consider the case where $q=2$. But $\mbox{Sp}_4({{\mathbb{F}}}_2)$ is isomorphic to the symmetric group $\mathfrak{S}_6$; an explicit isomorphism is described in [@HupI 9.21]. One checks that, under this isomorphism, $g_1=x_a(1) x_b(1)$ corresponds to an element of cycle type $(4,2)$ in $\mathfrak{S}_6$. We also need to identify $\rho_{r}$, $\rho_{\text{sgn}_a}$, $\rho_{\text{sgn}_b}$, $\rho_{x_0}$ in the character table of $\mathfrak{S}_6$. Now, $B({{\mathbb{F}}}_2)$ is a Sylow $2$-subgroup of $\mbox{Sp}_4
({{\mathbb{F}}}_2)$. Working out the character of the permutation representation on the cosets of this subgroup, one can identify the $5$ characters which are of the form $\rho_\epsilon$ for some $\epsilon\in {{\operatorname{Irr}}}(W)$. Looking also at character degrees, we can then identify $\rho_{\text{sgn}}$ and the sum $\rho_{\text{sgn}_a}+\rho_{\text{sgn}_b}$; finally, $\rho_{x_0}$ corresponds to the sign character of $\mathfrak{S}_6$. By inspection of the table of $\mathfrak{S}_6$, we find that $R_{x_0}(g_1)=2$ (for $q=2$) and, hence, $\zeta_{x_0}=1$ for any $q=2^f$ (using Proposition  \[p1\]). — Of course, this could also be deduced from the explicit knowledge of the “generic” character table of $G^F=\mbox{Sp}_4({{\mathbb{F}}}_q)$ for any $q=2^f$ (see Enomoto [@En]). But the point is that, once $\zeta_{x_0}$ is known in advance, the task of computing such a generic character table is considerably simplified!

\[so8\] Let $G=\mbox{SO}_8(k)$ be the $8$-dimensional special orthogonal group. The Weyl group $W$ 