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Here's a question from someone who's just found out what Serre duality (in the case of curves) is.

It occurs to me that the popular statement which can also be interpreted as the Riemann-Roch theorem corresponds to $q=1$ in the general statement as formulated here: Now what about the case $q=0$, namely $\Gamma(C,\mathcal F)\cong H^1(C, \Omega_C\otimes \mathcal F^*)^*$? Why is this obvious/wrong/unimportant?

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It is certainly not wrong. I think that in practice one can often compute global sections of $\mathcal F$ directly, so the challenge is to figure out $H^1(\mathcal F).$ – Andrew Dec 11 '12 at 20:26
@Igor You got two good answers. Why don't accept one of them? – user26857 Jan 4 '13 at 8:32

Depending on what you call obvious, it might be - but it's neither wrong nor unimportant. Choosing $\mathcal F=\Omega_C$, it yields the following nice corollary: $$h^0(\Omega_C)=h^1(C,\Omega_C\otimes\Omega_C^\ast) = h^1(\mathcal O_C) = g(c),$$ i.e. the dimension of $\Gamma(C,\Omega_C)$ equals the genus of the curve. What you also get out of Serre Duality then is $h^1(\Omega_C)=1$ by choosing $\mathcal F=\mathcal O_C$. Now substituting these two in RR, you get $$\deg\Omega_C = 2g(C)-2.$$ Doesn't seem obvious to me, and I am sure there are a couple more applications.

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Actually, this is used this all the time; it is just another way of describing the $q=1$ case! A symmetric way to state the situation for curves is that, for any locally free sheaf of $\mathcal O_C$-modules $\mathcal F$ on $C$, there is a natural perfect pairing $$H^0(C,\mathcal F) \times H^1(C, \Omega_C\otimes \mathcal F^{\vee}) \to k$$ (where $k$ is the ground field). Replacing $\mathcal F$ by $\Omega_C \otimes\mathcal F^{\vee}$ shows the symmetric role played by the $H^0$ and $H^1$, and shows that the $q=0$ and $q=1$ statements from Wikipedia are equivalent (in the case of a curve).

In fact even more canonically, the pairing is $$H^0(C,\mathcal F) \times H^1(C,\Omega_C \otimes \mathcal F^{\vee}) \to H^1(C,\Omega^1),$$ the pairing being induced by cup product from $H^0 \times H^1$ to $H^1$, together with (the tensor product with $\Omega^1$ of) the natural evaluation map $\mathcal F\otimes \mathcal F^{\vee} \to \mathcal O_C$. This is then combined with the fact that there is a canonical isomorphism (the trace map, or --- in more topological terms --- the fundamental class) $H^1(\Omega^1_C) \cong k$. (So the fact that $H^1(\Omega_C^1)$ --- which was noted in another answer --- is not just an incidental consequence of Serre duality; it is one of the linchpins.)

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