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!



Decomposition into irreducible of a representation of the wreath product $S_d\wr S_n$
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 $S^{\widetilde{\otimes n}}$.


