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 $P$?



families of curves on surfaces which are products of curves
$\mathrm{Pic}(\mathbb P^1\times \mathbb P^1)\simeq \mathbb Z\oplus\mathbb Z$, 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).




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*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.


