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?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – user41650
    Jul 27, 2020 at 5:14
  • $\begingroup$ 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 $\endgroup$
    – user41650
    Aug 4, 2020 at 15:33
  • $\begingroup$ 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 $Y$. $\endgroup$
    – user41650
    Aug 4, 2020 at 18:33

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