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Consider the Segre embedding $\mathbb{P}^n\times \mathbb{P}^n\rightarrow \mathbb{P}^N$, and let $S\subset\mathbb{P}^N$ be its image.

Then $rank(Z)\leq k$ implies that $Z\in Sec_k(S)$. Moreover if $Z\in Sec_k(S)$ is general then $rank(Z)\leq k$. Does this last statement hold for any $Z\in Sec_k(S)$ and not just for $Z\in Sec_k(S)$ general? What if we replace the Segre $S$ with the Veronese $V$ parametrizing rank $1$ symmetric matrices or the Grassmannian $\mathbb{G}(1,n)$ parametrizing rank $2$ anti-symmetric matrices?

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    $\begingroup$ What do you mean by rank ? Rank as a matrix or 'can be written as a sum of k rank 1 one matrices". I think you are asking about the latter which is border rank. As I recall the answer to the question "is border rank equal to rank?" is no. See J.M. Landsbergs book if this is what you are asking. $\endgroup$
    – meh
    Aug 27 '19 at 14:59
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    $\begingroup$ What I am asking is the following. Take any point $Z\in Sec_k(S)$. Is it true that $Z$ has rank at most $k$ or may it have rank $k+1$? $\endgroup$
    – user125056
    Aug 27 '19 at 15:14
  • $\begingroup$ But again, what do you call the rank of an element of $Sec_k(S)$? $\endgroup$
    – abx
    Aug 27 '19 at 17:09
  • $\begingroup$ The rank of $Z$ is the minimun numbers of points of $S$ such that we can write $Z$ as a linear combination of these points. $\endgroup$
    – user125056
    Aug 27 '19 at 18:09
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Your question is:

Let $S \subset \mathbb{P}^N$ be the image of the Segre map $\mathbb{P}^n \times \mathbb{P}^n \to \mathbb{P}^N$. Let $Z \in \operatorname{Sec}_k(S)$. Does $Z$ have rank at most $k$?

Yes. This is matrix rank. The elements of $\mathbb{P}^N$ are $(n+1) \times (n+1)$ matrices (up to a scalar factor) and rank with respect to $S$ is ordinary matrix rank. Every element of $\operatorname{Sec}_k(S)$ has rank at most $k$.

A modified question is:

Let $S \subset \mathbb{P}^N$ be the image of the Segre map $\mathbb{P}^{n_1} \times \dotsm \times \mathbb{P}^{n_s} \to \mathbb{P}^N$. Let $Z \in \operatorname{Sec}_k(S)$. Does $Z$ have rank at most $k$?

When $s \geq 3$, no. A tensor of border rank $k$ may have rank strictly greater than $k$ or $k+1$. An example is below.

In the symmetric case, the answer is yes for Veronese varieties of degree $2$ (corresponding to quadratic forms), and no for Veronese varieties of degree $d \geq 3$. For example, the homogeneous binary form $x y^{d-1}$ has rank $d$, but lies on $\operatorname{Sec}_2(V)$: $$ x y ^{d-1} = \lim_{t \to 0} \frac{(tx+y)^d - y^d}{dt}, $$ where each $(1/dt)((tx+y)^d - y^d) \in \operatorname{Sec}_2(V)$.

In the antisymmetric case I believe the answer is again no.

Landsberg's book is a good reference. See also more recently https://arxiv.org/abs/1811.12725 and https://arxiv.org/abs/1812.10267.

For an example of a tensor with border rank $k=2$ and rank greater than $2$ take $$ T = x y^{d-1} = x \otimes y \otimes \dotsm \otimes y + y \otimes x \otimes y \otimes \dotsm \otimes y + \dotsb + y \otimes \dotsm \otimes y \otimes x, $$ where $x,y \in \Bbbk^2$ are a basis. It visibly has rank at most $d$ and you can prove that it has rank equal to $d$. It has border rank $2$, since $T = \lim_{t \to 0} ((tx+y)^{\otimes d} - y^{\otimes d})/t$.

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    $\begingroup$ Well obviously I agree with your answer :) $\endgroup$
    – meh
    Aug 28 '19 at 1:37
  • $\begingroup$ Thank you very much. That is exaclty what I was asking. Is this true also for anti-symmetric matrices considering the Grassmannian of lines insted of the Segre? Do you have a reference for this? $\endgroup$
    – user125056
    Aug 28 '19 at 8:53

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