Maximum linear-flattening potency in even dimensions

Determine, for every even integer n, the largest integer r for which the linear flattening method can produce a nontrivial equation for the secant variety σ_r of n × n × n tensors.

Background

The paper defines the optimal potency Π(n) of linear flattenings and proves that 2n − 2 ≤ Π(n) ≤ 2n − 2 + 2/(n + 1). Consequently, the largest integer r accessible through the linear flattening method is either 2n − 3 or 2n − 2 when n is even.

The question remains unresolved for general even n. The paper records that n = 2 has answer r = 1, while for odd n the results establish r = 2n − 2, attained by the central Koszul flattening.

References

For an even integer $n$, what is the largest $r$ for which the linear flattening method can yield nontrivial equations for $\sigma_r$? By~\Cref{cor:sharpness}, $r \in {2n-3, 2n-2}$. For $n=2$, the answer is $r=1$ because $\sigma_2=(2){\otimes 3}$. For odd $n$, our results show that $r=2n-2$, and this is achieved by the central Koszul flattening.

— Explicit tensors beyond the linear flattening barrier  (2610.08504 - Lovitz, 6 Oct 2026) in Section 1, subsection “Open questions,” item 2