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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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 Kähler 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 Kähler forms.