Given a multi-affine map $f: \mathbb{R}^n \mapsto \mathbb{R}$, is it always possible to represent the function as a determinant? And is there a principled way to generate the matrix if it is possible?
To make a concrete examples, consider the following two functions in 4 variables $\mathbf{x} \in \mathbb{R}^4$: the cubic function $$ f(\mathbf{x}) = a_1x_2x_3x_4+a_2x_3x_4x_1+a_3x_4x_1x_2+a_4x_1x_2x_3 $$ and the quadratic form $$ g(\mathbf{x}) = b_{12}x_1x_2 + b_{13}x_1x_3 + b_{14}x_1x_4+b_{23}x_2x_3+b_{24}x_2x_4+b_{34}x_3x_4.\\ $$ Both of them are affine functions of each of the variables. Using educated guessing, I was able to find $$ f(\mathbf{x}) = \left| \begin{matrix} a_1 & -a_2 & a_3 & -a_4\\ x_1 & x_2 & 0 & 0\\ 0 & x_2 & x_3 & 0\\ 0 & 0 & x_3 & x_4 \end{matrix} \right|, $$ yet I am wondering what other solutions are possible, for example using a symmetric matrix. For the function $g$ I don't see a similarly simple solution. Any pointer to the right keywords would be appreciated.
The right keyword is, unsurprisingly, determinantal representation (of a polynomial). This has apparently been a topic of interest in algebraic geometry. A comparatively accessible account on the question of representing the zero set of any polynomial is the paper Polynomials with and without determinantal representations by Netzer and Thom.
The main points are: 1. Every polynomial in 2 variables has a representation as a determinant; 2. In general this is not true for more than 2 variables; 3. However, the zero set of any real quadratic polynomial can be represented as a determinant by taking a power of the polynomial.
This means I should be able to represent $g(\mathbf{x})=0$ as a determinantal equation by considering $\left( g(\mathbf{x}) \right)^n = 0$. The representation is based on Clifford algebras and concretely makes use of Weyl-Brauer matrices. This approach could be considered a principled way to arrive at a representation.