We present a detailed exposition of the theory of generalized Kac-Moody algebras associated to symmetrizable matrices. A proof of the character formula for a standard module is given, generalizing the argument of Garland and Lepowsky for the Kac-Moody case. A short proof of the theorem that any generalized Kac-Moody algebra can be decomposed into direct (vector space) sums of free subalgebras and a Kac-Moody subalgebra is given.
Generalized Kac-Moody algebras, called Borcherds algebras in [14], were investigated by R. Borcherds in [2]. We show that any generalized Kac-Moody algebra g that has no mutually orthogonal imaginary simple roots can be written as g = u + ⊕ (g J + h) ⊕ u − , where u + and u − are subalgebras isomorphic to free Lie algebras with given generators, and g J is a Kac-Moody algebra defined from a symmetrizable Cartan matrix (see Theorem 5.1). There is a formula due to Witt that computes the graded dimension of a free Lie algebra where all of the generators have been assigned degree one. It is known that Witt's formula can be extended to other gradings (e.g., [5]). We present a further generalization of the formula appearing in [5]. The denominator identity for g is obtained by using this generalization of Witt's formula and the denominator identity known for the Kac-Moody algebra g J . In this work, we are taking g to be the algebra defined by the appropriate generators and relations, rather than the quotient of this algebra by its radical. In particular, our main result and consequent proof of the denominator identity give a new proof that the radical of a generalized Kac-Moody algebra of the above type is zero. (We use the fact that the radical of g J is zero, which is Serre's theorem in the case that g J is finite-dimensional; this is the main case for us.)The most important application of our work is to the Monster Lie algebra m, defined by R. Borcherds [4]. In fact, we show that m = u + ⊕ gl 2 ⊕ u − , with u ± free Lie algebras. This result is obtained by applying the above Let k + J be the ideal of l generated by {(ad e i ) 1−2a ij /a ii e j | i, j ∈ J, i = j}, then U (n + J ) = U (l/k + J ) = U (l)/K, where K denotes the ideal of U (l) generated by k + J ⊂ U (l). Thus we can decompose the vector spaceApplying the elimination theorem once again, using the above decomposition we obtain:where c is the ideal in L( j∈I\J U (l)e j /R j ) generated by the sum of the Ke j . Each h e -module U (n + J )e j / i∈J U (n + J )(ad e i ) 1−2a ij /a ii e j is an integrable 20 lowest weight module for the Lie algebra g J , denoted by U (n + J ) · e j , with lowest weight α j . Thus we have a decomposition into semidirect products:It is clear that, as ideals of. Therefore, since all elements of k + 0 are zero in L( i∈I\J U (n + J ) · e j ),By the definition of g J , n + J = L({e i } i∈J )/k + J .Corollary 5.1 Let A be a matrix satisfying conditions C1-C3. Assume that the matrix A has only one positive diagonal entry, a ii > 0, and if a mj = 0 then m = i, or j = i or m = j. Let S = {(ad e i ) l e j } 0≤l≤−2a ij /a ii . The subalgebra n + ⊂ g(A) is the semidirect product of a one-dimensional Lie algebra and a free Lie algebra, n + = Re i ⊕ L(S). Similarly, n − = Rf i ⊕ L(η(S)). Thus g(A) = L(S) ⊕ (sl 2 + h) ⊕ L(η(S)).
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