Galois action and cyclic defect groups for \(\mathrm{Sp}_6 (2^a)\) (Q6543067)

Let \(G\) be a finite group of order \(n\), let \(\ell\) be a prime and let \(B\) be an \(\ell\)-block of \(G\) with nontrivial defect group \(D\). Then every complex character of \(G\) is defined over \(\mathbb{Q}(\xi_n)\) where \(\xi_n\) is a primitive \(n\)-th root of unity. The Galois group \(\mathrm{Gal}(\mathbb{Q}(\xi_n)/\mathbb{Q})\) acts on the set of complex characters of \(G\), inducing an action on the set \(\mathrm{Irr}_0(B)\) of characters of height \(0\).\N\NIn this situation, it is conjectured that for \(\ell \in \{2,3\}\), we have \(|\mathrm{Irr}_0(B)^{\sigma_1}| = \ell\) if and only if \(D\) is cyclic [\textit{N. Rizo} et al., Algebra Number Theory 14, No.\ 7, 1953--1979 (2020; Zbl 1511.20034)]. Here, \(\sigma_1 \in \mathrm{Gal}(\mathbb{Q}(\xi_n)/\mathbb{Q})\) is the unique element such that \(\sigma_1(\xi) = \xi\) whenever \(\xi\) is a root of unity of order coprime to \(\ell\), and \(\sigma_1(\xi) = \xi^{\ell + 1}\) whenever \(\xi\) is an \(\ell\)-power root of unity.\N\NThe present article proves this conjecture in the particular case \(G = \mathrm{Sp}_6(2^a)\) and \(\ell = 3\). The proof is largely computational, making detailed use of the known generic character tables for \(\mathrm{Sp}_6(2^a)\) to determine precisely which characters in each block are fixed by \(\sigma_1\) in each case.