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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.

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    $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented Feb 10, 2017 at 17:53
  • $\begingroup$ 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) $\endgroup$
    – mnr
    Commented Apr 11, 2017 at 11:33
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    $\begingroup$ @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 $\endgroup$
    – Ariyan Javanpeykar
    Commented Apr 11, 2017 at 13:58
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    $\begingroup$ @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). $\endgroup$
    – Ariyan Javanpeykar
    Commented Apr 11, 2017 at 14:06

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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 $G(k)$.

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  • $\begingroup$ 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. $\endgroup$
    – Joonas Ilmavirta
    Commented Feb 10, 2017 at 10:38
  • $\begingroup$ Indeed. I've clarified my answer. $\endgroup$
    – thierry stulemeijer
    Commented Feb 10, 2017 at 10:47
  • $\begingroup$ Great, thanks! I had tried looking in Borel's book but had missed this result. $\endgroup$
    – Daniel Loughran
    Commented Feb 10, 2017 at 10:56
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    $\begingroup$ 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 $\endgroup$
    – Jim Humphreys
    Commented Feb 10, 2017 at 17:44

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