Automorphisms and generalized involution models of finite complex reflection groups

Abstract

We prove that a finite complex reflection group has a generalized involution model, as defined by Bump and Ginzburg, if and only if each of its irreducible factors is either G(r, p, n) with gcd(p, n) = 1; G(r, p, 2) with r/p odd; or G 23 , the Coxeter group of type H 3 . We additionally provide explicit formulas for all automorphisms of G (r, p, n), and construct new Gelfand models for the groups G(r, p, n) with gcd(p, n) = 1.

Gelfand models and Robinson–Schensted correspondence

“…9 If q = 1, the result is straightforward since an S n -conjugacy class of absolute involutions in G(r, q, p, n) is also an S n -conjugacy class of absolute involutions in G(r, 1, p, n) and the definitions of the Gelfand models for G(r, p, q, n) and G(r, p, 1, n) are compatible with the projection G(r, p, 1, n) → G(r, p, q, n).…”

“…A Gelfand model of a finite group G is a G-module isomorphic to the multiplicityfree sum of all the irreducible complex representations of G. The study of Gelfand models originated from [3] and has found a wide interest in the case of reflection groups and other related groups (see, e.g., [1,2,[7][8][9][10]). In [5], a Gelfand model (M, ) was constructed (relying on the concept of duality in an essential way) for every involutory projective reflection group G(r, p, q, n) satisfying GCD(p, n) = 1, 2.…”

Isomorphisms, automorphisms, and generalized involution models of projective reflection groups

“…In [13,14], the second author classified which finite complex reflection groups have GIMs. Subsequently, the first author discovered an interesting reformulation of this classification, which suggests that these results are most naturally interpreted in the broader context of projective reflection groups.…”

“…Remark. Explicitly, G has a GIM if and only if (i) n = 2 and GCD(p, n) = 1 or (ii) n = 2 and either p or r/p is odd; this is the statement of [14,Theorem 5.2].…”

“…Deducing this theorem from [14,Theorem 5.2] is straightforward, given our next main result. Let r, n be positive integers and let p, p ′ , q, q ′ be positive divisors of r such that pq = p ′ q ′ divides rn.…”