2
$\begingroup$

This is a question following Decomposition into irreducible of a representation of the wreath product $S_d\wr S_n$

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?

$\endgroup$
  • $\begingroup$ My mistake I did not read the new version. Thanks a lot! $\endgroup$ – MarcO Dec 13 '18 at 12:05
  • $\begingroup$ @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? $\endgroup$ – MarcO Jan 23 at 12:28
  • $\begingroup$ 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. $\endgroup$ – Mark Wildon Jan 23 at 15:00
  • $\begingroup$ 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 $R_m$. $\endgroup$ – LSpice Jan 24 at 2:21
  • $\begingroup$ @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. $\endgroup$ – Mark Wildon Jan 27 at 3:22
2
$\begingroup$

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

$$ \begin{split} (U \boxtimes V)\bigl\uparrow_{S_d \wr S_{n-m} \times S_d \wr S_m}^{S_d \wr S_n} & \otimes \; \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \\ &= \Bigl( (U \boxtimes V) \otimes \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)}\bigl\downarrow_{S_d \wr S_{n-m} \times S_d \wr S_m} \bigr)\Bigr)\bigl\uparrow^{S_n} \end{split}$$

is reducible. Since restriction commutes with inflation,

$$\bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \bigl\downarrow_{S_d\wr S_{n-m} \times S_d \wr S_m} \, \cong \mathrm{Inf}^{S_d \wr S_{n-m} \times S_d \wr S_m}_{S_{n-m} \times S_m} \bigl( S^{(n-1,1)} \bigl\downarrow_{S_{n-m} \times S_{m}} \bigr). $$

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 $S^{(n-2,1,1)}$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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