Despite the title, this is probably actually a question in linear algebra or algebraic geometry. Let me write the question(s) first, before I explain the background.

**Problems**

Let $h^{\mu\nu}_{ij}$ represent a map from $\mathbb{R}^4\otimes\mathbb{R}^4$ to $\mathbb{R}^2\otimes\mathbb{R}^2$ (here $\mu\nu$ are indices in the $\mathbb{R}^4$ directions, and $ij$ are in $\mathbb{R}^2$ directions). We shall assume that $h$ is symmetric swapping the $\mu,\nu$ indices and also symmetric swapping the $i,j$ indices. Then for any $\xi_\mu$ in $\mathbb{R}^4$, the object $H(\xi):= h^{\mu\nu}_{ij}\xi_\mu\xi_\nu$ is a symmetric bilinear form on $\mathbb{R}^2$. We say that $\xi$ is characteristic if $H(\xi)$ is degenerate. In other words, $\xi$ is characteristic if $\det(H(\xi)) = 0$.

Since $H(\xi)$ is quadratic in $\xi$, the determinant is an 8th degree homogeneous polynomial in $\xi$. Furthermore, by definition if $\xi$ is characteristic, so is $-\xi$. Observe also that in general the characteristic set will have multiple sheets.

*Question 1, very specific*

~~Does there exist an $h$ such that the characteristic surface is given by $\xi_1^4 + \xi_2^4 + \xi_3^4 - \xi_4^4 = 0$? ~~

*Question 2, slightly more general*

In general are there any obstructions to having a sheet of the characteristic surface described by the zero set of an irreducible (over the reals) polynomial of degree strictly higher than 2?

*Question 3, even more general*

What if we relax the condition on $h$ so that it is a map from $\mathbb{R}^m\otimes\mathbb{R}^m$ to $\mathbb{R}^d\otimes\mathbb{R}^d$ with the same symmetric properties. Define $H(\xi)$ analogously. Can a sheet of the characteristic surface have algebraic degree more than 2?

I'm particularly interested in concrete examples.

**Motivation**

This comes from the study of hyperbolic systems partial differential equations. Recall that a second degree partial differential equation $$ h^{\mu\nu}_{ij} \partial_\mu\partial_\nu u^i = 0 $$ is said to be strictly hyperbolic in the direction of $e_\mu$ if the characteristic polynomial (a polynomial in $t$) $\det(H(x_\mu - te_\mu))$ is hyperbolic for any fixed $x_\mu$ linearly independent from $e_\mu$ and that the roots are distinct (it is enough that the second condition only holds for all by finitely many $x_\mu$ modulo $e_\mu$).

The classical examples for strictly hyperbolic systems (wave equation, crystal optics, etc) all have the sheets of the characteristic surfaces being linearly transformed versions of the standard quadratic double cone: in other words there exists a basis of $\mathbb{R}^m$ such that a sheet is given by $\sum_{i = 1}^{m-1} e_i^2 - e_m^2 = 0$.

I am guessing that for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity (though please let me know if I am wrong).

So my question is: is it possible for a non-strictly hyperbolic system (but one still hyperbolic) where some of the sheets have higher multiplicity to not come from "the square of a quadratic sheet" but from a genuinely quartic or higher polynomial?

**Postscript**

Please do let me know if you need any clarification on my question. Thanks.

**Update**

I struck out question 1 for the following reason: in view of my motivation from hyperbolic polynomials arising from second order PDEs, the answer is negative. The argument is thus: for a hyperbolic system of PDEs, the time-like direction $\xi_4$ should have its corresponding $h^{44}_{ij}$ negative definite, whereas the space-like directions $\xi_1,\xi_2,\xi_3$ should have their corresponding $h^{aa}_{ij}$ positive definite. A simple computation shows that the coefficient to the $\xi_a^4$ term in $\det H(\xi)$ must be $\det h^{aa}_{ij}$. If the target space is two dimensional, both positive definite and negative definite matrices have positive determinants. So for any hyperbolic polynomial arising from a second order system of PDEs, the coefficients for $\xi_a^4$ must be positive.