{align}
  Moreover, the counit map $\epsilon: U \to \KK$ makes $\KK$ into $U$-module, and it is the unit of the monoidal structure.
\end{proof}
\begin{remark}
  \Cref{UgHopfeqn1} makes the category of $U$-modules into a monoidal category for any bialgebra $U$.  One can define duals via \cref{UgHopfeqn2}, but if $U$ is not cocommutative, then $\anti$ may not be an involution, so a choices is required as to which variation of \cref{UgHopfeqn2} to take.  Moreover, when $U$ is not cocommutative, we do not, in general, have an isomorphism $V\otimes W \cong W\otimes V$.  See \cite{Qnotes} and references therein for more discussion of Hopf algebras.
\end{remark}

\begin{eg}
  When $U = \Uu\gg$ and $x\in \gg$, then $x$ acts on $V\otimes W$ by $v\otimes w \mapsto xv \otimes w + v\otimes w$, and on $\Hom(V,W)$ by $\phi \mapsto x\circ \phi - \phi \circ x$.
\end{eg}

\begin{defn}
  Let $(U,\mu,e,\epsilon)$ be a ``counital algebra'' over $\KK$, i.e.\ an algebra along with an algebra map $\epsilon: U\to \KK$; thus $\epsilon$ makes $\KK$ into a $U$-module.  Let $V$ be a $U$-module.  An element $v\in V$ is {\em $U$-invariant}\index{invariant} if the linear map $\KK\to V$ given by $1 \mapsto v$ is a $U$-module homomorphism.  We write $V^U$ for the vector space of $U$-invariant elements of $V$.
\end{defn}
\begin{lemma}
  When $U$ is a cocommutative Hopf algebra, the space $\Hom_\KK(V,W)^U$ of $U$-invariant linear maps is the same as the space $\Hom_U(V,W)$ of $U$-module homomorphisms.
\end{lemma}
\begin{eg}
  The $\Uu\gg$-invariant elements of a $\gg$-module $V$ is the set $V^\gg = \{v\in V \st x\cdot v = 0 \,\forall x\in \gg\}$.  We shorten the word ``$\Uu\gg$-invariant'' to ``$\gg$-invariant''.  A linear map $\phi \in \Hom_\KK(V,W)$ is $\gg$-invariant if and only if $x\circ \phi = \phi \circ x$ for every $x\in \gg$.
\end{eg}
\begin{defn}
  The \emindexing{center} of a Lie algebra $\gg$ is the space of $\gg$-invariant elements of $\gg$ under the adjoint action: $Z(\gg) \defined \gg^\gg = \{x\in \gg \st [\gg,x] = 0\}$.
\end{defn}

\section{Structure Theory of Lie Algebras}

\subsection{Many Definitions}

\cite[Lectures 17 and 18]{notes}

As always, we write ``$\gg$-module'' for ``$\Uu\gg$-module''.

\begin{defn}
  A $\gg$-module $V$ is \emindexing{simple} or \emindexing{irreducible} if there is no submodule $W\subseteq  V$ with $0 \neq W \neq V$.  A Lie algebra is \emindexing{simple} if it is simple as a $\gg$-module under the adjoint action.  An \emindexing{ideal}\index{Lie ideal} of $\gg$ is a $\gg$-submodule of $\gg$ under the adjoint action.
\end{defn}
\begin{prop} \label{derivedprop} If $\aa$ and $\bb$ are ideals in $\gg$, then so is $[\aa,\bb]$. \end{prop}

\begin{defn}
  The \emindexing{upper central series}\index{series!derived} of a Lie algebra $\gg$ is the series $\gg \geq \gg_1 \geq \gg_2 \geq \dots$ where $\gg_0 \defined \gg$ and $\gg_{n+1} \defined [\gg,\gg_n]$.  The Lie algebra $\gg$ is \emindexing{nilpotent} if $\gg_n = 0$ for some $n$.
\end{defn}

\begin{defn}
  The \emindexing{derived subalgebra}\index{subalgebra!derived} of a Lie algebra $\gg$ is the algebra $\gg' \defined [\gg,\gg]$.  The \emindexing{derived series}\index{series!derived} of $\gg$ is the series $\gg \geq \gg' \geq \gg'' \geq \dots$ given by $\gg^{(0)} \defined \gg$ and $\gg^{(n+1)} \defined [\gg^{(n)},\gg^{(n)}]$.  The Lie algebra $\gg$ is \emindexing{solvable} if $\gg^{(n)} = 0$ for some $n$.  An ideal $\rr$ in $\gg$ is solvable if it is solvable as a subalgebra.  By \cref{derivedprop}, if $\rr$ is an ideal of $\gg$, then so is $\rr^{(n)}$.
\end{defn}

\begin{eg}
  The Lie algebra of upper-triangular matrices in $\gl(n)$ is solvable.  A converse to this statement is \cref{uppertriangularconverse}.  The Lie algebra of strictly upper triangular matrices is nilpotent.  
\end{eg}

\begin{defn} \label{semisimpledef}
  A Lie algebra $\gg$ is \emindexing{semisimple} if its only solvable ideal is $0$.
\end{defn}
\begin{remark}
  If $\rr$ is a solvable ideal of $\gg$ with $\rr^{(n)} = 0$, then $\rr^{(n-1)}$ is abelian.  Conversely, any abelian ideal of $\gg$ is solvable.  Thus it is equivalent to replace the word ``solvable'' in \cref{semisimpledef} with the word ``abelian''.
\end{remark}

\begin{prop}
  Any nilpotent Lie algebra is solvable.
  A non-zero nilpotent Lie algebra has non-zero center.
\end{prop}

\begin{prop} \label{SolvNilProp}
  A subquotient of a solvable Lie algebra is solvable.  A subquotient of a nilpotent algebra is nilpotent.  If $\aa$ is an ideal of $\gg$ and if $\aa$ and $\gg/\aa$ are both solvable, then $\gg$ is solvable.  If $\aa$ is an ideal of $\gg$ and $\aa$ is nilpotent and if $\gg \acts \aa$ nilpotently, then $\gg$ is nilpotent.  Thus a central extension of a nilpotent Lie algebra is nilpotent.
\end{prop}

\begin{proof}The derived and upper central series of subquotients are subquotients of the derived and upper central series.  For the second statement, we start taking the derived series of $\gg$, eventually landing in $\aa$ (since $\gg/\aa \to 0$), which is solvable.  The nilpotent claim is similar.\end{proof}

\begin{eg} Let $\gg = \<x,y : [x,y]=y\>$ be the two-dimensional nonabelian Lie algebra.  Then $\gg^{(1)} = \<y\>$ and $\gg^{(2)} = 0$, but $\gg_2 = [\gg,\<y\>] = \<y\>$ so $\gg$ is solvable but not nilpotent.\end{eg}

\begin{defn}
  The \emindexing{lower central series}\index{series!lower central} of a Lie algebra $\gg$ is the series $0 \leq Z(\gg) \leq \zz_2 \leq \dots$ defined by $\zz_0 = 0$ and $\zz_{k+1} = \{x\in \gg \st [\gg,x] \subseteq \zz_k\}$.
\end{defn}

\begin{prop}
  For any of the derived series, the upper central series, and the lower central series, quotients of consecutive terms are abelian.
\end{prop}

\begin{prop}
  Let $\gg$ be a Lie algebra and $\{\zz_k\}$ its lower central series.  Then $\zz_n = \gg$ for some $n$ if and only if $\gg$ is nilpotent.
\end{prop}

\subsection{Nilpotency: Engel's Theorem and Corollaries}

\cite[Lecture 19]{notes}

\begin{lemmadef}
  A matrix $x\in \End(V)$ is \emindexing{nilpotent} if $x^n = 0$ for some $n$.  A Lie algebra $\gg$ {\em acts by nilpotents} on a vector space $V$ if for each $x\in \gg$, its image under $\gg \to \End(V)$ is nilpotent.  If $\gg\acts V,W$ by nilpotents, then $\gg\acts V\otimes W$ and $g\acts \Hom(V,W)$ by nilpotents.  If $v\in V$ and $\gg \acts V$, define the \emindexing{annihilator} of $v$ to be $\ann_\gg(v) = \{x\in \gg \st xv = 0\}$.  For any $v\in V$, $\ann_(v)$ is a Lie subalgebra of $\gg$.
\end{lemmadef}

\begin{thm}[Engel's Theorem] \label{EngelThm}
  If $\gg$ is a finite-dimensional Lie algebra acting on $V$ (possibly infinite-dimensional) by nilpotent endomorphisms, and $V \neq 0$, then there exists a non-zero vector $v\in V$ such that $\gg v = 0$.
\end{thm}
\begin{proof}
  It suffices to look at the image of $\gg$ in $\gl(V) = \Hom(V,V)$.  Then $\ad: \gg \acts \gg$ is by nilpotents.
  
  Pick $v_0\in V$ so that $\ann_\gg(v_0)$ has maximal dimension and let $\hh = \ann_\gg(v_0)$.  It suffices to show that $\hh = \gg$; suppose to the contrary that  $\hh \subsetneq \gg$.  By induction on dimension, the theorem holds for $\hh$.  Consider the vector space $\gg/\hh$; then $\hh \acts \gg/\hh$ by nilpotents, so we can find $x \in \gg/\hh$ nonzero with $\hh x = 0$.  Let $\hat x$ be a preimage of $x$ in $\gg$.  Then $\hat x \in \gg \sminus \hh$ and $[\hh,\hat x] \subseteq \hh$.  Then $\hh_1 \defined \<\hat x\> + \hh$ is a subalgebra of $\gg$.
  
  The space $U \defined \{ u\in V \st \hh u = 0\}$ is non-zero, since $v_0\in U$.  We see that $U$ is an $\hh_1$-submodule of $\hh_1 \acts V$: $hu = 0 \in U$ for $h\in \hh$, and $h(\hat xu) = [h,\hat x]u + \hat xhu = 0u + \hat x0 = 0$ so $\hat xu \in U$.  All of $\gg$ acts on all of $V$ by nilpotents, so in particular $x|_U$ is nilpotent, and so there is some vector $v_1 \in U$ with $xv_1 = 0$.  But then $\hh_1v_1 = 0$, contradicting the maximality of $\hh = \ann(v_0)$.
\end{proof}

\begin{cor} \begin{enumerate}
\item If $\gg \acts V$ by nilpotents and $V$ is finite dimension, then $V$ has a basis in which $\gg$ is strictly upper triangular. 
\item If $\ad x$ is nilpotent for all $x\in \gg$ finite-dimensional, then $\gg$ is a nilpotent Lie algebra.
\item Let $V$ be a simple $\gg$-module.  If an ideal $\aa \leq \gg$ acts nilpotently on $V$ then $\aa$ acts as $0$ on $V$.
 \end{enumerate}
\end{cor}

\begin{lemmadef}
  If $V$ is a finite-dimensional $\gg$ module, then there exists a \emindexing{Jordan-Holder series}\index{series!Jordan-Holder} $0 = M_0 < M_1 < M_2 < \dots < {\rm M}(n) = V$ such that each $M_i$ is a $\gg$-submodule and each $M_{i+1}/M_i$ is simple.
\end{lemmadef}

\begin{cor} \label{Zproof1}
  Let $V$ be a finite-dimensional $\gg$-module and $0 = M_0 < M_1 < M_2 < \dots < {\rm M}(n) = V$ a Jordan-Holder series for $V$.  An ideal $\aa \leq \gg$ acts by nilpotents on $V$ if and only if $\aa$ acts by $0$ on each $M_{i+1}/M_i$.  Thus there is a largest ideal of $\gg$ that acts by nilpotents on $V$.
\end{cor}

\begin{defn}
  The largest ideal of $\gg$ that acts by nilpotents on $V$ is the \emindexing{nilpotency ideal} of the action $\gg \acts V$.
\end{defn}

\begin{prop}
  Any nilpotent ideal $\aa \leq \gg$ acts nilpotently on $\gg$.
\end{prop}

\begin{cor} \label{AdNilCor}
  Any finite-dimensional Lie algebra has a largest nilpotent ideal: the nilpotency ideal of $\ad$.
\end{cor}

\begin{remark}
  Not every $\ad$-nilpotent element of a Lie algebra is necessarily in the nilpotency ideal of $\ad$.
\end{remark}

\begin{defn} \label{KillingDef1}
  Let $\gg$ be a Lie algebra and $V$ a finite-dimensional $\gg$-module.  Then $V$ defines a \emindexing{trace form} $\beta_V$: a symmetric bilinear form on $\gg$ given by $\beta_V(x,y) \defined \tr_V(x,y)$.  The \emindexing{radical} or \emindexing{kernel} of $\beta_V$ is the set $\ker \beta_V \defined \{x \in \gg \st \beta_V(x,\gg) = 0\}$.
\end{defn}
\begin{remark}
  The more standard notation seems to be $\rad \beta$ for what we call $\ker \beta$, c.f.\ \cite{notes}.  We prefer the term ``kernel'' largely to avoid the conflict of notation with \cref{radicaldef}.  Any bilinear form $\beta$ on $V$ defines two linear maps $V \to V^*$, where $V^*$ is the dual vector space to $V$, given by $x \mapsto \beta(x,-)$ and $x\mapsto \beta(-,x)$.  Of course, when $\beta$ is symmetric, these are the same map, and we can unambiguously call the map $\beta: V \to V^*$.  Then $\ker\beta_V$ defined above is precisely the kernel of the map $\beta_V : \gg \to \gg^*$.
\end{remark}

The following proposition follows from considering Jordan-Holder series:
\begin{prop}
  If and ideal $\aa \leq \gg$ of a finite-dimensional Lie algebra acts nilpotently on a finite-dimensional vector space $V$, then $\aa \leq \ker \beta_V$.
\end{prop}

\subsection{Solvability: Lie's Theorem and Corollaries}

\cite[Lectures 19 and 20]{notes}

\begin{lemmadef} \label{radicaldef}
  Let $\gg$ be a finite-dimensional Lie algebra.  Then $\gg$ has a largest solvable ideal, the \emindexing{radical} $\rad \gg$.
\end{lemmadef}
\begin{proof}
  If ideals $\aa,\bb \leq \gg$ are solvable, then $\aa + \bb$ is solvable, since we have an exact sequence of $\gg$-modules
\eqn{0 \to \aa \to \aa + \bb \to (\aa + \bb)/\aa \to 0}
which is also an extension of a solvable algebra (a quotient of $\bb$) by a solvable ideal.
\end{proof}

\begin{thm}[Lie's Theorem]  \label{LieThm}
  Let $\gg$ be a finite-dimensional solvable Lie algebra over $\KK$ of characteristic $0$, and $V$ a non-zero $\gg$-module.  Assume that $\KK$ contains eigenvalues of the actions of all $x\in \gg$.  Then $V$ has a one-dimensional $\gg$-submodule.
\end{thm}

\begin{proof}
  Without loss of generality $\gg \neq 0$; then $\gg' \neq \gg$ by solvability.  Pick any $\gg \geq \hh \geq \gg'$ a codimension-$1$ subspace.   Since $\hh \geq \gg'$, $\hh$ is an ideal of $\gg$.  Pick $x \in \gg \sminus \hh$, whence $\gg = \< x\> + \hh$.

Being a subalgebra of $\gg$, $\hh$ is solvable, and by induction on dimension $\hh\acts V$ has a one-dimensional $\hh$-submodule $\<w\>$.  Thus there is some linear map $\lambda: \hh \to \KK$ so that  $h \cdot w = \lambda(h)w$ for each $h\in \hh$.  Let $W = \KK[x]w$ for $x\in\gg\sminus \hh$ as above.  Then $W = \Uu(\gg)w$, as $\gg = \hh + \<x\>$ and $\hh w \subseteq \KK w$.

By induction on $m$, each $\<1,x,\dots,x^m\>w$ is an $\hh$-submodule of $W$:
\begin{align}
 \label{LieThmEqn1} h(x^mw) & = x^mhw + \sum_{k+l = m-1} x^k [h,x] x^lw\\ 
 \label{LieThmEqn2}  & = \lambda(h) x^m w + x^k h' x^l w
\end{align}
where $h' = [h,x] \in \hh$.  Thus $h'x^lw \in \<1,\dots,x^l\>w$ by induction, and so $x^k h' x^l w \in \<1,\dots,x^{k+l}\>w = \<1, \dots, x^{m-1}\>w$.

Moreover, we see that $W$ is a generalized eigenspace with eigenvalue $\lambda(h)$ for all $h\in \hh$, and so $\tr_W h = (\dim W)\lambda(h)$, by working in a basis where $h$ is upper triangular.  But for any $a,b$, $\tr[a,b] = 0$; thus $\tr_W[h,x] = 0$ so $\lambda([h,x]) = 0$.  Then \crefrange{LieThmEqn1}{LieThmEqn2} and induction on $m$ show that $W$ is an actual eigenspace.

Thus we can pick $v \in W$ an eigenvector of $x$, and then $v$ generates a one-dimensional eigenspace of $x + \hh = \gg$, i.e.\ a one-dimensional $\gg$-submodule. \end{proof}

\begin{cor} \label{uppertriangularconverse}
  Let $\gg$ and $V$ satisfy the conditions of \cref{LieThm}.  Then $V$ has a basis in which $\gg$ is upper-diagonal.
\end{cor}

\begin{cor}
  Let $\gg$ be a solvable finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$.  Then every simple finite-dimensional $\gg$-module is one-dimensional.
\end{cor}

\begin{cor} \label{LcorNilMod}
  Let $\gg$ be a solvable finite-dimensional Lie algebra over a field of characteristic $0$.  Then $\gg'$ acts nilpotently on any finite-dimensional $\gg$-module.
\end{cor}

\begin{remark}
  In spite of the condition on the ground field in \cref{LieThm}, \cref{LcorNilMod} is true over any field of characteristic $0$.  Indeed, let $\gg$ be a Lie algebra over $\KK$ and $\KK\leq \LL$ a field extension.  The upper central, lower central, and derived series are all preserved under $\LL\otimes_\KK$, so $\LL \otimes_\KK \gg$ is solvable if and only if $\gg$ is.  Moreover, $\gg\acts V$ nilpotently if and only if $\LL \otimes_\KK \gg \acts \LL \otimes_\KK V$ nilpotently.  Thus we may as well ``extend by scalars'' to an algebraically closed field.
\end{remark}

\begin{cor}
  \Cref{AdNilCor} asserts that any Lie algebra $\gg$ has a largest ideal that acts nilpotently on $\gg$.  When $\gg$ is solvable, then any element of $\gg'$ is $\ad$-nilpotent.  Hence the set of $\ad$-nilpotent elements of $\gg$ is an ideal.
\end{cor}

\subsection{The Killing Form}

\cite[Lecture 20]{notes}

We recall \cref{KillingDef1}.

\begin{prop}
  Let $\gg$ be a Lie algebra and $V$ a $\gg$-module.  The trace form $\beta_V: (x,y) \mapsto \tr_V(xy)$ on $\gg$ is \indexing{invariant} under the $\gg$-action:
  \eqn{ \beta_V\bigl([z,x],y\bigr) + \beta_V\bigl(x,[z,y]\bigr) = 0 }
\end{prop}

\begin{defn}
  Let $\gg$ be a Lie algebra.  The \emindexing{Killing form} $\beta \defined \beta_{(\gg,\ad)}$ on $\gg$ is the trace form of the adjoint representation $\gg \acts \gg$.
\end{defn}

\begin{prop}
  Let $\gg$ be a Lie algebra, $V$ a $\gg$-module, and $W \subseteq V$ a $\gg$-submodule.  Then $\beta_V = \beta_W + \beta_{V/W}$.
\end{prop}

\begin{cor}
  Let $\gg$ be a Lie algebra and $\aa\leq \gg$ an ideal.  Then $\beta_{(\gg/\aa,\ad)} |_{\aa \times \gg} = 0$, so $\beta|_{\aa \times \gg} = \beta_\aa|_{\aa \times \gg}$.  In particular, the Killing form of $\aa$ is $\beta|_{\aa \times \aa}$.
\end{cor}

\begin{prop}
  Let $V$ be a $\gg$-module of a Lie algebra $\gg$.  Then $\ker\beta_V$ is an ideal of $\gg$.
\end{prop}
\begin{proof}
  The invariance of $\beta_V$ implies that the map $\beta_V: \gg \to \gg^*$ given by $x \mapsto \beta_V(x,-)$ is a $\gg$-module homomorphism, whence $\ker \beta_V$ is a submodule a.k.a.\ and ideal of $\gg$.
\end{proof}

The following is a corollary to \cref{EngelThm}, using the Jordan-form decomposition of matrices:
\begin{prop}
  Let $\gg$ be a Lie algebra, $V$ a $\gg$-module, and $\aa$ an ideal of $\gg$ that acts nilpotently on $V$.  Then $\aa \subseteq \ker \beta_V$.
\end{prop}

\begin{cor} \label{Cartan2reverse}
  If the Killing form $\beta$ of a Lie algebra $\gg$ is \emindexing{nondegenerate} (i.e.\ if $\ker \beta = 0$), then $\gg$ is semisimple.
\end{cor}

\subsection{Jordan Form}

\cite[Lecture 20]{notes}, \cite[page 57]{anton}\footnote{My notes from MH's class \cite{notes} on the proof \cref{JordanThm} are sadly incomplete, and so I have quoted from \cite{anton}.  The statement of the Jordan decomposition in \cite[Proposition 20.3]{notes} is stronger than in \cite[Lemma 12.4]{anton}: in the former, not only do we assert that $s,n \in \KK[x]$, but that they are in $x\KK[x]$.  This is probably equivalent, but I haven't thought about it enough to be sure.

The two statements in this section are leading towards the statement and proof of \cref{CartanProp}.  It seems that this theorem requires an unmotivated piece of linear algebra; for us, this is \cref{LinAlgLemma}.  \cite[Lemma 12.3]{anton} states this result differently; only by inspecting the proofs are they obviously equivalent:
\begin{lemmann}
  Let $s$ be a diagonalizable linear operator on a vector space $V$ over $\KK$ algebraically closed of characteristic $0$.  If $s = a \diag(\lambda_1,\dots,\lambda_n) a^{-1}$ for $a$ an invertible matrix over $V$, and given $f : \KK \to \KK$ an arbitrary function, we define $f(x)$ as $a \diag(f(\lambda_1),\dots,f(\lambda_n))a^{-1}$.  Suppose that $\tr(xf(x)) = 0$ for any $\QQ$-linear map $f: \KK \to \KK$ that restricts to the identity on $\QQ \mono \KK$.  Then $s = 0$.
\end{lemmann}
}

\begin{thm}[Jordan decomposition] \label{JordanThm}
Let $V$ be a finite-dimensional vector space over an algebraically closed field $\KK$.  Then:
\begin{enumerate}
  \item Every $a \in \gl(V)$ has a unique \emindexing{Jordan decomposition} $a = s+n$, where $s$ is diagonalizable, $n$ is nilpotent, and they commute.
  \item $s,n \in \KK[a]$, in the sense that they are linear combinations of powers of $a$; as $a$ varies, $s$ and $n$ need not depend polynomially on $a$.
\end{enumerate}
\end{thm}
\begin{proof}
  We write $a$ in Jordan form; since strictly-upper-triangular matrices are nilpotent, existence of a Jordan decomposition of $a$ is guaranteed.  In particular, the diagonal part $s$ clearly commutes with $a$, and hence with $n = a - s$.  We say this again more specifically, showing that $s,n$ constructed this way are polynomials in $x$:

  Let the characteristic polynomial of $a$ be $\prod_i (x - \lambda_i)^{n_i}$.  In particular, $(x-\lambda_i)$ are relatively prime, so by the Chinese Remainder Theorem, there is a polynomial $f$ such that $f(x) = \lambda_i \mod (x-\lambda_i)^{n_i}$.  Choose a basis of $V$ in which $a$ is in Jordan form; since restricting to a Jordan block $b$ of $a$ is an algebra homomorphism $\KK[a] \onto \KK[b]$, we can compute $f(a)$ block-by-block.  Let $b$ be a block of $a$ with eigenvalue $\lambda_i$.  Then $(b - \lambda_i)^{n_i} = 0$, so $f(b) = \lambda_i$.  Thus $s = f(a)$ is diagonal in this basis, and $n = a - f(a)$ is nilpotent.
  
  For uniqueness in part 1., let $x = n' + s'$ be any other Jordan decomposition of $a$.  Then $n'$ and $s'$ commut with $a$ and hence with any polynomial in $a$, and in particular $n'$ commutes with $n$ and $s'$ commutes with $s$.  But $n' + s' = a = n + s$, so $n' - n = s' - s$.  Since everything commutes, $n' - n$ is nilpotent and $s' - s$ is diagonalizable, but  the only nilpotent diagonal is $0$.
\end{proof}

We now move to an entirely unmotivated piece of linear algebra:
\begin{lemma} \label{LinAlgLemma}
  Let $V$ be a finite-dimensional vector space over an algebraically closed field $\KK$ of characteristic $0$.  Let $B \subseteq A \subseteq \gl(V)$ be any subspaces, and define $T  = \{ x \in \gl(V) : [x,A] \subseteq B\}$.  Then if $t\in T$ satisfies $\tr_v(tu) = 0$ $\forall u\in T$, then $t$ is nilpotent.
\end{lemma}
We can express this as follows:   
  Let $\beta_V$ be the trace form on $\gl(V) \acts V$.  Then $\ker \beta_V |_{T\times T}$ consists of nilpotents.
\begin{proof}
  Let $t = s+n$ be the Jordan decomposition; we wish to show that $s = 0$.  We fix a basis $\{e_i\}$ in which $s$ is diagonal: $se_i = \lambda_ie_i$.  Let $\{e_{ij}\}$ be the corresponding basis of matrix units for $\gl(V)$.  Then $(\ad s)e_{ij} = (\lambda_i - \lambda_j)e_{ij}$.
  
  Now let $\Lambda = \QQ \{\lambda_i\}$ be the finite-dimensional $\QQ$-vector-subspace of $\KK$.  We consider an arbitrary $\QQ$-linear functional $f: \Lambda \to \QQ$; we will show that $f = 0$, and hence that $\Lambda = 0$.

By $\QQ$-linearity, $f(\lambda_i) - f(\lambda_j) = f(\lambda_i - \lambda_j)$, and we chose a polynomial $p(x) \in \KK[x]$ so that $p(\lambda_i - \lambda_j) = f(\lambda_i) - f(\lambda_j)$; in particular, $p(0) = 0$.

Now we define $u \in \gl(V)$ by $ue_i = f(\lambda_i)e_i$, and then $(\ad u) e_{ij} = (f(\lambda_i) - f(\lambda_j))e_{ij} = p(\ad s)e_{ij}$.  So $\ad u = p(\ad s)$.

Since $\ad: \gg \to \gl(\gg)$ is a Lie algebra homomorphism,  $\ad t = \ad s + \ad n$, and $\ad s,\ad n$ commute, and $\ad s$ is diagonalizable and $\ad n$ is nilpotent.  So $\ad s + \ad n$ is the Jordan decomposition of $\ad t$, and hence $\ad s = q(\ad t)$ for some polynomial $q \in \KK[x]$.  Then $\ad u = (p\circ q)(\ad t)$, and since every power of $t$ takes $A$ into $B$, we have $(\ad u)A \subseteq B$, so $u\in T$.

But by construction $u$ is diagonal in the $\{e_i\}$ basis and $t$ is upper-triangular, so $tu$ is upper-triangular with diagonal $\diag(\lambda_if(\lambda_i))$.   Thus $0 = \tr(tu) = \sum \lambda_i f(\lambda_i)$.  We apply $f$ to this: $0 = \sum (f(\lambda_i))^2 \in \QQ$, so $f(\lambda_i) = 0$ for each $i$.  Thus $f = 0$.
\end{proof}

\subsection{Cartan's Criteria}

\cite[Lecture 21]{notes}

\begin{prop} \label{CartanProp}
  Let $V$ be a finite-dimensional vector space over a field $\KK$ of characteristic $0$.  Then a subalgebra $\gg \leq \gl(V)$ is solvable if and only if $\beta_V(\gg,\gg') = 0$, i.e.\ $\gg' \leq \ker\beta_V$.
\end{prop}

\begin{proof}
  We can extend scalars and assume that $\KK$ is algebraically closed, thus we can use \cref{LinAlgLemma}.

The forward direction follows by Lie's theorem (\cref{LieThm}): we can find a basis of $V$ in which $\gg$ acts by upper-triangular matrices, and hence $\gg'$ acts by strictly upper-triangular matrices.

For the reverse, we'll show that $\gg'$ acts nilpotently, and hence is nilpotent by Engel's theorem (\cref{EngelThm}).  We use \cref{LinAlgLemma}, taking $V = V$, $A = \gg$, and $B = \gg'$.  Then $T = \{ t\in \gl(V) \st [t,\gg] \leq \gg'\}$, and in particular $\gg \leq T$, and so $\gg' \leq T$.

So if $[x,y]= t\in \gg'$, then $\tr_V(tu) = \tr_V([x,y]u) = \tr_V(y[x,u])$ by invariance, and $y\in \gg$ and $[x,u]\in \gg'$ so $\tr_V(y[x,u]) = 0$.  Hence $t$ is nilpotent.
\end{proof}

The following is a straightforward corollary:

\begin{thm}[Cartan's First Criterion] \label{CartanThm1}
  Let $\gg$ be a Lie algebra over a field of characteristic $0$.  Then $\gg$ is solvable if and only if $\gg' \leq \ker \beta$.
\end{thm}

\begin{proof}
  We have not yet proven \cref{AdoThm}, so we cannot assume that $\gg \mono \gl(V)$ for some $V$.  Rather, we let $V = \gg$ and $\tilde\gg = \gg / Z(\gg)$, whence $\tilde\gg \mono \gl(V)$ by the adjoint action.  Then $\gg$ is a central extension of $\tilde\gg$, so by \cref{SolvNilProp} $\gg$ is solvable if and only if $\tilde\gg$ is.  By \cref{CartanProp}, $\tilde\gg$ is solvable if and only if $\beta_{\tilde\gg}(\tilde\gg,\tilde\gg') = 0$.  But $\beta_\gg$ facrots through $\beta_{\tilde \gg}$:
  \eqn{ \beta_\gg = \{\gg \times \gg \overset{/Z(\gg)}{\longto} \tilde{\gg} \times \tilde{\gg} \overset{\beta_{\tilde{\gg}}}{\to} \KK \} }
  Moreover, $\gg' \overset{/Z(\gg)}\onto \tilde\gg'$, and so $\beta_\gg(\gg,\gg') = \beta_{\tilde\gg}(\tilde\gg,\tilde\gg')$.
\end{proof}

\begin{cor} \label{Cartan2precor}
  For any Lie algebra $\gg$ in characteristic zero with Killing form $\beta$, we have that $\ker\beta$ is solvable, i.e.\ $\ker\beta \leq \rad \gg$.
\end{cor}

The reverse direction of the following is true in any characteristic (\cref{Cartan2reverse}).  The forward direction is an immediate corollary of \cref{Cartan2precor}.

\begin{thm}[Cartan's Second Criterian] \label{CartanThm2}
  Let $\gg$ be a Lie algebra over characteristic $0$, and $\beta$ its Killing form.  Then $\gg$ is semisimple if and only if $\ker \beta = 0$.
\end{thm}

\begin{cor}
  Let $\gg$ be a Lie algebra over characteristic $0$.  The $\gg$ is semisimple if and only if any extension by scalars of $\gg$ is semisimple.
\end{cor}

\begin{remark}
  For any Lie algebra $\gg$, $\gg / \rad \gg$ is semisimple.  We will see latter (\cref{LeviThm}) that in characteristic $0$, $\rad \gg$ is a direct summand and $\gg$.
\end{remark}

\section{Examples: three-dimensional Lie algebras}

\cite[Lecture 21]{notes}

The classification of three-dimensional Lie algebras over $\RR$ or $\CC$ is long but can be done by hand, c.f.\ \url{http://en.wikipedia.org/wiki/Bianchi_classification}.  The classification of four-dimensional Lie algebras has been completed, but beyond this it is hopeless: there are too many extensions of one algebra by another.  In \cref{SemisimpleChapter} we will classify all semisimple Lie algebras.  For now we list two important Lie algebras:
\begin{lemmadef}
  The \emindexing{Heisenberg algebra} is a three-dimensional Lie algebra with a basis $x,y,z$, in which $z$ is central and $[x,y] = z$.  The Heisenberg algebra is nilpotent.
\end{lemmadef}
\begin{lemmadef}
  We define $\sl(2)$\index{sl2@$\sl(2)$} to be the three-dimensional Lie algebra with a basis $e,h,f$ and relations $[h,e] = 2e$, $[h,f] = -2f$, and $[e,f] = h$.  So long as we are not working over characteristic $2$, $\sl(2)$ is semisimple; simplicity follows from \cref{sl2simplecor}.
\end{lemmadef}
\begin{proof}  Just compute the Killing form $\beta_{\sl(2)}$. \end{proof}

We conclude this section with two propositions and two corollaries; these will play an important role in \cref{SemisimpleChapter}.

\begin{prop}
  Let $\gg$ be a Lie algebra such that every ideal $\aa$ of $\gg$ and every quotient $\gg/\aa$ of $\gg$ is semisimple.  Then $\gg$ is semisimple.  Conversely, let $\gg$ be a semisimple Lie algebra over characteristic $0$.  Then all ideals and all quotients of $\gg$ are semisimple.
\end{prop}

\begin{proof}
  We prove only the converse direction.  Let $\gg$ be semisimple, so that $\beta$ is nondegenerate.  Let $\alpha^\perp$ be the orthogonal subspace to $\aa$ with respet to $\beta$.  Then $\aa^\perp = \ker\{ x \mapsto \beta(-,x) : \gg \to \Hom(\aa,\gg) \}$, so $\aa^\perp$ is an ideal.  Then $\aa \cap \aa^\perp = \ker \beta|_\aa \leq \rad \aa$, and hence it's solvable and hence is $0$.  So $\aa$ is semisimple, and also $\aa^\perp$ is.  In particular, the projection $\aa^\perp \isom \gg/\aa$ is an isomorphism of Lie algebras, so $\gg/\aa$ is semisimple.
\end{proof}

\begin{cor} \label{extproof2}
  Every finite-dimensional semisimple Lie algebra $\gg$ over characteristic $0$ is a direct product $\gg = \gg_1 \times \dots \times \gg_m$ of simple nonabelian algebras.
\end{cor}
\begin{proof}
  Let $\aa$ be a minimal and hence simple idea.  Then $[\aa,\aa^\perp] \subseteq \aa \cap \aa^\perp = 0$.  Rinse and repeat.
\end{proof}

\begin{cor}  \label{sl2simplecor}
  $\sl(2)$ is simple.
\end{cor}

\section{Some Homological Algebra}

We will not need too much homological algebra; any standard textbook on the subject, e.g.\ \cite{Hom1,Hom2,Hom3}, will contain fancier versions of many of these constructions.

\subsection{The Casimir}

\cite[Lectures 21 and 22]{notes}

The following piece of linear algebra is a trivial exercise in definition-chasing, and is best checked in either the physicists' index notation or Penrose's graphical language:
\begin{prop} \label{CasimirProp}
  Let $\<,\>$ be a nondegenerate not-necessarily-symmetric bilinear form on finite-dimensional $V$.  Let $(x_i)$ and $(y_i)$ be dual bases, so $\<x_i,y_j\> = \delta_{ij}$.  Then $\theta = \sum x_i\otimes y_i \in V\otimes V$ depends only on the form $\<,\>$.  If $z \in \gl(V)$ leaves $\<,\>$ invariant, then $\theta$ is also invariant.
\end{prop}

\begin{cor} \label{CasimirCor}
  Let $\beta$ be a nondegenerate invariant (symmetric) form on a finite-dimensional Lie algebra $\gg$, and define $c_\beta = \sum x_iy_i$ to be the image of $\theta$ in \cref{CasimirProp} under the multiplication map $\gg \otimes \gg \to \Uu\gg$.  Then $c_\beta$ is a central element of $\Uu\gg$.
  \end{cor}

\begin{lemmadef}  \label{CasimirDefn}
  Let $\gg$ be a finite-dimensional Lie algebra and $V$ a $\gg$-module so that the trace form $\beta_V$ is nondegenerate.  Define the \emindexing{Casimir operator} $c_V = c_{\beta_V}$ as in \cref{CasimirCor}.  Then $c_V$ has the following properties:
  \begin{enumerate}
  \item $c_{V}$ only depends on $\beta_V$.
  \item $c_{V} \in Z(\Uu(\gg))$
  \item $c_{V} \in \Uu(\gg)\gg$, i.e.\ it acts as $0$ on $\KK$.
  \item $\tr_V(c_{V}) = \sum \tr_V(x_iy_i) = \dim \gg$.
\end{enumerate}
  In particular, $c_V$ distinguishes $V$ from the trivial representation.
\end{lemmadef}

\subsection{Review of $\Ext$}

\cite[Lectures 22 and 23]{notes}

\begin{defn}
  Let $\Cc$ be an abelian category.  A \emindexing{complex} (with \emindexing{homological indexing}\index{indexing!homological}) in $\Cc$ is a sequence $A_\bullet = \dots A_k \overset{d_k}\to A_{k-1} \to \dots$ of maps in $\Cc$ such that $d_k \circ d_{k+1} = 0$ for every $k$.  The \emindexing{homology} of $A_\bullet$ are the objects $H_k(A_\bullet) \defined \ker d_k / \im d_{k+1}$.  For each $k$, $\ker d_k$ is the object of {\em $k$-cycles}\index{cycle}, and $\im d_{k+1}$ is the object of {\em $k$-boundaries}\index{boundary}.
  
  We can write the same complex with \emindexing{cohomological indexing}\index{indexing!cohomological} by writing $A^k \defined A_{-k}$, whence the arrows go $\dots \to A^{k-1} \overset{\delta^k}\to A^k \to \dots$.  The \emindexing{cohomology} of a complex is $H^k(A^\bullet) \defined H_{-k}(A_\bullet) = \ker \delta^{k+1} / \im \delta^{k}$.  The {\em $k$-cocycles}\index{cocycle} are $\ker \delta^{k+1}$ and the {\em $k$-coboundaries}\index{coboundary} are $\im \delta^k$.
  
    A complex is \emindexing{exact} at $k$ if $H_k = 0$.  A \emindexing{long exact sequence} is a complex, usually infinite, that is exact everywhere.  A \emindexing{short exact sequence} is a three-term exact complex of the form $0 \to A \to B \to C \to 0$.  In particular, $A = \ker(B\to C)$ and $C = A / B$.
\end{defn}

\begin{defn}
  Let $U$ be an associative algebra and \cat{$U$-mod} its category of left modules.  A \emindexing{free module}\index{module!free} is a module $U\acts F$ that is isomorphic to a possibly-infinite direct sum of copies of $U\acts U$.  Let $M$ be a $U$-module.  A \emindexing{free resolution} of $M$ is a complex $F_\bullet = \dots \to F_1 \to F_0 \to 0$ that is exact everywhere except at $k = 0$, where $H_k(F_\bullet) = M$.  Equivalently, the \emindexing{augmented complex}\index{complex!augmented} $F_\bullet \to M \to 0$ is exact.
\end{defn}

\begin{lemma}
  Given any module $M$, a free resolution $F_\bullet$ of $M$ exists.
\end{lemma}
\begin{proof}
  Let $F_{-1} \defined M_0 \defined M$ and $M_{k+1} \defined \ker(F_k \to F_{k-1})$.  Define $F_k$ to be the free module on a generating set of $M_k$.
\end{proof}

\begin{lemmadef}
  Let $U$ be an associative algebra and $M,N$ two left $U$ modules.  Let $F_\bullet$ be a free resolution of $M$, and construct the complex
  \eqn{ \Hom_U(F_\bullet,N) = \Hom_U(F_0,N) \overset{\delta^1}\to \Hom_U(F_1,N) \overset{\delta^2}\to \dots }
  by applying the contravariant functor $\Hom_U(-,N)$ to the complex $F_\bullet$.  Define $\Ext_U^i(M,N) \defined H^i(\Hom_U(F_\bullet,N))$\index{Ext@$\Ext^\bullet$}.  Then $\Ext^0_U(M,N) = \Hom(M,N)$.  Moreover, $\Ext^i_U(M,N)$ does not depend on the choice of free resolution $F_\bullet$, and is functorial in $M$ and $N$.
\end{lemmadef}

\begin{proof}
  It's clear that for each choice of a free-resolution of $M$, we get a functor $\Ext^\bullet(M,-)$.
  
  Let $M\to M'$ be a $U$-morphism, and $F_\bullet'$ a free resolution of $M'$.  By freeness we can extend the morphism $M\to M'$ to a chain morphism, unique up to chain homotopy:
  \eqn{ \begin{tikzpicture}[baseline=(X.base)]
    \path coordinate (X) +(0,.5) node (M) {$M$} +(0,-.5) node (m) {$M'$};
    \path (M) ++(-1,0) node (F0) {$F_0$} ++(-1,0) node (F1) {$F_1$} ++(-1,0) node (D) {$\cdots$} ;
    \path (m) ++(-1,0) node (f0) {$F'_0$} ++(-1,0) node (f1) {$F'_1$} ++(-1,0) node (d) {$\cdots$} ;
    \draw[->] (M) -- (m);
    \draw[->] (D) -- (F1); \draw[->] (F1) -- (F0); \draw[->] (F0) -- (M);
    \draw[->] (d) -- (f1); \draw[->] (f1) -- (f0); \draw[->] (f0) -- (m);
    \draw[->,dotted] (F0) -- (f0); \draw[->,dotted] (F1) -- (f1);
  \end{tikzpicture} }
  Chain homotopies induce isomorphisms on $\Hom$, so $\Ext^\bullet(M,N)$ is functorial in $M$; in particular, letting $M' = M$ with a different free resolution shows that $\Ext^\bullet(M,N)$ is well-defined.
\end{proof}

\begin{lemmadef}
  The functor $\Hom(-,N)$ is \emindexing{left-exact} but not \emindexing{right-exact}, i.e.\ if $0 \to A \to B \to C \to 0$ is a short exact sequence then $\Hom(A,N) \from \Hom(B,N) \from \Hom(C,N) \from 0$ is exact, but $0 \from \Hom(A,N) \from \Hom(B,N)$ is not necessarily exact.  Rather, we get a {\em long exact sequence in $\Ext$}\index{long exact sequence!in $\Ext$}:
  \eqn{ \begin{tikzpicture}[xscale=3,yscale=1.5,baseline=(C1.base)]
    \path node (O) {$0$} ++(-1,0) node (C0) {$\Ext^0(C,N)$}  ++(-1,0) node (B0) {$\Ext^0(B,N)$}  ++(-1,0) node (A0) {$\Ext^0(A,N)$};
    \path (C0) ++(0,-1) node (C1) {$\Ext^1(C,N)$}  ++(-1,0) node (B1) {$\Ext^1(B,N)$}  ++(-1,0) node (A1) {$\Ext^1(A,N)$};
    \path (C1) ++(0,-1) node (C2) {$\Ext^2(C,N)$}  ++(-1,0) node (B2) {$\Ext^2(B,N)$}  ++(-1,0) node (A2) {$\Ext^2(A,N)$} ++(-1,0) node (d) {$\dots$};
    \foreach \n in {0,1,2} {\draw[->] (C\n) -- (B\n); \draw[->] (B\n) -- (A\n); }
    \draw[->] (O) -- (C0); \draw[->] (A2) -- (d);
    \path (C1.east) ++(0,.5) coordinate (c1);    \path (C2.east) ++(0,.5) coordinate (c2);
    \draw[->] (A0.west) .. controls +(-.25,0) and +(-.25,0) .. ++(0,-.5) -- (c1) .. controls +(.25,0) and +(.25,0) .. (C1.east);
    \draw[->] (A1.west) .. controls +(-.25,0) and +(-.25,0) .. ++(0,-.5) -- (c2) .. controls +(.25,0) and +(.25,0) .. (C2.east);
  \end{tikzpicture} }
  When $N = A$, the image of $1_A \in \Hom(A,A) = \Ext^0(A,A)$ in $\Ext^1(C,A)$ is the \emindexing{characteristic class} of the extension $0 \to A\to B \to C \to 0$.  The characteristic class determines $B$ up to equivalence; in particular, when $1_A \mapsto 0$, then $B \cong A\oplus C$.
\end{lemmadef}

\subsection{Complete Reducibility}

\begin{lemma}
  Let $\gg$ be a Lie algebra over $\KK$, $\Uu\gg$ its universal enveloping algebra, $N$ a $\gg$-module, and $F$ a free $\gg$-module.  Then $F\otimes_\KK N$ is free.
\end{lemma}
\begin{proof}
  Let $F = \bigoplus \Uu\gg$; then $F\otimes N = \bigl( \bigoplus \Uu\gg \bigr) \otimes_\KK N = \bigoplus (\Uu\gg \otimes_\KK N)$, so it suffices to show that $G \defined \Uu\gg\otimes_\KK N$ is free.
  
  We understand the $\Uu\gg$-action on $G$: let $x\in \gg$ and $u\otimes n\in G$, then $x\cdot (u\otimes n) = (xu)\otimes n + u\otimes(x\cdot n)$ as $\Delta x = x\otimes 1 + 1\otimes x$.  Here $xu$ is the product in $\Uu\gg$ and $x\cdot n$ is the action $\gg \acts N$.
  
  We can put a filtration on $G$ by $G_{\leq n} = \Uu\gg_{\leq n}\otimes_\KK N$. This makes $G$ into a \emindexing{filtered module}\index{module!filtered}:
\eqn{ \Uu(\gg)_{\leq k}G_{\leq l} \subseteq G_{\leq k+l}}
Thus $\gr G$ is a $\gr \Uu\gg$-module, but $\gr\Uu\gg = \Ss\gg$, and $S(\gg)$ acts through the first term, so $S(\gg) \otimes N$ is a free $S(\gg)$-module, by picking any basis of $N$.

let $\{n_\beta\}$ be a basis of $N$ and $\{x_\alpha\}$ a basis of $\gg$.  Then $\{x_{\vec{\alpha}}n_\beta\}$ is a basis of $\gr G = S(\gg) \otimes N$, hence also a basis of $\Uu(\gg)\otimes N$.  Thus $\Uu(\gg)\otimes N$ is free.  We have used \cref{PBWthm} implicitly multiple times.
\end{proof}

\begin{cor} \label{extproof1}
  If $M$ and $N$ are finite-dimensional $\gg$-modules, then: \eqn{ \Ext^i(M,N) \cong \Ext^i(\KK,\Hom(M,N)) \cong \Ext^i(\Hom(N,M),\KK) }
\end{cor}

\begin{proof}
  It suffices to prove the first equality.
  
  Let $F_\bullet \to M$ be a free resolution.  A $\Uu(\gg)$-module homomorphism is exactly a $\gg$-invariant linear map:
\begin{align} \Hom_{\Uu(\gg)}(F,N)^\bullet & = \Hom_\KK(F_\bullet,N)^\gg \\
 & = \Hom_\KK(F_\bullet \otimes_\KK N^*,\KK)^\gg \\
 & = \Ext^\bullet(M\otimes N^*,\KK) \end{align}
Using the finite-dimensionality of $N$ and the lemma that $F^\bullet \otimes N^*$ is a free resolution of $M\otimes N^*$.
\end{proof}

\begin{lemma}
  If $M,N$ are finite-dimensional $\gg$-modules and $c\in Z(\Uu\gg)$ such that the characteristic polynomials $f$ and $g$ of $c$ on $M$ and $N$ are relatively prime, then $\Ext^i(M,N) = 0$ for all $i$.
\end{lemma}
\begin{proof}
  By functoriality, $c$ acts $\Ext^i(M,N)$.  By centrality, the action of $c$ on $\Ext^i(M,N)$ must satisfy both the characteristic polynomials: $f(c), g(c)$ annihilate $\Ext^i(M,N)$.  If $f$ and $g$ are relatively prime, then $1 = af + bg$ for some polynomials $a,b$; thus $1$ annihilates $\Ext^i(M,N)$, which must therefore be $0$.
\end{proof}

\begin{thm}[Schur's Lemma] \label{SchurLemma}
  Let $U$ be an algebra and $N$ a simple non-zero $U$-module, and let $\alpha: N\to N$ a $U$-homomorphism; then $\alpha =0$ or $\alpha$ is an isomorphism.
\end{thm}
\begin{proof}
  The image of $\alpha$ is a submodule of $N$, hence either $0$ or $N$.  If $\im \alpha = 0$, then we're done.  If $\im \alpha = N$, then $\ker \alpha \neq 0$, so $\ker \alpha = N$ by simplicity, and $\alpha$ is an isom.
\end{proof}

\begin{cor} \label{extproof3}
  Let $M,N$ be finite-dimensional simple $U$-modules such that $c\in Z(U)$ annihilates $M$ but not $N$; then $\Ext^i(M,N) = 0$ for every $i$.
\end{cor}
\begin{proof}
  By \cref{SchurLemma}, $c$ acts invertibly on $N$, so all its eigenvalues (over the algebraic closure) are non-zero.  But the eigenvalues of $c$ on $M$ are all $0$, so the characteristic polynomials are relatively prime.
\end{proof}

\begin{thm}[$\Ext^1$ vanishes over a semisimple Lie algebra] \label{Ext1thm}
  Let $\gg$ be a semisimple Lie algebra over a field $\KK$ of characteristic $0$, and let $M$ and $N$ be finite-dimensional $\gg$-modules.  Then $\Ext^1(M,N) = 0$.
\end{thm}
\begin{proof}
  Using \cref{extproof1} we may assume that $M = \KK$.  Assume that $N$ is not a trivial module.  Then $\gg = \gg_1\times \dots \times \gg_k$ by \cref{extproof2} for $\gg_i$ simple, and some $\gg_i$ acts non-trivially on $N$.  Then $\beta_N$ does not vanish on $\gg_i$ by \cref{CartanThm2}, and so $\ker_{\gg_i} \beta_N = 0$ by simplicity.  Thus we can find a Casimir $c\in Z(\Uu\gg_i) \subseteq Z(\Uu\gg)$.  In particular, $\tr_N(c) = \dim \gg_i \neq 0$, but $c$ annihilates $\KK$, and so by \cref{extproof3} $\Ext^1(\KK,N) = 0$.
  
  If $N$ is a trivial module, then we use the fact that $\Ext^1(\KK,N)$ classifies extensions $0 \to N \to L \to \KK \to 0$, which we will classify directly.  (See \cref{extproof4} for a direct verification that $\Ext^1$ classifies extensions in the case of $\gg$-modules.)  Writing $L$ in block form (as a vector space, $L = N\oplus \KK$), we see that $\gg$ acts on $L$ like $\left[ \begin{matrix} 0 & * \\ 0 & 0 \end{matrix}\right]$.  Then $\gg$ acts by nilpotent matrices, but $\gg$ is semisimple, so $\gg$ annihilates $L$.  Thus the only extension is the trivial one, and $\Ext^1(\KK,N) = 0$.
\end{proof}

We list two corollaries, which are important enough to call theorems.
We recall the following definition:
\begin{defn}
  An object in an abelian category is \emindexing{simple} if it has no non-zero proper subobjects.  An object is \emindexing{completely reducible} if it is a direct sum of simple objects.
\end{defn}
\begin{thm}[Weyl's Complete Reducibility Theorem] \label{WeylThm}
  Every finite-dimensional representation of a semisimple Lie algebra over characteristic zero is completely reducible.
\end{thm}

\begin{thm}[Whitehead's Theorem] \label{WhiteheadThm}
  If $\gg$ is a semisimple Lie algebra over characteristic zero, and $M$ and $N$ are finite-dimensional non-isomorphic simple $\gg$-modules, then $\Ext^i(M,N)$ vanishes for all $i$.
\end{thm}

\subsection{Computing $\Ext^i(\KK,M)$}

\cite[Lecture 23 and 24]{notes}

\begin{prop}
  Let $\gg$ be a Lie algebra over $\KK$, and $\KK$ the trivial representation.  Then $\KK$ has a free $\Uu\gg$ resolution given by:
  \eqn[BRST1]{ \dots \to \Uu(\gg) \otimes_\KK \Wedge^2\gg \overset{d_2}\to \Uu(\gg) \otimes_\KK \gg \overset{d_1}\to \Uu(\gg) \overset{\epsilon}\to \KK \to 0}
The maps $d_k : \Uu(\gg) \otimes \Wedge^k\gg \to \Uu(\gg) \otimes \Wedge^{k-1}\gg$ for $k \leq 1$ are given by:
\begin{align}
  d_k(x_1\wedge \dots \wedge x_k) & = \sum_i (-1)^{i-1} x_i \otimes (x_1 \wedge \cdots \hat{x_i} \cdots \wedge x_k) \nonumber \\
  & \phantom{=}  - \sum_{i<j} (-1)^{i-j+1} 1 \otimes ([x_i,x_j] \wedge x_1 \wedge \cdots \hat{x_i} \cdots \hat{x_j} \cdots \wedge x_k)
\end{align}
\end{prop}
\begin{proof}
  That $d_k$ is well-defined requires only checking that it is antisymmetric.  That $d_{k-1}\circ d_k = 0$ is more or less obvious: terms cancel either by being sufficiently separated to appear twice with opposite signs (like in the free resolution of the symmetric polynomial ring), or by syzygy, or by Jacobi.
  
  For exactness, we quote a general principle:  Let $F_\bullet(t)$ be a $t$-varying complex of vector spaces, and choose a basis for each one.  Assume that the vector spaces do not change with $t$, but that the matrix coefficients of the differentials $d_k$ depend algebraically on $t$.  Then the dimension of $H^i$ can jump for special values of $t$, but does not fall at special values of $t$.  In particular, exactness is a Zariski open condition.
  
  Thus consider the complex with the vector spaces given by \cref{BRST1}, but with the differential given by
\begin{align}
  d_k(x_1\wedge \dots \wedge x_k) & = \sum_i (-1)^{i-1} x_i \otimes (x_1 \wedge \cdots \hat{x_i} \cdots \wedge x_k) \nonumber \\
  & \phantom{=}  - t\sum_{i<j} (-1)^{i-j+1} 1 \otimes ([x_i,x_j] \wedge x_1 \wedge \cdots \hat{x_i} \cdots \hat{x_j} \cdots \wedge x_k)
\end{align}  
  This corresponds to the Lie algebra $\gg_t = (\gg,[x,y]_t \defined t[x,y])$.  When $t \neq 0$, $\gg_t\cong \gg$, by $x \mapsto tx$, but when $t = 0$, $\gg_0$ is abelian, and the complex consists of polynomial rings and is obviously exact.
  
  Thus the $t$-varying complex is exact at $t = 0$ and hence in an open neighborhood of $0$.  If $\KK$ is not finite, then an open neighborhood of $0$ contains non-zero terms, and so the complex is exact for some $t\neq 0$ and hence for all $t$.  If $\KK$ is finite, we replace it by its algebraic closure.
\end{proof}

\begin{cor}
$\Ext^\bullet(\KK,M)$ is the cohomology of the \emindexing{Chevalley complex} with coefficients in $M$: 
\eqn{ 0 \to M \overset{\delta^1}{\to} \Hom_\KK(\gg,M) \overset{\delta^2}{\to} \Hom(\Wedge^2\gg,M) \to \dots}
If $g\in \Hom_\KK(\Wedge^{k-1}\gg, M)$, then the differential $\delta^kg$ is given by:
\begin{align}
  \delta^kg(x_1\wedge\dots\wedge x_k) & = \sum_i (-1)^{i-1}x_ig(x_1\wedge \cdots \hat{x_i} \cdots \wedge x_k)\nonumber \\
   & \phantom{=} - \sum_{i<j} (-1)^{i-j+1} g([x_i,x_j]\wedge x_1\wedge \cdots \hat{x_i} \cdots \hat{x_j} \cdots \wedge x_k)
\end{align}
\end{cor}

\begin{eg} \label{extproof4}
  Let $M$ and $N$ be finite-dimensional $\gg$-modules.  Then $\Ext^i(M,N)$ is the cohomology of $\cdots \overset{\delta^k}\to \Hom_\KK(\Wedge^l\gg,M^*\otimes N) \overset{\delta^{k+1}}\to \cdots$.  We compute $\Ext^1(M,N)$.
  
  If $\phi \in M^* \otimes N$ and $x\in \gg$, then the action of $x$ on $\phi$ is given by $x\cdot \phi = x_N \circ \phi - \phi \circ x_M = ``[x,\phi]$''.  A $1$-cocycle is a map $f: \gg \to M^*\otimes N$ such that $0 = \delta^1f(x\wedge y) = f([x,y]) - \bigl( (x\cdot f)(y) - (y\cdot f)(x) \bigr) = ``[x,f(y)] - [y,f(x)]$''.
  
  Let $0 \to N \to V \to M \to 0$ be a $\KK$-vector space, and choose a splitting $\sigma: M\to V$ as vector spaces.  Then $\gg$ acts on $M\oplus N$ by $x\mapsto \left[ \begin{matrix} x_N & f(x) \\ 0 & x_M \end{matrix}\right]$, and the cocycles $f$ exactly classify the possible ways to put something in the upper right corner.
  
  The ways to change the splitting $\sigma \mapsto \sigma' = \sigma + h$ correspond to $\KK$-linear maps $h : M \to N$.  This changes $f(x)$ by $x_N\circ h - h\circ x_M = \delta^1(h)$.
  
  We have seen that the 1-cocycles classify the splitting, and changing the 1-cocycle by a 1-coboundary changes the splitting but not the extension.  So $\Ext^1(M,N)$ classifies extensions up to isomorphism.
\end{eg}

\begin{eg} 
  Consider abelian extensions of Lie algebras $0 \to \mm \to \hat\gg \to \gg \to 0$ where $\mm$ is an abelian ideal of $\hat\gg$.  Since $\mm$ is abelian, the action $\hat\gg \acts \mm$ factors through $\gg = \hat\gg / \mm$.  Conversely, we can classify abelian extensions $0 \to \mm \to \hat\gg \to \gg \to 0$ given $\gg$ and a $\gg$-module $\mm$.
  
  We pick a $\KK$-linear splitting $\sigma: \gg\to \hat\gg$; then $\hat\gg  = \{ \sigma(x) + m\}$ as $x$ ranges over $\gg$ and $m$ over $\mm$, and the bracket is
\eqn{  [\sigma(x) + m, \sigma(y) + n] = \sigma([x,y]) + [\sigma(x),n] - [\sigma(y),m] + g(x,y) }
where $g$ is the error term measuring how far off $\sigma$ is from being a splitting of $\gg$-modules.  There is no $[m,n]$ term, because $\mm$ is assumed to be an abelian ideal of $\hat\gg$.

  Then $g$ is antisymmetric.  The Jacobi identity on $\hat\gg$ is equivalent to $g$ satisfying:
\begin{align}
 0 & = x\,g(y\wedge z) - y\,g(x\wedge z) + z\,g(x\wedge y) - g([x,y]\wedge z) + g([x,z]\wedge y) - g([y,z]\wedge x) \\
  & = x\,g(y\wedge z) - g([x,y]\wedge z) + \textrm{cycle permutations}
\end{align}
  I.e.\ $g$ is a $2$-cocycle in $\Hom_\KK(\gg,\mm)$.  In particular, the $2$-cocycles classify extensions of $\gg$ by $\mm$ along with a splitting.  If we change the splitting by $f: \gg \to \mm$, then $\gg$ changes by $(x\cdot f)(y) - (y\cdot f)(x) - f([x,y]) = \delta^2(f)$.  We have proved:
\end{eg}

\begin{prop}
  $\Ext^2_{\Uu\gg}(\KK,\mm)$ classifies abelian extensions $0 \to \mm \to \hat\gg \to \gg \to 0$ up to isomorphism.  The element $0\in \Ext^2$ corresponds to the semidirect product $\hat\gg = \gg \ltimes\mm$.
\end{prop}

\begin{cor} \label{Leviproof1}
  Abelian extensions of semisimple Lie groups are semidirect products.
\end{cor}

\begin{thm}[Levi's Theorem] \label{LeviThm}
  Let $\gg$ be a finite-dimensional Lie algebra over characteristic $0$, and let $\rr = \rad(\gg)$.  Then $\gg$ has a \emindexing{Levi decomposition}: semisimple \emindexing{Levi subalgebra}\index{subalgebra!Levi} $\ss \subseteq \gg$ such that $\gg = \ss\ltimes \rr$.
\end{thm}
\begin{proof}
  Without loss of generality, $\rr \neq 0$, as otherwise $\gg$ is already semisimple.
  
  Assume first that $\rr$ is not a minimal non-zero ideal.  In particular, let $\mm \neq 0$ be an ideal of $\gg$ with $\mm \subsetneq \rr$.  Then $\rr / \mm = \rad (\gg/\mm) \neq 0$, and by induction on dimension $\gg/\mm$ has a Levi subalgebra.  Let $\tilde\ss$ be the preimage of this subalgebra in $\gg/\mm$.  Then $\tilde\ss \cap \rr = \mm$ and $\tilde\ss / \mm \isom (\gg/\mm) / (\rr/\mm) = \gg/\rr$.    Hence $\mm = \rad(\tilde\ss)$.  Again by induction on dimension, $\tilde\ss$ has a Levi subalgebra $\ss$; then $\tilde\ss = \ss \oplus\mm$ and $\ss \cap \rr = 0$, so $\ss \isom \gg/\rr$.  Thus $\ss$ is a Levi subalgebra of $\gg$.
  
  We turn now to the case when $\rr$ is minimal.  Being a radical, $\rr$ is solvable, so $\rr' \neq \rr$, and by minimality $\rr' = 0$.  So $\rr$ is abelian.  In particular, the action $\gg \acts \rr$ factors through $\gg/\rr$, and so $0 \to \rr \to \gg \to \gg/\rr \to 0$ is an abelian extension of $\gg/\rr$, and thus must be semidirect by \cref{Leviproof1}.
\end{proof}
\begin{remark}  We always have $Z(\gg) \leq \rr$, and when $\rr$ is minimal, $Z(\gg)$ is either $0$ or $\rr$.  When $Z(\gg) = \rr$, then $0\to \rr \to \gg \to \gg/\rr \to 0$ is in fact an extension of $\gg/\rr$-modules, and so is a direct product by \cref{extproof4}.
\end{remark}

We do not prove the following, given as an exercise in \cite{notes}:
\begin{thm}[Malcev-Harish-Chandra Theorem]
  All Levi subalgebras of a given Lie algebra are conjugate under the action of the group $\exp \ad \nn \subseteq \GL(V)$, where $\nn$ is the largest nilpotent ideal of $\gg$.  (In particular, $\ad: \nn \acts \gg$ is nilpotent, so the power series for $\exp$ terminates.)
\end{thm}

We are now ready to complete the proof of \cref{fundthm}, with a theorem of Cartan:

\begin{thm}[Lie's Third Theorem]\label{LieThm3}
  Let $\gg$ be a Lie algebra over $\RR$.  Then $\gg = \Lie(G)$ for some analytic Lie group $G$.
\end{thm}
\begin{proof}
  Find a Levi decomposition $\gg = \ss \ltimes \rr$.  If $\ss = \Lie(S)$ and $\rr = \Lie(R)$ where $S$ and $R$ are connected and simply connected, then the action $\ss \acts \rr$ lifts to an action $S \acts R$.  Thus we can construct $G = S\ltimes R$, and it is a direct computation that $\gg = \Lie(G)$ in this case.
  
  So it suffices to find groups $S$ and $R$ with the desired Lie algebras.  We need not even assure that the groups we find are simply-connected; we can always take universal covers.  In any case, $\ss$ is simply connected, so the action $\ss \to \gl(\ss)$ is faithful, and thus we can find $S \subseteq \GL(\ss)$ with $\Lie(S) = \ss$.
  
  On the other hand, $\rr$ is solvable: the chain $\rr \geq \rr' \geq \rr'' \geq \dots$ eventually gets to $0$.  We can interpolate between $\rr$ and $\rr'$ by one-co-dimensional vector spaces, which are all necessarily ideals of some $\rr^{(k)}$, and the quotients are one-dimensional and hence abelian.  Thus any solvable Lie algebra is an extension by one-dimensional algebras, and this extension also lifts to the level of groups.  So $\rr = \Lie(R)$ for some Lie group $R$.
\end{proof}



\section{From Zassenhaus to Ado}

\cite[Lectures 25 and 26]{notes}

Ado's Theorem (\cref{AdoThm}) normally is not proven in a course in Lie Theory.  For example, \cite[page 8]{anton} mentions it only in a footnote, referring the reader to \cite[Appendix E]{FH}.  \cite{Knapp} also relegates Ado's Theorem to an appendix (B.3).  In fact, we will see that Ado's Theorem is a direct consequence of \cref{LeviThm}, although we will need to develop some preliminary facts.

\begin{lemmadef}
  A \emindexing{Lie derivation} of a Lie algebra $\aa$ is a linear map $f: \aa \to \aa$ such that $f([x,y]) = [f(x),y] + [x,f(y)]$.  Equivalently, a derivation is a one-cocycle in the Chevalley complex with coefficients in $\aa$.
  
  A \emindexing{derivation} of an associative algebra $A$ is a linear map $f: A\to A$ so that $f(xy) = f(x)\,y + x\,f(x)$.
  
  The product (composition) of (Lie) derivations is not necessarily a (Lie) derivation, but the commutator of derivations is a derivation.  We write $\Der \aa$ for the Lie algebra of Lie derivations of $\aa$, and $\Der A$ for the algebra of associative derivations of $A$.  Henceforth, we drop the adjective ``Lie'', talking about simply derivations of a Lie algebra.
  
  We say that $\hh \acts \aa$ {\em by derivations} \index{action!by derivations}\index{derivation!action by} if the map $\hh \to \gl(\aa)$ in fact lands in $\Der\aa$.
\end{lemmadef}

In very general language, let $A$ and $B$ be vector spaces, and $a:A^{\otimes n} \to A$ and $b: B^{\otimes n}\to B$ $n$-linear maps.  Then a \emindexing{homomorphism} from $(A,a)$ to $(B,b)$ is a linear map $\phi: A\to B$ so that $\phi \circ a = b \circ \phi^{\otimes n}$, and a \emindexing{derivation} from $(A,a)$ to $(B,b)$ is a linear map $\phi: A\to B$ such that $\phi \circ a = b \circ \left( \sum_{i=1}^n \phi_i\right)$, where $\phi_i \defined 1 \otimes \dots \otimes \phi \otimes \dots \otimes 1$, with the $\phi$ in the $i$th spot.  Then the space $\Hom(A,B)$ of homomorphisms is not generally a vector space, but the space $\Der(A,B)$ of derivations is.  If $(A,a) = (B,b)$, then $\Hom(A,A)$ is closed under composition and hence a monoid, whereas $\Der(A,A)$ is closed under the commutator and hence a Lie algebra.  The notions of ``derivation'' and ``homomorphism'' agree for $n=1$, whence the map $\phi$ must intertwine $a$ with $b$.  The difference between derivations and homomorphisms is the difference between grouplike and primitive elements of a bialgebra.

\begin{prop} \label{Zproof2}
Let $\aa$ be a Lie algebra.
\begin{enumerate}
  \item Every derivation of $\aa$ extends uniquely to a derivation of $\Uu(\aa)$.
  \item $\Der\aa \to \Der\Uu(\aa)$ is a Lie algebra homomorphism.
  \item If $\hh\acts\aa$ by derivations, then $\hh(\Uu(\aa)) \subseteq \Uu(\aa)\cdot \hh(\aa) \cdot \Uu(\aa)$ the two-sided ideal of $\Uu(\aa)$ generated by the image of the $\hh$ action in $\aa$.
  \item If $N\leq \Uu(\aa)$ is an $\hh$-stable two-sided ideal, so is $N^n$.
\end{enumerate}
\end{prop}
\begin{proof}
  \begin{enumerate}
  \item Let $d\in\Der\aa$, and define $\hat\aa \defined \KK d \oplus \aa$; then $\Uu(\aa) \subseteq \Uu(\hat\aa)$.  The commutative $[d,-]$ preserves $\Uu(\aa)$ and is the required derivation.  Uniqueness is immediate: once you've said how something acts on the generators, you've defined it on the whole algebra.
  \item This is an automatic consequence of the uniqueness: the commutator of two derivations is a derivation, so if it's unique, it must be the correct derivation.
  \item Let $a_1,\dots, a_k \in \aa$ and $h\in \hh$.  Then $h(a_1\cdots a_k) = \sum_{i=1}^k a_1 \cdots h(a_i)\cdots a_k \in \Uu(\aa)\,\hh(\aa)\,\Uu(\aa)$.
  \item $N^n$ is spanned by monomials $a_1\cdots a_n$ where all $a_i \in N$.  Assuming that $h(a_i) \in N$ for each $h\in \hh$, we see that $h(a_1\cdots a_k) = \sum_{i=1}^k a_1 \cdots h(a_i)\cdots a_k \in N^n$.
\end{enumerate}
\end{proof}

\begin{lemmadef}
  Let $\hh$ and $\aa$ be Lie algebras and
  let $\hh \acts \aa$ by derivations.  The \emindexing{semidirect product}\index{product!semidirect} $\hh \ltimes \aa$ is the vector space $\hh \oplus \aa$ with the bracket given by $[h_1+a_1,h_2+a_2] = [h_1,h_2]_\hh + [a_1,a_2]_\aa + h_1\cdot a_2 - h_2\cdot a_1$, where be $h\cdot a$ we mean the action of $h$ on $a$.  Then $\hh\ltimes \aa$ is a Lie algebra, and $0 \to \aa \to \hh\ltimes\aa \to \hh \to 0$ is a split short exact sequence in \cat{LieAlg}.
\end{lemmadef}

\begin{prop}
  Let $\hh \acts \aa$ by derivations, and let $\gg = \hh \ltimes \aa$ be the semidirect product.  Then the actions $\hh \acts \Uu\aa$ by derivations and $\aa \acts \Uu\aa$ by left-multiplication together make a $\gg$-action on $\Uu\aa$.
\end{prop}

\begin{proof}
  We need only check the commutator of $\hh$ with $\aa$.  Let $u\in\Uu(\aa)$, $h\in\hh$, and $a\in\aa$.  Then $(h\circ a)u = h(au) = h(a)\,u + a\,h(u) = [h,a]u + a\,h(u)$.  Thus $[h,a]\in \gg$ acts as the commutator of operators $h$ and $a$ on $\Uu(\aa)$.
\end{proof}

\begin{defn}
  An algebra $U$ is {\em left-noetherian}\index{noetherian} if left ideals of $U$ satisfy the \emindexing{ascending chain condition}.  I.e.\ if any chain of left ideals $I_1\leq I_2\leq \dots$ of $U$ stabilizes.
\end{defn}
We refer the reader to any standard algebra textbook for a discussion of noetherian rings.  For noncommutative ring theory see \cite{Lam,McCR,GW}.

\begin{prop} \label{Zproof3}
  Let $U$ be a filtered algebra.  If $\gr U$ is left-noetherian, then so is $U$.
\end{prop}
  In particular, $\Uu(\aa)$ is left-noetherian, since $\gr \Uu(\aa)$ is a polynomial ring on $\dim \aa$ generators.
  
\begin{proof}
  Let $I\leq U$ be a left ideal.  We define $I_{\leq n} = I\cap U_{\leq n}$, and hence $I = \bigcup I_{\leq n}$.  We define $\gr I = \bigoplus I_{\leq n}/I_{\leq n-1}$, and this is a left ideal in $\gr U$.  If $I \leq J$, then $\gr I \leq \gr J$, using the fact that $U$ injects into $\gr U$ as vector spaces.

So if we have an ascending chain $I_1 \leq I_2 \leq \dots$, then the corresponding chain $\gr I_1 \leq \gr I_2 \leq \dots$ eventually terminate by assumption: $\gr I_n = \gr I_{n_0}$ for $n\geq n_0$.  But if $\gr I = \gr J$, then by induction on $n$, $I_{\leq n} = J_{\leq n}$, and so $I = J$.  Hence the original sequence terminates.
\end{proof}

\begin{lemma} \label{Zproof4}
  Let $\jj = \hh + \nn$ be a finite-dimensional Lie algebra, where $\hh$ is a subalgebra of $\jj$ and $\nn$ an ideal.  Assume that $\gg \acts W$ is a finite-dimensional representation such that $\hh,\nn \acts W$ nilpotently.  Then $\gg \acts W$ nilpotently.
\end{lemma}
\begin{proof}
  If $W = 0$ there is nothing to prove.  Otherwise, by \cref{EngelThm} there is some $w\in W^\nn$, where $W^\nn$ is the subspace of $W$ annihilated by $\nn$.  Let $h\in \hh$ and $x\in \nn$.  Then:
  \eqn{ xh w = \underbrace{[x,h]}_{\in \nn}w + h\underbrace{xw}_{= 0} = 0 }
  Thus $hw\in V^\nn$, and so $w \in V^\gg$.  By modding out and iterating, we see that $\gg \acts V$ nilpotently.
\end{proof}

\begin{thm}[Zassenhaus's Extension Lemma] \label{ZThm}
  Let $\hh$ and $\aa$ be finite-dimensional Lie algebras so that $\hh \acts \aa$ by derivations, and let $\gg = \hh \ltimes \aa$.  Moreover, let $V$ be a finite-dimensional $\aa$-module, and let $\nn$ be the nilpotency ideal of $\aa \acts V$.  If $[\hh,\aa] \leq \nn$, then there exists a finite-dimensional $\gg$-module $W$ and a surjective $\aa$-module map $W\onto V$, and so that the nilpotency ideal $\mm$ of $\gg \acts W$ contains $\nn$.  If $\hh \acts \aa$ by nilpotents, then we can arrange for $\mm \subseteq \hh$ as well.
\end{thm}

\begin{proof}
  Consider a Jordan-Holder series $0 = M_0 \subseteq M_1 \subseteq \dots \subseteq {\rm M}(n) = V$.  Then $\nn = \bigcap \ker(M_i/M_{i-1})$ by \cref{Zproof1}.  We define $N = \bigcap \ker\bigl( \Uu\aa \to \End(M_i/M_{i-1})\bigr)$, an ideal of $\Uu\aa$. Then $N \supseteq \nn \supseteq [\hh,\aa]$, and so $N$ is an $\hh$-stable ideal of $\Uu\aa$ by the third part of \cref{Zproof2}, and $N^k$ is $\hh$-stable by the fourth part.
  
  Since $\Uu\aa$ is left-noetherian (\cref{Zproof3}), $N^k$ is finitely generated for each $k$, and hence $N^k / N^{k+1}$ is a finitely generated $\Uu\aa$ module.  But the action $\Uu\aa \acts (N^k / N^{k+1})$ factors through $\Uu\aa / N$, so in fact $N^k / N^{k+1}$ is a finitely generated $(\Uu\aa / N)$-module.  But $\Uu\aa/N \cong \bigoplus \im\bigl(\Uu\aa \to \End(M_i/M_{i-1})\bigr) \subseteq \bigoplus \End(M_i/M_{i-1})$, which is finite-dimensional.  So $\Uu\aa/N$ is finite-dimensional, $N^k / N^{k+1}$ a finitely-generated $(\Uu\aa/N)$-module, and hence $N^k/N^{k+1}$ is finite-dimensional.
  
  By construction, $N(M_k) \subseteq M_{k-1}$, so $N^n$ annihilates $V$, where $n$ is the length of the Jordan-Holder series $0 = M_0 \subseteq M_1 \subseteq \dots \subseteq {\rm M}(n) = V$.  Let $\dim V = d$, and define
  \eqn{ W \defined \bigoplus_{i=1}^d \Uu\aa / N^n }
  Then $W$ is finite-dimensional since $\Uu\aa / N^n \cong \Uu\aa / N \oplus N/N^2 \oplus \dots \oplus N^{n-1}/N^n$ as a vector space, and each summand is finite-dimensional.  To construct the map $W \onto V$, we pick a basis $\{v_i\}_{i=1}^d$ of $V$, and send $(0,\dots,1,\dots,0) \mapsto v_i$, where $1$ is the image of $1\in \Uu\aa$ in $\Uu\aa/N^n$, and it is in the $i$th spot.  By construction $\Uu\aa_{\leq 0}$ acts as scalars, and so $N$ does not contain $\Uu\aa_{\leq 0}$; thus the map is well-defined.  Moreover, $\gg \acts \Uu\aa$ by \cref{Zproof2}, and $N$ is $\hh$-stable and hence $\gg$-stable.  Thus $\gg \acts W$ naturally, and the action $\Uu\aa \acts V$ factors through $N^n$, and so $W \onto V$ is a map of $\gg$-modules.
  
  By construction, $N$ and hence $\nn$ acts nilpotently on $W$.  But $\nn$ is an ideal of $\gg$: a general element of $\gg$ is of the form $h+a$ for $h\in \hh$ and $a\in \aa$, and $[h + a,\nn] = [h,\nn] + [a,\nn] \subseteq [h,\aa] + [a,\nn] \subseteq \nn + \nn = \nn$.  So $\mm \supseteq \nn$, as $\mm$ is the largest nilpotency ideal of $\gg \acts W$.
  
  We finish by considering the case when $\hh \acts \aa$ nilpotently.  Then $\hh \acts W$ nilpotently, and since $[\hh,\aa] \subseteq \nn$, $\hh + \nn$ is an ideal of $\gg$.  By \cref{Zproof4}, $\hh + \nn$ acts nilpotently on $W$, and so is a subideal of $\mm$.
\end{proof}

\begin{cor} \label{AdoProof1}
  Let $\rr$ be a solvable Lie algebra over characteristic $0$, and let $\nn$ be its largest nilpotent ideal.  Then every derivation of $\rr$ has image in $\nn$.  In particular, if $\rr$ is an ideal of some larger Lie algebra $\gg$, then $[\gg,\rr] \subseteq \nn$.
\end{cor}

\begin{proof}
  Let $d$ be a derivation of $\rr$ and $\hh = \KK d \oplus \rr$.  Then $\hh$ is solvable by \cref{SolvNilProp}, and $\hh' \acts \hh$ nilpotently by \cref{LcorNilMod}.  But $d(\rr) \subseteq \hh'$ and $\rr$ is an ideal of $\hh$, and so $d(\rr)$ acts nilpotently on $\rr$, and is thus a subideal of $\nn$.
  
  The second statement follows from the fact that $[g,-]$ is a derivation; this follows ultimately from the Jacobi identity.
\end{proof}

\begin{thm}[Ado's Theorem] \label{AdoThm}
  Let $\gg$ be a finite-dimensional Lie algebra over characteristic $0$.  Then $\gg$ has a faithful representation $\gg \mono \gl(V)$, and this representation can be chosen so that the largest nilpotent ideal $\nn \leq \gg$ acts nilpotently on $V$.\footnote{It seems that Ado originally proved a weaker version of \cref{AdoThm} over $\RR$.  The version we present is due to \cite{HarishChandra}.  The dependence on characteristic is removed in \cite{Iwasawa}.}
\end{thm}

\begin{proof}
  We induct on $\dim \gg$.  The $\gg = 0$ case is trivial, and we break the induction step into cases:
\begin{description}
  \item[Case I: $\gg = \nn$ is nilpotent.]  Then $\gg \neq \gg'$, and so we choose a subspace $\aa \supseteq \gg'$ of codimensional $1$ in $\gg$.  This is automatically an ideal, and we pick $x\not\in \aa$, and $\hh = \<x\>$.  Any one-dimensional subspace is a subalgebra, and $\gg = \hh \ltimes \aa$.  By induction, $\aa$ has a faithful module $V'$ and acts nilpotently.

  The hypotheses of \cref{ZThm} are satisfied, and we get an $\aa$-module homomorphism $W \epi V'$ with $\gg \acts W$ nilpotently.  As yet, this might not be a faithful representation of $\gg$: certainly $\aa$ acts faithfully on $W$ because it does so on $V'$, but $x$ might kill $W$.  We pick a faithful nilpotent $\gg/\aa = \KK$-module $W_1$, e.g.\ $x \mapsto \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right] \in \gl(2)$.  Then $V = W\oplus W_1$ is a faithful nilpotent $\gg$ representation.

  \item[Case II: $\gg$ is solvable but not nilpotent.]  Then $\gg' \leq \nn \lneq \gg$.  We pick an ideal $\aa$ of codimension $1$ in $\gg$ such that $\nn \subseteq \aa$, and $x$ and $\hh = \KK x$ as before, so that $\gg = \hh \ltimes \aa$.  Then $\nn(\aa) \supseteq \nn$ --- if a matrix acts nilpotently on $\gg$, then certainly it does so on $\aa$, and by construction $\nn \subseteq \aa$ --- and we have a faithful module $\aa \acts V'$ by induction, with $\nn(\aa)\acts V'$ nilpotently.  Then $[\hh,\aa]\acts V'$ nilpotently, since $[\hh,\aa] \subseteq \nn(\aa)$ by \cref{AdoProof1}, so we use \cref{ZThm} to get $\gg \acts W$ and an $\aa$-module map $W \onto V'$, such that $\nn \acts W$ nilpotently.  We form $V = W \oplus W_1$ as before so that $\gg \acts V$ is faithful.  Since $\nn$ is contained in $\aa$ and $\aa$ acts as $0$ on $W_1$, $\nn$ acts nilpotently on $W$.

  \item[Case III: general.]  By \cref{LeviThm}, there is a splitting $\gg = \ss \ltimes \rr$ with $\ss$ semisimple and $\rr$ solvable.  By Case II, we have a faithful $\rr$-representation $V'$ with $\nn(\rr) \acts V'$ nilpotently.  By \cref{AdoProof1} the conditions of \cref{ZThm} apply, so we have $\gg \acts W$ and an $\rr$-module map $W\onto V'$, and since $\nn \leq \rr$ we have $\nn \leq \nn(\rr)$ so $\nn \acts W$ nilpotently.  We want to get a faithful representation, and we need to make sure it is faithful on $\ss$.  But $\ss = \gg/\rr$ is semisimple, so has no center, so $\ad: \ss \acts \ss$ is faithful.  So we let $W_1 = \ss = \gg/\rr$ as $\gg$-modules, and $\gg \acts V = W \oplus W_1$ is faithful with $\nn$ acting as $0$ on $W_1$ and nilpotently on $W$.
\end{description}
\end{proof}

\Section{Exercises}

\begin{enumerate}
  \item Classify the $3$-dimensional Lie algebras $\gg$ over an algebraically closed field $\KK$ of characteristic zero by showing that if $\gg$ is not a direct product of smaller Lie algebras, then either 
  \begin{itemize} \item $\gg \cong \sl(2,\KK)$, 
    \item $\gg$ is isomorphic to the nilpotent Heisenberg Lie algebra $\hh$ with basis $X , Y , Z$ such that $Z$ is central and $[X, Y ] = Z$, or 
    \item $\gg$ is isomorphic to a solvable algebra $\ss$ which is the semidirect product of the abelian algebra $\KK^2$ by an invertible derivation. In particular $\ss$ has basis $X , Y , Z$ such that $[Y , Z ] = 0$, and $\ad X$ acts on $\KK Y + \KK Z$ by an invertible matrix, which, up to change of basis in $\KK Y + \KK Z$ and rescaling $X$, can be taken to be either $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]$ or $\left[\begin{array}{cc} \lambda &  0 \\ 0 & 1 \end{array}\right]$ for some nonzero $\lambda \in \KK$.
  \end{itemize}

   \item \begin{enumerate} \item Show that the Heisenberg Lie algebra $\hh$ in Problem 1 has the property that $Z$ acts nilpotently in every finite-dimensional module, and as zero in every simple finite-dimensional module. 

    \item  Construct a simple infinite-dimensional $\hh$-module in which $Z$ acts as a non-zero scalar. [Hint: take $X$ and $Y$ to be the operators $d/dt$ and $t$ on $\KK[t]$.]

  \end{enumerate}
  \item  Construct a simple $2$-dimensional module for the Heisenberg algebra $\hh$ over any field $\KK$ of characteristic $2$. In particular, if $\KK = \bar\KK$, this gives a counterexample to Lie's theorem in non-zero characteristic. 

  \item  Let $\gg$ be a finite-dimensional Lie algebra over $\KK$. 
  \begin{enumerate}
    \item  Show that the intersection $\nn$ of the kernels of all finite-dimensional simple $\gg$-modules can be characterized as the largest ideal of $\gg$ which acts nilpotently in every finite-dimensional $\gg$-module. It is called the \emindexing{nilradical} of $\gg$. 

    \item  Show that the nilradical of $\gg$ is contained in $\gg' \cap \rad(\gg)$.

    \item  Let $\hh \subseteq \gg$ be a subalgebra and $V$ a $\gg$-module. Given a linear functional $\lambda : \hh \to \KK$, define the associated weight space to be $V_\lambda = \{v \in V : H v = \lambda(H )v \textrm{ for all } H \in \hh\}$.  Assuming $\textrm{char}(\KK) = 0$, adapt the proof of Lie's theorem to show that if $\hh$ is an ideal and $V$ is finite-dimensional, then $V_\lambda$ is a $\gg$-submodule of $V$. 

    \item  Show that if $\textrm{char}(\KK) =0$ then the nilradical of $\gg$ is equal to $\gg' \cap \rad(\gg)$. [Hint: assume without loss of generality that $\KK = \bar\KK$ and obtain from Lie's theorem that any finite-dimensional simple $\gg$-module $V$ has a non-zero weight space for some weight $\lambda$ on $\gg' \cap \rad(\gg)$.  Then use (c) to deduce that $\lambda = 0$ if $V$ is simple.] 

  \end{enumerate}
  \item  Let $\gg$ be a finite-dimensional Lie algebra over $\KK$, $\textrm{char}(\KK) =0$. Prove that the largest nilpotent ideal of $\gg$ is equal to the set of elements of $\rr = \rad \gg$ which act nilpotently in the adjoint action on $\gg$, or equivalently on $\rr$. In particular, it is equal to the largest nilpotent ideal of $\rr$.

  \item  Prove that the Lie algebra $\sl(2, \KK)$ of $2\times 2$ matrices with trace zero is simple, over a field $\KK$ of any characteristic $\neq 2$. In characteristic $2$, show that it is nilpotent. 

  \item  In this exercise, we�ll deduce from the standard functorial properties of $\Ext$ groups and their associated long exact sequences that $\Ext^1 (N, M )$ bijectively classifies extensions $0 \to M \to V \to N \to 0$ up to isomorphism, for modules over any associative ring with unity. 
  \begin{enumerate}
    \item  Let $F$ be a free module with a surjective homomorphism onto $N$, so we have an exact sequence $0\to K \to F \to N \to 0$. Use the long exact sequence to produce an isomorphism of $\Ext^1 (N, M )$ with the cokernel of $\Hom(F, M ) \to \Hom(K, M )$. 

    \item  Given $\phi \in \Hom(K, M )$, construct $V$ as the quotient of $F \oplus M$ by the graph of $-\phi$ (note that this graph is a submodule of $K \oplus  M \subseteq  F \oplus M$).

    \item  Use the functoriality of $\Ext$ and the long exact sequences to show that the characteristic class in $\Ext^1 (N, M )$ of the extension constructed in (b) is the element represented by the chosen $\phi$, and conversely, that if $\phi$ represents the characteristic class of a given extension, then the extension constructed in (b) is isomorphic to the given one. 

  \end{enumerate}
  \item Calculate $\Ext^i (\KK, \KK)$ for all $i$ for the trivial representation $\KK$ of $\sl(2, \KK)$, where $\textrm{char}(\KK) = 0$. Conclude that the theorem that $\Ext^i (M, N ) = 0$ for $i = 1, 2$ and all finite-dimensional representations $M , N$ of a semi-simple Lie algebra $\gg$ does not extend to $i > 2$. 

  \item  Let $\gg$ be a finite-dimensional Lie algebra. Show that $\Ext^1 (\KK, \KK)$ can be canonically identified with the dual space of $\gg/\gg'$, and therefore also with the set of $1$-dimensional $\gg$-modules, up to isomorphism. 

  \item  Let $\gg$ be a finite-dimensional Lie algebra. Show that $\Ext^1 (\KK, \gg)$ can be canonically identified with the quotient $\Der(\gg)/ \Inn(\gg)$, where $\Der(\gg)$ is the space of derivations of $\gg$, and $\Inn(\gg)$ is the subspace of inner derivations, that is, those of the form $d(x) = [y, x]$ for some $y \in  \gg$. Show that this also classifies Lie algebra extensions $\hat{\gg}$ containing $\gg$ as an ideal of codimension~$1$. 

  \item  Let $\gg$ be a finite-dimensional Lie algebra over $\KK$, $\textrm{char}(\KK) = 0$. The {Malcev-Harish-Chandra theorem} says that all Levi subalgebras $\ss \subseteq \gg$ are conjugate under the action of the group $\exp \ad \nn$, where $\nn$ is the largest nilpotent ideal of $\gg$ (note that $\nn$ acts nilpotently on $\gg$, so the power series expression for $\exp \ad X$ reduces to a finite sum when $X \in \nn$).
  \begin{enumerate}
    \item  Show that the reduction we used to prove Levi's theorem by induction in the case where the radical $\rr = \rad \gg$ is not a minimal ideal also works for the Malcev-Harish-Chandra theorem. More precisely, show that if $\rr$ is nilpotent, the reduction can be done using any nonzero ideal $\mm$ properly contained in $\rr$. If $\rr$ is not nilpotent, use Problem 4 to show that $[\gg,\rr] = \rr$, then make the reduction by taking $\mm$ to contain $[\gg, \rr]$. 
    \item  In general, given a semidirect product $\gg = \hh \ltimes \mm$, where $\mm$ is an abelian ideal, show that $\Ext^1_{\Uu (\hh)} (\KK, \mm)$ classifies subalgebras complementary to $\mm$, up to conjugacy by the action of $\exp \ad \mm$. Then use the vanishing of $\Ext^1 (M, N )$ for finite-dimensional modules over a semi-simple Lie algebra to complete the proof of the Malcev-Harish-Chandra theorem.
  \end{enumerate}
\end{enumerate}










\chapter{Classification of Semisimple Lie Algebras} \label{SemisimpleChapter}

\section{Classical Lie algebras over $\CC$}

\subsection{Reductive Lie algebras}

\cite[Lecture 26]{notes}

Henceforth every Lie algebra, except when otherwise marked, is finite-dimensional over a field of characteristic $0$.

\begin{lemmadef} \label{reductivedefn}
  A Lie algebra $\gg$ is \emindexing{reductive} if $(\gg,\ad)$ is completely reducible.
  
  A Lie algebra is reductive if and only if it is of the form $\gg = \ss \times \aa$ where $\ss$ is semisimple and $\aa$ is abelian.  Moreover, $\aa = Z(\gg)$ and $\ss = \gg'$.
\end{lemmadef}
\begin{proof}
  Let $\gg$ be a reductive Lie algebra; then $\gg = \bigoplus \aa_i$ as $\gg$-modules, where each $\aa_i$ is an ideal of $\gg$ and $[\aa_i,\aa_j] \subseteq \aa_i\cap \aa_j = 0$ for $i\neq j$.  Thus $\gg = \prod \aa_i$ as Lie algebras, and each $\aa_i$ is either simple non-abelian or one-dimensional.
\end{proof}

\begin{prop}
  Any $*$-closed subalgebra of $\gl(n,\CC)$ is reductive.
\end{prop}
\begin{proof}
  We define a symmetric real-valued bilinear form $(,)$ on $\gl(n,\CC)$ by $(x,y) = \real\bigl( \tr(xy^*)\bigr)$.  Then $(x,x) = \sum |x_{ij}|^2$, so $(,)$ is positive-definite.  Moreover:
  \eqn{ ([z,x],y) = -(x,[z^*,y]) }
  so $[z^*,-]$ is adjoint to $[-,z]$.
  
  Let $\gg \subseteq \gl(n,\CC)$ be any subalgebra and $\aa \leq \gg$ an ideal.  Then $\aa^\perp \subseteq \gg^*$ by invariance, where $\gg^*$ is the Lie algebra of Hermitian conjugates of elements of $\gg$.  If $\gg$ is $*$-closed, then $\gg^* = \gg$ and $\aa^\perp$ is an ideal of $\gg$.  By positive-definiteness, $\gg = \aa \oplus \aa^\perp$, and we rinse and repeat to write $\gg$ is a sum of irreducibles.
\end{proof}

\begin{eg}
  The \indexing{classical Lie algebras}\index{Lie algebra!classical} $\sl(n,\CC) \defined \{x \in \gl(n)\st \tr x = 0\}$\index{slnC@$\sl(n,\CC)$}, $\so(n,\CC) \defined \{x \in \gl(n)\st x + x^T = 0\}$\index{sonC@$\so(n,\CC)$}, and $\sp(n,\CC) \defined  \{x \in \gl({2n}) \st jx + x^Tj = 0\}$\index{spnC@$\sp(n,\CC)$} are reductive.  Indeed, since they have no center except in very low dimensions, they are all semisimple.  We will see later that they are all simple, except in a few low dimensions.
  
  Since a real Lie algebra $\gg$ is semisimple if $\gg \otimes_\RR \CC$ is, the real Lie algebras $\sl(n,\RR)$\index{slnR@$\sl(n,\RR)$}, $\so(n,\RR)$\index{sonR@$\so(n,\RR)$}, and $\sp(n,\RR)$\index{spnR@$\sp(n,\RR)$} also are semisimple.
\end{eg}

\subsection{Guiding examples: $\sl(n)$ and $\sp(n)$ over $\CC$}

\cite[Lectures 27 and 28]{notes}

Let $\gg = \sl(n)$ or $\sp(n)$.  We extract an abelian subalgebra $\hh \subseteq \gg$.  For $\sl_n$ we use the diagonal traceless matrices:
\eqn{ \hh \defined \left\{ \left[ \begin{array}{ccc} z_1 && \\ & \ddots & \\ && z_n \end{array} \right] \st \sum z_i = 0 \right\} }
For $\sp(n) \defined \{ x\in \gl(2n) \st jx + x^Tj = 0 \}$, it will be helpful to redefine $j$.  We can use any $j$ which defines a non-degenerate antisymmetric bilinear form, and we take:
\eqn{  j = \left[ \begin{array}{ccc:ccc} 
   &&& &&1 \\
   & 0 && & \adots & \\
   &&& 1 && \\ \hdashline
   && -1 &&& \\
   & \adots & && 0 & \\
   -1 && &&& 
   \end{array} \right] }
Let $a^R$ be the matrix $a$ reflected across the antidiagonal.  Then we can define $\sp(n)$ in block diagonal form:
\eqn[spnewdef]{ \sp(n) = \left\{ \left[ \begin{array}{c:c} a & b \\ \hdashline c & d \end{array} \right] \in {\rm M}(2n) = {\rm M}(2,{\rm M}(n)) \st d = -a^R, b = b^R, c = c^R \right\} }
In this basis, we take as our abelian subalgebra
\eqn{ \hh \defined \left\{ \left[ \begin{array}{ccc:ccc} z_1 &&&&& \\ & \ddots &&& 0 & \\ && z_n &&& \\ \hdashline  &&& -z_n && \\ & 0 &&& \ddots & \\ &&&&& -z_1 \end{array} \right] \right\}}

\begin{prop}
  Let $\gg = \sl(n)$ or $\sp(n)$.  For $\hh \leq \gg$ defined above, the adjoint action $\ad: \hh \acts \gg$ is diagonal.
\end{prop}
\begin{proof}
  We make explicit the basis of $\gg$.  For $\gg = \sl(n)$, the natural basis is $\{e_{ij}\}_{i\neq j} \cup \{e_{ii} - e_{i+1,i+1}\}_{i=1}^{n-1}$, where $e_{ij}$ is the matrix with a $1$ in the $(ij)$ spot and $0$s elsewhere.  In particular, $\{e_{ii} - e_{i+1,i+1}\}_{i=1}^{n-1}$ is a basis of $\hh$.  Let $h \in \hh$ be 
  \eqn{ h = \left[ \begin{array}{ccc} z_1 && \\ & \ddots & \\ && z_n \end{array} \right]}
  Then $[h,e_{ij}] = (z_i - z_j)e_{ij}$, and $[h,h'] = 0$ when $h' \in \hh$.
  
  For $\gg = \sp(n)$, the natural basis suggested by \cref{spnewdef} is
  \begin{align} \left\{ a_{ij} \defined \left[ \begin{array}{c:c} e_{ij} & 0 \\ \hdashline 0 & -e_{n+1-j,n+1-i} \end{array} \right] \right\} & \cup
  \left\{ b_{ij} \defined \left[ \begin{array}{c:c} 0 & e_{ij} + e_{n+1-j,n+1-i} \\ \hdashline 0 & 0 \end{array}\right] \st i+j \leq n+1 \right\} \nonumber \\
 & \cup  \left\{ c_{ij} \defined \left[ \begin{array}{c:c} 0& 0 \\ \hdashline  e_{ij} + e_{n+1-j,n+1-i}  & 0 \end{array}\right] \st i+j \leq n+1 \right\} \end{align}
  Of course, when $i=j$, then $ \left\{ \left[ \begin{array}{c:c} e_{ii} & 0 \\ \hdashline 0 & -e_{n+1-i,n+1-i} \end{array} \right] \right\}$ is a basis of $\hh$.  Let $h\in \hh$ be given by
  \eqn{ h = \left[ \begin{array}{ccc:ccc} z_1 &&&&& \\ & \ddots &&& 0 & \\ && z_n &&& \\ \hdashline  &&& -z_n && \\ & 0 &&& \ddots & \\ &&&&& -z_1 \end{array} \right] }
  Then $[h,a_{ij}] = (z_i - z_j)a_{ij}$, $[h,b_{ij}] = (z_i + z_j)b_{ij}$, and $[h,c_{ij}] = (-z_i - z_j)c_{ij}$.
\end{proof}
\begin{defn} \label{RootDefnTemporary}
  Let $\hh$ be a maximal abelian subalgebra of a finite-dimensional Lie algebra $\gg$ so that $\ad: \hh \acts\gg$ is diagonalizable, so diagonal in an eigenbasis.  Write $\hh^*$ for the vector space dual to $\hh$.  Each eigenbasis element of $\gg$ defines an eigenvalue to each $h\in \hh$, and this assignment is linear in $\hh$; thus, the eigenbasis of $\gg$ picks out a vector $\alpha \in \hh^*$.  The set of such vectors are the {\em roots}\index{root} of the pair $(\gg,\hh)$.
\end{defn}
We will refine this definition in \cref{RootsDefn}, and we will prove that the set of roots of a semisimple Lie algebra $\gg$ is determined up to isomorphism by $\gg$ (in particular, it does not depend on the subalgebra $\hh$).
\begin{eg}
  When $\gg = \sl(n)$ and $\hh$ is as above, the roots are $\{0\} \cup \{z_i - z_j\}_{i\neq j}$, where $\{z_i\}_{i=1}^n$ are the natural linear functionals $\hh \to \CC$.
  When $\gg = \sp(n)$ and $\hh$ is as above, the roots are $\{0\} \cup \{\pm 2z_i\} \cup \{\pm z_i \pm z_j\}_{i\neq j}$.
\end{eg}

\begin{lemmadef}
  Let $\gg$ and $\hh \leq \gg$ as in \cref{RootDefnTemporary}.  Then the roots break $\gg$ into eigenspaces:
  \eqn{ \gg = \bigoplus_{\alpha \textrm{ a root}} \gg_\alpha = \hh \oplus \bigoplus_{\alpha\neq 0\textrm{ a root}} \gg_\alpha}
  In particular, since $\hh$ is a maximal abelian subalgebra, the $0$-eigenspace of $\hh \acts \gg$ is precisely $\gg_0 = \hh$.  Then the spaces $\gg_\alpha$ are the {\em root spaces}\index{root space} of the pair $(\gg,\hh)$.  By the Jacobi identity, $[\gg_\alpha,\gg_\beta]\subseteq \gg_{\alpha + \beta}$.
\end{lemmadef}
\begin{lemma} \label{sl2lemmaslsp}
  When $\gg = \sl(n)$ or $\sp(n)$ and $\hh$ is as above, then for $\alpha \neq 0$ the root space $\gg_\alpha\subseteq \gg$ is one-dimensional.  Let $\hh_\alpha \defined [\gg_\alpha,\gg_{-\alpha}]$.  Then $\hh_\alpha = \hh_{-\alpha}$ is one-dimensional, and $\gg_\alpha \oplus \gg_{-\alpha} \oplus \hh_\alpha$ is a subalgebra of $\gg$ isomorphic to $\sl(2)$.
\end{lemma}
\begin{proof}
  For each root $\alpha$, pick a basis element $g_{\pm \alpha} \in \gg_{\pm \alpha}$ (in particular, we can use the eigenbasis of $\hh \acts \gg$ given above), and define $h_\alpha \defined [g_\alpha,g_{-\alpha}]$.  Define $\alpha(h_\alpha) = a$ so that $[h_\alpha,g_{\pm \alpha}] = \pm a g_{\alpha}$; one can check directly that $a \neq 0$.  For the isomorphism, we use the fact that $\CC$ is algebraically closed.
\end{proof}

\begin{defn}
  Let $\gg$ and $\hh \leq \gg$ as in \cref{RootDefnTemporary}.  The \emindexing{rank} of $\gg$ is the dimension of $\hh$, or equivalently the dimension of the dual space $\hh^*$.
\end{defn}

\begin{eg}
  The Lie algebras $\sl(3)$ and $\sp(2)$ are rank-two.  For $\gg = \sl(3)$, the dual space $\hh^*$ to $\hh$ spanned by the vectors $z_1-z_2$ and $z_2 - z_3$ naturally embeds in a three-dimensional vector space spanned by $\{z_1,z_2,z_3\}$, and we choose an inner product on this space in which $\{z_i\}$ is an orthonormal basis.  Let $\alpha_1 = z_1 - z_2$, $\alpha_2 = z_2 - z_3$, and $\alpha_3 = z_1 - z_3$.  Then the roots $\{0,\pm\alpha_i\}$ form a perfect hexagon:
  \eqn{ \begin{tikzpicture}[baseline=(O.base),scale=2]
    \node[label=right:$0$] (O) at (0,0) {$\bullet$};
    \node[label=above:$\alpha_2$] at (30:1) {$\bullet$};
    \node[label=above:$\alpha_3$] at (90:1) {$\bullet$};
    \node[label=above:$\alpha_1$] at (150:1) {$\bullet$};
    \node[label=below:$-\alpha_2$] at (210:1) {$\bullet$};
    \node[label=below:$-\alpha_3$] at (270:1) {$\bullet$};
    \node[label=below:$-\alpha_1$] at (330:1) {$\bullet$};
  \end{tikzpicture} }
  
  For $\gg = \sp(2)$, we have $\hh^*$ spanned by $\{z_1,z_2\}$, and we choose an inner product in which this is an orthonormal basis.  Let $\alpha_1 = z_1 - z_2$ and $\alpha_2 = 2z_2$.  The roots form a diamond:
  \eqn{ \begin{tikzpicture}[baseline=(O.base),scale=1.5]
    \node[label=below:$0$] (O) at (0,0) {$\bullet$};
    \node[label=below:$z_2$] at (1,0) {$\cdot$};
    \node[label=below:$z_1$] at (0,1) {$\cdot$};
    \node[label=above:$\alpha_2$] at (2,0) {$\bullet$};
    \node[label=above:$\alpha_1$] at (-1,1) {$\bullet$};
    \node[label=above:$\alpha_2 + \alpha_1$] at (1,1) {$\bullet$};
    \node[label=above:$\alpha_2 + 2\alpha_1$] at (0,2) {$\bullet$};
    \node at (-2,0) {$\bullet$};
    \node at (1,-1) {$\bullet$};
    \node at (-1,-1) {$\bullet$};
    \node at (0,-2) {$\bullet$};    
  \end{tikzpicture} }
\end{eg}
\begin{lemma}
  Let $\gg,\hh\leq \gg$ be as in \cref{RootDefnTemporary}.  Let $v\in \hh$ be chosen so that $\alpha(v) \neq 0$ for every root $\alpha$.  Then $v$ divides the roots into {\em positive roots}\index{root!positive}\index{positive root} and {\em negative roots} according to the sign of $\alpha(v)$.  A \emindexing{simple root}\index{root!simple} is any positive root that is not expressible as a sum of positive roots.
\end{lemma}
\begin{eg}
  Let $\gg = \sl(n)$ or $\sp(n)$, and choose $v\in \hh$ so that $z_1(v) > z_2(v) > \dots > z_n(v) > 0$.  The positive roots of $\sl(n)$ are $\{z_i - z_j\}_{i < j}$, and the positive roots of $\sp(n)$ are $\{z_i - z_j\}_{i < j} \cup \{z_i + z_i\} \cup \{2z_i\}$.  The simple roots of $\sl(n)$ are $\{ \alpha_i = z_i - z_{i+1}\}_{i=1}^{n-1}$, and the simple roots of $\sp(n)$ are $\{ \alpha_i = z_i - z_{i+1}\}_{i=1}^{n-1} \cup \{2z_n\}$.  In each case, the simple roots are a basis of $\hh^*$.  Moreover, the roots are in the $\ZZ$-span of the simple roots, i.e.\ the lattice generated by the simple roots, and the positive roots are in the intersection of this lattice with the positive cone, so that the positive roots are in the $\NN$-span of the simple roots.
  
  We partially order the positive roots by saying that $\alpha < \beta$ if $\beta - \alpha$ is a positive root.  Under this partial order there is a unique maximal positive root $\theta$, the \emindexing{highest root}\index{root!highest}; for $\sl(n)$ we have $\theta = z_1 - z_n = \alpha_1 + \dots + \alpha_{n-1}$, and for $\sp(n)$ we have $\theta = 2z_1 = 2(\alpha_1 + \dots + \alpha_{n-1}) + \alpha_n$.  We draw these partial orders:
  \[ \begin{tikzpicture}[baseline,scale=.5]
    \path (0,0) node[label=below:$\scriptstyle \alpha_1$] {$\bullet$} ++(2,0) node[label=below:$\scriptstyle \alpha_2$] {$\bullet$} ++(2,0) node[label=below:$\scriptstyle \alpha_3$] {$\bullet$} ++(3,0) node {$\cdots$}  ++(3,0) node[label=below:$\scriptstyle \alpha_{n-1}$] {$\bullet$} ;
    \foreach \x/\y in {3/1,5/1,2/2,4/2,6/2,8/2,3/3,5/3,7/3,4/4,6/4} {\node at (\x,\y) {$\bullet$}; \draw (\x,\y) -- +(-1,-1); \draw (\x,\y) -- +(1,-1); }
    \node[fill=white] at (7,1) {$\cdots$};
    \node[label=above:{$\scriptstyle \theta = \alpha_1 + \dots + \alpha_{n-1}$}] at (5,5) {$\bullet$}; \draw (5,5) -- +(-1,-1); \draw (5,5) -- +(1,-1);
    \node[label=above left:$\scriptstyle \alpha_1 + \alpha_2$] at (1,1) {$\bullet$}; \draw (1,1) -- +(-1,-1); \draw (1,1) -- +(1,-1);
    \node[label=above right:$\scriptstyle \alpha_{n-2} + \alpha_{n-1}$] at (9,1) {$\bullet$}; \draw (9,1) -- +(-1,-1); \draw (9,1) -- +(1,-1);
    \node at (5,-2) {$\sl(n)$};
  \end{tikzpicture}
      \hspace{.5in}
  \begin{tikzpicture}[baseline,scale=.5]
    \path (0,0) node[label=below:$\scriptstyle \alpha_1$] {$\bullet$} ++(2,0) node[label=below:$\scriptstyle \alpha_2$] {$\bullet$} ++(3,0) node {$\cdots$}  ++(3,0) node[label=below:$\scriptstyle \alpha_{n-1}$] {$\bullet$} ++(2,0)  node[label=below:$\scriptstyle \alpha_{n}$] {$\bullet$} ;
    \foreach \x/\y in {1/1,3/1,7/1,9/1,2/2,4/2,6/2,8/2,3/3,5/3,7/3,9/3,4/4,6/4,8/4,5/5,7/5,9/5,6/6,8/6,7/7,9/7,8/8,9/9} {\node at (\x,\y) {$\bullet$}; \draw (\x,\y) -- +(-1,-1); \draw (\x,\y) -- +(1,-1); }
    \foreach \y in {2,4,6,8} {\node at (10,\y) {$\bullet$}; \draw (10,\y) -- +(-1,-1);}
    \node[fill=white] at (5,1) {$\cdots$};
    \node[label=above:{$\scriptstyle \theta = 2(\alpha_1 + \dots + \alpha_{n-1}) + \alpha_n$}] at (10,10) {$\bullet$}; \draw (10,10) -- +(-1,-1); 
    \node at (5,-2) {$\sp(n)$};
  \end{tikzpicture}
  \]
\end{eg}
To make this very clear, we draw the rank-three pictures fully labeled (edges by the difference between consecutive nodes):
\[ \begin{tikzpicture}[baseline,scale=1.5]
  \node[label=below:{$\alpha_1 = z_1 - z_2$}] at (0,0) {$\bullet$};
  \node[label=below:{$\alpha_2 = z_2 - z_3$}] at (2,0) {$\bullet$};
  \node[label=below:{$\alpha_3 = z_3 - z_4$}] at (4,0) {$\bullet$};
  \node[label=above left:$z_1 - z_3$] at (1,1) {$\bullet$};
  \node[label=above right:$z_2 - z_4$] at (3,1) {$\bullet$};
  \node[label=above:{$\theta = z_1 - z_4$}] at (2,2) {$\bullet$};
  \draw (0,0) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,-1);
  \draw (2,0) -- node[auto] {$\scriptstyle \alpha_3$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,-1);
  \draw (1,1) -- node[auto] {$\scriptstyle \alpha_3$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,-1);
  \node at (2,-1) {$\sl(4)$};
\end{tikzpicture}\]
\[ \begin{tikzpicture}[baseline,scale=1.5]
  \node[label=below:{$\alpha_1 = z_1 - z_2$}] at (0,0) {$\bullet$};
  \node[label=below:{$\alpha_2 = z_2 - z_3$}] at (2,0) {$\bullet$};
  \node[label=below:{$\alpha_3 = 2z_3 $}] at (4,0) {$\bullet$};
  \node[label=above left:$z_1 - z_3$] at (1,1) {$\bullet$};
  \node[label=right:$z_2 + z_3$] at (3,1) {$\bullet$};
  \node[label=above left:{$z_1 + z_3$}] at (2,2) {$\bullet$};
  \node[label=right:$2z_2$] at (4,2) {$\bullet$};
  \node[label=above left:{$z_1 + z_2$}] at (3,3) {$\bullet$};
  \node[label=right:{$\theta = 2z_1$}] at (4,4) {$\bullet$};
  \draw (0,0) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,-1);
  \draw (2,0) -- node[auto] {$\scriptstyle \alpha_3$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,-1);
  \draw (1,1) -- node[auto] {$\scriptstyle \alpha_3$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,-1);
  \draw (2,2) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,1) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,-1);
  \draw (3,1) -- node[auto] {$\scriptstyle \alpha_2$} ++(1,1) ;
  \draw (3,3) -- node[auto] {$\scriptstyle \alpha_1$} ++(1,1) ;
  \node at (2.5,-1) {$\sp(3)$};
\end{tikzpicture}\]

Using these pictures of $\sl(n)$ and $\sp(n)$, we can directly prove the following:
\begin{prop}
  The Lie algebras $\sl(n,\CC)$ and $\sp(n,\CC)$ are simple.
\end{prop}
\begin{proof}
  Let $\gg = \sl(n)$ or $\sp(n)$, and $\hh$, the systems of positive and simple roots, and $\theta$ the highest root as above.  Recall that for each root $\alpha\neq 0$, the root space $\gg_\alpha$ is one-dimensional, and we pick an eigenbasis $\{g_\alpha\}_{\alpha \neq 0} \cup \{h_i\}_{i=1}^{\rank}$ for the action $\hh \acts \gg$.
  
  Let $x \in \gg$.  It is a standard exercise from linear algebra that $\hh x$ is the span of the eigenvectros $g_\alpha$, $\alpha \neq 0$, for which the coefficient of $x$ in the eigenbasis is non-zero.  In particular, if $x \in \gg\sminus \hh$, then $[\hh ,x]$ includes some $g_\alpha$.  By switching the roles of positive and negative roots if necessary, we can assure that $\alpha$ is positive; thus $[\hh, x] \supseteq \gg_\alpha$ for some positive $\alpha$.
  
  One can check directly that if $\alpha,\beta,\alpha+\beta$ are all nonzero roots, then $[\gg_\alpha,\gg_\beta] = \gg_{\alpha+\beta}$.  In particular, for any positive root $\alpha$, $\theta - \alpha$ is a positive root, and so $[\gg,\gg_\alpha] \supseteq [\gg_{\theta - \alpha},\gg_\alpha] = \gg_\theta$.  In particular, $g_\theta \in [\gg,x]$.
  
  But $[\gg_{\theta - \alpha},\gg_\theta] = \gg_{\alpha}$, and so $[\gg,g_\theta]$ generates all $g_\alpha$ for $\alpha$ a positive root.  We saw already (\cref{sl2lemmaslsp}) that $[\gg_\alpha,\gg_{-\alpha}] = \hh_{\alpha}$ is non-zero, and that $[\gg_{\pm\alpha},\hh_\alpha] = \gg_{\pm \alpha}$.  Thus $[\gg,\gg_\alpha] \supseteq \gg_{-\alpha}$, and in particular $g_\theta$ generates every $g_\alpha$ for $\alpha \neq 0$, and every $h_\alpha$.  Then $g_\theta$ generates all of $\gg$.
  
  Thus $x$ generates all of $\gg$ for any $x\in \gg \sminus \hh$.  If $x\in \hh$, then $\alpha(x) \neq 0$ for some $\alpha$, and then $[\gg_\alpha,x] = \gg_\alpha$, and we repeat the proof with some nonzero element of $\gg_\alpha$.  Hence $\gg$ is simple.
\end{proof}

When $\gg = \sl(n)$, let $\epsilon_i$ refer to the matrix $e_{ii}$, and when $\gg = \sp(n)$, let $\epsilon_i$ refer to the matrix $\left[ \begin{array}{c:c} e_{ii} & 0 \\ \hdashline 0 & -e_{n+1-i,n+1-i} \end{array} \right]$.  We construct a linear isomorphism $\hh^* \isom \hh$ by assigning an element $\alpha^\vee_i$ of $\hh$ to each simple root $\alpha_i$ as follows: to $\alpha_i = z_i - z_{i+1}$ for $1\leq i \leq n-1$ we assign $\alpha^\vee_i = \epsilon_i - \epsilon_{i+1}$, and to $\alpha_n = 2z_n$ a root of $\sp(n)$ we assign $\alpha^\vee_n = \epsilon_n$.  In particular, $\alpha_i(h_i) = 2$ for each simple root.  We define the \emindexing{Cartan matrix} $\aaa$ by $\aaa_{ij} \defined \alpha_i(h_j)$.
\begin{eg}
  For $\sl(n)$, we have the following $(n-1)\times (n-1)$ matrix:
  \eqn{ \aaa = \left[ \begin{array}{ccccc}
  2 & -1 & 0 & \dots & 0 \\
  -1 & 2 & -1 & & \vdots \\
  0 & -1 & \ddots & \ddots& 0 \\
  \vdots & & \ddots& 2 & -1 \\
  0 & \dots  & 0 & -1 & 2
\end{array} \right] }
  For $\sp(n)$, we have the following $n\times n$ matrix:
  \eqn{ \aaa = \left[ \begin{array}{ccccc:c}
  2 & -1 & 0 & \dots & 0 & 0 \\
  -1 & 2 & -1 & & \vdots & \vdots \\
  0 & -1 & \ddots & \ddots& 0 & \vdots \\
  \vdots & & \ddots& 2 & -1 & 0 \\
  0 & \dots  & 0 & -1 & 2 & -1 \\ \hdashline
  0 & \dots & \dots & 0 & -2 & 2
\end{array} \right]}
\end{eg}

To each of the above matrices we associate a \emindexing{Dynkin diagram}.  This is a graph with a node for each simple root, and edges assigned by: $i$ and $j$ are not connected if $\aaa_{ij} = 0$; they are singly connected if $\aaa_{ij}$ is a block $\left[ \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array} \right]$; we put a double arrow from $j$ to $i$ when the $(i,j)$-block is $\left[ \begin{array}{cc} 2 & -1 \\ -2 & 2 \end{array} \right]$.  So for $\sl(n)$ we get the graph \tikz[baseline=-1ex] \draw (0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$};, and for $\sp(n)$ we get \begin{tikzpicture}[baseline=-1ex] \draw (0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) ; \draw[doubly]  (4,0) node {$\bullet$} -- (3,0) node {$\bullet$};\end{tikzpicture}.
 
\begin{lemmadef}
The identification $\hh^* \isom \hh$ lets us construct reflections of $\hh^*$ by $s_i: \alpha \mapsto \alpha - \<\alpha,\alpha^\vee_i\>\alpha$, where $\<,\>$ is the pairing $\hh^* \otimes \hh \to \CC$ that we had earlier written as $\<\alpha,\beta\>  = \alpha(\beta)$.  These reflections generated the \emindexing{Weyl group} $W$.

  For each of $\sl(n)$ and $\sp(n)$, let $R\subseteq \hh^*$ be the set of roots and $W$ the Weyl group.  Then $W\acts R \sminus \{0\}$.  In particular, for $\sl(n)$, we have $W = S_n$ the symmetric group on $n$ letters, where the reflection $(i,i+1)$ acts as $s_i$; $W\acts R\sminus \{0\}$ is transitive.  For $\sp(n)$, we have $W = S_n \ltimes (\ZZ/2)^n$, the \emindexing{hyperoctahedral group}, generated by the reflections $s_i = (i,i+1) \in S_n$ and $s_n$ the sign change, and the action $W \acts R\sminus \{0\}$ has two orbits.
\end{lemmadef}

We will spend the rest of this chapter showing that the picture of $\sl(n)$ and $\sp(n)$ in this section is typical of simple Lie algebras over $\CC$.

\section{Representation Theory of {$\mathfrak {\lowercase{ s l}}(2)$}} \label{SL2RepThy}

\index{sl2@$\sl(2)$}\cite[Lectures 28 and 29]{notes}

Our hero for this section is the Lie algebra $\sl(2,\CC) \defined \< e,h,f : [e,f]=h, \,[h,e]=2e, \,[h,f]=-2f\> = \{ x\in {\rm M}(2,\CC) \st \tr x = 0\}$.
\begin{eg}
  As a subalgebra of ${\rm M}(2,\CC)$, $\sl(2)$ has a tautological representation on $\CC^2$, given by $E \mapsto \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$, $F \mapsto \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$, and $H \mapsto \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$.  Let $v_0$ and $v_1$ be the basis vectors of $\CC^2$.  Then the representation $\sl(2) \acts \CC^2$ has the following picture:
  \eqn{\begin{tikzpicture}[baseline]
    \path (0,.5) node[label=left:$v_0$] (v0) {$\bullet$} (0,-.5) node[label=left:$v_1$] (v1) {$\bullet$};
    \draw[->] (v0) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle H$} (v0);
    \draw[->] (v1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle H$} (v1);
    \draw[->] (v0) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle F$} (v1);
    \draw[->] (v1) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle E$} (v0);
  \end{tikzpicture}}
  This is the infinitesimal verion of the action $\SL(2) \acts \CC^2$ given by
  \begin{align}
  (\exp(-te)) \left[ \begin{array}{c} x \\ y \end{array} \right] & = \left[ \begin{array}{cc} 1 & -t \\ 0 & 1 \end{array} \right]\left[ \begin{array}{c} x \\ y \end{array} \right] \nonumber \\
   & = \left[ \begin{array}{c} x - ty \\ y \end{array} \right] \\
   \dzero \exp(-te) & = \left\{ \left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{c} -y \\ 0 \end{array} \right] \right\}\\
   \dzero \exp(-tf) & = \left\{\left[ \begin{array}{c} x \\ y \end{array} \right] \mapsto \left[ \begin{array}{c} 0 \\ -x \end{array} \right] \right\}
\end{align}
\end{eg}
\begin{eg} \label{VnDefn}
  Since $\SL(2)\acts \CC^2$, it acts also on the space of functions on $\CC^2$; by the previous calculations, we see that the action is:
  \eqn{ e = -y\partial_x,\;\;\; f = -x\partial_y,\;\;\; h = -x\partial_x + y\partial_y }
  These operations are homogenous --- they preserve the total degree of any polynomial --- and so the symmetric tensor product $\Ss^n(\CC^2) = \{\textrm{homogeneous polynomials of degree $n$ in $x$ and $y$}\}$ is a submodule of $\SL(2) \acts \{\textrm{functions}\}$.  Let $v_i \defined {n\choose i}x^iy^{n-i}$ be a basis vector in $\Ss^n(\CC x \oplus \CC y)$.  Then the action $\SL(2) \acts \Ss^2(\CC^2)$ has the following picture:
  \eqn[sl2vn]{ \begin{tikzpicture}[baseline=(v2.base)]
    \path (0,0) node[label=left:{$y^n = v_0$}] (v0) {$\bullet$} 
    ++(0,-1) node[label=left:{$nxy^{n-1} = v_1$}] (v1) {$\bullet$}
    ++(0,-1) node[label=left:{${n\choose 2}x^2y^{n-2} = v_2$}] (v2) {$\bullet$}
    ++(0,-1) node (dots) {$\vdots$}
    ++(0,-1) node[label=left:{${n\choose n-1} x^{n-1} y = v_{n-1}$}] (vn1) {$\bullet$}
    ++(0,-1) node[label=left:{$ x^n = v_n$}] (vn) {$\bullet$}
    ;
    \draw[->] (v0) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = n$} (v0);
    \draw[->] (v1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = n-2$} (v1);
    \draw[->] (v2) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = n-4$} (v2);
    \draw[->] (dots) .. controls +(330:1) and +(30:1) ..  (dots);
    \draw[->] (vn1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = 2-n$} (vn1);
    \draw[->] (vn) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = -n$} (vn);
    \draw[->] (v0) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 1$} (v1);
    \draw[->] (v1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 2$} (v2);
    \draw[->] (v2) .. controls +(300:.5) and +(60:.5) .. (dots);
    \draw[->] (dots) .. controls +(300:.5) and +(60:.5) ..  (vn1);
    \draw[->] (vn1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = n$} (vn);
    \draw[->] (v1) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle n = e$} (v0);
    \draw[->] (v2) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle n-1 =  e$} (v1);
    \draw[->] (dots) .. controls +(120:.5) and +(240:.5) ..  (v2);
    \draw[->] (vn1) .. controls +(120:.5) and +(240:.5) ..  (dots);
    \draw[->] (vn) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle 1 = e$} (vn1);
  \end{tikzpicture}}
  Let us call this module $V_n$.  Then $V_n$ is irreducible, because applying $e$ enough times to any non-zero element results in a multiple of $v_0$, and $v_0$ generated the module.
\end{eg}
\begin{prop}
  Let $V$ be any $(n+1)$-dimensional irreducible module over $\sl(2)$.  Then $V \cong V_n$.
\end{prop}
\begin{proof}
  Suppose that $v\in V$ is an eigenvector of $h$, so that $hv = \lambda v$.  Then $hev = [h,e]v + ehv = 2ev + \lambda ev$, so $ev$ is an $h$-eigenvector with eigenvalue $\lambda + 2$.  Similarly, $fv$ is an $h$-eigenvector with eigenvalue $\lambda - 2$.  So the space spanned by $h$-eigenvectors of $V$ is a submodule of $V$; by the irreducibility of $V$, and using the fact that $h$ has at least one eigenvector, this submodule must be the whole of $V$, and so $h$ acts diagonally.
  
  By finite-dimensionality, there is an eigenvector $v_0$ of $h$ with the highest eigenvalue, and so $ev_0 = 0$.  By \cref{PBWthm}, $\{f^ke^lh^m\}$ spans $\Uu\sl(2)$, and so $\{v_i \defined f^iv_0 / i!\}$ is a basis of $V$ (by irreducibility, $V$ is generated by $v_0$).  In particular, $v_n = f^nv_0/n!$, the $(n+1)$st member of the basis, has $fv_n = 0$, since $V$ is $(n+1)$-dimensional.
  
  We compute the action of $e$ by induction, using the fact that $hv_k = (\lambda_0 - 2k)v_k$:
  \begin{align}
    ev_0 & = 0 \\
    ev_1 & = efv_0 = [e,f]v_0 + fev_0 = hv_0 = \lambda_0v_0 \\
    ev_2 & = efv_1/2 = [e,f]v_1/2 + fev_1/2 = hv_1/2 + f\lambda_0v_0/2 \nonumber \\  & = (\lambda_0 - 2)v_1/2 + \lambda_0v_1/2 = (\lambda_0 - 1)v_1 \\ \dots & \nonumber \\
    ev_k & = efv_{k-1}/k = hv_{k-1}/k + fev_{k-1}/k = (\lambda_0 - 2k + 2) v_{k-1}/k + (\lambda_0 - k + 2)fv_{k-2}/k \nonumber \\
    & =  \bigl((\lambda_0 - 2k + 2)/k + (k-1)(\lambda_0 - k + 2)/k\bigr)v_{k-1}   = (\lambda_0 - k+1) v_{k-1}
  \end{align}
   But $fv_n = 0$, and so:
   \begin{align}
    0 & = efv_n = [e,f]v_n + fev_n = hv_n + (\lambda_0 - n + 1)fv_{n-1} \nonumber \\
     & = (\lambda_0 - 2n)v_n + (\lambda_0 - n + 1)nv_n = \bigl( (n+1)\lambda_0 - (n+1)n \bigr)v_n
   \end{align}
   Thus $\lambda_0 = n$ and $V$ is isomorphic to $V_n$ defined in \cref{sl2vn}.
\end{proof}

\section{Cartan subalgebras}

\subsection{Definition and Existence}

\cite[Lectures 29 and 30]{notes}

\begin{lemma}
  Let $\hh$ be a nilpotent Lie algebra over a field $\KK$, and $\hh \acts V$ a finite-dimensional representation.  For $h\in \hh$ and $\lambda \in \CC$, define $V_{\lambda,h} = \{v\in V \st \exists n \st (h-\lambda)^nv = 0\}$.  Then $V_{\lambda,h}$ is an $\hh$-submodule of $V$.
\end{lemma}
\begin{proof}
  Let $\ad : \hh \acts \hh$ be the adjoint action; since $\hh$ is nilpotent, $\ad h \in \End(\hh)$ is a nilpotent endomorphism.  Define $\hh_{(m)} \defined \ker\bigl((\ad h)^m\bigr)$; then $\hh_{(m)} = \hh$ for $m$ large enough.  We will show that $\hh_{(m)} V_{\lambda, h} \subseteq V_{\lambda,h}$ by induction on $m$; when $m=0$, $\hh_{(0)} = 0$ and the statement is trivial.
  
  Let $y\in \hh_{(m)}$, whence $[h,y] \in \hh_{(m-1)}$, and let $v\in V_{\lambda,h}$.  Then  $(h-\lambda)^n v = 0$ for $n$ large enough, and so
  \begin{align}
     (h-\lambda)^n y v & = y(h-\lambda)^nv + [(h-\lambda)^n,y]v \\
      & = 0 + [(h-\lambda)^n,y]v \\
      & = \sum_{k+1 = n-1} (h-\lambda)^k[h,y](h-\lambda)^lv
  \end{align}
  since $[\lambda,y] = 0$.  By increasing $n$, we can assure that for each term in the sum at least one of the following happens: $l$ is large enough that $(h-\lambda)^lv = 0$, or $k$ is large enough that $(h-\lambda)^k V_{\lambda,h} = 0$.  The large-$l$ terms vanish immediately; the large-$k$ terms vanish upon realizing that $(h-\lambda)V_{h,\lambda} \subseteq V_{h,\lambda}$ by definition and $[h,y]V_{h,\lambda} \subseteq \hh_{(m-1)}V_{h,\lambda} \subseteq V_{h,\lambda}$ by induction on $m$.
\end{proof}
\begin{cor}\label{vlambdadefcor}
  Let $\hh$ be a nilpotent Lie algebra over $\KK$, $\hh\acts V$ a  finite-dimensionalrepresentation, and $\lambda: \hh \to \KK$ a linear map.  Then $V_\lambda \defined \bigcap_{h\in \hh} V_{\lambda(h),h}$ is an $\hh$-submodule of $V$.
\end{cor}

\begin{prop}
  Let $\hh$ be a finite-dimensional nilpotent Lie algebra over an algebraically closed field $\KK$ of characteristic $0$, and $V$ a finite-dimensional $\hh$-module.  For each $\lambda \in \hh^*$, define $V_\lambda$ as in \cref{vlambdadefcor}.  Then $V = \bigoplus_{\lambda \in \hh^*} V_\lambda$.
\end{prop}
\begin{proof}
  Let $h_1,\dots,h_k \in \hh$, and let $H_k \subseteq \hh$ be the linear span of the $h_i$.  Let $W \defined \bigcap_{i=1}^k V_{\lambda(h_i),h_i}$.  It follows from \cref{LieThm} that $W = \bigcap_{h \in H} V_{\lambda(h),h}$, since we can choose a basis of $V$ in which $\hh \acts V$ by upper-triangular matrices.
  
  We have seen already that $W$ is a submodule of $V$.  Let $h_{k+1} \not\in H_k$; then we can decompose $W$ into generalized eigenspaces of $h_{k+1}$.  We proceed by induction on $k$ until we have a basis of $\hh$.
\end{proof}
\begin{defn}
  For $\lambda \in \hh^*$, the space $V_\lambda$ in \cref{vlambdadefcor} is a \emindexing{weight space} of $V$, and $V = \bigoplus_{\lambda \in \hh^*} V_\lambda$ the {\em weight space decomposition}.
\end{defn}
\begin{lemma} \label{TensorWeightLemma}
  Let $\hh$ be a finite-dimensional nilpotent Lie algebra over an algebraically closed field of characteristic $0$, and let $V$ and $W$ be two finite-dimensional $\hh$ modules.  Then the weight spaces of $V\otimes W$ are given by $(V\otimes W)_{\lambda} = \bigoplus_{\alpha + \beta = \lambda}V_\alpha \otimes W_\beta$.
\end{lemma}
\begin{proof}  $h(v\otimes w) = hv\otimes w + v\otimes hw$ \end{proof}
\begin{cor} \label{CartanExistsProof1}
  Let $\gg$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$, and $\hh \subseteq \gg$ a nilpotent subalgebra.  Then the weight spaces of $\ad: \hh \acts \gg$ satisfy $[\gg_\alpha,\gg_\beta] \subseteq \gg_{\alpha + \beta}$.
\end{cor}
\begin{prop} \label{Cartanprop}
  Let $\gg$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$, and let $\hh \subseteq \gg$ be a nilpotent subalgebra.  The following are equivalent:
  \begin{enumerate}
    \item $\hh = N(\hh) \defined \{x\in \gg \st [x,\hh]\subseteq \hh\}$, the \emindexing{normalizer} of $\hh$ in $\gg$.
    \item $\hh = \gg_0$ is the $0$-weight space of $\ad: \hh \acts \gg$.
  \end{enumerate}
\end{prop}
\begin{proof}
  Define $N^{(i)} \defined \{x\in \gg \st (\ad \hh)^ix \subseteq \hh\}$.  Then $N^{(0)} = \hh$ and $N^{(1)} = N(\hh)$, and $N^{(i)} \subseteq N^{(i+1)}$.  By finite-dimensionality, the sequence $N^{(0)} \subseteq N^{(1)} \subseteq \dots$ must eventually stabilize.  By definition $\bigcup N^{(i)} = \gg_0$, so 2.\ implies 1.  But $N^{(i+1)} = N(N^{(i)})$, and so 1.\ implies 2.
\end{proof}

\begin{defn}
  A subalgebra $\hh$ of $\gg$ satisfying the equivalent conditions of \cref{Cartanprop} is a \emindexing{Cartan subalgebra}\index{subalgebra!Cartan} of $\gg$.
\end{defn}

\begin{thm}[Existence of a Cartan Subalgebra]  \label{CartanExistsThm}
  Every finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$ has a Cartan subalgebra.
\end{thm}

Before we prove this theorem, we will need some definitions and lemmas.

\begin{defn}  \label{ZariskiDefn}
  Let $\KK$ be a field; we say that $X\subseteq \KK^n$ is \emindexing{Zariski closed} if $X = \{x\in \KK^n \st p_i(x) = 0\, \forall i\}$ for some possibly infinite set $\{p_i\}$ of polynomials in $\KK[x_1,\dots,x_n]$.  A subset $X \subseteq \KK^n$ is \emindexing{Zariski open} if $\KK^n \sminus X$ is Zariski closed.
\end{defn}
\begin{lemma} \label{CartanExistsProof2}
  If $\KK$ is infinite and $U,V \subseteq \KK^n$ are two non-empty Zariski open subsets, then $U\cap V$ is non-empty.
\end{lemma}
\begin{proof}
  Let $\bar U \defined \KK^n\sminus U$ and similarly for $\bar V$.  Let $u\in U$ and $v\in V$.  If $u = v$ we're done, and otherwise consider the line $L \subseteq \KK^n$ passing through $u$ and $v$, parameterized $\KK \isom \LL$ by $t \mapsto tu+(1-t)v$.  Then $L\cap \bar U$ and $L\cap \bar V$ are finite, as their preimages under $\KK \to \LL$ are loci of polynomials.  Since $\KK$ is infinite, $L$ contains infinitely many points in $U\cap V$.
\end{proof}

\begin{lemmadef} \label{CartanExistsProof3}
  Let $\gg$ be a  finite-dimensional Lie algebra over an algebraically closed field of characteristic $0$.  An element $x\in \gg$ is \emindexing{regular} if $\gg_{0,x}$ has minimal dimension.  If $x$ is regular, then $\gg_{0,x}$ is a nilpotent subalgebra of $\gg$.
\end{lemmadef}
\begin{proof}
  We will write $\hh$ for $\gg_{0,x}$.  That $\hh$ is a subalgebra follows from \cref{CartanExistsProof1}.  Suppose that $\hh$ is not nilpotent, and let $U \defined \{h \in \hh \st \ad h|_\hh \textrm{ if not nilpotent}\} \neq 0$.  Then $U = \{h\in \hh \st (\ad h|_\hh)^d \neq 0\}$ is a Zariski-open subset of $\hh$.  Moreover, $V \defined \{ h \in \hh \st h \textrm{ acts invertibly on }\gg/\hh\}$ is also a non-empty Zariski-open subset of $\hh$, where $V$ is the quotient of $\hh$-modules; it is non-empty because $x\in V$.  By \cref{CartanExistsProof2} (recall that any algebraically closed field is infinite), there exists $y \in U\cap V$.  Then $\ad y$ preserves $\gg_{\alpha,x}$ for every $\alpha$, as $y\in \hh = \gg_{0,x}$, and $y$ acts invertibly on every $\gg_{\alpha,x}$ for $\alpha \neq 0$.  Then$\gg_{0,y} \subseteq \gg_{0,x} = \hh$, but $y\in U$ and so $\gg_{0,y} \neq \hh$.  This contradicts the minimality of $\hh$.
\end{proof}

\begin{proof}[of \cref{CartanExistsThm}]
  We let $\gg$, $x\in \gg$, and $\hh = \gg_{0,x}$ be as in \cref{CartanExistsProof3}.  Then $\hh \subseteq \gg_{0,\hh}$ because $\hh$ is nilpotent, and $\gg_{0,\hh} \subseteq \gg_{0,x} = \hh$ because $x\in \hh$.  Thus $\hh$ is a Cartan subalgebra of $\gg$.
\end{proof}


We mention one more fact about the Zariski topology:
\begin{lemma}
  Let $U$ by a Zariski open set over $\CC$.  Then $U$ is path connected.
\end{lemma}
\begin{proof}
  Let $u,v\in U$ and construct the line $L$ as in the proof of \cref{CartanExistsProof2}.  Then $L \cap U$ is isomorphic to $\CC \sminus\{\textrm{finite}\}$, and therefore is path connected.
\end{proof}

\begin{prop} \label{CartanSubalgebraOverC}
  Let $\gg$ be a finite-dimensional Lie algebra over $\CC$.  Then all Cartan subalgebras of $\gg$ are conjugate by automophisms of $\gg$.
\end{prop}
\begin{proof}
  Consider $\ad: \gg \to \gl(\gg)$.  Then $\ad \gg \subseteq \gl(\gg)$ is a Lie subalgebra, and so corresponds to a connected Lie subgroup $\Int\gg \subseteq \GL(\gg)$ generated by $\exp(\ad\gg)$.  Since $\gg \acts \gg$ be derivations, $\exp(\ad\gg) \acts \gg$ by automorphisms, and so $\Int\gg \subseteq \Aut\gg$.
  
  Let $\hh \subseteq \gg$ be a Cartan subalgebra, and $\gg = \bigoplus \gg_{\alpha,\hh}$ the corresponding weight-space decomposition.  Since $\gg$ is finite-dimensional, the set
  \eqn{ R_\hh \defined \{h\in \hh \st \alpha(h) \neq 0 \textrm{ if }\alpha \neq 0 \textrm{ and }\gg_{\alpha,\hh} \neq 0\} = \{ h\in \hh \st \gg_{0,h} = \hh\} }
  is non-empty and open, since we can take $\alpha$ to range over a finite set (by finite-dimensionality).
  
  Let $\sigma: \Int\gg \times \gg \to \gg$ be the canonical action, and consider the restriction to $\sigma : \Int\gg \times R_\hh \to \gg$.  Pick $y\in R_\hh$ and let $e\in \Int\gg$ be the identity element.  We compute the image of the infinitesimal action $d\sigma\left( {\rm T}_{(e,y)}(\Int\gg \times R_\hh)\right) \subseteq {\rm T}_y \gg \cong \gg$.  By construction, varying the first component yields an action by conjugation: $x\mapsto [x,y]$.  Thus the image of ${\rm T}_e\Int\gg \times \{0\in {\rm T}_yR_\hh\}$ is $(\ad y)(\gg)$.  Since $y$ acts invertibly, $(\ad y)(\gg) \supseteq \bigoplus_{\alpha \neq 0}\gg_{\alpha,\hh}$.  By varying the second coordinate (recall that $R_\hh$ is open), we see that $d\sigma\left( {\rm T}_{(e,y)}(\Int\gg \times R_\hh)\right) \supseteq \hh = \gg_{0,\hh}$ also.  Thus $d\sigma\left( {\rm T}_{(e,y)}(\Int\gg \times R_\hh)\right) = \gg = {\rm T}_y\gg$, and so the image $(\Int \gg)(R_\hh)$ contains a neighborhood of $y$ and therefore is open.
  
  For each $y\in \gg$, consider the generalized nullspace $\gg_{0,y}$; the dimension of $\gg_{0,y}$ depends on the characteristic polynomial of $y$, and the coefficients of the characteristic polynomial depend polynomially on the matrix entries of $\ad y$.  In particular, $\dim \gg_{0,y} \geq r$ if and only if the last $r$ coefficients of the characteristic polynomial of $\ad y$ are $0$, and so $\{y \in \gg \st \dim \gg_{0,y}\geq r\}$ is Zariski closed.  Therefore $y \mapsto \dim \gg_{0,y}$ is upper semi-continuous in the Zariski topology.  In particular, let $r$ be the minimum value of $\dim \gg_{0,y}$, which exists since $\dim \gg_{0,y}$ takes values in integers.  Then $\textrm{Reg} \defined \{ y\in \gg \st \dim \gg_{0,y} = r\}$, the set of regular elements, is Zariski open and therefore dense.  In particular, $\textrm{Reg}$ intersects $(\Int \gg)(R_\hh)$.
  
  But if $y\in (\Int \gg)(R_\hh)$ then $\dim \gg_{0,y} = \dim \hh$.  Therefore $\dim \hh$ is the minimal value of $\dim \gg_{0,y}$ and in particular $R_\hh \subseteq \textrm{Reg}$.  Conversely, $\textrm{Reg} = \bigcup_{\hh' \textrm{ a Cartan}} R_{\hh'} = \bigcup_{\hh' \textrm{ a Cartan}} (\Int \gg)R_{\hh'}$.
  
  However, $\Int\gg$ is a connected group, $R_\hh$ is connected being $\CC^n$ minus some hyperplanes, and $\textrm{Reg}$ is connected on account of being Zariski open.  But the orbits of $(\Int\gg)R_\hh$ are disjoint, and their union is all of $\textrm{Reg}$, so $\textrm{Reg}$ must consist of a single orbit.
  
  To review: $\hh$ is Cartan and so contains regular elements of $\gg$, and any other regular element of $\gg$ is in the image under $\Int\gg$ of a regular element of $\hh$.  Thus every Cartan subalgebra is in $(\Int \gg)\hh$.
\end{proof}

\subsection{More on the Jordan Decomposition and Schur's Lemma}

\cite[Lectures 30 and 31]{notes}

Recall \cref{JordanThm} that every $x\in \End(V)$, where $V$ is a finite-dimensional vector spce over an algebraically closed field, has a unique decomposition $x = x_s + x_n$ where $x_s$ is diagonalizable and $x_n$ is nilpotent.  We will strengthen this result in the case when $x \in \gg \to \End(V)$ and $\gg$ is semisimple.

\begin{lemma}
  Let $\gg$ be a finite-dimensional Lie algebra over an algebraically closed field, and $\Der \gg \subseteq \End\gg$ the algebra of derivations of $\gg$.  If $x\in \Der\gg$, then $x_s,x_n \in \Der\gg$.
\end{lemma}
\begin{proof}
  For $x\in \Der\gg$, construct the weight-space decomposition $\gg = \bigoplus_{\lambda} \gg_{\lambda,x}$ of generalized eigenspaces of $x$.  Since $x$ is a derivation, the weight spaces add: $[\gg_{\mu,x},\gg_{\nu,x}] \subseteq \gg_{\mu + \nu,x}$.  Let $y \in \End\gg$ act as $\lambda$ on $\gg_{\lambda,x}$; then $y$ is a derivation by the additive property.
  But $y$ is diagonalizable and commutes with $x$, and $x-y$ is nilpotent because all its eigenvalues are $0$ so $y = x_s$.
\end{proof}
We have an immediate corollary:
\begin{lemmadef} \label{JordanDecompDef}
  If $\gg$ is a semisimple finite-dimensional Lie algebra over an algebraically closed field $\KK$, then every $x\in \gg$ has a unique \emindexing{Jordan decomposition} $x = x_s + x_n$ such that $[x_s,x_n] = 0$, $\ad x_s$ is diagonalizable, and $\ad x_n$ is nilpotent.
\end{lemmadef}
\begin{proof}
  If $\gg$ is semisimple then $\ad: \gg \to \Der\gg$ is injective as $Z(\gg) = 0$ and surjective because $\Der\gg / \ad\gg = \Ext^1(\gg,\KK) = 0$.
\end{proof}

\begin{thm}[Schur's Lemma over an algebraically closed field] \label{SchurLemma2}
  Let $U$ be an algebra over $\KK$ and algebraically closed field, and let $V$ be an irreducible $U$-module.  Then $\End_U(V) = \KK$.
\end{thm}
\begin{proof}
  Let $\phi \in \End_U(V)$ and $\lambda\in \KK$ an eigenvalue of $\phi$.  Then $\phi - \lambda$ is singular and hence $0$ by \cref{SchurLemma}.
\end{proof}
\begin{prop}
  Let $\gg$ be a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $0$, and let $\sigma: \gg \acts V$ be a finite-dimensional $\gg$ module.  For $x\in \gg$, write $x_s$ and $x_n$ as in \cref{JordanDecompDef}, and write $\sigma(x)_s$ and $\sigma(x)_n$ for the diagonalizable and nilpotent parts of $\sigma(x) \in \gl(\gg)$ as given by \cref{JordanThm}.  Then $\sigma(x)_s = \sigma(x_s)$ and $\sigma(x)_n = \sigma(x_n)$.
\end{prop}
\begin{proof}
  We reduce to the case when $V$ is an irreducible $\gg$-module using \cref{WeylThm}, and we write $\gg = \prod \gg_i$ a product of simples using \cref{extproof2}.  Then $\gg_i \acts V$ as $0$ for every $i$ except one, for which the action $\gg_i \acts V$ is faithful.  We replace $\gg$ by that $\gg_i$, whence $\sigma: \gg \mono \gl(V)$ with $\gg$ simple.
  
  It suffices to show that $\sigma(x)_s\in \sigma(\gg)$, since then $\sigma(x_s) = \sigma(s)$ for some $s\in \gg$, $\sigma(x)_n = \sigma(x) - \sigma(s) = \sigma(x-s)$, and $s$ and $n = x-s$ commute, sum to $x$, and act diagonalizably and nilpotently since the adjoint action $\ad: \gg \acts \gg$ is a submodule of $\gg \acts \gl(V)$, so $s = x_s$ and $n = x_n$.
  
  By semisimplicity, $\gg = \gg' \subseteq \sl(V)$.  By \cref{SchurLemma2}, the centralizer of $\gg$ in $\gl(V)$ consists of scalars.  In characteristic $0$, the only scalar in $\sl(V$ is $0$, so the centralizer of $\gg$ in $\sl(V)$ is $0$.  Define the normalizer $N(\gg) = \{x\in \sl(V) \st [x,\gg] \subseteq \gg\}$; then $N(\gg)$ is a Lie subalgebra of $\sl(V)$ containing $\gg$, and $N(\gg)$ acts faithfully on $\gg$ since the centralizer of $\gg$ in $\sl(V)$ is $0$, and this action is by derivations.  But all derivations are inner, as in the proof of \cref{JordanDecompDef}, and so $N(\gg)\acts \gg$ factors through $\gg \acts \gg$, and hence $N(\gg) = \gg$.
  
  So it suffices to show that $\sigma(x)_s \in N(\gg)$ for $x\in \gg$.  Since $\sigma(x)_n$ is nilpotent, it's traceless, and hence in $\sl(V)$; then $\sigma(x)_s\in \sl(V)$ as well.  We construct a generalized eigenspace decomposition of $V$ with respect to $\sigma(x): V = \bigoplus V_{\lambda,x}$.  Then $\sigma(x)_s$ acts on $V_{\lambda,x}$ by the scalar $\lambda$.  We also construct a generalized eigenspace decomposition $\gg = \bigoplus \gg_{\alpha,x}$ with respect to the adjoint action $\ad: \gg \acts \gg$.  Since $\gg \subseteq \gl(V)$, we have $\gg_{\alpha,x} = \gg \cap \End_\KK(V)_\alpha = \bigoplus \Hom_\KK(V_{\lambda,x},V_{\lambda + \alpha,x})$, by tracking the eigenvalues of the right and left actions of $\gg$ on $V$.
  
  Moreover, $\ad(\sigma(x)_s) = \ad(\sigma(x_s))$ because both act by $\alpha$ on $\Hom_\KK(V_{\lambda,x},V_{\lambda + \alpha,x})$ and hence on $\gg_\alpha$.  Thus $\sigma(x)_s$ fixes $\gg$ since $\sigma(x_s)$ does.  Therefore $\sigma(x)_s \in N(\gg)$.
\end{proof}

\subsection{Precise description of Cartan subalgebras}

\cite[Lecture 31]{notes}

\begin{lemma} \label{KillingNondegCartan}
  Let $\gg$ be a semisimple Lie algebra over characteristic $0$, $\hh \subseteq \gg$ a nilpotent subalgebra, and $\gg = \bigoplus \gg_\alpha$ the root space decomposition with respect to $\hh$.  Then the Killing form $\beta$ pairs $\gg_\alpha$ with $\gg_{-\alpha}$ nondegenerately, and $\beta(\gg_\alpha,\gg_{\alpha'}) = 0$ if $\alpha + \alpha' \neq 0$.
\end{lemma}
\begin{proof}
  Let $x\in\gg_\alpha$ and $y\in \gg_{\alpha'}$.  For any $h\in\hh$, $(\ad h - \alpha(h))^nx = 0$ for some $n$.  So 
\eqn{ 0 = \beta\bigl( \left(\ad h - \alpha(h)\right)^nx,y\bigr) = \beta\bigl(x, \left(-\ad h -\alpha (h)\right)^ny\bigl) }
but $\left(-\ad h -\alpha (h)\right)^n$ is invertible on $\gg_{\alpha'}$ unless $\alpha' = -\alpha$.  Nondegeneracy follows from nondegeneracy of $\beta$ on all of $\gg$.
\end{proof}

\begin{cor} \label{CartanIsAbelianProof1}
  If $\gg$ is a finite-dimensional semisimple Lie algebra over characteristic $0$, and let $\hh \subseteq \gg$ a nilpotent subalgebra, then the largest nilpotency ideal in $\gg_0$ of the action $\ad: \gg_0 \acts \gg$ is the $0$ ideal.
\end{cor}
\begin{proof}
  The Killing form $\beta$ pairs $\gg_0$ with itself nondegenerately.  As $\beta$ is the trace form of $\ad: \gg_0 \acts \gg$, and $\ad(\gg)$-nilpotent ideal of $\gg_0$ must be in $\ker \beta = 0$.
\end{proof}

\begin{prop} \label{CartanIsAbelianProp}
  Let $\gg$ be a finite-dimensional semisimple Lie algebra over an algebraically closed field $\KK$ of characteristic $0$, and let $\hh\subseteq \gg$ be a Cartan subalgebra.  Then $\hh$ is abelian and $\ad: \hh \acts \gg$ is diagonalizable.
\end{prop}
\begin{proof}
  By definition, $\hh$ is nilpotent and hence solvable, and by \cref{LieThm} we can find a basis of $\gg$ in which $\hh \acts \gg$ by upper triangular matrices.  Thus $\hh' = [\hh,\hh]$ acts by strictly upper triangular matrices and hence nilpotently on $\gg$.  But $\hh = \gg_0$, and so $\hh' = 0$ by \cref{CartanIsAbelianProof1}.  This proves that $\hh$ is abelian.
  
Let $x\in \hh$.  Then $\ad x_s = (\ad x)_s$ acts as $\alpha(x)$ on $\gg_\alpha$, and in particular $x_s$ centralizes $\hh$.  So $x_s \in \gg_0 = \hh$ and so $x_n = x - x_s \in \hh$.  But if  $n\in \hh$ acts nilpotently on $\gg$, then $\KK n$ is an ideal of $\hh$, since $\hh$ is abelian, and acts nilpotently on $\gg$, so $\KK n = 0$ by \cref{CartanIsAbelianProof1}.  Thus $x_n = 0$ and $x = x_s$.  In particular, $x$ acts diagonalizably on $\gg$.  To show that $\hh$ acts diagonalizably, we use finite-dimensionality and the classical fact that if $n$ diagonalizable matrices commute, then they can by simultaneously diagonalized.
\end{proof}
\begin{cor}
  Let $\gg$ be a finite-dimensional semisimple Lie algebra over an algebraically closed field  of characteristic $0$.  Then a subalgebra $\hh \subseteq \gg$ is Cartan\index{Cartan subalgebra} if and only if $\hh$ is a maximal diagonalizable abelian subalgebra.
\end{cor}
\begin{proof}
  We first show the maximality of a Cartan subalgebra.  Let $\hh$ be a Cartan subalgebra and $\hh_1 \supseteq \hh$ abelian.  Then $\hh_1 \subseteq \gg_0 = \hh$ because it normalizes $\hh$.
  
  Conversely, let $\hh$ be a maximal diagonalizable abelian subalgebra of $\gg$, and write $\gg = \bigoplus \gg_\alpha$ the weight space decomposition of $\hh \acts\gg$.  We want to show that $\hh = \gg_0$, the centralizer of $\hh$.  Pick $x \in \gg_0$;  then $x_s,x_n \in \gg_0$, and so $x_s \in \hh$ by maximality.  In particular, $\gg_0$  is spanned by $\hh$ and $\ad$-nilpotent elements.  Thus $\gg_0$ is nilpotent by \cref{EngelThm} and therefore solvable, so $\gg_0'$ acts nilpotently on $\gg$.  But $\gg_0'$ is an ideal of $\gg_0$ that acts nilpotently, so $\gg_0'=0$, so $\gg_0$ is abelian.  Then any one-dimensional subspace of $\gg_0$ is an ideal of $\gg_0$, and a subspace spanned by a nilpotent acts nilpotently, so $\gg_0$ doesn't have any nilpotents.  Therefore $\gg_0 = \hh$.
\end{proof}

\section{Root systems}

\subsection{Motivation and a Quick Computation}

\cite[Lectures 31 and 32]{notes}

In any semisimple Lie algebra over $\CC$ we can choose a Cartan subalgebra, to which we assign combinatorial data.  Since all Cartan subalgebras are conjugate, this data, called a \emindexing{root system}, will not depend on our choice.  Conversely, this data will uniquely describe the Lie algebra, based on the representation theory of $\sl(2)$.

\begin{defn}\label{RootsDefn}
Let $\gg$ be a semisimple Lie algebra over $\CC$, and $\hh$ a Cartan subalgebra.  The \emindexing{root space decomposition} of $\gg$ is the weight decomposition $\gg = \bigoplus_\alpha \gg_\alpha$ of $\ad:\hh \acts \gg$; each $\gg_\alpha$ is a \emindexing{root space}, and the set of weights $\alpha \in \hh^*$ that appear in the root space decomposition comprise the {\em roots}\index{root} of $\gg$.  By \cref{CartanSubalgebraOverC} the structure of the set of roots depends up to isomorphism only on $\gg$.
\end{defn}


\begin{lemmadef}
  Let $\gg$ be a semisimple Lie algebra over $\CC$ with Killing form $\beta$, $\hh$ a Cartan subalgebra, and $x_\alpha \in \gg_\alpha$ for $\alpha \neq 0$.  To $x_\alpha$ we can associate $y_\alpha \in \gg_{-\alpha}$ with $\beta(x_\alpha,y_\alpha) = -1$ and to the root $\alpha$ we associate a \emindexing{coroot} $h_\alpha$ with $\beta(h_\alpha,-) = \alpha$.  Then $\{x_\alpha,y_\alpha,h_\alpha\}$ span a  subalgebra $\sl(2)_\alpha$\index{sl2alpha@$\sl(2)_\alpha$} of $\gg$ isomorphic to $\sl(2)$.
\end{lemmadef}
\begin{proof}
  That $h_\alpha$ and $y_\alpha$ are well-defined follows from the nondegeneracy of $\beta$.  For any $h\in \hh$, $x\in \gg_\alpha$, and $y\in \gg_{-\alpha}$, we have
  \begin{align}
  \beta(h,[x,y]) & = \beta([x,h],y) \\
   & = -\alpha(h)\,\beta(x,y)
\end{align}
Thus $[x,y] = -\beta(x,y) h_\alpha$.  Moreover, since $x_\alpha \in \gg_\alpha$, $[h_\alpha,x_\alpha]   = \alpha(h_\alpha)x_\alpha$, and since $y_\alpha \in \gg_{-\alpha}$, $[h_\alpha,y_\alpha]   = -\alpha(h_\alpha)y_\alpha$.

Thus $x_\alpha,y_\alpha,h_\alpha$ span a three-dimensional Lie subalgebra of $\gg$, which is isomorphic to either $\sl(2)$ or or the Heisenberg algebra.  But in every finite-dimensional representation the Heisenberg algebra acts nilpotently, whereas $\ad(h_\alpha) \in \End(\gg)$ is diagonalizable.  Therefore this subalgebra is isomorphic to $\sl(2)$, and $\alpha(h_\alpha) \neq 0$.
\end{proof}

\begin{cor}
  Let $\alpha$ be a root of $\gg$.  Then $\pm \alpha$ are the only non-zero roots of $\gg$ in $\CC\alpha$, and $\dim \gg_{\alpha} = 1$.  In particular, $\sl(2)_\alpha = \gg_\alpha \oplus \gg_{-\alpha} \oplus \CC h_\alpha$.
\end{cor}
\begin{proof}
  We consider $\jj_\alpha \defined \bigoplus_{\alpha' \in \CC\alpha \sminus \{0\}} \gg_{\alpha'} \oplus \CC h_\alpha$; it is a subalgebra of $\gg$ and an $\sl(2)_\alpha$-submodule, since $[\gg_\alpha,\gg_{\alpha'}] \subseteq \gg_{\alpha + \alpha'}$, and $h_{c\alpha} = ch_\alpha$ for $c\in \CC$.  Let $\alpha' \in \CC \alpha\sminus \{0\}$ be a root; as a weight of the $\sl(2)_\alpha$ representation, we see that $\alpha' \in \ZZ\alpha/2$.  If any half-integer multiple of $\alpha$ actually appears, then $\alpha/2$ appears, and by switching $\alpha$ to $\alpha/2$ if necessary we can assure that $\jj_\alpha$ contains only representations $V_{2m}$.  But each $V_{2m}$ has contributes a basis vector in weight $0$, and the only part of $\jj_\alpha$ in weight $0$ is $\CC h_\alpha$.  Therefore $\jj_\alpha$ is irreducible as an $\sl(2)_\alpha$ module, contains $\sl(2)_\alpha$, and so equals $\sl(2)_\alpha$.
\end{proof}

\begin{cor}
  The roots $\alpha$ span $\hh^*$, and the coroots $h_\alpha$ span $\hh$.
\end{cor}
\begin{proof}
  We let $\alpha$ range over the non-zero roots.  Then
  \begin{align}
    \bigcap_{\alpha \neq 0} \ker \alpha & = Z(\gg) = 0 \\
    \sum_{\alpha \neq 0} \CC h_\alpha & = \gg' \cap \hh = \hh  
  \end{align}
  That $\beta^{-1}: \alpha \mapsto h_\alpha$ is a linear isomorphism $\hh^* \to \hh$ completes the proof.
\end{proof}

\begin{prop}
  Let $\gg$ be a finite-dimensional semisimple Lie algebra over $\CC$, and $\hh \subseteq \gg$ a Cartan subalgebra.  Let $R\subseteq \hh^*$ be the set of nonzero roots and $R^\vee\subseteq \hh$ the set of nonzero coroots.  Then $\alpha \mapsto \alpha^\vee \defined \frac{ 2h_\alpha}{\alpha(h_\alpha)}$ defines a bijection $\vee: R \to R^\vee$, and the triple $(R,R^\vee,\vee)$ comprise a \emindexing{root system} in $\hh$.
\end{prop}
We will define the words ``root system'' in the next section to generalize the data already computed.

\subsection{The Definition}

\cite[Lecture 33]{notes}

\begin{defn} \label{RootSystemDefn}
A \emindexing{root system} is a complex vector space $\hh$, a finite subset $R \subseteq \hh^*$, a subset $R^\vee \subseteq \hh$, a bijection $\vee: R \to R^\vee$, subject to
\begin{description}
  \item[RS1] $\<\alpha,\beta^\vee\> \in \ZZ$
  \item[RS2] $R = -R$ and $R^\vee = -R^\vee$, with $(-\alpha)^\vee = -(\alpha^\vee)$
  \item[RS3] $\<\alpha,\alpha^\vee\> = 2$
  \item[RS4] If $\alpha,\beta\in R$ are not proportional, then $(\beta + \CC\alpha)\cap R$ consists of a ``\indexing{string}'':
  \eqn{ \bigl< (\beta + \CC\alpha)\cap R, \alpha^\vee \bigr> = \{m,m-2,\dots,-m+2,-m\} }
   \item[Nondeg] $R$ spans $\hh^*$ and $R^\vee$ spans $\hh$
  \item[Reduced] $\CC\alpha \cap R = \{\pm \alpha\}$ for $\alpha\in R$.
\end{description}
Two root systems are {\em isomorphic}\index{root system!isomorphism} if there is a linear isomorphism of the underlying vector spaces, inducing an isomorphism on dual spaces, that carries each root system to the other.  The \emindexing{rank} of a root system is the dimension of $\hh$.
\end{defn}
\begin{defn}
  Given a root system $(R,R^\vee)$ on a vector space $\hh$, the \emindexing{Weyl group} $W \subseteq \GL(\hh^*)$ is the group generated by the reflections $s_\alpha: \lambda \mapsto \lambda - \<\lambda,\alpha^\vee\>\alpha$ as $\alpha$ ranges over $R$.
\end{defn}

\begin{prop}
  \begin{enumerate}
    \item It follows from {\bf RS3} that $s_\alpha^2 = e\in W$ for each root $\alpha$.
    \item It follows from {\bf RS4} that $WR = R$.  Thus $W$ is finite.  Moreover, $W$ preserves $\hh^*_\RR$, the $\RR$-span of $R$.
    \item The $W$-average of any positive-definite inner product on $\hh^*_\RR$ is a $W$-invariant positive-definite inner product.  Let $(,)$ be a $W$-invariant positive-definite inner product.  Then $s_\alpha$ is orthogonal with respect to $(,)$, and so $s_\alpha: \lambda \mapsto \lambda - \frac{2(\lambda,\alpha)}{(\alpha,\alpha)}\alpha$.  This inner product establishes an isomorphism $\hh \isom \hh^*$, under which $\alpha^\vee \mapsto 2\alpha/(\alpha,\alpha)$.
    \item Therefore {\bf Reduced} holds with $R$ replaced by $R^\vee$ if it holds at all.
    \item Let $W$ act on $\hh$ dual to its action on $\hh^*$.  Then $w(\alpha^\vee) = (w\alpha)^\vee$ for $w \in W$ and $\alpha \in R$.  Thus $s_{w\alpha} = ws_\alpha w^{-1}$.
    \item If $V\subseteq \hh^*$ is spanned by any subset of $R$, then $R\cap V$ and its image under $\vee$ form another nondegenerate root system.
    \item Two root systems with the same Weyl group and lattices are related by an isomorphism.
  \end{enumerate}
\end{prop}

\begin{defn}
  Let $R$ be a root system in $\hh^*$.  Define the \emindexing{weight lattice}\index{lattice!weight} to be $P \defined \{\lambda \in \hh^* \st \<\lambda,\alpha^\vee\> \in \ZZ\,\forall \alpha^\vee \in R^\vee\}$ and the \emindexing{root lattice}\index{lattice!root} $Q$ to be the $\ZZ$-span of $R$.  Then {\bf RS1} implies that $R \subseteq Q \subseteq P \subseteq \hh^*$; by {\bf Nondeg}, both $P$ and $Q$ are of full rank and so the index $P:Q$ is finite.  We define the \emindexing{coweight lattice}\index{lattice!coweight} to be $P^\vee$ and the \emindexing{coroot lattice}\index{lattice!coroot} to be $Q^\vee$.
\end{defn}

\subsection{Classification of rank-two root systems}

By {\bf Reduced}, there is a unique rank-one root system up to isomorphism, the root system of $\sl(2)$.

Let $R$ be a rank-two root system; then its Weyl group $W$ is a finite subgroup $W \subseteq \GL(2,\RR)$ generated by reflections.  The only finite subgroups of $\GL(2,\RR)$ are the cyclic and dihedral groups; only the dihedral groups are generated by reflections, and so $W \cong D_{2m}$ for some $m$.  Moreover, $W$ preserves the root lattice $Q$.
\begin{lemma}
  The only dihedral groups that preserve a lattice are $D_4,D_6,D_8,$ and $D_{12}$.
\end{lemma}
\begin{proof}
  Let $r_\theta$ be a rotation by $\theta$.  Its eigenvalues are $e^{\pm i\theta}$, and so $\tr(r_\theta) = 2\cos\theta$.  If $r_\theta$ preserves a lattice, its trace must be an integer, and so $2\cos\theta \in \{1,0,-1,-2\}$, as $2\cos\theta = 2$ corresponds to the identity rotation, and $|\cos\theta| \leq 1$.  Therefore $\theta \in \{\pi,2\pi/3,\pi/2,\pi/3\}$, i.e. $\theta = 2\pi/m$ for $m \in \{2,3,4,6\}$, and the only valid dihedral groups are $D_{2m}$ for these values of $m$.
\end{proof}
\begin{cor}
  There are four rank-two root systems, corresponding to the rectangular lattice, the square lattice, and the hexagonal lattice twice:
  \[ \begin{tikzpicture}[scale=.5]
  \draw[->,gray] (-3.5,0) -- (3.5,0);
  \draw[->,gray] (0,-2.5) -- (0,2.5);
  \foreach \X in {-3,-1.5,...,3} \foreach \Y in {-2,...,2}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \draw (1.5,1) -- (1.5,-1) -- (-1.5,-1) -- (-1.5,1) -- cycle;
  \node at (0,-3) {$\theta = \pi$};
\end{tikzpicture} \;\;\;\;\;
\begin{tikzpicture}[scale=.5]
  \draw[->,gray] (-2.5,0) -- (2.5,0);
  \draw[->,gray] (0,-2.5) -- (0,2.5);
  \foreach \X in {-2,...,2} \foreach \Y in {-1.732,0,1.732}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \foreach \X in {-1.5,...,1.5} \foreach \Y in {-.866,.866}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \draw (-1,0) -- (.5,.866) -- (.5,-.866) -- cycle;
  \node at (0,-3) {$\theta = 2\pi/3$};
\end{tikzpicture} \;\;\;\;\;
\begin{tikzpicture}[scale=.5]
  \draw[->,gray] (-2.5,0) -- (2.5,0);
  \draw[->,gray] (0,-2.5) -- (0,2.5);
  \foreach \X in {-2,...,2} \foreach \Y in {-2,...,2}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \draw (1,1) -- (-1,1) -- (-1,-1) -- (1,-1) -- cycle;
  \node at (0,-3) {$\theta = \pi/2$};
 \end{tikzpicture} \;\;\;\;\;
\begin{tikzpicture}[scale=.5]
  \draw[->,gray] (-2.5,0) -- (2.5,0);
  \draw[->,gray] (0,-2.5) -- (0,2.5);
  \foreach \X in {-2,...,2} \foreach \Y in {-1.732,0,1.732}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \foreach \X in {-1.5,...,1.5} \foreach \Y in {-.866,.866}
  {\fill (\X,\Y) circle (0.1cm);
  }
  \draw (-1,0) -- (-.5,.866) -- (.5,.866) -- (1,0) -- (.5,-.866) -- (-.5,-.866) -- cycle;
  \node at (0,-3) {$\theta = \pi/3$};
\end{tikzpicture}
 \]
\end{cor}

For each dihedral group, we can pick two reflections $\alpha_1,\alpha_2$ with a maximally obtuse angle; these generate $W$ and the lattice.  On the next page we list the four rank-two root systems with comments on their corresponding Lie groups:

\[\hspace{-1in} \vspace{-1in}
 \begin{tikzpicture}
  \matrix[column sep=6pt, row sep=6pt]{
    \node {\underline{$m$}}; & \node {\underline{Picture}}; & \node{\underline{Name}}; & \node{\underline{notes}}; \\ 
    \node {$2$}; & 
	  \draw[->,gray] (-2.5,0) -- (2.5,0);
	  \draw[->,gray] (0,-2) -- (0,2);
	  \foreach \X in {-2,...,2} \foreach \Y in {-1.4,-.7,...,1.4}
	     \fill (\X,\Y) circle (1pt);
	  \foreach \X/\Y in {1/0,-1/0,0/.7,0/-.7} \fill (\X,\Y) circle (2pt);
	  \node at (-.4,.3) {$\alpha_2$}; \node at (.5,-.3) {$\alpha_1$};
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (1,0);
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (0,.7);
   & \node{$A_1 \times A_1 = D_2$}; 
   & \node[text width=4in,text justified] {corresponding to the Lie algbera $\sl(2)\times \sl(2) = \so(4)$}; 
  \\
    \node {$3$}; &  
	  \draw[->,gray] (-2.5,0) -- (2.5,0);
	  \draw[->,gray] (0,-2.5) -- (0,2.5);
	  \foreach \X in {-2,...,2} \foreach \Y in {-1.732,0,1.732}
	     \fill (\X,\Y) circle (1pt);
	  \foreach \X in {-1.5,...,1.5} \foreach \Y in {-.866,.866}
	     \fill (\X,\Y) circle (1pt);
	   \foreach \X/\Y in {-1/0,-.5/.866,-.5/-.866,.5/.866,.5/-.866,1/0}
	     \fill (\X,\Y) circle (2pt);
	  \node at (-.5,.3) {$\alpha_2$}; \node at (.5,-.3) {$\alpha_1$};
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (1,0);
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (-.5, 0.866);
   & \node{$A_2$}; 
   & \node[text width=4in,text justified] {corresponding to $\sl(3)$ acting on the traceless diagonals}; 
 \\
   \node {$4$}; & 
	  \draw[->,gray] (-2.5,0) -- (2.5,0);
	  \draw[->,gray] (0,-2.5) -- (0,2.5);
	  \foreach \X in {-2,...,2} \foreach \Y in {-2,...,2}
	    \fill (\X,\Y) circle (1pt);
	  \foreach \X/\Y in {-1/-1,-1/0,-1/1,0/-1,0/1,1/-1,1/0,1/1}
	    \fill (\X,\Y) circle (2pt);
	  \node at (-.7,.3) {$\alpha_2$}; \node at (.5,-.3) {$\alpha_1$};
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (1,0);
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (-1, 1); 
   & \node{$B_2 = C_2$}; 
   & \node[text width=4in,text justified]{$\so(5) = \sp(4)$.  (When we get higher up, the $B$s and $C$s will separate, and we will have a new sequence of $D$s.)};  
  \\
    \node {$6$}; &  
	  \draw[->,gray] (-2.5,0) -- (2.5,0);
	  \draw[->,gray] (0,-2.5) -- (0,2.5);
	  \foreach \X in {-2,...,2} \foreach \Y in {-1.732,0,1.732}
	    \fill (\X,\Y) circle (1pt);
	  \foreach \X in {-1.5,...,1.5} \foreach \Y in {-.866,.866}
	    \fill (\X,\Y) circle (2pt);
	  \foreach \X/\Y in {-1/0,1/0,0/-1.732,0/1.732}
	    \fill (\X,\Y) circle (2pt);
	  \node at (-.85,.2) {$\alpha_2$}; \node at (.5,-.3) {$\alpha_1$};
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (1,0);
	  \draw[->,shorten >=3pt, ultra thick] (0,0) to (-1.5, 0.866);
   & \node{ $G_2$}; 
   & \node[text width=4in,text justified]{a new simple algebra of dimension $14 = $ number of roots plus dimension of root space.  We will see later that its smallest representation has dimension $7$.  There are many descriptions of this representation and the corresponding Lie algebra; the seven-dimensional representation comes from the \indexing{Octonians}, a non-associative, non-commutative ``field'', and $G_2$ is the automorphism group of the pure-imaginary part of the Octonians.}; 
  \\
  };
\end{tikzpicture} \hspace{-1in}\]

\begin{lemmadef}
  The axioms of a finite root system are symmetric under the interchange $R \tofrom R^\vee$.  This interchange assigns a \emindexing{dual} to each root system\index{root system!dual}.
\end{lemmadef}
\begin{proof}
  Only {\bf RS4} is not obviously symmetric.  We did not use {\bf RS4} to classify the two-dimensional root systems; we needed only a corollary:
  \begin{description}
    \item[RS4'] $W(R) = R$,
  \end{description}
  which is obviously symmetric.  But {\bf RS4} describes only the two-dimensional subspaces of a root system, and every rank-two root system with {\bf RS4'} replacing {\bf RS4} in fact satisfies {\bf RS4}.  This suffices to show that {\bf RS4'} implies {\bf RS4} for finite root systems.
\end{proof}

We remark that the statement is false for infinite root systems, and we presented the definition we did to accommodate the infinite case.  We will not discuss infinite root systems further.

\subsection{Positive roots}

\cite[Lecture 34]{notes}\footnote{I missed a few proofs from class.  In particular, the proofs of \cref{indeplemma1} and \cref{indeplemma2} are reproduced from \cite[page 156]{Knapp}.}

\begin{defn}
  A \emindexing{positive root system}\index{root system!positive} consists of a (finite) root system $R \subseteq \hh_\RR^*$ and a vector $v\in \hh_\RR$ so that $\alpha(v) \neq 0$ for every root $\alpha \in R$.  A root $\alpha \in R$ is {\em positive}\index{root!positive}\index{positive root} if $\alpha(v) > 0$, and {\em negative} otherwise.  Let $R_+$ be the set of positive roots and $R_-$ the set of negative ones; then $R = R_+ \sqcup R_-$, and by {\bf RS2}, $R_+ = - R_-$.
  
  The $\RR_{\geq 0}$-span of $R_+$ is a cone in $\hh_\RR^*$, and we let $\Delta$ be the set of extremal rays in this cone.  Since the root system is finite, extremal rays are generated by roots, and we use {\em Reduced} to identify extremal rays with positive roots.  Then $\Delta \subseteq R$ is the set of {\em simple roots}\index{simple root}\index{root!simple}.
\end{defn}
\begin{lemma} \label{indeplemma1}
  If $\alpha$ and $\beta$ are two simple roots, then $\alpha - \beta$ is not a root.  Moreover, $(\alpha,\beta) \leq 0$ for $\alpha \neq \beta$.
\end{lemma}
\begin{proof}
  If $\alpha - \beta$ is a positive root, then $\alpha = \beta + (\alpha - \beta)$ is not simple; if $\alpha - \beta$ is negative then $\beta$ is not simple.
  
  For the second statement, assume that $\alpha$ and $\beta$ are any two roots with $(\alpha,\beta) > 0$.  If $\alpha \neq \beta$, then they cannot be proportional, and we assume without loss of generality that $(\alpha,\alpha) \leq (\beta,\beta)$.  Then $s_\beta(\alpha) = \alpha - \frac{2(\alpha,\beta)}{(\beta,\beta)}\beta = \alpha - \beta$, because $2(\alpha,\beta) / (\beta,\beta) = \<\alpha,\beta^\vee\>$ is a positive integer strictly less than $2$.  Thus $\alpha - \beta$ is a root if $(\alpha,\beta) > 0$.
\end{proof}
\begin{lemma} \label{indeplemma2}
  Let $\RR^n$ have a positive definite inner product $(,)$, and suppose that $v_1,\dots,v_n \in \RR^n$ satisfy $(v_i,v_j) \leq 0$ if $i\neq j$, and such that there exists $v_0$ with $(v_0,v_i) > 0$ for every $i$.  Then $\{v_1,\dots,v_n\}$ is an independent set.
\end{lemma}
\begin{proof}
  Suppose that $0 = c_1v_1 + \dots + c_nv_n$.  Renumbering as necessary, we assume that $c_1,\dots,c_k \geq 0$, and $c_{k+1},\dots,c_{n} \leq 0$.  Let $v = c_1v_1 + \dots + c_kv_k = |c_{k+1}|v_{k+1} + \dots + |c_n|v_n$.  Then $0 \leq (v,v) = ( \sum_{i=1}^k c_iv_i,\sum_{j=k+1}^n -c_kv_k) = \sum_{j,k} |c_ic_j|(v_i,v_j) \leq 0$, which can happen only if $v = 0$.  But then $0 = (v,v_0) = \sum_{i=1}^k c_i(v_i,v_0) > 0$ unless all $c_i$ are $0$ for $i\leq k$.  Similarly we must have $c_j = 0$ for $j \geq k+1$, and so $\{v_i\}$ is independent.
\end{proof}

\begin{cor}
  In any positive root system, the set $\Delta$ of simple roots is a basis of $\hh^*$.
\end{cor}
\begin{proof}
  By \cref{indeplemma1}, $\Delta$ satisfies the conditions of \cref{indeplemma2} and so is independent.  But $\Delta$ generates $R_+$ and hence $R$, and therefore spans $\hh^*$.
\end{proof}

\begin{lemma} \label{HeightInductionLemma}
  Let $\Delta = \{\alpha_1,\dots,\alpha_n\}$ be a set of vectors in $\RR^m$ with inner product $(,)$, and assume that $\alpha_i$ are all on one side of a hyperplane: there exists $v$ such that  $(\alpha_i,v)>0$ $\forall i$.  Let $W$ be the group generated by reflections $S_{\alpha_i}$.  Let $R_+$ be any subset of $\RR_{\geq 0}\Delta \sminus \{0\}$ such that $s_i(R_+ \sminus \{\alpha_i\}) \subseteq R_+$ for each $i$, and such that the set of heights $\{(\alpha,v)\}_{\alpha\in R_+} \subseteq \RR_{\geq 0}$ is well-ordered.  Then $R_+ \subseteq W(\Delta)$.
\end{lemma}

\begin{proof}
  Let $\beta \in R_+$.  We proceed by induction on its height.
  
  There exists $i$ such that $(\alpha_i,\beta) > 0$, because if $(\beta,\alpha_i)\leq 0$ $\forall i$, then $(\beta,\beta) = 0$ since $\beta$ is a positive combination of the $\alpha_i$s.  Thus  $s_i(\beta) = \beta - (\textrm{positive})\alpha_i$; in particular, $\bigl(v, s_i(\beta)\bigr) < (v,\beta)$.
  
  If $\beta \neq \alpha_i$, then $s_i(\beta)\in R_+$ by hypothesis, so by induction $s_i(\beta) \in W(\Delta)$, and hence $\beta = s_i(s_i(\beta)) \in W(\Delta)$.  If $\beta = \alpha_i$, it's already in $W(\Delta)$.
\end{proof}

\begin{cor} \label{CartanMatrix2}
  Let $R$ be a finite root system, $R_+$ a choice of positive roots, and $\Delta$ the corresponding set of simple roots.  Then $R = W(\Delta)$, and the set $\{s_{\alpha_i}\}_{\alpha_i\in \Delta}$ generates $W$.
\end{cor}

\begin{cor}
  Let $R$ be a finite root system, $R_+$ a choice of positive roots, and $\Delta$ the corresponding set of simple roots.  Then $R \subseteq \ZZ\Delta$ and $R_+ \subseteq \ZZ_{\geq 0}\Delta$.
\end{cor}

\begin{prop} \label{CartanMatrix1}
  Let $R$ be a finite root system, and $R_+$ and $R_+'$ two choices of positive roots.  Then $R_+$ and $R_+'$ are $W$-conjugate.
\end{prop}
\begin{proof}
  Let $\Delta$ be the set of simple roots corresponding to $R_+$.
  If $\Delta \subseteq R_+'$, then $R_+ \subseteq R_+'$.  Then $R_- \subseteq R_-'$ by negating, and $R_+ \supseteq R_+'$ by taking complements, so $R_+ = R_+'$.
  
  Suppose $\alpha_i \in \Delta$ but $\alpha_i \not\in R_+'$, and consider the new system of positive roots $s_i(R_+')$, where $s_i = s_{\alpha_i}$ is the reflection corresponding to $\alpha_i$.  Then $s_i(R'_+) \cap R_+ \supseteq s_i(R_+' \cap R_+)$, because a system of roots that does not contain $\alpha_i$ does not lose anything under $s_i$.  But $\alpha_i \in R_-'$, so $-\alpha_i \in R_+'$, and so $\alpha_i \in s_i(R_+')$ and hence in $s_i(R_+') \cap R_+$.  Therefore $\bigl| s_i (R_+') \cap R_+ \bigr| > \bigl| R_+' \cap R_+ \bigr|$.
  
  If $s_i(R_+') \neq R_+$, then we can find $\alpha_j \in \Delta \sminus s_i(R_+')$.  We repeat the argument, at each step making the set $\bigl| w(R_+') \cap R_+ \bigr| $ strictly bigger, where $w = \cdots s_js_i \in W$.  Since $R_+$ is a finite set, eventually we cannot get any bigger; this can only happen when $\Delta \subseteq w(R_+')$, and so $R_+ = w( R_+')$.
\end{proof}

\section{Cartan Matrices and Dynkin Diagrams}

\subsection{Definitions}
\cite[Lecture 34]{notes}

\begin{defn}
  A finite-type \emindexing{Cartan matrix} of rank $n$ is an $n\times n$ matrix $\aaa_{ij}$ satisfying the following:
  \begin{itemize}
    \item $\aaa_{ii} = 2$ and $\aaa_{ij} \in \ZZ_{\leq 0}$ for $i\neq j$.
    \item $\aaa$ is \emindexing{symmetrizable}: there exists an invertible diagonal matrix $\ddd$ with $\ddd\aaa$ symmetric.
    \item $\aaa$ is {\em positive}\index{positive matrix}: all principle minors of $\aaa$ are positive.
  \end{itemize}
  An {\em isomorphism}\index{Cartan matrix!isomorphism of} between Cartan matrices $\aaa_{ij}$ and $\bbb_{ij}$ is a permutation $\sigma \in S_n$ such that $\aaa_{ij} = \bbb_{\sigma i,\sigma j}$.
\end{defn}
\begin{lemmadef}  \label{ClassificationLemma1}
  Let $R$ be a finite root system, $R_+$ a system of positive roots, and $\Delta = \{\alpha_1,\dots,\alpha_n\}$ the corresponding simple roots.  The {\em Cartan matrix}\index{Cartan matrix!of a root system} of $R$ is the matrix $\aaa_{ij} \defined \<\alpha_j,\alpha_i^\vee\> = 2(\alpha_i,\alpha_j)/(\alpha_i,\alpha_i)$.
  
  The Cartan matrix of a root system is a Cartan matrix.  It depends (up to isomorphism) only on the root system.  Conversely, a root system is determined up to isomorphism by its Cartan matrix.
\end{lemmadef}
\begin{proof}
  That the Cartan matrix depends only on the root system follows from \cref{CartanMatrix1}.  That the Cartan matrix determines the root system follows from \cref{CartanMatrix2}.
  
  Given a choice of root system and simple roots, let $d_i \defined (\alpha_i,\alpha_i)/2$, and let $\ddd_{ij} \defined d_i\delta_{ij}$ be the diagonal matrix with the $d_i$s on the diagonal.  Then $\ddd$ is invertible because $d_i > 0$, and $\ddd\aaa = (\alpha_i,\alpha_j)$ is obviously symmetric.  Let $I \subseteq \{1,\dots,n\}$; then the $I\times I$ principle minor of $\ddd\aaa$ is just $\prod_{i\in I} d_i$ times the corresponding principle minor of $\aaa$.  Since $d_i >0$ for each $i$ and $\ddd\aaa$ is the matrix of a positive-definite symmetric bilinear form, we see that $\aaa$ is positive.
\end{proof}

\subsection{Classification of finite-type Cartan matrices}

\cite[Lectures 34 and 35]{notes}

We classify (finite-type) Cartan matrices by encoding their information in graph-theoretic form (``Dynkin diagrams'') and then classifying (indecomposable) Dynkin diagrams.

\begin{defn}
  Let $\aaa$ be an integer matrix so that every principle $2\times 2$ sub-matrix has the form $\left[ \begin{matrix} 2 & -k \\ -l & 2 \end{matrix}\right]$ with $k,l \in \ZZ_{\geq 0}$ and either both $k$ and $l$ are $0$ or one of them is $1$.  Let us call such a matrix \emindexing{generalized-Cartan}.
\end{defn}
\begin{lemma}
  A Cartan matrix is generalized-Cartan.  A generalized-Cartan matrix is not Cartan if any entry is $-4$ or less.
\end{lemma}
\begin{proof}
  Consider a $2\times 2$ sub-matrix $\left[ \begin{matrix} 2 & -k \\ -l & 2 \end{matrix}\right]$.  Then if one of $k$ and $l$ is non-zero, the other must also be non-zero by symmetrizability.  Moreover, $kl < 4$ by positivity, and so one of $k$ and $l$ must be $1$.
\end{proof}
\begin{defn}
  Let $\aaa$ be a rank-$n$ generalized-Cartan matrix.  Its \emindexing{diagram} is a graph on $n$ vertices with (labeled, directed) edges determined as follows:  

  Let $1\leq i,j \leq n$, and consider the $\{i,j\}\times \{i,j\}$ submatrix of $\aaa$.  By definition, either $k$ and $l$ are both $0$, or one of them is $1$ and the other is a positive integer.  We do not draw an edge between vertices $i$ and $j$ if $k=l=0$.  We connect $i$ and $j$ with a single undirected edge if $k=l=1$.  For $k=2,3$, we draw an arrow with $k$ edges from vertex $i$ to vertex $j$ if the $\{i,j\}$ block is $\left[ \begin{matrix} 2 & -1 \\ -k & 2 \end{matrix}\right]$.
\end{defn}

\begin{defn}
  A diagram is {\em Dynkin}\index{Dynkin diagram}\index{diagram!Dynkin} if its corresponding generalized-Cartan matrix is in fact Cartan.
\end{defn}

\begin{lemmadef}
  The diagram of a generalized-Cartan matrix $\aaa$ is disconnected if and only if $\aaa$ is block diagonal, and connected components of the diagram correspond to the blocks of $\aaa$.  A block diagonal matrix $\aaa$ is Cartan if and only if each block is.  A connected diagram is \emindexing{indecomposable}.    We write ``$\times$'' for the disjoint union of Dynkian diagrams.
\end{lemmadef}
\begin{eg} \label{rank1diagrams}
  There is a unique indecomposable rank-$1$ diagram, and it is Dynkin: $A_1 = \bullet$.
  
  The indecomposable rank-$2$ Dynkin diagrams are:
  \begin{align*}
    A_2 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; \\
    B_2 = C_2 & = \tikz[baseline=-3pt] \draw[doubly] ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; \\
    G_2 & = \tikz[baseline=-3pt]{ \draw[doubly,double distance=2.5pt] ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; \draw (0,0) -- (1,0); }
  \end{align*}
\end{eg}

\begin{lemmadef}
  A \emindexing{subdiagram} of a diagram is a subset of the vertices, with edges induced from the parent diagram.  Subdiagrams of a Dynkin diagram correspond to principle submatrices of the corresponding Cartan matrix.  Any subdiagram of a Dynkin diagram is Dynkin.
\end{lemmadef}


By symmetrizability, if we have a triangle \tikz[baseline,thick]{
  \draw[singly] (0,.5) node {$\bullet$} -- node[auto,swap] {$k$} (.5,0) node {$\bullet$};
  \draw[singly] (.5,0) --  node[auto,swap] {$l$} (1,.5) node {$\bullet$};
  \draw[singly] (0,.5)  -- node[auto] {$m$} (1,.5) ;
}, then the multiplicities must be related: $m = kl$.  So $k$ or $l$ is $1$, and you can check that the three possibilities all have determinant $\leq 0$.  Moreover, a triple edge cannot attach to an edge, and two double edges cannot attach, again by positivity.  As such, we will never need to discuss the triple-edge again.

\begin{eg}
  The are three indecomposable rank-$3$ Dynkin diagrams:
  \begin{align*}
    A_3 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}  -- ++(1,0) node {$\bullet$}; \\
    B_3 & = \tikz[baseline=-3pt]{ \draw[doubly] (1,0) -- (2,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}   ++(1,0) node {$\bullet$}; } \\
    C_3 & = \tikz[baseline=-3pt]{ \draw[doubly] (2,0) -- (1,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}   ++(1,0) node {$\bullet$}; }
  \end{align*}
\end{eg}

\begin{defn} \label{neighbordefn}
Let $\aaa$ be a generalized-Cartan rank-$n$ matrix.  We can specify a vector in $\RR^n$ by assigning a ``\indexing{weight}'' to each vertex of the corresponding diagram.  The {\em neighbors}\index{neighbor} of a vertex are counted with multiplicity: an arrow leaving a vertex contributes only one neighbor to that vertex, but an arrow arriving contributes as many neighbors as the arrow has edges.  Naturally, each vertex of a weighted diagram has some number of ``weighted neighbors'': each neighbor is counted with multiplicity and multiplied by its weight, and these numbers are summed.
\end{defn}

\begin{lemma}
  Let $\aaa$ be a generalized-Cartan matrix, and think of a vector $\vec{x}$ as a weighting of the corresponding diagram.  With the weighted-neighbor conventions in \cref{neighbordefn}, the multiplication $\aaa\vec{x}$ can be achieved by subtracting the number of weighted neighbors of each vertex from twice the weight of that vertex.
  
  Thus, a generalized-Cartan matrix is singular if its corresponding diagram has a weighting such that each vertex has twice as many (weighted) neighbors as its own weight.
\end{lemma}

\begin{cor}
  A ring of single edges, and hence any diagram with a ring as a subdiagram, is not Dynkin.
\end{cor}
\begin{proof}
  We assign weight $1$ to each vertex; this shows that the determinant of the ring is $0$:
  \[ \begin{tikzpicture}[baseline]
  \draw  (330:2) node[fill=white] {$\cdot$}  \foreach \T in {0,40,80,...,280} { -- (\T:2) node {$\bullet$} } -- (310:2) node[fill=white] {$\cdot$} (320:2) node[fill=white] {$\cdot$};
  \draw[->, line width=2pt, decorate, decoration={snake,amplitude=.4mm,segment length=2mm,post length=1mm}] (3,0) -- (4,0);
  \draw[xshift=3in] (330:2) node[fill=white] {$\cdot$}  \foreach \T in {0,40,80,...,280} { -- (\T:2) node[fill=white] {$1$} } -- (310:2) node[fill=white] {$\cdot$} (320:2) node[fill=white] {$\cdot$} (180:2) node[anchor=east] {$\aaa \Bigg($}  (0:2) node[anchor=west] {$\Bigg) = 0$};
\end{tikzpicture} \]
\end{proof}

\begin{cor}
  The following diagrams correspond to singular matrices and hence are not Dynkin:
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a1) at (-.707,.707) 
  coordinate (a2) at (-.707,-.707) 
  coordinate (e1) at (3.707,.707) 
  coordinate (e2) at (3.707,-.707) 
  (a1) -- (b1) -- (a2) (e1) -- (d) -- (e2) (b1) -- (b2) -- (c) -- (d)
  ;
  \foreach \x in {b1,b2,d,a1,a2,e1,e2} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
  coordinate (a1) at (-.707,.707) 
  coordinate (a2) at (-.707,-.707) 
  coordinate (e1) at (.707,.707) 
  coordinate (e2) at (.707,-.707) 
  (a1) -- (b1) -- (a2) (e1) -- (b1) -- (e2) 
  ;
  \foreach \x in {b1,a1,a2,e1,e2} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a1) at (-.707,.707) 
  coordinate (a2) at (-.707,-.707) 
  coordinate (e) at (4,0) 
  (a1) -- (b1) -- (a2)  (b1) -- (b2) -- (c) -- (d)
  ;
  \draw[doubly] (d) -- (e);
  \foreach \x in {b1,b2,d,a1,a2,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a1) at (-.707,.707) 
  coordinate (a2) at (-.707,-.707) 
  coordinate (e) at (4,0) 
  (a1) -- (b1) -- (a2)  (b1) -- (b2) -- (c) -- (d)
  ;
  \draw[doubly] (e) -- (d);
  \foreach \x in {b1,b2,d,a1,a2,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a) at (-1,0) 
  coordinate (e) at (4,0) 
  (b1) -- (b2) -- (c) -- (d)
  ;
  \draw[doubly] (e) -- (d);
  \draw[doubly] (b1) -- (a);
  \foreach \x in {b1,b2,d,a,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a) at (-1,0) 
  coordinate (e) at (4,0) 
  (b1) -- (b2) -- (c) -- (d)
  ;
  \draw[doubly] (e) -- (d);
  \draw[doubly] (a) -- (b1);
  \foreach \x in {b1,b2,d,a,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (b1) at (0,0) 
    coordinate (b2) at (1,0) 
    node (c) at (2,0) {$\cdots$}
    coordinate (d) at (3,0) 
  coordinate (a) at (-1,0) 
  coordinate (e) at (4,0) 
  (b1) -- (b2) -- (c) -- (d)
  ;
  \draw[doubly] (d) -- (e);
  \draw[doubly] (b1) -- (a);
  \foreach \x in {b1,b2,d,a,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (a) at (0,0) 
    coordinate (b) at (1,0) 
    coordinate (c) at (2,0)
    coordinate (d) at (3,0) 
    coordinate (e) at (4,0) 
  (a) -- (b) (c)--(d)--(e)
  ;
  \draw[doubly] (b) -- (c);
  \foreach \x in {a,b,c,d,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (a) at (0,0) 
    coordinate (b) at (1,0) 
    coordinate (c) at (2,0)
    coordinate (d) at (3,0) 
    coordinate (e) at (4,0) 
  (a) -- (b) (c)--(d)--(e)
  ;
  \draw[doubly] (c) -- (b);
  \foreach \x in {a,b,c,d,e} \node at (\x) {$\bullet$};
\end{tikzpicture}\]
\end{cor}
\begin{proof}
  For example, we can show the last two as singular with the following weightings:
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (a) at (0,0) 
    coordinate (b) at (1,0) 
    coordinate (c) at (2,0)
    coordinate (d) at (3,0) 
    coordinate (e) at (4,0) 
  (a) -- (b) (c)--(d)--(e)
  ;
  \draw[doubly] (b) -- (c);
  \path (a) node[fill=white] {1}
  (b) node[fill=white] {2}
  (c) node[fill=white] {3}
  (d) node[fill=white] {2}
  (e) node[fill=white] {1};
\end{tikzpicture}\]
\[\begin{tikzpicture}
  \path[draw] 
    coordinate (a) at (0,0) 
    coordinate (b) at (1,0) 
    coordinate (c) at (2,0)
    coordinate (d) at (3,0) 
    coordinate (e) at (4,0) 
  (a) -- (b) (c)--(d)--(e)
  ;
  \draw[doubly] (c) -- (b);
  \path (a) node[fill=white] {2}
  (b) node[fill=white] {4}
  (c) node[fill=white] {3}
  (d) node[fill=white] {2}
  (e) node[fill=white] {1};
\end{tikzpicture}\]  
\end{proof}

\begin{lemma}
  The following diagrams are not Dynkin:
  \[ \begin{tikzpicture}
    \draw[doubly] (0,0) coordinate (a) -- ++(1,0);
    \draw (a) node {$\bullet$} -- ++(0,1) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(0,-1) node {$\bullet$};
    \path (a) ++(.5,-.5) node {$\det = -4$};
    
    \draw[doubly] (a) ++(3,0) coordinate (b) -- ++(1,0);
    \draw (b) node {$\bullet$} -- ++(108:1) node {$\bullet$} -- ++(36:1) node {$\bullet$} -- ++(324:1) node {$\bullet$} -- ++(252:1) node {$\bullet$};
    \path (b) ++(.5,-.5) node {$\det = -4$};
    
    \draw[doubly] (b) ++(3,0) coordinate (c) -- ++(1,0);
    \draw[doubly] (c) ++(0,1) -- ++(1,0);
    \draw (c) node {$\bullet$} -- ++(0,1) node {$\bullet$}  ++(1,0) node {$\bullet$} -- ++(0,-1) node {$\bullet$};
    \path (c) ++(.5,-.5) node {$\det = -8$};

    \draw[doubly] (c) ++(3,0) coordinate (d) -- ++(1,0);
    \draw[doubly] (d) ++(1,1) -- ++(-1,0);
    \draw (d) node {$\bullet$} -- ++(0,1) node {$\bullet$}  ++(1,0) node {$\bullet$} -- ++(0,-1) node {$\bullet$};
    \path (d) ++(.5,-.5) node {not symmetrizable};
  \end{tikzpicture} \]
\end{lemma}
\begin{cor}
  The indecomposable Dynkin diagrams with double edges are the following:
  \begin{align*}
    B_n & = \tikz[baseline=-3pt]{ \draw[doubly] (3,0) -- (4,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$}  ++(1,0) node {$\bullet$}; } \\
    C_n & = \tikz[baseline=-3pt]{ \draw[doubly] (4,0) -- (3,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$}  ++(1,0) node {$\bullet$}; } \\
    F_4 & = \tikz[baseline=-3pt]{ \draw[doubly] (1,0) -- (2,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; }
  \end{align*}  
\end{cor}
\begin{proof}
  Any indecomposable Dynkin diagram with a double edge is a chain.  The double edge must come at the end of the chain, unless the diagram has rank $4$.  
\end{proof}
\begin{lemma}
  Consider a $Y$-shaped indecomposable diagram.  Let the lengths of the three arms, including the middle vertex, be $k,l,m$.  Then the diagram is Dynkin if and only if $\frac1k + \frac1l + \frac1m > 1$.
\end{lemma}
\begin{proof}
  One can show directly that the determinant of such a diagram is $klm(\frac1k + \frac1l + \frac1m - 1)$.  We present null-vectors for the three ``\indexing{Egyptian fraction}'' decompositions of $1$ --- triples $k,l,m$ such that $\frac1k + \frac1l + \frac1m = 1$:
  \[ \begin{tikzpicture}[thick]
  \node (a0) at (0,0) {$3$};
  \node (a1) [left=.5 of a0] {$2$};
  \node (a2) [left=.5 of a1] {$1$};
  \node (a3) [above=.5 of a0] {$2$};
  \node (a4) [right=.5 of a3] {$1$};
  \node (a5) [right=.5 of a0] {$2$};
  \node (a6) [right=.5 of a5] {$1$};
  \draw (a0) -- (a1) -- (a2) (a0) -- (a3) -- (a4) (a0) -- (a5) -- (a6);
  \node (b0) at (6,0) {$4$};
  \node (b1) [above=.5 of b0] {$3$};
  \node (b2) [right=.5 of b1] {$2$};
  \node (b3) [left=.5 of b0] {$2$};
  \node (b4) [right=.5 of b2] {$1$};
  \node (b5) [right=.5 of b0] {$3$};
  \node (b6) [right=.5 of b5] {$2$};
  \node (b7) [right=.5 of b6] {$1$};
  \draw (b0) -- (b1) -- (b2) -- (b4) (b0) -- (b3) (b0) -- (b5) -- (b6) -- (b7);
\end{tikzpicture} \] 
\[ \begin{tikzpicture}[thick]
  \node (a0) at (0,0) {$6$};
  \node (a1) [above=.5 of a0] {$3$};
  \node (a3) [left=.5 of a0] {$4$};
  \node (a4) [left=.5 of a3] {$2$};
  \node (a5) [right=.5 of a0] {$5$};
  \node (a6) [right=.5 of a5] {$4$};
  \node (a7) [right=.5 of a6] {$3$};
  \node (a8) [right=.5 of a7] {$2$};
  \node (a9) [right=.5 of a8] {$1$};
  \draw (a0) -- (a1) (a0) -- (a3) -- (a4) (a0) -- (a5) -- (a6) -- (a7) -- (a8) -- (a9);
\end{tikzpicture} \]
\end{proof}
\begin{cor}
  The indecomposable Dynkin diagrams made entirely of single edges are:
  \begin{align*}
    A_n & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$} ;\\
    D_n & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$} -- +(45:1) node {$\bullet$} ++(0,0) -- +(315:1) node {$\bullet$} ; \\
    E_6 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; \\
    E_7 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; \\
    E_8 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; 
  \end{align*}
\end{cor}

All together, we have proven:
\begin{thm}[Classification of indecomposable Dynkin diagrams]  \label{DynkinClassificationThm}
  \index{Dynkin diagram!classification of}A diagram is Dynkin if and only if it is a disjoint union of indecomposable Dynkin diagrams.  The indecomposable Dynkin diagrams comprise four infinite families and five ``sporadic'' cases:
   \begin{align*}
    A_n & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$} ;\\
    D_n & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$} -- +(45:1) node {$\bullet$} ++(0,0) -- +(315:1) node {$\bullet$} ; \\
    B_n & = \tikz[baseline=-3pt]{ \draw[doubly] (3,0) -- (4,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$}  ++(1,0) node {$\bullet$}; } \\
    C_n & = \tikz[baseline=-3pt]{ \draw[doubly] (4,0) -- (3,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node[fill=white] {$\cdots$} -- ++(1,0) node {$\bullet$}  ++(1,0) node {$\bullet$}; } \\
    E_6 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; \\
    E_7 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; \\
    E_8 & = \tikz[baseline=-3pt] \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- +(0,1) node {$\bullet$} ++(0,0) -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ; \\
    F_4 & = \tikz[baseline=-3pt]{ \draw[doubly] (1,0) -- (2,0); \draw ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$} ++(1,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; } \\    
    G_2 & = \tikz[baseline=-3pt]{ \draw[doubly,double distance=2.5pt] ++(0,0) node {$\bullet$} -- ++(1,0) node {$\bullet$}; \draw (0,0) -- (1,0); }
  \end{align*} 
\end{thm}

\begin{eg} \label{DynkinCoincidences}
  We mention the small-rank coindidences\index{coincidences of Dynkin diagrams}\index{Dynkin diagram!coincidences of}.  We can continue the $E$ series for smaller $n$: $E_4 = A_4$, $E_5 = D_5$.  $E_3$ is not defined.  The $B$, $C$, and $D$ series make sense for $n\geq 2$, whence $B_2 = C_2$ and $D_2 = A_1 \times A_1$ and $D_3 = A_3$.  Some diagrams have nontrivial symmetries: for $n\geq 1$, the symmetry group of $A_n$ has order $2$, and similarly for $D_n$ for $n\neq 4$.  The diagram $D_4$ has an unexpected symmetry: its symmetry group is $S_3$, with order $6$.  The symmetry group of $E_6$ is order-$2$.
\end{eg}

\section{From Cartan Matrix to Lie Algebra}

\cite[Lectures 36 and 37]{notes}

In \cref{DynkinClassificationThm}, we classified indecomposable finite-type Cartan matrices, and therefore all finite-type Cartan matrices.  We can present generators and relations showing that each indecomposable Cartan matrix is the Cartan matrix of some simple Lie algebra --- indeed, the infinite families $A_n$, $B_n$, $C_n$, and $D_n$ correspond respectively to the classical Lie algebras $\sl(n,\CC)$\index{slnC@$\sl(n,\CC)$}, $\so(2n+1,\CC)$\index{sonC@$\so(n,\CC)$}, $\sp(n,\CC)$\index{spnC@$\sp(n,\CC)$}, and $\so(2n,\CC)$ --- and it is straightforward to show that a disjoint union of Cartan matrices corresponds to a direct product of Lie algebras.  

In this section, we explain how to construct a semisimple Lie algebra for any finite-type Cartan matrix, and we show that a semisimple Lie algebra is determined by its Cartan matrix.  This will complete the proof of the classification of semisimple Lie algebras.  Most, but not all, of the construction applies to generalized-Cartan matrices; the corresponding Lie algebras are {\em Kac-Moody}\index{Kac-Moody algebra}\index{algebra!Kac-Moody}, which are infinite-dimensional versions of semisimple Lie algebras.  We will not discuss Kac-Moody algebras here.

\begin{lemmadef} \label{tildegdef}
  Let $\Delta$ be a rank-$n$ Dynkin diagram with vertices labeled a basis $\{\alpha_1,\dots,\alpha_n\}$ of a vector space $\hh^*$, and let $\aaa_{ij}$ be the corresponding Cartan matrix.  Since $\aaa_{ij}$ is nondegenerate, it defines a map $\vee: \hh^* \to \hh$ by $\aaa_{ij} = \<\alpha_j,\alpha_i^\vee\>$.  We define $\tilde\gg = \tilde\gg_\Delta$ to be the Lie algebra generated by $\{e_i,f_i,h_i\}_{i=1}^n$ subject to the relations
  \begin{align}
\label{gtilderel1}  [h_i,e_j] & = \aaa_{ij} e_j \\
\label{gtilderel2}   [h_i,f_j] & = - \aaa_{ij} f_j \\
\label{gtilderel3}   [e_i,f_j] & = \delta_{ij} h_i \\
 \label{gtilderel4} [h_i,h_j] & = 0
\end{align}
  For each $i$, we write $\sl(2)_i$ for the subalgebra spanned by $\{e_i,f_i,h_i\}$; clearly $\sl(2)_i \cong \sl(2)$.
  
  Let $Q = \ZZ\Delta$ be the root lattice of $\Delta$.  Then the free Lie algebra generated by $\{e_i,f_i,h_i\}_{i=1}^n$ has a natural $Q$-grading, by $\deg e_i = \alpha_i$, $\deg f_i = -\alpha_i$, and $\deg h_i = 0$; under this grading, the relations are homogeneous, so the grading passes to the quotient $\tilde\gg_\Delta$.  
  
  Let $\tilde\hh \subseteq \tilde\gg$ be the subalgebra generated by $\{h_i\}_{i=1}^n$; then it is abelian and spanned by $\{h_i\}_{i=1}^n$.  The adjoint action $\ad: \hh \acts \tilde\gg$ is diagonalized by the grading: $h_i$ acts on anything of degree $q\in Q$ by $\<q,\alpha_i^\vee\>$.
  
  Let $\tilde\nn_+$ be the subalgebra of $\tilde\gg$ generated by $\{e_i\}_{i=1}^n$ and let $\tilde\nn_-$ be the subalgebra of $\tilde\gg$ generated by $\{f_i\}_{i=1}^n$; the algebras $\tilde\nn_{\pm}$ are called the ``upper-'' and ``lower-triangular'' subalgebras.
\end{lemmadef}
\begin{prop} \label{triangulardecompprop}
  Let $\Delta,\tilde\gg,\tilde\hh,\tilde\nn_\pm$ be as in \cref{tildegdef}.  Then $\tilde\gg = \tilde\nn_- \oplus \tilde\hh \oplus \tilde\nn_+$ as vector spaces; this is the ``\indexing{triangular decomposition}'' of $\tilde\gg$.
\end{prop}
\begin{proof}
  That $\tilde\nn_-,\tilde\hh,\tilde\nn_+$ intersect trivially follows form the grading, so it suffices to show that $\tilde\gg = \tilde\nn_- + \tilde\hh + \tilde\nn_+$.  By inspecting the relations, we see that $(\ad f_i)\tilde\nn_- \subseteq \tilde\nn_-$, $(\ad f_i) \tilde\hh \subseteq \<f_i\> \subseteq \tilde\nn_-$, and $(\ad f_i)\tilde\nn_+ \subseteq \tilde\hh + \tilde\nn_+$.  Therefore $\ad f_i$ preserves $\tilde\nn_- + \tilde\hh + \tilde\nn_+$, $\tilde\hh$ does so obviously, and $\ad e_i$ does so by the obvious symmetry $f_i \tofrom e_i$.  Therefore $\tilde\nn_- + \tilde\hh + \tilde\nn_+$ is an ideal of $\tilde\gg$ and therefore a subalgebra, but it contains all the generators of $\tilde\gg$.
\end{proof}
\begin{prop} \label{VermaProp}
  Let $\Delta,\tilde\gg$ be as in \cref{tildegdef}, and let $\lambda\in \hh^*$.  Write $\CC\< f_1,\dots,f_n\>$ for the free algebra generated by noncommuting symbols $f_1,\dots,f_n$ and $M_\lambda \defined \CC\< f_1,\dots,f_n\> v_\lambda$ for its free module generated by the symbol $v_\lambda$.  Then there exists  an action of $\tilde\gg$ on $M_\lambda$ such that:
  \begin{align}
  f_i \left( \prod f_{j_k} v_\lambda \right) & = \left( f_i\prod f_{j_k} \right) v_\lambda \\
  h_i \left( \prod f_{j_k} v_\lambda \right) & = \left( \lambda(h_i) - \sum_k \aaa_{i,j_k} \right) \left( \prod f_{j_k} v_\lambda \right) \\
  e_i  \left( \prod f_{j_k} v_\lambda \right) & = \sum_{k \st j_k = i} f_{j_1} \cdots f_{j_{k-1}} h_i f_{j_{k+1}} \cdots f_{j_l} \,v_\lambda
\end{align}
\end{prop}
\begin{proof}
  We have only to check that the action satisfies the relations \crefrange{gtilderel1}{gtilderel4}.  The $Q$-grading verifies \cref{gtilderel1,gtilderel2,gtilderel4}; we need only to check \cref{gtilderel3}.  When $i\neq j$, the action by $e_i$ ignores any action by $f_j$, and so we need only check that $[e_i,f_i]$ acts by $h_i$.  Write $\underline{f}$ for some monomial $f_{j_1} \cdots f_{j_n}$.  Then $e_if_i(\underline f v_\lambda) = e_i(f_i\underline f v_\lambda) = h_i\underline f v_\lambda + f_ie_i(\underline f v_\lambda)$, clear by the construction.
\end{proof}

\begin{defn}
  The $\tilde\gg$-module $M_\lambda$ defined in \cref{VermaProp} is the \emindexing{Verma module} of $\tilde\gg$ with weight\index{weight!of a Verma module} $\lambda$.
\end{defn}
\begin{cor}
  The map $\hh \to \tilde\hh$ is an isomorphism, so $\hh \mono \tilde\hh$.  The upper- and lower-triangular algebras $\tilde\nn_-$ and $\tilde\nn_+$ are free on $\{f_i\}$ and $\{e_i\}$ respectively.
\end{cor}

\begin{prop} \label{properidealsofg}
  Assume that $\Delta$ is an indecomposable system of simple roots, in the sense that the Dynkin diagram of the Cartan matrix of $\Delta$ is connected.  Construct $\tilde\gg$ as in \cref{tildegdef}.  Then any proper ideal of $\tilde\gg$ is graded, contained in $\tilde\nn_- + \tilde\nn_+$, and does not contain any $e_i$ or $f_i$.
\end{prop}
\begin{proof}
  The grading on $\tilde\gg$ is determined by the adjoint action of $\hh = \tilde\hh$.  Let $\aa$ be an ideal of $\tilde\gg$ and $a\in \aa$.  Let $a = \sum a_qg_q$ where $g_q$ are homogeneous of degree $q \in Q$.  Then $[h_i,a] = \sum \<q,\alpha_i^\vee\>a_qg_q$, and so $[\hh,a]$ has the same dimension as the number of non-zero coefficients $a_q$; in particular, $g_q\in [\hh,a]$.  Thus $\aa$ is graded.
  
  Suppose that $\aa$ has a degree-$0$ part, i.e.\ suppose that there is some $h\in \hh \cap \aa$.  Since the Cartan matrix $\aaa$ is nonsingular, there exists $\alpha_i\in\Delta$ with $\alpha_i(h) \neq 0$.  Then $[f_i,h] = \alpha_i(h)f_i \neq 0$, and so $f_i \in \aa$.
  
  Now let $\aa$ be any ideal with $f_i\in \aa$ for some $i$.  Then $h_i = [e_i,f_i] \in \aa$ and $e_i = -\frac12 [e_i,h_i] \in \aa$.  But let $\alpha_j$ be any neighbor of $\alpha_i$ in the Dynkin diagram.  Then $\aaa_{ij} \neq 0$, and so $[f_j,h_i] = \aaa_{ij}f_j \neq 0$; then $f_j \in \aa$.  Therefore, if the Dynkin diagram is connected, then any ideal of $\tilde\gg$ that contains some $f_i$ (or some $e_i$ by symmetry) contains every generator of $\tilde\gg$.
\end{proof}
\begin{cor}
  Under the conditions of \cref{properidealsofg}, $\tilde\gg$ has a unique maximal proper ideal.
\end{cor}
\begin{proof}
  Let $\aa$ and $\bb$ be any two proper ideals of $\tilde\gg$.  Then the ideal $\aa + \bb$ does not contain $\hh$ or any $e_i$ or $f_i$, and so is a proper ideal.
\end{proof}
\begin{defn} \label{gdef}
  Let $\Delta$ be a system of simple roots with connected Dynkin diagram, and let $\tilde\gg = \tilde\gg_\Delta$ be defined as in \cref{tildegdef}.  We define $\gg = \gg_\Delta$ as the quotient of $\tilde\gg$ by its unique maximal proper ideal.  Then $\<h_i,e_i,f_i\> \mono \gg$, where by $\<h_i,e_i,f_i\>$ we mean the linear span of the generators of $\tilde\gg$.  Since we quotiented by a maximal ideal, $\gg$ is simple.
\end{defn}

\begin{thm}[Serre Relations] \label{SerreThm}
  Let $\gg$ be as in \cref{gdef}, and $e_i,f_i$ the images of the corresponding generators of $\tilde\gg$.  Then:
  \begin{align}
  \label{SerreE} (\ad e_j)^{1 - \aaa_{ji}} e_i & = 0 \\
  \label{SerreF} (\ad f_j)^{1 - \aaa_{ji}} f_i & = 0
\end{align}
\end{thm}
\begin{proof}
  We will check \cref{SerreF}; \cref{SerreE} is exactly analogous.  Let $s$ be the left-hand-side of \cref{SerreF}, interpreted as an element of $\tilde\gg$.  We will show that the ideal generated by $s$ is proper.
  
  When $i=j$, $s = 0$, and when $i\neq j$, $\aaa_{ji} \leq 0$, and so the degree of $s$ is $-\alpha_i - (\geq 1)\alpha_j$.  In particular, bracketing with $f_k$ and $h_k$ only moves the degree further from $0$.  Therefore, the claim follows from the following equation:
  \eqn[SerreClaim]{ [e_k,s]_{\tilde\gg} = 0 \, \textrm{ for any }k}
  
  When $k \neq i,j$, $[e_k,f_i] = [e_k,f_j] = 0$.  So it suffices to check \cref{SerreClaim} when $k = i,j$.  Let $m = -\aaa_{ji}$.  When $k=j$, we compute:
  \begin{align}
    (\ad e_j)(\ad f_j)^{1 +m} f_i & = \bigl[ \ad e_j , (\ad f_j)^{1 +m} \bigr] f_i + (\ad f_j)^{1 +m }(\ad e_j)f_i \\
  \label{SerreClaim1}   & = \bigl[ \ad e_j , (\ad f_j)^{1 +m } \bigr] f_i + 0 \\
  \label{SerreClaim2} & = \sum_{l=0}^{m} (\ad f_j)^{m-l} \bigl(\ad[e_j,f_j]\bigr) (\ad f_j)^{l} f_i \\
   & = \sum_{l=0}^{m} (\ad f_j)^{m-l} (\ad h_j) (\ad f_j)^{l} f_i \\
   & = \sum_{l=0}^m \bigl( l(-\<\alpha_j,\alpha_j^\vee\>) - \<\alpha_i,\alpha_j^\vee\>\bigr) (\ad f_j)^m f_i \\
   & = \left( \sum_{l=0}^m \bigl( -2l + m\bigr) \right) (\ad f_j)^m f_i \\
   & = \left(  - 2\frac{m(m+1)}2 + (m+1)m\right)(\ad f_j)^m f_i = 0
  \end{align}
  where \cref{SerreClaim1} follows by $[e_i,f_j] = 0$, \cref{SerreClaim2} by the fact that $\ad$ is a Lie algebra homomorphism, and the rest is \cref{gtilderel2,gtilderel3}, that $m = -\aaa_{ji}$, and arithmetic.
  
  When $k = i$, $e_i$ and $f_j$ commute, and we have:
  \begin{align}
    (\ad e_i)(\ad f_j)^{1 +m} f_i & = \bigl[ \ad e_j , (\ad f_j)^{1 +m} \bigr] f_i + (\ad f_j)^{1 +m }(\ad e_i)f_i \\
    & = 0 +  (\ad f_j)^{1 +m }(\ad e_i)f_i \\
    & = (\ad f_j)^{1 +m }h_i  = 0
  \end{align}
  provided that $m\geq 1$.  When $m = 0$, we use the symmetrizability of the Cartan matrix: if $\aaa_{ji} = 0$ then $\aaa_{ij} = 0$.  Therefore
  \eqn{
    (\ad e_i)(\ad f_j)^{1 - \aaa_{ji}} f_i = (\ad e_i)[f_j,f_i] = -(\ad e_i) (\ad f_i)^{1 - \aaa_{ij}} f_j
  }
  which vanishes by the first computation.
\end{proof}

We have defined for each indecomposable Dynkin diagram $\Delta$ a simple Lie algebra $\gg_\Delta$.  If $\Delta = \Delta_1 \times \Delta_2$ is a disjoint union of Dynkin diagrams, we define $\gg_\Delta \defined \gg_{\Delta_1}\times \gg_{\Delta_2}$.

\begin{defn} \label{IntegrableDefn}
  Let $V$ be a (possibly-infinite-dimensional) $\gg$-module.  An element $v\in V$ is \emindexing{integrable} if for each $i$, the $\sl(2)_i$-submodule of $V$ generated by $v$ is finite-dimensional.  We write $I(V)$ for the set of integrable elements of $V$.
\end{defn}
\begin{lemma}
  Let $V$ be a $\gg$-module.  Then $I(V)$ is a $\gg$-submodule.
\end{lemma}
\begin{proof}
  Let $N \subseteq V$ be an $(n+1)$-dimensional irreducible representation of $\sl(2)_i$; then it is isomorphic to $V_n$ defined in \cref{VnDefn}.  It suffices to show that $e_jN$ is contained within some finite-dimensional $\sl(2)_i$ submodule of $V$ for $i\neq j$; the rest follows by switching $e\tofrom f$ and permuting the indices, using the fact that $\{e_j,f_j\}$ generate $\gg$.
  
  Then $N$ is spanned by $\{{f_i}^k v_0\}_{k=0}^n$ where $v_0 \in N$ is the vector annihilated by $e_i$; in particular, ${f_i}^{n+1}v_0 = 0$.  Since $e_j$ and $f_i$ commute, $e_jN$ is spanned by $\{{f_i}^k e_jv_0\}_{k=0}^n$.  It suffices to compute the $\sl(2)_i$ module generated by $e_jv_0$, or at least to show that it is finite-dimensional.  The action of $h_i$ on $e_jv_0$ is $h_ie_jv_0 = ([h_i,e_j] + e_jh_i)v_0 = (\aaa_{ij} + n)e_jv_0$.  For $k\neq n+1$, $f_i^k e_j v_0 = e_jf_i^k v_0 = 0$.  Moreover, by \cref{SerreThm}, $e_i^k e_j v_0  = [e_i^k,e_j]v_0 + e_je_i^kv_0 = (\ad e_i)^k(e_j) v_0 + 0$, which vanishes for large enough $k$.  Then the result follows by \cref{PBWthm} and the fact that $[e_i,f_i] = h_i$.
\end{proof}
\begin{cor}
  Let $\Delta$ be a Dynkin diagram and define $\gg$ as above.  Then $\gg$ is $\ad$-integrable.
\end{cor}
\begin{proof}
  Since $\{e_k,f_k\}$ generate $\gg$, it suffices to show that $e_k$ and $f_k$ are $\ad$-integrable for each $k$.  But the $\sl(2)_i$-module generated by $f_k$ has $f_k$ as its highest-weight vector, since $[e_i,f_k] = 0$, and is finite-dimensional, since $(\ad f_i)^nf_k = 0$ for large enough $n$ by \cref{SerreThm}.
\end{proof}

\begin{cor}
  The non-zero weights $R$ of $\ad: \gg \acts \gg$ form a root system.
\end{cor}
\begin{proof}
  Axioms {\bf RS1}, {\bf RS2}, {\bf RS3}, {\bf RS4}, and {\bf Nondeg} of \cref{RootSystemDefn} follow from the $\ad$-integrability.  Axiom {\bf Reduced} and that $R$ is finite follow from \cref{HeightInductionLemma}.
\end{proof}

\begin{thm}[Classification of finite-dimensional simple Lie algebras]
  The list given in \cref{DynkinClassificationThm} classifies the finite-dimensional simple Lie algebras over $\CC$.
\end{thm}
\begin{proof}
  A Lie algebra with an indecomposable root system is simple, because any such system has a highest root, and linear combination of roots generates the highest root, and the highest root generates the entire algebra.  So it suffices to show that two simple Lie algebras with isomorphic root systems are isomorphic.
  
  Let $\Delta$ be an indecomposable root system, and define $\tilde\gg$ and $\gg$ as above.  Let $\gg_1$ be a Lie algebra with root system $\Delta$.  Then the relations defining $\tilde\gg$ hold in $\gg_1$, and so there is a surjection $\tilde\gg \onto \gg_1$; if $\gg_1$ is simple, then the kernel of this surjection is a maximal ideal of $\tilde\gg$.  But $\tilde\gg$ has a unique maximal ideal, and $\gg$ is the quotient by this ideal; thus $\gg_1 \cong \gg$.
\end{proof}

\begin{eg}
  The families $ABCD$ correspond to the classical Lie algebras: $A_n \tofrom \sl(n+1)$\index{sln@$\sl(n)$}, $B_n \tofrom \so(2n+1)$\index{son@$\so(n)$}, $C_n \tofrom \sp(n)$\index{spn@$\sp(n)$}, and $D_n \tofrom \so(2n)$.  We recall that we have defined $\sp(n)$ as the Lie algebra that fixes the nondegenerate antisymmetric $2n\times 2n$ bilinear form: $\sp(n) \subseteq \gl(2n)$.  The EFG Lie algebras are new.

  The coincidences in \cref{DynkinCoincidences} correspond to coincidences of classical Lie algebras: $\so(6) \cong \sl(4)$, $\so(5) \cong \sp(2)$, and $\so(4) \cong \sl(2) \times \sl(2)$. The identity $\so(3) \cong \sl(2)$ suggests that we define $B_1 = A_1 = \bullet$, but $\sl(2)$ is not congruent to $\sp(1)$ or to $\so(2)$, so we do not assign meaning to $C_1$ or $D_1$, and justifying the name $B_1$ for $\bullet$ but not $C_1$ is ad hoc.
\end{eg}

\Section{Exercises}

\begin{enumerate}
  \item \begin{enumerate}
    \item  Show that $\SL(2, \RR)$\index{SL2R@$\SL(2,\RR)$} is topologically the product of a circle and two copies of $\RR$, hence it is not simply connected. 
    \item  Let $S$ be the simply connected cover of $\SL(2, \RR)$. Show that its finite-dimensional complex representations, i.e., real Lie group homomorphisms $S \to \GL(n, \CC)$, are determined by corresponding complex representations of the Lie algebra $\Lie(S )^\CC = \sl(2, \CC)$, and hence factor through $\SL(2, \RR)$. Thus $S$ is a simply connected real Lie group with no faithful finite-dimensional representation. 
   \end{enumerate}
  \item \begin{enumerate}
    \item Let $U$ be the group of $3\times 3$ upper-unitriangular complex matrices. Let $\Gamma \subseteq U$ be the cyclic subgroup of matrices 
    $$ \left[ \begin{array}{ccc} 1 & 0 & m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right], $$
    where $m\in \ZZ$. Show that $G = U/\Gamma$ is a (non-simply-connected) complex Lie group that has no faithful finite-dimensional representation. 
    \item  Adapt the solution to Set 4, Problem 2(b) to construct a faithful, irreducible infinite-dimensional linear representation $V$ of $G$. 
   \end{enumerate}
  \item  Following the outline below, prove that if $\hh \subseteq \gl(n,\CC)$  is a real Lie subalgebra with the property that every $X \in \hh$ is diagonalizable and has purely imaginary eigenvalues, then the corresponding connected Lie subgroup $H \subseteq  GL(n, \CC)$ has compact closure (this completes the solution to Set 1, Problem 7). \begin{enumerate}
    \item Show that $\ad X$ is diagonalizable with imaginary eigenvalues for every $X \in \hh$.
    \item Show that the Killing form of $\hh$ is negative semidefinite and its radical is the center of $\hh$. Deduce that $\hh$ is reductive and the Killing form of its semi-simple part is negative definite. Hence the Lie subgroup corresponding to the semi-simple part is compact. 
    \item Show that the Lie subgroup corresponding to the center of $\hh$ is a dense subgroup of a compact torus. Deduce that the closure of $H$ is compact. 
    \item Show that $H$ is compact --- that is, closed --- if and only if it further holds that the center of $\hh$ is spanned by matrices whose eigenvalues are rational multiples of $i$. 
  \end{enumerate}
  \item  Let $V_n = \Ss^n (\CC^2 )$ be the $(n + 1)$-dimensional irreducible representation of $\sl(2, \CC)$.
  \begin{enumerate}
    \item  Show that for $m \leq n$, $V_m \otimes V_n \cong V_{n-m} \oplus V_{n-m + 2} \oplus \dots \oplus V_{n+m}$, and deduce that the decomposition into irreducibles is unique. 
     \item Show that in any decomposition of $V^{\otimes n}_1$ into irreducibles, the multiplicity of $V_n$ is equal to $1$, the multiplicity of $V_{n-2k}$ is equal to ${n \choose k} - {n \choose k-1}$ for $k = 1, \dots , \lfloor n/2 \rfloor$, and all other irreducibles $V_m$ have multiplicity zero. 
  \end{enumerate}
  \item Let $a$ be a symmetric Cartan matrix, i.e.\ $a$ is symmetric with diagonal entries $2$ and off-diagonal entries $0$ or $-1$. Let $\Gamma$ be a subgroup of the automorphism group of the Dynkin diagram $D$ of $a$, such that every edge of $D$ has its endpoints in distinct $\Gamma$ orbits. Define the \emindexing{folding} $D'$ of $D$ to be the diagram with a node for every $\Gamma$ orbit $I$ of nodes in $D$, with edge weight $k$ from $I$ to $J$ if each node of $I$ is adjacent in $D$ to $k$ nodes of $J$. Denote by $a'$ the Cartan matrix with diagram $D'$.
  \begin{enumerate}
    \item Show that $a'$ is symmetrizable and that every symmetrizable generalized Cartan matrix (not assumed to be of finite type) can be obtained by folding from a symmetric one. 
    \item Show that every folding of a finite type symmetric Cartan matrix is of finite type. 
    \item  Verify that every non-symmetric finite type Cartan matrix is obtained by folding from a unique symmetric finite type Cartan matrix. 
  \end{enumerate}
  \item  An indecomposable symmetrizable generalized Cartan matrix $a$ is said to be of \emindexing{affine type} if $\det(a) = 0$ and all the proper principal minors of $a$ are positive. 
  \begin{enumerate}
    \item  Classify the affine Cartan matrices. 
    \item Show that every non-symmetric affine Cartan matrix is a folding, as in the previous problem, of a symmetric one. 
     \item Let $\hh$ be a vector space, $\alpha_i \in \hh^*$ and $\alpha_i^\vee \in \hh$ vectors such that 
$a$ is the matrix $\<\alpha_j,\alpha_i^\vee\>$.  Assume that this realization is non-degenerate in the sense that the vectors $\alpha_i$ are linearly independent. Define the \emindexing{affine Weyl group}\index{Weyl group!affine} $W$ to be generated by the reflections $s_{\alpha_i}$, as usual. Show that $W$ is isomorphic to the semidirect product $W_0 \ltimes Q$ where $Q$ and $W_0$ are the root lattice and Weyl group of a unique finite root system, and that every such $W_0 \ltimes Q$ occurs as an affine Weyl group. 
    \item Show that the affine and finite root systems related as in (c) have the property that 
the affine Dynkin diagram is obtained by adding a node to the finite one, in a unique way if 
the finite Cartan matrix is symmetric. 
  \end{enumerate}
  \item Work out the root systems of the orthogonal Lie algebras $\so(m, \CC)$ explicitly, thereby verifying that they correspond to the Dynkin diagrams $B_n$ if $m = 2n + 1$, or $D_n$ if $m = 2n$. Deduce the isomorphisms $\so(4, \CC) \cong \sl(2, \CC) \times \sl(2, \CC)$, $\so(5, \CC) \cong \sp(4, \CC)$, and $\so(6, \CC) \cong \sl(4, \CC)$.
    \item  Show that the Weyl group of type $B_n$ or $C_n$ (they are the same because these two root systems are dual to each other) is the group $S_n \ltimes (\ZZ/2\ZZ)^n$ of signed permutations, and that the Weyl group of type $D_n$ is its subgroup of index two consisting of signed permutations with an even number of sign changes, i.e., the semidirect factor $(\ZZ/2\ZZ)^n$ is replaced by the kernel of $S_n$-invariant summation homomorphism $(\ZZ/2\ZZ)^n \to \ZZ/2\ZZ$
  \item  Let $(\hh, R, R^\vee)$  be a finite root system, $\Delta = \{ \alpha_1,\dots,\alpha_n\}$  the set of simple roots with respect to a choice of positive roots $R_+$, $s_i = s_{\alpha_i}$  the corresponding generators of the Weyl group $W$. Given $w \in  W$, let $l(w)$ denote the minimum length of an expression for $w$ as a product of the generators $s_i$.
  \begin{enumerate}
    \item  If $w = s_{i_1} \dots s_{i_r}$ and $w(\alpha_j) \in R_-$, show that for some $k$ we have $\alpha_{i_k} = s_{i_{k+1}} \dots s_{i_r}(\alpha_j)$, and hence $s_{i_k}s_{i_{k+1}} \dots s_{i_r} = s_{i_{k+1}} \dots s_{i_r} s_j$. Deduce that $l(ws_j ) = l(w) -1$ if $w(\alpha_j ) \in R_-$. 
    \item Using the fact that the conclusion of (a) also holds for $v = ws_j$, deduce that $l(ws_j ) = l(w) + 1$ if $w(\alpha_j ) \not\in R_-$.
     \item  Conclude that $l(w) = |w(R_+) \cup R_-|$ for all $w \in W$. Characterize $l(w)$ in more explicit terms in the case of the Weyl groups of type $A$ and $B$/$C$. 
    \item Assuming that $\hh$ is over $\RR$, show that the dominant cone $X = \{ \lambda \in \hh : \<\lambda,\alpha_i^\vee\> \geq 0 \textrm{ for all } i \}$ is a fundamental domain for $W$, i.e., every vector in $\hh$ has a unique element of $X$ in its $W$ orbit. 
    \item  Deduce that $|W |$ is equal to the number of connected regions into which $\hh$ is separated by the removal of all the root hyperplanes $\<\lambda,\alpha^\vee\>$, $\alpha^\vee \in R^\vee$.
    \end{enumerate}
  \item Let $h_1, \dots, h_r$ be linear forms in variables $x_1, \dots, x_n$ with integer coefficients. Let $\FF_q$ denote the finite field with $q = p^e$ elements. Prove that except in a finite number of ``bad'' characteristics $p$, the number of vectors $v \in \FF_q^n$ such that $h_i (v) = 0$ for all $i$ is given for all $q$ by a polynomial $\chi(q)$ in $q$ with integer coefficients, and that $(-1)^n \chi(-1)$ is equal to the number of connected regions into which $\RR^n$ is separated by the removal of all the hyperplanes $h_i = 0$. 
  
  Pick your favorite finite root system and verify that in the case where the $h_i$ are the root hyperplanes, the polynomial $\chi(q)$ factors as $(q - e_1) \dots (q - e_n)$  for some positive integers $e_i$ called the \emindexing{exponents} of the root system. In particular, verify that the sum of the exponents is the number of positive roots, and that (by Problem 9(e)) the order of the Weyl group is $\prod_i (1 + e_i)$
  \item The \emindexing{height} of a positive root $\alpha$ is the sum of the coefficients $c_i$ in its expansion $\alpha = \sum_i c_i\alpha_i$ on the basis of simple roots. 
  
  Pick your favorite root system and verify that for each $k \geq 1$,  the number of roots of height $k$ is equal to the number of the exponents $e_i$ in Problem 10 for which $e_i \geq k$. 
  \item Pick your favorite root system and verify that if $h$ denotes the height of the highest root plus one, then the number of roots is equal to $h$ times the rank. This number $h$ is called the \emindexing{Coxeter number}. Verify that, moreover, the multiset of exponents (see Problem 10) is invariant with respect to the symmetry $e_i \mapsto  h - e_i$. 
    \item A \emindexing{Coxeter element} in the Weyl group $W$ is the product of all the simple reflections, once each, in any order. Prove that a Coxeter element is unique up to conjugacy. Pick your favorite root system and verify that the order of a Coxeter element is equal to the Coxeter number (see Problem 12). 
  \item The {\em fundamental weights}\index{fundamental weight}\index{weight!fundamental} $\lambda_i$ are defined to be the basis of the weight lattice $P$ dual to the basis of simple coroots in $Q^\vee$, i.e., $\<\lambda_i, \alpha^\vee_j\> = \delta_{ij}$.
  \begin{enumerate}
    \item Prove that the stabilizer in $W$ of $\lambda_i$ is the Weyl group of the root system whose Dynkin diagram is obtained by deleting node $i$ of the original Dynkin diagram.
     \item Show that each of the root systems $E_6$, $E_7$, and $E_8$ has the property that its highest root is a fundamental weight. Deduce that the order of the Weyl group $W(E_k )$ in each case is equal to the number of roots times the order of the Weyl group $W(E_{k-1})$, or $W (D_5 )$ for $k = 6$. Use this to calculate the orders of these Weyl groups. 
     \end{enumerate}
  \item Let $e_1, \dots, e_8$  be the usual orthonormal basis of coordinate vectors in Euclidean space $\RR^8$. The root system of type $E_8$ can be realized in $\RR^8$ with simple roots $\alpha_i = e_i - e_{i+1}$ for $i = 1,\dots, 7$  and 
  $$ \alpha_8 = \left( -\frac12, -\frac12, -\frac12, \frac12, \frac12, \frac12, \frac12, \frac12 \right). $$
  Show that the root lattice $Q$ is equal to the weight lattice $P$, and that in this realization, $Q$ consists of all vectors $\beta \in \ZZ^8$ such that $\sum_i \beta_i$  is even and all vectors $\beta \in  \left( \frac12, \frac12, \frac12, \frac12, \frac12, \frac12, \frac12, \frac12 \right) + \ZZ^8$ such that $\sum_i \beta_i$ is odd. Show that the root system consists of all vectors of squared length $2$ in $Q$,  namely, the vectors $\pm e_i \pm e_j$ for $i<j$, and all vectors with coordinates $\pm \frac12$ and an odd number of coordinates with each sign. 
  \item Show that the root system of type $F_4$ has $24$ long roots and $24$ short roots, and that the roots of each length form a root system of type $D_4$. Show that the highest root and the highest short root are the fundamental weights at the end nodes of the diagram. Then use Problem 14(a) to calculate the order of the Weyl group $W (F_4 )$. Show that $W (F_4 )$ acts on the set of short (resp.\ long roots) as the semidirect product $S_3 \ltimes W(D_4)$, where the symmetric group $S_3$ on three letters acts on $W (D_4 )$ as the automorphism group of its Dynkin diagram. 
  \item Pick your favorite root system and verify that the generating function $W(t) = \sum_{w\in W} t^{l(w)}$ is equal to $\prod_i \left(1 + t + \dots + t^{e_i}\right)$,  where $e_i$ are the exponents as in Problem 10. 
  \item Let $S$ be the subring of $W$-invariant elements in the ring of polynomial functions on $\hh$. Pick your favorite root system and verify that $S$ is a polynomial ring generated by homogeneous generators of degrees $e_i + 1$, where $e_i$ are the exponents as in Problem 10.
\end{enumerate}










\chapter{Representation Theory of Semisimple Lie Groups}

\section{Irreducible Lie-algebra representations}

\cite[Lectures 38 and 39]{notes} \cite[Lectures 18 and 19]{anton}\footnote{I was absent for Lecture 39 in Mark Haiman's class, and so those notes are missing from \cite{notes}.  To prepare this document, I thus turned to \cite{anton} for the proof of the Weyl Character Formula (\cref{WeylCharThm}).  The presentation of this material there is quite good, although differs from the presentation in \cite{notes}; in particular, \cite{anton} includes more motivation and examples, and discusses $\ch$ (\cref{chDefn}) in terms of group characters of the corresponding simply-connected Lie group.  See also Footnote 1 in \cite[page 106]{anton}.}

Any representation of a Lie group induces a representation of its Lie algebra, so we start our story there.  We recall \cref{WeylThm}: any finite-dimensional representation of a semisimple Lie algebra is the direct sum of simple representations.  In \cref{SL2RepThy} we computed the finite-dimensional simple representations of $\sl(2)$; we now generalize that theory to arbitrary finite-dimensional semisimple Lie algebras.  

\begin{lemmadef} \label{VermaDefn}
Let $\gg$ be a semisimple Lie algebra with Cartan subalgebra $\hh$ and root system $R$, and choose a system of positive roots $R_+$.  Let $\nn_\pm = \bigoplus_{\alpha \in R_{\pm}} \gg_\alpha$ be the upper- and lower-triangular subalgebras; then $\gg = \nn_- \oplus \hh \oplus \nn_+$ as a a vector space.  We define the \emindexing{Borel subalgebra} $\bb = \hh \oplus \nn_+$, and $\nn_+$ is an ideal of $\bb$ with $\hh = \bb/\nn_+$.

  Pick $\lambda \in \hh^*$; then $\bb$ has a one-dimensional module $\CC v_\lambda$, where $hv_\lambda = \lambda(h)\,v_\lambda$ for $h\in \hh$ and $\nn_+ v_\lambda = 0$.
  
  As a subalgebra, $\bb$ acts on $\gg$ from the right, and so we define the \emindexing{Verma module} of $\gg$ with weight $\lambda$ by:
  \eqn{ M_\lambda \defined \Uu\gg \otimes_{\Uu\bb} \CC v_\lambda }
  As a vector space, $M_\lambda \cong \Uu\nn_- \otimes_\CC \CC v_\lambda$.  It is generated as a $\gg$-module by $v_\lambda$ with the relations $hv_\lambda = \lambda(h)v_\lambda$, $\nn_+v_\lambda = 0$, and no relations on the action of $\nn_-$ except those from $\gg$.
\end{lemmadef}
\begin{proof}
  The explicit description of $M_\lambda$ follows from  \cref{PBWthm}: $\Uu\gg = \Uu\nn_- \otimes \Uu\hh \otimes \Uu\nn_+$ as vector spaces.
\end{proof}
\begin{cor}
  Any module with highest weight $\lambda$ is a quotient of $M_\lambda$.
\end{cor}

\begin{lemma}
  Let $\Delta = \{\alpha_1,\dots,\alpha_n\}$ be the simple roots of $\gg$, and let $Q \defined \ZZ\Delta$ be the root lattice and $Q_+ \defined \ZZ_{\geq 0}\Delta$.  Then the \emindexing{weight grading} given by the action of $\hh$ on the Verma module $M_\lambda$ is:
  \eqn{ M_\lambda = \bigoplus_{\beta \in Q_+} \left(M_\lambda\right)_{\lambda - \beta}}
  Moreover, let $N\subseteq M_\lambda$ be a proper submodule.  Then $N \subseteq  \bigoplus_{\beta \in Q_+ \sminus \{0\}} \left(M_\lambda\right)_{\lambda - \beta}$.
\end{lemma}
\begin{proof}
  The description of the weight grading follows directly from the description of $M_\lambda$ given in \cref{VermaDefn}.  Any submodule is graded by the action of $\hh$.  Since $(M_\lambda)_\lambda = \CC v_\lambda$ is one-dimensional and generates $M_\lambda$, a proper submodule cannot intersect $(M_\lambda)_\lambda$.
\end{proof}
\begin{cor} \label{LlambdaDefn}
  For any $\lambda \in \hh^*$, the Verma module $M_\lambda$ has a unique maximal proper submodule.  The quotient $M_\lambda \onto L_\lambda$ is an irreducible $\gg$-module.  Conversely, any irreducible $\gg$-module with highest weight $\lambda$ is isomorphic to $L_\lambda$, since it must be a quotient of $M_\lambda$ by a maximal ideal.
\end{cor}


\begin{defn}
Let $\gg$ be a semisimple Lie algebra and $\Delta = \{\alpha_1,\dots,\alpha_n\}$.  We recall the \emindexing{root lattice} $Q \defined \ZZ\Delta$ and  the \emindexing{weight lattice} $P \defined \{\lambda \in \hh^* \st \<\lambda,Q^\vee\> \subseteq \ZZ\}$.  A \emindexing{dominant integral weight}\index{weight!dominant integral} is an element of $P_+ \defined \{\lambda \in P \st \<\lambda,\alpha_i^\vee\> \geq 0\,\forall i\}$.
\end{defn}
We recall \cref{IntegrableDefn}.
\begin{prop} \label{IntegrableProp}
  If $\lambda \in P_+$, then $L_\lambda$ consists of integrable elements.
\end{prop}
\begin{proof}
  Since $L_\lambda$ is irreducible, its submodule of integrable elements is either $0$ or the whole module.  So it suffices to show that if $\lambda \in P_+$, then $v_\lambda$ is integrable.  Pick a simple root $\alpha_i$.  By construction, $e_iv_\lambda = 0$ and $h_iv_\lambda = \<\lambda,\alpha_i^\vee\>v_\lambda$.  Since $\lambda \in P_+$, $\<\lambda,\alpha_i^\vee\> = m\geq 0$ is an integer.  Consider the $\sl(2)_i$-submodule of $M_\lambda$ generated by $v_\lambda$; if $m$ is a nonnegative integer, from the representation theory of $\sl(2)$ we know that $e_if_i^{m+1}v_\lambda = 0$.  But if $j\neq i$, then $e_jf_i^{m+1}v_\lambda = f_i^{m+1}e_jv_\lambda = 0$.  Recalling the grading, we see then that $f_i^{m+1}v_\lambda$ generates a submodule of $M_\lambda$, and so $f_i^{m+1}v_\lambda \mapsto 0$ in $L_\lambda$.  Hence the $\sl(2)_i$-submodule of $L_\lambda$ generated by $v_\lambda$ is finite, and so $v_\lambda$ is integrable.
\end{proof}

\begin{defn}
  Let $\gg$ be a semisimple Lie algebra.  We define the category $\hat\Oo$\index{O@$\hat\Oo$} to be a full subcategory of the category $\cat{$\gg$-mod}$ of (possibly-infinite-dimensional) $\gg$ modules.  The objects $X\in \hat\Oo$ are required to satisfy the following conditions:
  \begin{itemize}
    \item The action $\hh \acts X$ is diagonalizable.
    \item For each $\lambda \in \hh^*$, the weight space $X_\lambda$ is finite-dimensional.
    \item There exists a finite set $S \subseteq \hh^*$ such that the weights of $X$ lie in $S + (-Q_+)$.
  \end{itemize}
\end{defn}
\begin{lemma}
  The category $\hat\Oo$ is closed under submodules, quotients, extensions, and tensor products.
\end{lemma}
\begin{proof}
  The $\hh$-action grades subquotients of any graded module, and acts diagonally.  An extension of graded modules is graded, with graded components extensions of the corresponding graded components.  Since $\gg$ is semisimple, any extension of finite-dimensional modules is a direct sum, and so the $\hh$-action is diagonal on any extension of objects in $\Oo$.  Finally, tensor products are handled by \cref{TensorWeightLemma}.
\end{proof}
\begin{defn}
  Write the additive group $\hh^*$ multiplicatively: $\lambda \mapsto x^\lambda$.  The \emindexing{group algebra} $\ZZ [\hh^*]$ is the algebra of ``polynomials'' $\sum c_ix^{\lambda_i}$, with the obvious addition and multiplication.  I.e.\ $\ZZ[\hh^*]$ is the free abelian group $\bigoplus_{\lambda \in \hh^*} \ZZ x^\lambda$, with multiplication given on a basis by $x^\lambda x^\mu = x^{\lambda + \mu}$.
  
  Let $\ZZ[-Q_+]$ be the subalgebra of $\ZZ[\hh^*]$ generated by $\{x^\lambda \st -\lambda \in Q_+\}$.  This has a natural topology given by setting $\| x^{-\alpha_i}\| = c^{-\alpha_i}$ for $\alpha_i$ a simple root and $c$ some real constant with $c > 1$.  We let $\ZZ[[-Q_+]]$ be the completion of $\ZZ[-Q_+]$ with respect to this topology.  Equivalently, $\ZZ[[-Q_+]]$ is the algebra of \emindexing{formal power series} in the variables $x^{-\alpha_1},\dots,x^{-\alpha_n}$ with integer coefficients.
  
  Then $\ZZ[-Q_+]$ is a subalgebra of both $\ZZ[\hh^*]$ and $\ZZ[[-Q_+]]$.  We will write $\ZZ[h^*,-Q_+]]$ for the algebra $\ZZ[\hh^*] \otimes_{\ZZ[-Q_+]} \ZZ[[-Q_+]]$.
\end{defn}
The algebra $\ZZ[h^*,-Q_+]]$ is a formal gadget, consisting of formal fractional Laurant series.  We use it as a space of generating functions.
\begin{defn}  \label{chDefn}
  Let $X \in \hat\Oo$.  We define $\ch(X) \in \ZZ[h^*,-Q_+]]$ by:
  \eqn{ \ch(X) \defined \sum_{\lambda \textrm{ a weight of }X} \dim(X_\lambda)x^\lambda }
\end{defn}
We remark that every coefficient of $\ch(X)$ is a nonnegative integer, and if $Y$ is a subquotient  
\begin{eg} \label{MChar}
  Let $M_\lambda$ be the Verma module with weight $\lambda$, and let $R_+$ be the set of positive roots of $\gg$.  Then
  \eqn{ \ch(M_\lambda) = \frac{x^\lambda}{\prod_{\alpha \in R_+} (1 - x^{-\alpha})}
   \defined x^\lambda \prod_{\alpha\in R_+} \sum_{l=0}^\infty x^{-l\alpha} 
  }
  This follows from \cref{PBWthm}, the explicit description of $M_\lambda \cong \Uu\nn_-\otimes \CC v_\lambda$, and some elementary combinatorics.
\end{eg}
\begin{prop} \label{WeylCharProof2}
  Let $\gg$ be simple Lie algebra, $P_+$ the set of dominant integral weights, and $W$ the Weyl group.
  Let $\lambda \in P_+$, and $L_\lambda$ the irreducible quotient of $M_\lambda$ given in \cref{LlambdaDefn}.  Then:
  \begin{enumerate}
    \item $\ch(L_\lambda)$ is $W$-invariant.
    \item If $\mu$ is a weight of $L_\lambda$, then $\mu \in W(\nu)$ for some $\nu \in P_+ \cap (\lambda - Q_+)$.
    \item $L_\lambda$ is finite-dimensional.
  \end{enumerate}
  Conversely, every finite-dimensional irreducible $\gg$-module is $L_\lambda$ for a unique $\lambda \in P_+$.
\end{prop}

\begin{proof}
\begin{enumerate}
  \item We use \cref{IntegrableProp}: $L_\lambda$ consists of integrable elements.  Let $\alpha_i$ be a root of $\gg$; then $L_\lambda$ splits as an $\sl(2)_i$ module: $L_\lambda = \bigoplus V_a$, where each $V_a$ is an irreducible $\sl(2)_i$ submodule.  In particular, $V_a = \CC v_{a,m} \oplus \CC v_{a,m-2} \dots \oplus \CC v_{a,-m}$ for some $m$ depending on $a$, where $h_i$ acts on $\CC v_{a,l}$ by $l$.  But $\ch(L_\lambda)  = \sum_a \ch(V_a) = \sum_a \sum_{j= -m,-m+2,\dots,m} \ch(\CC v_{a,m})$.  Let $\ch\bigl(\CC (v_a)_l\bigr) = x^{\mu_{a,l}}$; then $\<\mu_{a,l},\alpha_i^\vee\> = l$ by definition, and $v_{a,l - 2} \in f_i \CC v_{a,l}$, and so $s_i \mu_{a,l} = \mu_{a,-l}$.  This shows that $\ch(V_a)$ is fixed under the action of $s_i$, and so $\ch(L_\lambda)$ is also $s_i$-invariant.  But the reflections $s_i$ generate $W$, and so $\ch(L_\lambda)$ is $W$-invariant.
  \item We partially order $P$: $\nu \leq \mu$ if $\mu - \nu \in Q_+$.  In particular, the weights of $L_\lambda$ are all less than or equal to $\lambda$.

  Let $\lambda \in P$.  Then $s_i(\lambda) = \lambda - \<\lambda,\alpha_i^\vee\>\alpha_i$, and so $W(\lambda) \subseteq \lambda + Q$.  If $\lambda \in P_+$ then $\<\lambda,\alpha_i^\vee\>\geq 0$ for every $i$ and so $s_i \leq \lambda$; if $\lambda \in P \sminus P_+$ then there is some $i$ with $\<\lambda,\alpha_i^\vee\>< 0$, i.e.\ some $i$ with $s_i(\lambda) > \lambda$.  But $W$ is finite, so for any $\lambda \in P$, $W(\lambda)$ has a maximal element, which must be in $P_+$.  This proves that $P = W(P_+)$.
  
  Thus, if $\mu$ is a weight of $L_\lambda$, then $\mu \in W(\nu)$ for some $\nu \in P_+$.  But by 1., $\nu$ is a weight of $L_\lambda$, and so $\nu \leq \lambda$.  This proves statements 2.
  
  Moreover, the $W$-invariance of $\ch(L_\lambda)$ shows that if $\lambda \in  P_+$, then $W(\lambda) \subseteq \lambda - Q_+$, and moreover that $P_+$ is a fundamental domain of $W$.
  \item The Weyl group $W$ is finite.  Consider the two cones $\RR_{\geq 0} P_+$ and $-\RR_{\geq 0}Q_+$.  Since the inner product (the symmetrization of the Cartan matrix) is positive definite and by construction the inner product of anything in $\RR_{\geq 0}P_+$ with anything in $-\RR_{\geq 0}Q_+$ is negative, the two cones intersect only at $0$.  Thus there is a hyperplane separating the cones: i.e.\ there exists a linear functional $\eta: \hh^*_\RR \to \RR$ such that its value is positive on $P_+$ but negative on $-Q_+$.  Then $\lambda - Q_+$ is below the $\eta = \eta(\lambda)$ hyperplane.  But $-Q_+$ is generated by $-\alpha_i$, each of which has a negative value under $\eta$, and so $\lambda - Q_+$ contains only finitely many points $\mu$ with $\eta(\mu) \geq 0$.  Thus $P_+ \cap (\lambda - Q_+)$ is finite, and hence so is its image under $W$.
\end{enumerate}

  For the converse statement, let $L$ be a finite-dimensional irreducible $\gg$-module, and let $v \in L$ be any vector.  Then consider $\nn_+v$, the image of $v$ under repeated application of various $e_i$s.  By finite-dimensionality, $\nn_+v$ must contain a vector $l \in \nn_+v$ so that $e_il = 0$ for every $i$.  By the $\sl(2)$ representation theory, $l$ must be homogeneous, and indeed a top-weight vector of $L$, and by the irreducibility $l$ generates $L$.  Let the weight of $l$ be $\lambda$; then the map $v_\lambda \to l$ generates a map $M_\lambda \onto L$.  But $M_\lambda$ has a unique maximal submodule, and since $L$ is irreducible, this maximal submodule must be the kernel of the map $M_\lambda \onto L$.  Thus $L \cong L_\lambda$.
\end{proof}


\begin{lemmadef}
  Let $\gg$ be a semisimple Lie algebra, $\hh$ its Cartan subalgebra, and $\Delta = \{\alpha_1,\dots,\alpha_n\}$ its simple root system.  For each $i = 1,\dots,n$, we define a \emindexing{fundamental weight}\index{weight!fundamental} $\Lambda_i \in \hh^*$ by $\<\Lambda_i,\alpha_j^\vee\> = \delta_{ij}$.  Then $P_+ = \ZZ_{\geq 0}\{\Lambda_1,\dots,\Lambda_n\}$.
  
  The following are equivalent, and define the \emindexing{Weyl vector} $\rho$:
  \begin{enumerate}
    \item $\rho = \sum_{i=1}^n \Lambda_i$.  I.e.\ $\<\rho,\alpha_j^\vee\> = 1$ for every $j$.
    \item $\rho = \frac12 \sum_{\alpha \in R_+} \alpha$.
  \end{enumerate}
\end{lemmadef}
\begin{proof}
  Let $\rho_2 = \frac12 \sum_{\alpha \in R_+} \alpha$.  Since $s_i(R_+\sminus \{\alpha_i\}) = R_+$ but $s_i(\alpha_i) = -\alpha_i$, we see that $s_i(\tilde\rho) = \tilde\rho - \alpha_i$, and so $\<\tilde\rho,\alpha_i^\vee\> = 1$ for every $i$.  The rest is elementary linear algebra.
\end{proof}
\begin{defn}
  Let $\lambda \in P_+$.  We define the \emindexing{character} $\chi^\lambda$ of $\lambda$ to be $\ch(L_\lambda)$.
\end{defn}
\begin{thm}[Weyl Character Formula] \label{WeylCharThm}
  Let $\epsilon: W \to \{\pm 1\}$ be given by $\epsilon(w) = \det_\hh w$; i.e.\ $\epsilon$ is the group homomorphism generated by $s_i \mapsto -1$ for each $i$.  Let $\lambda \in P_+$.  Then $\chi^\lambda$ can be computed as follows:
  \eqn[WeylCharEqn]{ \chi^\lambda = \frac{\displaystyle \sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) - \rho}}{\displaystyle \prod_{\alpha \in R_+} (1 - x^{-\alpha})} =  \frac{\displaystyle \sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) }}{\displaystyle \prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} }
\end{thm}
The equality of fractions follows simply from the description $\rho = \frac12 \sum_{\alpha \in R_+} \alpha$.

\begin{remark}
  The sum in \cref{WeylCharEqn} is finite.

  Indeed,  the numerator and denominator on the right-hand-side fraction are obiously antisymmetric in $W$, and so the whole expression is $W$-invariant.  The numerator on the left-hand-side fraction is a polynomial, and each $(1 - x^{-\alpha})$ is invertible as a power series: $(1-x^{-\alpha})^{-1} = \sum_{n=0}^\infty x^{-n\alpha}$.  So the fraction is a $W$-invariant power series, and hence a polynomial.
\end{remark}

To prove \cref{WeylCharThm} we will need a number of lemmas.  In \cref{MChar} we computed the character of the Verma module $M_\lambda$.  Then \cref{WeylCharThm} asserts that:
  \eqn{ \ch(L_\lambda) = \sum_{w\in W} \epsilon(w) \,\ch\bigl(M_{w(\lambda + \rho) - \rho}\bigr) }
As such, we will begin by understanding $M_\lambda$ better.  
We recall \cref{CasimirDefn}: Let $(,)$ be the Killing form on $\gg$, and $\{x_i\}$ any basis of $\gg$ with dual basis $\{y_j\}$, i.e.\ $(x_i,y_j) = \delta_{ij}$ for every $i,j$; then $c = \sum x_iy_i \in \Uu\gg$ is central, and does not depend on the choice of basis. 
  
  \begin{lemma}
    Let $\lambda \in \hh^*$ and $M_\lambda$ the Verma module with weight $\lambda$.  Let $c \in \Uu\gg$ be the Casimir, corresponding to the Killing form on $\gg$.  Then $c$ acts on $M_\lambda$ by multiplication by $(\lambda,\lambda + 2\rho)$.
  \end{lemma}
  \begin{proof}
    Let $\gg$ have rank $n$.  Write $R$ for the set of roots of $\gg$, $R_+$ for the positive roots, and $\Delta$ for the simple roots, as we have previously.
    
  Recall \cref{KillingNondegCartan}.  We construct a basis of $\gg$ as follows: we pick an orthonormal basis $\{u_i\}_{i=1}^n$ of $\hh$.  For each $\alpha$ a non-zero root of $\gg$, the space $\gg_\alpha$ is one-dimensional; let $x_\alpha$ be a basis vector in $\gg_\alpha$.  Then the dual basis to $\{u_i\}_{i = 1}^n \cup \{x_\alpha\}_{\alpha \in R \sminus \{0\}}$ is $\{u_i\}_{i = 1}^n \cup \{ y_\alpha\}_{\alpha \in R \sminus \{0\}}$, where $y_\alpha = \frac{x_{-\alpha} }{ (x_\alpha,x_{-\alpha})}$.  Then:
  \eqn{ c = \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R\sminus \{0\}} x_\alpha y_\alpha =  \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R\sminus \{0\}} \frac{x_\alpha x_{-\alpha}}{(x_\alpha,x_{-\alpha})} = \sum_{i=1}^n u_i^2 + \sum_{\alpha \in R_+} \frac{x_\alpha x_{-\alpha} + x_{-\alpha}x_\alpha}{(x_\alpha,x_{-\alpha})} }
  
  Since $M_\lambda$ is generated by its highest weight vector $v_\lambda$, and $c$ is central, to understand the action of $c$ on $M_\lambda$ it suffices to compute $cv_\lambda$.  We use the fact that for $\alpha \in R_+$, $x_\alpha v_\lambda = 0$; then 
  \eqn[WeylCharProof2]{ x_\alpha x_{-\alpha} v_\lambda = h_\alpha v_\lambda + x_{-\alpha} x_\alpha v_\lambda = h_\alpha v_\lambda = \lambda(h_\alpha) \, v_\lambda }
  where for each $\alpha \in R_+$ we have defines $h_\alpha \in \hh$ by $h_\alpha = [x_\alpha,x_{-\alpha}]$.  
  Moreover, $(,)$ is $\gg$-invariant, and $[h_\alpha,x_\alpha] = \alpha(h_\alpha)x_\alpha$, where $\alpha(h_\alpha) \neq 0$.   So:
  \eqn{ (x_\alpha,x_{-\alpha}) = \frac1{\alpha(h_\alpha)} \bigl([h_\alpha,x_\alpha], x_{-\alpha}\bigr)  = \frac1{\alpha(h_\alpha)} \bigl(h_\alpha,[x_\alpha, x_{-\alpha}]\bigr) = \frac{(h_\alpha,h_\alpha)}{\alpha(h_\alpha)} }
  We also have that $u_iv_\lambda = \lambda(u_i)v_\lambda$, and since $\{u_i\}$ is an orthonormal basis, $(\lambda,\lambda) = \sum_{i=1}^n \bigl(\lambda(u_i)\bigr)^2$.  Thus:
  \eqn{ cv_\lambda = \sum_{i=1}^n \bigl(\lambda(u_i)\bigr)^2 v_\lambda + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)}{\frac{(h_\alpha,h_\alpha)}{\alpha(h_\alpha)} } v_\lambda = \left( (\lambda,\lambda) + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} \right) v_\lambda}
  We recall that $h_\alpha$ is proportional to $\alpha^\vee$, that $(\alpha,\alpha) = 4/(\alpha^\vee,\alpha^\vee)$, and that $\lambda(\alpha^\vee) = (\lambda,\alpha)/(\alpha,\alpha)$.  Then $\frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} = (\lambda,\alpha)$, and so:
  \eqn{ (\lambda,\lambda) + \sum_{\alpha \in R_+} \frac{\lambda(h_\alpha)\,\alpha(h_\alpha)}{(h_\alpha,h_\alpha)} = (\lambda,\lambda) + \sum_{\alpha \in R_+} (\lambda,\alpha) = (\lambda,\lambda + 2\rho) }
  Thus $c$ acts on $M_\lambda$ by multiplication by $(\lambda,\lambda + 2\rho)$.
  \end{proof}
  
  \begin{lemmadef}
    Let $X$ be a $\gg$-module.  We weight vector $v\in X$ is {\em singular}\index{singular vector} if $\nn_+ v = 0$.  In particular, any highest-weight vector is singular, and conversely any singular vector is the highest weight vector in the submodule it generates.  
  \end{lemmadef}
  
  \begin{cor}
    Let $\lambda \in P$, and $M_\lambda$ the Verma module with weight $\lambda$.  Then $M_\lambda$ contains finitely many singular vectors, in the sense that their span is finite-dimensional.
  \end{cor}
  \begin{proof}
  Let $C^\lambda$ be the set $C^\lambda \defined  \{\mu \in P \st (\mu + \rho,\mu + \rho) = (\lambda + \rho,\lambda + \rho)\}$.  Then $C^\lambda$ is a sphere in $P$ centered at $-\rho$, and in particular it is a finite set.  On the other hand, since $(\mu + \rho,\mu + \rho) = (\mu ,\mu + 2\rho) + (\rho,\rho)$, we see that:
  \eqn{ C^\lambda = \{\mu \in P \st c\textrm{ acts on }M_\mu \textrm{ by } (\lambda,\lambda + 2\rho) \} }
  Recall that any module with highest weight $\mu$ is a quotient of $M_\mu$.  Let $v \in M_\lambda$ be a non-zero singular vector with weight $\mu$.  Then on the one hand $c v = (\lambda,\lambda + 2\rho)v$, since $v\in M_\lambda$, and on the other hand $c v = (\mu,\mu + 2\rho)$, since $v$ is in a quotient of $M_\mu$.  In particular, $\mu \in C^\lambda$.  But the weight spaces $(M_\lambda)_\mu$ of $M_\lambda$ are finite-dimensional, and so the dimension of the space of singular vectors is at most $\sum_{\mu \in C^\lambda} \dim\bigl((M_\lambda)_\mu\bigr) < \infty$.
  \end{proof}
  
  \begin{cor} \label{WeylCharProof1}
    Let $\lambda \in P$.  Then there are nonnegative integers $b_{\lambda,\mu}$ such that
    \eqn[KLeqn]{ \ch M_\lambda = \sum  b_{\lambda,\mu} \ch L_\mu }
    and $b_{\lambda,\mu} = 0$ unless $\mu \leq \lambda$ and $\mu \in C^\lambda$.  Moreover, $b_{\lambda,\lambda} = 0$.
  \end{cor}
  \begin{proof}
    We construct a filtration on $M_\lambda$.  Since $M_\lambda$ has only finitely many non-zero singular vectors, we choose $w_1$ a singular vector of minimal weight $\mu_1$, and let $F_1M_\lambda$ be the submodule of $M_\lambda$ generated by $w_1$.  Then $F_1M_\lambda$ is irreducible with highest weight $\mu_1$.  We proceed by induction, letting $w_i$ be a singular vector of minimal weight in $M_\lambda / F_{i-1}M_\lambda$, and $F_iM_\lambda$ the primage of the subrepresentation generated by $w_i$.  This filters $M_\lambda$:
  \eqn[MlambdaFiltered]{ 0 = F_0M_\lambda \subseteq F_1M_\lambda \subseteq \dots }
  Moreover, since $M_\lambda$ has only finitely many weight vectors all together, the filtration must terminate:
  \eqn{ 0 = F_0M_\lambda \subseteq F_1M_\lambda \subseteq \dots \subseteq F_kM_\lambda = M_\lambda }
  By construction, the quotients are all irreducible: $F_iM_\lambda / F_{i-1}M_\lambda = L_{\mu_i}$ for some $\mu_i \in C^\lambda$, $\mu_i \leq \lambda$.
  
  We recall that $\ch$ is additive for extensions.  Therefore
  \eqn{ \ch M_\lambda = \sum_{i=1}^k \ch(F_iM_\lambda / F_{i-1}M_\lambda) = \sum_{i=1}^k \ch L_{\mu_i}}
  Then $b_{\lambda,\mu}$ is the multiplicity of $\mu$ appearing as the weight of a singular vector of $M_\lambda$, and we have \cref{KLeqn}.  The conditions stated about $b_{\lambda,\mu}$ are immediate: we saw that $\mu$ can only appear as a weight of $M_\lambda$ if $\mu \in C^\lambda$ and $\mu \leq \lambda$; moreover, $L_\lambda$ appears as a subquotient of $M_\lambda$ exactly once, so $b_{\lambda,\lambda} = 1$.
  \end{proof}
  \begin{defn}
  The coefficients $b_{\lambda,\mu}$ in \cref{KLeqn} are the {\em Kazhdan-Luztig multiplicities}\index{Kazhdan-Luztig multiplicity}.
  \end{defn}
  
\begin{lemma}  \label{WeylCharProof3}
  If $\lambda \in P_+$, $\mu \leq \lambda$, $\mu \in C^\lambda$, and $\mu + \rho \geq 0$, then $\mu = \lambda$.
\end{lemma}
\begin{proof}
  We have that $(\mu + \rho,\mu + \rho) = (\lambda + \rho,\lambda + \rho)$ and that $\lambda - \mu = \sum_{i=1}^n k_i\alpha_i$, where all $k_i$ are nonnegative.  Then
  \begin{align*}
    0 & = (\lambda + \rho,\lambda + \rho) - (\mu + \rho,\mu + \rho) \\
      & = \bigl( (\lambda + \rho) - (\mu + \rho), (\lambda + \rho) + (\mu + \rho) \bigr) \\
      & = ( \lambda - \mu, \lambda + \mu + 2\rho) \\
      & = \sum_{i=1}^n k_i(\alpha,\lambda + \mu + 2\rho)
  \end{align*}
  But $\lambda, \mu+\rho \geq 0$, and $(\alpha_i,\rho) > 0$, so $(\alpha,\lambda + \mu + 2\rho) > 0$, and so all $k_i = 0$ since they are nonnegative.
\end{proof}

\begin{proof}[of \cref{WeylCharThm}]
  We have shown (\cref{WeylCharProof1}) that $\ch M_\lambda = \sum b_{\lambda,\mu} \ch L_\mu$, were $b_{\lambda,\mu}$ is a lower-triangular matrix on $C^\lambda = C^\mu$ with ones on the diagonal.  Thus it has a lower-triangular inverse with ones on the diagonal:
  \eqn{ \ch L_\lambda = \sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} \ch M_\mu }
  But by \cref{WeylCharProof2} statement 1., $\ch L_\lambda$ is $W$-invariant, provided that $\lambda \in P_+$, thus so is $\sum c_{\lambda,\mu} \ch M_\mu$.  We recall \cref{MChar}:
  \eqn{ \ch M_\mu = \frac{ x^\mu}{\prod_{\alpha \in R_+} (1 - x^{-\alpha})} = \frac{ x^{\mu+\rho}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})}  }
  Therefore
  \eqn{ \ch L_\lambda = \frac{\sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} x^{\mu + \rho}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})}  }
  But the denominator if $W$-antisymmetric, and so the numerator must be as well:
  \eqn{ \sum_{\mu\leq \lambda, \mu \in C^\lambda}  c_{\lambda,\mu} x^{w(\mu + \rho)} = \sum_{\mu\leq \lambda, \mu \in C^\lambda}  \epsilon(w) c_{\lambda,\mu} x^{\mu + \rho} \textrm{ for every $w\in W$}}
  This is equivalent to the condition that $c_{\lambda,\mu} = \epsilon(w)c_{\lambda,w(\mu +\rho) - \rho}$.  By the proof of \cref{WeylCharProof2} statement 2., we know that $P_+$ is a fundamental domain of $W$; since $c_{\lambda,\lambda} = 1$, if $\mu + \rho \in W(\lambda +\rho)$, then $c_{\lambda,\mu} = \epsilon(w)$, and so:
  \eqn{ \sum_{\mu\leq \lambda, \mu \in C^\lambda} c_{\lambda,\mu} x^{\mu + \rho} = \sum_{w\in W} \left( x^{w(\lambda + \rho)} + \mathop{\sum_{\mu < \lambda, \mu \in C^\lambda}}_{\mu + \rho \in P^+} c_{\lambda,\mu} x^{w(\mu+\rho)} \right)}
  But the rightmost sum is empty by \cref{WeylCharProof3}.
\end{proof}

\begin{remark}
  Specializing to the trivial representation $L_0$, \cref{WeylCharThm} says that  
  \eqn[dimform1]{ 1 = \frac{ \sum_{w\in W} \epsilon(w) x^{w(\rho)}}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} }
  So we can rewrite \cref{WeylCharEqn} as
  \eqn{ \chi^\lambda = \frac{ \sum_{w\in W} \epsilon(w) x^{w(\lambda + \rho)} }{\sum_{w\in W} \epsilon(w) x^{w(\rho)} } }
\end{remark}

The following is an important corollary:
\begin{thm}[Weyl Dimension Formula]
  Let $\lambda \in P_+$.  Then $\displaystyle \dim L_\lambda = \prod_{\alpha \in R_+} \frac{(\alpha,\lambda + \rho)}{(\alpha,\rho)}$.
\end{thm}
\begin{proof}
  The formula $\ch(L_\lambda) = \frac{\sum_{w\in W} \epsilon(w)\,x^{w(\lambda + \rho) }}{\prod_{\alpha \in R_+} (x^{\alpha/2} - x^{-\alpha/2})} $ is a polynomial in $x$.  In particular, it defines a real-valued function on $\RR_{>0}\times \hh$ given by $x^\alpha \mapsto a^{\alpha(h)}$ --- when $a = 1$ or $h = 0$, the formula as written is the indeterminate form $\frac00$, but the function clearly returns $\sum_{\mu} \dim\bigl((L_\lambda)_\mu\bigr) = \dim L_\lambda$.  We will calculate this value of the function by taking a limit, using l'H\^opital's rule.
  
  In particular, letting $x^\alpha \mapsto e^{t(\alpha,\lambda + \rho)}$ in \cref{dimform1} gives
  \eqn{ \prod_{\alpha \in R_+}  \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) = \sum_{w\in W} \epsilon(w) e^{t(w(\rho),\lambda + \rho)} = \sum_{w\in W} \epsilon(w) e^{t(\rho,w(\lambda + \rho)) }}
  where the second equality comes from $w\mapsto w^{-1}$ and $ (w^{-1}x,y) = (x,wy) $.  On the other hand, we let $x^\alpha \mapsto e^{t(\alpha,\rho)}$ in \cref{WeylCharEqn}.  Then
  \begin{align}
     \ch L_\lambda |_{x = e^{t\rho}} & = \frac{ \sum_{w\in W}\epsilon(w) e^{t (w(\lambda + \rho),\rho)} }{\prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right) } \\
     & = \frac{\prod_{\alpha \in R_+}  \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{\prod_{\alpha \in R_+} \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \\
     & = \prod_{\alpha \in R_+} \frac{  \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)}   
   \end{align}
 Therefore
 \begin{align}
   \dim L_\lambda & = \lim_{t\to 0} \prod_{\alpha \in R_+} \frac{  \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)}   \\
    & = \prod_{\alpha \in R_+}\lim_{t\to 0} \frac{  \left( e^{t(\alpha/2,\lambda + \rho)} - e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( e^{t(\alpha/2,\rho)} - e^{-t(\alpha/2,\rho)} \right)} \\
    & \overset{\!\! \textrm{l'H}\!\!}= \prod_{\alpha \in R_+}  \lim_{t\to 0} \frac{  \left( (\alpha/2,\lambda + \rho)e^{t(\alpha/2,\lambda + \rho)} + (\alpha/2,\lambda + \rho)e^{-t(\alpha/2,\lambda + \rho)} \right) }{ \left( (\alpha/2,\rho)e^{t(\alpha/2,\rho)} +(\alpha/2,\rho) e^{-t(\alpha/2,\rho)} \right)} \\
    & = \prod_{\alpha \in R_+} \frac{(\alpha,\lambda + \rho)}{(\alpha,\rho)}
 \end{align}
  \end{proof}
\begin{eg}
  Let us compute the dimensions of the irreducible representations of $\gg = \sl(n+1)$.  We work with the standard the simple roots be $\Delta = \{\alpha_1,\dots,\alpha_{n}\}$, whence $R_+ = \{\alpha_i + \alpha_{i+1} + \cdots + \alpha_j \}_{1 \leq i < j \leq n}$.  Let us write $\lambda$ and $\rho$ in terms of the fundamental weights $\Lambda_i$, defined by $(\Lambda_i,\alpha_j) = \delta_{ij}$: $\rho = \sum_{i=1}^n \Lambda_i$ and $\lambda + \rho = \sum_{i=1}^n a_i\Lambda_i$.  Then:
  \begin{align}
    \dim L_\lambda & = \prod_{\alpha \in R_+} \frac{(\lambda + \rho,\alpha)}{(\rho,\alpha)} \\
     & = \prod_{1 \leq i \leq j \leq n} \frac{a_i + a_{i+1} + \dots + a_{j-1} + a_j}{j-i+1} \\
     & = \frac1{n!!} \prod_{1 \leq i \leq j \leq n} \sum_{k=i}^j a_k
  \end{align}
  where we have defined $n!! \defined n! \,(n-1)! \cdots 3!\,2!\,1!$.  For example, the irrep of $\sl(3)$ with weight $\lambda + \rho = 3\Lambda_1 + 2\Lambda_2$ has dimension $\frac1{2!!} 2 \cdot 3 \cdot (2+3) = 15$.
\end{eg}

\section{Algebraic Lie Groups}

We have classified the representations of any semisimple Lie algebra, and therefore the representations of its simply connected Lie group.  But a Lie algebra corresponds to many Lie groups, quotients of the simply connected group by (necessarily central) discrete subgroups, and a representation of the Lie algebra is a representation of one of these groups only if the corresponding discrete normal subgroup acts trivially in the representation.  We will see that the simply connected Lie group of any semisimple Lie algebra is algebraic, and that its algebraic quotients are determined by the finite-dimensional representation theory of the Lie algebra.

\subsection{Guiding example: $\SL(n)$ and $\PSL(n)$}

\cite[Lecture 40]{notes}

Our primary example, as always, is the Lie algebra $\sl(2,\CC)$\index{sl2C@$\sl(2,\CC)$}, consisting of traceless $2\times 2$ complex matrices.  It is the Lie algebra of $\SL(2,\CC)$\index{SL2C@$\SL(2,\CC)$}, the group of $2\times 2$ complex matrices with determinant $1$.
\begin{lemmadef}
  The group $\SL(2,\CC)$ has a non-trivial center: $Z(\SL(2,\CC)) = \{\pm 1\}$.  We define the \emindexing{projective special linear group}\index{special linear group!projective} to be $\PSL(2,\CC) \defined \SL(2,\CC)/\{\pm 1\}$\index{PSL2C@$\PSL(2,\CC)$}.  Equivalently, $\PSL(2,\CC) = \PGL(2,\CC) \defined \GL(2,\CC) / \{\textrm{scalars}\}$, the \emindexing{projective general linear group}\index{general linear group!projective}\index{PGL2C@$\PGL(2,\CC)$}.
\end{lemmadef}
\begin{prop}
  The group $\SL(2,\CC)$ is connected and simply connected.  The kernel of the map $\ad: \SL(2,\CC) \to \GL(\sl(2,\CC))$ is precisely the center, and so $\PSL(2,\CC)$ is the connected component of the group of automorophisms of $\sl(2,\CC)$.  The groups $\SL(2,\CC)$ and $\PSL(2,\CC)$ are the only connected Lie groups with Lie algebra $\sl(2,\CC)$.
\end{prop}
\begin{proof}
  The only nontrivial statement is that $\SL(2,\CC)$ is simply connected.  Consider the subgroup $U \defined \left\{ \left[ \begin{matrix} 1 & * \\ 0 & 1\end{matrix}\right] \in \SL(2,\CC) \right\}$.  Then $U$ is the stabilizer of the vector $\left[ \begin{matrix} 1 \\ 0 \end{matrix}\right] \in \CC^2 \sminus \{0\}$, and $\SL(2,\CC)$ acts transitively on $ \CC^2 \sminus \{0\}$.  Thus the space of left cosets $\SL(2,\CC) / U$ is isomorphic to the space $ \CC^2 \sminus \{0\} \cong \RR^4 \sminus \{0\}$ as a real manifold.  But $U \cong \CC$, so $\SL(2,\CC)$ is connected and simply connected.
\end{proof}
\begin{lemma}
  The groups $\SL(2,\CC)$ and $\PSL(2,\CC)$ are algebraic.
\end{lemma}
\begin{proof}
  The determinant of a matrix is a polynomial in the coefficients, so $\{x\in {\rm M}(2,\CC) \st \det x = 1\}$ is an algebraic group.  Any automorphism of $\sl(2,\CC)$ preserves the Killing form, a nondegenerate symmetric pairing on the three-dimensional vector space $\sl(2,\CC)$.  Thus $\PSL(2,\CC)$ is a subgroup of ${\rm O}(3,\CC)$.  It is connected, and so a subgroup of $\SO(3,\CC)$, and three-dimensional, and so is all of $\SO(3,\CC)$.  Moreover, $\SO(3,\CC)$ is algebraic: it consists of matrices $x\in {\rm M}(3,\CC)$ that preserve the nondegenerate form (a system of quadratic equations in the coefficients) and have unit determinant (a cubic equation in the coefficients).
\end{proof}

Recall that any irreducible representation of $\sl(2,\CC)$ looks like a chain: $e$ moves up the chain, $f$ down, and $h$ acts diagonally with eigenvalues changing by $2$ from $m$ at the top to $-m$ at the bottom:
  \eqn{ \begin{tikzpicture}[baseline=(v2.base)]
    \path (0,0) node[label=left:{$v_0$}] (v0) {$\bullet$} 
    ++(0,-1) node[label=left:{$v_1$}] (v1) {$\bullet$}
    ++(0,-1) node[label=left:{$v_2$}] (v2) {$\bullet$}
    ++(0,-1) node (dots) {$\vdots$}
    ++(0,-1) node[label=left:{$ v_{m-1}$}] (vn1) {$\bullet$}
    ++(0,-1) node[label=left:{$ v_m$}] (vn) {$\bullet$}
    ;
    \draw[->] (v0) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m$} (v0);
    \draw[->] (v1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m-2$} (v1);
    \draw[->] (v2) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = m-4$} (v2);
    \draw[->] (dots) .. controls +(330:1) and +(30:1) ..  (dots);
    \draw[->] (vn1) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = 2-m$} (vn1);
    \draw[->] (vn) .. controls +(330:1) and +(30:1) .. node[auto,swap] {$\scriptstyle h = -m$} (vn);
    \draw[->] (v0) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 1$} (v1);
    \draw[->] (v1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = 2$} (v2);
    \draw[->] (v2) .. controls +(300:.5) and +(60:.5) .. (dots);
    \draw[->] (dots) .. controls +(300:.5) and +(60:.5) ..  (vn1);
    \draw[->] (vn1) .. controls +(300:.5) and +(60:.5) .. node[auto] {$\scriptstyle f = m$} (vn);
    \draw[->] (v1) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle m = e$} (v0);
    \draw[->] (v2) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle m-1 =  e$} (v1);
    \draw[->] (dots) .. controls +(120:.5) and +(240:.5) ..  (v2);
    \draw[->] (vn1) .. controls +(120:.5) and +(240:.5) ..  (dots);
    \draw[->] (vn) .. controls +(120:.5) and +(240:.5) .. node[auto] {$\scriptstyle 1 = e$} (vn1);
  \end{tikzpicture}}
The exponential map $\exp: \sl(2,\CC) \to \SL(2,\CC)$ acts on the Cartan by $th = \left[ \begin{matrix} t & \\ & -t \end{matrix}\right] \mapsto \left[ \begin{matrix} e^t & \\ & e^{-t} \end{matrix}\right] $.  Let $T = \exp(\hh)$; then the kernel of $\exp: \hh \to T$ is $2\pi i \ZZ h$.  On the other hand, when $t = \pi i$, $\exp(th) = -1$, which maps to $1$ under $\SL(2,\CC) \onto \PSL(2,\CC)$; therefore the kernel of the exponential map $\hh \to \PSL(2,\CC)$ is just $\pi i \ZZ h$.

In particular, the $(m+1)$-dimensional representation $V_m$ of $\sl(2,\CC)$ is a representation of $\PSL(2,\CC)$ if and only if $m$ is even, because $-1 \in \SL(2,\CC)$ acts on $V_m$ as $(-1)^m$.  We remark that $\ker\{\exp: \hh \to \SL(2,\CC)\}$ is precisely $2\pi i Q^\vee$, where $Q^\vee$ is the coroot lattice of $\sl(2)$, and $\ker\{\exp: \hh \to \PSL(2,\CC)\}$ is precisely the coweight lattice $2\pi i P^\vee$

\begin{remark}
This will be the model for any semisimple Lie algebra $\gg$ with Cartan subalgebra $\hh$.   We will understand the exponential map from $\hh$ to the simply connected Lie group $G$ corresponding to $\gg$, and we will also understand the map to $G / Z(G)$, the simplest quotient.  Every group with Lie algebra $\gg$ is a quotient of $G$, and hence lies between $G$ and $G/Z(G)$.  The kernels of the maps $\hh \to G$ and $\hh \to G/Z(G)$ will be precisely $2\pi i Q^\vee$ and $2\pi i P^\vee$, respectively, and every other group will correspond to a lattice between these two.
\end{remark}

Let us consider one further example: $\SL(n,\CC)$\index{SLnC@$\SL(n,\CC)$}\index{slnC@$\sl(n,\CC)$}.  It is simply-connected, and its center is $Z(\SL(n,\CC)) = \{n\textrm{th roots of unity}\}$.  We define the \emindexing{projective special linear group}\index{special linear group!projective} to be $\PSL(n,\CC) \defined \SL(n,\CC) / Z(\SL(n,\CC))$\index{PSLnC@$\PSL(n,\CC)$}; the groups with Lie algebra $\sl(n,\CC)$ live between these two, and so correspond to subgroups of $Z(\SL(n,\CC)) \cong \ZZ/n$, the cyclic group with $n$ elements.

We now consider the Cartan $\hh \subseteq \sl(n,\CC)$, thought of as the space of traceless diagonal matrices: $\hh = \{ \<z_1,\dots,z_n\> \in \CC^n \st \sum z_i = 0\}$.  In particular, $\sl(n,\CC)$ is of $A$-type, and so we can identify roots and coroots: $\alpha_i =  \alpha_i^\vee =  \<0,\dots,0,1,-1,0,\dots,0\>$, where the non-zero terms are in the $(i,i+1)$th spots.  Then the coroot lattice $Q^\vee$ is the span of $\alpha_i^\vee$: if $\sum z_i = 0$, then we can write $\<z_1,\dots,z_n\> \in \ZZ^n$ as $z_1 \alpha_1 + (z_1 + z_2)\alpha_2 + \dots + (z_1 + \dots + z_{n-1})\alpha_{n-1}$, since $z_n = - (z_1 + \dots + z_{n-1})$.  The coweight lattice $P^\vee$, on the other hand, is the lattice of vectors $\<z_1,\dots,z_n\>$ with $\sum z_i = 0$ and with $z_i - z_{i+1}$ an integer for each $i \in \{1,\dots,n-1\}$.  In particular, $\sum z_i = z_1+ \bigl( z_1 + (z_2 - z_1)\bigr) + \dots + \bigl( z_1 + (z_2 - z_1) + \dots + (z_n - z_{n-1}) \bigr) = nz_1 + \textrm{integer}$.  Therefore $z_1 \in \ZZ \frac1n$, and $z_i \in z_1 + \ZZ$.  So $P^\vee = Q^\vee \sqcup (\<\frac1n,\dots,\frac1n\> + Q^\vee) \sqcup \dots \sqcup (\< \frac{n-1}n ,\dots,\frac{n-1}n\> + Q^\vee)$.  In this way, $P^\vee / Q^\vee$ is precisely $\ZZ/n$, in agreement with the center of $\SL(n,\CC)$.


\subsection{Definition and General Properties of Algebraic Groups}

\cite[Lecture 41]{notes}

We have mentioned already (\cref{AGroupDefn}) the notion of an ``\indexing{algebraic group}\index{group!algebraic}'', and we have occasionally used some algebraic geometry (notably in the proof of \cref{CartanExistsThm}), but we have not developed that story.  We do so now.

\begin{defn}
  A subset $X \subseteq \CC^n$ is an \emindexing{affine variety}\index{variety} if it is the vanishing set of a set $P\subseteq \CC[x_1,\dots,x_n]$ of polynomials:
  \eqn{ X = V(P) \defined \{ x \in \CC^n \st p(x) = 0 \, \forall p \in P\} }
  Equivalently, $X$ is \emindexing{Zariski closed} (see \cref{ZariskiDefn}).  To any affine variety $X$ with associate an ideal $I(X) \defined \{p\in \CC[x_1,\dots,x_n] \st p|_X = 0\}$.  The \emindexing{coordinate ring} of, or the {\em ring of polynomial functions}\index{polynomial function} on, $X$ is the ring $\Ooo(X) \defined \CC[x]/I(X)$.
\end{defn}
\begin{lemma}
  If $X$ is an affine variety, then $I(X)$ is a radical ideal.  If $I \subseteq J$, then $V(I) \supseteq V(J)$, and conversely if $X \subseteq Y$ then $I(X) \supseteq I(Y)$.  It is clear from the definition that if $X$ is an affine varienty, then $V(I(X)) = X$; more generally, we can define $I(X)$ for any subset $X\supseteq \CC^n$, whence $V(I(X))$ is the Zariski closure of $X$.
\end{lemma}
\begin{defn}
  A {\em morphism of affine varieties}\index{morphism!of affine varieties}\index{affine variety!morphism} is a function $f:X \to Y$ such that the coordinates on $Y$ are polynomials in the coordinates of $X$.  Equivalently, any function $f: X\to Y$ gives a homomorphism of algebras $f^\# : \Fun(Y) \to \Fun(X)$, where $\Fun(X)$ is the space of all $\CC$-valued functions on $X$.  A function $f: X\to Y$ is a morphism of affine varieties if $f^\#$ restricts to a map $f^\#: \Ooo(Y) \to \Ooo(X)$.
\end{defn}
\begin{lemmadef}
  Any point $a\in \CC^n$ gives an \emindexing{evaluation map} $\ev_a: p \mapsto p(a): \CC[x_1,\dots,x_n] \to \CC$.  If $X$ is an affine variety, then $a\in X$ if and only if $I(X) \subseteq \ker \ev_a$ if and only if $\ev_a: \Ooo(X) \to \CC$ is a morphism of affine varieties.
\end{lemmadef}
\begin{cor}
  The algebra $\Ooo(X)$ determines the set of evaluation maps $\Ooo(X) \to \CC$, and if $\Ooo(X)$ is presented as a quotient of $\CC[x_1,\dots,x_n]$, then it determines $X \subseteq \CC^n$.  A morphism $f$ of affine varieties is determined by the algebra homomorphism $f^\#$ of coordinate rings, and conversely any such algebra homomorphism determines a morphism of affine varieties.  Thus the category of affine varieties is precisely the opposite category to the category of finitely generated commutative algebras over $\CC$.
\end{cor}

\begin{lemmadef}
  The category of affine varieties contains all finite products.  The {\em product}\index{product!of affine varieties} of affine varieties $X \subseteq \CC^m$ and $Y \subseteq \CC^l$ is $X\times Y \subseteq \CC^{m+l}$ with $\Ooo(X \times Y) \cong \Ooo(X) \otimes_\CC \Ooo(Y)$.
\end{lemmadef}
\begin{proof}
  The maps $\Ooo(X),\Ooo(Y) \to \Ooo(X\times Y)$ are given by the projections $X\times Y \to X,Y$.  The map $\Ooo(X) \otimes \Ooo(Y) \to \Ooo(X\times Y)$ is an isomorphism because all three algebras are finitely generated and the evaluation maps separate functions.
\end{proof}

We recall \cref{AGroupDefn}:
\begin{defn}
  An {\em affine algebraic group}\index{algebraic group} is a group object in the category of affine varieties.  We will henceforth drop the adjective ``affine'' from the term ``algebraic group'', as we will never consider non-affine algebraic groups.
  
  Equivalently, an algebraic group is a finitely generated commutative algebra $\Ooo(G)$\index{OG@$\Ooo(G)$} along with algebra maps
  \begin{description}
   \item[comultiplication] $\Delta : \Ooo(G) \to \Ooo(G) \otimes_\CC \Ooo(G)$ dual to the multiplication $G\times G \to G$
   \item[antipode] $\anti: \Ooo(G) \to \Ooo(G)$ dual to the inverse map $G\to G$
   \item[counit] $\epsilon = \ev_e : \Ooo(G) \to \CC$
  \end{description}
  The group axioms \cref{GroupAssoc,GroupId,GroupInv} are equivalent to the axioms of a commutative Hopf algebra\index{Hopf algebra!commutative} (\cref{HopfDefn}).
\end{defn}
\begin{lemmadef}
  Let $A$ be a Hopf algebra.  An algebra ideal $B \subseteq A$ is a \emindexing{Hopf ideal}\index{ideal!Hopf} if $\Delta(B) \subseteq B \otimes A + A\otimes B \subseteq A\otimes A$.  An ideal $B \subseteq A$ is Hopf if and only if the Hopf algebra structure on $A$ makes the quotient $B/A$ into a Hopf algebra.
\end{lemmadef}
\begin{defn}
  A commutative but not necessarily reduced Hopf algebra is a \emindexing{group scheme}.
\end{defn}
\begin{defn}
  An affine variety $X$ over $\CC$ is \emindexing{smooth} if $X$ is a manifold.
\end{defn}
\begin{prop}
  An algebraic group is smooth.
\end{prop}
\begin{proof}
  Let $E = \ker \epsilon$.  Since $e \cdot e = e$, we see that the following diagram commutes:
  \eqn{ \begin{tikzpicture}[baseline=(X.base),scale=1.5]
    \path coordinate (X) +(0,.5) node (G) {$\Ooo(G)$} +(2,.5) node (GG) {$\Ooo(G) \otimes \Ooo(G)$} +(0,-.5) node (C) {$\CC$} +(2,-.5) node (CC) {$\CC \otimes \CC$};
    \draw[->] (G) -- node[auto] {$\scriptstyle \Delta$} (GG);
    \draw[->] (G) -- node[auto] {$\scriptstyle \epsilon$} (C);
    \draw[->] (GG) -- node[auto] {$\scriptstyle \epsilon \otimes \epsilon$} (CC);
    \draw[<->] (CC) -- node[auto,swap] {$\sim$} (C);
  \end{tikzpicture} }
  In particular, 
  \eqn[HopfIdeal]{\Delta E \subseteq E \otimes \Ooo(G) + \Ooo(G) \otimes E,} and so $E$ is a Hopf ideal, and $\Ooo(G) / E$ is a Hopf algebra.   Moreover, \cref{HopfIdeal} implies that $\Delta(E^n) \subseteq \sum_{k+l = n} E^k \otimes E^l$, and so $\Delta$ and $\anti$ induce maps $\tilde\Delta$ and $\tilde\anti$ on $R = \gr_E\Ooo(G) \defined \bigoplus_{k\in \NN}E^k / E^{k+1}$.  In particular, $R$ is a \emindexing{graded Hopf algebra}\index{Hopf algebra!graded}, and is generated as an algebra by $R_1 = E / E^2$.  Moreover, if $x\in R_1$, then $x$ is primitive: $\Delta x = x\otimes 1 + 1\otimes x$.
  
  Since $R_1 = E/E^2$ is finitely dimensional, $R$ is finitely generated; let $R = \CC[y_1,\dots,y_n] / J$ where $n = \dim G$ and $J$ is a Hopf ideal of the Hopf algebra $\CC[y_1,\dots,y_n]$ with the generators $y_i$ all primitive.  We can take the $y_i$s to be a basis of $R_1$, and so $J_1 = 0$.  We use the fact that $\CC[y_1,\dots,y_n] \otimes \CC[y_1,\dots,y_n] = \CC[y_1,\dots,y_n,z_1,\dots,z_n]$, and that the antipode $\Delta$ is given by $\Delta: f(y) \mapsto f(y + z)$.  Then a minimal-degree homogeneous element of $J$ must be primitive, so $f(y + z) = f(y) + f(z)$, which in characteristic zero forces $f$ to be homogeneous of degree $1$.  A similar calculation with the antipode forces the minimal-degree homogeneous elements $f\in J$ to satisfy $\anti f = -f$.
  
  In particular, $\gr_E\Ooo(G)$ is a polynomial ring.  We leave out the fact from algebraic geometry that this is equivalent to $G$ being smooth at $e$.  But we have shown that the Hopf algebra maps are smooth, whence $G$ is smooth at every point.
\end{proof}
\begin{cor}
  An algebraic group over $\CC$ is a Lie group.
\end{cor}

Recall that if $G$ is a Lie group with $\Ccc(G)$ the algebra of smooth functions on $G$, and if $\gg = \Lie(G)$, then $\Uu\gg$ acts on $\Ccc(G)$ by left-invariant differential operators, and indeed is isomorphic to the algebra of left-invariant differential operators.

\begin{defn}
  Let $G$ be a group.
  A subalgebra $S \subseteq \Fun(G)$ is {\em left-invariant}\index{left-invariant algebra}\index{algebra!left-invariant} if for any $s\in S$ and any $g\in G$, the function $h\mapsto s(g^{-1}h)$ is an element of $S$.  Equivalently, we define the action $G \acts \Fun(G)$ by $gs = s\circ g^{-1}$; then a subalgebra is left-invariant if it is fixed by this action.
\end{defn}
\begin{lemma} \label{LieAlgAlgLemma}
  Let $S \subseteq \Fun(G)$ be a left-invariant subalgebra, and let $s\in S$ be a function such that $\Delta s = \{ (x,y) \mapsto s(xy) \} \subseteq \Fun(G\times G)$ is in fact an element of $S \otimes S \subseteq \Fun(G) \otimes \Fun(G) \mono \Fun(G\times G)$.  Then let $\Delta s = \sum s_1 \otimes s_2$, where we suppress the indices of the sum.  The action $G\acts S$ is given by
  \eqn{ g: s \mapsto \sum s_1(g^{-1}) s_2 }
\end{lemma}
\begin{cor}
  Let $u$ be a left-invariant differential operator and $s\in S$ as in \cref{LieAlgAlgLemma}, where $S \subseteq \Ccc(G)$ is a left-invariant algebra of smooth functions.  Then $us \in S$.
\end{cor}
\begin{proof}
  The left-invariance of $u$ implies that $u(gs) = g\,u(s)$.  Since $s(g^{-1}) \in \CC$, we have:
  \eqn{ u(gs)(h) = u\left( \sum s_1(g^{-1})s_2\right)(h) = \sum s_1(g^{-1}) \,u(s_2)(h) }
  Let $h = e$.  Then $\sum s_1(g^{-1})\, u(s_2)(e) = u(gs)(e) = g(us)(e) = (us)(g^{-1})$.  In particular:
  \eqn[LieAlgAldEqn1]{ (us)(g) = \sum s_1(g)\,u(s_2)(e)}
  But $(us_2)(e)$ are numbers.  Thus $us \in S$.
\end{proof}
\begin{cor}
  Let $G$ be an algebraic group, with Lie algebra $\gg= \Lie(G)$.  Then $\Uu\gg$ acts on $\Ooo(G)$ by left-invariant differential operators.
  
  Since a differential operator is determined by its action on polynomials, we have a natural embedding $\Uu\gg \mono \Ooo(G)^*$ of vector spaces.
\end{cor}

\begin{lemma}
  Let $u,v \in \Uu\gg$, where $G$ is an algebraic group.  Then $uv\,s = \sum u(s_1)\,v(s_2)(e)$.
\end{lemma}
\begin{proof}  This follows from \cref{LieAlgAldEqn1}.  \end{proof}
\begin{cor} \label{LieAlgAldLemma2}
  For each differential operator $u\in \Uu\gg$, let $\lambda_u \in \Ooo(G)^*$ be the map $\lambda_u:s \mapsto u(s)(e)$.  Then $\lambda_{uv}(s) = \sum \lambda_u(s_1) \,\lambda_v(s_2)$.
\end{cor}

\begin{lemma}
  Let $A$ be any (counital) coalgebra, for example a Hopf algebra.  Then $A^*$ is naturally an algebra: the map $A^* \otimes A^* \to A^*$ is given by $\<\mu \nu,a \> \defined \<\mu \otimes \nu,\Delta a\>$, and $\epsilon: A \to \CC$ is the unit $\epsilon \in A^*$.
\end{lemma}
\begin{remark}
  The dual to an algebra is not necessarily a coalgebra; if $A$ is an algebra, then it defines a map $\Delta: A^* \to (A\otimes A)^*$, but if $A$ is infinite-dimensional, then $(A\otimes A)^*$ properly contains $A^* \otimes A^*$.
\end{remark}
\begin{remark}
  Following the historical precedent, we take the pairing $(A^* \otimes A^*) \otimes (A\otimes A)$ to be $\<\mu \otimes \nu,a\otimes b\> = \<\mu,a\>\<\nu,b\>$.  This is in some sense the wrong pairing --- it corresponds to writing $(A\otimes B)^* = A^* \otimes B^*$ for finite-dimensional vector spaces $A,B$, whereas $B^* \otimes A^*$ would be more natural --- and is ``wrong'' in exactly the same way that the ``$-1$'' in the definition of the left action of $G$ on $\Fun(G)$ is wrong.
\end{remark}

\begin{prop}
  The embedding $\Uu\gg \mono \Ooo(G)^*$ is given by the map $u \mapsto \lambda_u$ in \cref{LieAlgAldLemma2}, and is an algebra homomorphism.
\end{prop}

\begin{defn}
  Let $G$ be any group; then we define the \emindexing{group algebra} $\CC[G]$ of $G$ to be the free vector space on the set $G$, with the multiplication given on the basis by the multiplication in $G$.  The unit $e\in G$ becomes the unit $1 \cdot e \in \CC[G]$.
\end{defn}
\begin{lemma}  
  If $G$ is an algebraic group, then $\CC[G] \mono \Ooo(G)^*$ is an algebra homomorphism given on the basis $g \mapsto \ev_g$.
\end{lemma}

\subsection{Constructing $G$ from $\gg$}

\cite[Lectures 42 and 43]{notes}

A Lie algebra $\gg$ does not determine the group $G$ with $\gg = \Lie(G)$.  We will see that the correct extra data consists of prescribed representation theory.  Throughout the discussion, we gloss the details, merely waving at the proofs of various statements.

\begin{lemmadef}
  Let $G$ be an algebraic group.  A finite-dimensional module $G\acts V$ is {\em algebraic}\index{algebraic module}\index{module!algebraic} if the map $G \to \GL(V)$ is a morphism of affine varieties.
  
  Any finite-dimensional algebraic (left) action $G\acts V$ of an algebraic group $G$ gives rise to a (left) \emindexing{coaction} $V^* \to \Ooo(G) \otimes V^*$:
  \eqn{ \begin{tikzpicture}[auto,baseline=(X.base)]
  \coordinate (X) at (0,1);
  \node  (v) at (0,2) {$V^*$};  \node (ov1) at (5,2) {$\Oo(G) \otimes V^*$}; 
  \node (ov2) at (0,0)  {$\Oo(G) \otimes V^*$};  \node (oov) at (5,0) {$\Oo(G) \otimes \Oo(G) \otimes V^*$}; \\
  \draw[->] (v) to node {\scriptsize \rm coact} (ov1); \draw[->] (v) to node {\scriptsize \rm coact} (ov2);
  \draw[->] (ov2) to node {\scriptsize \rm id $\otimes$ coact} (oov);
  \draw[->] (ov1) to node {\scriptsize \rm comult $\otimes$ id} (oov);
\end{tikzpicture} }
  This in turn gives rise to a (left) action $\Ooo(G)^* \acts V$, which specializes to the actions $G \acts V$ and $\Uu\gg \acts V$ under $G \mono \CC[G] \mono \Ooo(G)^*$ and $\Uu\gg \mono \Ooo(G)^*$.
\end{lemmadef}

We will take the following definition, referring the reader to \cite{ES} for the connections between rigid categories and Hopf algebras, and \cite{BK} and references therein for a thorough category-theoretic discussion.
\begin{defn}
  A \emindexing{rigid category} is an abelian category $\Mm$ with a (unital) monoidal product and duals.  We will write the monoidal product as $\otimes$.
  
  A {\em rigid subcategory} of $\Mm$ is a full subcategory that is a tensor category with the induced abelian and tensor structures.  I.e.\ it is a full subcategory containing the zero object and the monoidal unit, and closed under extensions, tensor products, and duals.
\end{defn}
\begin{defn}
  A rigid category $\Mm$ is \emindexing{finitely generated} if for some finite set of objects $V_1,\dots,V_n \in \Mm$, any object is a subquotient of some tensor product of $V_i$s (possibly with multiplicities).  Of course, by letting $V_0 = V_1 \oplus \dots \oplus V_n$, we see that any finitely generated rigid category is in fact generated by a single object.
\end{defn}
\begin{eg}
  For any Lie algebra $\gg$, the category \cat{$\gg$-mod} of finite-dimensional representations of $\gg$ is a tensor category; indeed, if $U$ is any Hopf algebra, then \cat{$U$-mod} is a tensor category.
\end{eg}
\begin{defn}
  Let $\gg$ be a finite-dimensional Lie algebra over $\CC$, and let $\Mm$ be a rigid subcategory of \cat{$\gg$-mod}.  By definition, for each $V \in \Mm$, we have a linear map $\Uu\gg \to \End V$.  Thus for each linear map $\phi: \End V \to \CC$ we can construct a map $\{\Uu\gg \to \End V \overset{\phi}\to \CC\} \in \Uu\gg^*$; we let $A_\Mm \subseteq \Uu\gg^*$ be the set of all such maps.  Then $A_\Mm$ is the set of {\em matrix coefficients}\index{matrix coefficient} of $\Mm$.  Indeed, for each $V$, the maps $\Uu\gg \to \End V \to \CC$ are the matrix coefficients of the action $\gg \acts V$.  In particular, for each $V\in \Mm$, the space $(\End V)^*$ is naturally a subspace of $A_\Mm$, and $A_\Mm$ is the union of such subspaces.
\end{defn}
\begin{lemma}
  If $\Mm$ is a rigid subcategory of \cat{$\gg$-mod}, then $A_\Mm$ is a subalgebra of the commutative algebra $\Uu\gg^*$.  Moreover, $A_\Mm$ is a Hopf algebra, with comultiplication dual to the multiplication in $\Uu\gg \otimes \Uu\gg \to \Uu\gg$.
\end{lemma}
\begin{proof}
  The algebra structure on $A = A_\Mm$ is straightforward: the multiplication and addition stem from the rigidity of $\Mm$, the unit is $\epsilon: \Uu\gg \to \CC \isom \Cc$, and the subtraction is not obvious but is straightforward; it relies on the fact that $\Mm$ is abelian, and so contains all subquotients.
  
  We will explain where the Hopf structure on $A$ comes from --- since $\Uu\gg$ is infinite-dimensional, $\Uu\gg^*$ does not have a comultiplication in general.  But $\Mm$ consists of finite-dimensional representations; if $V \in \Mm$, then we send $\{\Uu\gg \to \End V \overset{\phi}\to \CC\} \in A$ to $\{(\End V \otimes \End V) \overset{\textrm{multiply}}{\longto} \End V \overset{\phi} \to \CC\} \in (\End V)^* \otimes (\End V)^*\subseteq A \otimes A$.
  
  That this is dual to the multiplication in $\Uu\gg$ comes from the fact that $\Uu\gg \to \End V$ is an algebra homomorphism.
\end{proof}
\begin{cor}
  The map $\Uu\gg \to A^*$ dual to $A \mono \Uu\gg^*$ is an algebra homomorphism.
\end{cor}
\begin{prop} \label{thepropabove}
  Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod}.  Then $A_\Mm = \Ooo(G)$ for some algebraic group $G$.
\end{prop}
\begin{proof}
  If $\Mm$ is finitely generated, then there is some finite-dimensional representation $V_0 \in \Mm$ so that $(\End V_0)^*$ generates $A_\Mm$.  Then $A_\Mm$ is a finitely generated commutative Hopf algebra, and so $\Ooo(G)$ for some algebraic group $G$.
\end{proof}

\begin{lemma} \label{thepreviouslemma}
  Let $\gg$ be a finite-dimensional Lie algebra, $\Mm$ a finitely-generated rigid subcategory of \cat{$\gg$-mod} , and $G$ the algebraic group corresponding to the algebra $A_\Mm$ of matrix coefficients of $\Mm$.  We will henceforth write $\Ooo(G)$ for $A_\Mm$.  Then $G$ acts naturally on each $V \in \Mm$.
\end{lemma}
\begin{proof}
  Let $\{v^1,\dots,v^n\}$ be a basis of $V$ and $\{\xi_1,\dots,\xi_n\}$ the dual basis of $V^*$.  For each $i$, we define $\lambda_i: V \to \Ooo(G)$ by $v \mapsto \{ u \mapsto \<\xi_i,uv\>\}$ where $v\in V$ and $u\in \End V$.  \
  Then we define $\sigma: V \to V \otimes \Ooo(G)$ a right coaction of $\Ooo(G)$ on $V$ by $v \mapsto \sum_{i=1}^n v^i \otimes \lambda_i(v)$.  It is a coaction because  $uv = \sum_{i=1}^n v^i\lambda_i(v)(u)$ by construction.  In particular, it induces an action $G \acts V$.
\end{proof}

\begin{prop}
 Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod} that contains a faithful representation of $\gg$.  Then the map $\Uu\gg \to A_\Mm^*$ is an injection.
\end{prop}

\begin{proof}
  Let $\sigma: G \acts V$ as in the proof of \cref{thepreviouslemma}.  Then the induced representation $\Lie(G) \acts V$ is by contracting $\sigma$ with point derivations.  But $\gg \acts V$ and the map $\Uu\gg \to \Ooo(G)^*$ maps $x\in \gg$ to a point derivation since $x\in \gg$ is primitive.  Thus the following diagram commutes for each $V \in \Mm$:
  \eqn{ \begin{tikzpicture}[baseline=(X.base), scale=1.5]
    \path coordinate (X) +(0,.5) node (g) {$\gg$} +(1.5,.5) node (G) {$\Ooo(G)^*$} +(0,-.5) node (L) {$\Lie(G)$} +(1.5,-.5) node (V) {$\gl(V)$};
    \draw[->] (g) -- (G);
    \draw[->] (g) -- (L);
    \draw[->] (L) -- (V);
    \draw[->] (G) -- (V);
  \end{tikzpicture}}
  The map $\gg \to \Lie(G)$ does not depend on $V$.  Thus, if $\Mm$ contains a faithful $\gg$-module, then $\Uu\gg \mono \Uu\Lie(G) \mono \Ooo(G)^*$.
\end{proof}

\begin{eg}
  Let $\gg = \CC$ be one-dimensional, and let $\Mm$ be generated by one-dimensional representations $V_\alpha$ and $V_\beta$, where the generator $x\in \gg$ acts on $V_\alpha$ by multiplication by $\alpha$, and on $V_\beta$ by $\beta$.  Then $\Mm$ is generated by $V_\alpha \oplus V_\beta$, and $x$ acts as the diagonal matrix $\left[ \begin{matrix} \alpha & \\ & \beta \end{matrix}\right]$.  Let $\alpha,\beta \neq 0$, and let $\alpha \not\in \QQ\beta$.  Then $\Lie(G)$ will contain all diagonal matrices, since $\alpha / \beta \not\in \QQ$, but $\gg \mono \Lie(G)$ as a one-dimensional subalgebra.  The group $G$ is the complex torus, and the subgroup corresponding to $\gg \subseteq \Lie(G)$ is the irrational line.
 \end{eg}

\begin{prop}
  Let $V_0$ be the generator of $\Mm$ satisfying the conditions of \cref{thepropabove}, and let $W$ be a neighborhood of $0\in \gg$.  Then the image of $\exp(W)$ is Zariski dense in $G$.
\end{prop}
\begin{proof}
  Assume that $\Mm$ contains a faithful representation of $\gg$; otherwise, mod out $\gg$ by the kernel of the map $\gg \to \Lie(G)$.  Thus, we may consider $\gg \subseteq \Lie(G)$, and let $H \subseteq G$ be a Lie subgroup with $\gg = \Lie(H)$.  Let $f \in \Ooo(G)$ and $u \in \Uu\gg$; then the pairing $\Uu\gg \otimes \Ooo(G) \to \CC$ sends $u\otimes f \mapsto u(f|_H)(e)$.  In particular, the pairing depends only on a neighborhood of $e\in H$, and hence only on a neighborhood $W \ni 0$ in $\gg$.  But the pairing is nondegenerate; if the Zariski closure of $\exp W$ in $G$ were not all of $G$, then we could find $f,g \in \Ooo(G)$ that agree on $\exp W$ but that have different behaviors under the pairing.
\end{proof}

\begin{defn}
  Let $\Mm$ be a finitely-generated rigid subcategory of \cat{$\gg$-mod}, and let $G$ be the corresponding algebraic group as in \cref{thepropabove}.  Then $\gg$ is \emindexing{algebraically integrable} with respect to $\Mm$ if the map $\gg \to \Lie(G)$ is an isomorphism.  In particular, $\Mm$ must contain a faithful representation of $\gg$.
\end{defn}

\begin{eg} \label{abelianhenceintegrable}
  Let $\gg$ be a finite-dimensional abelian Lie algebra over $\CC$, and let $X \subseteq \gg^*$ be a lattice of full rank, so that $X \otimes_\ZZ \CC = \gg^*$.  Let $\{\xi_1,\dots,\xi_n\}$ be a $\ZZ$-basis of $X$ and hence a $\CC$-basis of $\gg^*$, and let $\Mm = \{\bigoplus \CC_\lambda \st \lambda \in X\}$, where $\gg \acts \CC_\lambda$ by $z \mapsto \lambda(z)\times$.  Then $V_0 = \bigoplus \CC_{\xi_i}$ is a faithful representation of $\gg$ in $\Mm$ and generates $\Mm$.
  
  Then $G \subseteq \GL(V_0)$ is the Zariski closure if $\exp \gg$, and for $z\in \gg$, $\exp(z_1,\dots,z_n)$ is the diagonal matrix whose $(i,i)$th entry is $e^{\xi_i(z)}$.  Thus $G$ is a torus $T \cong (\CC^\times)^n$, with $\Oo(T) = \CC[t_1^{\pm 1},\dots,t_n^{\pm 1}]$.  In particular, $\gg$ is algebraically integrable with respect to $\Mm$, since $X$ is a lattice.
\end{eg}

\begin{prop} \label{sumhenceintegrable}
  Let $\gg$ be a finite-dimensional Lie algebra and $\Mm$ a finitely generated rigid subcategory of \cat{$\gg$-mod} containing a faithful representation.  Suppose   that $\gg = \bigoplus_{i=1}^r \gg_i$ as a vector space, where each $\gg_i$ is a Lie subalgebra of $\gg$; then $\Mm$ embeds in \cat{$\gg_i$-mod} for each $i$.  If each $\gg_i$ is algebraically integrable with respect to (the image of) $\Mm$, then so is $\gg$.
\end{prop}
\begin{proof}
  Let $G$, $G_i$ be the algebraic groups corresponding to $\gg \acts \Mm$ and to $\gg_i \acts \Mm$.  Then for each $i$ we have a map $G_i \to G$.  Let $H\subseteq G$ be the subgroup of $G$ corresponding to $\gg \subseteq \Lie(G)$.  Consider the map $m: G_1 \times \cdots \times G_r \to G$ be the function that multiplies in the given order; it is not a group homomorphism, but it is a morphism of affine varieties.  Since each $G_i \to G$ factors through $H$, and since $H$ is a subgroup of $G$, the map $m$ factors through $H$.  Indeed, the differential of $m$ at the identity is the sum map $\bigoplus \gg_i \to \gg$.
  
  Thus we have an algebraic map $m$, with Zariski dense image.  But it is a general fact that any such map (a \emindexing{dominant morphism}\index{morphism!dominant}) is dimension non-increasing.  Therefore $\dim G \leq \dim (G_1 \times \cdots \times G_r) = \dim \gg$, and so $\gg = \Lie(G)$.
\end{proof}

\begin{thm}[Semisimple Lie algebras are algebraically integrable] \label{algint1thm}
  Let $\gg$ be a semisimple finite-dimensional Lie algebra over $\CC$, and let $\hh \subseteq \gg$ be its Cartan subalgebra and $Q$ and $P$ the root and weight lattices.  Let $X$ be any lattice between these: $Q \subseteq X \subseteq P$.  Let $\Mm$ be the category of finite-dimensional $\gg$-modules with highest weights in $X$.  Then $\Mm$ is finitely generated rigid and contains a faithful representation of $\gg$, and $\gg$ is algebraically integrable with respect to $\Mm$.
\end{thm}
\begin{proof}
  Let $V \in \Mm$; then its highest weights are all in $X$, and so all its weights are in $X$ since $X \supseteq Q$.  Moreover, the decomposition of $V$ into irreducible $\gg$-modules writes $V = \bigoplus L_\lambda$, where each $\lambda \in P_+ \cap X$.  This shows that $\Mm$ is rigid.  It contains a faithful representation because the representation of $\gg$ corresponding to the highest root is the adjoint representation, and the highest root is an element of $Q$ and hence of $X$.  Moreover, $\Mm = \{ \bigoplus V_\lambda \st \lambda \in P_+ \cap X\}$ is finitely generated: $P_+ \cap X$ is $\ZZ_{\geq 0}$-generated by finitely many weights.
  
  We recall the triangular decomposition (c.f.\ \cref{triangulardecompprop}) of $\gg$: $\gg = \nn_- \oplus \hh \oplus \nn_+$.  Then $\hh$ is abelian and acts on modules in $\Mm$ diagonally; in particular, $\hh$ is algebraically integrable by \cref{abelianhenceintegrable}.  On the other hand, on any $\gg$-module, $\nn_+$ and $\nn_-$ act by strict upper- and strict lower-triangular matrices, and the matrix exponential restricted to strict upper- (lower-) triangular matrices is a polynomial.  In particular, by finding a faithful generator of $\Mm$ (for example, the sum of the generators plus the adjoint representation), we see that $\nn_\pm$ are algebraically integrable.  The conclusion follows by \cref{sumhenceintegrable}.
\end{proof}
\begin{thm}[Classification of Semisimple Lie Groups over $\CC$]
  Let $\gg$ be a finite-dimensional semisimple Lie algebra over $\CC$.  Any connected Lie group $G$ with $\Lie(G)$ is semisimple; in particular, the algebraic groups constructed in \cref{algint1thm} comprise all integrals of $\gg$.
\end{thm}
\begin{proof}
  Let $\tilde G$ be the connected and simply connected Lie group with $\Lie(G) = \gg$; then any integral of $\gg$ is a quotient of $\tilde G$ be a discrete and hence central subgroup of $\tilde G$, and the integrals are classified by the kernels of these quotients and hence by the subgroups of the center $Z(\tilde G)$.  Let $G_X$ be the algebraic group corresponding to $X$.  Since $Z(G_P) = P/Q$, it suffices to show that $G_P$ is connected and simply connected.
  
  We show first that $G_X$ is connected.  It is an affine variety; $G_X$ is connected if and only if $\Ooo(G_X)$ is an integral domain.  Since $G_X$ is the Zariski closure of $\exp W$ for a neighborhood $W$ of $0\in \gg$, and $\exp W$ is connected, so is $G_X$.
  
  Let $U_\pm$ be the image of $\exp(\nn_\pm)$ in $G_X$, and let $T = \exp(\hh)$.  But $\exp: \nn_\pm \to U_\pm$ is the matrix exponential on strict triangular matrices, and hence polynomial with polynomial inverse; thus $U_\pm$ are simply connected.  
  
  We quote a fact from algebraic geometry: the image of an algebraic map contains a set Zariski open in its Zariski closure.  In particular, since the image of $U_- \times T \times U_+$ is Zariski dense, it contains a Zariski open set, and so the complement of the image must live inside some closed subvariety of $G_X$ with complex codimension at least 1, and hence real codimension at least 2, since locally this subvariety is the vanishing set of some polynomials in $\CC^n$.  So in any one-complex-dimensional slice transverse to this subvariety, the subvariety consists of just some points.  Therefore any path in $G_X$ can be moved off this subvariety and hence into the image of $U_- \times T \times U_+$.
  
  It suffices to consider paths in $G_X$ from $e$ to $e$, and by choosing for each such path a nearby path in $U_-TU_+$, we get a map $\pi_1(U_-TU_+) \onto \pi_1(G_X)$.  On the other hand, by the LU decomposition (see any standard Linear Algebra textbook, e.g.\ \cite{WikiLU}), the map $U_- \times T \times U_+ \to U_-TU_+$ is an isomorphism.  Since $U_\pm$ are isomorphic as affine varieties to $\nn_\pm$, we have:
   \eqn{ \pi_1(U_-TU_+) = \pi_1(U_- \times T \times U_+) = \pi_1(T) }
   And $\pi_1(T) = X^*$, the co-lattice to $X$, i.e.\ the points in $\gg$ on which all of $X$ takes integral values.
   
   Thus, it suffices to show that the map $\pi_1(T) \onto \pi_1(G_P)$ collapses loops in $T$ when $X = P$.  But then $\pi_1(T) = P^* = Q^\vee$ is generated by the simple coroots $\alpha_i^\vee$.  For each generator $\alpha_i^\vee = h_i$, we take $\sl(2)_i \subseteq \gg$ and exponentiate to a map $\SL(2,\CC) \to G$.  Then the loops in $\exp(\RR h_i)$, which generate $\pi_1(T)$, go to loops in $\SL(2,\CC)$ before going to $G$.  But $\SL(2,\CC)$ is simply connected.  This shows that the map $\pi_1(T) \onto \pi_1(G_P)$ collapses all such loops, and $G_P$ is simply connected.
\end{proof}

\section{Conclusion}

\cite[Lecture 44]{notes}

In Chapter 5, we put semisimple Lie algebras over $\CC$ into bijection with possibly-disconnected Dynkin diagrams, and classified all such diagrams.  In the current chapter, we described how to compute the representation theory of any such diagram: the finite-dimensional irreducible representations correspond to elements of the weight lattice of the diagram.  Then we showed that the integrals of any semisimple Lie algebra are algebraic, and correspond to lattices between the weight and root lattice: the simply connected group corresponds to the weight lattice, and the adjoint group corresponds to the root lattice.  The category of representations of the Lie group corresponding to a given lattice is precisely the category of representations whose highest (and hence all) weights lie in the lattice.  We remark that the index of the weight lattice on the root lattice --- the size of the quotient --- is precisely the determinant of the Cartan matrix.

The story we have told can be generalized.  We will not justify it, but only sketch how it goes.

We recall \cref{reductivedefn}.  We say that a complex Lie group $G$ is \emindexing{reductive} if it is linear and if its finite-dimensional representations are completely reducible.  We demand linearity to assure that the group have finite-dimensional representations: for example, an elliptic curve is a complex Lie group but it is compact, so any holomorphic function is constant, so its only finite-dimensional representations are trivial.
Let $G$ be a reductive Lie group and $\gg = \Lie(G)$ its Lie algebra.  Then we have a Levi decomposition $\gg = \ss \oplus \rr$, and if $G$ is reductive, then $\rr = \zz$ is abelian and is the center of $\gg$, so $\gg$ is reductive as a Lie algebra.

Let $\gg$ be a reductive Lie algebra.   Then the Cartan subalgebra $\hh$ of $\gg$ is $\hh_{\ss} \oplus \zz$, where $\hh_\ss$ is the Cartan subalgebra $\ss$.  We have a coroot lattice $Q^\vee \subseteq \hh_\ss$, and we pick a lattice $X^* \supseteq Q^\vee$ such that $X \supseteq Q$ and $X^*$ spans $\hh$.  Then let $\Mm$ be the category of finite-dimensional representations of $\gg$ with weights in $X$ and on which $\hh$ acts diagonally --- this is an extra condition, because $\zz$ need not act diagonally on a representation of $\gg$.  Then the entire story of algebraic integrals applies.  For any \emindexing{root datum} --- a choice of roots $\alpha$ and coroots $\alpha^\vee$ satisfying natural conditions --- we get a Cartan matrix, the Cartan matrix of $\ss$.

In particular, as we will not explain here, the root data $(X,\alpha_i\in X,\alpha_i^\vee\in X^*)$ classify:
\begin{enumerate}
  \item reductive Lie groups over $\CC$
  \item reductive Lie algebras over $\CC$
  \item compact real Lie groups
  \item reductive algebraic groups over any algebraically closed field in any characteristic
  \item {\em group schemes}\index{group scheme}, or ``algebraic groups over $\ZZ$'', and therefore {\em Chevalley groups}\index{Chevalley group}\index{group!Chevalley}, the tensor products of group schemes with finite fields
  \item finite groups of Lie type, essentially the source of the finite simple groups
\end{enumerate}

Some of this story is outlined but mostly not proved in \cite{anton}.  It would make a good second semester for this one-semester course, but in fact the second semester will tell a different story of quantum groups \cite{Qnotes}.

\Section{Exercises}

\begin{enumerate}
  \item Show that the simple complex Lie algebra $\gg$ with root system $G_2$ has a $7$-dimensional matrix representation with the generators shown below. 
  \eqn[hw61gen]{ \begin{array}{ccc}
    e_1 = \left[ \begin{array}{ccccccc}
      0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0
    \end{array} \right] & &
    f_1 = \left[ \begin{array}{ccccccc}
      0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0
    \end{array}\right] \\ \\
    e_2 = \left[ \begin{array}{ccccccc}
      0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0
    \end{array}\right] & &
    f_2 = \left[ \begin{array}{ccccccc}
      0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0
    \end{array}\right] 
  \end{array} }
  \item \begin{enumerate}
    \item Show that there is a unique Lie group $G$ over $\CC$ with Lie algebra of type $G_2$. 
     \item  Find explicit equations of $G$ realized as the algebraic subgroup of $\GL(7, \CC)$ whose Lie algebra is the image of the matrix representation in Problem 1.

  \end{enumerate}
  \item Show that the simply connected complex Lie group with Lie algebra $\so(2n, \CC)$ is a double cover $\Spin(2n, \CC)$ of $\SO(2n, \CC)$, whose center $Z$ has order four. Show that if $n$ is odd, then $Z$ is cyclic, and there are three connected Lie groups with this Lie algebra: $\Spin(2n, \CC)$, $\SO(2n, \CC)$ and $\SO(2n, \CC)/ \{\pm I \}$. If $n$ is even, then $Z \cong (\ZZ/2\ZZ)^2$, and there are two more Lie groups with the same Lie algebra. 

  \item If $G$ is an affine algebraic group, and $\gg$ its Lie algebra, show that the canonical algebra homormorphism $\Uu\gg \to \Ooo(G)^*$  identi�es $\Uu\gg$ with the set of linear functionals on $\Ooo(G)$ whose kernel contains a power of the maximal ideal $\mm = \ker(\ev_e)$.
  \item Show that there is a unique Lie group over $\CC$ with Lie algebra of type $E_8$. Find the dimension of its smallest matrix representation.

  \item Construct a finite dimensional Lie algebra over $\CC$ which is not the Lie algebra of any algebraic group over $\CC$. [Hint: the adjoint representation of an algebraic group on its Lie algebra is algebraic.]

\end{enumerate}

















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\end{thebibliography}


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