$\lambda$ be the non-trivial character of $A(g_1)$. As above, we obtain: $$\chi_{g_1,\lambda}(g)=\left\{\begin{array}{cl} q^2 &\quad\mbox{if 
$g=g_1$},\\ -q^2 & \quad \mbox{if $g=g_1'$},\\ 0 & \quad \mbox{if 
$g\not\in C^F$},
\end{array}\right.$$ where $g_1'\in C^F$ corresponds to the non-trivial element of $A(g_1)$. Now we have $R_{x_0}=\zeta_{x_0}\chi_{g_1,\lambda}$. In order to show that $\zeta_{x_0}=1$, we can use the known character table of $\mbox{SO}_8^+({{\mathbb{F}}}_2)$; see the [ATLAS]{} [@atl p. 85]. In fact, using an explicit realization in terms of orthogonal $8\times 8$-matrices, one can create $\mbox{SO}_8^+({{\mathbb{F}}}_2)$ as a matrix group in [GAP]{} and simply re-calculate that table using the [CharacterTable]{} function. The advantage of this re-calculation is that [GAP]{} also computes a list of representatives of the conjugacy classes of $\mbox{SO}_8^+({{\mathbb{F}}}_2)$. So one can identify the class to which $g_1$ belongs. Arguing as in the previous example, one can identify the characters $\rho_{x_0}$, $\rho_{(21,1)}$ and the sum $\rho_{(22,\varnothing)}+\rho_{(2,11)}$ in the table of $\mbox{SO}_8^+({{\mathbb{F}}}_2)$. (We omit the details.) In this way, one finds that $R_{x_0} (g_1)=4$ (for $q=2$), as required.

\[rsplit\] Assume that $Z(G)=\{1\}$, as above. Let $x\in 
X(W)$ be such that $A_x$ is cuspidal and let $C$ be the $F$-stable conjugacy class of $G$ such that $\mbox{supp}(A_x)=\overline{C}$. The above examples highlight the importance of singling out a specific 