TITLE: Interpretations and models of permanent
QUESTION [25 upvotes]: The standard interpretation of permanent of a $0/1$ matrix if considered as a biadjacency matrix of a bipartite graph is number of perfect matchings of the graph or if considered as a adjacency matrix of a directed graph is number of vertex-disjoint cycle covers and these are also the standard models.
Given the importance of permanent as a $\#P$ complete problem that arises in various contexts such as lattices and polyhedra are there any other non-trivial interpretations of the permanent?
I would be interested in any interpretations to number theory (I know none and if there is one it would be very interesting) or algebraic geometry (however tenuous it may be) particularly since the latter is contemplated to be useful in studying the Permanent-Determinant problem.
I would even consider this as an interpretation for special matrices. Permanent represents natural numbers in a canonical way through special matrices Is every positive integer the permanent of some 0-1 matrix?. However if there is any multiplicative and additive property that comes along with it that would be very great and so far I do not see it but it looks like it should be manageable to concoct interesting classes of matrices which have specific interpretations.

REPLY [7 votes]: The permanent appears in various "weighted" versions of Bézout's theorem (on number of solutions of systems of polynomials). Recall that Bézout's theorem says that the number of isolated solutions of $n$ polynomials $f_1, \ldots, f_n$ in $n$ variables is bounded by $\prod_i \deg(f_i)$. The weighted version of Bézout's theorem generalizes this bound to $\prod_i (\omega(f_i)/\omega(x_i)$, where $\omega$ is any "weighted degree" with positive weights. This is further generalized by the "weighted multi-homogeneous" version which lets you partition the variables into a bunch of (disjoint) groups and consider separately the weights on these groups, and this is where permanent appears naturally.
Indeed, let $I_1, \ldots, I_s$ be a partition of $[n] := \{1, \ldots n\}$, i.e. $[n] = \bigcup_j I_j$ and $\sum_j |I_j| = n$. For each $j$, fix a weighted degree $\omega_j$ on $(x_k: k \in I_j)$ such that

$\omega_{j,k} := \omega_j(x_k)$ is positive for each $j,k$, and
the $\gcd$ of the weights $\omega_{j,k}$ is $1$ for each $j$.

Then the weighted multi-homogeneous Bézout bound for the number of isolated zeroes of $f_1, \ldots, f_n$ is
$$\frac{\text{perm}(D)}{(\prod_j n_j!)(\prod_{j,k} \omega_{j,k})}$$
where $n_j := |I_j|$ and $D$ is the following matrix:
$$  \begin{pmatrix}
      d_{1,1}  & \cdots & d_{1,1} 
        & \cdots & \cdots 
        & d_{1,s}  & \cdots & d_{1,s} \\
        \vdots & & \vdots & & & \vdots & & \vdots \\
        d_{n, 1}  & \cdots & d_{n,1} & \cdots & \cdots & d_{n,s}  & \cdots & d_{n,s}
       \end{pmatrix}
$$
where each $d_{i,j} := \omega_j(f_i)$ is repeated $n_j$ times in the $i$-th row.
One can also try to generalize the weighted bound and the weighted multi-homogeneous bound to incorporate non-positive weights, see Theorems X.36 and X.39 of arxiv:1806.05346 for my attempts - permanent appears naturally in both places. The classical versions described above are also treated in that document (see Chapter VIII).
As asked by OP in a comment below, all these are related to mixed volumes of polytopes via the Bernstein-Kushnirenko theorem. More precisely:

The Bézout bound $\prod(\deg(f_i))$ is the mixed volume of simplices $\{\alpha \in \mathbb{R}_{\geq 0}^n: \alpha_1 + \cdots + \alpha_n \leq \deg(f_i)\}$.

More generally, the weighted Bézout bound is the mixed volume of simplices $\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i)\}$.

More generally, the weighted multi-homogeneous Bézout bound is the mixed volume of products of the above simplices.

If $\omega = (\omega_1, \ldots, \omega_n)$ has some negative/zero components, then the problem is that the polyhedron $\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i)\}$ is unbounded, and one has to decide on a polytope that has a facet on the hyperplane $\{\alpha:\langle \omega, \alpha \rangle = \omega(f_i)\}$. One natural choice is to bound the body by hyperplanes parallel to coordinate hyperplanes, i.e. to consider the polytopes
$$\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i),\alpha_j \leq \deg_{x_j}(f_i)\ \text{for all}\ j\ \text{such that}\ \omega_j \leq 0\}$$
The general weighted (respectively, weighted multi-homogeneous) Bézout bound cited above corresponds to the mixed volume of these (respectively, products of these) polytopes.