More specifically, what facts do you need to know to conclude $H^2(V) = H^{1,1}$? In general, are there hypersurfaces in $CP^n$ without holomorphic $k$-forms for some $k$?
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3$\begingroup$ All you need for a nonsingular hypersurface in $\mathbb{P}^n$ is that $\deg V\le n$. For $0<k<\dim V$, there's never a $k$-form. $\endgroup$– Alex DegtyarevMar 18, 2016 at 8:39
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$\begingroup$ This is probably very trivial but I am not seeing it for some reason. How is the adjunction formula enforcing the conditions you wrote above? $\endgroup$– mokimMar 18, 2016 at 16:06
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1 Answer
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Nothing more than adjunction formula, see here, knowing that $\omega_{\mathbb{P}^n} \cong \mathcal{O}_{\mathbb{P}^n}(-n-1)$.
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$\begingroup$ 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. $\endgroup$– EnricoMar 18, 2016 at 15:43