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.

  • 1
    $\begingroup$ 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 $5$. $\endgroup$ – 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 $X$.


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