# 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 $([n],\emptyset)$?

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 $$p_n(x)+p_n(y)= \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(x) + \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(y).$$
• To clarify then, the problem simply reduces to matching these schur functions. In particular, if we call a bi-partition with one empty set and one hook a unihook, then $\chi^{\alpha,\beta}([n],0)=0$ on all non-unihooks, and $(-1)^m$ on uni-hooks $[(n-i,1^i),\emptyset]$ and $[\emptyset,(n-i,1^i)]$? – Max Hopkins Nov 27 '17 at 19:55
• What is an example of an element in the conjugacy class of $(n, \emptyset)$ here? Previously I had thought it was something like (say for $n=3$), $(1,0,0, (123))$, since this corresponds to a $2n$-cycle $(1, -2, -3, -1, 2, 3),$ and is what is suggested in the notation of Richard Bayley's thesis. However, I just did something I should have done months ago and checked the character computations on GAP, and it appears to correspond to an element $(0,0,0,(123))$. Which is correct? – Max Hopkins Apr 4 '18 at 18:52
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.)