Characteristic surface for systems of PDE

Question

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.

Answer 1

There is a natural example where an irreducible component of the symbol, a hyperbolic polynomial, is of degree higher than $2$. It occurs in compressible magnetohydrodynamics (MHD). It is a coupling of the Euler system of compressible, inviscid gas dynamics, with Maxwell's equations, when the magnetic part dominates the EM field. It consists of $8$ conservation laws, governing mass density ($1$), momentum ($3$), energy ($1$) and magnetic field ($3$). When linearizing around a uniform state, we obtain a linear, first-order hyperbolic system. The determinant of the symbol splits into the square of a linear factor (pure transport), a quadratic factor (Alfvén mode) and an irreducible factor of degree $4$. The roots of the latter are the velocities in direction $\xi$ of the fast/slow backward/forward waves.

You can also get a complex structure in nonlinear models of EM fields. See here.

Answer 2

I dont't see why "for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity" should hold. Your questions appear to be about the concept of second order symmetric hyperbolic systems ("symmetric" here refers to the symmetry in $\mu$ and $\nu$). First order symmetric hyperbolic systems definitely don't necessarily have a characteristic surface that consists only of quadratic sheets. Since there are even more degrees of freedom for second order systems, I don't see why they would have more constraints (but maybe I missed something). The simplest non-quadratic example is a diagonal system where each diagonal entry is a first order operator squared (instead of a scalar wave operator).

As always, I am extremely curious about why you want to study nonstandard examples of hyperbolic systems. The only place I know where such things have arisen in nature is in the isoemetric embedding problem.