First let me state the problem, then I'll explain its origin and finally, I'll ask the main question..

**Problem S.** Fix a positive integer $n$. Find all the pairs $(V, S)$, whith the following properties.

**1.** $V$ is a finite dimensional complex vector space equipped with a Hermitian metric. $\DeclareMathOperator{\Sym}{Sym}$ We denote by $\Sym(V)$ the space of symmetric complex linear operators $V\to V$.

**2.** $\newcommand{\bR}{\mathbb{R}}$ $S$ is a linear map $S:\bR^n\to\Sym(V)$ such that for any $\xi\in\bR^n\setminus 0$ the symmetric operator $S(\xi)$ is invertible.

Readers familiar with the basics of p.d.e.-s will surely recognize $S(\xi)$ as the principal **symbol** of an elliptic, first order partial differential operator with constant coefficients that acts on $C^\infty(\bR^n, V)$. That explains the letter $S$ in the name of the problem.

We denote by $\newcommand{\eS}{\mathscr{S}}$ $\eS_n$ the space of solutions of **Problem S** for a given positive integer $n$.

Observe that $\eS_n$ is equipped with two basic algebraic operations $\oplus,\otimes$ $\newcommand{\one}{\boldsymbol{1}}$

$$(V_1, S_1)\oplus (V_2, S_2):= ( V_1\oplus V_2, S_1\oplus S_2), $$

$$ (V_1, S_1)\otimes (V_2, S_2):= ( V_1\otimes V_2, S_1\otimes\one_{V_2}+ \one_{V_1}\otimes S_2). $$

The group $\DeclareMathOperator{\GL}{GL}$ $\GL(n,\bR)$ acts in an obvious way on $\eS_n$. More precisely if $S:\bR^n\to\Sym(V)$ is a solution $S\in\eS_n$, and $T\in \GL(n,\bR)$, then $S\circ T\in \eS_n$.

Let us also observe that for each $n$, the set $\eS_n$ is not empty. We can obtain maps $S: \bR^n\to\Sym(V)$ with the desired properties by using complex representations of the Clifford algebra generated by an Euclidean inner product on the space $\bR^n$. I will refer to such examples as Clifford examples and I will denote by $\newcommand{\eC}{\mathscr{C}}$ $\eC_n$ the subset of $\eS_n$ constructed as above using representations of Clifford algebras. Observe that $\eC_n$ is also closed under the operations $\oplus,\otimes$ and invariant under the above action of $\GL(n,\bR)$

**Main Question.** Fix $n$ Are there non Clifford solutions to Problem S? In other words, is the set $\eS_n\setminus \eC_n$ non-empty?

**Addendum.** Apparently this question is related to a classical question discussed by Porteous in his book *Topological Geometry*. For a given *real* vector space $V$ find the *largest* $n$ find the maximal subspaces $\DeclareMathOperator{\Endo}{End}$ $S\subset \Endo(V)$ such that $S\setminus 0 \subset \GL(V)$. The answer has to do with Radon-Hurwitz numbers, and it basically says that if $S$ is such a subspace, maximal or not, then $V$ s a module over the Clifford algebra generated by an inner product on $S$.