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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!

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  • $\begingroup$ Is there any reasonable sense in which your notion is independent of the choice of embedding in a symmetric group? Of course there is the canonical realisation of a finite group as a group of permutations of its underlying set. $\endgroup$
    – LSpice
    Commented Jun 6, 2020 at 19:29
  • $\begingroup$ @LSpice. I mean the Cayley embedding: en.wikipedia.org/wiki/Cayley%27s_theorem. Which you consider the canonical realisation. I am not sure if it is independent for different embeddings. $\endgroup$
    – user6671
    Commented Jun 6, 2020 at 19:37
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    $\begingroup$ The matrix in the question differs from the Cayley embedding, no? In the Cayley embedding the row corresponding to the identity should be all zeros off the diagonal, if I understand correctly (since only the identity has any fixed points in this action). It seems the matrix above comes from the action of $D_4$ on vertices of a square. $\endgroup$ Commented Jun 6, 2020 at 19:59
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    $\begingroup$ @orgesleka the code used above doesn't use the cayley embedding, since sage implements DihedralGroup(4) as a permutation group acting on 4 elements (subgroup of $S_4$), while the cayley embedding expresses $D_8$ as subgroup of $S_8$ $\endgroup$ Commented Jun 6, 2020 at 21:16
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    $\begingroup$ The matrix $M$ can be interpreted as a $G×G$-invariant map from the group algebra of $G$ to itself. Since the group algebra has a multiplicity free decomposition, $M$ acts by a scalar times the identity on each isotypic component. Hence the characteristic polynomial of $M$ has the form $(x−a_1)^{d_1^2}...(x−a_r)^{d_r^2}$ where $r$ is the number of irreps of $G$ and $d_i$ is the dim of the $i$th irrep. However, as in the $D_8$ case, there may be coincidences among the eigenvalues $a_i$. It looks like a nice problem to figure out when these coincidences happen. $\endgroup$ Commented Jun 6, 2020 at 21:35

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