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Let $C$ be a projective curve (over an algebraically closed field) of genus $\geq 1$. Let $S = C \times C$. By normalisation we have a ramified cover $C \to \mathbb{P}^1$ and so a map $p: S \to \mathbb{P}^1 \times \mathbb{P}^1$. I am interested in finding families of curves on $S$ passing through a given point $P \in S$. One obvious way to get such families is to consider a family in $\mathbb{P}^1 \times \mathbb{P}^1$ passing through $f(P)$ and then take its preimage under $p$.

My question is: are there any other families of curves that pass through $P$?

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To give a very simple-minded reason why the answer is to your question is "yes, there are other families", observe that if we have a family of curves on P^1 X P^1 through f(P) and take the preimage on S, then all the curves in the family will contain not only P, but also the points P_1, P_2, ...P_N in the fibre $p^{-1}(P)$. But a simple argument involving linear sections of a projective embedding of S says there are families of curves on S, all containing P, but not all containing any other point P'. –  Artie Prendergast-Smith Mar 13 '13 at 19:06
    
Of course, this is just really just a more elementary paraphrase of Mohan's answer. –  Artie Prendergast-Smith Mar 13 '13 at 19:25
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3 Answers

up vote 4 down vote accepted

$\mathrm{Pic}(\mathbb P^1\times \mathbb P^1)\simeq \mathbb Z\oplus\mathbb Z$, while $\mathrm{rank}\ \mathrm{Pic} (C\times C)\geq 3$ (the diagonal has self-intersection $2-2g$ and its intersection with both $\{q\}\times C$ and $C\times \{q\}$ is $1$, so the determinant of the intersection matrix of these three curves is $2g$ and hence they are linearly independent) and hence $$ p^*\mathrm{Pic}(\mathbb P^1\times \mathbb P^1)\subsetneq \mathrm{rank}\ \mathrm{Pic} (C\times C). $$

This implies that there are plenty (in fact most) ample line bundles and hence very ample ones on $C\times C$ that do not come from $\mathbb P^1\times \mathbb P^1$. These (at least the very ample ones) give you families that contain any given point (Just take hyperplane sections through that point for the corresponding embedding).


This is an answer to Dima's question in the comments below I need the more sophisticated LaTeX capabilities of an answer than a comment to do this right....

So, how do you decide that there is a positive dimensional sublinear system going through your point? If you're taking a very ample or at least a basepoint-free system of dimension at least $2$, then the answer is yes. Somewhat less is enough, but this is probably the easiest to check. Here is why.

The following short exact sequence tells you which elements of the linear system of $L$ pass through the point $P\in S$:

$$ 0\to \mathscr O_S(L)\otimes \mathfrak m_P\to \mathscr O_S(L) \to \kappa(P)\to 0 $$

where $\mathfrak m_P$ is the maximal ideal and $\kappa(P)$ is the residue field of $P\in S$.

So, you get that you have a long exact sequence: $$ 0\to H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)\to H^0(S,\mathscr O_S(L)) \to \kappa(P)\to \dots $$

The group $H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)$ represents those sections of $\mathscr O_S(L)$ that vanish at $P$, so there is a positive dimensional sublinear system going through your point if and only if $\dim H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)>1$.

If $L$ is basepoint-free, or more generally $P$ is not a basepoint of $L$, then the map $H^0(S,\mathscr O_S(L)) \to \kappa(P)$ is surjective and then the above condition is equivalent to $$\dim H^0(S,\mathscr O_S(L))>2.$$

If $P$ is a basepoint of $L$, then that map is zero and $H^0(S,\mathscr O_S(L)\otimes \mathfrak m_P)= H^0(S,\mathscr O_S(L))$, so it is enough that $\dim H^0(S,\mathscr O_S(L))>1$. But this actually seems harder to guarantee.

So, the short answer is this: If you take a generic projective embedding of $S$ and choose $L$ to be the corresponding very ample divisor, then there is a positive dimensional sublinear system of $L$ going through any given point.

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Even if one takes a very ample line bundle $L$ that is a pull back from $\mathbb P^1\times \mathbb P^1$, the general section of $L$ is not a pull back of a section on $\mathbb P^1\times \mathbb P^1$. –  rita Mar 13 '13 at 21:37
    
@rita: you're right, but that does not mean that this proof is wrong. actually, I think this is simpler than that.... –  Sándor Kovács Mar 13 '13 at 22:50
    
@Sandor: of course the proof is correct. I just wanted to point out that if you have a finite morphism $X\to Y$ of degree $>1$ then you can find divisors on $X$ that are not pull backs also when the Picard numbers of $X$ and $Y$ are equal. –  rita Mar 14 '13 at 8:38
    
Yes, I understand and agree. Ciao! –  Sándor Kovács Mar 14 '13 at 21:31
    
Dear Sandor, so basically the solution is to take a generic enough projective embedding. Is there a way to decide for a given line bundle $L$ (which is not a pullback of a line bundle form $\mathbb{P}^1 \times \mathbb{P}^1$) if there is a positive-dimensional linear subsystem of divisors corresponding to L passing through a given point? I guess this question might be too general. –  Dima Sustretov Mar 16 '13 at 9:36
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There are plenty of such families. For any smooth surface $S$ (not necessarily as in your question) and $p\in S$, if $L$ is a sufficiently ample line bundle and $I$ is the ideal sheaf defining $p$, the sections of $H^0(I\otimes L)$ will give curves passing through $p$. Thus taking such an $L$ in your case, no multiple of which comes from $\mathbb{P}^1\times\mathbb{P}^1$ (and there are plenty such), you can construct families which do not come from $\mathbb{P}^1\times\mathbb{P}^1$.

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Dear Mohan, how can one see that there exist line budles on $S$ that are not pullbacks of line bundles on $\mathbb{P}^1 \times \mathbb{P}^1$? Note that we are also not interested in line bundles that are pullbacks from the factors $C$ of $S$ as there are only finitely many sections of such that vanish at a given point of $S$. –  Dima Sustretov Mar 13 '13 at 18:48
    
In general, I do not see how one can estimate the dimension of $H^0(I \otimes L)$. One would need to show that it is bigger than one to answer the question. –  Dima Sustretov Mar 13 '13 at 18:51
    
Dima, you first question is answered in my answer bellow –  Sándor Kovács Mar 13 '13 at 19:07
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OK, here is an alternative more concrete solution that shows that there exist (several sets of) two families of curves such that with respect to any morphism $q:C\times C\to \mathbb P^1\times \mathbb P^1$ one of the families is not a pull-back via $q$. This is a partial answer to a follow up question asked in the comments to my other answer by Dima.


Let $\Delta\subset C\times C$ be the diagonal, $R_1=C\times \{q_1\}$, and $R_2=\{q_2\}\times C$ for some $q_1,q_2\in C$. Now let $L_a=a\Delta+nR_1+mR_2$ for some $a,n,m\in\mathbb N$, $a$ fixed and $n,m\gg 0$. For any given $b\in\mathbb N$, choose $n,m\gg 0$ such that they work for both $a=2b$ and $a=2b+1$ and fix these choices for those $a$'s

Using the intersection matrix of $\Delta, R_1,R_2$, which is $$ \left[\begin{matrix} 2-2g & 1 & 1\\ 1 & 0 & 1\\ 1& 1 & 0 \end{matrix}\right], $$ it is easy to check that $L_a$ is ample (for $n,m\gg 0$). Taking a multiple of $L_a$ (or $L_a$ for $a\gg 0$, in which case we need $n,m\gg a\gg 0$) gives you a desired family of curves as explained in my other answer.

Claim For any given morphism $q:C\times C\to \mathbb P^1\times \mathbb P^1$ and any given $a\in\mathbb N$

  1. $\Delta$ is not a pull-back of a divisor via $q$
  2. For any $a\in\mathbb N$, $L_a$ is not a pull-back of a divisor via (the original) $p$
  3. For any $b\in\mathbb N$, at most one of $L_{2b}$ and $L_{2b+1}$ may be the pull-back of a divisor via $q$

Proof
1. follows from the fact that if $\Delta$ were the pull-back of a divisor via $q$, then it would have to be the pull-back of an effective divisor, but the self-intersection of any effective divisor on $\mathbb P^1\times\mathbb P^1$ is non-negative.
2. follows from 1. and that $L_a-a\Delta$ is a pull-back of a divisor via $p$.
3. follows from 1. and that $L_{2b+1}-L_{2b}=\Delta$. $\square$

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