# Is anything known about the eigenspectrum of the regular representation of the permutation group?

I am looking for information like upper bounds on how many times any eigenvalue can occur or something like how many eigenvalues can be there in some given range. Is anything like this known?

The only thing I know is this may be trivial thing : Fix a group element $g \in S_n$. Let $R(g)$ be the matrix of $g$ in the regular representation. If for an irrep say $\pi$ of $S_n$ if $\pi(g)$ (the matrix of $g$ in the irrep $\pi$) has an eigenvalue $\lambda$ with multiplicity $m_{\pi(g)}(\lambda)$ and $\pi$ has dimension $d_\pi$ then $\lambda$ will occur as an eigenvalue of the $R(g)$ with multiplicity at least $d_\pi m_{\pi(g) }(\lambda )$

It would be great if this same question can be answered for any other arbitrary representation of $S_n$.

• 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. – ARupinski Apr 25 '15 at 1:49
• 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? – user6818 Apr 25 '15 at 2:02
• Then what is the idea in the other answers? – user6818 Apr 25 '15 at 2:06
• @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. – Qiaochu Yuan Apr 25 '15 at 5:49
• 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. – Neil Strickland Apr 25 '15 at 10:27

For the eigenvalues of matrices in arbitrary irreducible representations of $S_n$, see the paper by John Stembridge at http://msp.org/pjm/1989/140-2/pjm-v140-n2-p06-s.pdf.

• Any particular part of this paper that you would like to specifically point to? – user6818 Apr 25 '15 at 1:37
• But now see what the comment above by ARupinksi says! I am now confused! – user6818 Apr 25 '15 at 2:57
• 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... – Dima Pasechnik Apr 25 '15 at 5:38
• If the eugenvalues are so trivial then what is Stembridge calculating? – Anirbit Apr 25 '15 at 7:50
• @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. – Dima Pasechnik Apr 25 '15 at 14:40

the data you like to know is encoded in the character value; once the character is known, the eigenvalues of each group element can be found, see e.g. https://math.stackexchange.com/questions/427529/computing-eigenvalues-from-characters

GAP can compute these using the function EigenvaluesChar, see Sect 72.8-19 of its manual: http://gap-system.org/Manuals/doc/ref/chap72.html

• Yes. But from knowing an irreducible representation of $S_n$ (say given as the partition of $n$ to which it corresponds to) how do I infer its eigenspectrum from it? – user6818 Apr 25 '15 at 1:36
• But now see what the comment above by ARupinksi says! I am now confused! – user6818 Apr 25 '15 at 2:57
• for the elements of order 2, it's trivial that the number of eigenvalues equal to 1 equals the number of eigenvalues equal to -1, as the trace of each non-identity element equals to 0. – Dima Pasechnik Apr 25 '15 at 14:47
• a similar argument works for elements of order 3; as an eigenvalue $\lambda$ and its conjugate must occur with the same multiplicity, we see that the number of eigenvalues equal to 1 must be equal to twice the number of eigenvalues $\zeta$, for $\zeta$ a fixed primitive cubic root of unity. – Dima Pasechnik Apr 25 '15 at 14:57