I'm working through Audrey Terras' Fourier Analysis on Finite Groups with Applications, and at the end of the chapter on induced representations, there is an exercise which reads:

Let $$S = \bigcup_{i=1}^s C_i$$ where $C_i$ is a conjugacy class in a finite group $G$. Suppose $S$ is symmetric; that is, $x\in S$ implies $x^{-1}\in S$. Consider the Cayley graph $X(G,S)$. Show that the eigenvalues of the adjacency matrix of this graph have the form $$\lambda_\pi = \frac{1}{d_\pi}\sum_{s\in S}\chi_\pi(s),$$ where $\pi \in \widehat G$, and $d_\pi=$ the degree of $\pi$.

I tried to mimic the proof for the abelian case, which is straightforward as you can use the fact that $\chi_\pi$ is always a homomorphism in that case. So I'd like to know if there is a way to compute it directly for the nonabelian case, taking $A$ to be the adjacency matrix of the Cayley graph, and $v_\pi=(\chi_\pi(g_1),\cdots,\chi_\pi(g_n))$ for $G=\{g_i\}_{i=1}^n$. Then where I get stuck in the computation is

$$ \begin{split} (Av_\pi)_x & = \sum_{y\in G}A_{xy}(v_\pi)_y \\ & = \sum_{x^{-1}y\in S}\chi_\pi(y) \\ & = \sum_{s\in S} \chi_\pi(xs) \\ & = \sum_{s\in S} \sum_{i=1}^{d_\pi} \pi_{ii}(xs) \end{split} $$

I did try this for the two-dimensional character of $S_3$, with $S$={(123),(132)} to try and get some intuition as to why this is true and maybe a way to prove it, but that didn't seem to provide any insight.


Hints: Define $$ a=\sum_{s\in S}s\in\mathbb C[G] $$ and let $L_a:\mathbb C[G]\to\mathbb C[G]$ denote left multiplication by $a$. Then the matrix of $L_a$ with respect to the basis $G$ of $\mathbb C[G]$ is $A$. Now use the isomorphism $$ \mathbb C[G]\cong\bigoplus_V\mathrm{End}_{\mathbb C}(V) $$ where the sum is over all irreducible representations.

  • $\begingroup$ Thanks for the reply, one of the things I was looking at can be found at: tcs.hut.fi/Studies/T-79.300/2002S/esitelmat/…, theorem (5.2) and corollary (5.4) However, their notation was dense and defined throughout the paper, so the scavenger hunt of doing so was frustrating at 1 am last night. This was a very clear presentation on how it starts, I'll try and work through the rest of the details now. $\endgroup$ – kholli Jul 4 '17 at 13:34
  • $\begingroup$ Also, this proof doesn't seem to use the fact that $S$ is closed under inverses--nor does the one in the link. It seems like, given that this is in the section on induced representations and we have this slightly stronger condition, we should be able to take a subgroup and a representation (I would guess the trivial) and use Frobenius in a straightforward manner (for one thing, Terras never mentions the Endomorphism ring). $\endgroup$ – kholli Jul 4 '17 at 14:20
  • $\begingroup$ @kholli I guess you're right, it should hold if $S$ is not closed under inverses with the right interpretation (the Cayley graph would be directed and $A$ would be non-symmetric). $\endgroup$ – stewbasic Jul 4 '17 at 21:57
  • $\begingroup$ As long as the connection set is a union of conjugacy classes, the formula is correct - you do not need it to be symmetric but, as noted, you now have a directed Cayley graph. $\endgroup$ – Chris Godsil Jul 5 '17 at 6:21

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