Kac-Moody Lie Algebras
Chapter I

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

Last update: 16 August 2012

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

Construction of the algebras

The construction of the Kac-Moody algebras works for any matrix over $K$:

$A={\left(a_{ij}\right)}_{1\le i,j\le n}$

– the $a_{ij}$ needn't even be integers.

We begin with a lemma of linear algebra. Given a matrix $A$ as above, a realization of $A$ is a triple $(𝔥,B,B^{\vee })$ where:

$\begin{array}{c}𝔥\text{is a finite-dimensional vector space over}k;\\ B=({\alpha }_1,\dots ,{\alpha }_n)\text{is a linearly independent set of vectors in}{𝔥}^{*}=\text{dual of}𝔥;\\ B^{\vee }=(h_1,\dots ,h_n)\text{is a linearly independent set of vectors in}𝔥;\text{such that}\\ {\alpha }_j\left(h_i\right)=a_{ij}(1\le i,j\le n).\end{array}$

A realization of $A$ will be called minimal if dim $𝔥$ is as small as possible (evidently it must be $\ge n$).

Let $l=\text{rank}\left(A\right)$

(1.1)

1. If $(𝔥,B,B^{\vee })$ is a realization of $A$, then dim $𝔥\ge 2n-l$.
2. $A$ has a minimal realization, of dimension $2n-l$, which is unique up to isomorphism (but the isomorphism is not unique if $l<n$).

		Proof.
	Extend $B^{\vee }$ to a basis $h_1,\dots ,h_N$ of $𝔥$ (so that dim $𝔥=N$). The $N\times n$ matrix $M={\left({\alpha }_j\left(h_i\right)\right)}_{1\le i\le N,1\le j\le n}$ is of the form $M=\left(\begin{array}{c}A\\ B\end{array}\right)$ and has rank $n$ (because its columns are linearly independent). Let $V_A,V_b,V_M$ denote the spaces spanned by the rows of $A,B,M$ respectively. Then we have $V_M=V_A+V_B$, and $\text{dim}{V}_A=\text{rank}\left(A\right)=l,\text{dim}{V}_M=\text{rank}\left(M\right)=l$. Hence $N-n\ge \text{dim}{V}_B\ge n-l$ i.e., $N\ge 2n-l$. By reordering the rows and the columns of $A$ we may assume that the $l\times l$ minor of $A$ in the top left-hand corner is nonsingular, say $A=\left(\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\end{array}\right)$ with $A_1$ a nonsingular $l\times l$ matrix. Let $C=\left(\begin{array}{ccc}{A}_1& A_2& 0\\ A_3& A_4& 1_{n-l}\\ 0& 1_{n-l}& 0\end{array}\right)$, then $\text{det}C=\pm \text{det}{A}_1\ne 0$; hence the rows of $C$ are linearly independent. Take $𝔥=k^{2n-l}$ (row vectors); ${\alpha }_j$ the $j$th coordinate function on $𝔥,h_i$ the $i$th row of WHAT GOES HERE?. Then ${\alpha }_j\left(h_i\right)=a_{ij}$ and we have a realization of $A$. Conversely, let $(𝔥,B,B^{\vee })$ be a minimal realization of $A$ $(\text{dim}𝔥=2n-l)$. Extend $B^{\vee }$ to a basis $h_1,\dots ,h_{2n-l}$ of $𝔥$, and define ${\alpha }_{n+1},\dots ,{\alpha }_{2n-l}\in {𝔥}^{*}$ so that the matrix $D={\left({\alpha }_j\left(h_i\right)\right)}_{1\le i,j\le 2n-l}$ has the form $D=\left(\begin{array}{ccc}{A}_1& A_2& 0\\ A_3& A_4& 1_{n-l}\\ B_1& B_2& 0\end{array}\right)$ $B_1,B_2$ at present unspecified. Then $\text{det}D=\pm \text{det}\left(\begin{array}{cc}{A}_1& A_2\\ B_1& B_2\end{array}\right)$, and I claim that this matrix is nonsingular. For the submatrix $M=\left(\begin{array}{cc}{A}_1& A_2\\ A_3& A_4\\ B_1& B_2\end{array}\right)=\left(\begin{array}{c}A\\ B\end{array}\right)$ of $D$ has rank $n$, as before, and now we have $V_M=V_A\oplus V_B$; but $V_A$ has the first $l$ rows of $A$ as a basis, and the rows of $B$ form a basis of $V_B$; hence the rows of $\left(\begin{array}{cc}{A}_1& A_2\\ B_1& B_2\end{array}\right)$ are linearly independent, as claimed. It follows that $D$ is nonsingular, hence that ${\alpha }_1,\dots ,{\alpha }_{2n-l}$ are a basis of ${𝔥}^{*}$. By adding to $h_{n+1},\dots ,h_{2n-l}$ suitable linear combinations of $h_1,\dots ,h_l$, we can make $B_1=0$. But then $\text{det}{B}_2\ne 0$, and we can choose another basis of the subspace of $𝔥$ spanned by $h_{n+1},\dots ,h_{2n-l}$ so as to make $B_2=1_{n-l}$, i.e. $D=C$. This completes the proof. $\square$

A matrix $A$ as above will be said to be decomposable if we can partition the index set $\{1,\dots ,n\}$ into two non-empty disjoint subsets $I,J$ such that $a_{ij}=a_{ji}=0$ whenever $i\in I$ and $j\in J$. In other words, if after simultaneous permutation of rows and columns $A$ becomes a nontrivial direct sum $A_1\oplus A_2$.

Clearly, if $({𝔥}_i,B_i,B_i^{\vee })$ is a minimal realization of $A_i(i=1,2)$, then $(𝔥,B,B^{\vee })$ is a minimal realization of $A$, where

$\begin{array}{ccc}𝔥& =& {𝔥}_1\times {𝔥}_2{𝔥}^{*}={𝔥}_1^{*}\times {𝔥}_2^{*}\text{(direct sum)}\\ B^{\vee }& =& (B_1^{\vee }\times 0)\cup (0\times B_2^{\vee })\\ B& =& (B_1\times 0)\cup (0\times B_2)\end{array}$

Again, if $(𝔥,B,B^{\vee })$ is a minimal realization of $A$, then $({𝔥}^{*},B^{\vee },B)$ is a minimal realization of $A^t$.

Let $A$ be any $n\times n$ matrix over $k$, as before; let $(𝔥,B,B^{\vee })$ be a minimal realization of $A(B=({\alpha }_1,\dots ,{\alpha }_n);B^{\vee }=(h_1,\dots ,h_n))$. Let $\stackrel{\sim }{𝔤}\left(A\right)$ denote the Lie algebra generated by $𝔥$ and $2n$ elements $e_i,f_i(1\le i\le n)$ subject to the relations

$\begin{array}{cc}\text{(1.2)}& \{\begin{array}{ccc}[h,h^{\prime }]=0& & (\text{all}h,h^{\prime }\in 𝔥)\\ [e_i,f_j]={\delta }_{ij}{h}_i& & (1\le i,j\le n)\\ \begin{array}{c}[h,e_i]={\alpha }_i\left(h\right)e_i\\ [h,f_i]=-{\alpha }_i\left(h\right)f_i\end{array}\}& & (1\le i\le n;h\in 𝔥)\end{array}\end{array}$

By (1.1), $\stackrel{\sim }{𝔤}\left(A\right)$ depends (up to isomorphism) only on the matrix $A$.

Objects defined by generators and relations are often not easy to handle directly. To get a group on $\stackrel{\sim }{𝔤}\left(A\right)$ we shall construct a family of representations ${\rho }_{\lambda }$ of $\stackrel{\sim }{𝔤}\left(A\right)$, one for each $\lambda \in {𝔥}^{*}$check the wedge?. These representations will act on the same vector space $X:X$ is the free associative algebra over $k$ on $n$ generators $x_1,\dots ,x_n$, and we define an action of the generators of $\stackrel{\sim }{𝔤}\left(A\right)$ on $X$ as follows: Let $\lambda \in {𝔥}^{*}$ and define

$\begin{array}{cccc}\left(a\right)& h\left(1\right)=\lambda \left(h\right)·1;& h\left(x_{j}x\right)=x_{j}h\left(x\right)-{\alpha }_j\left(h\right)x_{j}x& (x\in X,h\in 𝔥,1\le j\le n)\\ \left(b\right)& e_i\left(1\right)=0;& e_i\left(x_{j}x\right)=x_j{e}_i\left(x\right)+{\delta }_{ij}{h}_i\left(x\right)& (x\in X,1\le i,j\le n)\\ \left(c\right)& f_i\left(x\right)=x_{i}x& (x\in X,1\le i\le n).\end{array}$

I claim that these formulas define a representation ${\rho }_{\lambda }$ of $\stackrel{\sim }{𝔤}\left(A\right)$ on $X$. To verify this, we have to check the defining relations (1.2).

First, it follows from ($a$) that

$h\left(x_{j_1}\dots x_{j_r}\right)=(\lambda -{\alpha }_{j_1}-\dots -{\alpha }_{j_r})\left(h\right)x_{j_1}\dots x_{j_r}$

by induction on $r$; hence each $h\in 𝔥$ acts diagonally on $X$ (relative to the basis of $X$ formed by the monomials) and therefore $[h,h^{\prime }]=0$ for all $h,h^{\prime }\in 𝔥$.

Next, we have

$\begin{array}{cc}[e_i,f_j]={\delta }_{ij}{h}_j& \text{from}\left(b\right)\text{and}\left(c\right)\\ [h,f_j]=-{\alpha }_j\left(h\right)f_j& \text{from}\left(a\right)\text{and}\left(c\right)\end{array}$

and therefore it remains to show that $[h,e_i={\alpha }_i\left(h\right)e_i]$ (as linear transformations of the vector space $X$). So let $u=[h,e_i]-{\alpha }_i\left(h\right)e_i$, then

$\begin{array}{ccc}[u,f_j]& =& [[h,e_i],f_j]-{\alpha }_i\left(h\right)[e_i,f_j]\\ & =& [h,[e_i,f_j]]-[e_i,[h,f_j]]-{\alpha }_i\left(h\right)[e_i,f_j]\\ & =& [h,{\delta }_{ij}{h}_i]+{\alpha }_j\left(h\right){\delta }_{ij}{h}_i-{\alpha }_i\left(h\right){\delta }_{ij}{h}_i\\ & =& 0\end{array}$

Hence $u\left(x_{j}x\right)=x_{j}u\left(x\right)$ for all $x\in X$ and $1\le j\le n$; hence (induction on $r$) $u\left(x_{j_1}\dots x_{j_r}\right)=x_{j_1}\dots x_{j_r}u\left(1\right)$; but

$\begin{array}{ccc}u\left(1\right)& =& h\left(e_i\left(1\right)\right)=-e_i\left(h\left(1\right)\right)-{\alpha }_i\left(h\right)e_i\left(1\right)\\ & =& -\lambda \left(h\right)e_i\left(1\right)=0\end{array}$

and therefore $u=0$ as required.

Thus for each $\lambda \in {𝔥}^{*}$ the formulas ($a$) - ($c$) define a representation ${\rho }_{\lambda }$ of $\stackrel{\sim }{𝔤}\left(A\right)$ on $X$.

This may look like a rabbit pulled out of a hat: in fact it is a standard construction (Verma module).

We may remark straightaway that the canonical mapping $𝔥\to \stackrel{\sim }{𝔤}\left(A\right)$ is injective. For if $h\in 𝔥$ because zero in $\stackrel{\sim }{𝔤}\left(A\right)$, then from ($a$) we have $\lambda \left(h\right)·1=0$ for all $\lambda \in {𝔥}^{*}$, and hence $h=0$.

Let ${\stackrel{\sim }{𝔫}}_{+}$ (resp. ${\stackrel{\sim }{𝔫}}_{-}$) denote the subalgebra of $\stackrel{\sim }{𝔤}\left(A\right)$ generated by $e_1,\dots ,e_n$ (resp. $f_1,\dots ,f_n$).

(1.3)

1. $\stackrel{\sim }{𝔤}\left(A\right)={\stackrel{\sim }{𝔫}}_{-}\oplus 𝔥\oplus {\stackrel{\sim }{𝔫}}_{+}$ (direct sum of vector spaces)
2. ${\stackrel{\sim }{𝔫}}_{+}$ (resp. ${\stackrel{\sim }{𝔫}}_{-}$) is the free Lie algebra generated by $e_1,\dots ,e_n$ (resp. $f_1,\dots ,f_n$)
3. $\exists$ unique involutory automorphsim $\stackrel{\sim }{\omega }$ of $\stackrel{\sim }{𝔤}\left(A\right)$ such that

   $\stackrel{\sim }{\omega }\left(e_i\right)=-f_i,\stackrel{\sim }{\omega }\left(f_i\right)=-e_i,\stackrel{\sim }{\omega }\left(h\right)=-h,(h\in 𝔥)$.

		Proof.
	We shall take these in reverse order. is clear, since the relations (1.2) are stable under $\stackrel{\sim }{\omega }$. Since $X$ is the free associative algebra on $x_1,\dots ,x_n$, L($X$) is the free Lie algebra on the same generators. Now the mapping $\phi :{\stackrel{\sim }{𝔫}}_{-}\to \text{L}\left(X\right)$ defined by $\phi \left(f\right)=f\left(1\right)$ takes $f_i$ to $x_i$ and is a Lie algebra homomorphism. Since L($X$) is free, $\phi$ must be an isomorphism, and ${\stackrel{\sim }{𝔫}}_{-}$ is the free Lie algebra on $f_1,\dots ,f_n$, and $U\left({\stackrel{\sim }{𝔫}}_{-}\right)\cong X$. By applying $\stackrel{\sim }{\omega }$, we see that ${\stackrel{\sim }{𝔫}}_{+}$ is the free Lie algebra on $e_1,\dots ,e_n$. Let $𝔞={\stackrel{\sim }{𝔫}}_{-}+𝔥+{\stackrel{\sim }{𝔫}}_{+}$. It follows easily from the defining relations (1.2) that $𝔞$ is stable under ad $e_i$, ad $f_i$ and ad $h$ $h\in 𝔥$. Hence it is an ideal in $\stackrel{\sim }{𝔤}\left(A\right)$, and since it contains the generators $e_i,f_i,h\in 𝔥$ it is the whole of $\stackrel{\sim }{𝔤}\left(A\right)$. Remains to prove that the sum is direct. Suppose then that we have ${𝔫}_{-}\in {\stackrel{\sim }{𝔫}}_{-},h\in 𝔥,{𝔫}_{+}\in {\stackrel{\sim }{𝔫}}_{+}$ such that ${𝔫}_{-}+h+{𝔫}_{+}=0$ Apply ${\rho }_{\lambda }$ and evaluate at $1\in X$. We have ${𝔫}_{+}\left(1\right)=0$ (because $e_i\left(1\right)=0$), hence ${𝔫}_{-}\left(1\right)+\lambda \left(h\right)1=0\text{in}X$ whence $\lambda \left(h\right)=0$ and ${𝔫}_{-}\left(1\right)=0$. Since this is true for all $\lambda \in {𝔥}^{*}$, it follows first that $h=0$; next, as we have seen, ${𝔫}_{-}\mapsto {𝔫}_{-}\left(1\right):{\stackrel{\sim }{𝔫}}_{-}\to X$ is the embedding of ${\stackrel{\sim }{𝔫}}_{-}$ in its universal enveloping algebra $X\cong U\left({\stackrel{\sim }{𝔫}}_{-}\right)$; hence ${𝔫}_{-}=0$, whence finally ${𝔫}_{+}=0$ and the proof is complete. $\square$

Gradings

In general, if $A$ is a $k$–algebra and $G$ an abelian group, a $G$–grading of $A$ is a decomposition

$\begin{array}{cc}{\displaystyle A=\underset{\alpha \in G}{\oplus }{A}_{\alpha }}& \left(1\right)\end{array}$

of $A$ into a direct sum of $k$–subspaces $A_{\alpha }$, indexed by $G$, such that

$A_{\alpha }{A}_{\beta }\subset A_{\alpha +\beta }(\alpha ,\beta \in G)$.

The elements of $A_{\alpha }$ are said to be homogeneous of degree $\alpha$; the decomposition (1) says that any $x\in A$ can be written uniquely as the sum $x=\sum _{\alpha }{x}_{\alpha }$ of its homogeneous components (only finitely many of which can be $\ne 0$).

An ideal $𝔞$ in $A$ is a graded ideal if

${\displaystyle 𝔞=\underset{\alpha }{\oplus }{𝔞}_{\alpha }}$

where ${𝔞}_{\alpha }=𝔞\cap A_{\alpha }$, that is to say if whenever $x\in 𝔞$ all the homogeneous components $x_{\alpha }$ of $x$ lie in $𝔞$. Any sum of graded ideals is graded; any ideal generated by homogeneous elements is graded.

If $𝔞$ is a graded (two-sided) ideal, then $A/𝔞$ is a $G$–graded algebra:

${\displaystyle A/𝔞=\underset{\alpha \in G}{\oplus }{A}_{\alpha }/{𝔞}_{\alpha }}$.

In the present context, let

${\displaystyle Q=\sum _{i=1}^n\mathbb{Z}{\alpha }_i(\cong {\mathbb{Z}}^n)}$

denote the Lattice generated by $B$ in ${𝔥}^{*}$ (the root lattice). Also let

${\displaystyle Q^{+}=\sum _{i=1}^n\mathbb{N}{\alpha }_i}$

For $\alpha \in Q$ we write $\alpha \ge 0$ to mean $\alpha \in Q^{+}$, i.e. $\alpha =\sum _1^n{m}_i{\alpha }_i$ with all $m_i=0$; also $\alpha >0$ to mean $\alpha \in G^{+}$ and $\alpha \ne 0$. Likewise $\alpha \le 0,\alpha <0$. If $\alpha =\sum _1^n{m}_i{\alpha }_i\in Q$, we define the height of $\alpha$ to be

$\text{ht}\left(\alpha \right)=\sum _1^n{m}_i$.

Now the free Lie algebra generated by $𝔥$ and $e_1,\dots ,e_n,f_1,\dots ,f_n$ is $G$–graded by assigning degree 0 to each $h\in 𝔥$, degree ${\alpha }_i$ to $e_i$ and degree $-{\alpha }_i$ to $f_i$ $(1\le i\le n)$.

The relations (1.2) are homogeneous, hence $\stackrel{\sim }{𝔤}\left(A\right)$ is a $Q$–graded Lie algebra:

$\stackrel{\sim }{𝔤}\left(A\right)=\underset{\alpha \in Q}{\oplus }{\stackrel{\sim }{𝔤}}_{\alpha },[{\stackrel{\sim }{𝔤}}_{\alpha },{\stackrel{\sim }{𝔤}}_{\beta }]\subset {\stackrel{\sim }{𝔤}}_{\alpha +\beta }$

where ${\stackrel{\sim }{𝔤}}_{\alpha }$ consists of the homogeneous elements of degree $\alpha$ in $\stackrel{\sim }{𝔤}\left(A\right)$. By (1.3) we have ${\stackrel{\sim }{𝔤}}_0=𝔥$ and (for $\alpha \ne 0$) ${\stackrel{\sim }{𝔤}}_{\alpha }=0$ unless either $\alpha >0$ or $\alpha <0$, because

${\stackrel{\sim }{𝔫}}_{+}=\underset{\alpha >0}{\oplus }{\stackrel{\sim }{𝔤}}_{\alpha },{\stackrel{\sim }{𝔫}}_{-}=\underset{\alpha <0}{\oplus }{\stackrel{\sim }{𝔤}}_{\alpha }$.

We can introduce other gradings on $\stackrel{\sim }{𝔤}\left(A\right)$. Let $s:Q\to \mathbb{Z}$ be any homomorphism of abelian groups, and for each $m\in \mathbb{Z}$ define

${\stackrel{\sim }{𝔤}}_m\left(s\right)=\sum _{s\left(\alpha \right)=m}{\stackrel{\sim }{𝔤}}_{\alpha }$

Then $\stackrel{\sim }{𝔤}\left(A\right)=\underset{m\in \mathbb{Z}}{\oplus }{\stackrel{\sim }{𝔤}}_m\left(s\right)$, and $[{\stackrel{\sim }{g}}_m\left(s\right),{\stackrel{\sim }{g}}_{m^{\prime }}\left(s\right)]\subset {\stackrel{\sim }{𝔤}}_{m+m^{\prime }}\left(s\right)$, giving a $\mathbb{Z}$–grading of $\stackrel{\sim }{𝔤}\left(A\right)$. The most important case of this is the principal grading, defined by $s\left({\alpha }_i\right)=1(1\le i\le n)$. For this choice of $s$ we have $s\left(\alpha \right)=\text{ht}\left(\alpha \right)$, and we therefore define

${\stackrel{\sim }{𝔤}}_m=\sum _{\text{ht}\left(\alpha \right)=m}{\stackrel{\sim }{𝔤}}_{\alpha }(m\in \mathbb{Z})$

We have ${\stackrel{\sim }{𝔤}}_0=𝔥$; ${\stackrel{\sim }{𝔤}}_1$ is spanned by $e_1,\dots ,e_n;{\stackrel{\sim }{𝔤}}_{-1}$ by $f_1,\dots ,f_n$; and ${\stackrel{\sim }{𝔫}}_{+}=\underset{m\ge 1}{\oplus }{\stackrel{\sim }{𝔤}}_m,{\stackrel{\sim }{𝔫}}_{-}=\underset{m\ge 1}{\oplus }{\stackrel{\sim }{𝔤}}_{-m}$.

Let $\alpha >0$. Then ${\stackrel{\sim }{𝔤}}_{\alpha }$ is the $\alpha$–component of the free Lie algebra ${\stackrel{\sim }{𝔫}}_{+}$ generated by $e_1,\dots ,e_n$, hence it is spanned by all commutators

$x_{\alpha }=\left[e_{i_1}\dots e_{i_r}\right]$

such that ${\alpha }_{i_1}+\dots +{\alpha }_{i_r}=\alpha$ (notation: $\left[u_1\dots u_r\right]$ means $[u_1,[u_2,\dots u_r]]$). There are only finitely many of these (at most ${𝔫}^{h+\alpha }),$ hence ${\stackrel{\sim }{𝔤}}_{\alpha }$ is finite-dimensional. Likewise when $\alpha <0$. In particular, ${\stackrel{\sim }{𝔤}}_{\pm {\alpha }_i}$ are 1-dimensional.

For $x_{\alpha }$ as above we have, for each $h\in 𝔥$,

$\begin{array}{ccc}[h,x_{\alpha }]& =& {\displaystyle \sum _{p=1}^r}\left[e_{i_1}\dots [h,e_{i_p}]\dots e_{i_r}\right]\\ \multicolumn{3}{c}{\text{(because ad}h\text{is a derivation)}}\\ & =& {\displaystyle \sum _{p=1}^r}{\alpha }_{i_p}\left(h\right)\left[e_{i_1}\dots e_{i_p}\dots e_{i_r}\right]\text{by}\text{(1.2)}\\ & =& \alpha \left(h\right)x_d\end{array}$

Hence if temporarily we write

$M_{\alpha }=\{x\in \stackrel{\sim }{𝔤}\left(A\right):[h,x]=\alpha \left(h\right)x\text{for all}h\in 𝔥\}$

for each $\alpha \in Q$, then we have

${\stackrel{\sim }{𝔤}}_{\alpha }\subset M_{\alpha },\text{all}\alpha \in Q$.

(For the calculation above shows this is true when $\alpha >0$; similarly when $\alpha <0$ or $\alpha =0$; and in other cases ${\stackrel{\sim }{𝔤}}_{\alpha }=0$).

Moreover a standard argument shows that the sum $\sum _{\alpha \in Q}{M}_{\alpha }$ is direct. For if there exist non-trivial relations

$\begin{array}{cc}{\displaystyle \sum _{\alpha }}{x}_{\alpha }=0& \left(1\right)\end{array}$

with $x_{\alpha }\in M_{\alpha }$ (and only finitely many $x_{\alpha }\ne 0$), choose such a relation with as few non-zero terms as possible; by applying $\text{ad}h$ we conclude that

$\begin{array}{cc}\sum _{\alpha }\alpha \left(h\right)x_{\alpha }=0& \left(2\right)\end{array}$

for each $h\in 𝔥$. We can then subtract a multiple of (1) from (2) to obtain a shorter relation: contradiction. Hence we have

$\stackrel{\sim }{𝔤}\left(A\right)=\underset{\alpha \in Q}{\oplus }{\stackrel{\sim }{𝔤}}_{\alpha }\subset \underset{\alpha \in Q}{\oplus }{M}_{\alpha }\subset \stackrel{\sim }{𝔤}\left(A\right)$

from which it follows that $M_{\alpha }={\stackrel{\sim }{𝔤}}_{\alpha }$ for all $\alpha \in Q$. To summarize:

(1.4) For each $\alpha \in Q$, let ${\stackrel{\sim }{𝔤}}_{\alpha }$ denote the component of degree $\alpha$ in $\stackrel{\sim }{𝔤}\left(A\right)$. Then

1. ${\stackrel{\sim }{𝔤}}_{\alpha }=\{x\in \stackrel{\sim }{g}\left(A\right):[h,x]=\alpha \left(h\right)x\text{for all}h\in 𝔥\}$
2. ${\stackrel{\sim }{𝔤}}_{\alpha }=0$ unless $\alpha >0,\alpha <0$ or $\alpha =0$; moreover ${\stackrel{\sim }{𝔤}}_0=𝔥$.
3. each ${\stackrel{\sim }{𝔤}}_{\alpha }$ is finite-dimensional over $k$, and $\text{dim}{\stackrel{\sim }{𝔤}}_{\pm {\alpha }_i}=1$.

Ideals in $\stackrel{\sim }{𝔤}\left(A\right)$

We shall next prove that all ideals in $\stackrel{\sim }{𝔤}\left(A\right)$ are $Q$–graded ideals. This will be a consequence of the following lemma:

(1.5) Let $𝔥$ be an abelian Lie Algebra, $M$ an $𝔥$–module. For each $\lambda \in {𝔥}^{*}$ let

$M_{\lambda }=\{x\in M:h·x=\lambda \left(h\right)x\text{for all}h\in 𝔥\}$.

Suppose that $M=\underset{\lambda \in {𝔥}^{*}}{\oplus }{M}_{\lambda }$, and let $M^{\prime }$ be a submodule of $M$. Then

$M^{\prime }=\underset{\lambda \in {𝔥}^{*}}{\oplus }{M^{\prime }}_{\lambda },\text{where}{M^{\prime }}_{\lambda }=M^{\prime }\cap M_{\lambda }$.

		Proof.
	Each $x\in M^{\prime }$ can be written in the form $x={\displaystyle \sum _{i=1}^m}{x}_{{\lambda }_i}$ where ${\lambda }_1,\dots ,{\lambda }_m$ are district elements of ${𝔥}^{*}$, and $x_{{\lambda }_i}\in M_{{\lambda }_i}$. We have to show that each $x_{{\lambda }_i}\in M^{\prime }$. The polynomial function $\prod _{i<j}({\lambda }_i-{\lambda }_j)$ on $𝔥$ is not zero, hence $\exists h\in 𝔥$ such that ${\lambda }_1\left(h\right),\dots ,{\lambda }_m\left(h\right)$ are all distinct. We have $h^j·x={\displaystyle \sum _{i=1}^m}{\lambda }_i{\left(h\right)}^j{x}_{{\lambda }_i}(0\le j\le m-1)$ and we can solve these equations for $x_{{\lambda }_1},\dots ,x_{{\lambda }_m}$ by Cramer's rule, since $\text{det}{\left({\lambda }_i{\left(h\right)}^j\right)}_{\genfrac{}{}{0ex}{}{1\le j\le m}{0\le j\le m-1}}=\prod _{i<j}({\lambda }_i\left(h\right)-{\lambda }_j\left(h\right))\ne 0$. Hence each $x_{{\lambda }_i}$ is a linear combination of the $h^j·x$, hence lies in $M^{\prime }$. $\square$

We apply this lemma with $M=\stackrel{\sim }{𝔤}\left(A\right)$ and $M^{\prime }$ an ideal $𝔞$ in $\stackrel{\sim }{𝔤}\left(A\right)$. Then $𝔞$ is a $𝔥$–submodule of $\stackrel{\sim }{𝔤}\left(A\right)$ under the adjoint action, hence by (1.4) and (1.5) we have

$𝔞=\underset{\alpha \in Q}{\oplus }{𝔞}_{\alpha }$

where ${𝔞}_{\alpha }=𝔞\cap {\stackrel{\sim }{𝔤}}_{\alpha }$ i.e. $𝔞$ is a $Q$–graded ideal. Hence also

$𝔞=\underset{m\in \mathbb{Z}}{\oplus }{𝔞}_m$

where ${𝔞}_m=𝔞\cap {\stackrel{\sim }{g}}_m$ (principal grading).

Consider now ideals $𝔞$ in $\stackrel{\sim }{𝔤}\left(h\right)$ such that ${𝔞}_0=0$, i.e. $𝔞\cap 𝔥=0$. Any sum of such ideals has the same property, hence there is a unique largest ideal $𝔯$ in $\stackrel{\sim }{𝔤}\left(A\right)$ such that $𝔯\cap 𝔥=0$. We have

$𝔯={𝔯}_{+}\oplus {𝔯}_{-}$

where

${𝔯}_{+}=\underset{m>0}{\oplus }{𝔯}_m=𝔯\cap {\stackrel{\sim }{𝔫}}_{+}$
${𝔯}_{-}=\underset{m<0}{\oplus }{𝔯}_m=𝔯\cap {\stackrel{\sim }{𝔫}}_{-}$

We have $[f_i,{𝔯}_{+}]=\underset{m>0}{\oplus }[f_i,{𝔯}_m]\subset \underset{m>0}{\oplus }{𝔯}_{m-1}={𝔯}_{+}$; and since clearly $[𝔥,{𝔯}_{+}]\subset {𝔯}_{+},[e_i,{𝔯}_{+}]\subset {𝔯}_{+}$ it follows that ${𝔯}_{+}$ is an ideal in $\stackrel{\sim }{𝔤}\left(A\right)$. Similarly, of course, for ${𝔯}_{-}$.

Next I claim that ${𝔯}_1={𝔯}_{-1}=0$. For ${𝔯}_1=\underset{i=1}{\overset{n}{\oplus }}{𝔯}_{{\alpha }_i}$; if ${𝔯}_{{\alpha }_i}\ne 0$ then ${𝔯}_{{\alpha }_i}={𝔤}_{{\alpha }_i}$ (because ${𝔤}_{{\alpha }_i}$ is 1-dimensional, spanned by $e_i$), hence $e_i\in 𝔯$; but then $h_i=[e_i,f_i]\in 𝔯\cap 𝔥$, contradiction. Hence ${𝔯}_1=0$, and similarly ${𝔯}_{-1}=0$.

Finally, we must have $\stackrel{\sim }{\omega }\left(𝔯\right)=𝔯$ (for $\stackrel{\sim }{\omega }\left(𝔯\right)$ has the same properties as $𝔯$).

To summarize:

(1.6)

1. All ideals in $\stackrel{\sim }{𝔤}\left(A\right)$ are $Q$–graded.
2. The set of ideals $𝔞$ in $\stackrel{\sim }{𝔤}\left(A\right)$ such that $𝔞\cap 𝔥=0$ has a unique maximal element of $𝔯$.
3. ${𝔯}_{+}=𝔯\cap {\stackrel{\sim }{𝔫}}_{+}$ and ${𝔯}_{-}=𝔯\cap {\stackrel{\sim }{𝔫}}_{-}$ are ideals in $\stackrel{\sim }{𝔤}\left(A\right)$, and $𝔯={𝔯}_{+}\oplus {𝔯}_{-}$ (direct sum)
4. ${𝔯}_1={𝔯}_{-1}=0$
5. $\stackrel{\sim }{\omega }\left(𝔯\right)=𝔯$

Now define

$𝔤\left(A\right)=\stackrel{\sim }{𝔤}\left(A\right)/𝔯$.

It is this algebra which is the object of our investigations. If $A$ is a Cartan matrix, $𝔤\left(A\right)$ is the Kac-Moody algebra defined by the matrix $A$.

(1.7) Remarks
Since the ideal $𝔯$ is $Q$–graded (1.6), $𝔤\left(A\right)$ is a $Q$–graded Lie algebra:

$𝔤\left(A\right)=\underset{\alpha \in Q}{\oplus }{𝔤}_{\alpha }$

where ${𝔤}_{\alpha }={\stackrel{\sim }{𝔤}}_{\alpha }/{𝔯}_{\alpha }$; thus

1. ${𝔤}_0={\stackrel{\sim }{𝔤}}_0=𝔥$
2. Since ${𝔯}_1={𝔯}_{-1}=0$ (1.6), ${𝔤}_1={\stackrel{\sim }{𝔤}}_1=\sum _{i=1}^{n}k{e}_i;{𝔤}_{-1}={\stackrel{\sim }{𝔤}}_{-1}=\sum _{i=1}^{n}k{f}_i;{𝔤}_{{\alpha }_i}=k{e}_i,{𝔤}_{{-\alpha }_i}=k{f}_i,(1\le i\le n)$
   (Since the images of $e_1,\dots ,e_n,f_1,\dots ,f_n$ in $𝔤\left(A\right)$ remain linearly independent, we continue to denote them by the same symbols).
3. ${𝔤}_{\alpha }=0$ unless $\alpha -0$ or $\alpha >0$ or $\alpha <0$, by (1.4)
4. ${𝔤}_{\alpha }=\{x\in 𝔤\left(A\right):[h,x]=\alpha \left(h\right)x\text{for all}h\in 𝔥\}$ (same proof as in (1.4)) and each ${𝔤}_{\alpha }$ is finite-dimensional. If $\alpha \ne 0$ and ${𝔤}_{\alpha }=0$ we say $\alpha$ is a root of $𝔤\left(A\right)$ with multiplicity $m_{\alpha }=\text{dim}{𝔤}_{\alpha }$. (In the classical case, all $m_{\alpha }$ are 1).

5. Let ${𝔫}_{+}={\stackrel{\sim }{𝔫}}_{+}/{𝔯}_{+}=\underset{\alpha >0}{\oplus }{𝔤}_{\alpha },{𝔫}_{-}={\stackrel{\sim }{𝔫}}_{-}/{𝔯}_{-}=\underset{\alpha <0}{\oplus }{𝔤}_{\alpha }$

   These are subalgebras of $𝔤\left(A\right)$, generated by $e_1,\dots ,e_n$ and by $f_1,\dots ,f_n$ respectively, and

   $𝔤\left(A\right)={𝔫}_{-}\oplus 𝔥\oplus {𝔫}_{+}\text{(vector space direct sum)}$

6. All ideals $𝔞$ in $𝔤\left(A\right)$ are $Q$–graded (1.6): $𝔞=\underset{\alpha \in Q}{\oplus }{𝔞}_{\alpha }$; and $𝔤\left(A\right)$ has no ideal $𝔞\ne 0$ such that $𝔞\cap 𝔥=0$ (by construction)
7. Since $𝔯$ is stable under the involution $\stackrel{\sim }{\omega }$ (1.6) we have an involution $\omega :𝔤\left(A\right)\to 𝔤\left(A\right)$ under which $e_i\mapsto -f_i,f_i\mapsto -e_i,h\mapsto -h(h\in 𝔥)$ we have $\omega \left({𝔤}_{\alpha }\right)={𝔤}_{-\alpha }$ for all $\alpha \in Q$, hence $\omega$ interchanges ${𝔫}_{+}$ and ${𝔫}_{-}$.

The following lemma is frequently useful:

(1.8)

1. Let $x\in {𝔫}_{+}$ be such that $[x,f_i]=0(1\le i\le n)$. Then $x=0$.
2. Let $x\in {𝔫}_{-}$ be such that $[x,e_i]=0(1\le i\le n)$. Then $x=0$.

		Proof.
	We shall prove (i), and then (ii) will follow by use of the involution $\omega$. Write $x=\sum _{\alpha >0}{x}_{\alpha }$, then we have $\sum _{\alpha >0}[x_{\alpha },f_i]=0$ for $1\le i\le n$. Hence $[x_{\alpha },f_i]=0$ for each $\alpha$ and each $i$; in other words we may assume $x$ homogeneous. Consider the ideal $U\left(𝔤\right)·x=𝔞$ say, generated by $x$. ($U\left(𝔤\right)$ acting via ad). Since $𝔤={𝔫}_{-}\oplus 𝔥\oplus {𝔫}_{+}$ (1.7) we have (corollary of P-B-W) $U\left(𝔤\right)=U\left({𝔫}_{+}\right)U\left(𝔥\right)U\left({𝔫}_{-}\right)$ By assumption, $U\left({𝔫}_{-}\right)·x=kx;U\left(𝔥\right)·x=kx$, hence $𝔞=U\left(𝔤\right)·x=U\left({𝔫}_{+}\right)·x$ has only positive components, hence $𝔞\cap 𝔥=0$. Hence ((1.7)(vi)) $𝔞=0$, i.e. $x=$0? $\square$

Example Suppose $A=0$ (the $n\times n$ zero matrix). What does $𝔤\left(0\right)$ look like?

We have ${\alpha }_j\left(h_i\right)=a_{ij}=0$ for all $i,j$, hence the relations (1.2) give

$[h_i,e_j]={\alpha }_j\left(h_i\right)e_j=0(\text{all}i,j)$

and likewise

$[h_i,f_j]=0(\text{all}i,j)$

Consider now $[e_i,e_j]$. We have

$\begin{array}{ccc}[[e_i,e_j],f_k]& =& [e_i,[e_j,f_k]]-[e_j,[e_i,f_k]]\\ & =& [e_i,{\delta }_{jk}{h}_k]-[e_j,{\delta }_{ij}{h}_k]=0\end{array}$

whence by (1.8) $[e_i,e_j]=0$. Hence ${𝔫}_{+}$ is abelian, i.e. ${𝔫}_{+}={𝔤}_1=\sum _1^{n}k{e}_i$. Similarly ${𝔫}_{-}={𝔤}_{-1}=\sum _1^{n}k{f}_i$, and

$𝔤\left(0\right)={𝔤}_{-1}\oplus {𝔤}_0\oplus {𝔤}_1$

Note that ${𝔤}_0=𝔥$ has dimension $2n$ (because here $l=0$).

The algebra ${𝔤}^{\prime }\left(A\right)$

Let ${𝔤}^{\prime }=D𝔤\left(A\right)$ be the derived algebra of $𝔤\left(A\right)$.

(1.9) ${𝔤}^{\prime }\left(A\right)$ is the subalgebra of $𝔤\left(A\right)$ generated by $e_1,\dots ,e_n,f_1,\dots ,f_n,$ and

${𝔤}^{\prime }\left(A\right)={𝔫}_{-}\oplus {𝔥}^{\prime }\oplus {𝔫}_{+}$

where ${𝔥}^{\prime }$ is the subspace of $𝔥$ generated by $h_1,\dots ,h_n$.

Thus ${𝔤}^{\prime }\left(A\right)=𝔤\left(A\right)$ iff det $\left(A\right)\ne 0$.

		Proof.
	Let $𝔞$ denote the subalgebra of $𝔤\left(A\right)$ generated by $e_1,\dots ,f_n$, and let $𝔟={𝔫}_{-}\oplus {𝔥}^{\prime }\oplus {𝔫}_{+}$. Since $[h,e_i]={\alpha }_i\left(h\right)e_i$ for all $h\in 𝔥$, and since $\exists h\in 𝔥$ such that ${\alpha }_i\left(h\right)\ne 0$, it follows that $e_i\in {𝔤}^{\prime }\left(A\right)$; similarly $f_i\in {𝔤}^{\prime }\left(A\right)$, and therefore $\begin{array}{cc}𝔞\subset {𝔤}^{\prime }\left(A\right)& \left(1\right)\end{array}$ Next, by (1.7)(v), ${𝔫}_{+}$ and ${𝔫}_{-}$ are subalgebras of $𝔞$; and since $h_i=[e_i,f_i]\in 𝔞$, it follows that ${𝔥}^{\prime }\subset 𝔞$, whence $\begin{array}{cc}𝔟\subset 𝔞& \left(2\right)\end{array}$ Finally, I claim that $𝔟$ is an ideal in $𝔤\left(A\right)$. We have to check that $[h,b]\subset 𝔟(h\in 𝔥);[e_i,𝔟]\subset 𝔟;[f_i,b]\subset 𝔟$ The first of these is obvious. As to the second, we have $[e_i,{𝔫}_{-}]\subset {𝔫}_{-}+{𝔥}^{\prime }$ (because $[e_i,f_j]={\delta }_{ij}{h}_i\text{(1.2))};[e_i,{𝔥}^{\prime }]\subset {𝔫}_{+};[e_i,{𝔫}_{+}]\subset {𝔫}_{+};\text{and}[f_i,𝔟]\subset 𝔟$ is proved similarly. Since $𝔤\left(A\right)/𝔟\cong 𝔥/{𝔥}^{\prime }$ is abelian, it follows that $\begin{array}{cc}{𝔤}^{\prime }\left(A\right)\subset 𝔟\text{.}& \left(3\right)\end{array}$ (1), (2), (3) complete the proof. $\square$

Remark The algebra ${𝔤}^{\prime }\left(A\right)$ is also sometimes called the Kac-Moody algebra associated to the matrix $A$ (if $A$ is a Cartan matrix). We can give a more direct construction of ${𝔤}^{\prime }\left(A\right)$, as follows: Let ${\stackrel{\sim }{𝔤}}^{\prime }\left(A\right)$ denote the Lie algebra with $3n$ generators $e_i,f_i,h_i(1\le i\le n)$ subject to the relations

$\begin{array}{cc}(1.2^{\prime })& \{\begin{array}{c}[h_i,h_j]=0\\ [e_i,f_j]={\delta }_{ij}{h}_i\\ [h_i,e_j]=a_{ij}{e}_j\\ [h_i,f_j]=-a_{ij}{f}_j& & (1\le i,j\le n)\end{array}\end{array}$

Let $Q(\cong {\mathbb{Z}}^n)$ be a free abelian group on generators ${\alpha }_1,\dots ,{\alpha }_n$; then ${\stackrel{\sim }{𝔤}}^{\prime }\left(A\right)$ is $Q$–graded by assigning degrees 0, ${\alpha }_i$, $-{\alpha }_i$ to $h_i,e_i,f_i$ respectively $(1\le i\le n)$, and there exists a unique maximal $Q$–graded ideal ${𝔯}^{\prime }$ subject to ${𝔯}^{\prime }\cap {𝔥}^{\prime }=0$ (where ${𝔥}^{\prime }=\sum _1^{n}k{h}_i$ as above). Then ${𝔤}^{\prime }\left(A\right)={\stackrel{\sim }{𝔤}}^{\prime }\left(A\right)/{𝔯}^{\prime }$.

We can then construct $𝔤\left(A\right)$ as a semidirect product of ${𝔤}^{\prime }\left(A\right)$ by a suitable algebra of derivations.

Semidirect products

In general, let $𝔤$ be a Lie algebra, $𝔫$ an ideal in $𝔤$, $𝔞$ a subalgebra of $𝔤$, such that $𝔤=𝔫\oplus 𝔞$ (vector space direct sum). If $x_1,x_2\in 𝔤$, say $x_i=n_i+a_i(n_i\in 𝔫,a_i\in 𝔞)$ then

$[x_1,x_2]=[n_1,n_2]+[a_1.n_2]-[a_2,n_2]+[a_1,a_2]$

in which the first 3 terms on the right lie in $𝔫$ (because $𝔫$ is an ideal in $𝔤$).

The Lie algebra $𝔞$ acts (via ad) on $𝔫$ as an algebra of derivations:

$\text{ad}:𝔞\to \text{Der}\left(𝔫\right)$

Conversely, if we are given Lie algebras $𝔫,𝔞$ and a Lie algebra homomorphism $\phi :𝔞\to \text{Der}\left(𝔫\right)$, we construct the semidirect product $𝔤=𝔫⋊𝔞$ as follows: $𝔤=𝔫\oplus 𝔞$ as a vector space, and the Lie bracket in $𝔤$ is defined by

$[n_1+a_1,n_2+a_2]=[n_1,n_2]+\phi \left(a_1\right)n_2-\phi \left(a_2\right)n_1+[a_1,a_2]$.

One has of course to check the Jacobi identity, which is tedious but straightforward.

In the present case, let $𝔞$ be a vector space complement of ${𝔥}^{\prime }$ in $𝔥$: then

$𝔤\left(A\right)={𝔤}^{\prime }\left(A\right)\oplus 𝔞$

with ${𝔤}^{\prime }\left(A\right)$ an ideal (1.9) and $𝔞$ a subalgebra. Hence $𝔤\left(A\right)$ may be constructed as the semidirect product ${𝔤}^{\prime }\left(A\right)⋊𝔞$, with $𝔞$ acting as an (abelian) algebra of derivations.

The centre of $𝔤\left(A\right)$

(1.10) The algebras $𝔤\left(A\right),{𝔤}^{\prime }\left(A\right)$ have the same centre $𝔠$:

${𝔥}_0=𝔠=\bigcap _{i=1}^n\text{Ker}\left({\alpha }_i\right)\subset {𝔥}^{\prime }$.

We have dim $𝔠=n-l$, hence $𝔠=0$ iff $A$ is nonsingular.

		Proof.
	Suppose $x\in 𝔤\left(A\right)$ commutes with $e_1,\dots ,f_n$. Say $x=\sum _{r\in \mathbb{Z}}{x}_r$ (principal grading); then $0=[x,f_i]=\sum _r[x_r,f_i]$, so that $[x_r,f_i]=0$ for $1\le i\le n$ and all $r\in \mathbb{Z}$. By (1.8) it follows that $x_r=0$ if $r\ge 1$, and similarly $x_r=0$ for $r\le$what goes here?. Hence $x=x_0\in 𝔥$. But then (1.2) $0=[x,e_i]=x_i\left(x\right)e_i$ so that $x_i\left(x\right)=0$ $(1\le i\le n)$, whence $x\in \bigcap _1^n\text{Ker}{\alpha }_i$. Conversely, if ${\alpha }_i\left(x\right)=0$ for $1\le i\le n$, then by (1.2) we have $[x,e_i]=[x,f_i]=0$, and of course $[x,𝔥]=0$. This shows that the centre of $𝔤\left(A\right)$ is $𝔠=\bigcap _{i=1}^n\text{Ker}\left({\alpha }_i\right)$; since the ${\alpha }_i$ are independent linear forms on $𝔥$, we have dim $𝔠=$dim $𝔥-𝔫=n-l$. Finally, I claim that $𝔠\in {𝔥}^{\prime }$. For ${\displaystyle \begin{array}{ccc}\sum _1^n{\mu }_i{h}_i\in 𝔠\cap {𝔥}^{\prime }& \iff & \sum _1^n{\mu }_i{\alpha }_i\left(h_i\right)=0(1\le j\le n)\\ & \iff & \sum _1^n{\mu }_i{a}_{ij}=0(1\le j\le n)\end{array}}$ Since $A=\left(a_{ij}\right)$ has rank $l$, it follows that $𝔠\cap {𝔥}^{\prime }$ has dimension $n-l=\text{dim}𝔠$. Hence $𝔠\in {𝔥}^{\prime }$ as claimed and therefore $𝔠$ is also the centre of ${𝔤}^{\prime }\left(A\right)$. $\square$

Decomposability

Let us say that two $n\times n$ matrices $A=\left(a_{ij}\right)$ and $A^{\prime }=a_{ij}^{\prime }$ are equivalent:

$A\sim A^{\prime }$

if $\exists w\in s_n$ such that

$a_{ij}^{\prime }=a_{w\left(i\right),w\left(j\right)}(1\le i,j\le n)$

i.e. if $A^{\prime }$ is obtained from $A$ by applying the same permutation to rows and columns. Clearly $A\sim A^{\prime }\Rightarrow 𝔤\left(A\right)\cong 𝔤\left(A^{\prime }\right)$: we have merely reindexed the generators. Now suppose that $A$ satisfies the condition

$\begin{array}{cc}{a}_{ij}=0\iff a_{ji}=0& \left(✶\right)\end{array}$

We associate with $A$ a graph $𝔯\left(A\right)$, as follows: the vertices of $A$ are the indices $1,2,\dots ,n$, and distinct vertices $i$ and $j$ are joined by an edge iff $a_{ij}\ne 0$ or $a_{ji}\ne$something

(1.11) Assume that $A$ satisfies ($✶$). Then the following conditions on $A$ are equivalent:

1. $A$ is equivalent to a nontrivial diagonal sum $\left(\begin{array}{cc}{A}_1& 0\\ 0& A_2\end{array}\right)$;
2. There exist non-empty complimentary subsets $I,J$ of $\{1,2,\dots ,n\}$ such that $a_{ij}=0$ for $i\in I$ and $j\in J$;
3. $𝔯\left(A\right)$ is not connected.

		Proof.
	Obvious. $\square$

If these equivalent conditions are satisfied, we say that $A$ is decomposable.

(1.12) If $A$ is decomposable, say $A\sim \left(\begin{array}{cc}{A}_1& 0\\ 0& A_2\end{array}\right)$, then

$𝔤\left(A\right)\cong 𝔤\left(A_1\right)\times 𝔤\left(A_2\right)$

(direct product).

		Proof.
	Consider $𝔤\left(A\right)=𝔤\left(A_1\right)\times 𝔤\left(A_2\right)$, which is a Lie algebra generated by $e_1,\dots ,e_n,f_1,\dots ,f_n$ and $𝔥={𝔥}_1\oplus {𝔥}_2$. Check that these generators satisfy relations (1.2) (for the $Q$–graded matrix $A$); hence $𝔤$ is a homomorphic image of $\stackrel{\sim }{𝔤}\left(A\right)$, i.e. we have a surjective homomorphism $\stackrel{\sim }{\phi }:\stackrel{\sim }{𝔤}\left(A\right)\to 𝔤$. Then $\stackrel{\sim }{\phi }\left(r\right)=𝔞$ say is an ideal of $𝔤$ such that $𝔞\cap 𝔥=0$. But $𝔤$ is a direct product, hence $𝔞={𝔞}_1\times {𝔞}_2$, where ${𝔞}_i$ is an ideal in $𝔤\left(A_i\right)$ which intersects ${𝔥}_i$ trivially $(i=1,2)$. Hence (1.7) ${𝔞}_1={𝔞}_2=0$ and therefore $𝔞=0$, consequently $\stackrel{\sim }{\phi }$ induces a surjective homomorphism $\phi :𝔤\left(A\right)\to 𝔤$. The kernel of $\phi$ is an ideal $𝔟$ such that $𝔟\cap 𝔥=0$ (because $\phi /𝔥$ is injective), hence $𝔟=0$ (1.7). Hence $\phi$ is an isomorphism. $\square$

$\Rightarrow \alpha \in R$, then Supp($\alpha$) connected.

Ideals in $𝔤\left(A\right)$

Assume that $A$ satisfies the condition

$a_{ij}=0\iff a_{ji}=0$.

(1.13)

1. Suppose $A$ is indecomposable. Then every ideal in $𝔤\left(A\right)$ either contains ${𝔤}^{\prime }\left(A\right)$ or is contained in the centre $𝔠$.
2. $𝔤\left(A\right)$ is simple iff $A$ is indecomposable and nonsingular.

		Proof.
	Let $𝔞$ be an ideal in $𝔤\left(A\right)$. By (1.7)(vi), $𝔞$ is $Q$–graded, hence we may write $𝔞=\underset{r\in \mathbb{Z}}{\oplus }{𝔞}_r$ (principal grading). Suppose first that ${𝔞}_0\subset 𝔠$. If ${𝔞}_1\ne 0$, then $e_i\in {𝔞}_1$ for some $i$. But then $h_i=[e_i,f_i]\in {𝔞}_0$, hence $h_i\in 𝔠$ and therefore (1.10) $a_{ij}={\alpha }_j\left(h_i\right)=0$ for $1\le j\le$something. This contradicts the assumption that $A$ is indecomposable. Hence ${𝔞}_1=0$. It now follows by induction on $r$ that ${𝔞}_r=0$ for all $r\ge 1$. For if $x\in {𝔞}_r$ where $r\ge 2$, then $[x,f_i]\in {𝔞}_{r-1}=0$ by ind. hyp., hence $x=0$ by (1.8). Likewise ${𝔞}_r=0$ for $r\le -1$ and therefore $𝔞={𝔞}_0\subset c$. Now suppose that ${𝔞}_0\not\subset 𝔠$. Let $h\in {𝔞}_0$, $h\notin 𝔠$. By (1.10) we have ${\alpha }_i\left(h\right)\ne 0$ for some $i$, hence $e_i={\alpha }_i{\left(h\right)}^{-1}[h_i,e_i]\in 𝔞$; similarly $f_i\in 𝔞$ (for this value of $i$), and $h_i=[e_i,f_i]\in 𝔞$. Since $𝔯\left(A\right)$ is connected (1.11) $\exists j\in [1,n]$ such that $a_{ij}\ne 0$; since $[h_i,e_j]=a_{ij}{e}_j$, it follows that $e_j\in 𝔞$, and likewise $f_j\in 𝔞$. It now follows that $e_j,f_j\in 𝔞$ for every index $j$ connected to $i$ by a path in the graph $𝔯\left(A\right)$ – i.e. $e_1,\dots ,f_n\in 𝔞$, hence (1.9) $𝔞\supset {𝔤}^{\prime }\left(A\right)$. $\Rightarrow$ If $A$ is decomposable, $𝔤\left(A\right)$ is not simple, by (1.12). Again, if $A$ is singular, i.e. $l<n$, then $𝔠\ne 0$ (1.10) and again $𝔤\left(A\right)$ is not simple. $\Leftarrow$ If $A$ is nonsingular ($l=n$) then ${𝔤}^{\prime }\left(A\right)=𝔤\left(A\right)$ and $𝔠=0$. Now use (i). $\square$

Note for later use the following corollary of (1.13):

(1.13$\frac{1}{2}$). Assume $A$ indecomposable. Then the following conditions are equivalent:

1. $𝔤\left(A\right)$ is infinite-dimensional
2. $R$ is infinite
3. For each $\alpha \in R^{+}$ there exists $i$ such that $\alpha +{\alpha }_i\in R^{+}$.

		Proof.
	(i) $\Rightarrow$ (ii) is clear from the root space decomposition $𝔤\left(A\right)=𝔥\oplus \underset{\alpha \in R}{\oplus }{𝔤}_{\alpha }$ (iii) $\Rightarrow$ (ii) is clear (ii) $\Rightarrow$ (iii) If (iii) is false, there exists a positive root $\alpha$ such that $\alpha +{\alpha }_i\notin R$ $(1\le i\le n)$. Let $x\in {𝔤}_{\alpha },x\ne 0$. Then we have $[x,e_i]=0(1\le i\le n)$, from which it follows that $U\left({𝔫}_{+}\right)·x=kx$ and therefore the ideal $𝔞=U\left(𝔤\right)·x$ generated by $x$ in $𝔤\left(A\right)$ is $𝔞=U\left({𝔫}_{-}\right)U\left(𝔥\right)U\left({𝔫}_{+}\right)·x=U\left(𝔫\right)·x$ Hence ${𝔞}_{\beta }=0$ unless $\beta \le \alpha$. But by (1.13) we have $𝔞\supset {𝔤}^{\prime }\left(A\right)$ (because clearly $𝔞\not\subset 𝔠$), hence in particular $𝔞\supset {𝔫}_{+}$. It follows that all roots $\beta$ are $\le \alpha$, whence $R^{}$ and therefore $R$ is finite. $\square$

Thus if $R$ is finite there is a unique highest root $q$ such that $\alpha \le \phi$ for all $\alpha \in R$.

The algebra ${\overline{𝔤}}^{\prime }={𝔤}^{\prime }\left(A\right)/𝔠$

We have

${𝔤}^{\prime }\left(A\right)={𝔫}_{-}\oplus {𝔥}^{\prime }\oplus {𝔫}_{+}$

by (1.9), and $𝔠\subset {𝔥}^{\prime }$ (1.10), hence

${\overline{𝔤}}^{\prime }\left(A\right)={𝔫}_{-}\oplus {\overline{𝔥}}^{\prime }\oplus {𝔫}_{+}$

where ${\overline{𝔥}}^{\prime }={𝔥}^{\prime }/𝔠$ (so that dim ${\overline{𝔥}}^{\prime }=n-(n-l)=l$).

Assume that $A$ is indecomposable (and that $a_{ij}=0\iff a_{ji}=0$). The proof of (1.13)(i) shows that any $Q$–graded ideal $𝔞$ in ${𝔤}^{\prime }\left(A\right)$ such that ${𝔞}_0(=𝔞\cap {𝔥}^{\prime })\subset 𝔠$ is contained in $𝔠$. Hence ${\overline{𝔤}}^{\prime }\left(A\right)$ has no nontrivial $Q$–graded ideal $\overline{𝔞}$ such that ${\overline{𝔞}}_0=0$. From this it follows that (1.8) is valid for the algebra ${\overline{𝔤}}^{\prime }\left(A\right)$.

We shall make use of this remark in the proof of the following proposition:

(1.14) Assume that $A$ is indecomposable and that each root $\alpha$ has a nonzero restriction to ${𝔥}^{\prime }$. Then the algebra ${\overline{𝔤}}^{\prime }\left(A\right)$ is simple.

		Proof.
	Let $𝔞\ne 0$ be an ideal in ${\overline{𝔤}}^{\prime }\left(A\right)$. Each $x\in 𝔞$ is of the form ${\displaystyle x=\sum _{\alpha \in S}{x}_{\alpha }}$ where $S$ is some finite subset of $Q$, each $x_{\alpha }\ne 0$ and $x_{\alpha }\in {𝔤}_{\alpha }$ for $\alpha \ne 0,x_0\in {\overline{𝔥}}^{\prime }$. Call $\left|S\right|$ the length of $x$, and the number ${\displaystyle \underset{\alpha \in S}{\text{max}}\text{ht}\left(\alpha \right)}$ the height of $x$. Choose $x\ne 0$ in $𝔞$ of minimal length. Suppose that the chosen $x$ has height $r\ge 1$, so that $x=x_{\alpha }+\dots$ where $\text{ht}\left(\alpha \right)=r$, $x_{\alpha }\ne 0$. Since $x_{\alpha }\ne 0$ we have $[x_{\alpha },f_i]\ne 0$ for some $i$, by the remark above; hence $[x,f_i]$ is a nonzero element of $𝔞$, of minimal length and height $r-1$. By proceeding in this way we shall obtain an element $y\ne 0$ of $𝔞$ of minimal length and height 0, say ${\displaystyle \begin{array}{cc}y={\bar{h}}_0+\sum _{\alpha \in S^{\prime }}{y}_{\alpha }& \left(1\right)\end{array}}$ where ${\bar{h}}_0\in {\overline{𝔥}}^{\prime }$ (and $\ne 0$), and $\left|S^{\prime }\right|=\left|S\right|-1$. Similarly, if $x$ has height $<0$, we use the $e_i$ rather than the $f_i$ to achieve the same result. From (1) we have, for all $\bar{h}\in {\overline{𝔥}}^{\prime }$, ${\displaystyle [\bar{h},y]=\sum _{\alpha \in S^{\prime }}\alpha \left(\bar{h}\right)y_{\alpha }}$ which is an element of $𝔞$ of length $<\left|S\right|$, hence is 0. Hence $\alpha \left(\bar{h}\right)=0$ for all $\alpha \in S^{\prime }$ and all $\bar{h}\in {\bar{h}}^{\prime }$, i.e. $\alpha \in S^{\prime }\Rightarrow \alpha |{𝔥}^{\prime }=0$. By hypothesis, therefore $S^{\prime }$ is empty and therefore $y={\bar{h}}_0\in {\overline{𝔥}}^{\prime }$. Since ${\bar{h}}_0\ne 0$ we have ${\alpha }_i\left({\bar{h}}_0\right)\ne 0$ (by (1.something)) for some $i$, and therefore $e_i,f_i\in 𝔞$ for this value of $i$ (because ${\alpha }_i\left({\bar{h}}_0\right)e_i=[{\bar{h}}_0,e_i]\in 𝔞$). But now it follows as in the proof of (1.13) that $e_1,\dots ,f_n$ all lie in $𝔞$, whence $𝔞={\overline{𝔤}}^{\prime }\left(A\right)$. $\square$

Remark: The converse of (1.14) is true if $A$ is a Cartan matrix (proof later) I do not know whether it is so in general.

I shall conclude this chapter with some properties of $𝔤\left(A\right)$ that are valid only when $A$ is a Cartan matrix. So assume now that the matrix $A$ satisfies the condition $\left(C\right)$:

$\left(C\right)$

$a_{ij}\in \mathbb{Z};a_{ii}=2;a_{ij}\le 0\text{if}i\ne j;a_{ij}=0\iff a_{ji}=0$.

For each $i=1,\dots ,n$ let $s_i$ denote the subspace of $𝔤\left(A\right)$ spanned by $e_i,f_i,h_i$. From (1.2) we have

$[e_i,f_i]=h_i,[h_i,e_i]=2e_i,[h_i,f_i]=-2f_i$

(1.15) $s_i$ is a subalgebra of $𝔤\left(A\right)$, isomorphic to ${𝔰𝔩}_2\left(k\right)$.

		Proof.
	The relations just written show that $s_i$ is a 3-dimensional subalgebra of $𝔤\left(A\right)$. The mapping ${\displaystyle e_i\mapsto \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f_i\mapsto \left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h_i\mapsto \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)}$ is an isomorphism of $s_i$ onto ${𝔰𝔩}_2\left(k\right)$ $\square$

Next we require the following lemma:

(1.16) Let $x,y$ be elements of an associative ring $R$. Then for each positive integer $N$ we have

1. ${\displaystyle x^{N}y=\sum _{r=0}^N\left(\genfrac{}{}{0ex}{}{N}{r}\right){\left(\text{ad}x\right)}^{r}y{x}^{N-r}}$.
2. ${\displaystyle x{y}^N=\sum _{r=0}^N{(-1)}^r\left(\genfrac{}{}{0ex}{}{N}{r}\right)y^{N-r}{\left(\text{ad}y\right)}^{r}x}$.

		Proof.
	We shall prove (ii); the proof of (i) is analogous. Let ${\lambda }_y;p_y:R\to$something denote respectively left and right multiplication by $y$ in $R$. Since $R$ is associative they commute with each other and hence also with $\text{ad}y={\lambda }_y-p_y$. Hence ${\displaystyle \begin{array}{ccc}x{y}^N=p_y^N\left(x\right)& =& {({\lambda }_y-\text{ad}y)}^{N}x\\ & =& \sum _{r=0}^N{(-1)}^r\left(\genfrac{}{}{0ex}{}{N}{r}\right){\lambda }_y^{N-r}{\left(\text{ad}y\right)}^{r}x\text{.}\end{array}}$ $\square$

Let us apply this formula with $x=e_i,y=f_i,R=U\left(s_i\right)$: we have

$\begin{array}{c}\left(\text{ad}{f}_i\right)e_i=[f_i,e_i]=-h_i\\ {\left(\text{ad}{f}_i\right)}^2{e}_i=-[f_i,h_i]=-2f_i\\ {\left(\text{ad}{f}_i\right)}^3{e}_i=-2(f_i,f_i)=0\end{array}$

and therefore

$\begin{array}{cc}\text{(1.17)}& {\displaystyle \begin{array}{ccc}{e}_i{f}_i^N& =& f_i^N{e}_i+N{f}_i^{N-1}{h}_i+\left(\genfrac{}{}{0ex}{}{N}{2}\right)f_i^{N-2}·-2f_i\\ & =& f_i^N{e}_i+N{f}_i^{N-1}(h_i-N+1)\text{.}\end{array}}\end{array}$

(1.18) In $𝔤\left(A\right)$ we have

${\left(\text{ad}{e}_i\right)}^{1-a_{ij}}{e}_j=0$
${\left(\text{ad}{f}_i\right)}^{1-a_{ij}}{f}_j=0$

whenever $i\ne j$.

		Proof.
	It is enough to prove one of these relations, because the other then follows by applying the involution $\omega$. Let $f_{ij}={\left(\text{ad}{f}_i\right)}^{1-a_{ij}}{f}_j$. By (1.8), in order to show that $f_{ij}=0$ it is enough to show that $[e_k,f_{ij}]=0(1\le k\le n)$. There are 3 cases to consider: $k\ne i,k\ne j$. Then $e_k$ commutes with $f_i$ and $f_j$ (1.2), hence with $f_{ij}$. $k=j,k\ne i$. Then $e_j$ commutes with $f_i$, hence $\begin{array}{ccc}[e_j,f_{ij}]& =& {\left(\text{ad}{f}_i\right)}^{1-a_{ij}}[e_j,f_j]\\ & =& {\left(\text{ad}{f}_i\right)}^{-a_{ij}}[f_i,h_j]\\ & =& a_{ji}{\left(\text{ad}{f}_i\right)}^{-a_{ij}}{f}_i\text{.}\end{array}$ If $a_{ij}\ne 0$ this is zero, whilst if $z_{ij}=0$ then $a_{ji}=0$ (by (c)), so again it is 0. $k=i,k\ne j$. We have, using the formula (1.17) $\begin{array}{ccc}[e_i,f_{ij}]& =& \left(\text{ad}{e}_i\right){\left(\text{ad}{f}_i\right)}^{1-a_{ij}}{f}_j\\ & =& \left(\text{ad}{f}_i^{1-a_{ji}}\right)\left(\text{ad}{e}_i\right)f_j+(1-a_{ij}){\left(\text{ad}{f}_i\right)}^{-a_{ji}}(\left(h_i\right)f_j+a_{ij}{f}_j)\\ & =& 0\text{by (1.2).}\end{array}$ $\square$

Remark For an arbitrary Cartan matrix $A$, it is still an open question whether the relations (1.2) together with (1.17) are a complete set of defining relations for the algebra $𝔤\left(A\right)$: or, equivalently, whether the left sides of the relations (1.18) generate the ideal $𝔯$ in $\stackrel{\sim }{𝔤}\left(A\right)$. At any rate this is known to be true (proof later, perhaps) if $A$ is symmetrizable.

In general, a derivation $d$ of a Lie algebra $𝔤$ is said to be locally nilpotent if for each $x\in 𝔤$ there exists a positive integer $N\left(x\right)$ such that $d^{N\left(x\right)}x=0$: i.e. if for each $x\in 𝔤$ is killed by some power of $d$. In that case $e^d:𝔤\to 𝔤$ is well defined, because the series

${\displaystyle e^d\left(x\right)x\sum _{n\ge 0}\frac{d^{n}x}{n!}}$

terminates for each $x\in 𝔤$. The Leibniz formula shows that $e^d[x,y]=[e^{d}x,e^{d}y]$ and hence that $e^d$ is an automorphism of the Lie algebra $𝔤$ (with inverse what does this say??).

(1.19) $\text{ad}{e}_i$ and $\text{ad}{f}_i(1\le i\le n)$ are locally nilpotent derivations of $𝔤\left(A\right)$ (and of ${𝔤}^{\prime }\left(A\right)$). (Consequently $e^{\text{ad}{e}_i},e^{\text{ad}{f}_i}$ are automorphisms of $𝔤\left(A\right)$ and of ${𝔤}^{\prime }\left(A\right)$.)

		Proof.
	It is enough to consider $\text{ad}{e}_i=\phi$, say. Let $\alpha$ be the subspace of $𝔤\left(A\right)$ consisting of all $x\in 𝔤\left(A\right)$ killed by some power of $\phi$. Since $\phi$ is a derivation, $𝔞$ is a subalgebra of $𝔤$, by virtue of the Leibnitz formula: if ${\phi }^{r}x=0$ and ${\phi }^{s}y=0$, then ${\phi }^{r+s-1}[x,y]=0$. Hence to show that $𝔞=𝔤\left(A\right)$ it is enough to show that the generators $e_j,f_j,h\in 𝔥$ belong to $𝔞$. For $e_j$ this follows from (1.18). for $h\in 𝔥$ we have $\phi \left(h\right)=[e_i,h]=-{\alpha }_i\left(h\right)e^i\text{(1.2)}$ whence ${\phi }^2\left(h\right)=0$. Finally, for $f_j$ we have $\phi \left(f_j\right)=[e_i,f_j]=0$ if $j\ne i$, and $\phi \left(f_i\right)=h_i,{\phi }^2\left(f_i\right)[e_i,h_i]=-2e_i,{\phi }^3\left(f_i\right)=-2[e_i,e_i]=0$. $\square$

The Lie algebra defined by a principal submatrix

Let $A={\left(a_{ij}\right)}_{1\le i,j\le n}$ be any $n\times n$ matrix with entries in $k$. For any non-empty subset $J$ of $\{1,2,\dots ,n\}$ let

$A_J={\left(a_{ij}\right)}_{i,j\in J}$

be the principal submatrix defined by the subset $J$. Write

$n_J=\text{Card}\left(J\right),l_J=\text{rank}\left(A_J\right)$.

We wish to see how $𝔤\left(A_J\right)$ is related to $𝔤\left(A\right)$.

Let $(𝔥,B,B^{\vee })$ be a minimal realization of $A$, so that $\text{dim}𝔥=2n-l=N$ say. Let

${\displaystyle {𝔥}_J^{\prime }=\sum _{j\in J}k{h}_j}$
${\displaystyle {𝔠}_J=\bigcap _{j\in J}\text{Ker}\left({\alpha }_j\right)}$

which are subspaces of $𝔥$.

(a) Let $V$ be a vector subspace of $𝔥$. Then the restrictions ${\alpha }_j|V(j\in J)$ are linearly independent (as linear forms on $V$) iff $V+{𝔠}_J=𝔥$.

		Proof.
	Take annihilators: $V+{𝔠}_J=𝔥\iff V^0\cap {𝔠}_J^0=0$ (in ${𝔥}^{*}$). But ${𝔠}_J^0$ is the subspace of ${𝔥}^{*}$ spanned by the ${\alpha }_j(j\in J)$, whence the result. Note that ${𝔠}_J={⟨h{\alpha }_j;j\in J⟩}^0$. $\square$

(b) Let ${𝔥}_J$ be minimal among subspaces $V$ of $𝔥$ satisfying (i) $V\supset {𝔥}_J^{\prime }$; (ii) $V+{𝔠}_J=𝔥$. Let check the following

${\displaystyle B_J=\left\{{\alpha }_j\right|{𝔥}_J:j\in J\},B_J^{\vee }=\{h_j:j\in J\}}$

Then $({𝔥}_J,B_J,B_J^{\vee })$ is a minimal realization of $A_J$.

		Proof.
	By (a), the elements of $B_J$ are linearly independent in ${𝔥}_J^{*}$. It remains to show that $\text{dim}{𝔥}_J=2n_J-l_J$. Let $\pi :𝔥\to 𝔥/{𝔥}_J^{\prime }$ be the projection. We have $\pi \left({𝔠}_J\right)=({𝔠}_J+{𝔥}_J^{\prime })/{𝔥}_J^{\prime }\cong {𝔠}_J/({𝔠}_J\cap {𝔥}_J^{\prime })$ and $\text{dim}({𝔠}_J\cap {𝔥}_J^{\prime })=n_K-l_J$ just as in the proof of (1.10); also $\text{dim}{𝔠}_J=N-n_J$, so that $\text{dim}\pi \left({𝔠}_J\right)=N-2n_J+l_J=N-N_J$ say where $N_J=2n_J-l_J$. Clearly ${𝔥}_J$ must be such that ${𝔥}_J/{𝔥}_J^{\prime }=\pi \left({𝔥}_J\right)$ is a vector space complement of $\pi \left({𝔠}_J\right)$ in $𝔥/{𝔥}_J^{\prime }$, and therefore $\begin{array}{ccc}\text{dim}\pi \left({𝔥}_J\right)& =& \text{dim}(𝔥/{𝔥}_J^{\prime })-\text{dim}\pi \left({𝔠}_J\right)\\ & =& (N-n_J)-(N-N_J)=N_J-n_J\end{array}$ and finally $\text{dim}{𝔥}_J=n_J+(N_J-n_J)=N_J$. $\square$

(c) From (1.3) we have

$\stackrel{\sim }{𝔤}\left(A_J\right)={\stackrel{\sim }{𝔫}}_{J,-}\oplus {𝔥}_J\oplus {\stackrel{\sim }{𝔫}}_{J,+}$

where ${\stackrel{\sim }{𝔫}}_{J,+}$ (resp. ${\stackrel{\sim }{𝔫}}_{J,-}$) is the free Lie algebra generated by the $e_j,j\in J$ (resp. by the $f_j,j\in J$). Hence $\stackrel{\sim }{𝔤}\left(A_J\right)$ is a subalgebra of $\stackrel{\sim }{𝔤}\left(A\right)$, and if we put $Q_J=\sum _{j\in J}\mathbb{Z}{\alpha }_j$ we have

${\displaystyle \stackrel{\sim }{𝔤}\left(A_J\right)={𝔥}_J+\sum _{\genfrac{}{}{0ex}{}{\beta \in Q_J}{\beta \ne 0}}{\stackrel{\sim }{𝔤}}_{\beta }}$

with components ${\stackrel{\sim }{𝔤}}_{\beta }(\beta \in Q_J,\beta \ne 0)$ the same as those in $\stackrel{\sim }{𝔤}\left(A\right)$.

(d) Let ${𝔯}_J$ be the unique largest ideal in $\stackrel{\sim }{𝔤}\left(A_J\right)$ satisfiying ${𝔯}_J\cap {𝔥}_J=0$ (so that $𝔤\left(A_J\right)=\stackrel{\sim }{𝔤}\left(A_J\right)/{𝔯}_J$). Then

${𝔯}_J=\stackrel{\sim }{𝔤}\left(A_J\right)\cap 𝔯$

and hence

${\displaystyle {𝔯}_J=\underset{\beta \in Q_J}{\oplus }{𝔯}_{\beta }}$

		Proof.
	Let ${𝔯}_J^{\prime }=\stackrel{\sim }{𝔤}\left(A_J\right)\cap 𝔯=\oplus {𝔯}_{\beta }$. this is an ideal in $\stackrel{\sim }{𝔤}\left(A_J\right)$ which intersects ${𝔥}_J$ trivially, hence certainly ${𝔯}_J^{\prime }\subset {𝔯}_J$. Conversely, let $\phi :\stackrel{\sim }{𝔤}\left(A_J\right)\hookrightarrow \stackrel{\sim }{𝔤}\left(A\right)\to 𝔤\left(A\right)$, so that $\text{Ker}\left(\phi \right)={𝔯}_J^{\prime }$. Let $x\in {𝔯}_J$, where $\beta \in Q_J$; I claim that $\phi \left(x\right)=0$ in $𝔤\left(A\right)$. Suppose for example $\beta >0$, and proceed by induction on $m=\text{ht}\left(\beta \right)$. If $m=1$ then $x=0$ (1.6), so certainly $\phi \left(x\right)=0$. If $m>1$, consider $[\phi \left(x\right),f_i]$. There are two cases: if $i\in J$, then $[\phi \left(x\right),f_i]=\phi \left([x,f_i]\right)=0$ by the inductive hypothesis, because $[x,f_i]\in {𝔯}_{J,\beta -{\alpha }_i}$ and $\text{ht}(\beta -{\alpha }_i)=m-1$; if $i\ne J$, then since $\phi \left(x\right)\in {𝔤}_{\beta }$ we have $[\phi \left(x\right),f_i]\in {𝔤}_{\beta -{\alpha }_i}$; but $\beta -{\alpha }_i$ is not a root, because ${\displaystyle \beta =\sum _{j\in J}{m}_j{\alpha }_j}$ (say) and $i\notin J$. Hence $[\phi \left(x\right),f_i]=0$ in both cases, and therefore by (1.8) $\phi \left(x\right)=0$. Likewise if $\beta <0$. It follows that ${𝔯}_J\subset \text{Ker}\left(\phi \right)={𝔯}_J^{\prime }$. $\square$

(e) From (d) it follows that the embedding of $\stackrel{\sim }{𝔤}\left(A_J\right)$ in $\stackrel{\sim }{𝔤}\left(A\right)$ induces an embedding of $𝔤\left(A_J\right)$ in $𝔤\left(A\right)$. We have

${\displaystyle 𝔤\left(A_J\right)={𝔥}_J+\sum _{\genfrac{}{}{0ex}{}{\beta \in Q_J}{\beta \ne 0}}{\stackrel{\sim }{𝔤}}_{\beta }/{𝔯}_{\beta }}$

from (c) and (d); but ${\stackrel{\sim }{𝔤}}_{\beta }/{𝔯}_{\beta }={𝔤}_{\beta }$ (summand of $𝔤\left(A\right)$). Hence

$\begin{array}{cc}\text{(1.20)}& {\displaystyle 𝔤\left(A_J\right)={𝔥}_J+\sum _{\genfrac{}{}{0ex}{}{\beta \in Q_J}{\beta \ne 0}}{𝔤}_{\beta }}\end{array}$

Hence is $R$ (resp. $R_J$) is the set of roots of $𝔤\left(A\right)$ (resp. $𝔤\left(A_J\right)$) we have $RJ=R\cap Q_J$, and the multiplicity of $\beta \in R_J$ a root of $𝔤\left(A_J\right)$ is the same as its multiplicity as a root of $R$.

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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