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Apr 27, 2015 at 23:37 comment added ARupinski @Anirbit: You are correct that for $g$ of finite order $n$, all eigenvalues of $\rho(g)$ are indeed $n^{th}$ roots of unity for any representation $\rho$. However, for an arbitrary representation the roots need not be evenly distributed around the unit circle; indeed if they were there wouldn't anything interesting to study about character theory. The regular representation turns out to be a very special case in which all eigenvalue multiplicities are equal.
Apr 25, 2015 at 19:36 vote accept user6818
Apr 25, 2015 at 16:10 comment added Student ^Isn't the argument of ARupinsky more generic? g^(its order) = e and hence in any representation the matrix of g will have the characteristic equation x^(its order) = 1 and then all the eigenvalues become roots of unity? What am I missing?
Apr 25, 2015 at 14:40 comment added Dima Pasechnik @Anirbit: ARupinski claims something for the regular representation (of any finite group) only; Stembridge gives formulae for the irreducible representations of $S_n$, which is a completely different story.
Apr 25, 2015 at 13:40 comment added Richard Stanley @Anirbit: Stembridge's Theorem 3.3 gives a combinatorial formula for the eigenvalues. The characters of $S_n$ are not involved.
Apr 25, 2015 at 7:50 comment added Student If the eugenvalues are so trivial then what is Stembridge calculating?
Apr 25, 2015 at 5:38 comment added Dima Pasechnik there is no contradiction; well, I don't see immediately how to (dis)prove ARupinski's statement, but it was probably already known to Frobenuis...
Apr 25, 2015 at 2:57 comment added user6818 But now see what the comment above by ARupinksi says! I am now confused!
Apr 25, 2015 at 1:37 comment added user6818 Any particular part of this paper that you would like to specifically point to?
Apr 25, 2015 at 0:44 history answered Richard Stanley CC BY-SA 3.0