What are the higher homotopy groups of Spec Z ?

Question

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?

Answer 1

$Spec(\mathbb{Z})$ should only be considered as $S^3$, if you "compactify" that is add the point at the real place. This is demonstrated by taking cohomology with compact support. The étale homotopy type of $Spec(\mathbb{Z})$ is however contractible (indeed what do you get by removing a point form a sphere?) to see this (all results apper in Milne's Arithmetic Dualities Book)

(let $X=Spec(\mathbb{Z})$ )

1. $H^r_c(X_{fl},\mathbb{G}_m)=H^r_{c}(X_{et},\mathbb{G}_m) = 0$ for $r \neq 3$.

2. $H^3_c(X_{fl},\mathbb{G}_m)=H^3_{c}(X_{et},\mathbb{G}_m) = \mathbb{Q}/\mathbb{Z}$

3. by 2+1, we have:

$H^3_c(X_{fl},\mu_n)= \mathbb{Z}/n$

$H^r_c(X_{fl},\mu_n)= 0$ for $r \neq 3$.

4. since we have a duality $$H^r(X_{fl},\mathbb{Z}/n)\times H^{3-r}_c(X_{fl},\mu_n) \to \mathbb{Q}/\mathbb{Z} $$ we have

5. $H^0(X_{fl},\mathbb{Z}/n) = H^0(X_{et},\mathbb{Z}/n) = \mathbb{Z}/n,$

$H^r(X_{fl},\mathbb{Z}/n) = H^r(X_{et},\mathbb{Z}/n) = 0$, $r >0$

6. Now since $\pi_1$ is trivial we have by the Universal Coefficient Theorem, the Hurewicz Theorem and the profiniteness theorem for 'etale homotopy that all homotopy groups are zero.

Answer 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.

Answer 3

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.