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\proclaim Central extensions of generalized Kac-Moody algebras.
J. Algebra Vol 140, No. 2, July 1991, 330--335.
Richard E. Borcherds, %author
D.P.M.M.S.,
Cambridge University,
16 Mill Lane,
Cambridge,
CB2 1SB,
England.
\bigskip
The main result of Borcherds [1] states that graded Lie algebras with
an ``almost positive definite'' contravariant bilinear form are
essentially the same as central extensions of generalized Kac-Moody
algebras. In this paper we calculate these central
extensions. Ordinary Kac-Moody algebras have nontrivial centers when
the Cartan matrix is singular; generalized Kac-Moody algebras turn out
to have some ``extra'' center in their universal central extensions
whenever they have simple roots of multiplicity greater than 1, and in
particular the dimension of the Cartan subalgebra can be larger than
the number of rows of the Cartan matrix.
\proclaim 1.~Statement of results.
The main result of this paper states that, roughly speaking, graded
Lie algebras with an ``almost positive definite'' contravariant
bilinear form are the same as a sort of generalization of Kac-Moody
algebras. More precisely,
\proclaim Theorem 1.
Suppose that $G$ is a real Lie algebra. Then condition (2) implies
(1), and (1) implies (2) if $\dim G_i <\infty$ for all $i$.
\item{(1)} $G$ has a grading $G=\oplus_i G_i$ with $G_0\subset [G,G]$,
an involution $\omega$ which maps $G_i$ to $G_{-i}$ and is $-1$ on
$G_0$, and an invariant bilinear form $(,)$ such that $(a,b)=0$ if $a$
and $b$ have different degrees and such that $(a,-\omega(a))>0$ if $a$
is a nonzero element of $G_i$ for $i\ne 0$. In addition $G$ is a sum
of one dimensional eigenspaces of $G_0$; this follows from the other
conditions if the spaces $G_i$, $i\ne 0$, are finite dimensional. We
can summarize these conditions roughly by saying that $G$ is a graded
Lie algebra with an almost positive definite contravariant bilinear
form.
\item{(2)} There is a symmetric matrix $a_{ij}$, $i,j\in I$, such that
$a_{ij}\le 0$ if $i\ne j$, and such that if $a_{ii}>0$ then
$2a_{ij}/a_{ii}$ is an integer for any $j$. Some central extension of
$G$ is given by the following generators and relations.
\itemitem{} Generators:
\itemitem{} Elements $e_i$, $f_i$, $h_{ij}$ for $i,j\in I$.
\itemitem{} Relations:
\itemitem{} $[e_i,f_j]=h_{ij}$.
\itemitem{} $[h_{ij},e_k]=\delta_i^ja_{ik}e_k$,
$[h_{ij},f_k]=-\delta_i^ja_{ik}f_k$.
\itemitem{} If $a_{ii}>0$ then $\Ad(e_i)^ne_j=0=\Ad(f_i)^nf_j$, where
$n=1-2a_{ij}/a_{ii}$.
\itemitem{} If $a_{ii}\le0$, $a_{jj}\le 0$ and $a_{ij} = 0 $
then $[e_i,e_j]=0=[f_i,f_j]$.
The subalgebra $G_0$ must be abelian because the automorphism $\omega$
acts as $-1$ on it. If $a_{ii}>0$ for all $i$ then the conditions of
(2) are equivalent to the defining relations for the Kac-Moody algebra
with symmetrized Cartan matrix $a$ (with $h_{ii}=h_i$). If we allow
$a_{ii}$ to be non-positive but add the condition that $h_{ij}=0$ if
$i\ne j$ then we obtain the relations for ``generalized Kac-Moody
algebras'' as in Borcherds [1]. In any case, $h_{ij}$ is 0 unless the
$i$th and $j$th columns of $a$ are equal, and is in the center of
$G$ unless $i=j$, so the Lie algebra above is a central extension of a
generalized Kac-Moody algebra.
Theorem 3.1 of Borcherds [1] implies that any Lie algebra satisfying
the conditions of (1) above is a central extension of a generalized
Kac-Moody algebra. Because of this, Theorem 1 follows almost
immediately from the following theorem, which we prove in Section 2 of
this paper.
\proclaim Theorem 2.
Suppose that $G$ is the Lie algebra defined in part (2) of Theorem
1. The subalgebra of $G$ generated by the elements $h_{ij}$ is abelian
and has a basis consisting of the elements $h_{ij}$ for $i,j\in I$
such that the $i$th and $j$th columns of $A$ are the same. If $a$
has no zero columns then $G$ is perfect and equal to its own universal
central extension.
In particular we recover the well-known result that ordinary Kac-Moody
algebras are their own universal central extensions, because in this
case it is not possible for two columns of the symmetrized Cartan
matrix to be equal.
It is easy to extend most of the results of Borcherds [1] about
generalized Kac-Moody algebras, such as the generalized Kac-Weyl
formula for characters of highest weight modules, to these central
extensions of generalized Kac-Moody algebras.
Example. The monstrous generalized Kac-Moody algebra $G$ [2] has as
root lattice the even 26 dimensional unimodular Lorentzian lattice
$L=II_{25,1}$, and if $r$ is a nonzero element of $L$ then $r$ has
multiplicity $p_{24}(1-r^2/2)$, which is the number of partitions of
$1-r^2/2$ into parts of 24 colors. This lattice has a certain norm 0
``Weyl vector'' $\rho$, and the simple roots of $G$ are given by
\item{(1)} All norm 2 vectors of $L$ which have inner product $-1$
with $\rho$. These are the real simple roots of $G$.
\item{(2)} All positive integral multiples of $\rho$, each with multiplicity
24. These are the norm 0 simple roots of $G$.
Therefore the universal central extension $\hat G$ of $G$ has a Cartan
subalgebra which is the sum of
\item{(1)} A one dimensional space for each real simple root of $G$.
\item{(2)} A $24^2=576$ dimensional space for each positive integer.
The center of $\hat G$ has index 26 in the Cartan subalgebra, and the quotient
of $\hat G$ by its center is simple.
{\it Remark.} It is possible to define generalized Kac-Moody
superalgebras, and many theorems about generalized Kac-Moody algebras
can be generalized to generalized Kac-Moody superalgebras, provided we
make a few changes such as replacing the involution $\omega$ by an
element of order 4 whose square is 1 on even elements of the
superalgebra and $-1$ on odd elements. Theorem 1 generalizes to
superalgebras provided we make all the ``usual'' changes to get from
algebras to superalgebras, and also add to part (2) of Theorem 1 the
condition that $G$ has no odd real simple roots. There are many finite
dimensional simple superalgebras satisfying this condition.
\proclaim 2.~Proofs.
In this section we give the proof of Theorem 2, which consists mainly
of checking that the Lie algebra in (2) of Theorem 1 is its own
universal central extension if it is perfect.
We start by showing that a simpler Lie algebra is its own universal
central extension.
\proclaim Theorem 3.
Let $a_{ij}$ be any real matrix with no zero columns and let $G$ be
the Lie algebra given by the following generators and relations.
\item{}Generators:
\itemitem{}$e_i$, $f_i$, $h_{ij}$, $i,j\in I$.
\item{}Relations:
\itemitem{}$[e_i,f_j]=h_{ij}$,
\itemitem{}$[h_{ij}e_k]=\delta_i^ja_{ik}e_k$,
$[h_{ij}f_k]=-\delta_i^ja_{ik}f_k$.
\item{}Then $G$ is perfect and is equal to its own universal central extension.
{\it Proof.} It is obvious that $G$ is perfect and that $h_{ij}$ is
in the center of $G$ if $i\ne j$. Also,
$$\eqalign{
a_{ij}h_{jk} &= [a_{ij}e_j,f_k] = [[h_{ii},e_j],f_k]=
[h_{ii},[e_j,f_k]] - [e_j,[h_{ii},f_k]] \cr
&= [h_{ii},h_{jk}]+ [e_j,a_{ik}f_k] = [h_{ii}, h_{jk}] +a_{ik}h_{jk}.\cr
}$$
If $j=k$ this implies that $[h_{ii},h_{jj}]=0$, so all the $h$'s
commute with each other. If $j\ne k$ this then implies that $h_{jk}=0$
unless the $j$th and $k$th columns of $a$ are equal.
Let $e_i', f_j', h_{ij}'$ be elements of the universal central
extension $\hat G$ of $G$ mapping onto $e_i,f_j, h_{ij}$. We
repeatedly use the fact that if the images of two elements $x'$ and
$y'$ commute in $G$, then $[x',y']$ is in the center of $\hat G$. Then
$[h_{ii}',[h_{jj}'e_k']]=[h_{jj}',[h_{ii}',e_k']]$ because
$[h_{ii}',h_{jj}']$ is in the center of $\hat G$. Therefore
$$a_{jk}[h_{ii}',e_k']=[h_{ii}',[h_{jj}',e_k']]
=[h_{jj}',[h_{ii}',e_k']] = a_{ik}[h_{jj}',e_k'].
$$
For any $i$ with $a_{ik}$ nonzero we can form the element
$[h_{ii}',e_k']/a_{ik}$ of $\hat G$; this does not depend on $i$ by
the equality above and maps onto $e_k$ in $G$. We may therefore
assume that $e_k'$ is equal to this element so that
$a_{ik}e_k'=[h_{ii}',e_k']$ for all $i$ and may likewise assume that
$a_{ik}f_k'=-[h_{ii}',f_k']$. Finally we may redefine $h_{ij}'=[e_i',
f_j']$. We wish to show that these new elements satisfy the relations
of $G$, and the only relations which are not part of their definition
are that the elements $h_{jl}'$ commute with $e_k'$ and $f_k'$ if $j\ne
l$. If we choose $i$ with $a_{ik}\ne 0$ then this follows from
$$a_{ik}[h_{jl}',e_k'] = [h_{jl}',[h_{ii}',e_k']]=[h_{ii}',[h_{jl}',e_k']]=0,$$
the last equality holding because $[h_{jl}', e_k']$ is in the center of
$\hat G$. This shows that $\hat G$ contains elements satisfying the relations
of $G$, so that the extension $\hat G$ splits.
Therefore $G$ is its own universal central extension and Theorem 3 is proved.
{\it Remark.} The Lie algebra in Theorem 3 is the universal central
extension of a Lie algebra associated to the matrix $a$ in Kac [3],
and is equal to this Lie algebra when all columns of $a$ are
different, which is always the case for ordinary Kac-Moody algebras.
\proclaim Corollary.
Let $G$ be the Lie algebra of Theorem 3 and let $H$ be any ideal of
$G$. The the universal central extension of $G/H$ is $G/[G,H]$, and
in particular if $H=[G,H]$ then $G/H$ is its own universal central
extension.
{\it Proof.} This is true for any perfect Lie algebra $G$ which is its own
universal central extension. The pullback of $G\rightarrow G/H$ and
the universal central extension of $G/H$ is a perfect central
extension of $G$ and hence equal to $G$, so the map from $G$ to $G/H$
factors through the universal central extension of $G/H$. It is easy
to see that this implies that the universal central extension of $G/H$
is $G/[G,H]$. This proves the corollary.
We can now give the proof of the second part of Theorem 2. If $G$ is
the Lie algebra of Theorem 1 and $H$ is its ideal generated by some
elements of the form $\Ad(e_i)^ne_j$, $\Ad(f_i)^nf_j$, $[e_i,e_j]$,
and $[f_i,f_j]$ as in Theorem 1, then by the corollary we only have to
check that $H=[G,H]$. The element $\Ad(e_i)^n(e_j)$ is in $[G,H]$
because $[h_{ii},\Ad(e_i)^ne_j]=(na_{ii}-a_{ij})\Ad(e_i)^ne_j$ and
$na_{ii}-a_{ij}$ is nonzero because $a_{ii}>0$, $a_{ij}\le 0$. The
element $[e_i,e_j]$ is in $[G,H]$ when $a_{ii}\le 0$, $a_{jj}\le 0$,
$a_{ij}=0$ because if we choose any $k$ with $a_{ki}\ne 0$ then
$[h_{kk}, [e_i,e_j]]=(a_{ki}+a_{kj})[e_i,e_j]$ and $a_{ki}+a_{kj}$ is
nonzero because $a_{ki}<0$, $a_{kj}\le 0$. Similarly the elements of
the forms $\Ad(f_i)^nf_j$ and $[f_i,f_j]$ are in $[G,H]$, and this
proves the second part of Theorem 2.
Finally we have to prove the first part of Theorem 2, the only
nontrivial part of which is to prove that the $h$'s are linearly
independent. In particular we show that the elements $h_{ij}$ are
nonzero if the $i$th and $j$th columns of $a$ are the same, so the
central extensions we have constructed are nontrivial.
\proclaim Theorem 4.
Let $a$ be any matrix (possibly with some zero columns) and let $G$ be
the Lie algebra with the generators and relations of Theorem 3. Then
the subalgebra of $G$ generated by the $h$'s is abelian and has a
basis consisting of the elements $h_{ij}$ for $i,j\in I$ such that the
$i$th and $j$th columns of $a$ are equal. If these columns are not
equal then $h_{ij}=0$.
{\it Proof.} This proof is a slight modification of the one in Kac [3]
for the case of Kac-Moody algebras. The only nontrivial thing to check
is that the $h$'s are linearly independent, which we show by
constructing sufficiently many ``lowest weight'' representations of
$G$. We let the space $V$ be the universal associative algebra
generated by elements $e_i$, and let $e_i$ act on $V$ by left
multiplication. We choose any real numbers $b_{ij}$ with $b_{ij}=0$
unless the $i$th and $j$th columns of $a$ are equal, and define
operators $h_{ij} $ on $V$ such that $h_{ij}(1)=b_{ij}1$ and
$$[h_{ij},e_k] = \delta_i^ja_{ik}e_k\qquad\hbox{for all $i,j,k$.}\eqno {(1)}$$
Similarly we can define operators $f_j$ on $V$ such that $f_j(1)=0$ and
$$[f_j,e_i] = -h_{ij}\qquad\hbox{for all $i,j,k$.}\eqno{(2)}$$
This will give a representation of $G$ provided that the relation
$[h_{ij}f_k]=-\delta_i^ka_{ik}f_k$ is satisfied, and if it is this
will prove the theorem because we will have constructed enough
representations of $G$ on which elements of the Cartan subalgebra act
non-trivially. The operator $[h_{ij},h_{kl}]$ commutes with all the
$e$'s and vanishes on 1, so it is 0 on $V$, and for the same reason
$h_{ij}$ is 0 when the $i$th and $j$th columns of $a$ are different
because $b_{ij}=0$. From the relations (1) and (2) we find that
$$\eqalign{
[[h_{ij},f_k]+\delta_i^ja_{ik}f_k,e_l]
&= [h_{ij},[f_k,e_l]]- [f_k,[h_{ij},e_l]] - \delta_i^ja_{ik}h_{lk}\cr
&=-[h_{ij},h_{lk}] +\delta_i^ja_{il}h_{lk} - \delta_i^ja_{ik}h_{lk},\cr
}$$
which is 0 because $h_{lk}$ is 0 unless $a_{ik}=a_{il}$. Hence
$[h_{ij},f_k] +\delta_i^ja_{ik}f_k$ is 0 on $V$ because it vanishes on
1 and commutes with the $e$'s. This proves Theorem 4.
Theorem 2 follows from this in the same way that the corresponding theorem is proved for Kac-Moody algebras, as in Kac [3].
{\it Remark.} If the $i$th and $j$th columns of $a$ are equal then the
Lie algebra $G$ of Theorem 1 has an outer derivation $h$ defined by
$[h,e_i]=e_j$, $[h,e_j]=-e_i$, $[h,e_k]=0$ if $k\ne i,j$,
$[h,f_i]=f_j$, $[h,f_j]=-f_i$, $[h,f_k]=0$ if $k\ne i,j$. If $a$ has
$n$ equal columns then these derivations generate an action of the
orthogonal group $O_n(R)$ on $G$. These outer derivations do not
always commute with the elements of the Cartan subalgebra.
{\it Errata.} There is a mistake in the statement of Theorem 5.1 of
Borcherds [1]: the phrase ``nonsingular SCM'' should twice be replaced
by ``nonsingular bilinear form on the Cartan subalgebra''. Line 15 on
page 502 should read ``follows from (4).'' Line $-4$ on page 502 should
read ``... with all real simple roots.'' Proposition 2.2 should read
``A positive root...''.
\proclaim References.
\item {1.} R. E. Borcherds, Generalized Kac-Moody algebras,
J. Algebra 115 (1988), 501--512.
\item{2.} R. E. Borcherds, The monster Lie algebra,
Adv. Math. 83 (1990), 30--47.
\item{3.} V. G. Kac, ``Infinite dimensional Lie algebras'',
Birkh\"auser, Basel, 1983.
\bye