# Decomposition into irreducible of a representation of the wreath product $S_d \wr S_m$ (2)

Let $$F$$ be the trivial and $$S$$ be the standard representations of $$S_d$$ (of dimension $$1$$ and $$d-1$$).

Let: $$R_m= \bigl( F^{\widetilde{\otimes n-m}} \boxtimes S^{\widetilde{\otimes m}} \bigr)\bigl\uparrow_{S_d\wr S_{n-m} \times S_d\wr S_{m}}^{S_d\wr S_n} :$$ This is an irreducible representation of $$S_d \wr S_n$$.

Let $$S^{(n-1,1)}$$ and $$S^{(n-2,1^2)}$$ be irreducible representations (Specht modules) of the symmetric group $$S_n$$. $$S_d \wr S_n$$ naturally acts on it (just considering $$S_n\subset S_d \wr S_n$$).

Are $$R_m\otimes S^{(n-1,1)}$$ and $$R_m\otimes S^{(n-2,1^2)}$$ irreducibles?

If not, how difficult is this problem, would you have some reference to advice to takle this kind of problem?

• My mistake I did not read the new version. Thanks a lot! – MarcO Dec 13 '18 at 12:05
• @MarkWildon Sorry to come back to this, I have the impression now it can be reduced. Maybe because we are not talking about the same object? Here is an example: For d=2, I write $t=e_1+e_2$ and $\delta=e_1-e_2$. For n=3, I write $\delta_{ij}=e_i-e_j$. Then in $R_1\otimes S^{(2,1)}$, the span of $\{tt\delta\otimes\delta_{12}, t\delta t\otimes\delta_{13}, \delta tt\otimes\delta_{23}\}$ is stable? – MarcO Jan 23 '19 at 12:28
• I misread your module as one of the form $(F^{\widetilde{\otimes (n-m)}} \otimes U) \boxtimes (\mathrm{sgn}^{\widetilde{\otimes m}} \otimes V) \uparrow_{S_{n-m} \times S_m}^{S_n}$, where $U$ is an irreducible $FS_{n-m}$-module (inflated to $S_d \wr S_{n-m}$) and $V$ is an irreducible $FS_m$-module (inflated to $S_d \wr S_m)$. These are irreducible (by the classification – you could take any two distinct simple modules in place of $F$ and $\mathrm{sgn}$), but I now see your question is different. I'm sorry to have mislead you. – Mark Wildon Jan 23 '19 at 15:00
• What does any of the notation mean ($F$, $S$, and the tilde accent)? Is it $S_d \wr S_n$ as in the body or $S_d \wr S_m$ as in the title? You use both $m$ and $n$, but not $d$, in the definition of $R_m$. – LSpice Jan 24 '19 at 2:21
• @LSpice The tilde notation is defined in the linked question mathoverflow.net/q/317485/7709; $F$ and $S$ are representations of $S_d$ . In the definition of $R_m$ the induction should be from $S_d \wr S_{n-m} \times S_d \wr S_m$ to $S_d \wr S_n$. I'm not sure what $S$ is: maybe it is the sign representation of $S_d$, or some other simple module: it doesn't matter for my answer. The first paragraph of my answer tries to make clear the action of $S_n$: it is a bit 'loose' to say $S_n \subset S_d \wr S_n$, when what's really needed is a quotient and the inflation map. – Mark Wildon Jan 27 '19 at 3:22

Contrary to a comment I've now deleted, the representation is reducible. It's important to be clear about the action of $$S_n$$: given a representation $$\rho: S_n \rightarrow \mathrm{GL}(V)$$, composing with the quotient homomorphism $$S_d \wr S_n \rightarrow S_n$$ gives a representation of $$S_d \wr S_n$$, denoted $$\mathrm{Inf}_{S_n}^{S_d \wr S_n} V$$.
Using this, I'll show more generally that if $$U$$ is any representation of $$S_d \wr S_{n-m}$$ and $$V$$ is any representation of $$S_d \wr S_m$$ where $$1 \le m < n$$, then
$$\begin{split} (U \boxtimes V)\bigl\uparrow_{S_d \wr S_{n-m} \times S_d \wr S_m}^{S_d \wr S_n} & \otimes \; \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \\ &= \Bigl( (U \boxtimes V) \otimes \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)}\bigl\downarrow_{S_d \wr S_{n-m} \times S_d \wr S_m} \bigr)\Bigr)\bigl\uparrow^{S_n} \end{split}$$
$$\bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \bigl\downarrow_{S_d\wr S_{n-m} \times S_d \wr S_m} \, \cong \mathrm{Inf}^{S_d \wr S_{n-m} \times S_d \wr S_m}_{S_{n-m} \times S_m} \bigl( S^{(n-1,1)} \bigl\downarrow_{S_{n-m} \times S_{m}} \bigr).$$
Since $$S^{(n-1,1)}\!\!\downarrow_{S_{n-m} \times S_m}\, \cong \bigl( F \boxtimes F \bigr) \oplus \bigl( F \boxtimes S^{(m-1,1)} \bigr) \oplus \bigl( S^{(n-m-1,1)} \boxtimes F\bigr)$$ is reducible (this follows from Young's rule using Frobenius reciprocity, or the more general Littlewood–Richardson rule, or could be proved using an explicit basis for $$S^{(n-1,1)}$$), so is the right-hand side immediately above, and hence so is the original module. A similar argument will work replacing $$S^{(n-1,1)}$$ with $$S^{(n-2,1,1)}$$.