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

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!

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