2
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.