# Can we move curves which are members of very ample systems?

Let us take the second degree Hirzebruch surface F_2 which is a holomorphic CP^1 bundle over CP^1 having sections of self intersections +2 and -2. Let me denote the class of the -2 section by C and class of the sphere fiber by F. Picard group of F_2 is generated by C and F.

1) It is known that the linear system |2C+5F| is very ample and it contains an irreducible curve, let me denote it by 2C+5F. On the other hand, I take the linear system |C+2F|. By the same proposition in Hartshorne, this system is not very ample but contains an irreducible curve, denoted by C+2F.

The curves (2C+5F) and (C+2F) intersect at 5 points generically. My question is : Can we move 2C+5F so that it intersects C+2F at one point with multiplicity 5? But while moving I do not want to change the genera and self intersections. Or let me put it another way; is there a member of the very ample system |2C+5F| which intersects a member of |C+2F| at one point with multiplicity 5?

As I know in very ample systems there is more freedom, since they are not rigid. So I was hoping the curve 2C+5F in the way that I want, but I could not find any proof in relation to these.

2) My second question: The fiber F and the -2 section C of F_2 intersect once, say, at the point p. Can we find a member of the very ample linear system |2C+5F| which passes through this point p such that, at p, C and 2C+5F intersect once, and F and 2C+5F intersect once but with multiplicity 2?

I would appreciate any suggestions or resources. Thanks.

• It looks to me like one positive solution of this problem boils down to an enumerative computation: for a general divisor class $D$ of degree $4$ on a genus $2$ curve $X$, there are $10$ ordered pairs $(p,q)\in X\times X$ such that $D\sim -\underline{p} +5 \underline{q}$ (if I computed correctly). For $D$ equal to $2K_X$, $6$ of the $10$ solutions come from Weierstrass points, $(p,q)=(r,r)$. However, the $4$ remaining solutions give a $g^3_5$ of $2K_X+\underline{p}$ lying on a singular quadric cone and having an osculating hyperplane with contact order $5$. – Jason Starr Feb 5 at 15:19

I am just writing up my comment as an answer. For every smooth, genus $$2$$ curve $$X$$, the image of the Abel map, $$u_1:X\to \text{Pic}^1(X), \ \ p\mapsto [\mathcal{O}_X(\underline{p})],$$ is a Cartier divisor in the algebraic equivalence class of the theta divisor $$\theta$$. Similarly, for every integer $$n\neq 0$$, the composite morphism $$u_n$$, $$X\to \text{Pic}^1(X) \xrightarrow{m_n} \text{Pic}^n(X), \ \ p\mapsto [\mathcal{O}_X(n\underline{p})],$$ is a Cartier divisor in the algebraic equivalence class of $$|n|\cdot \theta.$$ In particular, for every Cartier divisor class $$D$$ on $$X$$ of degree $$4$$, the two divisor classes $$D+[u_1(X)]$$ and $$[u_5(X)]$$ on $$\text{Pic}^5(X)$$ are $$\theta$$ and $$5\theta$$. Thus, the intersection number of these divisor classes is $$(D+[u_1(X)])\cdot [u_5(X)] = 5\theta^2 = 10[\{*\}].$$ Thus, for every degree $$4$$ divisor class $$D$$, when counted with the appropriate multiplicities, the following scheme has length $$10$$, $$Z_D:=\{(p,q)\in X\times X | 5\underline{q} \sim \underline{p} + D \}.$$
When $$D$$ is the divisor class $$2K_X$$, then $$Z_D$$ contains $$(r,r)$$ for each of the $$6$$ Weierstrass points $$r$$ of $$X$$. Since $$10>6$$, there exists $$(p,q)\in Z_D$$ with $$p\neq q$$. Now consider the complete linear system of the divisor class $$\underline{p} + 2K_X$$. By Riemann-Roch and Serre duality, this complete linear system is a $$g^3_5$$ on $$X$$ that embeds $$X$$ as a degree $$5$$ curves in $$\mathbb{P}^3$$.
Moreover, the composition of this embedding with linear projection away from $$p$$ is a $$g^2_4$$ on $$X$$ given by the complete linear system of $$2K_X$$, i.e., it is the composition of the canonical $$g^1_2$$ from $$X$$ to $$\mathbb{P}^1$$ and the Veronese $$2$$-uple map from $$\mathbb{P}^1$$ to a smooth conic $$\overline{Q}$$ in $$\mathbb{P}^2$$. Since linear projection away from $$p$$ factors through the conic, the image of $$X$$ in $$\mathbb{P}^3$$ is contained in the singular quadric cone $$Q$$ over the smooth plane conic $$\overline{Q}$$.
Of course the minimal desingularization of $$Q$$ is a Hirzebruch surface $$\mathbb{F}_2$$. Also, the divisor class of the image of $$X$$ in $$\mathbb{F}_2$$ equals $$2C+5F$$. Finally, by construction, the osculating hyperplane $$H_q$$ to $$X$$ in $$\mathbb{P}^3$$ at $$q$$ has contact order $$5$$ with $$X$$. The hyperplane section $$H_q\cap Q$$ is a plane conic contained in $$Q$$ whose divisor class in $$\mathbb{F}_2$$ equals $$C+2F$$. Since $$q$$ does not equal $$p$$, and since $$X$$ does contain $$p$$, the hyperplane section is not reducible (every reducible hyperplane section of $$Q$$ contains the vertex $$p$$ of the cone). Thus, the hyperplane section is a smooth curve $$D$$. Altogether, the strict transforms of $$X$$ and $$D$$ in $$\mathbb{F}_2$$ are smooth curves with respective divisor classes $$2C+5F$$ and $$C+2F$$ such that $$X\cdot D$$ equals $$q$$ with multiplicity $$5$$.
The answer to the second question is similar and easier: just use the linear system $$\underline{r}+2K_X$$ on $$X$$ where $$r$$ is one of the $$6$$ Weierstrass points of $$X$$.