# What are the higher homotopy groups of Spec Z ?

The homotopy groups of the étale topos of a scheme were defined by Artin and Mazur. Are these known for Spec Z? Certainly π1 is trivial because Spec Z has no unramified étale covers, but what is known about the higher homotopy groups?

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Since nobody else is answering this question I will give a few vague thoughts. Caveat: this is a completely speculative answer for various reasons, not least of which because I don't know Artin-Mazur's definition of pi_i.

Since X = Spec(Z) is "simply connected", one can pretend that the Hurewicz theorem applies.

I believe that H^i(X,Z/nZ) is trivial for all i, as a consequence of class field theory and the fact that Z has neither many units nor n-th roots of unity. (I'm not completely sure about i = 3 and n = 2 here.) One can then squint and imagine that the higher homotopy groups of X are trivial. This seems a little dodgy. Another direction one could go is to note that the groups H^i(X,G_m) vanish unless i = 3, and H^3(X,G_m) = Q/Z. From this (and other) facts it has been argued that X is analogous to the 3-sphere.

For what it is worth, both computations suggest that pi_2(X) is trivial. If one wanted to turn this comment into mathematics, one should try to define an algebraic Hurewicz map.

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Indeed H^i(X) is supposed to point out that X looks like a 3-sphere and it's not like homotopy groups of S^3 are easy. –  Ilya Nikokoshev Nov 3 '09 at 9:03
My memory is vague, but does H^3(X,G_m) = Q/Z imply that there are nontrivial elements in H^3(X,Z/2) arising from the kernel of the squaring map G_m -> G_m, since Z/2 and the 2nd roots of unity represent the same etale sheaf on this site? –  Tyler Lawson Nov 3 '09 at 12:23
\mu_2 is not etale over Spec(Z). –  moonface Nov 3 '09 at 13:52
The word "represent" may not have been the best to use. No, it's not etale, but it's a sheaf on the etale site as the kernel of the map G_m -> G_m. (The cokernel, if I remember correctly, is isomorphic the direct image of the additive group on the etale site of Z/2.) –  Tyler Lawson Nov 3 '09 at 15:02
Oh, I see what you mean. Thanks for clarifying! –  moonface Nov 3 '09 at 17:20