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Assume we have a smooth algebraic curve $C$ over complex numbers. Consider then all possible embeddings of $C$ into a smooth algebraic surface $X$.

Here are two questions:

1) How many are there different such embeddings for fixed $C$ and $X$? (in moduli-theoretic sense)

(I understand that this must heavily depend on $X$: type, algebraic structure etc. as well as on $C$).

2) Is there a way to characterize the divisor classes on $C$ corresponding to normal bundles of $C$ for various such embeddings?

Here $X$ is still fixed. A generalized version of this question for $C$ fixed and $X$ varying in the moduli space of the type to which $X$ belongs is partially answered by Sasha below, where he chose $X$ to vary in the class of geometrically ruled surfaces.

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  • $\begingroup$ For any line bundle $L$ on $C$ you can embed $C$ into $P_C(O \oplus L)$ as a section in such a way that the normal bundle is $L$, so the normal bundle can be arbitrary. $\endgroup$
    – Sasha
    Commented Oct 6, 2013 at 5:28
  • $\begingroup$ Sasha, thank you. This is elementary, of course, but somehow I was not able to get it at the moment of writing my comment. $\endgroup$
    – N B
    Commented Oct 6, 2013 at 14:23
  • $\begingroup$ N B and @Sasha: I thought the problem was for fixed $C$ and $X$? Those projective bundles certainly won't be isomorphic for all $L$... $\endgroup$
    – user5117
    Commented Oct 6, 2013 at 16:58
  • $\begingroup$ Artie, thank you. I apologize for not formulating it clearly. Actually, Sasha answered the version of the question stated in parentheses which allows to vary $X$ (he chose $X$ to be geometrically ruled). When one fixes $X$, it is still not answered here. $\endgroup$
    – N B
    Commented Oct 6, 2013 at 17:12
  • $\begingroup$ By the way, Sasha, I think your argument gives the answer for the divisors of the same degree, this is when you vary the algebraic structure of $X$, while topologically $X$ is the same. One can do it for every fixed degree in this way, of course. $\endgroup$
    – N B
    Commented Oct 6, 2013 at 17:27

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