# The locus of lines intersecting with another fixed line on a Fano threefold

Let $$Y$$ be an index $$2$$, degree $$5$$, Picard number $$1$$ Fano threefold, i.e $$Y$$ is a linear section of Grassmannian $$\operatorname{Gr}(2,5)$$. Let $$\Sigma(Y)$$ be the Hilbert scheme of lines on $$Y$$, it is isomorphic to $$\mathbb{P}^2$$. Let $$\mathcal{B}\in \lvert\mathcal{O}_Y(2)\rvert$$ be a smooth quadric hyersurface, it is a degree $$10$$ K3 surface. Now, I consider the following two situations:

1. I fix a line $$L_1\in Y$$, consider all lines $$L_t$$ intersects with $$L_1$$. Since the intersection with a fixed line is a codimension $$1$$ condition, I think such a family of lines is parametrized by $$\mathbb{P}^1$$? Or at least, can I choose a pencil of lines intersecting with the fixed $$L_1$$?

2. I consider a family of lines $$L_t$$ tangent to $$\mathcal{B}$$, is this family also a $$\mathbb{P}^1$$ or just a smooth curve?

Maybe the general question is how to describe those families rigorously?

Question 1. Let $$I(Y) \subset \Sigma(Y) \times \Sigma(Y) \cong \mathbb{P}^2 \times \mathbb{P}^2$$ be the incidence scheme (parameterizing pairs of intersecting lines). Then $$I(Y) \cong \mathrm{Fl}(1,2;3) \subset \mathbb{P}^2 \times \mathbb{P}^2$$; I think you can find this in Sanna, Giangiacomo. Small charge instantons and jumping lines on the quintic del Pezzo threefold. Int. Math. Res. Not. IMRN 2017, no. 21, 6523-6583. In particular, lines intersecting a given line $$L$$ are parameterized by $$p_1(p_2^{-1}([L])) \subset \Sigma(Y)$$ which is indeed a line on $$\mathbb{P}^2$$ (here $$p_i$$ denote the projections of $$I(Y)$$ to the factors).

Question 2. Recall that $$Y \subset \mathbb{P}^6 = \mathbb{P}(V)$$. In particular, every line on $$Y$$ is a line in $$\mathbb{P}(V)$$. This defines an embedding $$\Sigma(Y) \to \mathrm{Gr}(2,V).$$ It is defined by a rank-2 vector bundle $$\mathcal{U}$$ on $$\Sigma(Y)$$. A description of this bundle can be found in the same reference, for now it is important that $$\det(\mathcal{U}) \cong \mathcal{O}(-3)$$. A quadric in $$Y$$ is cut out by a quadric in $$\mathbb{P}(V)$$; its equation is in $$S^2V^\vee$$, and it induces a global section of $$S^2\mathcal{U}^\vee$$. The tangency locus is the degeneracy locus of the corresponding section of $$S^2\mathcal{U}^\vee$$, or equivalently of the induced morphism $$q \colon \mathcal{U} \to \mathcal{U}^\vee.$$ Its equation is $$\det(q) \colon \mathcal{O}(-3) \cong \det(\mathcal{U}) \to \det(\mathcal{U}^\vee) \cong \mathcal{O}(3)$$; thus the the tangency locus is a sextic curve in $$\Sigma(Y) \cong \mathbb{P}^2$$. For general $$q$$ it is smooth, but it is not true that it is smooth for any smooth quadric divisor --- if, for instance, a divisor contains a line, this line is contained in the tangency locus and gives a singular point on it.

• Thanks so much! In my case, the real thing I am considering is the curve $\Gamma$ as a double cover of this sextic curve on X_{10}, and the ruled surface $S$ over this $\Gamma$ and the fibers are exactly coming from all the lines intersecting $\Gamma$. And your comment on if a line is in this branch divisor , then it will give a singular point on $\Gamma$ also clears my confusion Jul 27 '20 at 5:14
• Hi Sasha, I have a confusion. The vector space V is dimension 7 right? So I think Gr(2,V) and $\mathbb{P}^2=\Sigma(Y)$ is embedding to $\mathbb{P}^{20}$, so this is Veronese degree 5 embedding, so deg($\mathcal{U}$) should be $O(-5)$? Otherwise, I think the dimension of $S^2V$ and $H^0(Y,S^2\mathcal{U}^{\vee})$ does not match, the previous one is 28 and the latter seems is 10 Aug 4 '20 at 15:33
• Sorry, I think the dimension of $H^0(Y,S^2\mathcal{U}^{\vee})$ is of dimension 24. By the way, the computation of $ch(S^2\mathcal{U}^{\vee})$ in the paper you mentioned should be wrong, the correct number should be $3+3H+\frac{27}{2}L-\frac{1}{2}P$, thus $\chi(S^2\mathcal{U}^{\vee})=3+\frac{8}{3}\times 3+\frac{27}{2}-\frac{1}{2}=24$. I think higher cohomology would vanish. Of course, I think the bundle $\mathcal{U}^{\vee}$ should be on $\Sigma(Y)$ instead of $Y$, it seems that the vector bundle $\mathcal{U}$ in that paper is the tautological bundle on $Y$. Aug 4 '20 at 18:33