Given two three order tensors $T$ and $S$ in $F^{m\times n\times p}$ and $F^{a\times b\times c}$. Clearly $\operatorname{rk}(T\otimes S)\le \operatorname{rk}(T)\operatorname{rk}(S)$. Does the equality holds for generic $T$ in $F^{m\times n\times p}$ and generic $S$ in $F^{a\times b\times c}$ (tensors are viewed as projective varieties via Segre embedding).


1 Answer 1


We don't know.

There are formats in which the equality is false for generic tensors: take $F^{n \times n \times 1}$ and $F^{n \times n \times n^2}$. Generic tensors in both formats are isomorphic to matrix multiplication tensors $\left<n,1,1\right>$ and $\left<1,n,n\right>$ with ranks $n$ and $n^2$ respectively, but the rank of their tensor product $\left<n,n,n\right>$ is less than $n^3$ for $n \geq 2$. If you don't like dimension $1$, the same happens for $F^{n \times n^2 \times n}$ and $F^{n \times n \times n^2}$.

For formats in which the subgeneric rank tensors form a hypersurface, the techniques of the paper "Border Rank Is Not Multiplicative under the Tensor Product" by Christandl, Gesmundo and Jensen, I believe, can be used to prove that for a generic tensor $R(T^{\otimes 2}) < R(T)^2$, but this does not extend to products of different tensors.

On the other hand, for generic $2 \times 2 \times 2$ tensors multiplicativity is true.

My conjecture would be that rank multiplicativity does not hold in most cases.

  • 2
    $\begingroup$ Thanks a lot for your answer and counterexample $\endgroup$
    – Nick Chen
    Dec 16, 2023 at 11:33

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