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 Post subject: Geometric GenusPosted: Wed May 11, 2016 11:06 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
Let $X$ be a non-singular, projective, rational variety. Show that its geometric genus $p_{g}(X)$ equals $0$.

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 Post subject: Re: Geometric GenusPosted: Sun Nov 12, 2017 1:01 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
Let $n = \dim X$. As $X$ is rational, (by definition) $X$ is birationally equivalent to $\mathbb{P}^{n}$, and since the geometric genus $p_{g}(X) (= P_{1}(X) = \dim H^{0} (X, \omega_{X} ) )$ is a birational invariant, we have that $p_{g} (X) = p_{g} (\mathbb{P}^{n})$. But the canonical sheaf of $\mathbb{P}^{n}$ is
$\omega_{\mathbb{P}^{n}} \cong \mathscr{O}_{\mathbb{P}^{n}} (-n-1) ,$thus it has no (non-trivial) global sections. Hence
$p_{g} (X) = p_{g} (\mathbb{P}^{n}) = \dim H^{0} ( \mathbb{P}^{n}, \omega_{\mathbb{P}^{n}} ) = 0.$

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