Evaluation of irreducible representations of the hyperoctahedral group at bipartition $(\lambda,\mu)=([n],\emptyset)$ There is a very simple formulation for the character of irreducible representations of $S_n$ evaluated on an n-cycle, i.e. that it is 0 on all non-hook partitions, and $(-1)^m$ on hooks. Is there an analogous computation for irreducible characters of $B_n$, the hyperoctahedral group, evaluated on signed 2n-cycles? That is, on the conjugacy class indexed by the bipartition $([n],\emptyset)$?
 A: In general, if $(\lambda,\mu)$ is a bipartition of $n$, then
  $$ \prod_i(p_{\lambda_i}(x)+p_{\lambda_i}(y))\cdot\prod_j
       (p_{\mu_j}(x)-p_{\mu_j}(y)) = \sum_{(\alpha,\beta)}
           \chi^{\alpha,\beta}(\lambda,\mu)s_\alpha(x)s_\beta(y), $$
where $(\alpha,\beta)$ ranges over all bipartitions of $n$ and                                $\chi^{\alpha,\beta}$ is the irreducible character of $B_n$ indexed                         by $(\alpha,\beta)$. Setting $(\lambda,\mu)=
(n,\emptyset)$ gives
  $$ p_n(x)+p_n(y) = \sum_{(\alpha,\beta)}
     \chi^{\alpha,\beta}(n,\emptyset)s_\alpha(x)s_\beta(y). $$
But (as alluded to in the question)
  $$ p_n = \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}, $$
so
 $$ p_n(x)+p_n(y)= \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(x) +
     \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(y). $$
A: Did you try small values of $n$? A quick examination of the character table (computed using GAP, say) for $5\leq n\leq 9$ shows that in this range there always is a $2n$-class with character values $0,\pm 1$.
(For $n$ odd it is unique, so one does not even need to check that this is the class you are looking at in this case.)
Surely it should not be too hard to make a conjecture about these values, based on such experimental info.
