Kac-Moody Lie AlgebrasIntroduction

Last update: 13 August 2012

Abstract.
This is a typed version of I.G. Macdonald's lecture notes on Kac-Moody Lie algebras from 1983.

Introduction

Before we get down to serious business, let me begin by telling you a little in general terms about this subject, and what I propose to do (and what I do not propose to do) is this course of lectures. Basically it is an outgrowth of the theory of finite-dimensional complex simple (or semisimple) Lie algebras developed by Killing and E. Cartan nearly 100 years ago. If $𝔤$ is a complex finite-dimensional semisimple Lie algebra, one can associate canonically with $𝔤$ (I will go into details later) a certain matrix $C=\left({C}_{ij}\right)$ of integers, called the Cartan matrix of $𝔤$, which determines $𝔤$ up to isomorphism. This matrix satisfies the following conditions:

 $\left(C\right)$ ${c}_{ii}=2\phantom{\rule{0.2em}{0ex}};\phantom{\rule{1em}{0ex}}{c}_{ij}\le 0\phantom{\rule{0.5em}{0ex}}\text{if}\phantom{\rule{0.5em}{0ex}}i\ne j\phantom{\rule{0.2em}{0ex}};\phantom{\rule{1em}{0ex}}{c}_{ij}=0⇒{c}_{ji}=0$ $\left(P\right)$ All principal minors of $C$ are positive.

Conversely, any matrix of integers satisfying these two conditions is the Cartan matrix of some semisimple Lie algebra $𝔤$, and one can write down generators and relations for $𝔤$ which involve only the integers ${c}_{ij}$. (SOMETHING GOES HERE) (I will write them down explicitly a little later.)

In the late 1960's, V. Kac and R. Moody more or less simultaneously and independently, had the idea of starting with a "generalized Cartan matrix" (satisfying $\left(C\right)$ but not $\left(P\right)$), writing down the same set of generators and relations as in Serr's theorem, and looking at the resulting Lie algebra (which is now infinite-dimensional). These are the so-called Kac-Moody Lie algebras, or Lie algebras defined by generalized Cartain matrices.

Of course, nearly always when one takes an attractive and elegant piece of mathematics, such as the classical theory of finite-dimensional semisimple Lie algebras over $ℂ$, and starts tinkering with it by weakening the axioms, the result is usually of no interest at all. The surprising thing in this case is that one does continue to get a coherent theory (which of course includes the classical theory as a special case) in which all the main features of the classical theory have their counterparts: root system, Weyl group, representations and characters – culminating in a generalization (due to V. Kac) of Weyl's character formula.

Moreover it has become clear during the last 10 years or so that these Kac-Moody Lie algebras impinge on many other areas of mathematics:

 Number Theory (Modular forms) Combinatorics (Partitions, Rogers-Ramanujan identities) Topology (Loop spaces and Loop groups) Linear Algebra (Representations of Quivers) Singularities Completely integrable systems Mechanics and Particle Physics

My aim in this course of lectures is to cover the basic structure and representation theory of Kac-Moody Lie algebras, but not any of the applications listed above. Also I don't propose to assume any particular knowledge of Lie algebras, so I will begin by briefly reviewing some of the basic notions. We shall work always over a field $k$ of characteristic 0 (and fairly soon $k$ will be $ℂ$, the field of complex numbers). A Lie algebra $𝔤$ is then a vector space over $k$ endowed with a bilinear multiplication

 $\left(x,y\right)↦\left[x,y\right]$ (Lie bracket) $𝔤×𝔤\to 𝔤$

satisfying

 (1) $\left[x,x\right]=0$ for all $x\in 𝔤$ (2) $\left[x,\left[y,z\right]\right]+\left[y,\left[z,x\right]\right]+\left[z,\left[x,y\right]\right]=0$

for all $x,y,z\in 𝔤\phantom{\rule{0.5em}{0ex}}$ (Jacobi identity).

By applying (1) to $x+y$ and using the bilinearity of the bracket we have

 (1$\prime$) $\left[x,y\right]=-\left[y,x\right]$

and conversely (1$\prime$) $⇒$ (1) (take $x=y$).

Examples

1. $V$ any vector space over $k$, define $\left[x,y\right]=0$ for all $x,y\in V$. This is an abelian Lie algebra.
2. $A$ any associative $k$–algebra, define $\left[x,y\right]=xy-yx$. Check that (2) holds. We have a Lie algebra L($A$).
3. $A$ any $k$–algebra – i.e. $A$ is a $k$–vector space endowed with a bilinear multiplication $\left(x,y\right)↦xy\phantom{\rule{0.4em}{0ex}}:\phantom{\rule{0.4em}{0ex}}A×A\to A$.
A derivation $d:A\to A$ is a $k$–linear mapping which satisfies
$\left(✶\right)$
$d\left(xy\right)=\left(dx\right)y+x\left(dy\right)\phantom{\rule{3em}{0ex}}\left(x,y\in A\right)$
If ${d}_{1},{d}_{2}$ are derivations, so is $\left[{d}_{1},{d}_{2}\right]={d}_{1}{d}_{2}-{d}_{2}{d}_{1}$ (check this). Again the Jacobi identity is satisfied (the verification is the same as in Ex. 2), so we have a Lie algebra Der($A$).
It follows from $\left(✶\right)$ that
${d}^{n}\left(xy\right)=\sum _{p+q=n}\frac{n!}{p!q!}\left({d}^{p}x\right)\left({d}^{q}y\right)\phantom{\rule{3em}{0ex}}\text{(Leibniz)}$
by induction on $n$; hence if $d$ is nilpotent $\left({d}^{N}=0\phantom{\rule{0.4em}{0ex}}\text{for some}\phantom{\rule{0.4em}{0ex}}N>0\right)$
${e}^{d}=\sum _{n\ge 0}\frac{{d}^{n}}{n!}$
is well-defined (because the sum is finite) and is an automorphism of $A$:
${e}^{d}\left(xy\right)=\sum _{n\ge 0}\frac{{d}^{n}\left(xy\right)}{n!}=\sum _{p,q\ge 0}\frac{{d}^{p}x}{p!}·\frac{{d}^{q}y}{q!}={e}^{d}\left(x\right){e}^{d}\left(y\right)$
so that ${e}^{d}:A\to A$ is a $k$–algebra homomorphism, hence an automorphism because ${e}^{d}·{e}^{-d}=1$.
4. Let $V$ be a finite-dimensional $k$–vector space, $A=$End($V$). Then L($A$) (Ex. 2) is a Lie algebra denoted by $𝔤𝔩\left(V\right)$. If $V={k}^{n}$, so that $A={M}_{n}\left(k\right)$ is the algebra of $n×n$ matrices over $k$, we write ${𝔤𝔩}_{n}\left(k\right)$ in place of $𝔤𝔩\left({k}^{n}\right)$. ${𝔰𝔩}_{n}\left(k\right)=\left\{X\in {𝔤𝔩}_{n}\left(k\right)\phantom{\rule{0.4em}{0ex}}:\phantom{\rule{0.4em}{0ex}}\text{trace}\phantom{\rule{0.2em}{0ex}}X=0\right\}$ is a subalgebra of ${𝔤𝔩}_{n}\left(k\right)$, because $\text{trace}\phantom{\rule{0.2em}{0ex}}\left[X,Y\right]=\text{trace}\phantom{\rule{0.2em}{0ex}}XY-\text{trace}\phantom{\rule{0.2em}{0ex}}YX=0$.
5. $G$ a (real or complex) Lie group, $𝔤=$tangent space ${T}_{e}\left(G\right)$ to $G$ at the identity element $e$ ($k=ℝ$ or $ℂ$ here). $𝔤$ inherits from the group $G$ a Lie algebra structure: roughly speaking, addition $X+Y$ in $𝔤$ corresponds to multiplication in $G$ near the identity, and the bracket $\left[X,Y\right]$ ERROR HERE? to formation of the commutator $xy{x}^{-1}{y}^{-1}$ (for $x,y$ near $e$). Here of course $𝔤$ is finite-dimensional: $\text{dim}\phantom{\rule{0.2em}{0ex}}𝔤=\text{dim}\phantom{\rule{0.2em}{0ex}}G$. This is the origin of the subject: Lie algebra $𝔤=$linear approximation to $G$ at $e$.

Basic concepts

Many notions for groups have counterparts for Lie algebras. This is hardly surprising, given the origin of the subject.

Let $𝔤$ be a Lie algebra. If $𝔞,𝔟$ are vector subspaces (or just subsets) of $𝔤$, let $\left[𝔞,𝔟\right]$ denote the subspace of $𝔤$ spanned by all $\left[x,y\right]$ with $x\in 𝔞,y\in 𝔟$. Observe that $\left[𝔞,𝔟\right]=\left[𝔟,𝔞\right]$ (because $\left[x,y\right]=-\left[y,x\right]$ ).

Subalgebra: a vector subspace $𝔞$ of $𝔤$ is a subalgebra if $\left[𝔞,𝔞\right]\subset 𝔞$ (so that $𝔞$ is a Lie algebra in it's own right).

Ideal: a vector subspace $𝔫$ of $𝔤$ is an ideal of $𝔤$ if $\left[𝔤,𝔫\right]\subset 𝔫$ (normal subgroup).

Quotient algebra: Let $𝔫$ be an ideal in $𝔤$, and form the vector space quotient $𝔤/𝔫$, whose elements are the cosets $\stackrel{‾}{x}=x+𝔫$. Define $\left[\stackrel{‾}{x},\stackrel{‾}{y}\right]=\stackrel{‾}{\left[x,y\right]}$, this does not depend on the choice of representations, and makes $𝔤/𝔫$ into a Lie Algebra. $\left(G/N\right)$

Homomorphism: a homomorphism from $𝔤$ to $𝔥$ is a $k$–linear map $f:𝔤\to 𝔥$ such that $f\left(\left[x,y\right]\right)=\left[f\left(x\right),f\left(y\right)\right]$ for all $x,y\in 𝔤$.
It's kernel $𝔫={f}^{-1}\left(0\right)$ is an ideal in $𝔤$, it's image $𝔞=f\left(𝔤\right)$ is a subalgebra of $𝔥$, and $f$ induces an isomorphism $𝔤/𝔫\stackrel{\sim }{\to }𝔞$

If $x,y\in 𝔤$ are such that $\left[x,y\right]=0$, we say that $x,y$ commute. In particular, if any two elements of $𝔤$ commute, ie if $\left[𝔤,𝔤\right]=0,$ we say that $𝔤$ is abelian (Ex. 1 above).

Centre of $𝔤=z=\left\{x\in 𝔤\phantom{\rule{0.4em}{0ex}}:\phantom{\rule{0.4em}{0ex}}\left[x,𝔤\right]=0\right\}$. It is an ideal in $𝔤$.

Derived algebra $\phantom{\rule{0.5em}{0ex}}𝔤\prime =D𝔤=\left[𝔤,𝔤\right]$ consists of all linear combinations of brackets $\left[x,y\right]$. $D𝔤$ is an ideal in $𝔤$ (by virtue of the Jacobi identity):

$\left[\left[x,y\right],z\right]=-\left[\left[y,z\right],x\right]-\left[\left[z,x\right],y\right]\in D𝔤,$

hence $\left[D𝔤,𝔤\right]\subset D𝔤$. Moreover $𝔤/D𝔤$ is abelian (and $D𝔤$ is the smallest ideal with abelian quotient).
Derived series, upper and lower central series; nilpotent, solvable Lie algebras.

For each $x\in 𝔤$ we define $\text{ad}\left(x\right):𝔤\to 𝔤$ by $\text{ad}\left(x\right)y=\left[x,y\right]$. Then

$\text{ad}:𝔤\to 𝔤𝔩\left(𝔤\right)$

is a homomorphism of Lie algebras, because for all $x,y,z\in 𝔤$ we have

$\begin{array}{ccc}\text{ad}\phantom{\rule{0.2em}{0ex}}\left[x,y\right]·z& =& \left[\left[x,y\right],z\right]\\ & =& -\left[z,\left[x,y\right]\right]\\ & =& \left[x,\left[y,z\right]\right]-\left[y,\left[x,z\right]\right]\phantom{\rule{3em}{0ex}}\text{(Jacobi)}\\ & =& \left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)\left(\text{ad}\phantom{\rule{0.2em}{0ex}}y\right)z-\left(\text{ad}\phantom{\rule{0.2em}{0ex}}y\right)\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)z\\ & =& \left[\text{ad}\phantom{\rule{0.2em}{0ex}}x,\text{ad}\phantom{\rule{0.2em}{0ex}}y\right]·z\end{array}$

Moreover each $\text{ad}\phantom{\rule{0.2em}{0ex}}x$ is a derivation of $𝔤$:

$\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)\left[y,z\right]=\left[\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)y,z\right]+\left[y,\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)z\right]$

This is another equivalent form of the Jacobi identity.

The kernel of ad is the centre $z$ of $𝔤$.

Inner automorphisms

If $x\in 𝔤$ is such that $\text{ad}\phantom{\rule{0.2em}{0ex}}x$ is nilpotent $\left({\left(x\right)}^{N}=0\phantom{\rule{0.2em}{0ex}}\text{for some}\phantom{\rule{0.2em}{0ex}}N>0\right)$ then we can form ${e}^{\text{ad}\phantom{\rule{0.2em}{0ex}}x}$ (Ex. 3 above) which is an automorphism of $𝔤$. The subgroup Int($𝔤$) of Aut($𝔤$) generated by these ${e}^{\text{ad}\phantom{\rule{0.2em}{0ex}}x}$ is the group of inner automorphisms of $𝔤$. It is a normal subgroup of Aut($𝔤$), because if $\phi \in \text{Aut}\left(𝔤\right)$ we have $\phi \left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right){\phi }^{-1}=\text{ad}\phantom{\rule{0.2em}{0ex}}\phi \left(x\right)$ and therefore also

$\phi \left({e}^{\text{ad}\phantom{\rule{0.2em}{0ex}}x}\right){\phi }^{-1}={e}^{\text{ad}\phantom{\rule{0.2em}{0ex}}\phi x}$.

Representations

Let $𝔤$ be a Lie algebra. A representation $\rho$ of $𝔤$ on a $k$–vector space $V$ is by definition a Lie algebra homomorphism $\rho :𝔤\to 𝔤𝔩\left(V\right)$. In other words, for each $x\in 𝔤$ we have a linear transformation $\rho \left(x\right):V\to V$ depending on linearity on $x$:

$\rho \left(\alpha x+\beta y\right)=\alpha \rho \left(x\right)+\beta \rho \left(y\right)\phantom{\rule{3em}{0ex}}\left(x,y\in 𝔤;\phantom{\rule{0.2em}{0ex}}\alpha ,\beta \in k\right)$

and satisfying

$\rho \left(\left[x,y\right]\right)=\rho \left(x\right)\rho \left(y\right)-\rho \left(y\right)\rho \left(x\right)$.

An equivalent notion is that of a $𝔤$module, which is a vector space $V$ on which $𝔤$ acts linearly, i.e. we are given a bilinear mapping

$\left(x,v\right)↦x·v:\phantom{\rule{0.5em}{0ex}}𝔤×V\to V$

satisfying

$\left[x,y\right]·v=x·y·v-y·x·v\phantom{\rule{3em}{0ex}}\left(x,y\in 𝔤;\phantom{\rule{0.2em}{0ex}}v\in V\right)$

To connect the two notions, define $x·v=\rho \left(x\right)v$.

Usual notions of irreducibility, direct sums etc.

Universal enveloping algebra of a Lie algebra

If $G$ is a group, a $G$–module (or representation of $G$) is the same thing as a $kG$–module, where $kG$ is the group algebra of $G$ over $k$. The analogue of this for Lie algebras is the universal enveloping algebra $U\left(𝔤\right)$ of a Lie algebra $𝔤$, which may be defined as follows: for the tensor algebra of the vector space $𝔤$

$T\left(𝔤\right)=\underset{n\ge 0}{\oplus }{T}^{n}\left(𝔤\right)$

where ${T}^{0}\left(𝔤\right)=k,\phantom{\rule{0.5em}{0ex}}{T}^{1}\left(𝔤\right)=𝔤,\phantom{\rule{0.5em}{0ex}}{T}^{n}\left(𝔤\right)=𝔤\otimes \dots \otimes 𝔤\phantom{\rule{0.5em}{0ex}}\left(n\phantom{\rule{0.2em}{0ex}}\text{factors}\right)\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}n\ge 2$.
Let ${J}_{𝔤}$ be the two-sided ideal of IS THIS A T? $U\left(𝔤\right)$ generated by all

$x\otimes y-y\otimes x-\left[x,y\right]\phantom{\rule{3em}{0ex}}\left(x,y\in 𝔤\right)$

and define

$U\left(𝔤\right)=T\left(𝔤\right)/{J}_{𝔤}$.

$U\left(𝔤\right)$ is functional in $𝔤:\phantom{\rule{0.2em}{0ex}}$ if $\phi :𝔤\to 𝔥$ is a homomorphism of Lie algebras, it induces $T\left(\phi \right):T\left(𝔤\right)\to T\left(𝔥\right)$, and

$T\left(\phi \right)\left(x\otimes y-y\otimes x-\left[x,y\right]\right)$
$\phi x\otimes \phi y-\phi y\otimes \phi x-\left[\phi x,\phi y\right]\in {J}_{𝔥}$

so that T($\phi$) maps ${J}_{𝔤}$ into ${J}_{𝔥}$ and hence induces

$U\left(\phi \right):U\left(𝔤\right)\to \text{}\left(𝔥\right)$

U is the left adjoint of the functor L (Ex. 2) from associative algebras to Lie algebras (over $k$): for each Lie algebra $𝔤$ and each associative algebra A, there is a canonical bijection

${\text{Hom}}_{\text{assoc. alg}}\left(U\left(𝔤\right),\text{A}\right)\stackrel{\sim }{\to }{\text{Hom}}_{\text{Lie alg.}}\left(𝔤,\text{L(A)}\right)$

For if $\phi :𝔤\to \text{L(A)}$ is a Lie algebra homomorphism, it is a $k$–linear mapping $𝔤\to \text{A}$ such that

$\phi \left[x,y\right]=\phi \left(x\right)\phi \left(y\right)-\phi \left(y\right)\phi \left(x\right)$. $\phantom{\rule{3em}{0ex}}$(1)

Extend $\phi$ to a homomorphism $\phi :T\left(𝔤\right)\to \text{A}$ by defining

$\phi \left({x}_{1}\otimes \dots \otimes {x}_{n}\right)=\phi \left({x}_{1}\right)\dots \phi \left({x}_{n}\right),$

then the kernel of $\phi$ contains the ideal ${J}_{𝔤}$, by virtue of (1). Hence $\phi$ induces a homomorphism of associative algebras ${\phi }^{#}:U\left(𝔤\right)\to \text{A}$.

In the other direction, let $\theta :U\left(𝔤\right)\to \text{A}$ be a homomorphism of associative algebras, and form the linear mapping

${\theta }^{b}:g↪T\left(𝔤\right)\to U\left(𝔤\right)\stackrel{\theta }{\to }\text{A}$

which is a Lie algebra homomorphism $𝔤\to \text{L(A)}$. Finally verify that the mappings $\phi \to {\phi }^{#},\theta \to {\theta }^{b}$ are inverses of each other.

In particular, if $\rho :𝔤\to 𝔤𝔩\left(V\right)=\text{L(End(}V\text{))}$ is a representation, we have ${\rho }^{#}:U\left(𝔤\right)\to \text{End(}V\text{)}$, i.e. $V$ is a $U\left(𝔤\right)$–module.

The Poincaré-Birkhoff-Witt Theorem

Recall that $U\left(𝔤\right)=T\left(𝔤\right)/{J}_{𝔤};\phantom{\rule{0.2em}{0ex}}T\left(𝔤\right)=\underset{n\ge 0}{\oplus }{T}^{n}\left(𝔤\right)$ is a graded algebra, but ${J}_{𝔤}$ is not a graded ideal, because the generators $x\otimes y-y\otimes x-\left[x,y\right]$ are not homogeneous: $x\otimes y-y\otimes x\in {T}^{2}\left(𝔤\right),\left[x,y\right]\in {T}^{1}\left(𝔤\right)$. So $U\left(𝔤\right)$ is not a graded algebra; but it does carry a filtration, defined as follows: Let

${T}_{n}=\underset{i=0}{\overset{n}{\oplus }}{T}^{i}\left(𝔤\right)$

Let $\pi :T\left(𝔤\right)\to U\left(𝔤\right)$ be the canonical homomorphism, and let

${U}_{n}=\pi \left({T}_{n}\right)$

The ${U}_{n}$ are vector subspaces of $U$:

$k={U}_{0}\subset {U}_{1}\subset \dots \phantom{\rule{2em}{0ex}};\phantom{\rule{0.5em}{0ex}}U=\bigcup _{n\ge 0}{U}_{n}$

and since ${T}_{m}\otimes {T}_{n}\subset {T}_{m+n}$ we have ${U}_{m}·{U}_{n}\subset {U}_{m+n}$, ie $U\left(𝔤\right)$ is a strong filtered associative $k$–algebra. Now form the associated graded algebra: define

${G}^{n}={U}_{n}/{U}_{n-1}\phantom{\rule{2em}{0ex}}\left(n\ge 0;{U}_{-1}=0\right)$

then the multiplication in $U\left(𝔤\right)$ induces bilinear mappings

${G}^{m}×{G}^{n}\to {G}^{m+n}$

namely (for $x\in {U}_{m},y\in {U}_{n}$ )

$\left(x+{U}_{m-1}\right)\left(y+{U}_{n-1}\right)=xy+{U}_{m+n-1}$

which make $G=\text{Gr}\left(U\left(𝔤\right)\right)=\underset{n\ge 0}{\oplus }{G}^{n}$ into a graded associated $k$–algebra. For each $n\ge 1$ we have

commutative with exact rows; $\phi :{T}^{n}\left(𝔤\right)\to {G}^{n}$ is surjective, hence we have a surjective algebra homomorphism

$\phi :T\left(𝔤\right)\to G$

defined by

$\phi \left(x\right)=\pi \left(x\right)+{U}_{n-1}\phantom{\rule{3em}{0ex}}\left(x\in {T}_{n}\right)$

In particular, if $x,y\in 𝔤$ we have

$\pi \left(x\otimes y-y\otimes x\right)=\pi \left(\left[x,y\right]\right)\in {U}_{1}$

and therefore

$\phi \left(x\otimes y-y\otimes x\right)=0\phantom{\rule{1em}{0ex}}\text{in}\phantom{\rule{0.5em}{0ex}}{G}_{2}$

Hence the kernel of $\phi$ contains the two-sides ideal $I$ of $T\left(𝔤\right)$ generated by all $x\otimes y-y\otimes x\phantom{\rule{0.5em}{0ex}}\left(x,y\in 𝔤\right)$; now $T\left(𝔤\right)/I$ is by definition the symmetric algebra $S=S\left(𝔤\right)$ of the vector space $𝔤$; hence $\phi$ induces a surjective homomorphism

$\omega :S\left(𝔤\right)\to \text{Gr}\left(U\left(𝔤\right)\right)$

(P – B – W) $\omega$ is an isomorphism.

For the proof, see the standard texts (Bourbaki (Ch. I), Humphreys, Jacoborn)

Let $\sigma :T\left(𝔤\right)\to S\left(𝔤\right)$ be the canonical homomorphism.

Let $V$ be a vector subspace of ${T}^{n}\left(𝔤\right)$ which is mapped isomorphically by $\sigma$ onto ${S}^{n}\left(𝔤\right)$. Then $\pi \left(V\right)\left(\cong V\right)$ is a complement of ${U}_{n-1}$ in ${U}_{n}$.

 Proof. The diagram $Un$ $Sn$ $Gn$ $V⊂Tn$ $θ$ $ω$ $π$ $σ$ commutes, hence $\theta$ maps $V$ isomorphically onto ${G}^{n}={U}_{n}/{U}_{n-1}$ (because $\omega$ is an isomorphism, by P-B-W). Hence the result. $\square$

The canonical map $g↪T\left(𝔤\right)\stackrel{\pi }{\to }U\left(𝔤\right)$ is injective.

We may therefore identify $𝔤$ with it's image in $U\left(𝔤\right)$.

Let ${\left({x}_{\lambda }\right)}_{\lambda \in L}$ be a totally ordered $k$–basis of $𝔤$. Then the elements

${x}_{{\lambda }_{1}}\dots {x}_{{\lambda }_{n}}=\pi \left({x}_{{\lambda }_{1}}\otimes \dots \otimes {x}_{{\lambda }_{n}}\right)$

such that ${\lambda }_{1}\le \dots \le {\lambda }_{n}\phantom{\rule{0.2em}{0ex}}\left(\text{for all}\phantom{\rule{0.2em}{0ex}}n\ge 0\right)$ form a $k$–basis of $U\left(𝔤\right)$.

 Proof. Let ${V}_{n}$ be the subspace of ${V}^{n}\left(𝔤\right)$ spanned by all ${x}_{{\lambda }_{1}}\otimes \dots \otimes {x}_{{\lambda }_{n}}$ with ${\lambda }_{1}\le \dots \le {\lambda }_{n}$. Clearly $\sigma$ maps ${V}_{n}$ isomorphically onto ${S}^{n}\left(𝔤\right)$, hence by Corollary 7.2 $\pi \left({V}_{n}\right)$ is a complement of ${U}_{n-1}$ in ${U}_{n}$. By induction on $n$ it follows that $U\left(𝔤\right)$ is the direct sum of the $\pi \left({V}_{n}\right)$ for all $n\ge 0$. (Corollary 7.4 is also known as the P-B-W theorem). $\square$

Recap

To recapitulate from last time:– to each Lie algebra $𝔤$ we associate $U\left(𝔤\right)$, its universal enveloping algebra: $U\left(𝔤\right)=T\left(𝔤\right)/{J}_{𝔤}$ where $T\left(𝔤\right)$ generated by all $x\otimes y-y\otimes x-\left[x,y\right]\phantom{\rule{0.5em}{0ex}}\left(x,y\in 𝔤\right)$. $𝔤$ embeds in $U\left(𝔤\right)$ (by virtue of the P-B-W theorem) and we identify 𝔤 with its image in $U\left(𝔤\right)$. In $U\left(𝔤\right)$ we have $\left[x,y\right]=xy-yx\phantom{\rule{0.5em}{0ex}}\left(x,y\in 𝔤\right)$. Moreover $U\left(𝔤\right)$ is universal in the following sense: if $\phi :𝔤\to A$ is any $k$–linear mapping of $𝔤$ into an associative $k$–algebra $A$ such that $\phi \left[x,y\right]=\phi \left(x\right)\phi \left(y\right)-\phi \left(y\right)\phi \left(x\right)$ – i.e. if $\phi :𝔤\to \text{L}\left(A\right)$ is a Lie algebra homomorphism, then $\phi$ extends uniquely to a homomorphism ${\phi }^{#}:U\left(𝔤\right)\to A$, as follows :– first extend $\phi$ to $\stackrel{\sim }{\phi }:T\left(𝔤\right)\to A$ in the obvious way:

$\stackrel{\sim }{\phi }\left({x}_{1}\otimes \dots \otimes {x}_{n}\right)=\phi \left({x}_{1}\right)\dots \phi \left({x}_{n}\right)\phantom{\rule{3em}{0ex}}\left({x}_{1},\dots ,{x}_{n}\in 𝔤\right)$

and then observe that ${J}_{𝔤}\subset \text{Ker}\phantom{\rule{0.2em}{0ex}}\stackrel{\sim }{\phi }$, so that $\stackrel{\sim }{\phi }$ induces ${\phi }^{#}:U\left(𝔤\right)\to A$ as desired. Thus $\phi ↦{\phi }^{#}$ is a mapping

Hom Lie alg ( 𝔤, L(A) ) Hom assoc. alg ( U(𝔤)A )

which one easily verifies to be bijective (i.e. $U$ is a left adjoint of the functor L, as I said last time).

Recall also (Corollary 7.4 of P-B-W th.) that if ${\left({x}_{\lambda }\right)}_{\lambda \in \text{L}}$ is an ordered $k$–basis of $𝔤$, then the monomials ${x}_{{\lambda }_{1}}\dots {x}_{{\lambda }_{n}}$ with ${\lambda }_{1}\le \dots \le {\lambda }_{n}$ and $n\ge 0$ (if $n=0$, the product is empy and conventionally is said to be read as 1, the identity element of $U\left(𝔤\right)$) form a $k$–basis of $U\left(𝔤\right)$. This has the following consequence: if

$𝔤=𝔞\oplus 𝔟$

where $𝔞,𝔟$ are subalgebras and $𝔤$ is the direct sum of the vector spaces $𝔞,𝔟$, then $U\left(𝔞\right),U\left(𝔟\right)$ are subalgebras of $U\left(𝔤\right)$ and

$U\left(𝔤\right)=U\left(𝔞\right)U\left(𝔟\right)=U\left(𝔟\right)U\left(𝔞\right)$

(Take ordered bases $\left({y}_{\mu }\right),\left({z}_{\nu }\right)$ of $𝔞,𝔟$ respectively; the monomials ${y}_{{\mu }_{1}}\dots {y}_{{\mu }_{m}}$ with ${\mu }_{1}\le \dots \le {\mu }_{m}$ form a $k$–basis of $U\left(𝔞\right)$, the monomials ${z}_{{\nu }_{1}}\dots {z}_{{\nu }_{n}}$ with ${\nu }_{1}\le \dots \le {\nu }_{n}$ form a $k$–basis of $U\left(𝔟\right)$, and the monomials ${y}_{{\mu }_{1}}\dots {y}_{{\mu }_{m}}{z}_{{\nu }_{1}}\dots {z}_{{\nu }_{n}}$ with ${\mu }_{1}\le \dots \le {\mu }_{m}$ and ${\nu }_{1}\le \dots \le {\nu }_{n}$ form a $k$–basis of $U\left(𝔤\right)$.)

Free Lie algebras

Let $X$ be a set, $k$ a field. We want to define the free Lie algebra Lie($X$) on the set $X$. There are two ways of proceeding: one involves P-B-W, the other doesn't.

(1) Form the free non-associative algebra F($X$) on $X$. How does one do this?

Define inductively sets ${X}_{n},n\ge 1$ by

$\begin{array}{ccc}{X}_{1}& =& X\\ {X}_{2}& =& {X}_{1}×{X}_{1}\\ {X}_{3}& =& \left({X}_{2}×{X}_{1}\right)\bigsqcup \left({X}_{1}×{X}_{2}\right)\\ \multicolumn{3}{c}{\text{and in general}}\\ {X}_{n}& =& \coprod _{p=1}^{n-1}{X}_{p}×{X}_{n-p}\end{array}$

and put $M\left(X\right)=\coprod _{n\ge 1}{X}_{n}$ (disjoint union) (the "free magma" on $X$).

If $a,b\in M\left(X\right)$, say $a\in {X}_{p}$ and $b\in {X}_{q}$, then $\left(a,b\right)\in {X}_{p}×{X}_{q}\subset {X}_{p+q}\subset M$ so we have a multiplication $ab=\left(a,b\right)$ in $M\left(X\right)$. Then $F\left(X\right)$ is the $k$–algebra with $M\left(X\right)$ as basis, i.e. it consists of all finite linear combinations $\sum {\lambda }_{i}{a}_{i}$ with ${\lambda }_{i}\in k$ and ${a}_{i}\in M\left(X\right)$, and multiplication defined in the obvious way:

$\left(\sum {\lambda }_{i}{a}_{i}\right)\left(\sum {\mu }_{j}{b}_{j}\right)=\sum _{i,j}{\lambda }_{i}{\mu }_{j}{a}_{i}{b}_{j}$.

Now let $J$ be the 2-sided ideal in $F\left(X\right)$ generated by all

$xx,\phantom{\rule{2em}{0ex}}x\left(yz\right)+y\left(zx\right)+z\left(xy\right)\phantom{\rule{3em}{0ex}}\left(x,y,z\in F\left(X\right)\right)$

and define

$\text{Lie}\left(X\right)=F\left(X\right)/J$

where $F\left(X\right)$ is a graded algebra and $J$ is a homogeneous ideal. It is clear that Lie($X$) is a Lie algebra and that if

$j:X↪F\left(X\right)\to \text{Lie}\left(X\right)$

is the canonical embedding, then any mapping $\phi$ of the set $X$ into a Lie algebra $𝔤$ extends uniquely to a Lie algebra homomorphism

${\phi }^{#}:\text{Lie}\left(X\right)\to 𝔤$

(extend $\phi$ in the obvious way to $\stackrel{\sim }{\phi }:F\left(X\right)\to 𝔤$ and observe that the generators of $J$ in the kernel of $\stackrel{\sim }{\phi }$, by definition).

In other words, we have a bijection

${\text{Hom}}_{\text{sets}}\left(X,𝔤\right)\stackrel{\sim }{\to }{\text{Hom}}_{\text{Lie alg.}}\left(\text{Lie}\left(X\right),𝔤\right)$

i.e. the functor Lie (from sets to Lie algebras) is a left adjoint of the forgetful functor $\Phi$ (from Lie algebras to sets).

(2) Let $A\left(X\right)$ be the free associative algebra on $X$ ($=F\left(X\right)/I$, where $I$ is the 2-sided ideal generated by all $\left(xy\right)z-x\left(yz\right)\phantom{\rule{0.5em}{0ex}}\left(x,y,z\in F\left(X\right)\right)$). The embedding

$X↪A\left(X\right)=\text{L}\left(A\left(X\right)\right)$

induces, as we have just seen, a Lie algebra homomorphism

$\alpha :\text{Lie}\left(X\right)\to \text{L}\left(A\left(X\right)\right)$

hence also

$\beta :U\left(\text{Lie}\left(X\right)\right)\to A\left(X\right)$

But also we have a mapping $X↪\text{Lie}\left(X\right)\to U\left(\text{Lie}\left(X\right)\right)$, hence (by the universal property of $A\left(X\right)$) a homomorphism of associative algebras

$\gamma :A\left(X\right)\to U\left(\text{Lie}\left(X\right)\right)$

Check that these two homomorphisms $\beta ,\gamma$ are inverses of each other, hence that

$U\left(\text{Lie}\left(X\right)\right)\cong A\left(X\right)$

By P-B-W, Lie($X$) embeds in $U\left(\text{Lie}\left(X\right)\right)$, hence in $A\left(X\right)$, so that the mapping $\alpha$ above is injective. In other words, the free Lie algebra Lie($X$) may be described as the subalgebra of $\text{L}\left(A\left(X\right)\right)$ generated by $X$.

We have $\beta k=alpha$ and $\gamma \alpha =k$, also $\alpha j=i$, hence $\gamma \beta k=\gamma \alpha =k$, hence $\gamma \beta =1$ (because $k$ injective); $\beta \gamma i=\beta \gamma \alpha j=\beta kj=\alpha j=i$, hence $\beta \gamma =1$ (because $i$ is injective).

Finally, if $R$ is any subject of Lie($X$), the Lie algebra generated by $X$ subject to the relations $R$ is by definition $\text{Lie}\left(X\right)/𝔞$, where $𝔞$ is the ideal of the Lie($X$) generated by $R$ (i.e. the intersection of all ideals of Lie($X$) which contain $R$).

Finite-dimensional simple Lie algebras / $ℂ$

This is the classical theory we intend to generalise. A Lie algebra $𝔤$ is said to be simple if its only ideals are 0 and $𝔤$, and if also $𝔤$ is non-abelian (thus deliberately excluding the 1-dimensional abelian Lie algebra). Take $k=ℂ$, and dim $𝔤<\infty$.

An element $x\in 𝔤$ is semisimple if ad $x:𝔤\to 𝔤$ is a semisimple (i.e., diagonalizable) linear transformation.

Let $𝔤$ be simple, finite-dimensional. Then $𝔤$ has nonzero subalgebras consisting of semisimple elements (toral subalgebras); they are necessarily abelian.

Let $𝔥$ be a maximal toral subalgebra (or Cartan subalgebra) of $𝔤$. Certainly such exist, for dimensional reasons. Moreover (a non-trivial fact) any two such are conjugate in $𝔤$ (i.e. transforms of each other under the group Int($𝔤$) of inner automorphisms). Fix $𝔥$ once for all. $\phantom{\rule{0.5em}{0ex}}l=\text{dim}\phantom{\rule{0.2em}{0ex}}𝔥$ is called the rank of $𝔤$.

Let ${𝔥}^{*}$ be the vector space dual of $𝔥$. Introduce the killing form

$⟨x,y⟩=\text{trace}\phantom{\rule{0.2em}{0ex}}\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)\left(\text{ad}\phantom{\rule{0.2em}{0ex}}y\right)\phantom{\rule{3em}{0ex}}\left(x,y\in 𝔤\right)$

This is symmetric, nondegenerate and invariant, ie

$⟨\left[x,z\right],y⟩=⟨x,\left[z,y\right]⟩\phantom{\rule{3em}{0ex}}\left(x,y,z\in 𝔤\right)$

Moreover its restriction to $𝔥$ is nondegenerate, hence defines an isomorphism $\omega :𝔥\stackrel{\sim }{\to }{𝔥}^{*}\phantom{\rule{0.5em}{0ex}}\left(\omega \left(x\right)\left(y\right)=⟨x,y⟩\right)$ and a symmetric bilinear form $⟨\lambda ,\mu ⟩$ on ${𝔥}^{*}\phantom{\rule{0.5em}{0ex}}\left(⟨\lambda ,\mu ⟩=⟨{\omega }^{-1}\lambda ,{\omega }^{-1}\mu ⟩\right)$.

Example $𝔤={𝔰𝔩}_{n}\left(ℂ\right)=$ Lie algebras of $n×n$ matrices with trace 0. Here we may take $𝔥$ to consist of the diagonal matrices

$k=\left(\begin{array}{ccc}{h}_{1}& & \\ & \ddots & \\ & & {h}_{n}\end{array}\right)\phantom{\rule{1em}{0ex}}\text{with}\phantom{\rule{1em}{0ex}}\sum _{1}^{n}{h}_{i}=0$

(so that $l=\text{rank}\left(g\right)=n-1$).

Roots

Consider the adjoint representation ${\text{ad}}_{𝔤}$ of $𝔤$, restricted to $𝔥$: this is a representation of $𝔥$ on $𝔤$. Since $𝔥$ is abelian, all its irreducible representations are 1-dimensional, so that $𝔤$ splits up into a direct sum of 1-dimensional $𝔥$–modules. explicitly, for each $\alpha \in {𝔥}^{*}$ define

${𝔤}_{\alpha }=\left\{x\in 𝔤\phantom{\rule{0.2em}{0ex}}:\phantom{\rule{0.2em}{0ex}}\left(\text{ad}\phantom{\rule{0.2em}{0ex}}x\right)=\alpha \left(h\right)x\phantom{\rule{0.5em}{0ex}}\text{for all}\phantom{\rule{0.2em}{0ex}}h\in 𝔥\right\}$

Then it turns out that ${𝔤}_{0}=𝔥$; the nonzero $\alpha \in {𝔥}^{*}$ such that ${𝔤}_{\alpha }\ne 0$ are called the roots of $𝔤$ (relative to $𝔥$). They form a finite subject $R$ of ${𝔥}^{*}$, called the root system of $\left(𝔤,𝔥\right)$, and we have

$𝔤=𝔥+\sum _{\alpha \in R}{𝔤}_{\alpha }\phantom{\rule{3em}{0ex}}\text{(direct sum)}$

Moreover each ${𝔤}_{\alpha }\left(\alpha \in R\right)$ is 1-dimensional, and $\left[{𝔤}_{\alpha },{𝔤}_{\beta }\right]\subset {𝔤}_{\alpha +\beta }$ (hence is 0 if $\alpha +\beta \notin R\bigcup \left\{0\right\}$). If $\alpha$ is a root, so is $-\alpha$.

In the case of ${𝔰𝔩}_{n}\left(ℂ\right)$, let ${e}_{ij}\phantom{\rule{0.2em}{0ex}}\left(1\le i,j\le n\right)$ be the matrix units, and for $1\le i\le n$ let ${u}_{i}:𝔥\to ℂ$ be the $i$th projection: ${u}_{i}\left(h\right)={h}_{i}$. Then

$\left(\text{ad}\phantom{\rule{0.2em}{0ex}}h\right){e}_{ij}=\left[h,{e}_{ij}\right]=h{e}_{ij}-{e}_{ij}h=\left({h}_{i}-{h}_{j}\right){e}_{ij}=\left({u}_{i}-{u}_{j}\right)\left(h\right)·{e}_{ij}\phantom{\rule{2em}{0ex}}\left(h\in 𝔥\right)$

which shows that the roots are $\alpha ={u}_{i}-{u}_{j}\left(i\ne j\right)\phantom{\rule{0.2em}{0ex}};\phantom{\rule{0.2em}{0ex}}{𝔤}_{\alpha }=ℂ{e}_{ij}$, and the root space decomposition is clear. We compute the Killing form on $𝔥$ as follows from above

$\left(\text{ad}\phantom{\rule{0.2em}{0ex}}h\right)\left(\text{ad}\phantom{\rule{0.2em}{0ex}}h\prime \right){e}_{ij}=\left({h}_{i}-{h}_{j}\right)\left({h\prime }_{i}-{h\prime }_{j}\right){e}_{ij}$

so that

$\begin{array}{ccc}⟨h,h\prime ⟩& =& \sum _{i,j}\left({h}_{i}-{h}_{j}\right)\left({h\prime }_{i}-{h\prime }_{j}\right)\\ & =& 2n\sum _{1}^{n}{h}_{i}{h\prime }_{i}\end{array}$

(remember that $\sum {h}_{i}\sum {h\prime }_{i}=0$). So it is a multiple of the obvious scalar product.

It is possible to choose roots ${\alpha }_{1},\dots ,{\alpha }_{l}\phantom{\rule{0.2em}{0ex}}\left(l=\text{dim}\phantom{\rule{0.2em}{0ex}}𝔥\right)$ such that each root $\alpha \in R$ is of the form $\alpha =\sum _{1}^{l}{n}_{i}{\alpha }_{i}$ with coefficients ${n}_{i}\in ℤ$ and either ${n}_{i}\ge 0$ (positive roots) or all ${n}_{i}\le 0$ (negative roots). The ${\alpha }_{i}$ are called a set of simple roots or a basis $B$ of $R$ (they are also a basis of ${𝔥}^{*}$). Choose such a basis once for all. There is then a unique highest root, for which $\sum {n}_{i}$ is a maximum; and a unique lowest root, for which $\sum {n}_{i}$ is a minimum.

Weyl group

For each $\alpha \in R$, let ${w}_{\alpha }$ denote the reflection in the hyperplane orthogonal to $\alpha$ in ${𝔥}^{*}$, so that

${w}_{\alpha }\left(\lambda \right)=\lambda -⟨\lambda ,{\alpha }^{\vee }⟩\alpha \phantom{\rule{3em}{0ex}}\left(\lambda \in {𝔥}^{*}\right)$

where ${\alpha }^{\vee }=2\alpha /{||\alpha ||}^{2}$ is the coroot of $\alpha$. The reflections ${w}_{{\alpha }_{i}}$ corresponding to the simple roots generate a finite group of isometries of ${𝔥}^{*}$, called the Weyl group $W$ of (𝔤,𝔥). Each reflection ${w}_{\alpha }$ lies in $W$; $R$ is stable under $W$; and each root $\alpha \in R$ is of the form ${w}_{{\alpha }_{i}}$ for some $w\in W$ and some simple root ${\alpha }_{i}$. Moreover, any other basis of $R$ is of the form $\left({w}_{{\alpha }_{1}},\dots ,{w}_{{\alpha }_{l}}\right)=wB$ for some (unique) $w\in W$.

Cartan matrix

The numbers

${\alpha }_{ij}=⟨{\alpha }_{i}^{\vee },{\alpha }_{j}⟩=\frac{2⟨{\alpha }_{i},{\alpha }_{j}⟩}{⟨{\alpha }_{i},{\alpha }_{j}⟩}$

are integers, and the matrix ${\left({a}_{ij}\right)}_{1\le i,j\le l}$ is called the Cartan matrix of. It is independent of the choices of $𝔥$ and of basis of $R$. It satisfies the following conditions:

 $\left(C\right)$ ${a}_{ii}=2\phantom{\rule{0.2em}{0ex}}\left(1\le i\le l\right);\phantom{\rule{1em}{0ex}}{a}_{ij}\le 0\phantom{\rule{0.5em}{0ex}}\text{if}\phantom{\rule{0.5em}{0ex}}i\ne j;\phantom{\rule{1em}{0ex}}{a}_{ij}=0⇔{a}_{ji}=0$ $\left(P\right)$ All principal minors of $A$ are positive.

In the case of ${𝔰𝔩}_{n}\left(ℂ\right)$ we may take ${\alpha }_{i}={u}_{i}-{u}_{i+1}\phantom{\rule{0.5em}{0ex}}\left(1\le i\le n-1\right)$. The Weyl group $W$ is the symmetric group ${S}_{n}$ (for ${w}_{{\alpha }_{i}}$ interchanges ${u}_{i}$ and ${u}_{i+1}$ and leaves the other ${u}_{j}$ fixed). Here the Cartan matrix is

$A=\left(\begin{array}{ccccc}2& -1& & & \\ -1& 2& -1& & \\ & -1& 2& & \\ & & & \ddots & -1\\ & & & -1& 2\end{array}\right)$

(with $l=n-1$ rows and columns).

Generators and relations

The Cartan matrix $A$ determines $𝔤$ up to isomorphism.

Choose generators ${e}_{i}\in {𝔤}_{{\alpha }_{i}},\phantom{\rule{0.2em}{0ex}}{f}_{i}\in {𝔤}_{{-\alpha }_{i}}\phantom{\rule{0.5em}{0ex}}\left(1\le i\le l\right)$ such that $⟨{e}_{i},{f}_{i}⟩=1,$ and elements ${h}_{1},\dots ,{h}_{l}\in 𝔥$ such that $⟨{h}_{i},h⟩={\alpha }_{i}^{\vee }\left(h\right)$, so that

${\alpha }_{j}\left({h}_{i}\right)=⟨{\alpha }_{i}^{\vee },{\alpha }_{j}⟩={a}_{ij}$

Then the $3l$ elements ${e}_{i},{f}_{i},{h}_{i}$ generate $𝔤$ subject to the following relations (Serre):

$\begin{array}{c}\left[{h}_{i},{h}_{j}\right]=0\phantom{\rule{0.5em}{0ex}}\text{for all}\phantom{\rule{0.2em}{0ex}}i,j\\ \left[{e}_{i},{f}_{j}\right]={\delta }_{ij}{h}_{i}\\ \left[{h}_{i},{e}_{j}\right]={a}_{ij}{e}_{j}\phantom{\rule{0.5em}{0ex}};\phantom{\rule{0.5em}{0ex}}\left[{h}_{i},{f}_{j}\right]=-{a}_{ij}{f}_{j}\\ {\left(\text{ad}\phantom{\rule{0.2em}{0ex}}{e}_{i}\right)}^{1-{a}_{ij}}{e}_{j}={\left(\text{ad}\phantom{\rule{0.2em}{0ex}}{f}_{i}\right)}^{1-{a}_{ij}}{f}_{j}=0\phantom{\rule{2em}{0ex}}\left(i\ne j\right).\end{array}$

The idea now is (roughly) the following: start with any matrix $A$ of integers satisfying ($C$), and form the Lie algebra with the above generators and relations.

However there is one remark that should be made at this point. In the classical set up (which I have just been describing) the Cartan matrix $A$ is nonsingular, the ${h}_{i}\phantom{\rule{0.5em}{0ex}}\left(1\le i\le l\right)$ form a basis of the Cartan subalgebra $𝔥$, and the simple roots ${\alpha }_{j}\in {𝔥}^{*}$. Now a generalized Cartan matrix may well be singular (and it would be foolish to exclude this possibility, because for the affine Lie algebras the Cartan matrix is singular).