# The direct product of the geometric fundamental group and the absolute Galois group

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?

• Does complex conjugation ever act trivially on a nontrivial fundamental group of a variety? Dec 26 '19 at 4:04
• @KevinCasto Maybe there is a variety with $\pi_1 = \mathbb{Z}/2$? Dec 26 '19 at 4:27
• @TheoJohnson-Freyd Fair enough -- I guess I really had in mind the setting of the question, with an infinite fundamental group Dec 26 '19 at 4:37
• @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?
– user145520
Dec 26 '19 at 10:57
• @TheoJohnson-Freyd An Enriques surface has geometric fundamental group of order $2$.
– Jef
Dec 26 '19 at 18:06

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.