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?

$\begingroup$ Does complex conjugation ever act trivially on a nontrivial fundamental group of a variety? $\endgroup$ – Kevin Casto Dec 26 '19 at 4:04

1$\begingroup$ @KevinCasto Maybe there is a variety with $\pi_1 = \mathbb{Z}/2$? $\endgroup$ – Theo JohnsonFreyd Dec 26 '19 at 4:27

$\begingroup$ @TheoJohnsonFreyd Fair enough  I guess I really had in mind the setting of the question, with an infinite fundamental group $\endgroup$ – Kevin Casto Dec 26 '19 at 4:37

$\begingroup$ @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? $\endgroup$ – user145520 Dec 26 '19 at 10:57

1$\begingroup$ @TheoJohnsonFreyd An Enriques surface has geometric fundamental group of order $2$. $\endgroup$ – Jef Dec 26 '19 at 18:06