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Denote by $\mathbb{F}_q$ the finite field with $q$ elements, and denote by $\bar{\mathbb{F}}_q$ its algebraic closure. Let $G$ be an affine algebraic group over $\bar{\mathbb{F}}_q$, and let $F$ be a Frobenius morphism of $G$ which defines a $\mathbb{F}_q$ structure on $G$.

It is known that each $F$-stable Borel subgroup of $G$ contains a $F$-stable maximal torus of $G$. However, is any $F$-stable maximal torus always contained in some $F$-stable Borel subgroup?

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  • $\begingroup$ I think you should really read Chapter 4 of Geck's book. It has all the answers to your questions. The answer here is no. There always exists an $F$-stable maximal torus and Borel subgroup $T_0\leqslant B_0$ in $G$. If you take $T = gT_0g^{-1}$ with $g^{-1}F(g) = n \in N_G(T_0)$ then this is always $F$-stable. For convenience, assume $F$ acts trivially on the Weyl group $N_G(T_0)/T_0$. Then the maximal torus $T$ is contained in an $F$-stable Borel subgroup if and only if $n \in T_0$. Hence there are lots of $F$-stable maximal tori that are not contained in $F$-stable Borels. $\endgroup$
    – Jay Taylor
    Commented May 19, 2015 at 8:57
  • $\begingroup$ Look at Example 4.3.3 in Geck's "An Introduction to Algebraic Geometry and Algebraic Groups" and the results around there. $\endgroup$
    – Jay Taylor
    Commented May 19, 2015 at 8:58
  • $\begingroup$ @JayTaylor I see. Thank you. Actually I am reading the book "Representations of finite groups of Lie type" by Francois DIGNE and Jean MICHEL. Now I think that I had better turn to read the book recommended by you. $\endgroup$
    – Hebe
    Commented May 19, 2015 at 9:03
  • $\begingroup$ It's very hard to understand what's going on just by reading Digne--Michel, I wouldn't advise it. You might want to look at my answer to this question mathoverflow.net/questions/203602/…. Good luck. $\endgroup$
    – Jay Taylor
    Commented May 19, 2015 at 9:09
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    $\begingroup$ I'm voting to close this question as it has been answered in the comments $\endgroup$
    – Dima Pasechnik
    Commented May 19, 2015 at 11:25

2 Answers 2

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I think it's instructive to think about these problems on the level of $\mathbb{F}_q$ points; then you can see counterexamples even in the simplest cases.

For instance, if $G = GL_2$ defined over $\mathbb{F}_q$ and $B$ is an $F$-stable torus, then $B$ is conjugate (in $GL_2(\mathbb{F}_q)$) to the Borel subgroup of upper triangular matrices. Now consider the torus $T$ with $\mathbb{F}_q$ points

$$T(\mathbb{F}_q) = \bigg\{t = \begin{pmatrix} x & y \\ \alpha y & x\end{pmatrix}\bigg\}$$ where $\alpha\in \mathbb{F}_q$ is not a square. Then we can check that $T$ does not conjugate into $B$. For instance, if $y \neq 0$ then the characteristic polynomial of $t$ does not split in $\mathbb{F}_q$, whereas every element of $B(\mathbb{F}_q)$ has a characteristic polynomial that splits.

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The answer is sort of maximally negative. For a (smooth) connected linear algebraic group $G$ over any field $k$ whatsoever, if $G$ admits a Borel $k$-subgroup (as happens for finite $k$) then all such $B$ constitute a single $G(k)$-conjugacy class and all maximal $k$-tori $T$ in any such $B$ map isomorphically onto the quotient $B/U$ by the unipotent radical (which is always "defined over $k$" in such $B$), whence there is only one $k$-isomorphism class of such $T$.

Pick a single such $B_0$ and a single maximal $k$-torus $T_0$ in such a $B_0$. There will generally be lots of maximal $k$-tori $T$ in $G$ that are not $k$-isomorphic to $T_0$ as $k$-groups (e.g., for split $G$ we can generally find lots of non-split $T$, such as usually happens for $G= {\rm{GL}}_n$ when $n > 1$, as in John Binder's example for $n=2$) provided that $k$ isn't too close to being separably closed. All such $T$ are then counterexamples.

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