$G$ is quasisplit at almost all places

Let $G$ be a connected, reductive group over a global field $k$. I am trying to understand why $G_v = G \times_k k_v$ is quasisplit for almost all places $v$ of $k$. There are several equivalent formulations of quasi-split, but none of them seem to be easy to work with:

1 . $G_v$ is quasisplit if and only if every/at least one minimal parabolic subgroup is solvable.

2 . Let $S_v$ be a maximal $k_v$-split torus of $G_v$, and let $T_v$ be a maximal torus of $G_v$ which is defined over $k_v$ containing $S_v$. There exist systems of positive roots for $T_v$ and $S_v$ such that positive roots of $T_v$ restrict to nonnegative characters. Let $\Delta$ be a base for a system of positive roots on $T_v$. Then $G_v$ is quasisplit if and only if none of the $\alpha \in \Delta$ are trivial on $S_v$.

3 . $G_v$ is quasisplit if and only if the following holds: whenever $P$ is a parabolic subgroup of $G_v \times_{k_v} \overline{k_v}$, $\gamma \in \textrm{Gal}((k_v)_s/k_v)$, and $\gamma(P)$ and $P$ are conjugate, then $P$ is conjugate to a parabolic subgroup which is defined over $k_v$.

The first two formulations seem especially difficult to work with. For example, in the second formulation, I see no way to control how the $S_v$ are chosen. For the third one, I thought one might be able to proceed by choosing in the beginning a Galois extension $E$ of $k$ over which $G$ splits, and for each finite place $v$ of $k$, choosing a place $w$ of $E$ lying over $v$. Then $G_v$ splits over $E_w$, and for the Galois action one can work with the subgroup

$$\textrm{Gal}(E_w/k_v) = \{ \sigma \in \textrm{Gal}(E/k) : \sigma \mathfrak P_w = \mathfrak P_w \}$$

where $\mathfrak P_w$ is the prime of $\mathcal O_E$ corresponding to $w$.

I would appreciate any hint or references. Thank you.

• Another characterisation: there is a cocycle $c \in Z^1(k, G_{\text{ad}})$ such that the twist $G_c$ is quasisplit, and $G$ itself is quasisplit if and only if $c$ is a coboundary. Maybe one can show such a cocycle must be a coboundary almost everywhere? – LSpice Jun 5 '17 at 1:15
• You need to make contact with integral structure. A connected reductive group over a finite field is quasi-split; one has to bootstrap from that. By "spreading out" formalism (entails real work), there exists a non-empty finite set $S$ of places of $k$ (containing the archimedean places) such that $G$ is the generic fiber of an $O_{k,S}$-group $\mathbf{G}$ that is reductive in the sense of SGA3: smooth affine with connected reductive fibers. The scheme of Borel subgroups of $\mathbf{G}$ is $O_{k,S}$-smooth, so a $\kappa(v)$-point lifts to an $O_{k_v}$-point; the associated $k_v$-point does it. – nfdc23 Jun 5 '17 at 2:35
• None of your 3 approaches "remembers" that you began with a group over $k$ rather than over $k_v$; that is another reason such approaches get stuck. For details on my suggested argument, see XIX 2.6 in SGA3 for the spreading-out over some $O_{k,S}$ and XXII 5.8.3(i) for the existence and smoothness of the scheme of Borel subgroups (or instead in the article Reductive Group Schemes in the proceedings of the 2011 Luminy summer school on SGA3 see 3.1.9(1) and 3.1.12 for spreading-out over some $O_{k,S}$ and 5.2.11(3) for the existence and smoothness of the scheme of Borel subgroups). – nfdc23 Jun 5 '17 at 2:49
• Building on the suggestion of @LSpice, here's another approach. We have $\xi \in {\rm{H}}^1(k, G^{\rm{ad}})$ whose triviality over an extension field $k'/k$ ensures being quasi-split over $k'$. Geometrically, $\xi$ corresponds to a $G^{\rm{ad}}$-torsor $E$ over $k$. Taking $O_{k,S}$ as in my previous comments and increasing $S$ by a finite amount, $E$ spreads out to a $\mathbf{G}^{\rm{ad}}$-torsor $\mathbf{E}$. By Lang's theorem, this torsor has a $\kappa(v)$-point for all $v\not\in S$, so has an $O_{k_v}$-point ($\mathbf{E}$ is $O_{k,S}$-smooth) and thus a $k_v$-point for all $v\not\in S$. – nfdc23 Jun 5 '17 at 4:09
• Now I realize that my suggestion is what nfdc23 is saying as well. The definition of the scheme parameterizing Borel subgroups (the "flag variety") uses SGA 3. – Jason Starr Jun 5 '17 at 10:13