TITLE: Evaluation of irreducible representations of the hyperoctahedral group at bipartition $(\lambda,\mu)=([n],\emptyset)$
QUESTION [7 upvotes]: 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 $([n],\emptyset)$?

REPLY [4 votes]: 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
 $$ p_n(x)+p_n(y)= \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(x) +
     \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(y). $$