3
$\begingroup$

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.

$\endgroup$
  • 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
0
$\begingroup$

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 $X$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.