# 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. – Jim Humphreys Feb 10 '17 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 Apr 11 '17 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 – Ariyan Javanpeykar Apr 11 '17 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). – Ariyan Javanpeykar Apr 11 '17 at 14:06

## 1 Answer

Yes. This follows directly from Theorem 20.9 (i) in "Armand Borel, Linear Algebraic Groups, Second enlarged edition, 1991" which goes like this:

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

• Can you add the statement of the theorem (or a rough version) to your answer? Currently it's not clear from your answer what the answer to the question is. – Joonas Ilmavirta Feb 10 '17 at 10:38
• Indeed. I've clarified my answer. – thierry stulemeijer Feb 10 '17 at 10:47
• Great, thanks! I had tried looking in Borel's book but had missed this result. – Daniel Loughran Feb 10 '17 at 10:56
• Note that this basic conjugacy result is due to Borel and Tits: Thm. 4.13 in their foundational 1965 paper Groupes reductifs. For online access see numdam.org/numdam-bin/item?id=PMIHES_1965__27__55_0 – Jim Humphreys Feb 10 '17 at 17:44