Timeline for Representations of products of symmetric groups
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 15, 2021 at 15:32 | comment | added | Maxime Ramzi | @BenjaminSteinberg ah that's a good point ! thanks ! (no worries about the name ;) ) | |
| Apr 15, 2021 at 15:23 | comment | added | Benjamin Steinberg | @MaximeRamzi, I just noticed I had a typo in your name when I tried to respond to your comment. Sorry about that. | |
| Apr 12, 2021 at 21:23 | comment | added | Benjamin Steinberg | @MaximeRanzi, the number of irreps over Q is the number of conjugacy classes of cyclic subgroups. So Z/pxZ/p has p+1 non trivial plus the trivial adult Z/p has two irreps so you don't come close to getting them all this way either. Also all the non trivial ones are p-1 dimensional. | |
| Apr 12, 2021 at 21:15 | comment | added | Benjamin Steinberg | @MaximeRanzi, you should get examples working over $\mathbb Q$ for $Z/pxZ/p$ with p a prime bigger than 2. Then there is a unique non trivial rep of Z/p over Q corresponding to L=Q(primitive p-root of 1). The tensor product of these is not irreducible because L\otimes_Q L splits into a product of fields. | |
| Apr 12, 2021 at 15:34 | comment | added | LSpice | @MaximeRamzi, re, I think you meant to link @XandiTuni's answer. | |
| Apr 12, 2021 at 14:32 | comment | added | Maxime Ramzi | Xandi here (mathoverflow.net/questions/29919/…) seems to be saying that the tensor product need not be irreducible in general. | |
| Apr 12, 2021 at 2:56 | comment | added | Benjamin Steinberg | Andy your proof just needs the representations are absolutely irreducible which means that it works for direct products of symmetric groups over Q. It’s also essentially the semi simple version of @Mare’s argument without talking about idempotents because you are going one simple component at a time | |
| Apr 12, 2021 at 1:29 | comment | added | Andy Putman | @JeremyRickard: Whoops, you’re right! I should have used the rotation action of $\mathbb{Z}/15$ on $\mathbb{R}^2$ instead. I will fix this next time I am in my office. Thanks! | |
| Apr 11, 2021 at 23:12 | comment | added | Jeremy Rickard | @Andy Is the example in your note not the tensor product of the nontrivial irreducible representations of $\mathbb{Z}/3$ and $\mathbb{Z}/2$ over $\mathbb{R}$? | |
| Apr 11, 2021 at 21:23 | comment | added | Maxime Ramzi | Right no I think I went too quickly and just made a mistake (that's why I should write things down !) | |
| Apr 11, 2021 at 21:14 | comment | added | Andy Putman | @MaximeRamzi: I could be missing something very obvious, but why can we assume that the $v_i$ are linearly independent over the endomorphism ring? And as a related question (I suspect that I will understand this once I understand the answer to my previous one), why can you then turn around that do the same to the $w_i$ after you've reduced to the case where there is only one $v_i$? | |
| Apr 11, 2021 at 21:05 | comment | added | Maxime Ramzi | I don't know if it's appropriate to do that here, but here's a sketch of what I have in mind : take an arbitrary tensor $\sum_i (v_i\otimes w_i)$. Up to rearranging the terms one may assume the $v_i$'s and $w_i$'s are all nonzero, and the $v_i$'s are linearly independent over $End_G(V)$ . The latter is a skew field, so by Jacobson's density theorem we can find $a\in k[G]$ sending all the $v_i$'s except for one to $0$. Thus any stable subspace must contain a pure tensor, and hence the whole space | |
| Apr 11, 2021 at 21:01 | comment | added | Andy Putman | @MaximeRamzi: I haven't thought about that, and am not entirely sure. | |
| Apr 11, 2021 at 20:47 | comment | added | Maxime Ramzi | That's a nice document ! am I right to believe that the claim 1 also works over a non algebraically closed field, although the proof needs to be changed slightly ? (I have in mind a proof using the Jacobson density theorem as well but in a slightly subtler way) | |
| Apr 11, 2021 at 19:47 | history | answered | Andy Putman | CC BY-SA 4.0 |