$I_n$ is the $n \times n$ or order $n$ identity matrix, $J_n$ is the order $n$ matrix of all ones, and $n \in \mathbb{Z}^+$.
We define a Latin square $\mathcal{L_n}$ to be a set of $n$ permutation matrices $P_i$ each of order $n$ with the following condition:
$\sum_{i=1}^{n} P_i = J_n$
This isn’t the usual definition, but it helps set up the structure I’m looking for. Also, if the condition holds, then we can always get the “normal” Latin square $\bar{\mathcal{L_n}}$ as the following:
$\bar{\mathcal{L_n}} = \sum_{i=1}^{n} iP_i$
For the rest of the post, we will only consider $\mathcal{L_n}$ such that $P_1 = I_n$. Therefore each $P_{i > 1}$ is a derangement.
Any Latin square can be taken as the multiplication or Cayley table of a quasigroup. This is a closed binary operation $(Q_n, \cdot)$ over the set $Q_n = \{x \leq n : x \in \mathbb{Z}^+\}$ such that for any $a, b \in Q_n$, there exists a unique $x, y \in Q_n$ such that $a \cdot x = b$ and $y \cdot a = b$.
With the bolded condition imposed above, we’ve restricted ourselves to unipotent quasigroups, where there exists some unique unipotent element $e$, such that $a \cdot a = e$ for any element $a$ (since $e \cdot e = e$, it is also an idempotent element - the only idempotent element - giving us a pique, or a pointed idempotent quasigroup). In terms of $\bar{\mathcal{L_n}}$, $e$ will always be $1$, but it’s important to make clear that neither $e$ nor $1$ should be taken as some sort of identity element.
With all of that covered, we now define quasi-commutativity.
Starting with a partition of the elements of $Q_n$ into disjoint subsets or blocks, we say elements $a$ and $b$ are equivalent, or $a \sim b$, if they are in the same block in the partition. A quasi-commutative binary operation satisfies the condition that for every pair of elements $a$ and $b$, there exists equivalent elements $c \sim d$ such that $a \cdot b = c$ and $b \cdot a = d$.
The equivalence relation allows us to partition $J_n$ - alternatively a partition over $Q_n \times Q_n$ with the natural order over $Q_n$ - into $1 \leq m \leq n$ classes $R_j$ such that $\sum_{j=1}^{m} R_j = J_n$. We use this terminology as a reference to the connection with association schemes.
We say some order $n$ permutation matrix $P$ is embedded into $R_j$ if their Hadamard product $P \circ R_j = P$. A resulting property of quasi-commutativity is that for any $P_i \in \mathcal{L_n}$ embedded into $R_j$, its transpose $P^T_i$ must also be embedded into $R_j$ (if $P_i = P^T_i$, then the permutation is either the identity, or a symmetric derangement i.e. a product of disjoint transpositions). Even stronger, for any general $P$ embedded into $R_j$, $P^T$ must also be embedded into $R_j$. This means that $R_j$ must be a symmetric matrix.
If $I_n$ can’t be embedded into some class, then we say that it is a deranged class.
For the rest of the post, we assume the following equivalent conditions:
- $R_1 = I_n$
- $e \not\sim a$ for all $a \in Q_n$\$\{e\}$
- any $R_{j > 1}$ is a deranged class
*The set of classes now satisfy the first 3 axioms of a symmetric association scheme (aside from class indexing since we start at 1). However, I don’t know if the classes in general commute, or if we can characterize when they do. I also don’t know if in general we can define the structural constants
Given that $R_j$ is a symmetric matrix which can be decomposed into the sum of $k_j$ disjoint permutation matrices of order $n$ (due to the Latin square property of $\mathcal{L_n}$), the sum of each column and the sum of each row must be equal to $k_j$ and $\frac{1}{k_j}R_j$ is a doubly stochastic matrix (transition matrix for a random walk on a $k$-regular graph).
*I’d be interested to know of any interesting properties the associated Markov chain would have. It’s not a strong area of mine, so I haven’t dug in too much here
Given the above construction of a unipotent quasi-commutative quasigroup, my question is if it’s possible to give a characterization of the classes as a specific type of combinatorial design?
The motivation for the question is to understand the coefficient orbit matrices (COMs) of a square matrix $A$. Given the set of inner automorphisms of $A$, the COM of some $A_{ij}$ is a square binary matrix which has a $1$ in each cell that $A_{ij}$ can be sent to through an automorphism.
I’ve only been able to analyze small graphs, $|V| \leq 8$, but if $A$ is the adjacency matrix of an undirected vertex transitive simple graph, then the COMs (of edges and non-edges) seem to be exactly the classes of some unipotent quasi-commutative quasigroup $(Q_{|V|}, \cdot)$.
In the examples I’ve worked out, each COM is permutation similar to a circulant matrix, and is therefore the adjacency matrix of a circulant graph, but I’m not sure if every COM defined from the adjacency matrix of a vertex transitive graph has the same circulant property (or more generally if $A$ has a row/column transitive automorphism group).
