# Rank of matrices and secant varieties

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?

• 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.
– meh
Aug 27 '19 at 14:59
• 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 $k+1$? Aug 27 '19 at 15:14
• But again, what do you call the rank of an element of $Sec_k(S)$?
– abx
Aug 27 '19 at 17:09
• 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. Aug 27 '19 at 18:09

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 $$T = \lim_{t \to 0} ((tx+y)^{\otimes d} - y^{\otimes d})/t$$.

• Well obviously I agree with your answer :)
– meh
Aug 28 '19 at 1:37
• 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? Aug 28 '19 at 8:53