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 $S^{\widetilde{\otimes n}}$.

  • $\begingroup$ 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 $S_d$. $\endgroup$ – MarcO Dec 12 '18 at 13:31
  • $\begingroup$ ... 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? $\endgroup$ – MarcO Dec 12 '18 at 14:55
  • $\begingroup$ I have corrected my rough statement of the classification of irreducible representations to answer your question (in the affirmative). $\endgroup$ – Mark Wildon Dec 12 '18 at 22:22
  • $\begingroup$ Thanks a lot! As requested, I posted a new question here: mathoverflow.net/questions/317585/… $\endgroup$ – MarcO Dec 13 '18 at 9:32
  • $\begingroup$ 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. $\endgroup$ – MarcO Dec 13 '18 at 9:34

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