Kac-Moody Lie Algebras
Chapter IV: Affine Lie algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 7 October 2012

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

Loop algebras

Let $A$ be an indecomposable Cartan matrix of finite type, so that $𝔤\left(A\right)$ is finite-dimensional and simple.

Let $L=k[t,t^{-1}]$ denote the algebra of Laurent polynomials in one variable over $k$.

The loop algebra of $𝔤$ is defined to be

$$L\left(𝔤\right)=L{\otimes }_k𝔤=\underset{m\in \mathbb{Z}}{⨁}{t}^m\otimes 𝔤$$

i.e. it is constructed from $𝔤$ by extension of scalars from $k$ to $L\text{.}$ I shall drop the tensor product notation and write $t^{m}x$ in place of $t^m\otimes x$ $(m\in \mathbb{Z},x\in 𝔤)\text{.}$ Then the Lie bracket in $L\left(𝔤\right)$ is defined by

$$\begin{array}{cc}{[t^{m}x,t^{n}y]}_0=t^{m+n}[x,y]& \text{(1)}\end{array}$$

$(m,n\in \mathbb{Z};x,y\in 𝔤)\text{.}$

Recall that $A$ is symmetrizable and hence that $𝔤$ carries an invariant scalar product $⟨x,y⟩\text{.}$ We extend this to $L\left(𝔤\right)$ as follows:

$$\begin{array}{cc}⟨t^{m}x,t^{n}y⟩=\{\begin{array}{cc}⟨x,y⟩,& \text{if}m+n=0,\\ 0,& \text{otherwise.}\end{array}& \text{(2)}\end{array}$$

One verifies immediately that this scalar product on $L\left(𝔤\right)$ is still invariant.

Finally we define a derivation $d$ of $L\left(𝔤\right)$ by

$$\begin{array}{cc}d\left(t^{m}x\right)=m{t}^{m}x(m\in \mathbb{Z},x\in 𝔤)& \text{(3)}\end{array}$$

i.e., $d=t\frac{d}{dt}\text{.}$ That $d$ is a derivation is immediate from (1).

The next stage is to construct a 1-dimensional central extension of $L\left(𝔤\right)\text{.}$

Central extensions of Lie algebras

In general let

$$0⟶𝔞⟶{𝔤}_1\stackrel{p}{⟶}⟶𝔤⟶0$$

be an exact sequence of Lie algebras with $𝔞=\text{Ker}\left(p\right)$ contained in the centre of $𝔤$ i.e. $[𝔞,{𝔤}_1]=0\text{.}$ Choose a section $s:𝔤\to {𝔤}_1,$ i.e. a $k\text{-linear}$ map such that $p\circ s=1_{𝔤}\text{.}$ Then for $x,y\in 𝔤$

$$\begin{array}{cc}\psi (x,y)=[sx,sy]-s[x,y]\in 𝔞& \text{(1)}\end{array}$$

(because it is killed by $p\text{);}$ the function $\psi :𝔤\times 𝔤\to 𝔞$ is bilinear, skew-symmetric and satisfies $\delta \psi =0,$ where

$$\begin{array}{cc}\begin{array}{ccc}\delta \psi (x,y,z)& =& \psi ([x,y],z)+\psi ([y,z],x)+\psi ([z,x],y)\end{array}& \text{(2)}\end{array}$$

For we have

$$\begin{array}{ccc}\psi ([x,y],z)& =& [s[x,y],sz]-s[[x,y],z]\\ & =& [[sx,sy],sz]-s[[x,y],z]\end{array}$$

by (1) and the centrality of $𝔞;$ now apply the Jacobi identity.

In other words, $\psi$ is a 2-cocycle on $𝔤$ with values in $𝔞$ (with trivial $𝔤\text{-action).}$

Conversely, given a 2-cocycle $\psi :𝔤\times 𝔤\to 𝔞,$ define ${𝔤}_1=𝔤\times 𝔞$ (direct product of vector spaces) with Lie bracket given by

$$\begin{array}{cc}[(x,a),(y,b)]=([x,y],\psi (x,y))& \text{(3)}\end{array}$$

$(x,y\in 𝔤;a,b\in 𝔞)\text{.}$ Then the Jacobi identity holds in ${𝔤}_1$ by virtue of (2): for we have

$$\begin{array}{ccc}[[(x,a),(y,b)],(z,c)]& =& [([x,y],\psi (x,y)),(z,c)]\\ & =& ([[x,y],z],\psi ([x,y],z))\end{array}$$

so that cyclic summation gives 0 by virtue of (2) and the Jacobi identity in $𝔤\text{.}$ (The motive for the definition (3) is that in the original context we have $[sx+a,sy+b]=[sx,sy]=s[x,y]+\psi (x,y)\text{.)}$

Thus ${𝔤}_1$ is a Lie algebra, $𝔞$ is a central ideal in $𝔤,$ and $p:{𝔤}_1/𝔞\to 𝔤$ is an isomorphism of Lie algebras $\text{(}𝔤$ is not a subalgebra of ${𝔤}_1\text{).}$

In the present context we define $\psi :L\left(𝔤\right)\times L\left(𝔤\right)\to k$ by

$$\psi (\xi ,\eta )=⟨d\xi ,\eta ⟩(\xi ,\eta \in L\left(𝔤\right))$$

Explicitly, if $\xi =t^{m}x,$ $\eta =t^{n}y\text{.}$ $(m,n\in \mathbb{Z};x,y\in 𝔤)$ then

$$\begin{array}{ccc}\psi (\xi ,\eta )& =& ⟨m{t}^{m}x,t^{n}y⟩\\ & =& \{\begin{array}{cc}m⟨x,y⟩,& \text{if}m+n=0,\\ 0,& \text{otherwise,}\end{array}\end{array}$$

from which it follows that $\psi (\eta ,\xi )=-\psi (\xi ,\eta )\text{.}$

Next we verify that $\delta \psi (\xi ,\eta ,\zeta )=0\text{.}$ By linearity we may assume that $\xi =t^{p}x,$ $\eta =t^{q}y,$ $\zeta =t^{r}z$ $(p,q,r\in \mathbb{Z};x,y,z\in 𝔤)\text{.}$ If $p+q+r\ne 0$ then certainly $\delta \psi =0,$ and if $p+q+r=0$ we have

$$\begin{array}{ccc}\delta \psi (\xi ,\eta ,\zeta )& =& (p+q)⟨[x,y],z⟩+(q+r)⟨[y,z],x⟩+(r+p)⟨[z,x],y⟩\\ & =& p⟨x,[y,z]⟩+q⟨[x,y],z⟩+(q+r)⟨x,[y,z]⟩+(r+p)⟨z,[x,y]⟩\\ & =& 0\text{(since}p+q+r=0\text{).}\end{array}$$

So we construct a 1-dimensional central covering $\stackrel{\sim }{L}\left(𝔤\right)$ of $L\left(𝔤\right)$ as follows:

$$\begin{array}{ccc}\stackrel{\sim }{L}\left(𝔤\right)& =& L\left(𝔤\right)\oplus kc\end{array}$$

with Lie bracket defined by

$$\begin{array}{ccc}[\xi +\lambda c,\eta +\mu c]& =& {[\xi ,\eta ]}_0+\psi (\xi ,\eta )c\\ & =& {[\xi ,\eta ]}_0+⟨d\xi ,\eta ⟩c\end{array}$$

Notice that $𝔤$ is a subalgebra of $\stackrel{\sim }{L}\left(𝔤\right),$ since $d\xi =0$ for $\xi \in 𝔤\text{.}$

We extend the derivation $d$ to $\stackrel{\sim }{L}\left(𝔤\right)$ by requiring that $dc=0\text{.}$ We ought to verify that $d,$ extended in this way, is still a derivation. On the one hand we have

$$\begin{array}{ccc}d[\xi +\lambda c,\eta +\mu c]& =& d{[\xi ,\eta ]}_0\end{array}$$

and on the other hand

$$\begin{array}{ccc}[d(\xi +\lambda c),\eta +\mu c]+[\xi +\lambda c,d(\eta +\mu c)]& =& [d\xi ,\eta +\mu c]+[\xi +\lambda c,d\eta ]\\ & =& {[d\xi ,\eta ]}_0+\psi (d\xi ,\eta )+{[\xi ,d\eta ]}_0+\psi (\xi ,d\eta )\\ & =& d{[\xi ,\eta ]}_0+\psi (\xi ,d\eta )-\psi (\eta ,d\xi )\\ & =& d{[\xi ,\eta ]}_0+⟨d\xi ,d\eta ⟩-⟨d\eta ,d\xi ⟩\\ & =& d{[\xi ,\eta ]}_0\text{.}\end{array}$$

Finally we construct the semidirect product

$$\begin{array}{ccc}\hat{L}\left(𝔤\right)& =& \stackrel{\sim }{L}\left(𝔤\right)⋊kd=\stackrel{\sim }{L}\left(𝔤\right)\oplus kd\end{array}$$

with bracket

$$\begin{array}{ccc}[\xi +{\lambda }_{1}d,\eta +{\mu }_{1}d]& =& [\xi ,\eta ]+{\lambda }_{1}d\eta -{\mu }_{1}d\xi \end{array}$$

$\text{(}\xi ,\eta \in \stackrel{\sim }{L}\left(𝔤\right);{\lambda }_1,{\mu }_1\in k\text{).}$ So altogether

$$\begin{array}{ccc}\hat{L}\left(𝔤\right)& =& L\left(𝔤\right)\oplus kc\oplus kd\end{array}$$

and

$$\begin{array}{ccc}[\xi +\lambda c+{\lambda }_{1}d,\eta +\mu c+{\mu }_{1}d]& =& {[\xi ,\eta ]}_0+{\lambda }_{1}d\eta -{\mu }_{1}d\xi +⟨d\xi ,\eta ⟩c\text{.}\end{array}$$

Our aim is to show that $\hat{L}\left(𝔤\right)\cong 𝔤\left(A^{\left(1\right)}\right),$ where $A^{\left(1\right)}$ is an indecomposable Cartan matrix of affine type. To construct $A^{\left(1\right)}$ we need the following lemma:

(4.1) The root system $R$ of $𝔤\left(A\right)$ has a unique highest root $\phi$ such that $\phi \ge \alpha$ for all $\alpha \in R$ (i.e. $\phi -\alpha \in Q^{+}\text{).}$ We have $\phi ={\displaystyle \sum _{i=1}^{\ell }}{a}_i{\alpha }_i$ with each coefficient $a_i\ge 1;$ $\phi \left(h_i\right)\ge 0$ for all $i,$ and $\phi \left(h_i\right)>0$ for some $i\text{.}$

		Proof.
	Since $A$ is of finite type, $R$ is finite. Let $\phi =\sum a_i{\alpha }_i$ be a maximal element of $R$ (with respect to the partial ordering $\ge \text{).}$ Since $w_i\phi =\phi -\phi \left(h_i\right){\alpha }_i,$ it follows that $\phi \left(h_i\right)\ge 0$ for all $i\text{.}$ If $\phi \left(h_i\right)=0$ for all $i,$ then $⟨\phi ,{\alpha }_i⟩=0$ for all $i$ and therefore $⟨\phi ,\phi ⟩=0,$ whence $\phi =0\text{.}$ Hence $\phi \left(h_i\right)>0$ for at least one value of $i\text{.}$ Clearly $\phi \in R^{+}$ (otherwise $-\phi >\phi \text{),}$ hence the coefficients $a_i$ are all $\ge 0\text{.}$ Let $J=\text{supp}\left(\phi \right)=\{i:a_i\ne 0\}\text{.}$ If $s\ne [1,\ell ]$ then by connectedness there exists $j\in J$ and $k\notin J$ such that $a_{kj}<0,$ whence $$\phi \left(h_k\right)=\sum _{i\in J}{a}_i{\alpha }_i\left(h_k\right)=\sum _{i\in J}{a}_i{a}_{ki}<0$$ (because all the terms are $\le 0,$ and at least one is $<0\text{).}$ This is a contradiction hence all the $a_i$ are $\ge 1\text{.}$ Finally let ${\phi }^{\prime }\ne \phi$ be another maximal root. Then from above (with $\phi$ replaces by ${\phi }^{\prime }\text{)}$ we have $⟨{\phi }^{\prime },{\alpha }_j⟩\ge 0$ for all $j,$ and $⟨{\phi }^{\prime },{\alpha }_j⟩>0$ for some $j,$ whence $⟨{\phi }^{\prime },\phi ⟩=\sum a_j⟨{\phi }^{\prime },{\alpha }_j⟩>0$ and therefore also ${\phi }^{\prime }\left(h_{\phi }\right)>0\text{.}$ By (2.31) (root strings) ${\phi }^{\prime }-\phi \in R\text{.}$ Hence either ${\phi }^{\prime }>\phi$ or $\phi >{\phi }^{\prime },$ neither of which is possible, and therefore $\phi$ is unique. $\square$

Now let $e_i,f_i$ $(1\le i\le \ell )$ as usual be the generators of $𝔤=𝔤\left(A\right),$ and let $𝔥$ be the Cartan subalgebra. Normalize the scalar product on $𝔤$ so that $⟨\phi ,\phi ⟩=2,$ and choose $e_{\phi }\in {𝔤}_{\phi },$ $f_{\phi }\in {𝔤}_{-\phi }$ such that

$$\begin{array}{cc}\begin{array}{ccc}[e_{\phi },{𝔤}_{\phi }]& =& h_{\phi }\end{array}& \text{(1)}\end{array}$$

(or equivalently) such that $⟨e_{\phi },f_{\phi }⟩=1\text{.}$ (Recall that $[e_{\phi },f_{\phi }]=⟨e_{\phi },f_{\phi }⟩h_{\phi }^{\vee }$ and that $h_{\phi }^{\vee }=h_{\phi }$ because $\phi ={\phi }^{\vee },$ by our choice of scalar product.)

Define

$$\begin{array}{cc}{e}_0=t{f}_{\phi },f_0=t^{-1}{e}_{\phi },h_0=-h_{\phi }+c& \text{(2)}\end{array}$$

and let $\widehat{𝔥}=𝔥\oplus kc\oplus kd\text{.}$ We extend each root $\alpha \in R$ to a linear form (also denoted by $\alpha \text{)}$ on $\widehat{𝔥}$ by setting $\alpha \left(c\right)=\alpha \left(d\right)=0;$ also define $\delta \in {\widehat{𝔥}}^{*}$ by

$$\delta \mid (𝔥\oplus kc)=0,\delta \left(d\right)=1$$

Finally set

$$\begin{array}{cc}{\alpha }_0=\delta -\phi & \text{(4)}\end{array}$$

so that

$$\begin{array}{cc}\sum _{i=0}^{\ell }{a}_i{\alpha }_i=\delta & \text{(5)}\end{array}$$

where $a_0=1,$ and $a_1,\dots ,a_{\ell }$ are the coefficients of $\phi$ (4.1).

Let

$$\begin{array}{cc}{a}_{ij}={\alpha }_j\left(h_i\right)(0\le i,j\le \ell )& \text{(6)}\end{array}$$

and let $A^{\text{(1)}}={\left(a_{ij}\right)}_{0\le i,j\le \ell }\text{.}$ The matrix $A^{\text{(1)}}$ has $A$ as a principal submatrix

(4.2) $A^{\text{(1)}}$ is an indecomposable Cartan matrix of affine type.

		Proof.
	We calculate: $$\begin{array}{c}{\alpha }_0\left(h_0\right)=(\delta -\phi )(c-h_{\phi })=\phi \left(h_{\phi }\right)=2;\\ {\alpha }_0\left(h_i\right)=(\delta -\phi )\left(h_i\right)=-\phi \left(h_i\right)\le 0\text{by (4.1)}\\ {\alpha }_j\left(h_0\right)={\alpha }_j(c-h_{\phi })=-{\alpha }_j\left(h_{\phi }\right)=-⟨\phi ,{\alpha }_j⟩\end{array}$$ which is a positive scalar multiple of $-⟨\phi ,{\alpha }_j^{\vee }⟩=-\phi \left(h_j\right),$ hence also $\le 0\text{.}$ Hence $A^{\text{(1)}}$ is a Cartan matrix, and is indecomposable because $A$ is indecomposable and $a_{0i}=-\phi \left(h_i\right)<0$ for some $i\ne 0,$ again by (4.1). Also from (3) and (5) we have $$\sum _{j=0}^{\ell }{a}_{ij}{a}_j=\sum _{j=0}^{\ell }{a}_j{\alpha }_j\left(h_i\right)=\delta \left(h_i\right)=0$$ for $0\le i\le \ell ,$ so that by (2.17) $A^{\text{(1)}}$ is of affine type. $\square$

If $A$ is of type $X$ $\text{(}=A_n,\dots ,G_2:$ see table F), then $A^{\text{(1)}}$ is of type $\stackrel{\sim }{X}$ (see table A; the integers ${\alpha }_i$ are the labels attached to the vertices there).

(4.3) Theorem

$$\begin{array}{ccc}\hat{L}\left(𝔤\right)& \cong & 𝔤\left(A^{\text{(1)}}\right)\text{(with}{e}_i,f_i,\widehat{𝔥}\text{as generators)}\\ \stackrel{\sim }{L}\left(𝔤\right)& \cong & {𝔤}^{\prime }\left(A^{\text{(1)}}\right)\\ L\left(𝔤\right)& \cong & \overline{{𝔤}^{\prime }}\left(A^{\text{(1)}}\right)\text{.}\end{array}$$

		Proof.
	The proof is a sequence of verifications: $(\widehat{𝔥},{\left(h_i\right)}_{0\le i\le \ell },{\left({\alpha }_i\right)}_{0\le i\le \ell })$ is a minimal realization of the Cartan matrix $A^{\text{(1)}}\text{.}$ Well, $\text{dim}\widehat{𝔥}=\ell +2=2n-\ell$ where $n=\ell +1$ is the number of rows of $A^{\text{(1)}},$ and $\ell$ is its rank. Clearly the $h_i$ are linearly independent in $\widehat{𝔥},$ and the ${\alpha }_i$ are $\ell \text{.}i\text{.}$ in ${\widehat{𝔥}}^{*}\text{.}$ The $e_i,f_i$ and $\widehat{𝔥}$ satisfy the defining relations (1.2). Since $𝔤$ is a subalgebra of $\hat{L}\left(𝔤\right),$ we have $[e_i,f_j]={\delta }_{ij}{h}_i$ for $1\le i,j\le \ell ;$ moreover $$\begin{array}{c}[e_0,f_j]=[t{f}_{\phi },f_j]=t[f_{\phi },f_j]+⟨t{f}_{\phi },f_j⟩c=0,(1\le j\le \ell )\\ [e_0,f_0]=[t{f}_{\phi },t^{-1}{e}_{\phi }]=[f_{\phi },e_{\phi }]+⟨f_{\phi },e_{\phi }⟩c=-h_{\phi }+c=h_0;\end{array}$$ next, if $\hat{h}\in \widehat{𝔥},$ say $\hat{h}=h+\lambda c+\mu d$ $(h\in 𝔥;\lambda ,\mu \in k)$ then for $i=1,2,\dots ,\ell$ we calculate $$\begin{array}{ccc}[\hat{h},e_i]& =& [h+\lambda c+\mu d,e_i]=[h,e_i]+\mu d\left(e_i\right)=[h,e_i]\\ & =& {\alpha }_i\left(h\right)e_i={\alpha }_i\left(\hat{h}\right)e_i;\\ [\hat{h},e_0]& =& [h+\lambda c+\mu d,t{f}_{\phi }]=t[h,f_{\phi }]+\mu d\left(t{f}_{\phi }\right)\\ & =& -\phi \left(h\right)t{f}_{\phi }+\mu t{f}_{\phi }=(\mu -\phi \left(h\right))e_0={\alpha }_0\left(\hat{h}\right)e_0\end{array}$$ (because ${\alpha }_0\left(\hat{h}\right)=(\delta -\phi )(h+\lambda c+\mu d)=\mu \phi \left(h\right)\text{).}$ Finally, it is clear that $\widehat{𝔥}$ is abelian. Let $\alpha \in R\cup \left\{0\right\},$ $m\in \mathbb{Z}\text{.}$ Then $t^m{𝔤}_{\alpha }$ is a weight space for the adjoint action of $\widehat{𝔥}$ on $\hat{L}\left(𝔤\right),$ with weight $\alpha +m\delta :$ for with $\hat{h}$ as above, $$\begin{array}{ccccc}[\hat{h},t^{m}x]& =& [h+\lambda c+\mu d,t^{m}x]& =& t^m[h,x]+\mu d\left(t^{m}x\right)\\ & & & =& (\alpha \left(h\right)+m\mu )t^{m}x\\ & & & =& (\alpha +m\delta )\left(\hat{h}\right)t^{m}x,\end{array}$$ Thus the $\alpha +m\delta$ $(\alpha \in R\cup \left\{0\right\},m\in \mathbb{Z})$ are the roots of $\hat{L}\left(𝔤\right)$ relative to $\widehat{𝔥}\text{.}$ Now let $𝔞$ be an ideal of $\hat{L}\left(𝔤\right)$ such that $𝔞\cap \widehat{𝔥}=0\text{.}$ By (1.5) $𝔞$ is the direct sum of its weight spaces $𝔞\cap t^m{𝔤}_{\alpha }$ (where ${𝔤}_0$ is to be interpreted as $\widehat{𝔥}\text{.}$ Hence if $𝔞\ne 0$ there exists $\alpha \in R\cup \left\{0\right\},$ $x\ne 0$ in ${𝔤}_{\alpha }$ and $m\in \mathbb{Z}$ such that $t^{m}x\in 𝔞\text{.}$ Choose $y\in {𝔤}_{-\alpha }$ such that $⟨x,y⟩=1,$ then $[x,y]=h_{\alpha }^{\vee }\ne 0$ and $$\begin{array}{ccccc}z& =& [t^{m}x,t^{-m}y]& =& [x,y]+⟨d\left(t^{m}x\right),t^{-m}y⟩c\\ & & & =& [x,y]+mc\ne 0\end{array}$$ lies in $𝔞\cap \widehat{𝔥}:$ contradiction. Hence $\hat{L}\left(𝔤\right)$ has no ideals $𝔞\ne 0$ such that $𝔞\cap \widehat{𝔥}=0\text{.}$ To complete the proof, it remains to be shown that $\hat{L}\left(𝔤\right)$ is generated by the $e_i,f_i$ $(0\le i\le \ell )$ and $\stackrel{}{𝔥}\text{.}$ Let $L_1$ be the subalgebra generated by these elements. Then certainly $𝔤\subset L_1;$ also $t{f}_{\phi }=e_0\in L_1,$ and since $𝔤$ is simple it is generated (as a $𝔤$-module) by $f_{\phi },$ i.e. $[f_{\phi },𝔤]=𝔤$ and therefore $[e_0,𝔤]=t𝔤\text{.}$ Thus $t𝔤\subset L_1\text{.}$ Now assume that $t^k𝔤\subset L_1$ for some $k\ge 1\text{.}$ Since $𝔤=[𝔤,𝔤]$ we have $t^{k+1}𝔤=[t𝔤,t^k𝔤]\subset L_1,$ and hence $t^k𝔤\subset L_1$ for all $k\ge 0\text{.}$ In the same way we prove that $t^k𝔤\subset L_1$ for all $k\le 0,$ and hence that $L_1=\hat{L}\left(𝔤\right)\text{.}$ $\square$

(4.4) Corollary If $S$ is the root system of $𝔤\left(A^{\text{(1)}}\right)$ then

$$\begin{array}{ccc}{S}_{\text{re}}& =& \{\alpha +m\delta :\alpha \in R,m\in \mathbb{Z}\}\\ S_{\text{im}}& =& \{m\delta :m\in \mathbb{Z},m\ne 0\}\end{array}$$

each imaginary root $m\delta$ has multiplicity $\ell =\text{rank}\left(A^{\text{(1)}}\right)\text{.}$

The positive real roots are

$$\alpha +m\delta (\alpha \in R,m\ge 1\text{and}\alpha \in R^{+},m=0)$$

The bilinear form $⟨\xi ,\eta ⟩$ on $L\left(𝔤\right)$ we extend to $\hat{L}\left(𝔤\right)$ as follows:

$$⟨c,L\left(𝔤\right)⟩=⟨d,L\left(𝔤\right)⟩=0;⟨c,c⟩=⟨d,d⟩=0;⟨c,d⟩=1$$

It is still invariant: the only nontrivial case to be checked is that

$$\begin{array}{ccc}⟨[d,\xi ],\eta ⟩& =& ⟨d,[\xi ,\eta ]⟩(\xi ,\eta \in L\left(𝔤\right))\end{array}$$

which is true because

$$\begin{array}{ccc}⟨[d,\xi ],\eta ⟩& =& ⟨d\xi ,\eta ⟩\end{array}$$

and

$$\begin{array}{ccccc}⟨d,[\xi ,\eta ]⟩& =& ⟨d,{[\xi ,\eta ]}_0+⟨d\xi ,\eta ⟩c⟩& =& ⟨d\xi ,\eta ⟩\text{.}\end{array}$$

Remark. $L\left(𝔤\right)$ is certainly not simple: it has lots of ideals. For example, let $a\in k^{*}$ and let $u_a:L\left(𝔤\right)\to 𝔤$ be the homomorphism defined by $u_a\left(t^{m}x\right)=a^{m}x$ $(m\in \mathbb{Z};x\in 𝔤)\text{.}$ Then $u_a$ is a Lie algebra homomorphism, and its kernel is a nontrivial ideal, indeed a maximal ideal.

Construction of the remaining affine Lie algebras

Let $A={\left(a_{ij}\right)}_{1\le i,j\le n}$ be an indecomposable symmetric Cartan matrix of finite type, i.e. of type $A,$ $D$ or $E\text{.}$ As usual, let $𝔥,R,Q,W$ denote the Cartan subalgebra, the root system, the root lattice and the Weyl group of the algebra $𝔤\left(A\right)=𝔤\text{.}$ The invariant bilinear form $⟨x,y⟩$ on $𝔤,$ constructed as in (3.12), is such that $⟨h_i,h_j⟩=a_{ij}$ and also (on ${𝔥}^{*}\text{)}$ $⟨{\alpha }_i,{\alpha }_j⟩=a_{ij}\text{.}$

In particular, $$ {\mid {\alpha }_i\mid }^2=a_{ii}=2;$$ since the scalar product on ${𝔥}^{*}$ is $W$-invariant and all the roots are real, we have ${\mid \alpha \mid }^2=2$ for all roots $\alpha \in R\text{.}$ Conversely, by (2.34), if $\alpha \in Q$ is such that ${\mid \alpha \mid }^2=2,$ then $\alpha \in R\text{.}$

Let $\alpha ,\beta \in R\text{.}$ Then (Cauchy-Schwarz)

$$\mid ⟨\alpha ,\beta ⟩\mid \le \mid \alpha \mid ·\mid \beta \mid =2$$

and since $⟨\alpha ,\beta ⟩$ is an integer, it can therefore take only the values $0,\pm 1,\pm 2\text{.}$

(4.5) We have $⟨\alpha ,\beta ⟩=2,1,0,-1,-2$ respectively if and only if

$$\alpha =\beta ;\alpha -\beta \in R;\alpha \pm \beta \notin R\cup \left\{0\right\};\alpha +\beta \in R;\alpha =-\beta$$

		Proof.
	For example, since ${\mid \alpha -\beta \mid }^2={\mid \alpha \mid }^2+{\mid \beta \mid }^2-2⟨\alpha ,\beta ⟩=4-2⟨\alpha ,\beta ⟩$ it follows that $$⟨\alpha ,\beta ⟩=\left\{\begin{array}{c}1\\ 2\end{array}\right\}\iff {\mid \alpha -\beta \mid }^2=\left\{\begin{array}{c}2\\ 0\end{array}\right\}\iff \alpha -\beta \left\{\begin{array}{c}\in R\\ =0\end{array}\right\}\iff$$ Similarly with $\beta$ replaced by $-\beta \text{.}$ $\square$

Now let $\Delta$ be the Dynkin diagram of $A,$ and let $s$ be an automorphism of $\Delta \text{.}$ In terms of the matrix $A,$ this means that $s$ is a permutation of $\{1,2,\dots ,n\}$ such that

$$a_{si,sj}=a_{i,j}$$

for all $i,j\text{.}$ Let $k$ be the order of $s\text{.}$ If $k\ne 1$ (i.e. if $s\ne 1\text{)}$ there are just 5 possibilities:

$$\begin{array}{ccc}{A}_{2\ell },\ell \ge 1& & k=2\\ A_{2\ell -1},\ell \ge 2& & k=2\\ D_{\ell +1},\ell \ge 3& & k=2\\ E_6& & k=2\\ D_4& & k=3\end{array}$$

(In each case, vertices of $\Delta$ in the same vertical line are in the same $s$-orbit.) Thus $k=2$ or $3$ in every case.

The graph automorphism $s$ determines an automorphism (also denoted by $s\text{)}$ of period $k$ of the Lie algebra $𝔤=𝔤\left(A\right)$ by the rule

$$s\left(e_i\right)=e_{si},s\left(f_i\right)=f_{si},s\left(h_i\right)=h_{si}(1\le i\le n)$$

This is clear from the construction of $𝔤\left(A\right)$ in Chapter I, since the relations (1.2) are stable under $s\text{.}$

By transposition, $s$ also acts on ${𝔥}^{*}:$ $\left(s\lambda \right)\left(h\right)=\lambda \left(s^{-1}h\right)(h\in 𝔥,\lambda \in {𝔥}^{*})$ We have $s{\alpha }_j={\alpha }_{sj},$ because

$$\left(s{\alpha }_j\right)\left(h_i\right)={\alpha }_j\left(s^{-1}{h}_i\right)={\alpha }_j\left(h_{s^{-1}i}\right)=a_{s^{-1}i,j}=a_{i,sj}={\alpha }_{sj}\left(h_i\right)\text{.}$$

The scalar product on $𝔤$ (hence on $𝔥$ and ${𝔥}^{*}\text{)}$ is $s$-invariant.

Since $𝔥$ is stable under $s,$ it follows that $s$ permutes the root-spaces ${𝔤}_{\alpha }$ and hence also the roots $\alpha \in R:s\left({𝔤}_{\alpha }\right)={𝔤}_{s\alpha }:$ and this action agrees with that already described on ${𝔥}^{*},$ because if $x\in {𝔤}_{\alpha }$ and $h\in 𝔥$ we have

$$[h,sx]=s[s^{-1}h,x]=s\left(\alpha \left(s^{-1}h\right)x\right)=\left(s\alpha \right)\left(h\right)sx\text{.}$$

Moreover, since $s$ permutes the simple roots ${\alpha }_i,$ it follows that $s$ permutes $R^{+}\text{.}$ Hence $\alpha +s\alpha$ is never zero, and $\alpha -s\alpha$ is never a root. This observation, together with (4.5), proves the first part of

(4.6) Let $\alpha \in R$ and assume $\alpha \ne s\alpha \text{.}$ Then

1. $⟨\alpha ,s\alpha ⟩=0$ or $-1;$
2. If $⟨\alpha ,s\alpha ⟩=-1$ (so that $\beta =\alpha +s\alpha \in R\text{)}$ then $k=2$ and $s$ acts as $-1$ on ${𝔤}_{\beta }\text{.}$

		Proof of (ii).
	If $k=3$ then $⟨\alpha ,s^2\alpha ⟩=⟨\alpha ,s^{-1}\alpha ⟩=⟨s\alpha ,\alpha ⟩=-1,$ and $⟨s\alpha ,s^2\alpha ⟩=⟨\alpha ,s\alpha ⟩=-1,$ hence ${\mid \alpha +s\alpha +s^2\alpha \mid }^2=6-2·3=0$ and therefore $\alpha +s\alpha +s^2\alpha =0;$ which is plainly impossible. Hence $k=2\text{.}$ Let $e_{\alpha }$ generate ${𝔤}_{\alpha },$ then $s{e}_{\alpha }$ generates ${𝔤}_{s\alpha }$ and $x=[e_{\alpha },s{e}_{\alpha }]$ is a nonzero element of ${𝔤}_{\beta }\text{.}$ Hence $x$ generates ${𝔤}_{\beta },$ and $sx=[s{e}_{\alpha },e_{\alpha }]=-x\text{.}$ $\square$

Let $\omega$ be a primitive $k$th roots of unity (assumed to lie in the ground field $K$ if $k=3\text{).}$ For each integer $r$ define

$$\begin{array}{ccc}{𝔤}^{\left(r\right)}& =& \{x\in 𝔤:sx={\omega }^{r}x\}\end{array}$$

so that ${𝔤}^{\left(r\right)}$ is the ${\omega }^r$-eigenspace of $s$ in $𝔤,$ and depends only on $r$ and $k\text{.}$ We have

$$\begin{array}{cc}𝔤=\underset{r=0}{\overset{k-1}{⨁}}{𝔤}^{\left(r\right)}& \text{(1)}\end{array}$$

the decomposition being

$$x=\sum _{r=0}^{k-1}{x}^{\left(r\right)}$$

where

$$x^{\left(r\right)}=\frac{1}{k}\sum _{i=0}^{k-1}{\omega }^{-ir}{s}^{i}x$$

Also

$$\begin{array}{cc}[{𝔤}^{\left(p\right)},{𝔤}^{\left(q\right)}]\subset {𝔤}^{(p+q)}& \text{(2)}\end{array}$$

for all $p,q\in \mathbb{Z},$ so that (1) is a $\mathbb{Z}/k\mathbb{Z}$-grading of $𝔤\text{.}$

In particular, ${𝔤}^{\left(0\right)}$ is the Lie algebra of $s$-invariants of $𝔤,$ and each ${𝔤}^{\left(r\right)}$ is a ${𝔤}^{\left(0\right)}$-module under the adjoint action.

Next we have

(4.7) The restriction of the bilinear form $⟨x,y⟩$ to ${𝔤}^{\left(p\right)}\times {𝔤}^{\left(q\right)}$ is

1. zero if $p+q\not\equiv 0\text{(mod}k\text{)}$
2. nondegenerate if $p+q\equiv 0\text{(mod}k\text{).}$

		Proof.
	Let $x\in {𝔤}^{\left(p\right)},y\in {𝔤}^{\left(q\right)}\text{.}$ Then $$⟨x,y⟩=⟨sx,sy⟩={\omega }^{p+q}⟨x,y⟩$$ which proves (a); then (b) follows because $⟨x,y⟩$ is nondegenerate on $𝔤\text{.}$ $\square$

Now let

$$\begin{array}{ccc}L(𝔤,s)& =& \sum _{r\in \mathbb{Z}}{t}^r{𝔤}^{\left(r\right)}\subset L\left(𝔤\right)\\ \stackrel{\sim }{L}(𝔤,s)& =& L(𝔤,s)\oplus Kc\subset \stackrel{\sim }{L}\left(𝔤\right)\\ \hat{L}(𝔤,s)& =& \stackrel{\sim }{L}(𝔤,s)\oplus Kd\subset \hat{L}\left(𝔤\right)\end{array}$$

It follows from (2) that $L(𝔤,s)$ is a subalgebra of $L\left(𝔤\right),$ and then that $\stackrel{\sim }{L}(𝔤,s)$ (resp. $\hat{L}(𝔤,s)\text{)}$ is a subalgebra of $\hat{L}\left(𝔤\right)$ (resp. $\hat{L}\left(𝔤\right)\text{).}$

Our aim is now to show that $\hat{L}(𝔤,s)\cong 𝔤\left(A^{\left(k\right)}\right)$ where $A^{\left(k\right)}$ is an indecomposable Cartan matrix of affine type, to be defined presently.

Let ${\Delta }_i$ $(1\le i\le \ell )$ be the orbits of $s$ in $\Delta ,$ and number the vertices of $\Delta$ so that $i\in {\Delta }_i\text{.}$ With one exception (case $A_{2\ell }\text{)}$ ${\Delta }_i$ is discrete (no joining edges). In the exceptional case, ${\Delta }_i$ is connected (of type $A_2\text{).}$ Define

$$u_i=\{\begin{array}{cc}1,& \text{if}{\Delta }_i\text{is discrete},\\ 2,& \text{if}{\Delta }_i\text{is connected},\end{array}$$

and put

$$u=\underset{1\le i\le \ell }{\text{max}}{u}_i$$

(so that $u=1$ except in case $A_{2\ell },$ where $u=2\text{.)}$

Let

$$E_i=u_i^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sum _{j\in {\Delta }_i}{e}_j,F_i=u_i^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sum _{j\in {\Delta }_i}{f}_j,H_i=u_i^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sum _{j\in {\Delta }_i}{h}_j$$

for $1\le i\le \ell \text{.}$ These elements are all fixed by $s,$ hence they generate a subalgebra $\overline{𝔤}$ of ${𝔤}^{\left(0\right)}\text{.}$ Let $\overline{𝔥}$ be the subspace of $𝔥$ spanned by the $H_i,$ and note that $\overline{𝔥}={𝔥}^{\left(0\right)}=\{h\in 𝔥:sh=h\}\text{.}$

Next define

$${\bar{a}}_{ij}=u_i\sum _{p\in {\Delta }_i}{a}_{pj}(1\le i,j\le \ell )$$

and let $\bar{A}={\left(a_{ij}\right)}_{1\le i,j\le \ell }\text{.}$

We have then

$$\begin{array}{ccc}⟨H_i,H_j⟩& =& u_i{u}_j\sum _{\underset{q\in {\Delta }_j}{p\in {\Delta }_i}}{a}_{pq}\\ & =& u_j\mid {\Delta }_j\mid {\bar{a}}_{ij}\text{.}\end{array}$$

(4.8)

1. $\bar{A}$ is an indecomposable Cartan matrix of finite type, given by the following table $$\begin{array}{|ccccccc|}\hline k& 2& 2& 2& 2& 2& 3\\ A& A_2& A_{2\ell }(\ell \ge 2)& A_{2\ell -1}(\ell \ge 2)& D_{\ell H}(\ell \ge 3)& E_6& D_4\\ \bar{A}& A_1& D_{\ell }& C_{\ell }& B_{\ell }& F_4& G_2\\ \hline\end{array}$$
2. $\overline{𝔤}\cong 𝔤\left(\bar{A}\right)\text{.}$

		Proof.
	It is straightforward to verify from the definition (4) that $\bar{A}$ is a Cartan matrix, and that $\mid {\bar{a}}_{ij}\mid \le 3\text{.}$ Let $\Delta ,\bar{\Delta }$ be the Dynkin diagrams of $A$ and $\bar{A}\text{.}$ Then $\bar{\Delta }$ is derived from $\Delta$ by the following rules (which are a restatement of (4)): $$\begin{array}{ccc} & \text{is replaced by}& \\ & \text{is replaced by}& \\ & \text{is replaced by}& \\ & \text{is replaced by}& \\ & \text{is replaced by}& \end{array}$$ (In the left-handed column, vertices in the same vertical line are in the same $s$-orbit). Hence the type of $\bar{A}$ is as stated in the table above. It is straightforward to verify that the generators $E_i,F_i,H_i$ of $\overline{𝔤}$ satisfy the relations $\text{(1.2}\prime \text{)}$ for the matrix $\bar{A}\text{.}$ To complete the proof, it will be enough to verify that they satisfy Serre's relations $$\begin{array}{cc}{\left(\text{ad}{E}_i\right)}^{1-{\bar{a}}_{ij}}{E}_j={\left(\text{ad}{F}_i\right)}^{1-{\bar{a}}_{ij}}{F}_j=0(i\ne j)\text{.}& \text{(}✶\text{)}\end{array}$$ For it will then follow from (2.???) that $\overline{𝔤}$ is a homomorphic image of $𝔤\left(\bar{A}\right);$ but $𝔤\left(\bar{A}\right)$ is simple, hence $\overline{𝔤}\cong 𝔤\left(A\right)\text{.}$ To prove $\left(✶\right),$ there are two cases to consider, according as the orbit ${\Delta }_i\subset \Delta$ is discrete or connected. Suppose ${\Delta }_i$ discrete. If it consists of the single element $i,$ then we have ${\bar{a}}_{ij}=a_{ij},$ $E_i=e_i,$ and $\left(✶\right)$ follows from the corresponding relation for $𝔤\text{.}$ If ${\Delta }_i$ consists of $k$ elements, then for $p\ne q$ in ${\Delta }_i$ we have $⟨{\alpha }_p,{\alpha }_q⟩=0$ and therefore by (4.6) ${\alpha }_p+{\alpha }_q$ is not a root, so that $[e_p,e_q]=0$ and hence $\text{ad}{e}_p,$ $\text{ad}{e}_q$ commute. Hence $${\left(\text{ad}{E}_i\right)}^{1-{\bar{a}}_{ij}}{E}_j={\left(\sum _{p\in {\Delta }_i}\text{ad}{e}_p\right)}^{1-{\bar{a}}_{ij}}\left(\sum _{q\in {\Delta }_j}{e}_q\right)$$ is a sum of terms $\prod _{p\in {\Delta }_i}{\left(\text{ad}{e}_p\right)}^{n_p}{e}_q,$ where $q\in {\Delta }_j$ and $$\sum _{p\in {\Delta }_i}{n}_p=1-{\bar{a}}_{ij}=1-\sum _{p\in {\Delta }_i}{a}_{pq},$$ so that $n_p\ge 1-a_{pq}$ for at least one $p\in {\Delta }_i,$ and therefore ${\left(\text{ad}{e}_p\right)}^{n_p}{e}_q=0\text{.}$ It follows that ${\left(\text{ad}{E}_i\right)}^{1-{\bar{a}}_{ij}}{E}_j=0,$ and likewise with the $E$'s replaced by ??? Suppose ${\Delta }_i$ connected. Then $k=2$ and ${\Delta }_i=\{i,si\}\text{.}$ Since $\Delta$ contains no cycles, at least one of $a_{ij},$ $a_{si,j}$ is zero. If both are zero, then $[E_i,e_j]=0$ and therefore $[E_i,E_j]=0\text{.}$ If say $a_{ij}=0,$ $a_{si,j}=-1,$ then ${\bar{a}}_{ij}=-2,$ and we have to show that ${(\text{ad}{e}_i+\text{ad}{e}_{si})}^3{e}_j=0\text{.}$ Let $x=\text{ad}{e}_i,$ $y=\text{ad}{e}_{si}\text{.}$ Then $x{e}_j=y^2{e}_j=0;$ Moreover $z=[x,y]=\text{ad}[e_i,e_{si}]$ commutes with $x$ and $y$ (because $2{\alpha }_i+{\alpha }_{si}$ and ${\alpha }_i+2{\alpha }_{si}$ are not roots), hence $x^{2}y{e}_j=xz{e}_j=zx{e}_j=0$ and $yxy{e}_j=yz{e}_j=zy{e}_j=-yxy{e}_j,$ so that we have altogether $$x{e}_j=y^2{e}_j=x^{2}y{e}_j=yxy{e}_j=0$$ and therefore $${(x+y)}^3{e}_j={(x+y)}^{2}y{e}_j=(x+y)xy{e}_j=0\text{.}$$ $\square$

$\overline{𝔥}$ is a Cartan subalgebra of $\overline{𝔤}\text{.}$ Let $p:{𝔥}^{*}\to {\overline{𝔥}}^{*}$ be the restriction map, and let ${\bar{\alpha }}_i=p\left({\alpha }_i\right)$ $(1\le i\le \ell )\text{.}$ Then

$${\bar{\alpha }}_j\left(H_i\right)=e_j\left(u_i\sum _{p\in {\Delta }_i}{h}_p\right)={\bar{a}}_{ij}$$

so that the ${\bar{\alpha }}_i$ are the simple roots of $\overline{𝔤}$ (relative to $\overline{𝔥}\text{).}$ If $\alpha \in R,$ say $\alpha =\sum _1^n{m}_i{\alpha }_i$ the $p\left(\alpha \right)=\text{????}\in \bar{Q}$ in particular $p\left(\alpha \right)\ne 0\text{.}$

Let $\bar{R}$ be the root system of $\overline{𝔤}$ and let $\bar{Q}=\sum _1^{\ell }\mathbb{Z}{\bar{\alpha }}_i$ be the root lattice.

We have $⟨H_i,H_j⟩=u_j\mid {\Delta }_j\mid {\bar{a}}_{ij},$ from which it follows that

$$\begin{array}{ccc}⟨{\bar{\alpha }}_i,{\bar{\alpha }}_j⟩& =& {\left(u_i\mid {\Delta }_i\mid \right)}^{-1}{\bar{a}}_{ij}\\ & =& {\mid {\Delta }_i\mid }^{-1}\sum _{k\in {\Delta }_i}⟨{\alpha }_k,{\alpha }_j⟩\\ & =& ⟨\pi {\alpha }_i,{\alpha }_j⟩\end{array}$$

where ${\displaystyle \pi =\frac{1}{k}\sum _{i=0}^{k-1}{s}^i:{𝔥}^{*}\to {𝔥}^{*}\text{.}}$ Hence, by linearity, we have

$$⟨p\left(\lambda \right),p\left(\mu \right)⟩=⟨\pi \left(\lambda \right),\mu ⟩$$

for all $\lambda ,\mu \in {𝔥}^{*}\text{.}$

In particular

$${\mid {\bar{\alpha }}_i\mid }^2=2{\left(u_i\mid {\Delta }_i\mid \right)}^{-1}$$

and therefore ${\mid {\bar{\alpha }}_i\mid }^2=2u^{-1}$ or $2{\left(ku\right)}^{-1}$ for all $\bar{\alpha }\in \bar{R}\text{.}$

(4.9) Let $\alpha ,\beta \in R$ be such that $p\left(\alpha \right)=p\left(\beta \right)\text{.}$ Then $\alpha ,\beta$ are in the same $s$-orbit in $R\text{.}$

		Proof.
	Suppose not, i.e. $\beta \ne s^i\alpha$ for $0\le i\le k-1\text{.}$ Since $p(\beta -s^i\alpha )=p(\beta -\alpha )=0,$ it follows that $\beta -s^i\alpha \notin R,$ hence (4.5) $⟨\beta ,s^i\alpha ⟩\le 0\text{.}$ But then $${\mid p\left(\alpha \right)\mid }^2=⟨p\left(\alpha \right),p\left(\beta \right)⟩=⟨\beta ,\pi \alpha ⟩\le 0$$ so that $p\left(\alpha \right)=0,$ which is impossible. $\square$

Since $\overline{𝔤}$ is subalgebra of ${𝔤}^{\left(0\right)},$ each ${𝔤}^{\left(r\right)}$ (and $𝔤)$ is a $\overline{𝔤}$-module. Let $S^{\left(r\right)}$ (resp. $S\text{)}$ be the set of nonzero weights of ${𝔤}^{\left(r\right)}$ (resp. $𝔤\text{)}$ as $\overline{𝔤}$-module (or $\overline{𝔥}$-module).

Clearly

$$\bar{R}\subset S^{\left(0\right)};S=\bigcup _{r=0}^{k-1}{S}^{\left(r\right)};$$

also

$$p\left(R\right)=S\subset \bar{Q}$$

because $R$ is the set of weights of $𝔤$ as $𝔥$-module. By (4.9) the fibres of $p:R\to S$ are the orbits of $S$ in $R,$ hence have 1 or $k$ elements.

1. If $\alpha =s\alpha$ then $s\left({𝔤}_{\alpha }\right)={𝔤}_{\alpha },$ hence $s{e}_{\alpha }={\omega }^r{e}_{\alpha }$ for some $r=0,\dots ,k-1,$ and correspondingly $e_{\alpha }\in {𝔤}^{\left(r\right)},$ hence $p\left(\alpha \right)\in s^{\left(r\right)}$ with multiplicity 1, for this value of $r\text{.}$
2. If $\alpha \ne s\alpha$ then $e_{\alpha },s{e}_{\alpha },\dots ,s^{k-1}{e}_{\alpha }$ are linearly independent, hence $e_{\alpha }^{\left(r\right)}\ne 0$ for each $r=0,1,\dots ,k-1\text{.}$ It follows that $p\left(\alpha \right)\in S^{\left(r\right)}$ with multiplicity 1, for each $r=0,1,\dots ,k-1\text{.}$

Thus all nonzero weights of each ${𝔤}^{\left(r\right)}$ occur with multiplicity 1.

For $\alpha \in R$ there are three (mutually exclusive) possibilities:

1. $\alpha =s\alpha ;$
2. $\alpha \ne s\alpha ;$ $⟨\alpha ,s\alpha ⟩=0;$
3. $\alpha \ne s\alpha ;$ $⟨\alpha ,s\alpha ⟩=-1\text{.}$

		Proof.
	We have $$\begin{array}{ccccc}{\mid p\left(\alpha \right)\mid }^2& =& ⟨\alpha ,\pi \alpha ⟩& =& \{\begin{array}{cc}⟨\alpha ,\alpha ⟩=2,& \text{in case (i),}\\ \frac{1}{k}⟨\alpha ,\alpha ⟩=\frac{2}{k},& \text{in case (ii),}\\ \frac{1}{2}(⟨\alpha ,\alpha ⟩+⟨\alpha ,s\alpha ⟩)=\frac{1}{2},& \text{in case (iii),}\end{array}\end{array}$$ (by (4.6), since $k=2$ in case (iii)). Suppose first that $u=1\text{.}$ Then case (iii) does not occur. For by (2.34) we have $$\text{min}\{{\mid \lambda \mid }^2:\lambda \in \bar{Q},\lambda \ne 0\}=\frac{2}{k}$$ and $p\left(\alpha \right)\in \bar{Q}\text{.}$ Again by (2.34), if ${\mid p\left(\alpha \right)\mid }^2=\frac{2}{k}$ (case (ii)) then $p\left(\alpha \right)\in \bar{R};$ whilst if ${\mid p\left(\alpha \right)\mid }^2=2,$ i.e. if $\alpha =s\alpha ,$ then if $\alpha =\sum _{i=1}^n{m}_i{\alpha }_i$ we have $m_i=m_{si}$ and therefore $p\left(\alpha \right)=\sum _{i=1}^{\ell }\mid {\Delta }_i\mid m_i{\bar{\alpha }}_i,$ whence again by (2.34) we have $p\left(\alpha \right)\in \bar{R}\text{.}$ So if $u=1$ we have $$S=\bar{R}$$ and therefore (since $\bar{R}\subset S^{\left(0\right)}\subset S\text{)}$ $$S^{\left(0\right)}=\bar{R},S^{\left(r\right)}={\bar{R}}_{\text{short}}(1\le r\le k-1)$$ where ${\bar{R}}_{\text{short}}$ is the set of short roots $\bar{\alpha }\in \bar{R}$ (i.e. with ${\mid \bar{\alpha }\mid }^2=\frac{2}{k}\text{).}$ There remains the case $A_{2\ell }^{\left(2\right)},$ where $u=2\text{.}$ In this case ${\mid \bar{\alpha }\mid }^2=1$ or $\frac{1}{2}$ for $\bar{\alpha }\in \bar{R},$ and we find by direct calculation that $$S^{\left(o\right)}=\bar{R},S^{\left(1\right)}=\bar{R}\cup 2{\bar{R}}_{\text{short}}\text{.}$$ Thus in all cases $\begin{array}{|c|}\hline {𝔤}^{\left(0\right)}=\overline{𝔤}\text{.}\\ \hline\end{array}$ Now let $\bar{\psi }$ be the highest short root of $\bar{R}$ (i.e., ${\bar{\psi }}^{\vee }$ is the highest root of ${\bar{R}}^{\vee }\text{),}$ and put $$\bar{\phi }=u\bar{\phi }\text{.}$$ Then $\bar{\phi }$ is the highest weight of ${\overline{𝔤}}^{\left(r\right)}$ $(1\le r\le k-1),$ and is the only weight of ${\overline{𝔤}}^{\left(r\right)}$ such that $\bar{\phi }+{\bar{\alpha }}_i\notin S^{\left(r\right)},$ for $i=1,\dots ,\ell \text{.}$ It follows that ${\overline{𝔤}}^{\left(r\right)}$ is simple $(1\le r\le k-1)\text{.}$ as a $\overline{𝔤}$-module. For in any case, by complete reducibility, ${\overline{𝔤}}^{\left(r\right)}$ is a direct sum of simple $\overline{𝔤}$-modules. Let $M$ be one of these, with highest weight $\lambda \in S^{\left(r\right)}\text{.}$ One sees easily that $\lambda \ne 0,$ hence $M_{\lambda }={𝔤}_{\lambda }^{\left(r\right)}$ (because this space is 1-dimensional). But then $[E_i,{𝔤}_{\lambda }^{\left(r\right)}]=[E_i,M_{\lambda }]=0$ for $1\le i\le \ell ,$ and therefore (Lemma) $\lambda +{\bar{\alpha }}_i\notin S^{\left(r\right)}$ and consequently $\lambda =\bar{\phi }\text{.}$ $\square$

(Alternatively: instead of using completed reducibility, use the dimension formula to compute $\text{dim}L\left(\bar{\phi }\right)$ in each case.)

We now proceed as in the case considered previously.

Normalize the scalar product on $\overline{𝔤}$ so that we have

$$⟨\bar{\phi },\bar{\phi }⟩=2$$

(no renormalization in case $A_{2\ell }^{\left(2\right)},$ because then ${\mid \bar{\psi }\mid }^2=\frac{1}{2},$ $u=2,$ hence ${\mid \bar{\phi }\mid }^2=2\text{).}$ Choose $E_{\bar{\phi }}\in {𝔤}_{\bar{\phi }}^{(-1)},$ $F_{\bar{\phi }}={𝔤}_{-\bar{\phi }}^{\left(1\right)}$ such that

$$⟨E_{\bar{\phi }},F_{\bar{\phi }}⟩=1$$

(this is possible by (4.7)). Then we have

$$[E_{\bar{\phi }},F_{\bar{\phi }}]=H_{\bar{\phi }}$$

(where $H_{\bar{\phi }}$ is the image of $\bar{\phi }$ in $\overline{𝔥}\text{),}$ because if $H\in \overline{𝔥}$ we have

Then define

$$E_0=t{F}_{\bar{\phi }}\in t{𝔤}^{\left(1\right)},F_0=t^{-1}{E}_{\bar{\phi }}\in t^{-1}{𝔤}^{(-1)},H_0=-H_{\bar{\phi }}+c$$

and set $\stackrel{\Delta }{𝔥}=\overline{𝔥}\oplus kc\oplus kd\text{.}$ Extend each $\bar{\alpha }\in S$ to a linear form (also denoted by $\bar{\alpha }\text{)}$ on $\stackrel{\Delta }{𝔥}$ by setting $\bar{\alpha }\left(c\right)=\bar{\alpha }\left(d\right)=0;$ also define $\bar{\delta }\in {\stackrel{\Delta }{𝔥}}^{*}$ by

$$\bar{\delta }\mid (\overline{𝔥}\oplus kc)=0,\bar{\delta }\left(d\right)=1\text{(}\bar{\delta }=\text{restriction of}\delta \text{for}\stackrel{\Delta }{𝔥}\text{)}$$

Finally set

$${\bar{\alpha }}_0=\bar{\delta }-\bar{\phi };$$

we have

$$\bar{\phi }=\sum _{i=1}^{\ell }{\bar{a}}_i{\bar{\alpha }}_i$$

with coefficients ${\bar{a}}_i\ge 1,$ so that if we define ${\bar{a}}_0=1$ we have

$$\sum _{i=0}^{\ell }{\bar{a}}_i{\bar{\alpha }}_i=\bar{\delta }$$

Let

$${\bar{a}}_{ij}={\bar{\alpha }}_j\left(H_i\right)(0\le i,j\le \ell )$$

and let $A^{\left(k\right)}={\left({\bar{a}}_{ij}\right)}_{0\le i,j\le \ell }\text{.}$ The matrix $A^{\left(k\right)}$ has $\bar{A}$ as a principal submatrix. One then verifies that

1. $A^{\left(k\right)}$ is an indecomposable Cartan matrix of affine type;
2. $(\stackrel{\Delta }{𝔥},{\left(H_i\right)}_{0\le i\le \ell },{\left({\bar{\alpha }}_i\right)}_{0\le i\le \ell })$ is a minimal realization of $A\left(k\right);$
3. the $E_i,F_i,\stackrel{\Delta }{𝔥}$ satisfy the defining relations (1.2)
4. the weights of $\stackrel{\Delta }{𝔥}$ in $\hat{L}(𝔤,s)$ are $\bar{\alpha }+r\bar{\delta }$ where $\bar{\alpha }\in S^{\left(r\right)},$ $r\in \mathbb{Z},$ and $r\bar{\delta }$ with multiplicity $=\text{dim}{𝔥}^{\left(r\right)}=$ multiplicity of ${\omega }^m$ as eigenvalues of $s$ on $𝔥\text{.}$ Thus $$m_{r\bar{\delta }}=\{\begin{array}{cc}\ell ,& \text{if}r\equiv 0\text{(mod}k\text{),}\\ \frac{n-\ell }{k-1},& \text{otherwise,}\end{array}$$ (because if the latter multiplicity is ${\ell }^{\prime }$ then $\ell +(k-1){\ell }^{\prime }=n\text{)}$
5. $\begin{array}{ccc}\hat{L}(𝔤,s)& \cong & 𝔤\left(A^{\left(k\right)}\right)\\ \stackrel{\sim }{L}(𝔤,s)& \cong & {𝔤}^{\prime }\left(A^{\left(k\right)}\right)\\ L(𝔤,s)& \cong & {\overline{𝔤}}^{\prime }\left(A^{\left(k\right)}\right)\end{array}$

References

I.G. Macdonald
Issac Newton Institute for the Mathematical Sciences
20 Clarkson Road
Cambridge CB3 OEH U.K.

Version: October 30, 2001

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