This question is related to my last [question](https://mathoverflow.net/questions/423223/the-existence-of-big-incompatible-families-of-weight-supports) 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](https://www.fernuni-hagen.de/stochastik/docs/pub/voting.pdf)) stating that a voting system is weighted iff it is trade robust. 

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 $V(v)=\{u\in V: vu\in \mathcal S\}$. For any distinct vertices $v,v'\in V$ we define an auxiliary graph $H=H(vv')$ with the vertex set $V(v)\cap V(v’)$ 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. $\square$

My last condition involves linear algebra. 

For each set $F\subset E$ let $\overline{F}$ be its characteristic function,  that an element of the linear space $\mathbb R^{E}$, such that for any $e\in F$, the $e$-th coordinate of $\overline{F}$ equals $1$, if $e\in F$, and $0$, otherwise. 

**Condition 4.** Let $v$ be any vertex of $V$, $u_1,u_2,u_3$ be any distinct vertices of $V(v)$, and $F=\{u_1u_2,u_1u_3,u_2u_3\}$. Then $\{v,u_1,u_2,u_3\}\in\mathcal K$ provided there exist a natural number $m$, distinct vertices $u_{1i}, u_{2i}, u_{3i}$ of $V(v)$ such that $\{v, u_{1i}, u_{2i}, u_{3i}\}\in\mathcal K$, $F_i=\{u_{1i}u_{2i},u_{1i}u_{3i},u_{2i}u_{3i}\}$ for each positive integer $i\le m$, and real numbers $\lambda_1,\dots,\lambda_m$ such that $\overline{F}=\sum_{i=1}^m \lambda_i\overline{F_i}$. 

*Proof.* For each positive integer $i\le m$ we have $$0=w(\{v, u_{1i}, u_{2i}, u_{3i}\})=w(vu_{1i})w(u_{2i} u_{3i})+ w(vu_{2i})w(u_{1i} u_{3i})+ w(vu_{3i})w(u_{1i} u_{3i}).$$ Also $$w(\{v,u_1,u_2,u_3\})=w(vu_1)w(u_2u_3)+ w(vu_2)w(u_1u_3)+w(vu_3)w(u_1u_2).$$ The equality $\overline{F}=\sum_{i=1}^m \lambda_i\overline{F_i}$ implies that 
$$0=\sum_{i=1}^m \lambda_i\frac{w(\{v, u_{1i}, u_{2i}, u_{3i}\})}{w(vu_{1i})w(vu_{2i})w(vu_{3i})}=$$ $$\sum_{i=1}^m \lambda_i \left(\frac{w(u_{2i} u_{3i})}{w(vu_{2i})w(vu_{3i})}+\frac{w(u_{1i} u_{3i})}{w(vu_{1i})w(vu_{3i})}+\frac{w(u_{1i} u_{2i})}{w(vu_{1i})w(vu_{2i})}\right)=$$ $$\frac{w(u_{2} u_{3})}{w(vu_{2})w(vu_{3})}+\frac{w(u_{1} u_{3})}{w(vu_{1})w(vu_{3})}+\frac{w(u_{1}u_{2})}{w(vu_{1})w(vu_{2})}=$$
$$\frac{w(\{v,u_1,u_2,u_3\})}{w(vu_{1})w(vu_{2})w(vu_{3})}. \square$$

Thanks.