The hyperoctahedral group $H_n$ can be seen as the centralizer of the permutation $(12)(34)\cdots (2n-1\,2n)$ in $S_{2n}$. It has $2^nn!$ elements.

The quantities $$ \omega_\lambda(\pi)=\frac{1}{2^nn!}\sum_{h\in H_n}\chi_{2\lambda}(h\pi)$$ are called the zonal spherical functions of the Gelfand pair $(S_{2n},H_n)$. Here $2\lambda=(2\lambda_1,2\lambda_2,...)$ and $\chi$ are irreducible characters of $S$.

I have observed the following very nice result: $$ \frac{2^nn!}{(2n)!}\sum_{\lambda\vdash n}\chi_{2\lambda}(1^{2n})\omega_\lambda(\pi)=\begin{cases} 1, &\pi\in H_n\\0, &\pi\notin H_n\end{cases}$$

Does anyone know how to prove this? (Orthogonality of characters is not straightforwardly useful) If so, can it be generalized to $\sum_{\lambda\vdash n}\chi_{2\lambda}(\mu)\omega_\lambda(\pi)$?


1 Answer 1


The essential thing here is that the characters $\chi_{2\lambda}$ are exactly the irreducible constituents of the induced character $(1_{H_n})^{S_n}$. The result generalizes to an arbitrary subgroup $H\leq G$ of a finite group $G$ as follows: For $x$, $y\in G$, we have $\DeclareMathOperator{\Irr}{Irr}$ $$ \frac{ |H| }{ |G| } \sum_{ \chi\in \Irr(G \mid 1_H) } \chi(x^{-1} ) \omega_{\chi}(y) = \frac{ |x^G \cap Hy| }{ |x^G| }. \tag{*} $$ Here $\Irr(G \mid 1_H)$ denotes the set of irreducible constituents of $(1_H)^G$, which by Frobenius reciprocity is the set of $\chi\in \Irr(G)$ such that $1_H$ is a constituent of the restriction $\chi_H$. As above, $$ \omega_{\chi}(y) = \frac{1}{|H|} \sum_{h\in H} \chi(hy), $$ but we do not have to assume that $(G,H)$ is a Gelfand pair. ($x=1$ and $y=\pi$ is your result.)

Proof: Let
$$ e_H = \frac{1}{|H|} \sum_{h\in H} h \in \mathbb{C} H, $$ the central primitive idempotent of the group algebra belonging to $1_H$. Then $\omega_{\chi}(g) = \chi(e_H g)$ by defintion. If $1_H$ is not a constituent of the restriction $\chi_H$ for $\chi\in \Irr(G)$, then $\chi(e_Hg) = 0$ for all $g\in G$. Thus we can let run $\chi$ over all of $\Irr(G)$ in the sum in (*). We get $$ \frac{ |H| }{ |G| } \sum_{ \chi\in \Irr G } \chi( x^{-1} ) \omega_{\chi}(y) = \frac{ 1 }{ |G| } \sum_{ h\in H } \sum_{ \chi\in \Irr G } \chi(x^{-1})\chi(hy), $$ and the second orthogonality relation for characters yields the result.

  • $\begingroup$ By $x^G$ you mean $\{gxg^{-1},g\in G\}$? $\endgroup$
    – Marcel
    Feb 24, 2016 at 17:09
  • $\begingroup$ @Marcel: Yes, exactly: $x^G$ is the conjugacy class of $x$ in $G$. $\endgroup$ Feb 24, 2016 at 17:15

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