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Let $X$ be a (edit: nonsingular) projective complex algebraic curve. The genus of $X$ can be defined as the dimension of the space of holomorphic $1$-forms on $X$, which in turn can be defined either analytically or algebraically in terms of Kahler differentials. It can also be defined as the topological genus of $X$ considered as a surface, which in turn can be defined either topologically as the number of tori in a connected sum decomposition of $X$ or homologically in terms of the Betti numbers of $X$. Does anyone know of a reasonably self-contained reference where some or all of these equivalences are proven?

(There is a related question about computing the genus of a curve from its function field as well as a nice post by Danny Calegari explaining the relationship to the Newton polygon, but I am mostly interested in the algebraic-to-topological step of going from Kahler differentials to the number of tori in a connected sum decomposition.)

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For $\mathrm{dim} H^0(X, \Omega^1_X) = \dim H^1(X, \mathbb{Q})$ see http://en.wikipedia.org/wiki/Hodge_theory. For $\dim H^1(X, \mathbb{Q}) =$ number of tori use induction and the Mayer-Vietoris sequence.

(And for $\mathrm{dim} H^0(X, \Omega^1_X) = \mathrm{dim} H^1(X, \mathcal{O}_X)$ see http://en.wikipedia.org/wiki/Serre_duality.)

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  • $\begingroup$ All right, help me out here; which one of the things I said is the sheaf cohomology? $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2010 at 18:33
  • $\begingroup$ "the dimension of the space of holomorphic 1-forms on X" is $\mathrm{dim} H^0(X,\Omega^1_X)$. $\mathrm{dim} H^1(X,\mathbb{Q})$ is sheaf or singular cohomology. $\endgroup$
    – user19475
    Commented Jan 14, 2010 at 18:39
  • $\begingroup$ Sure, but which one of the things I said is $\dim H^1(X, \mathcal{O}_X)$? $\endgroup$
    – Qiaochu Yuan
    Commented Jan 14, 2010 at 18:44
  • $\begingroup$ None, I added it just for reference. $\endgroup$
    – user19475
    Commented Jan 14, 2010 at 18:46
  • $\begingroup$ @Qiaochu: There's a theorem called Serre Duality which tells you that for an Algebraic Curve (smooth, projective) you have $\dim H^1(X,O_X)=\dim H^0(X,\Omega^1_X)$, reference is Fulton's "Algebraic Curves" book, or Hartshorne, or Griffiths and Harris, or...any book on algebraic geometry, really. $\endgroup$
    – Charles Siegel
    Commented Jan 14, 2010 at 20:12

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