Timeline for Is it easy to see that a cubic surface $V$ in $CP^3$ has no holomorphic 2-forms?
Current License: CC BY-SA 3.0
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| Mar 18, 2016 at 15:43 | comment | added | Enrico | In terms of canonical class, if $X$ is a smooth surface of (to stay in your example), degree 3 in $\mathbb{P}^3$, you have $$ K_X= (K_{\mathbb{P}^3}+[X])|_X,$$ and this gives to you $K_X \cong \mathcal{O}_X(-1)$. Then you use $$ H^2(X) \cong H^0(X, K_X) \oplus H^{1,1}(X) \oplus H^2(X, \mathcal{O}_X)$$ with the first (and then the third) term being zero, since $\mathcal{O}_X(-1)$ is negative. | |
| Mar 18, 2016 at 15:07 | comment | added | mokim | Could you elaborate a little bit more? | |
| Mar 18, 2016 at 11:42 | history | answered | Enrico | CC BY-SA 3.0 |