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!

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?

Here is some SAGEMATH code to play around with:

def J(A,B):
    XA = set([ (x,A[x]) for x in range(len(A))])
    XB = set([ (x,B[x]) for x in range(len(A))])
    print(XA)
    print(XB)
    return QQ(len(XA.intersection(XB)))/QQ(len(XA.union(XB)))

G = DihedralGroup(4)

M = matrix([[ J(Permutation(x),Permutation(y)) for x in list(G)] for y in list(G)])

print(factor(M.charpoly())

Thanks for your help!