Timeline for Irreducible representation of $S_n$: contained in tensor powers of the standard representation?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2020 at 19:42 | comment | added | Mark Wildon | There is a nice proof in Alperin's book Local Representation Theory that works in any characteristic and for any faithful module: see Theorem 1 on page 45. | |
| Jul 31, 2020 at 18:56 | answer | added | Will Sawin | timeline score: 11 | |
| Jul 31, 2020 at 18:01 | answer | added | Benjamin Steinberg | timeline score: 13 | |
| Jul 31, 2020 at 17:50 | comment | added | Benjamin Steinberg | @LSpice for instance of you take the monoid of all maps on n letters then the trivial representation is a quotient of the natural representation on $\mathbb C^n$ but not a subobject of any Tensor power. | |
| Jul 31, 2020 at 17:47 | comment | added | Benjamin Steinberg | @LSpice, for finite groups every irreducible appears as subrepresentation of the regular representation because the group algebra is Frobenius. R. Steinberg proves the regular representation appears inside of a direct sum of the tensor powers and so you get them as subobjects. But Steinberg also proves his theorem for finite semigroups and in that case you do not get them all as subobjects. | |
| Jul 31, 2020 at 14:14 | comment | added | LSpice | (Just to have the name here, @BenjaminSteinberg's reference is to Robert Steinberg - Complete sets of representations of algebras.) | |
| Jul 31, 2020 at 14:12 | comment | added | LSpice | @BenjaminSteinberg, wow, yet another illustration of the beauty that's to be found in (that other) Steinberg's works! As @lambda and @DavidESpeyer mention, the sense of 'constituent' there is 'subquotient'. Is there an easy example of a representation that can't be realised as a subobject of a tensor power $V^{\otimes m}$? | |
| Jul 31, 2020 at 14:10 | comment | added | user108998 | @Dianbin, the OP mentioned that the result is standard in this case, it is not an assumption of the question that $\mathbb{K}$ is of zero characteristic. | |
| Jul 31, 2020 at 14:09 | comment | added | LSpice | @DianbinBao, re your comment, I think that the post mentions that this is a known fact when $n! \ne 0$, but does not actually assume that $n! \ne 0$. | |
| Jul 31, 2020 at 14:08 | comment | added | David E Speyer | Does "occurs" mean as a subfactor, as a subrepresentation, as a summand, ... ? | |
| Jul 31, 2020 at 14:06 | comment | added | Dianbin Bao | @LSpice, yes, it works for characteristic 0 case. As the OP mentioned $n!\neq 0$ in $\mathbb{K}$. The inner product makes sense and the argument still works by Schur's lemma. | |
| Jul 31, 2020 at 14:01 | history | edited | YCor | CC BY-SA 4.0 |
expanded title to be specific
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| Jul 31, 2020 at 14:01 | comment | added | Benjamin Steinberg | See ams.org/journals/proc/1962-013-05/S0002-9939-1962-0141710-X/… | |
| Jul 31, 2020 at 13:58 | comment | added | lambda | @LSpice I believe that if the word "occurs" is interpreted correctly for the non-semisimple setting (i.e. a composition factor, not necessarily a summand) then it is true all characteristics, but I guess I did glance past the fact that the OP wanted this for arbitrary fields. | |
| Jul 31, 2020 at 13:56 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, changed tag
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| Jul 31, 2020 at 13:51 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading
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| Jul 31, 2020 at 13:46 | review | Close votes | |||
| Jul 31, 2020 at 14:01 | |||||
| Jul 31, 2020 at 13:45 | comment | added | Dianbin Bao | @LSpice, yes, it only works for characteristic 0 case. | |
| Jul 31, 2020 at 13:40 | comment | added | LSpice | Not to keep beating the same (possibly wrong, to mix metaphors) horse, but, @DianbinBao, isn't that a characteristic-$0$ argument? | |
| Jul 31, 2020 at 13:34 | comment | added | Dianbin Bao | Use the orthogonality relation and show that the inner product of any irreducible character of $S_n$ with character of $V^{\otimes m}$ does not vanish for large $m$. | |
| Jul 31, 2020 at 13:33 | comment | added | LSpice | @lambda, is it true for any faithful representation in any characteristic? I don't know that it isn't, but I thought that there might be some problems in case the characteristic of $\mathbb k$ divides $n!$. | |
| Jul 31, 2020 at 13:28 | comment | added | vidyarthi | how about using schur functions here? | |
| Jul 31, 2020 at 13:28 | review | First posts | |||
| Jul 31, 2020 at 13:37 | |||||
| Jul 31, 2020 at 13:27 | comment | added | lambda | This is a standard fact that holds for any faithful representation of any finite group. | |
| Jul 31, 2020 at 13:23 | history | asked | Eggon Viana | CC BY-SA 4.0 |