3
$\begingroup$

Suppose $T \in V_1 \otimes \cdots \otimes V_k$ is a tensor, where each $V_i$ is a finite dimensional complex vector space. A $1$-flattening (or a flattening) is a realization of $T$ as a matrix in the space of matrices in $k$ essentially different ways as follows: \begin{equation} V^*_{i} \rightarrow V_{1} \otimes \cdots \otimes V_{k-1}. \end{equation} Is it true that the rank of a flattening of $T$ is always a lower bound for the border rank of $T$? If not in general, is it true when $T$ is a symmetric tensor?

$\endgroup$
6
$\begingroup$

Yes, the flattening rank is a lower bound for border rank.

First note that flattening rank is a lower bound for rank. If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every flattening of $T$ has rank one. If $T$ has rank $r$, then $T$ is a sum of $r$ decomposable tensors, and so every flattening of $T$ is a sum of the corresponding $r$ flattenings of decomposable tensors. Each of those flattenings has rank $1$. A sum of $r$ linear maps of rank $1$ has rank at most $r$, so every flattening of $T$ has rank at most $r$.

Now, the set of tensors of border rank less than or equal to $r$ is the Zariski closure of the set of tensors of rank $r$.

(A digression: This is the $r$th secant variety of the variety of decomposable tensors, i.e., a Segre variety—at least if you set the $k$th secant variety of a variety $X$ to be the closure of the union of $(k-1)$-planes spanned by $k$ points of $X$. In this numbering the "ordinary" secant variety of lines is the $2$nd secant variety. ... At some point in writing this I have rather glossed over any distinction between affine varieties, such as sets of tensors, and projective varieties. But it doesn't matter much for discussions of rank—all the sets of tensors here are cones, so they might as well be projective varieties.)

On the other hand, the set of tensors with flattening rank less than or equal to an integer $k$ is also a Zariski closed set. Indeed it is defined precisely by the vanishing of the $(k+1) \times (k+1)$ minors of the flattenings. (I am being intentionally vague as to which flattenings. You may decide if you want to consider $1$-flattenings only, or all flattenings $V_I^* \to V_{[n]\setminus I}$ for $I \subset [n]$, where $V_I = \bigotimes_{i \in I} V_i$, etc.)

Putting it all together, the set of tensors of (ordinary) rank less than or equal to $r$ is contained in the set of tensors of flattening rank less than or equal to $r$. The latter is Zariski closed. Therefore the Zariski closure of the first set, the set of tensors of border rank less than or equal to $r$, is also contained in the set of tensors of flattening rank less than or equal to $r$. This is just to say that if $T$ has border rank $r$ (or less), then the flattening rank of $T$ is also less than or equal to $r$. Hence yes, as you asked, the flattening rank of $T$ is less than or equal to the border rank of $T$.

It is also true for symmetric tensors with (symmetric) border rank, either by the same argument, or by an inequality between symmetric border rank and (ordinary, non-symmetric) tensor border rank.

Some introductions include Landsberg, Tensors: Geometry and Applications (bonus: has a ridiculous drawing of a pickle) or Carlini, Grieve, Oeding, Four Lectures on Secant Varieties (no pickles).

$\endgroup$
  • $\begingroup$ Thank you Zach! The $n$-tuple of $1$-flattening ranks is the multi-linear rank $r_i$ of the tensor $T \in \bigotimes_{i=1}^n V_i$. In Landsberg's book, the Zariski closure of all tensors of multi-linear rank $(r_1, \cdots , r_n)$ is called the subspace variety $\text{Sub}_{r_1, \cdots , r_n}$. In exercise (5.11) of your 2nd reference (here), the $r$-th secant variety $\sigma_r(X)$ of a Segre variety $X$ (the space of tensors with border rank no greater than $r$) is contained in the Sub$_{r_1, \cdots , r_n}$, IF $r \leq r_i$. Am I missing something? $\endgroup$ – SMD Sep 7 '17 at 12:43
  • $\begingroup$ No, I don't think you're missing anything. Say $T$ is a tensor of border rank $r$. Then the $1$-flattening ranks $r_i(T) \leq r$. If $r_1,\dots,r_n$ are integers that are $\geq r$, then $r_i(T) \leq r_i$ for each $i$. So $T \in \operatorname{Sub}_{r_1,\dots,r_n}$. $\endgroup$ – Zach Teitler Sep 7 '17 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.