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arXiv · 2408.07550

Exact values of generic subrank

Abstract

In this article we prove the subrank of a generic tensor in $\mathbb{C}^{n,n,n}$ to be $Q(n) = \lfloor\sqrt{3n - 2}\rfloor$ by providing a lower bound to the known upper bound. More generally, we find the generic subrank of tensors of all orders and dimensions. This answers two open questions posed in arXiv:2205.15168v2. Finally, we compute dimensions of varieties of tensors of subrank at least $r$.

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BibTeXRIS

Paweł Pielasa, Matouš Šafránek, Anatoli Shatsila. 2024-08-14. Exact values of generic subrank. https://arxiv.org/abs/2408.07550

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