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?



What are the higher homotopy groups of Spec Z ?
$Spec(\mathbb{Z})$ 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.


