Does every $\mathbb{P}^{19}\subset \mathbb{P}(\mathbb{C}^5\otimes\mathbb{C}^5)$ intersect the Segre variety of rank one matrices in at least a $\mathbb{P}^1$? A naive dimension count suggests this is possible. The intersection is a $3$fold and I would be happy for any qualitative information about it.

1$\begingroup$ I think arxiv.org/abs/1410.7803v2 should tell you something about this 3fold; I believe that corollary 5.5 tells you that it's a CalabiYau birational to a quintic threefold? I'm very unsure of this (I haven't read the paper in depth). $\endgroup$ – dhy Dec 15 '14 at 0:57

4$\begingroup$ Correction: It's a bit more complicated than that (the quintic threefold will be singular, and this 3fold will be a resolution (see arxiv.org/abs/math/0508127v2). In any case it looks like the BPS numbers of this CalabiYau threefold have been calculated in arxiv.org/abs/1101.2746v2. In particular, I believe that 41.1 shows that there are 100 lines on a generic such threefold (and in particular that there are at least 100 on any such threefold). $\endgroup$ – dhy Dec 15 '14 at 1:29

$\begingroup$ @dhy: I don't know much about this stuff, but are you sure these "virtual" counts imply anything about the number of actual lines? $\endgroup$ – Daniel Litt Dec 15 '14 at 1:57

1$\begingroup$ @DanielLitt: "... are you sure these 'virtual' counts imply anything ...?" Yes, the virtual count equals the actual number if the set is "transversal". Since the empty set is transversal, a nonzero virtual count implies that there exists at least one line. Since that is all that Landsberg asked, dhy has answered Landsberg's question. $\endgroup$ – Jason Starr Dec 15 '14 at 2:11
Although dhy's comment does solve this problem, I thought I would point out another approach. Let $U$, $V$ and $W$ be $5$dimensional $k$vector spaces. Let $$B: U\times V \to W,$$ be a bilinear pairing. Landsberg's question, essentially, asks whether there exists a $2$dimensional subspace, $S\subset U$, and a $1$dimensional subspace, $T\subset V$, such that $B$ is zero on $S\times T$. On the parameter space $\text{Grass}(2,U)\times \mathbb{P}V$ for pairs $([S],[T])$, this is a straightforward "degeneracy locus" problem. Denoting by $$ \sigma: U^\vee\otimes_k \mathcal{O}_{\text{Grass}} \to \mathcal{S}^\vee$$ the universal rank $2$ locally free quotient on $\text{Grass}(2,U)$, and denoting by $$ \tau: V^\vee \otimes_k \mathcal{O}_{\mathbb{P}V} \to \mathcal{T}^\vee$$ the universal rank $1$ locally free quotient on $\mathbb{P}V$, then $B$ induces a global section, say $\widetilde{B}$, of the rank $10$ locally free sheaf on $\text{Grass}(2,U)\times \mathbb{P}V$, $$\textit{Hom}_{\mathcal{O}}(\text{pr}_{\text{Grass}}^*\mathcal{S} \otimes_{\mathcal{O}} \text{pr}_{\mathbb{P}V} \mathcal{T}, W\otimes_k \mathcal{O}).$$ This rank $10$ locally free sheaf on the $10$dimensional projective scheme $\text{Grass}(2,U)\times \mathbb{P}V$ has top Chern class $$5\text{pr}_{\text{Grass}}^*[c_1(\mathcal{S}^\vee)^4\cap c_2(\mathcal{S}^\vee)] \cap \text{pr}_{\mathbb{P}V}^*[c_1(\mathcal{T}^\vee)^4].$$ The total degree of this Chern class is $10$. Since it is nonzero, the zero scheme of $\widetilde{B}$ is nonempty, i.e., there exists a line $\mathbb{P}S$ in $\mathbb{P}U$ and a singleton $\mathbb{P}T$ in $\mathbb{P}V$ such that the image under the Segre morphism of $\mathbb{P}S\times \mathbb{P}T$ is a line in $\mathbb{P}(U\otimes_k V)$ that is contained in the zero scheme of the linear form $B$.
Of course, by a symmetric argument, there also exists a $1$dimensional subset $S$ of $U$ and a $2$dimensional subspace $T$ of $V$ such that $\mathbb{P}S\times \mathbb{P}T$ is in the zero locus of $B$. Thus, we know that there are at least $2$ lines. If $B$ is general, there should be precisely $20$ lines contained in the intersection of the Segre locus and the zero locus of $B$.
Edit. As Wajcha correctly points out, I had the wrong computation for $c_2(\text{pr}_{\text{Grass}}^*\mathcal{S}^\vee\otimes_{\mathcal{O}}\text{pr}_{\mathbb{P}V}^*\mathcal{T}^\vee)$. The correct formula is $$ \alpha = \text{pr}_{\text{Grass}}^*c_2(\mathcal{S}^\vee) + \text{pr}_{\text{Grass}}^*c_1(\mathcal{S}^\vee)\cap\text{pr}_{\mathbb{P}V}^*c_1(\mathcal{T}^\vee) + \text{pr}_{\mathbb{P}V}^*c_1(\mathcal{T}^\vee)^2. $$ Therefore, the top Chern class of the rank 10 locally free sheaf equals $$ \alpha^5 = 5\text{pr}_{\text{Grass}}^*[c_1(\mathcal{S}^\vee)^4\cap c_2(\mathcal{S}^\vee) +6c_1(\mathcal{S}^\vee)^2\cap c_2(\mathcal{S}^\vee)^2+ 2c_2(\mathcal{S}^\vee)^3]\cap \text{pr}_{\mathbb{P}V}^*c_1(\mathcal{T})^2. $$ Via Schubert calculus, the term in square brackets is $10$ times the class of a point (rather than just $2$, as I computed before). Therefore $\alpha^5$ has total degree $50$. So, if $B$ is generic, there are $50$ lines in the intersection lying in the first "ruling" of the Segre variety by linear $\mathbb{P}^4$s. By a symmetric argument, there are also $50$ lines lying in the second ruling. Therefore, altogether, there should be $100$ lines, just as in the article cited by dhy.

$\begingroup$ Thanks! This works and generalizes from 5 to m, 19 to m^2m1 and 1=21 to sqrt{m1}1 when this is an integer $\endgroup$ – JM Landsberg Dec 15 '14 at 19:25
This is a comment, but I cannot comment as my reputation is too low :/
I'm very sorry: why is this the top class? I.e. why we do not get additionally $pr^*_{Grass}[c_2(S^\vee)^2c_1(S^\vee)^2]$ (giving $30=5*6$) and $c_2(S^\vee)^3$ (giving $10$)? Did I miscompute that the top class is $(c_2(S^\vee)+c_1(T^\vee)c_1(S^\vee)+c_1(T^\vee)^2)^5$ or I have an error elswhere?

1$\begingroup$ You are correct. I made a mistake. I will fix it now. $\endgroup$ – Jason Starr Dec 15 '14 at 11:52