Ennola duality conjecture for character degrees of general linear and unitary congruence quotients

Let nNn\in\mathbb{N}. For ε{1,1}\varepsilon\in\{-1,1\}, let o\mathfrak{o} be a compact discrete valuation ring with residue cardinality qq, let G\mathsf{G} be one of the o\mathfrak{o}-group schemes GLn\mathsf{GL}_n or GUn\mathsf{GU}_n, and write ε=εG\varepsilon=\varepsilon_{\mathsf{G}} accordingly. Ennola duality conjecture. There exist a finite index set II, polynomials fi(ε)Z[1/n!][t]f_i^{(\varepsilon)}\in\mathbb{Z}[1/n!][t] and gi(ε)Z[t]g_i^{(\varepsilon)}\in\mathbb{Z}[t], ascending chains of finite sets Bi,1Bi,2N\mathcal{B}_{i,1}\subset\mathcal{B}_{i,2}\subset\cdots\subset\mathbb{N}, and non-negative integers Aij(ε),BijA_{ij}^{(\varepsilon)},B_{ij} such that, for every N\ell\in\mathbb{N}, the character degrees and representation zeta function of G(o)\mathsf{G}(\mathfrak{o}_\ell) are respectively

cd(G(o))={gi(ε)(q)qBijiI,jBi,},\operatorname{cd}(\mathsf{G}(\mathfrak{o}_\ell))=\{g_i^{(\varepsilon)}(q)q^{B_{ij}\mid i\in I,j\in\mathcal{B}_{i,\ell}}\},

and

ζG(o)(s)=iIjBi,fi(ε)(q)qAij(ε)(gi(ε)(q)qBij)s.\zeta_{\mathsf{G}(\mathfrak{o}_\ell)}(s)=\sum_{i\in I}\sum_{j\in\mathcal{B}_{i,\ell}}f_i^{(\varepsilon)}(q)q^{A_{ij}^{(\varepsilon)}}\left(g_i^{(\varepsilon)}(q)q^{B_{ij}}\right)^{-s}.

Moreover, for every iIi\in I, Ennola duality holds for character degrees:

gi(1)(t)=(1)deg(gi(1))gi(1)(t).g_i^{(-1)}(t)=(-1)^{\deg(g_i^{(1)})}g_i^{(1)}(-t).

The conjecture extends the observed uniformity of character degrees and zeta functions for finite congruence quotients of general linear and unitary groups; the cited low-level and low-rank cases are known, while the asserted uniform description for all nn and levels remains open.

Sources & referencesView supporting material

Primary source

Nir Avni, Benjamin Klopsch, Uri Onn and Christopher Voll, “Similarity classes of integral p-adic matrices and representation zeta functions of groups of type A_2”, arXiv:1410.4533 (2015).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.