Etale $\pi_1$ of Grassmannian I am forced to know the etale fundamental group of the grassmannian over the rational field. I searched it but couldn't find any hint. I am wondering whether there are some positive results or recipe to compute it.
Thank you!
 A: The answer is that any Grassmannian is geometrically simply connected, so the etale fundamental group over $\mathbb{Q}$ is simply [edit: !!] the absolute Galois group $\operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$.
In more detail: let $X$ be a geometrically integral variety defined over $\mathbb{Q}$, let 
$\overline{X}$ be its basechange to an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $\mathfrak{g}_{\mathbb{Q}} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ be the absolute Galois group of $\mathbb{Q}$.  
1) Via a choice of geometric points (which we suppress), we get a short exact sequence of profinite groups
$1 \rightarrow \pi_1(\overline{X}) \rightarrow \pi_1(X) \rightarrow \mathfrak{g}_{\mathbb{Q}} \rightarrow 1$.  
2) Let $K$ be an algebraically closed field of characteristic $0$.  Suppose that either $X$ is complete or that $X$ is nonsingular [in our application, both hold].  Then the natural map
$\pi_1(\overline{X}) \rightarrow \pi_1(\overline{X} \otimes K)$ is an isomorphism.  
[Comment: if instead of $\mathbb{Q}$, our ground field was a field $k$ of positive characteristic and $K$ is an algebraically closed field containing $\overline{k}$, this map is still an isomorphism for complete varieties but not necessarily for all smooth varieties and indeed, not even for the affine line!]
3) Take $K = \mathbb{C}$.  Then the analytification functor induces an isomorphism from the 
profinite completion of the topological fundamental group of $X(\mathbb{C})$ (with the $\mathbb{C}$-analytic topology) to the etale fundamental group $\pi_1(X \otimes \mathbb{C})$.  
4) If $X/\mathbb{Q}$ is any Grassmannian, then it is nonsingular, complete and its analytification is the usual complex Grassmannian, which is simply connected: see e.g.
http://books.google.com/books?id=WHjO9K6xEm4C&pg=PA748&lpg=PA748&dq=simply+connected+Grassmannian&source=bl&ots=waYTv_whVx&sig=ErlPHYKL5FdQPUdBlIrIgYQGhbE&hl=en&ei=9FiITJuNNYjW9ASP_p3fDg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CC8Q6AEwBTgK#v=onepage&q&f=false
Therefore putting the previous parts together the geometric fundamental group $\pi_1(\overline{X})$ is trivial, so we naturally have $\pi_1(X) \cong \mathfrak{g}_{\mathbb{Q}}$.
References for these facts may be found, for instance, in Chapter 5 of Szamuely's Galois Groups and Fundamental Groups.  
