6
$\begingroup$

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 $\sum_{\lambda\vdash n}\chi_{2\lambda}(\mu)\omega_\lambda(\pi)$?

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ By $x^G$ you mean $\{gxg^{-1},g\in G\}$? $\endgroup$
    – Marcel
    Feb 24 '16 at 17:09
  • $\begingroup$ @Marcel: Yes, exactly: $x^G$ is the conjugacy class of $x$ in $G$. $\endgroup$ Feb 24 '16 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.