Timeline for Factorization in the group algebra of symmetric groups
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2016 at 4:15 | answer | added | Igor Pak | timeline score: 3 | |
| Sep 17, 2016 at 6:31 | vote | accept | Jianrong Li | ||
| Sep 16, 2016 at 19:19 | answer | added | Richard Stanley | timeline score: 11 | |
| Sep 16, 2016 at 13:43 | comment | added | YCor | OK. But as noticed by several people it's of limited interest then: you're asking about a factorization of $(n!,0,\dots,0)$ in a product of matrix algebras $\mathbb{C}\times\prod_{i=2}^nM_{n_i}(\mathbb{C)}$, so on each matrix component $M_{n_i}(\mathbb{C})$ the question boils down to solve the equation $XY=0$. There are many solutions... | |
| Sep 16, 2016 at 13:40 | comment | added | Jianrong Li | @YCor, thank you very much. The group algebra is $\mathbb{C}S_n$. The factorization will be something like factor a polynomial into a product of irreducible polynomials. I edited the post. | |
| Sep 16, 2016 at 13:37 | history | edited | Jianrong Li | CC BY-SA 3.0 |
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| Sep 16, 2016 at 13:02 | answer | added | Geoff Robinson | timeline score: 3 | |
| Sep 16, 2016 at 9:07 | comment | added | Nick Gill | Related (a little bit): mathoverflow.net/questions/177747/… | |
| Sep 16, 2016 at 9:00 | comment | added | Nick Gill | Put another way, you're trying to write the group as a product: $S_n=A.B$, where $|A| \cdot |B|=n!$ (so there are no repetitions). This seems kind of tricky to me -- it's like a Zappa-Szep product for subsets, rather than subgroups... en.wikipedia.org/wiki/Zappa%E2%80%93Sz%C3%A9p_product | |
| Sep 16, 2016 at 8:57 | comment | added | Nick Gill | You'll get a load of factorizations for every subgroup (all elements of the subgroup multiplied by a bunch of coset reps). The set of all such factorizations is conjugation invariant, to reference the comment of @YCor... But, now, I guess it'd be interesting to know what other reps can occur. | |
| Sep 16, 2016 at 8:57 | comment | added | Wolfgang | Take a set of generators. Then all information about factorings (in your sense, which may need to be made more precise) involving one of those generators is encoded in the corresponding Cayley graph, by looking at its symmetries/automorphsms. Question: how easy is it to retrieve that info from the Cayley graph? And are all the Cayley graphs for, say, non redundant sets of generators of $S_n$ known? | |
| Sep 16, 2016 at 8:50 | comment | added | YCor | And could you specify you consider the group algebra over which ring? the ring of integers? | |
| Sep 16, 2016 at 8:34 | answer | added | Venkataramana | timeline score: 3 | |
| Sep 16, 2016 at 8:23 | history | edited | YCor |
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| Sep 16, 2016 at 8:22 | comment | added | YCor | What do you call "the factorization of $T$"? All the ways to write it as a product of two elements? more than two? some ways only? Note that since $T$ is conjugation-invariant, whenever you have a factorization, all conjugates yield a conjugation (in your example, you conjugate by the transposition $(13)$). | |
| Sep 16, 2016 at 8:05 | history | asked | Jianrong Li | CC BY-SA 3.0 |