Given a geometrically connected variety $X$ over $\mathbb{Q}$ we have a short exact sequence $$ 1\to \pi_1(X_{\overline{\mathbb{Q}}})\to \pi_1(X)\to Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to 1. $$ A rational point on $X$ provides a splitting of this sequence i.e. it exhibits $\pi_1(X)$ as a semidirect product of $\pi_1(X_{\overline{\mathbb{Q}}})$ and the absolute Galois group of $\mathbb{Q}$. Does there exist a smooth and proper variety (to avoid trivialities with an infinite geometric fundamental group) such that $\pi_1(X)$ is actually the direct product of these two groups? Is there an example where such a splitting comes from a rational point?
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$\begingroup$ Does complex conjugation ever act trivially on a nontrivial fundamental group of a variety? $\endgroup$– Kevin CastoCommented Dec 26, 2019 at 4:04
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1$\begingroup$ @KevinCasto Maybe there is a variety with $\pi_1 = \mathbb{Z}/2$? $\endgroup$– Theo Johnson-FreydCommented Dec 26, 2019 at 4:27
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$\begingroup$ @TheoJohnson-Freyd Fair enough -- I guess I really had in mind the setting of the question, with an infinite fundamental group $\endgroup$– Kevin CastoCommented Dec 26, 2019 at 4:37
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$\begingroup$ @KevinCasto is your objection that conjugation has to interchange $H^{1, 0}$ and $H^{0, 1}$? Could the fundamental group be perfect to avoid that? $\endgroup$– user145520Commented Dec 26, 2019 at 10:57
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1$\begingroup$ @TheoJohnson-Freyd An Enriques surface has geometric fundamental group of order $2$. $\endgroup$– JefCommented Dec 26, 2019 at 18:06
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1 Answer
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If $X$ is an abelian variety, then it is not a direct product. For example, because the absolute galois group cannot act trivially on the torsion points (e.g. by Mordell-Weil). It follows that it is not a direct product for any $X$ with a non-trivial Albanese.
answered Dec 26, 2019 at 21:26