A Short note on automorphisms of generalized Kac-Moody superalgebras

Urmie RAY^{1}

Max-Planck-Institut Für Mathematik, Vivatsgasse 7, D-53111 Bonn and

Dept de Math., UFR Sciences, Univ. de Reims, Moulin de la Housse, F-51062 Reims Cedex

The aim of this note is to give the decomposition of an arbitrary automorphism of a generalized Kac-Moody superalgebra. In order to do so, we show that Cartan subalgebras of generalized Kac-Moody superalgebras are conjugate under the action of inner automorphisms. This well-known result of Peterson-Kac [PK] for Kac-Moody algebras therefore remains valid in the more general context. This implies that not only the rank of a Cartan subalgebra of a generalized Kac-Moody superalgebra, but also the rank of of the Kac-Moody subalgebra generated by all the simple real root spaces, are invariants of the superalgebra. This result therefore puts an end to the speculation according to which the Monster vertex algebra is the tensor product of the vertex algebra of a Niemeier lattice and of the vertex algebra of the rank two even unimodular Lorentzian lattice.

1. A generalization of a Gabber-Kac result

The definitions of Kac-Moody algebras given independently by Kac [K1] and Moody [M] are different but were shown to be equivalent by Gabber and Kac [GK]. Both Kac’s definition and Moody’s definition can be generalized to define generalized Kac-Moody superalgebras [Ra]. We start by showing that in this more general context, both definitions remain equivalent. This was assumed to be true in [Ra] and we use a result from [Ra] in this paper.

We first fix some notation used throughout this paper. As usual G\overline{0} and G\overline{1} will denote the even and odd part of the Lie superalgebra G.

1 Supported by the MPI

The base field F will be either R or C. Let I be a finite set {1, ..., n} or a countable one identified with Z_{+} and S be a subset of I. The set S indexes the generators contained in G\overline{1}. Let H be a F-vector space with a non-degenerate symmetric bilinear form (., .), containing vectors h_{i} indexed by the set I. Set a_{ij} := (h_{i}, h_{j}) and assume that the matrix A = (a_{ij})_{i,j∈I} is a generalized symmetric Cartan matrix, i.e.

a_{ij} = a_{ji};

a_{ij} ≤ 0 if i ≠ j; \frac{2a_{ij}}{a_{ii}} ∈ Z if a_{ii} > 0; \frac{a_{ij}}{a_{ii}} ∈ Z if a_{ii} > 0 and i ∈ S. The Lie superalgebra G̃ = G̃(A, S, H) is generated by the vector space

H considered as an even abelian subalgebra and elements e_{i}, f_{i}, i ∈ I, satisfying the defining relations:

(i) [e_{i}, f_{j}] = δ_{ij}h_{i}; (ii) [h, e_{j}] = (h, h_{i})e_{j}, h ∈ H; (iii) [h, f_{j}] = −(h, h_{j})f_{j} (iv) deg e_{i} = deg f_{i} = \overline{0} if i ∉ S

(v) deg e_{i} = deg f_{i} = \overline{1} if i ∈ S

Following Moody’s definition [M], we define the generalized Kac-Moody algebra G = G(A, S, H) by generators and relations.

Definition. The generalized Kac-Moody algebra G = G(A, S, H) with Cartan matrix A and Cartan subalgebra H is the Lie superalgebra generated by H, e_{i}, and f_{i}, i ∈ I and with defining relations (i)-(v) and with the following extra ones:

(vi) (ad e_{i})^{\frac{−2a_{ij}}{a_{ii}} +1}e_{j} = 0 = (ad f_{i})^{\frac{−2a_{ij}}{a_{ii}} +1}f_{j} for all i ∈ I such that a_{ii} > 0 and j ∈ I (vii) (ad e_{i})^{\frac{−a_{ij}}{a_{ii}} +1}e_{j} = 0 = (ad f_{i})^{\frac{−a_{ij}}{a_{ii}} +1}f_{j} for all i ∈ S such that a_{ii} > 0 and j ∈ I (viii) [e_{i}, e_{j}] = 0 = [f_{i}, f_{j}] if a_{ij} = 0.

The Lie superalgebra G̃ has a unique maximal ideal R̃ which intersects the Cartan subalgebra H trivially [K4]. Following Kac’s definition in the Lie algebra case [K1], the generalized Kac-Moody algebra G = G(A, S, H) is the quotient G̃/R̃. We show that both definitions are equivalent.

We first need a few more notations. Let G be the generalized Kac Moody superalgebra according to the above Definition. let Δ denote the set of roots, Δ^{+} the set of positive roots, Δ_{0} the set of even roots and Δ_{0}^{+} the set of even positive roots. Let N^{+} := ⊕_{α∈Δ_{0}^{+}}G_{α} and N_{0}^{+} = N^{+} ∩ G\overline{0}. Let W be the Weyl group (see [Ra] for details).

Lemma 1. The even subalgebra G\overline{0} of G is a generalized Kac-Moody algebra.

Proof. The Lie subalgebra N_{0}^{+} is generated by root spaces G_{α} with α ∈ Δ_{0}^{+} for which there does not exist any positive even roots β_{1}, β_{2} ∈ Δ_{0}^{+} satisfying α = β_{1} + β_{2}. Let Π^{0} be the set of all such positive even roots. Consider α, β ∈ Π^{0}.

We first show that (α, β) ≤ 0.

If α is real then (α, β) ≤ 0. Otherwise, (α, β) > 0, and so α − β is an even root. Hence β = α + (β − α) or α = β + (α − β), and so either α or β is not in Π^{0}, contradicting assumptions.

If α is imaginary, then by what precedes, (α, α_{i}) ≤ 0 for all real roots α_{i}. If α_{i} is even, its a direct consequence of the above. Otherwise take 2α_{i}. Hence (α, α_{i}) ≤ 0 for all i ∈ I. Therefore, (α, β) ≤ 0.

It remains to show that the roots α, β satisfy relation of the above type (vi) and (viii).

If the root α is real, then there exists w ∈ W such that w(α) = α_{i}. Therefore for any e_{α} ∈ G_{α} and e_{β} ∈ G_{β},

ade_{α}^{\frac{−2(α,β)}{(α,α)} +1}(e_{β}) = 0.

Finally suppose that α and β are both imaginary orthogonal roots and that [e_{α}, e_{β}] ≠ 0. Then as (α, α_{i}) ≤ 0 for all i ∈ I, (α, α_{i}) = 0 for every root α_{i} in the support of β. Suppose that α_{i} is in the support of β but not in the support of α. Then (α_{j}, α_{i}) = 0 for every root α_{j} in the support of α. Hence α + β does not have a connected support which contradicts the fact that all roots have connected support [K4]. Hence α and β have the same support. Hence as the support of α is connected, all the roots in its support are real, and (α, α_{i}) = 0 for all α_{i} in the support of α. Hence the support of α generates an affine Lie superalgebra. It therefore follows from the structure of the affine Kac-Moody Lie superalgebras [K3] that [e_{α}, e_{β}] = 0, contradicting assumptions.

Hence, we can conclude that G\overline{0} is a GKM algebra with simple root set Π^{0}.

Proposition 2. Any ideal of G intersecting the Cartan subalgebra H trivially is trivial.

Proof. Without loss of generality, we assume that the Cartan matrix A is indecomposable. Let R be the unique maximal ideal of G having trivial intersection with the Cartan subalgebra H.

Since G\overline{0} is a generalized Kac-Moody algebra, the kernel of the invariant bilinear symmetric form (., .)_{0} induced by the Cartan matrix of G\overline{0} is contained in H [K4]. Let (., .) be the invariant bilinear symmetric form on G induced by the Cartan matrix A [K4] (with respect to the simple root set Π^{0} of G_{0}) Then (., .) is consistent [K2] and non-trivial on G\overline{0}. Hence its restriction to G\overline{0} is a non-trivial multiple of the form (., .)_{0}. Therefore (., .) is non-degenerate on G\overline{0}.

Let x_{α} ∈ G_{α}, where α is a positive root of G be an element generating the ideal R. Then [x_{α}, f_{i}] = 0 for all i ∈ I. In particular, [x_{α}, x_{−α}] = 0 for all x_{−α} ∈ G_{−α}. It follows that (x_{α}, x_{−α}) = 0 since [x_{α}, x_{−α}] = (x_{α}, x_{−α})h_{α}, where (h, h_{α}) = α(h) for all h ∈ H [K4]. As (x_{α}, y) = 0 for any y ∈ G_{β} for all β ≠ −α, x_{α} is therefore in the kernel K of (., .). Since K ∩ G\overline{0} = 0, x_{α} ∈ G\overline{1}. As (., .) is invariant, [x_{α}, x_{α}] ∈ K. Hence as [x_{α}, x_{α}] ∈ G\overline{0}, [x_{α}, x_{α}] = 0. So (α, α) = 0. As [x_{α}, f_{i}] = 0 for all i ∈ I, (α, α_{i}) ≤ 0 for all i ∈ I. As x_{α} generates R, α cannot be a simple root. Thus the support of the root α being connected, all the simple roots in its support are real. Therefore the structure of affine Kac-Moody Lie superalgebras [K3] implies that there exits i ∈ I such that [x_{α}, f_{i}] ≠ 0, contradicting assumptions. This proves that R = 0.

Therefore the generalized Kac-Moody superalgebra G is the quotient of the Lie superalgebra G̃ by the ideal R̃.

2. Conjugacy of Cartan subalgebras of the Lie superalgebra G

Let J = {i ∈ I|a_{ii} > 0} be the set indexing the real simple roots of G and G_{J} be the Kac-Moody sub-superalgebra of G generated by the elements e_{i} and f_{i} for i ∈ J.

Define U^{+} (resp. U^{−}) as the Lie sub-superalgebra generated by the root spaces G_{α} where the support of the root α ∈ Δ^{+} (resp. −α ∈ Δ^{+}) contains at least one imaginary (non-positive norm) simple root. Then:

Lemma 3. Keeping the above notation,

G = U^{+} ⊕ (G_{J} + H) ⊕ U^{−}

as a direct sum of vector spaces.

Proof Using arguments similar to [J], this is a direct consequence of the elimination Theorem [Bou].

Proposition 4. All Cartan subalgebras of G are contained in the sub-superalgebra G_{J} + H.

Proof. Without loss of generality, we may assume that the Cartan matrix of G is indecomposable. Suppose that K is a Cartan subalgebra of G not contained in G_{J} + H. By definition, K is an even subalgebra of G.

Since K is a Cartan subalgebra, it acts on G in a locally finite way. Let

x = u + v + y

be an element of K, where u ∈ U^{+}, v ∈ U^{−} and y ∈ G_{J} + H, which is not in G_{J} +H. We first assume that 0 ≠ u. So there exists positive roots β_{1}, ..., β_{r} ∈ Δ containing imaginary simple roots in their support and root vectors u_{i} ∈ G_{β_{i}} such that u = ∑_{i=1}^{r} u_{i}.

Suppose that there exists a simple root α_{i} such that for some j = 1, ...r, (β_{j}, α_{i}) < 0.

Since K is an even subalgebra, all roots β_{j} are even. So the roots β_{j}, 1 ≤ j ≤ r are all imaginary since they contain imaginary simple roots in their support and cannot thus be conjugate to a simple root under the action of the Weyl group. This implies that (adu_{j})^{n}e_{i} ≠ 0 for all n ∈ Z_{+} [Ra].

We claim that there are infinitely many vectors (adx)^{n}(e_{i}), n ∈ Z_{+} which are linearly independent. Let k ∈ {1, ..., r} be such that β_{k} is of maximum height with the property that β_{k} + α_{i} is a root. Then for each positive integer n, (adu)^{n}(e_{i}) is the sum of root vectors, exactly one of which corresponds to the root nβ_{i} + α_{i}.

This contradicts the fact that K acts in a locally finite way.

Next suppose that (β_{j}, α_{i}) ≥ 0 for all j = 1, ..., r and all i ∈ I. If α_{i} is imaginary, then this forces (β_{j}, α_{i}) = 0. Hence if α_{i} is in the support of β_{j}, then α_{i} is orthogonal to all the simple roots distinct from α_{i} in the support of β_{j}. Since the support of a root is connected, we can deduce that the roots β_{j} are simple imaginary roots for all j = 1, ..., r. Furthermore they are orthogonal to all the imaginary simple roots. Since β_{j} is simple, (β_{j}, α_{i}) ≤ 0 for any simple root α_{i}. Hence the simple roots β_{j} are orthogonal to all simple roots. As the Cartan matrix is assumed to be indecomposable, it follows that j = 1 and |I| = 1 and G is a Heisenberg algebra of rank 1. Therefore the Cartan subalgebra H is unique and in particular K = H, contradicting assumptions.

So u = 0. If v ≠ 0, then similar arguments lead to a contradiction. It follows that x ∈ G_{J} +H and this forces the Cartan subalgebra K to be contained in G_{J} + H, contradicting assumptions and proving the result.

Theorem 5. All Cartan subalgebras of G are conjugate under the action of the derived Lie group of G_{J}.

Proof. As all Cartan subalgebras are contained in G_{J} +H by Proposition 4 and G_{J} is a Kac-Moody algebra, this is a consequence of the conjugacy Theorem of Peterson-Kac [PK] (which can easily be generalized to Kac-Moody superalgebras).

The following is an immediate consequence of this result.

Corollary 6. The rank of the Lie superalgebra is invariant and there is a uniquely defined generalized Cartan matrix for G.

Remark. Since the Cartan matrix uniquely defines the generalized Kac-Moody superalgebra G, it follows that the rank of the Kac-Moody sub-superalgebra G_{J} generated by the elements e_{i}, f_{i}, i ∈ J is also an invariant of the Lie superalgebra G.

We next give two applications of Theorem 5.

a) The Monster vertex algebra

The crucial first step towards the proof of the Moonshine Theorem was the construction of a Z-graded module V for the Monster simple group [FLM]. It has a rich but complex algebraic structure. One of its most important characteristics is that it is a vertex algebra [FLM], [Bor1]. However the structure of a vertex algebra on it is very hard to see. It would therefore be useful to have a different construction of the Monster vertex algebra, which is the tensor vertex algebra of V and the lattice vertex algebra corresponding to the even unimodular Lorentzian lattice of rank 2 II_{1,1}. In particular this may help to find an adequate integral form on V for modular moonshine questions [Ry] [BR] [Bor4].

Unlike the Monster vertex algebra, the Fake Monster vertex algebra has a nice construction. It is the lattice vertex algebra associated to the even unimodular non-degenerate Lorentzian lattice II_{25,1} of rank 26 [Bor1], [Bor2]. Now, an important property of the Monster vertex algebra is that the Virasoro algebra acts on it with central charge 26. Similarly, an important aspect of a lattice vertex algebra constructed from an even non-degenerate lattice is that the Virasoro algebra acts on it with central charge equal to the rank of the lattice. Therefore there were speculations that the Monster vertex algebra would also have a construction derived from lattices in the following manner [Bor5]: For a lattice L, let V_{L} be the corresponding vertex algebra [Bor1]. Let L be a Niemeier lattice [CS]. So by definition the lattice L is positive definite of rank 24. Then the Virasoro algebra acts on the vertex algebra V_{L} ⊗ V_{II_{1,1}} with central charge 26 [Bor1].

Due to Theorem 5, the Monster vertex algebra cannot be constructed as a tensor product vertex algebra of this form, contrary to what had been thought.

Corollary 7. Let II_{1,1} be the even unimodular rank 2 Lorentzian lattice and L a Niemeier lattice. Then V_{L} ⊗ V_{II_{1,1}} is not isomorphic to the Monster vertex algebra.

Proof. Let L be a Niemeier lattice, the direct sum lattice L ⊕ II_{1,1} is even unimodular Lorentzian of rank 26 . Up to isomorphism there is a unique lattice II_{25,1} with this property [CS] and V_{II_{25,1}} is the root lattice of the Fake Monster Lie algebra [Bor2]. Therefore V = V_{L} ⊗V_{II_{1,1}} is isomorphic to the Fake Monster vertex algebra. Let G be the generalized Kac-Moody algebra which can be constructed from as a sub-quotient of V according to the method in [Bor1]. Then G is the Monster Lie algebra [Bor3] and is isomorphic to the Fake Monster Lie algebra. Hence by Corollary 6, they both have the same Cartan matrix (up to reordering). Therefore G cannot be the Monster Lie algebra.

b) Automorphisms of generalized Kac-Moody superalgebras

We finally give the decomposition of an arbitrary automorphism of the generalized kac-Moody superalgebra G. Let ω be the Chevalley automorphism of period 4 of G [K3]. It is the Chevalley involution if G is a generalized Kac-Moody algebra. We remind the reader that a diagram automorphism θ of G satisfies by θe_{i} = e_{θ̇(i)} and θf_{i} = f_{θ̇(i)} for all i ∈ I, where θ̇ is a bijection on the indexing set I such that a_{θ̇i,θ̇j} = a_{i,j} and θ̇(i) ∈ S if and only if i ∈ S.

Corollary 8. Let ϕ be an automorphism of the Lie superalgebra G. Then there exists an inner automorphism ϕ_{1} and a diagram automorphism ϕ_{2} of G such that ϕ = ω^{i}ϕ_{1}ϕ_{2} for some integer 0 ≤ i ≤ 3.

Proof. Since ϕ(H) is a Cartan subalgebra of G, there exists an inner automorphism ϕ_{1} such that ψ_{1} = ϕ_{1}^{−1}ϕ fixes H. Let Π be a base of H. Then ψ_{1}(Π) is a base of H. All bases of H are conjugate under the action of ±W [K4]. The Weyl group elements are given by inner automorphisms ([FH]: the proof is the same in the infinite dimension and superalgebra case). Therefore for some integer 0 ≤ i ≤ 3, the automorphism ω^{i}ψ_{1} fixes the base Π. Hence as this automorphism leaves the bilinear form invariant, it is a diagram automorphism.

Acknowledgements. The author is very grateful to the Max-Planck Institute for its financial support and hospitality during this work.

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