# Conjugacy of Borel subgroups over arbitrary fields

Let $k$ be a field and $G$ a connected semisimple algebraic group over $k$.

If $k$ is algebraically closed, then it is well known that all Borel subgroups of $G$ are conjugate by the action of $G(k)$. I would like to know whether this is also true over non-closed fields.

Are all Borel subgroups of $G$ over $k$ conjugate by the action of $G(k)$?

Recall that a Borel subgroup $B$ of $G$ over $k$ is a closed subgroup of $G$ over $k$ such that the base change of $B$ to the algebraic closure $\bar{k}$ is a Borel subgroup of $G \times_k \bar{k}$.

Of course Borel subgroups need not exist in general; when they do one says that $G$ is quasi-split. If $G$ is not quasi-split then the statement is vacuously true.

• As you note, Borel subgroups over $k$ need not exist in general. In fact they rarely exist over the typical fields people work with which are not algebraically closed: real, $p$-adic, number fields, etc. But it's a standard fact that a connected semisimple group defined over a finite field (or more generally over a field of cohomological dimension 1) is quasi-split over that field. Borel and Tits developed their more elaborate theory involving relative roots and $k$-parabolic subgroups to deal with the general ($k$-isotropic) case. Commented Feb 10, 2017 at 17:53
• I have a similar question. Suppose that G / Zp is a split group. Is it true that in G all Borel subgroups are conjugate? (Zp being the p-adic integers)
– mnr
Commented Apr 11, 2017 at 11:33
• @mnr It might be better to ask a separate question. In any case, have a look at Thm. 5.2.11 in math.stanford.edu/~conrad/papers/luminysga3.pdf Commented Apr 11, 2017 at 13:58
• @mnr Wait, actually, have a look at Prop. 6.2.11 (last sentence). That says that (G,T,B) is isom to (G,T,B') and that the isom can be given by conjugation with an element in G(S). Commented Apr 11, 2017 at 14:06

Theorem: Let $G$ be a connected reductive group over a field $k$. The minimal parabolic $k$-subgroups of $G$ are conjugate under $G(k)$.