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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 $S$.

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Do you allow for the various Clifford actions entering into $\mathscr{C}_n$ to come from differing inner products (viz, Riemannian metrics) on $\mathbb{R}^n$? – Branimir Ćaćić Jan 4 '13 at 14:11
Good question. No, the only inner products allowed on $\mathbb{R}^n$ are constant inner products, i.e., the induced Riemann metric is constant. I'll have to update my question. – Liviu Nicolaescu Jan 4 '13 at 14:46
up vote 11 down vote accepted

Maybe I'm misunderstanding something, but it seems that the answer is probably 'no', at least if $d = \dim_\mathbb{C} V$ is large enough.

What really matters is the $n$-dimensional real subspace $\mathsf{S}=\mathrm{Im}(S)\subset\mathrm{Sym}(V)$. What you need is that this space not meet the cone of singular matrices in $\mathrm{Sym}(V)$ anywhere but at the origin. This is an open condition on the element $\mathsf{S}\in\mathrm{Gr}_n\bigl(\mathrm{Sym}(V),\mathbb{R}\bigr)$, and this latter space has dimension $n(d^2{-}n)$. Thus, the set of such subspaces with your property has this latter dimension, but the group of unitary transformations on $V$ (which, I gather, is the space of symmetries of the problem) is only of dimension $d^2$, so there must be at least an $(n{-}1)d^2-n^2$ parameter family of 'inequivalent' subspaces that meet your criteria.

However, there are only a finite number of inequivalent $d$-dimensional complex representations of the Clifford algebra on $\mathbb{R}^n$ endowed with a definite inner product.

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I think that the argument above shows that non-Clifford examples must exist. Take $n=d=3$. – Robert Bryant Jan 4 '13 at 16:28
Thanks! This does it. – Liviu Nicolaescu Jan 4 '13 at 16:38

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