I have thought about the connection of the mixed volume and the mixed discriminant for a while now but I got no satisfying answer out of the process. In detail:
Mixed Volume: Let $\mathcal{K}^n$ be the space of convex bodies (i.e., nonempty, compact, convex sets) in $\mathbb{R}^n$. With elementwise addition and scalar operation $$A+B := \{a+b \mid a \in A, b \in B\}, \quad \lambda A := \{\lambda x \mid x \in A\} \qquad\Big(A,B \in\mathcal{K}^n, \lambda \in \mathbb{R}\Big) $$ the space $\mathcal{K^n}$ possesses the structure of an $\mathbb{R}_{\geq 0}$-semivector space, i.e., it suffices the properties of a vector space under these operations with the difference that we only allow non-negative scalars.
We know that the volume function $\text{vol}$ that maps each convex body to its volume is an $n$-homogeneous function, i.e., $$ \text{vol}(\lambda A) = \lambda ^n \text{vol}(A) $$ and we know that there exists a unique symmetric $\mathbb{R}_{\geq 0}$-multilinear map $V: (\mathcal{K}^n)^n \to \mathbb{R}_{\geq0}$ with $$ V(A, \dots, A) = \text{vol}(A), \qquad \Big(\forall A \in \mathcal{K}^n \Big) \qquad (1)$$ which just means that $V$ is the polarization of the volume function (see "Schneider, R. (2013). Convex Bodies: The Brunn–Minkowski Theory, Ch. 5" or another question for a proof; the fact that we don't work with vector spaces does not matter). We call this $V$ the mixed volume.
Mixed Discriminant: There is a similar relationship for the determinant: As the determinant $\det$ is a $n$-homogeneous function on $\mathbb{R}^{n,n}$, there also exists a unique symmetric $\mathbb{R}$-multilinear map $D: (\mathbb{R}^{n,n})^n \to \mathbb{R}$ with \begin{equation} \label{detref} D(M, \dots, M) = \det(M),\qquad \Big(\forall M \in \mathbb{R}^{n,n} \Big) \qquad (2)\end{equation} called the mixed discriminant.
My question begins here: It is commonly known that the (oriented) volume of a parallelotope $\text{Par}(M)$ given in terms of a matrix $M \in \mathbb{R}^{n,n}$ is just the determinant of $M$, i.e., \begin{equation} \label{voldet} \text{vol}\Big(\text{Par}(M)\Big)=|\det(M)|. \qquad (3)\end{equation} My central question is now: Does the connection of volumes and determinants somehow extend to the mixed versions? This can be expected as the mixed volume is determined by the standard volume function by $(1)$ and the mixed discriminant is determined by the determinant by $(2)$ and both the volume function (only for parallelotopes) and the determinant (with absolute value) are closely related by $(3)$.
My approaches: My first idea was that the volume function is completely determined by parallelltopes, which would define the volume function completely in terms of determinant. If we dive into measure theory, then the Lebesgue measure and therefore the volume function $\text{vol}$ are uniquely determined by their box sets which are a special case of parallelotopes. This means that by (3), the mixed volume is (somehow) given uniquely by the determinant but very inexplicit. As we work only with convex bodies, I have hope to make a more explicit and direct construction from parallelotopes to convex bodies, maybe by finding a way to exhaust all convex bodies by operations on parallelotopes. Maybe some density arguments could work, as the volume function is continuous under the Hausdorff metric, but I could not find anything that would provide such a relationship.
The other way around is easier, as by $(3)$ the determinant is of course defined in terms of volumes, so that the mixed discriminant $D$ is completely uniquely given in terms of the function $\text{vol} \circ \text{Par}$, which is not much new information. Therefore, it would be interesting to establish a more explicit (or even geometric) connection, but there are two problems that occur with linearity, as neither the orientation of matrices is closed under addition nor $\text{Par}$ is linear.
Other ideas: I have stumbled upon another formula (see $(1.4)$ ) that gives a direct integral formula in terms of other convex-geometric concepts such as support function. It works only if we assume smooth convex bodies, but approximation via density theorems to general convex bodies is possible. However, it seems a bit complex (although reasonable for smooth bodies) as in my mind (3) should give a more direct and intuitive connection.
I have also wondered if there is something more intrinsical that I miss hear that can be explained by multilinear algebra and tensors.
Thank you all for reading that and I appreciate any comment.