Since each group $G$ can be considered as a subgroup of the symmetric group, we might see the elements of $G$ as permutations $\pi$.
Consider for each $\pi \in G$ the set:
$$X(\pi) := \{ (i,\pi(i)) | 1 \le i \le n \}$$
Then the Jaccard kernel, which is positive definite, is:
$$J(\pi,\pi'):= \frac{ |X(\pi) \cap X(\pi')|}{|X(\pi) \cup X(\pi')|}$$
We can consider the matrix
$$M = (J(g,h)_{g,h \in G})$$
ordered somehow by an ordering of $G$.
My question is if the characteristic polynomial $\chi_{M}(t)$ when factored in irreducible factors over $\mathbb{Q}$ gives some insight in the degrees of irreducible representations of $G$.
For example: $G = D_8$ = Dihedral Group with 8 elements:
Then $M$ is given by:
$$\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\ 0 & 1 & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & 1 & 0 & 0 & 0 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{1}{3} & \frac{1}{3} & 0 & 0 \\ \frac{1}{3} & 0 & 0 & \frac{1}{3} & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & 0 & \frac{1}{3} & 0 & 1 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 1 & \frac{1}{3} \\ 0 & 0 & \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3} & 1 \end{array}\right) $$
with characteristic polynomial:
$$\chi_M(t) = (x - \frac{5}{3})^{2} \cdot (x - \frac{1}{3})^{2} \cdot (x - 1)^{4}$$
For $D_8$ we have:
$$(1^2+1^2)+(1^2+1^2)+2^2 = 8 = 2+2+4$$
Is this just a coincidence or can it be proven?
Thanks for your help!