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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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2 Answers 2

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

If etale pi_1 classifies obstructions to trivializing finite flat unramified Z-algebras, it would be nice if the whole etale homotopy type classified obstructions to trivializing simplicial commutative Z-algebras that were finite, flat, and unramified in a homotopy sense. I think all of these notions make sense: "finite" means that the homotopy groups vanish in high degrees and are finitely generated, "flat" means that these homotopy groups have no torsion, and "unramified" means that the cotangent complex is zero. Is that right?

Presumably algebraic topologists have thought about the sphere spectrum version of this question. Are there any connective E-infinity ring spectra that are finite, flat, and unramified over the sphere?

After Tyler's comments, I see that this is a bad analogy. The dictionary between etale locally constant sheaves of sets and finite flat unramified algebras (which in one direction takes an algebra and associates the sheaf of sections of its spectrum over Spec Z) just doesn't extend to a dictionary between homotopy-style locally constant sheaves and homotopy-style finite flat unramified algebras.

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With regards to your second question: This appears in John Rognes' paper on Galois theory of structured ring spectra, at least in part. There are none if you ask that the generators all live in pi_0. If you allow new generators in positive degrees, there are square-zero extensions and their ilk, and more following those that are harder to classify. I am not sure about your first question. –  Tyler Lawson Nov 3 '09 at 21:52
Are you really saying that there are square-zero extensions of the sphere spectrum that are unramified? This isn't possible with plain rings. –  David Treumann Nov 3 '09 at 22:55
I guess I was thinking by analogy with your "homotopy vanishing in high degrees" examples above. No, those can't ever have trivial cotangent complex, and thickenings of a commutative Z-algebra to something with positive homotopy groups usually can't either; e.g. if R -> S is an isomorphism on pi_0 then the first nonvanishing relative homotopy group coincides with the first nonzero homotopy group of the cotangent complex. –  Tyler Lawson Nov 4 '09 at 0:35

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