According to wikipedia, it is unknown what the maximal rank of a tensor is in the space $F^{n_{1}} \otimes ... \otimes F^{n_{k}}$. It is quite easy to show that it is at most $$\prod _{i=1} ^{k} n_{i}/\max\left(n_{i}\right)$$ Since $F^{n_{1}} \otimes ... \otimes F^{n_{k}}=F^{n_{1}} \otimes V $ where $V=F^{n_{2}} \otimes ... \otimes F^{n_{k}}$ and supposing $n_{1}$ is maximal this gives the bound above. I am interested in the case where $n_{1}=...=n_{k}$, simply call it $n$. Can we do better in this case. That is, can we show whether
- The bound $n^{k-1}$ is tight,
- if not, is there a stronger bound?
*Edit: The bound above follows since the rank in $U\otimes V$ is at most $\dim\left(U\right)$ and at most $\dim\left(V\right)$.