Do there exist two smooth projective schemes over $\mathbb{Q}$ that are etale homotopy equivalent and only one of them has a $\mathbb{Q}$point?
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1$\begingroup$ I'm having a hard time visualizing the etale homotopy type of $C = \{ x^2 + y^2 + z^2 = 0 \}\subseteq \mathbb{P}^2_{\mathbb{Q}}$. It looks like it has the same homotopy groups as $\mathbb{P}^1_\mathbb{Q}$, but I'm not sure if they are homotopy equivalent. $\endgroup$ – Piotr Achinger Apr 13 at 4:14

3$\begingroup$ Some work by Tomer Schlank might be of interest here. He proves that the exitence of a rational point "up to homotopy" (i.e. a section of the map induced by $X\to\mathrm{Spec}\,\mathbb{Q}$ on étale homotopy type) is equivalent to the vanishing of various descent obstructions, but there are schemes without actual rational points where these obstructions vanish. $\endgroup$ – Denis Nardin Apr 13 at 9:00

2$\begingroup$ @PiotrAchinger I think that there is no section for the map $Et(C)\to B\Gamma_\mathbb{Q}$ (by computing the basechange to $\mathbb{R}$ of the third homotopy obstruction class modulo 2). Since this map is canonical (being the first postnikov piece), this property distinguishes the two varueties you consider. You can try to pass to imaginary quadratic field, though. $\endgroup$ – S. carmeli Apr 13 at 11:06

$\begingroup$ related, via the section conjecture : Anabelian geometry with étale homotopy types Jakob Stix, Alexander Schmidt Annals of Mathematics 184 (2016), no. 3, 817868, to be found here math.unifrankfurt.de/~stix/publikationen.html $\endgroup$ – Niels Apr 15 at 7:39