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Let $A,B,C,D,E,F$ be vector spaces over a field.

Let $x\in A \otimes B \otimes C$ and $y \in D \otimes E \otimes F$ be tensors that are semistable with respect to the natural actions of $\text{SL}(A) \times \text{SL}(B) \times \text{SL}(C)$ and $\text{SL}(D) \times \text{SL}(E) \times \text{SL}(F)$ respectively.

Then $x \otimes y \in A \otimes B \otimes C \otimes D \otimes E \otimes F$ is certainly semistable with respect to the natural action of $\text{SL}(A) \times \text{SL}(B) \times \text{SL}(C)\times \text{SL}(D) \times \text{SL}(E) \times \text{SL}(F)$.

Is $x \otimes y \in A \otimes B \otimes C \otimes D \otimes E \otimes F$ also necessarily semistable with respect to the natural action of $\text{SL}(A \otimes D) \times \text{SL}(B \otimes E) \times \text{SL}(C \otimes F)$?

This seems "too good to be true", especially given how large $\text{SL}(A \otimes D)$ can be relative to $\text{SL}(A) \otimes \text{SL}(D)$, so my current inclination is to believe the answer is no.

On the other hand, the obvious place to look for a counterexample won't work. We might try to use the Hilbert-Mumford criterion to show that $x \otimes y$ is unstable. The obvious one-parameter subgroups of $SL(A \otimes B)$ to try are those acting diagonally on some basis of $A \otimes B$ obtained as a product of some basis of $A$ and some basis of $B$, and similarly for the other groups. If the Hilbert-Mumford criterion for instability is verified for such a one-parameter subgroup, then then $x$ or $y$ is necessarily unstable.

For some very special $x$ and $y$ (e.g. diagonal tensors), one can check that the only possible one-parameter subgroups are of this form, so we can show the answer is yes. In particular, if $\dim A = \dim B = \dim C=2$ then there is only one possibility for $x$, and one can check in this case that the answer is yes. I'm pretty sure I can check it also if the dimensions are 2,2,3 and 2,3,3, but 3,3,3 is already a mystery to me.


Let me explain some of my motivation for this question.

Terry Tao reformulated Ellenberg and Gijswijt’s variant of the Croot-Lev-Pach breakthrough method for bounding the size of progression-free sets in terms of a new concept, slice rank.

His argument showed that the size of a capset in $\mathbb F_3^n$ was bounded by the slice rank of a $3^n \times 3^n \times 3^n$ tensor, constructed by tensor product from a $3 \times 3 \times 3$ tensor, and that the slice rank was bounded by a certain explicit function, exponentially smaller than $3^n$.

Since reading that I’ve been trying to understand structural properties of the slice rank and sharpness properties of these slice rank bounds. In particular, there is a tensor product constructions that takes a $k \times k \times k$ tensor to a $k^n\times k^n \times k^n$ tensor. It would be combinatirally useful to find the asymptotics of the slice rank of the $k^n \times k^n \times k^n$ tensor in terms of the original $k \times k \times k$ tensor. I also think this problem is algebraically interesting.

It would be especially useful to find necessary and sufficient condition for the $k^n \times k^n \times k^n$ tensor to have slice rank grow exponentially slower than $k^n$. In fact it is sufficient for the original tensor to be GIT-unstable (Theorem 4.10).

Is this condition necessary?

I tried to encapsulate the difficulty of that question, in a way as algebraic-geometrically necessary as possible, in this Mathoverflow question. A positive answer to this MO question would imply that this condition is necessary, and a negative answer would either show it isn’t or else improve slice rank theory in a different way.

Either has the potential to prove new combinatorial bounds, because both upper and lower bounds for slice rank are used in the solution to the capset problem, and both would be needed in generalizations of this problem. However, new upper bounds seem more promising than new lower bounds because past generalizations have mostly been successful using the same lower bound step. Conversely, a positive answer to this question would be more fruitful for impossibility results.

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  • $\begingroup$ Hi @WillSawin. I have a question regarding the asymptotic slice rank of tensors: which upper bound - theorem (4.10) in arxiv.org/pdf/1605.06702.pdf or the inequality (9) in terrytao.wordpress.com/2016/08/24/… - is tight for a GIT-unstable tensor? or is there a relationship between the instability of a GIT-unstable tensor and the entropy of marginal distributions on its support? Thanks in advance! $\endgroup$
    – SMD
    Aug 19, 2017 at 12:43
  • $\begingroup$ @SMD (9) is the sharpest upper bound we have, and in particular is always at least as tight as theorem (4.10). So yes, there is a relationship, that the first bound is always at least as big as the second bound, which you should be able to prove directly by examining the proofs. (9) is often tight, and you can sometimes verify this using the techniques in that blog post. $\endgroup$
    – Will Sawin
    Aug 19, 2017 at 13:29
  • $\begingroup$ Thank you @WillSawin! One more question: does being GIT-unstable for a tensor have anything to do with its support being an anti-chain? $\endgroup$
    – SMD
    Aug 19, 2017 at 19:23
  • $\begingroup$ @SMD No, being GIT-unstable still gives you a natural basis with an ordering, via the Hilbert-Mumford criterion, but there is no reason for the support to be an antichain in this basis. For instance, you could sum a stable tensor with an unstable tensor. $\endgroup$
    – Will Sawin
    Aug 19, 2017 at 19:45

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