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Geometric Genus

Algebraic Geometry
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Tsakanikas Nickos
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Geometric Genus

#1

Post by Tsakanikas Nickos » Wed May 11, 2016 11:06 am

Let $X$ be a non-singular, projective, rational variety. Show that its geometric genus $p_{g}(X)$ equals $0$.
Tsakanikas Nickos
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Re: Geometric Genus

#2

Post by Tsakanikas Nickos » Sun Nov 12, 2017 1:01 am

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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