Let $G$ be a finite group with $n = |G|$ elements. By Cayley's theorem for finite groups, we have an injective homomorphism of groups: $$ \pi : G \rightarrow S_n, \quad g \mapsto \pi(g) $$ where each group element $g$ is mapped to the permutation of $G$ by left multiplication: $$ \pi(g): G \rightarrow G, \quad x \mapsto g \cdot x. $$
We want to associate to each group element $g$ a matrix and then use the Frobenius inner product to define a positive semi-definite function $k$ on the group $G$ as follows:
It is known, see for instance "The Kendall and Mallows Kernels for Permutations" by Yunlong Jiao and Jean-Philippe Vert, that the Kendall-tau function can be made into a positive semi-definite kernel $k$ for permutations.
Let $\mathbf{1}_{\{x\}}$ be the indicator function, which is $=1$ if the boolean variable $x$ is true and $0$ if the boolean variable $x$ is false, and let: $$ \phi: S_n \rightarrow M_n(\mathbb{R}), \quad \sigma \mapsto \left(\mathbf{1}_{\{\sigma(i)>\sigma(j)\}}-\mathbf{1}_{\{\sigma(i)<\sigma(j)\}}\right)_{1\le i,j \le n} $$ where $M_n(\mathbb{R})$ denotes $n \times n$ matrices over $\mathbb{R}$.
The embedding from the finite group $G$ to $M_n(\mathbb{R})$ is then given by: $$ \psi: G \rightarrow M_n(\mathbb{R}), \quad g \mapsto \frac{1}{\sqrt{n(n-1)}} \cdot \phi(\pi(g)). $$
We use the Frobenius inner product on $M_n(\mathbb{R})$, which is given by: $$ \left \langle A, B \right \rangle := \operatorname{tr}(A \cdot B^T) $$ to define a positive semi-definite, symmetric function $k$ on $G$: $$ k: G \times G \rightarrow \mathbb{R}, \quad (g,h) \mapsto \operatorname{tr}(\psi(g) \cdot \psi(h)^T). $$
Question: Does the Gram-Matrix determine the group?
Here are a few Gram-matrices not normalized:
C3
[ 6 -2 -2]
[-2 6 -2]
[-2 -2 6]
C4
[12 0 -4 0]
[ 0 12 0 -4]
[-4 0 12 0]
[ 0 -4 0 12]
KleinFourGroup
[ 12 4 -4 -12]
[ 4 12 -12 -4]
[ -4 -12 12 4]
[-12 -4 4 12]
C5
[20 4 -4 -4 4]
[ 4 20 4 -4 -4]
[-4 4 20 4 -4]
[-4 -4 4 20 4]
[ 4 -4 -4 4 20]
C6
[30 10 -2 -6 -2 10]
[10 30 10 -2 -6 -2]
[-2 10 30 10 -2 -6]
[-6 -2 10 30 10 -2]
[-2 -6 -2 10 30 10]
[10 -2 -6 -2 10 30]
S3
[ 30 18 2 -10 -10 -22]
[ 18 30 -10 -22 2 -10]
[ 2 -10 30 18 -22 -10]
[-10 -22 18 30 -10 2]
[-10 2 -22 -10 30 18]
[-22 -10 -10 2 18 30]
C7
[42 18 2 -6 -6 2 18]
[18 42 18 2 -6 -6 2]
[ 2 18 42 18 2 -6 -6]
[-6 2 18 42 18 2 -6]
[-6 -6 2 18 42 18 2]
[ 2 -6 -6 2 18 42 18]
[18 2 -6 -6 2 18 42]
C8
[56 28 8 -4 -8 -4 8 28]
[28 56 28 8 -4 -8 -4 8]
[ 8 28 56 28 8 -4 -8 -4]
[-4 8 28 56 28 8 -4 -8]
[-8 -4 8 28 56 28 8 -4]
[-4 -8 -4 8 28 56 28 8]
[ 8 -4 -8 -4 8 28 56 28]
[28 8 -4 -8 -4 8 28 56]
C2xC4
[ 56 40 8 -8 -8 -24 8 -8]
[ 40 56 -8 8 -24 -8 -8 8]
[ 8 -8 56 40 8 -8 -8 -24]
[ -8 8 40 56 -8 8 -24 -8]
[ -8 -24 8 -8 56 40 8 -8]
[-24 -8 -8 8 40 56 -8 8]
[ 8 -8 -8 -24 8 -8 56 40]
[ -8 8 -24 -8 -8 8 40 56]
C2xC2xC2
[ 56 40 24 8 -8 -24 -40 -56]
[ 40 56 8 24 -24 -8 -56 -40]
[ 24 8 56 40 -40 -56 -8 -24]
[ 8 24 40 56 -56 -40 -24 -8]
[ -8 -24 -40 -56 56 40 24 8]
[-24 -8 -56 -40 40 56 8 24]
[-40 -56 -8 -24 24 8 56 40]
[-56 -40 -24 -8 8 24 40 56]
D4
[ 56 40 8 -8 -8 -24 -8 -24]
[ 40 56 -8 -24 -24 -8 8 -8]
[ 8 -8 56 40 -8 -24 -8 -24]
[ -8 -24 40 56 8 -8 -24 -8]
[ -8 -24 -8 8 56 40 -24 -8]
[-24 -8 -24 -8 40 56 -8 8]
[ -8 8 -8 -24 -24 -8 56 40]
[-24 -8 -24 -8 -8 8 40 56]
Quaternions
[ 56 32 24 32 -24 -32 -24 -32]
[ 32 56 32 24 -32 -24 -32 -24]
[ 24 32 56 32 -24 -32 -24 -32]
[ 32 24 32 56 -32 -24 -32 -24]
[-24 -32 -24 -32 56 32 24 32]
[-32 -24 -32 -24 32 56 32 24]
[-24 -32 -24 -32 24 32 56 32]
[-32 -24 -32 -24 32 24 32 56]
Edit for clarification:
Conjectured properties:
- Is $k(xg,xh) = k(gx,hx) = k(g,h) \text{ for all } x,g,h \in G$?
- Is $\phi$ an injective mapping?
- Let $\chi_t(g) =$ the characteristic polynomial of the matrix $\phi(g)$ in variable $t$. Is $\chi_t(g) = \chi_t(h)$ for all $g,h \in G$?
- What properties does $\psi(g)$ inherit from properties of skew-symmetric matrices?
- Is $\det(\phi(\pi(g))) = 1$ if $n \equiv 0 \mod 2$ and $0$ otherwise?
- What properties can be deduced from the Pfaffian of $\phi(\pi(g))$?
- The eigenvalues of $\phi(\pi(g))$ are, since the matrix is skew-symmetric, either $=0$ in the $n =$ odd case and the others come in pairs of purely imaginary eigenvalues, or are all purely imaginary: $\lambda i, -\lambda i$ with $\lambda \in \mathbb{R}$?
The method to generate a closed parametric curve for $g \in G$ is given by for individual group elements:
$$f_g(t):= \sum_{m=1 }^n m\exp(k(g,h_m) \cdot t \cdot i)$$
Or it could also be defined as (for the whole group): $$F(t):= \sum_{g \in G } \exp(k(1,g) \cdot t \cdot i)$$
The resulting images looked different, so I suspected the reason was, that the Gram matrix / first row in the Gram matrix if the 1. point in the conjectured properties is true, determined the group.
Here are two images illustrating the point:
Another method of visualization would be, to scale the kernel and use the Fubini-Study metric:
$$d(g,h)=\operatorname{arccos}(\frac{|k(g,h)|}{\sqrt{k(g,g)k(h,h)}})$$
Having this metric between two points $g,h$ I divide the circle in $n$ pieces and to each piece I draw the remaining circles which have radius $d(g,h)$ from the circle corresponding to $g$.
The intersection points of the circles give a visually perceived distinct pattern for the group. It has been pointed to me, that maybe it would be better not to plot the whole group, but only conjugacy classes using the same method. I will try this in the future. Here are two images illustrating this approach:
Klein-Four-Group:
As can be seen, the complexity of the pattern increases with the size of the group, so maybe I should try rather sooner then later the approach with the conjugacy classes.
I have tried for about 40 small groups the approach and the resulting Gram-Matrix was unique for these groups.
I have not tried the whole SmallGroups of GAP.
Second edit:
As for visualization, one might use the Fiedler graph of a simplex, which is defined in the book of Miroslav Fiedler "Matrices and graphs in geometry" (Definition 3.1.2 "acute angles", singed graph $G^+$). Here some graphs are shown for some small groups with properties of graphs:
C4
properties of the Fiedler graph of C4 computed with SageMath:
['is_antipodal', 'is_apex', 'is_arc_transitive', 'is_asteroidal_triple_free', 'is_biconnected', 'is_bipartite', 'is_cactus', 'is_cayley', 'is_circulant', 'is_circular_planar', 'is_cograph', 'is_comparability', 'is_connected', 'is_cycle', 'is_distance_regular', 'is_edge_transitive', 'is_eulerian', 'is_hamiltonian', 'is_line_graph', 'is_long_antihole_free', 'is_long_hole_free', 'is_odd_hole_free', 'is_partial_cube', 'is_perfect', 'is_permutation', 'is_planar', 'is_regular', 'is_strongly_regular', 'is_triangle_free', 'is_vertex_transitive', 'is_weakly_chordal']
Klein Four group:
properties of graph:
['is_antipodal', 'is_apex', 'is_arc_transitive', 'is_asteroidal_triple_free', 'is_biconnected', 'is_bipartite', 'is_cactus', 'is_cayley', 'is_circulant', 'is_circular_planar', 'is_cograph', 'is_comparability', 'is_connected', 'is_cycle', 'is_distance_regular', 'is_edge_transitive', 'is_eulerian', 'is_hamiltonian', 'is_line_graph', 'is_long_antihole_free', 'is_long_hole_free', 'is_odd_hole_free', 'is_partial_cube', 'is_perfect', 'is_permutation', 'is_planar', 'is_regular', 'is_strongly_regular', 'is_triangle_free', 'is_vertex_transitive', 'is_weakly_chordal']
properties:
['is_antipodal', 'is_apex', 'is_arc_transitive', 'is_asteroidal_triple_free', 'is_biconnected', 'is_cayley', 'is_circulant', 'is_cograph', 'is_comparability', 'is_connected', 'is_distance_regular', 'is_edge_transitive', 'is_eulerian', 'is_hamiltonian', 'is_line_graph', 'is_long_antihole_free', 'is_long_hole_free', 'is_odd_hole_free', 'is_perfect', 'is_permutation', 'is_planar', 'is_polyhedral', 'is_regular', 'is_strongly_regular', 'is_triconnected', 'is_vertex_transitive', 'is_weakly_chordal']
As for the Gram-Matrix $\mathbf{G}(G)$, it is defined as:
$$\mathbf{G}(G) := (k(g_i,g_j))_{1 \le i,j \le n}$$
where $G=\{g_1,g_2,\cdots,g_n\}$ are the elements of the finite group $G$ in some order.
The question then is:
If there exists a permutation matrix $P$, such that $$\mathbf{G}(G) = \mathbf{G}(H) \cdot P$$,
is then $G \cong H$?
I think this is true, because of conjectured property $1$.