This question is related to my last question and is originally motivated by recent advances in quantum physics.
I am looking for combinatorial characterizations of some algebraically defined families of sets, like the well-known Taylor and Zwicker theorem (see, for instance, here) stating that a voting system is weighted iff it is trade robust. In particular, this approach allows us to use SAT solvers to deal with the families satisfying the respective combinatorial conditions.
Thus, we are given an even number $n$, a complete graph $K_n$, $K_n=(V,E)$, and a weight function $w:E\to\mathbb C$. Then we can extend the function $w$ to any nonempty even-sized subset $U$ of $V$ putting $w(U)=\sum w(e_1)...w(e_k)$, where the sum is taken over all perfect matchings $\{e_1,\dots, e_k\}$ of the subgraph of $K_n$ induced by $U$. It easily implies the sum decomposition $w(U)=\sum_{u\in U\setminus\{v\}} w(\{v,u\})w(U\setminus\{v,u\})$ for any vertex $v\in U$. It is also additionally required that $w(V)\ne 0$. Let $\mathcal S=\{U\subset V: w(U)\ne 0\}$ and $\mathcal K=\{U\subset V: w(U)=0\}$ be the support and the kernel of the weight $w$, respectively.
Question. What are combinatorial characterizations of families $\mathcal S$ of nonempty even-sized subsets of $V$ which are weight supports?
As a partial answer, I found the following necessary combinatorial conditions for a weight support $\mathcal S$.
The first two of them directly follow from the sum decomposition.
Condition 1. For any set $U\in\mathcal S$ with $|U|\ge 4$ and any element $v\in U$ there exists an element $u\in U$ such that both sets $\{v,u\}$ and $U\setminus\{v,u\}$ belong to $\mathcal S$. $\square$
Condition 2. An even-size subset $U$ of $V$ belongs to $\mathcal S$ provided for some element $v\in U$ there exists a unique element $u\in U$ such that both sets $\{v,u\}$ and $U\setminus\{v,u\}$ belong to $\mathcal S$. $\square$
To formulate a more refined condition, for any vertex $v\in V$ put $\mathcal S^2_v=\{u\in V: vu\in \mathcal S\}$. For any distinct vertices $v,v'\in V$ we define an auxiliary graph $H=H(vv')$ with the vertex set $\mathcal S^2_v\cap \mathcal S^2_{v'}$ such that two vertices $u$ and $u'$ of $H$ are adjacent in $H$, if $\{v,v',u,u'\}\in\mathcal K$.
Condition 3. Each odd cycle in $H$ contains an edge $e\in\mathcal S$. Moreover, if $vv'\in\mathcal K$ then the graph $H$ is bipartite.
Proof. For each vertex $u$ of $H$ put $w'(u)=w(vu)/w(v'u)$. Let $u_1\dots u_k$ be any odd cycle in $H$. Put $u_{k+1}=u_1$. Since for any positive integer $i$ with $1\le i\le k$, the vertices $u_i$ and $u_{i+1}$ are adjacent in $H$, we have $\{v,v',u_{i},u_{i+1}\}\in\mathcal K$, that is $w(vv'u_{i}u_{i+1})=0$. But $$w(vv'u_{i}u_{i+1})=w(vu_{i})w(v'u_{i+1})+w(vu_{i+1})w(v'u_{i})+w(vv')w(u_{i}u_{i+1}).$$
If either $vv'\in\mathcal K$ or $u_{i}u_{i+1}\in\mathcal K$ then the latter summand is zero. It follows $w'(u_{i+1})=-w'(u_i)$. Since $k$ is odd, we obtain a contradiction. $\square$
Since the weight function $w:E\to \mathbb C$ directly determines members of $\mathcal K$ of size $2$, the next step is to describe members of $\mathcal K$ of size $4$. For this purpose, given any vertex $v\in V$ put $$\mathcal K^4_v=\{U\subset V\setminus\{v\}: |U|=3 \mbox{ and } \{v\}\cup U\in\mathcal K\}.$$
We need some linear algebra for the next condition. For each triple $T\subset V$, $T=\{u,u',u''\}$, let $\widetilde T$ be a vector with the entries indexed by $E$, such that for any $e\in E$, the $e$-th entry of $\widetilde T$ equals $1$, if $e\in \{uu',uu'',u'u''\}$, and equals $0$, otherwise. For each $v\in V$ let $\widetilde v$ be a vector with the entries indexed by $E$, such that for any $\{u,u'\}\in E$, the $\{u,u'\}$-th entry of $\widetilde v$ equals $\frac{w(uu')}{w(vu)w(vu')}$, if $\{u,u'\}\subset\mathcal S^2_v$, and equals $0$, otherwise. Then for any triple $T\subset \mathcal S^2_v$, $T=\{u,u',u''\}$, we have $\frac{w(\{v,u,u',u''\})}{w(vu)w(vu')w(vu'')}=(\widetilde T, \widetilde v)$. For each family $\mathcal T$ of triples of $V$, let $[\mathcal T]$ be the matrix with the entries indexed by $\mathcal T\times E$, such that for each $T\in\mathcal T$ the $T$-th column of the matrix is $\widetilde T$.
Condition 4. Let $v$ be any vertex of $V$. Then $\operatorname{rank}\,[\mathcal K^4_v]<\operatorname{rank}\,[\mathcal T]$ for each subfamily $\mathcal T$ of triples of $\mathcal S^2_v$ such that $\mathcal T$ properly contains $\mathcal K^4_v$.
Proof. Suppose for a contradiction that $\operatorname{rank}\,[\mathcal K^4_v]=\operatorname{rank}\,[\mathcal T]$. Pick any triple $S\in\mathcal T\setminus\mathcal K^4_v$, $S=\{s,s',s''\}$. The equality of the ranks implies that there exist real numbers $\lambda_T, T\in\mathcal T$ such that $\widetilde S=\sum_{T\in\mathcal T}\lambda_T\widetilde T$. Then $$0=\sum_{\{u,u',u''\}\in\mathcal T}\lambda_{\{u,u',u''\}}\frac{w(\{v,u,u',u''\})}{w(vu)w(vu')w(vu'')}= \sum_{T\in\mathcal T}\lambda_{T}(\widetilde T, \widetilde v)=$$ $$\left(\sum_{T\in\mathcal T}\lambda_{T}\widetilde T, \widetilde v\right)=(\widetilde S,\widetilde v)= \frac{w(\{v,s,s',s''\})}{w(vs)w(vs')w(vu'')},$$ so $\{s,s',s''\}\in\mathcal K^4_v,$ a contradiction. $\square$
A special case of conditions is when we provide some even-sized subsets of $V$ to belong to $\mathcal K$, that is when some polylinear forms of the weights $w(e)$, $e\in E$ are zeroes. When we study the set of solutions of the respective system of equations, we can find that some weights (or other polylinear forms of them) are zeroes, for instance, as in the condition below.
Given a subset $U$ of $V$, let $G_\mathcal S[U]$ be the graph with the vertex set $U$ and the edge set consisting of all pairs $\{v,u\}$ of distinct elements of $U$ such that $\{v,u\}\in\mathcal S$, that is $w(\{v,u\})\ne 0$.
Condition 5. Let $U$ be a subset of $V$ such that each four-element subset of $U$ belongs to $\mathcal K$. Let $G$ be the graph $G_\mathcal S[U]$ with all isolated vertices removed. If $G$ has at least four vertices then $G$ is isomorphic to one of the following graphs: the star $K_{1,m}$ for some natural $m\ge 3$, the cycle $C_4$, the complete graph $K_4$, the complete graph $K_5$, and the complete graph $K_4$ without an edge.
Proof. Let $\overline{G}$ be the complement of $G$. Then each component of $\overline{G}$ is a clique, because for any its distinct vertices $v$, $v'$, and $v''$ such that $v'$ is adjacent to $v$ and $v''$ is adjacent to $v'$, we have $v''$ is adjacent to $v$. To show the latter pick any vertex $u$ adjacent in $G$ to its nonisolated vertex $v'$. Thus we have $w(vv')=w(v'v'')=0$ and $w(v'u)\ne 0$. Then the equality $$0=w(\{v,v',v'',u\})=w(vv')w(v''u)+w(vv'')w(v'u)+w(v'v'')w(vu)$$ implies that $w(vv'')=0$, that is $v$ is adjacent to $v''$ in $\overline{G}$.
Let $n'$ be the order of $G$. If $n'=4$ then it is easy to check that the graph $G$ is isomorphic to one of the following graphs: the star $K_{1,3}$, the cycle $C_4$, the complete graph $K_4$ without an edge, and the complete graph $K_4$.
Suppose that $n'\ge 5$. No component of $\overline{G}$ consists of two vertices $v$ and $v'$, because otherwise the graph $H(vv')$ is $K_{n'-2}$, which contradicts the second part of Condition 3. On the other hand, by the first part of Condition 3, there is no component of $\overline{G}$ whose order is strictly between $2$ and $n'-1$. Thus either $\overline{G}$ is an empty graph or the edges of $\overline{G}$ form a clique or order $n'-1$. Thus $G$ is either the complete graph or the star.
Moreover, if $n'\ge 6$ then $G$ is not a complete graph. Indeed, otherwise pick any set $U'$ of five vertices of $G$ and any vertex $v$ of $G$, which is not in $U'$. Then $U'\subset\mathcal S^2_v$. Let $\mathcal T$ be the family of all triples of $U'$. Then for any triple $T\in\mathcal T$, $T=\{u,u',u''\}$, we have $$0=\frac{w(\{v,u,u',u''\})}{w(vu)w(vu')w(vu'')}=(\widetilde T, \widetilde v),$$ so $[\mathcal T]\widetilde v=0$. It can be checked that the matrix $[\mathcal T]$ is nonsingular, so $\widetilde v=0$, a contradiction. $\square$
