# Decomposition into irreducible of a representation of the wreath product $S_d\wr S_n$

Let $$S_d, S_n$$ be the permutation groups of $$d,n$$ elements.

An intuitive representation of the wreath product $$S_d\wr S_n$$ is $$V_1\otimes...\otimes V_n$$, where each $$V_i$$ is of dimension $$d$$. Writing $$e_{i_1}\otimes...\otimes e_{i_n}$$ the canonical basis (where $$i_j=1..d$$), $$S_n$$ permutes the $$j$$ and each copy $$k$$ of $$S_d$$ in the wreath product permutes $$i_k$$.

How does this representation decomposes into irreducible representations of the wreath product?

Thanks a lot!

Given a representation $$U$$ of $$S_d$$ and $$m \in \mathbb{N}$$, we can extend the action of $$S_d \times \cdots \times S_d$$ on $$U \otimes \cdots \otimes U = U^{\otimes m}$$ to the wreath product $$S_d \wr S_m$$ by making $$S_m$$ act on the $$m$$ factors by place permutation. Let $$U^{\widetilde{\otimes m}}$$ denote this representation of $$S_d \wr S_m$$. The representation in the question is then $$V^{\widetilde{\otimes n}}$$ where $$V$$ is the natural $$d$$-dimensional representation of $$S_d$$.

Conjugacy classes and irreducible representations of wreath products are classified in Chapter 4 of The representation theory of the symmetric groups by James and Kerber. In particular any representation induced from a tensor product of irreducibles $$U^{\widetilde{\otimes n_i}}$$ for subgroups $$S_d \wr S_{n_i} \le S_d \wr S_n$$, where $$S_{n_1} \times \cdots \times S_{n_\ell}$$ is a Young subgroup of $$S_n$$, is irreducible, and all irreducibles are obtained by tensoring these modules with the inflations to $$S_d \wr S_n$$ of irreducible representations of $$S_n$$.

Given all this theory, questions like the one above become routine. Over any field $$F$$ of characteristic zero, $$V$$ decomposes as $$F\oplus S$$ where $$F$$ is the trivial representation and $$S = \langle e_i - e_j : 1 \le i < j \le d \rangle$$ is a $$(d-1)$$-dimensional irreducible representation (isomorphic to the Specht module $$S^{(n-1,1)}$$). Now

$$V^{\widetilde{\otimes n}} \cong \bigoplus_{m=0}^{n} \bigl( F^{\widetilde{\otimes n-m}} \boxtimes S^{\widetilde{\otimes m}} \bigr)\bigl\uparrow_{S_{n-m} \times S_{m}}^{S_m}$$

where each summand is irreducible. To see this, use basic Clifford theory to show that each summand on the right-hand side occurs in the left-hand side, and finish by counting dimensions:

$$d^n = \sum_{m=0}^n \binom{n}{m} (d-1)^m.$$

In particular there are summands $$F^{\widetilde{\otimes n}}$$ and $$S^{\widetilde{\otimes n}}$$.

• Thanks a lot! I'm not used to those notations but I think I guessed what it means: Let $t=\sum e_i$ a basis of $F$ and $\delta_p$ a basis of $S$. The term $(F^{\otimes\tilde{n-m}}\boxtimes S^{\otimes\tilde{m}}\uparrow_{S_{n-m}\times S_m}^{S_m})$ is of dimension $(d-1)^m\binom{n}{m}$, of basis all the possible reordering (with $S_n$) of $\delta_{p_1}\otimes...\otimes\delta_{p_m}\otimes t\otimes ...\otimes t$ and the action of $S_d\wr S_n$ over this is a reordering using the permutation in $S_n$ and when there is a $\delta$ in the $k$-th position, an action of the element in the $k$-th $S_d$. – MarcO Dec 12 '18 at 13:31
• ... I have a new question now. I call $R_m$ the irreducible summands. Let $S^{(n-1,1)}$ and $S^{(n-2,1^2)}$ be irreducible representations (Specht modules) of the symmetric group $S_n$. Are $R_m\otimes S^{(n-1,1)}$ and $R_m\otimes S^{(n-2,1^2)}$ irreducibles (I guess the action under $S_d\wr S_n$ is clear)? If not, how difficult is this problem, would you have some reference to advice to takle this kind of problem? – MarcO Dec 12 '18 at 14:55
• I have corrected my rough statement of the classification of irreducible representations to answer your question (in the affirmative). – Mark Wildon Dec 12 '18 at 22:22
• Thanks a lot! As requested, I posted a new question here: mathoverflow.net/questions/317585/… – MarcO Dec 13 '18 at 9:32
• I think the representation I consider is something like the $S_n$-imprimitive representation of the $S_d$ natural representation? Maybe I can put it in the title. – MarcO Dec 13 '18 at 9:34