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I noticed that the holomorphic Euler characteristic $\chi(C,\mathcal{O}_C)=1-g$ of a smooth complex curve $C$ of genus $g$ is just a half of the topological Euler characteristic $\chi_{top}(C)=2-2g$. Do they coincide by accident or is there any good explanation of this?

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I think of this as a special case of a massive and beautiful theory called Hodge theory. Hodge theory builds all sorts of connections between holomorphic and topological data.

Define $h^{p,q}=\dim_\mathbb{C} H^q(C,\Omega^p)$ the dimension of the $q$-th cohomology of the sheaf of holomorphic $p$-forms. One of the big theorems in Hodge theory says that these so-called Hodge numbers satisfy a symmetry: $h^{p,q}=h^{q,p}$.

Another theorem in Hodge theory says that the cohomology breaks up (has a "Hodge decomposition") as follows $$H^m(C, \mathbb{C})\simeq \bigoplus_{p+q=m}H^q(C,\Omega^p).$$

Thus the Betti numbers are related to this holomorphic data.

The above two theorems tell us $\chi(C)=b_2(C)-b_1(C)+b_0(C)=2-(h^{0,1}+h^{1,0})=2-h^{1,0}-h^{1,0}=2-2g$.

On the other hand $\chi(\mathcal{O}_C)=h^{0,0}-h^{0,1}=1-g$.

This probably isn't what you are looking for because you already know a way to calculate these things which led you to the question in the first place. I guess the takeaway I'm going for is that both can be expressed in terms of the Hodge numbers using Hodge theory.

Another interesting and related phenomena is that for any smooth, compact, holomorphic manifold, the odd Betti numbers must be even by the above symmetry.

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Matt, thank you for sharing your idea with me. Yes, maybe Hodge theory play a role behind this coincidence. I feel that there may be more than Hodge theory. Anyways, thank you for the wonderful post. – M. K. Nov 27 '12 at 4:06

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