Lie-algebra-like relation for totally symmetric 4-tensors

Question

There are many totally symmetric real 4-tensors, $T_{ijkl}$, which satisfy the relation

$$T_{ijmn}T_{mnkl} + T_{ikmn}T_{mnjl} + T_{ilmn}T_{mnjk} = c T_{ijkl}$$

with some constant $c$. By the way this relationship reminds me one of the properties of the totally anti-symmetric structure constants, $f_{abc}$, of simple Lie-algebras, $f_{ade}f_{beg}f_{cgd} = c f_{abc}$ but that's not really important :)

What I was wondering if anything is known about some special solutions of this quadratic equation. Specifically, is there a solution such that the associated quartic monomial is positive semi-definite and also it has non-trivial zeros:

$$T_{ijkl}x_i x_j x_k x_l \geq 0 \qquad {\rm for\;\;all\;\;}x$$ $$T_{ijkl}x_{0i} x_{0j} x_{0k} x_{0l} = 0 \qquad {\rm for\;\;some\;\;}x_0 \neq 0$$

I've been trying to find an example but couldn't so was thinking that maybe no such 4-tensor exists. As I was trying to prove it, it became clear that the second condition above means $T_{ijkl} x_{0i} x_{0j} = 0$ for all $k,l$ which means that $T_{ijkl}$ for any fixed index pair must be an indefinite matrix in the remaining 2 indices. This might mean that there are ``too many'' directions for $x$ in which $T_{ijkl}x_ix_jx_kx_l$ can be negative, violating the first condition.

However I couldn't make this precise, any pointers would be great!

Answer 1

It turns out the most general solution with the given requirements is

$$T_{ijkl} = \sum_A t_A\; e^{(A)}_i e^{(A)}_j e^{(A)}_k e^{(A)}_l$$

with some vectors $e^{(A)}$ and constants $t_A$, $A=1 \ldots N$, where if the dimension of the original vector space is $n$ i.e. $i,j,k,l = 1 \ldots n$ then $N < n$. Clearly this tensor satisfies the quadratic equation, its quartic monomial is positive semi-definite and if $x_0$ is orthogonal to all $e^{(A)}$ then it's a non-trivial zero.