1
$\begingroup$

Let $W_n$ be a Coxeter group of type $B_n$ with $n\geq 1$. Concretely, it is generated by a set of simple reflexions $S = \{s_1,\ldots ,s_n\}$ which satisfy the relations $s_i^2 = 1, s_is_j=s_js_i$ as long as $|i-j| \geq 2$, $(s_is_{i-1})^3 = 1$ for $2\leq i \leq n-1$ and $(s_ns_{n-1})^4 = 1$.

Given integers $a,b \geq 0$ such that $a+b = n$, consider the subgroup $H_{a,b} \simeq \mathfrak S_a \times W_b \subset W_n$ which is generated by all the simple reflexions except $s_a$.

Complex characters of the symmetric group $\mathfrak S_n$ (resp. of the group $W_n$) are classically classified by partitions of $n$ (resp. by bipartitions of $n$). Any such representation will be denoted by the letter $\rho$ with index a (bi)partition. Given a partition $\lambda \vdash a$ and a bipartition $(\alpha,\beta) \vdash b$, is there any paper or textbook reference describing the Littlewood-Richardson type formula to compute the induced character $$\mathrm{Ind}_{H_{a,b}}^{W_n} \, \rho_{\lambda}\otimes \rho_{\alpha,\beta}?$$ In the special case where $\lambda = (a)$ is the trivial partition, this induction corresponds to Pieri's rule for groups of type $B_n$, which is described in paragraph 6.1.9 of Geck and Pfeiffer's book Characters of finite Coxeter groups and Iwahori-Hecke algebras (2000). In terms of Young diagrams, the rule says that we obtain the multiplicity-free sum of all the irreducible characters $\rho_{\alpha',\beta'}$ where, for some $0 \leq d \leq a$, $\alpha'$ (resp. $\beta'$) can be obtained from $\alpha$ (resp. $\beta$) by successively adding $d$ boxes (resp. $a-d$ boxes) in the Young diagram, with no two boxes in the same column.

If anything, I am particularly interested in the case where $\lambda$ is a hook partition, ie. $\lambda = (x,1^{a-x})$ for some $1\leq x \leq a$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Note that $\newcommand{\Sa}{\mathfrak{S}_a} \DeclareMathOperator{Ind}{Ind} \DeclareMathOperator{Res}{Res}$ $$\Sa \times W_b \subseteq W_a \times W_b \subseteq W_{a+b}.$$ Hence

$$ \Ind_{\Sa \times W_b}^{W_{a+b}} \rho_\lambda \boxtimes \rho_{\alpha,\beta} = \Ind^{W_{a+b}}_{W_a \times W_b} (\Ind^{W_a}_{\Sa} \rho_\lambda) \boxtimes \rho_{\alpha,\beta}.$$ The problem breaks into two parts: decomposing $\Ind^{W_a}_{\Sa} \rho_\lambda$, and then describing the functor $\Ind^{W_{a+b}}_{W_a \times W_b}$.

Induction from $\Sa$: recall that the irreducible representation $\rho_{\alpha',\beta'}$ of $W_{a}$ is given by $$ \rho_{\alpha',\beta'} = \Ind^{W_a}_{W_{|\alpha'|} \times W_{|\beta'|} } \rho_{\alpha'} \boxtimes ((-\mathbf{1})^{\otimes |\beta'|}\rho_{\beta'}) $$ where $(-\mathbf{1})$ means the nontrivial representation of $\mathbb Z/2$, and where we view $W_n = \mathfrak{S}_n \wr (\mathbb Z/2)$. Hence $$ \langle \rho_{\alpha',\beta'}, \Ind_{\Sa}^{W_a} \rho_{\lambda}\rangle = \langle \Res_{\Sa}^{W_a} \Ind^{W_a}_{W_{|\alpha'|} \times W_{|\beta'|} } \rho_{\alpha'} \boxtimes ((-\mathbf{1})^{\otimes |\beta'|}\rho_{\beta'}),\rho_{\lambda}\rangle.$$ Now we apply the Mackey theorem. Note that $\Sa \backslash W_a /( W_{|\alpha'|} \times W_{|\beta'|})$ has only one element, and $\Sa \cap (W_{|\alpha'|} \times W_{|\beta'|}) = \mathfrak{S}_{|\alpha'|} \times \mathfrak{S}_{|\beta'|}$, so we obtain $$ \Res_{\Sa}^{W_a} \Ind^{W_a}_{W_{|\alpha'|} \times W_{|\beta'|} } \rho_{\alpha'} \boxtimes ((-\mathbf{1})^{\otimes |\beta'|}\rho_{\beta'}) \cong \Ind_{\mathfrak{S}_{|\alpha'|}\times \mathfrak{S}_{|\beta'|}}^{\Sa} \rho_{\alpha'} \boxtimes \rho_{\beta'}.$$ Thus $$ \langle \rho_{\alpha',\beta'} , \Ind_{\Sa}^{W_a}\rho_{\lambda}\rangle = c_{\alpha',\beta'}^{\lambda},$$ where $c_{*,*}^*$ are the usual Littlewood-Richardson coefficients.

Induction from $W_a \times W_b$: This is the type B Littlewood-Richardson rule. We have $$ \Ind_{W_a \times W_b}^{W_{a+b}} \rho_{\alpha',\beta'} \boxtimes \rho_{\alpha,\beta} = \sum_{\gamma,\delta} c_{\alpha',\alpha}^{\gamma} c_{\beta',\beta}^{\delta} \rho_{\gamma,\delta}.$$ This analysis holds for general wreath products of finite groups with $\mathfrak{S}_n$, see e.g. Zelevinsky's book.

Conclusion: $$\Ind_{\Sa \times W_b}^{W_{a+b}} \rho_\lambda \boxtimes \rho_{\alpha,\beta} = \sum_{\alpha',\beta',\delta,\gamma} c^{\lambda}_{\alpha',\beta'} c^{\gamma}_{\alpha',\alpha} c^{\delta}_{\beta',\beta} \rho_{\gamma,\delta}.$$

$\endgroup$
1
  • $\begingroup$ Thank you very much, it's all perfectly clear! $\endgroup$
    – Suzet
    Oct 23, 2023 at 5:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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