Timeline for An $n!\times n!$ determinant
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2018 at 21:32 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
title
|
| Feb 19, 2012 at 5:04 | answer | added | darij grinberg | timeline score: 9 | |
| Feb 14, 2012 at 19:10 | answer | added | Jonathan Novak | timeline score: 9 | |
| Feb 14, 2012 at 16:43 | comment | added | Richard Stanley | The polynomial $\sum_{g\in S_n} \chi(g)x^{c(g)}$ is evaluated in Exercise 7.50 of Enumerative Combinatorics, vol. 2 (solution on page 515). | |
| Feb 14, 2012 at 16:03 | vote | accept | Igor Makhlin | ||
| Feb 14, 2012 at 11:23 | answer | added | David E Speyer | timeline score: 16 | |
| Feb 14, 2012 at 9:31 | answer | added | Gjergji Zaimi | timeline score: 14 | |
| Feb 14, 2012 at 5:34 | comment | added | darij grinberg | I might make the above into a readable answer tomorrow (although I'd prefer a more combinatorial proof). Today I'll definitely not have the time for it (the only reason I am not in bed yet are two unsolved homework problems). | |
| Feb 14, 2012 at 5:29 | comment | added | darij grinberg | Anyway, let me correct myself: $\sum\limits_{g\in S_n} \chi\left(g\right)x^{c\left(g\right)}$ is not "the trace of the sum of all $g\in S_n$ acting on $V^{\otimes n}_{\chi}$" but the inner product of the character $\chi$ with the character of the $S_n$-module $V^{\otimes n}$. In other words, it is the number of $\chi$'s in $V^{\otimes n}$. This is (by Schur-Weyl duality) the dimension of the Schur functor corresponding to the partition corresponding to $\chi$, evaluated at the vector space $V$. Now use the Weyl character formula to obtain a polynomial formula for this. | |
| Feb 14, 2012 at 5:18 | comment | added | Gjergji Zaimi | See also theorem 110 here qspace.library.queensu.ca/bitstream/1974/5235/1/… | |
| Feb 14, 2012 at 5:17 | comment | added | darij grinberg | Ah! $x^{c\left(g\right)} = p_{C\left(g\right)}\left(1,1,...,1\right)$, where $C\left(g\right)$ denotes the cycle type of $g$ (this is a partition), $p$ stands for "power sum", and there are $x$ $1$'s in the bracket (which only makes sense for $x\in\mathbb N$ if interpreted directly, but probably with the notion of virtual alphabet we can get rid of this restriction). And yes, we can use Schur-Weyl duality: $\sum\limits_{g\in S_n} \chi\left(g\right)x^{c\left(g\right)}$ is the trace of the sum of all $g\in S_n$ acting on $V^{\otimes n}_{\chi}$, where $V$ is an $x$-dimensional vector space. | |
| Feb 14, 2012 at 5:11 | comment | added | darij grinberg | Apparently the map $g\mapsto x^{c\left(g\right)}$ is called the Polya character of $S_n$ (adjoined to $x$ or something like that). | |
| Feb 14, 2012 at 5:00 | comment | added | darij grinberg | Something that might work: Your determinant is a particular case of a group determinant (a.k.a. generalized circulant, a.k.a. Frobenius determinant; see math.uconn.edu/~kconrad/articles/groupdet.pdf for details), and thus equals $\prod\limits_{\chi\text{ is an irreducible character of }S_n} \left(\sum\limits_{g\in S_n} \chi\left(g\right) x^{c\left(g\right)} \right)^{\deg\chi}$. The sum $\sum\limits_{g\in S_n} \chi\left(g\right) x^{c\left(g\right)}$ looks like something we get out of Schur-Weyl duality, but I don't see how exactly to get it. | |
| Feb 14, 2012 at 4:47 | comment | added | darij grinberg | Anyway the topic of determinants of $n!\times n!$ matrices (which are usually very hard to evaluate; here is a conjectural one: mathematik.uni-marburg.de/~welker/preprints/inv.pdf ) cries for some interpretation. When expanded as a sum over permutations, they have $\left(n!\right)!$ addends. Is there something monadic about the symmetric group? | |
| Feb 14, 2012 at 4:40 | comment | added | darij grinberg | Very interesting!! (But it's not the characteristic polynomial of the $n$-th YJM idempotent...) | |
| Feb 14, 2012 at 3:49 | history | asked | Igor Makhlin | CC BY-SA 3.0 |