simplicity of complete Kac-Moody groups over finite fields Communicated by C.A. Weibel
MSC:Primary: 20E42 secondary: 20E32 17B67 20E18 22F50 a b s t r a c t Let G be a Kac-Moody group over a finite field corresponding to a generalized Cartan matrix A, as constructed by Tits. It is known that G admits the structure of a BN-pair, and acts on its corresponding building. We study the complete Kac-Moody group G which is defined to be the closure of G in the automorphism group of its building. Our main goal is to determine when complete Kac-Moody groups are abstractly simple, that is have no proper non-trivial normal subgroups. Abstract simplicity of G was previously known to hold when A is of affine type. We extend this result to many indefinite cases, including all hyperbolic generalized Cartan matrices A of rank at least four. Our proof uses Tits' simplicity theorem for groups with a BN-pair and methods from the theory of pro-p groups.Remark. In the hypotheses of Theorem 1.1, a submatrix is not necessarily proper.Abstract simplicity of G(A) was previously known only when A is of finite type (in which case G(A) is a finite group) or A is of affine type, in which case G(A) is isomorphic to the group of K-points of a simple algebraic group defined over K = k((t))[18]. The class of matrices covered by Theorem 1.1 includes many indefinite examples, including all hyperbolic matrices of rank at least four -see Proposition 2.1.
Remark. Recently, Caprace and Remy [3] proved (abstract) simplicity of the incomplete group G(A) modulo its finite centerin the case when the associated Coxeter group is not affine and assuming that |k| is sufficiently large. If A is affine, the incomplete group G(A) (modulo its center) is infinite and residually finite, and hence cannot be simple.Briefly, our approach to proving Theorem 1.1 will be as follows. A celebrated theorem of Tits [2] gives sufficient conditions (called "simplicity axioms") for a group with a BN-pair to be simple. The group G(A) routinely satisfies most of these axioms if |k| > 3 (for arbitrary A), which already implies topological simplicity of G(A). It is not clear if the remaining axioms. We have already defined the group G J . Note that by relations (R3), G J ⊃ U α for every α ∈ Φ J . Also introduce the subgroupsIn view of this proposition, we can identify U J (resp. G J ) with U(A J ) (resp. G(A J )). Let U J be the closure of U(A J ) in G(A J ), as before, and let U J be the closure of U J in G.Theorem 5.2. Assume that A is symmetric. The groups U J and U J are (topologically) isomorphic.Proof of Proposition 5.1. As before, identify Φ(A J ) with the subset Φ J of Φ. To distinguish between generators of G(A J ) and G(A), we use the symbols {x α (u) | α ∈ Φ(A J ), u ∈ k} for the generators of G(A J ) (the generators of G(A) are denoted {χ α (u)} as usual). From the defining presentation of Kac-Moody groups, it is clear that there exists a map ϕ :Clearly, ϕ(G(A J )) = G J , ϕ(U(A J )) = U J and ϕ(B(A J )) = B J , so we only need to show that ϕ is injective. We proceed in ...