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What is the relation between the dimension of the space of Kähler differentials of a projective curve and the genus of the curve?

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How do you define the genus ? Have you heard about Riemann-Roch theorem ? – QiL'8 Apr 20 '12 at 14:45
Via Riemann-Roch. – averageman Apr 20 '12 at 14:59
Then input some particular divisor in RR. – QiL'8 Apr 20 '12 at 15:32
I guess I know the "wrong" version of R-R... (I know the version for rational functions). Can you point me to the "good" version. – averageman Apr 20 '12 at 15:56
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Ah. If $K$ is a canonical divisor, then for any divisor $D$, we have $$l(D)-l(K-D)=\deg D+1-g.$$ Note that $l(K)$ is the dimension of the space of the regular differential forms. – QiL'8 Apr 21 '12 at 0:43

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As QiL pointed out, you can use Riemann-Roch on a curve $C$ with $L=\mathcal O_C$ to get that $l(K_C)=g-1$, where $g=g(C)$ is the genus of the curve. But you don't really need R-R to see that. The geometric genus of the curve is defined to be $h^0(C,\mathcal O_C(K_C))$ which is exactly the dimension of the vector space of Kahler differentials on $C$. The arithmetic genus of the curve is defined to be $h^1(C,\mathcal O_C)$, and in the case of a curve these are the same by Serre duality. Here $l(K_C)=h^0(C,\mathcal O_C(K_C)) - 1$ by definition, which is the dimension of the projective space of Kahler forms.

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