Let $A,B,C,D$ be $n$-dimensional vector spaces over a field $k$.
There is a natural homomorphism from the $mn^m$th tensor power $A^{\otimes (m n^m)} $ of $A$ to $k$ given by the determinant map $A^{\otimes (m n^m)} = \left(A^{\otimes m}\right)^{\otimes n^m} \to k$, becasue $A^{\otimes m}$ has dimension $n^m$. Hence there is a natural homomorphism $$\left( A \otimes B \otimes C \otimes D\right)^{\otimes (mn^m)} = A ^{\otimes (mn^m)} \otimes B^{\otimes (mn^m)}\otimes C^{\otimes (mn^m)} \otimes D^{\otimes (mn^m)} \to k$$
We can evaluate this on a tensor in $A \otimes B \otimes C \otimes D$, producing for each $m$ a $SL(A) \times SL(B) \times SL(C) \times SL(D)$ invariant $i_m$ of these tensors.
What is the set of tensors $x \in A \otimes B \otimes C \otimes D$ such that, for all but finitely many $m$, $i_m(x)=0$?
I can only say three things about this set:
It is a countable union of closed varieties (this follows immediately from the definition).
It contains the GIT-unstable tensors (because all nontrivial homogeneous invariants vanish on GIT-unstable tensors).
It does not contain the diagonal tensor (by explicit evaluation, these invariants are all nonvanishing on the diagonal tensor).
This question is already interesting in the case $n=2$.
Given a specific tensor in $A \otimes B \otimes C \otimes D$, this is a combinatorial question about an explicit sequence of sums. But the sums seem complicated enough that I'm not very optimistic about the hope of a direct combinatorial attack. Here is an example special case:
Consider permutations $\sigma: \mathbb F_2^m \to \mathbb F_2^m$ satisfying the condition that, if $x \neq y$, then $\sigma(x)- x \neq \sigma(y)- y$. In other words $x \mapsto \sigma(x)-x$ is also a permutation of $\mathbb F_2^m$.
Let $r_m$ be the sum of the signs of all permutations satisfying this condition.
Is $r_m$ nonzero for infinitely many $n$?
This corresponds to the $2 \times 2 \times 2 \times 2$ tensor $e_1 \otimes e_1 \otimes e_1 \otimes e_1 + e_2 \otimes e_2 \otimes e_1 \otimes e_1 + e_2 \otimes e_1 \otimes e_2 \otimes e_2 + e_1 \otimes e_2 \otimes e_2 \otimes e_2$.
Here is the general version of the problem:
Let $\alpha_0,\dots,\alpha_{15}$ be fixed. For each $4$-tuple of permutations of the numbers $0$ through $2^m-1$, write the numbers of each permuatation in order, in binary, forming a string of $m2^m$ binary digits. Place these strings on top of each other, so the columns are strings of $4$ binary digits. Let $r_m(\alpha_0,\dots,\alpha_{15})$ be the sum over all $4$-tuples of permutations of the product over $i$ of $\alpha_i$ raised to the number of columns in this figure equal to the binary expansion of $i$, times the product of the signs of the four permutations.
For which $\alpha_0,\dots,\alpha_{15}$ is $r_m(\alpha_0,\dots,\alpha_{15})=0$ for all but finitely many $m$?
One can restrict to very special cases (e.g. $\alpha_i=1$ for $i=0,3,5,9,14$ and $0$ otherwise) but the complexity of the problem is too high.
Let me explain some of the motivation for this problem. It is an attempt to answer a variant of this question using the invariant suggested by Abdelmalek Abdesselam in his answer to this question. Hence a sufficiently good answer to this question would tell me new things about slice rank. Because an abstract approach to this family of questions failed, I've been trying to make it as concrete an combinatorial as possible. Unfortunately the concrete approach doesn't look any easier to me. But maybe someone has an idea?
