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 $([n],\emptyset)$?
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3$\begingroup$ there are lots of references in mathoverflow.net/questions/25625/… so perhaps it is an exercise in one of them. $\endgroup$– Dima PasechnikNov 26, 2017 at 11:47
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2 Answers
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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 $$ p_n(x)+p_n(y)= \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(x) + \sum_{i=0}^{n-1} (-1)^i s_{n-i,1^i}(y). $$
answered Nov 26, 2017 at 17:47
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$\begingroup$ 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 $[\emptyset,(n-i,1^i)]$? $\endgroup$ Nov 27, 2017 at 19:55
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$\begingroup$ 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? $\endgroup$ Apr 4, 2018 at 18:52
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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.
answered Nov 26, 2017 at 11:30