2
$\begingroup$

In some papers about the cayley graphs of finite groups the behaviour of the eigenvalues and eigenvectors of $\phi$ were discussed when $\phi=\sum_{g\in G} \lambda_G(g)$ and $\lambda_G(g)$ is defined by $\lambda_G(g)\delta_v=\delta_{gv}$ where $\delta_g(a)=1$ if $g=a$ and for other elements of $G$ the value of $\delta_g(a)=0$. It is proved that $\chi\in Irr(G)$ satisfies the following relation and so $\chi$ is an eigenvector of $\phi$:

$$\phi(\chi)=(\sum_{i=1}^h {|K_i|\chi(k_i)\over \chi(1)})\chi$$ when $K_i$ is a conjugacy class of $G$ and $k_i\in K_i$.

It is stated that this result can proved by showing that $\theta=\phi(\chi)$ satisfies 5.4 in "Character theory: Feit" as follows: $$\theta(g)\theta(h)={\theta(1)\over |G|}\sum_{t\in G} \theta(gt^{-1}ht)$$ for each $g,h\in G$. But I have no idea how we can show this result.

Could you kindly help me for proving this result.

$\endgroup$
8
  • $\begingroup$ Which "some papers"? $\endgroup$
    – Igor Rivin
    Commented Nov 13, 2017 at 20:44
  • 1
    $\begingroup$ total perfect codes in cayley graph by Zhou, Des. Codes and Cryptogr. 2016 $\endgroup$
    – Maja
    Commented Nov 13, 2017 at 20:46
  • $\begingroup$ Also in Etienne, Perfect codes and regular partitions in graphs and groups, Eur. J. Combin. 1987 $\endgroup$
    – Maja
    Commented Nov 13, 2017 at 20:48
  • $\begingroup$ Does my correction get your intent right? $\endgroup$ Commented Nov 14, 2017 at 0:19
  • $\begingroup$ Okay, let me prove your first formula independently of whatever Feit has done. First of all, the sum on the right hand side is simply $\dfrac{\sum_{g \in G}\chi\left(g\right)}{\chi\left(1\right)}$. But character orthogonality shows that $\sum_{g \in G}\chi\left(g\right)$ equals $\left|G\right|$ when $\chi$ is the trivial character, and equals $0$ otherwise (see, e.g., math.stackexchange.com/questions/908947/… ). The case when $\chi$ is the trivial character is easy (indeed, in this case we have $\lambda_G \left(g\right) \chi = \chi$ for ... $\endgroup$ Commented Nov 14, 2017 at 2:37

0

You must log in to answer this question.