Timeline for Is anything known about the eigenspectrum of the regular representation of the permutation group?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2015 at 19:36 | vote | accept | user6818 | ||
| Apr 25, 2015 at 19:15 | comment | added | Dima Pasechnik | @QiaochuYuan - right, I missed this point. | |
| Apr 25, 2015 at 19:10 | comment | added | Qiaochu Yuan | @Dima: the behavior of the permutation matrix describing multiplication by $g$ is completely determined by the order of $g$; the only possible cycle length is the order of $g$. | |
| Apr 25, 2015 at 19:07 | comment | added | Dima Pasechnik | @QiaochuYuan - the behaviour of a permutation matrix is not determined by the order alone; it is determined by the cyclic structure of the permutation. E.g. the multiplicity of eigenvalue 1 is the number of cycles. | |
| Apr 25, 2015 at 18:50 | comment | added | user6818 | And why wouldn't the same argument as of ARupinky not go through for any irrep? | |
| Apr 25, 2015 at 10:27 | comment | added | Neil Strickland | The two answers appear to assume that you want to know the spectrum for the action on an arbitrary irreducible. If you are really only interested in the regular representation, then @ARupinski's comment is all you need. | |
| Apr 25, 2015 at 7:48 | comment | added | Student | I am missing something here. Why are the character formulas so much more difficult for this permutation group if the eigenvalues are so trivial? | |
| Apr 25, 2015 at 5:49 | comment | added | Qiaochu Yuan | @user6818: thinking about the irrep decomposition actually makes this harder. You already know what $R(g)$ looks like in the regular representation: it's the permutation matrix of the permutation corresponding to multiplication by $g$. The behavior of this permutation matrix is completely determined by the order of $g$ and that's all there is to it. A similarly easy result which again does not require you to know anything about the irreps of $G$ is that the trace of $R(g)$ is $0$ if $g$ is not the identity. | |
| Apr 25, 2015 at 2:06 | comment | added | user6818 | Then what is the idea in the other answers? | |
| Apr 25, 2015 at 2:02 | comment | added | user6818 | How!? So you are saying this spectrum is completely blind to the irreps of $G$!? That what the irreducibles of $G$ look like is completely irrelevant? | |
| Apr 25, 2015 at 1:49 | comment | added | ARupinski | Unless I'm completely misreading your question, in any (finite) group $G$ if $g$ is an element of order $n$, the eigenspectrum of $R(g)$ is $|G|/n$ copies of $\zeta^k$ for each $k = 0\ldots {n-1}$ and $\zeta$ a primitive $n^{th}$ root of unity. | |
| Apr 25, 2015 at 0:44 | answer | added | Richard Stanley | timeline score: 4 | |
| Apr 25, 2015 at 0:07 | answer | added | Dima Pasechnik | timeline score: 3 | |
| Apr 24, 2015 at 23:35 | history | asked | user6818 | CC BY-SA 3.0 |