I have the following set of equation systems, and I would like to find a short, formal way to write it down. My main difficulty is that I cannot find a good way to write the indices of the variables $\omega$. Any suggestion is highly apprechiated. 


**n=4:**
$$\omega\_{A,B,a,b}\cdot\omega\_{C,D,c,d}+\omega\_{A,C,a,c}\cdot\omega\_{B,D,b,d}+\omega\_{A,D,a,d}\cdot\omega\_{B,C,b,c}=\delta\_{a,b}\cdot\delta\_{b,c}\cdot\delta\_{c,d}$$
with $\omega\_{X,Y,x,y} \in \mathbb{C}$, and $x,y \in \{0,1\}$. This leads to 24 independent variables ($\omega\_{A,B,0,0}$, $\omega\_{A,B,0,1}$, $\omega\_{A,B,1,0}$, $\omega\_{A,B,1,1}$, $\omega\_{A,C,0,0}$, $\omega\_{A,C,0,1}$, $\omega\_{A,C,1,0}$, $\omega\_{A,C,1,1}$, $\omega\_{A,D,0,0}$, $\omega\_{A,D,0,1}$, $\omega\_{A,D,1,0}$, $\omega\_{A,D,1,1}$, $\omega\_{B,C,0,0}$, $\omega\_{B,C,0,1}$, $\omega\_{B,C,1,0}$, $\omega\_{B,C,1,1}$, $\omega\_{B,D,0,0}$, $\omega\_{B,D,0,1}$, $\omega\_{B,D,1,0}$, $\omega\_{B,D,1,1}$, $\omega\_{C,D,0,0}$, $\omega\_{C,D,0,1}$, $\omega\_{C,D,1,0}$, $\omega\_{C,D,1,1}$), and 16 equations:


$$
\omega\_{A,B,0,0}\cdot\omega\_{C,D,0,0}+\omega\_{A,C,0,0}\cdot\omega\_{B,D,0,0}+\omega\_{A,D,0,0}\cdot\omega\_{B,C,0,0}=1\\
\omega\_{A,B,0,0}\cdot\omega\_{C,D,0,1}+\omega\_{A,C,0,0}\cdot\omega\_{B,D,0,1}+\omega\_{A,D,0,1}\cdot\omega\_{B,C,0,0}=0\\
\omega\_{A,B,0,0}\cdot\omega\_{C,D,1,0}+\omega\_{A,C,0,1}\cdot\omega\_{B,D,0,0}+\omega\_{A,D,0,0}\cdot\omega\_{B,C,0,1}=0\\
\omega\_{A,B,0,0}\cdot\omega\_{C,D,1,1}+\omega\_{A,C,0,1}\cdot\omega\_{B,D,0,1}+\omega\_{A,D,0,1}\cdot\omega\_{B,C,0,1}=0\\
$$
$$
\omega\_{A,B,0,1}\cdot\omega\_{C,D,0,0}+\omega\_{A,C,0,0}\cdot\omega\_{B,D,1,0}+\omega\_{A,D,0,0}\cdot\omega\_{B,C,1,0}=0\\
\omega\_{A,B,0,1}\cdot\omega\_{C,D,0,1}+\omega\_{A,C,0,0}\cdot\omega\_{B,D,1,1}+\omega\_{A,D,0,1}\cdot\omega\_{B,C,1,0}=0\\
\omega\_{A,B,0,1}\cdot\omega\_{C,D,1,0}+\omega\_{A,C,0,1}\cdot\omega\_{B,D,1,0}+\omega\_{A,D,0,0}\cdot\omega\_{B,C,1,1}=0\\
\omega\_{A,B,0,1}\cdot\omega\_{C,D,1,1}+\omega\_{A,C,0,1}\cdot\omega\_{B,D,1,1}+\omega\_{A,D,0,1}\cdot\omega\_{B,C,1,1}=0\\
$$
$$
\omega\_{A,B,1,0}\cdot\omega\_{C,D,0,0}+\omega\_{A,C,1,0}\cdot\omega\_{B,D,0,0}+\omega\_{A,D,1,0}\cdot\omega\_{B,C,0,0}=0\\
\omega\_{A,B,1,0}\cdot\omega\_{C,D,0,1}+\omega\_{A,C,1,0}\cdot\omega\_{B,D,0,1}+\omega\_{A,D,1,1}\cdot\omega\_{B,C,0,0}=0\\
\omega\_{A,B,1,0}\cdot\omega\_{C,D,1,0}+\omega\_{A,C,1,1}\cdot\omega\_{B,D,0,0}+\omega\_{A,D,1,0}\cdot\omega\_{B,C,0,1}=0\\
\omega\_{A,B,1,0}\cdot\omega\_{C,D,1,1}+\omega\_{A,C,1,1}\cdot\omega\_{B,D,0,1}+\omega\_{A,D,1,1}\cdot\omega\_{B,C,0,1}=0\\
$$
$$
\omega\_{A,B,1,1}\cdot\omega\_{C,D,0,0}+\omega\_{A,C,1,0}\cdot\omega\_{B,D,1,0}+\omega\_{A,D,1,0}\cdot\omega\_{B,C,1,0}=0\\
\omega\_{A,B,1,1}\cdot\omega\_{C,D,0,1}+\omega\_{A,C,1,0}\cdot\omega\_{B,D,1,1}+\omega\_{A,D,1,1}\cdot\omega\_{B,C,1,0}=0\\
\omega\_{A,B,1,1}\cdot\omega\_{C,D,1,0}+\omega\_{A,C,1,1}\cdot\omega\_{B,D,1,0}+\omega\_{A,D,1,0}\cdot\omega\_{B,C,1,1}=0\\
\omega\_{A,B,1,1}\cdot\omega\_{C,D,1,1}+\omega\_{A,C,1,1}\cdot\omega\_{B,D,1,1}+\omega\_{A,D,1,1}\cdot\omega\_{B,C,1,1}=1
$$




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**n=6:**
$$\omega\_{A,B,a,b}\cdot\omega\_{C,D,c,d}\cdot\omega\_{E,F,e,f}+\omega\_{A,B,a,b}\cdot\omega\_{C,E,c,e}\cdot\omega\_{D,F,d,f}+\omega\_{A,B,a,b}\cdot\omega\_{C,F,c,f}\cdot\omega\_{D,E,d,e}\\
+\omega\_{A,C,a,c}\cdot\omega\_{B,D,b,d}\cdot\omega\_{E,F,e,f}+\omega\_{A,C,a,c}\cdot\omega\_{B,E,b,e}\cdot\omega\_{D,F,d,f}+\omega\_{A,C,a,c}\cdot\omega\_{B,F,b,f}\cdot\omega\_{D,E,d,e}\\
+\omega\_{A,D,a,d}\cdot\omega\_{B,C,b,c}\cdot\omega\_{E,F,e,f}+\omega\_{A,D,a,d}\cdot\omega\_{B,E,b,e}\cdot\omega\_{C,F,c,f}+\omega\_{A,D,a,d}\cdot\omega\_{B,F,b,f}\cdot\omega\_{C,E,c,e}\\
+\omega\_{A,E,a,e}\cdot\omega\_{B,C,b,c}\cdot\omega\_{D,F,d,f}+\omega\_{A,E,a,e}\cdot\omega\_{B,D,b,d}\cdot\omega\_{C,F,c,f}+\omega\_{A,E,a,e}\cdot\omega\_{B,F,b,f}\cdot\omega\_{C,D,c,d}\\
+\omega\_{A,F,a,f}\cdot\omega\_{B,C,b,c}\cdot\omega\_{D,E,d,e}+\omega\_{A,F,a,f}\cdot\omega\_{B,D,b,d}\cdot\omega\_{C,E,c,e}+\omega\_{A,F,a,f}\cdot\omega\_{B,E,b,e}\cdot\omega\_{C,D,c,d}=\delta\_{a,b}\cdot\delta\_{b,c}\cdot\delta\_{c,d}\cdot\delta\_{d,e}\cdot\delta\_{e,f}
$$




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* The rule is, that I multiply $m=\left(\frac{n}{2}\right)$ variables $\omega\_{X\_i,Y\_i,x\_i,y\_i}$ ($\omega\_{X\_1,Y\_1,x\_1,y\_1}\cdot\omega\_{X\_2,Y\_2,x\_2,y\_2}\cdot\dots\cdot\omega\_{X\_m,Y\_m,x\_m,y\_m}$), such that $X\_0,Y\_0,X\_1,Y\_1,\dots,X\_m,Y\_m$ contains each of the first $n$ letters in the alphabet exactly once.
* There are $|\omega|=4\frac{n(n-1)}{2}$ variables and $|Q|=2^n$ equations.
* The indices could be generalized to $x,y \in \{0,1,...,c-1\}$. Then $|\omega|=c^2\frac{n(n-1)}{2}$, $|Q|=c^n$.




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> 
> **Question.** How can one write this infinite set of equation systems in a concise, formal way?
> 
> 
> 


There are infinite equation systems because $n$ can be an arbitrary even integer, and $c$ can contain arbitrarily many terms.


PS: The current question is a reformulation of this question on [graph theory](https://mathoverflow.net/questions/311325/vertex-coloring-inherited-from-perfect-matchings-motivated-by-quantum-physics), independent of graphs and perfect matchings.


.....



Concise formulation of set of equation systems
Let's use an alphabet $X\_i $ indexed by an integer $1\leq i\leq n$, and also accompanying variables $x\_i \in \{ 0,1,\ldots ,c-1 \} $. Up to now, it seems you've only defined the variables $\omega\_{X\_i, X\_j, x\_i, x\_j} $ for $i<j$. For convenience, define also the auxiliary variables $\omega\_{X\_j, X\_i, x\_j, x\_i} = \omega\_{X\_i, X\_j, x\_i, x\_j} $ for $i<j$. Then, it seems to me your equations are
$$
\frac{1}{(n/2)! 2^{n/2} } \sum\_{m=1}^{n!} \prod\_{j=1}^{n/2} \omega\_{X\_{P^{(m)}(2j-1)}, X\_{P^{(m)}(2j)}, x\_{P^{(m)}(2j-1)}, x\_{P^{(m)}(2j)}} = \prod\_{i=1}^{n-1} \delta\_{x\_i,x\_{i+1} }
$$
where $P^{(m)}(k)$ denotes the $k$-th entry in the $m$-th permutation of $1,2,\ldots ,n$. The prefactor $1/2^{n/2} $ arises because the sum over permutations generates also terms containing $\omega\_{X\_j, X\_i, x\_j, x\_i} $ for $i<j$ in addition to your original $\omega\_{X\_i, X\_j, x\_i, x\_j} $; the products contain $n/2$ factors $\omega $, and in each factor we can exchange $X\_i \leftrightarrow X\_j $, so we're overcounting each distinct term $2^{n/2} $ times. The prefactor $1/(n/2)!$ arises because we're generating all possible orderings of the factors $\omega $ in each product, so we're generating each distinct term $(n/2)!$ times. As you already indicate, there are $c^n $ equations because we can fix each of the variables $x\_i $ to $c$ different choices.


