Hyperdeterminants and real solutions to polynomials in multiple indeterminates

Question

I am interested in real solutions to homogeneous polynomials in $\mathbb{R}[x_1, ..., x_n]$.

For the case $n = 2$, we can compute the determinant of the symmetric $2 \times 2$ matrix $[a_{ij}]$, $a_{ij} \in \mathbb{R}$. If this determinant is positive, then the corresponding homogeneous polynomial $a_{11}x_1^2 + 2a_{12} x_1 x_2 + a_{22} x_2^2$ factorises into $(c_1 x_1 + c_2 x_2)(d_1 x_1 + d_2 x_2)$, with $c_k, d_k \in \mathbb{R}$.

Can we make a similar statement in $n = 3$ using the hyperdeterminant of the $3 \times 3 \times 3$ tensor corresponding to a homogeneous polynomial of degree 3? Specifically, if this hyperdeterminant is positive, can we say that it factorises into $(c_1 x_1 + c_2 x_2 + c_3 x_3)(d_1 x_1 + d_2 x_2 + d_3 x_3)(e_1 x_1 + e_2 x_2 + e_3 x_3)$ with $c_k, d_k, e_k \in \mathbb{R}$? And if the hyperdeterminant is negative, is there a sense in which the two zeroes of this polynomial are "complex conjugates" like in the single variable cubic case?

I'm just beginning to learn algebraic geometry and have skimmed through Discriminants, Resultants, and Multidimensional Determinants, but haven't found an answer (probably because I don't know enough algebraic geometry!).

Answer 1

The notion of hyperdeterminant (in the technical sense of GKZ) is not the right one here, because it is meant for not necessarily symmetric tensors, whereas the question is about homogeneous polynomials, i.e., symmetric tensors.

Moreover, there are two different issues to be treated separately.

If $F\in \mathbb{R}[x_1,\ldots,x_n]$ is homogeneous of degree $d$, it is not in general true that it factors as a product of linear forms $$ F(x)=\prod_{i=1}^{d}\left(\sum_{j=1}^{n}c_{ij} x_j\right) $$ even if the coefficients are allowed to be complex.

This factorization occurs if and only if the polynomial satisfies the Brill equations. For references on the latter see my answer to

which homogeneous polynomials split into linear factors?

Then, once that is settled, one needs a way to check if one can arrange for the coefficients to be real. A necessary condition is that if you restrict $F$ to an arbitrary line $x=t_1 a+t_2 b$ where the $n$-component vectors $a,b$ are real, the resulting binary form, in the pair of variables $(t_1,t_2)$, factorizes as a product of linear forms with real coefficients. Equivalently the associated nonhomogeneous polynomial should only have real roots, i.e., the polynomial should be real stable. See my answer to

real symmetric matrix has real eigenvalues - elementary proof

for a link to the book by Basu, Pollack and Roy on real algebraic geometry where one can find a characterization of univariate polynomials with only real roots in terms of subresultants and subdiscriminants.

I suspect the above is not only a necessary condition but also a sufficient condition, but didn't have time to think about it.

In the mentioned example, Brill equations should not be hard to write, and the real root condition is just that the discriminant of the cubic is nonnegative.