Let $G=SL_n(\mathbb C), $ or $SO_n(\mathbb C)$ or $Sp_{2n}(\mathbb C)$.

Then it is known that $B\cap G$ is a Borel subgroup of $G$ where $B$ is the Borel subgroup of $GL_n$ (for the right $n$) of upper triangular matrices; hence it follows that given any Borel subgroup $B$ of $G$, there is a Borel subgroup $B'$ of $GL_n$ (for the right $n$) such that $B=B'\cap G$.

My question is: Is this true for any general kind of connected affine algebraic groups other than those listed above ?

Please help . Thanks in advance


1 Answer 1


$\newcommand{\GL}{\mathrm{GL}}$As stated the answer is it's always true if $G$ is connected and has a Borel subgroup $B$, over any field $k$ (characteristic $0$ so I don't have to worry about smoothness hypotheses).

Recall that if $G$ is a connected linear algebraic group then a smooth closed subgroup $P$ of $G$ is called parabolic if $G/P$ is proper (if you just want to work over $\mathbb{C}$, you can replace every instance of 'proper' with 'compact in the analytic topology'). For example, a Borel is a parabolic subgroup. We then have the following theorem of Chevalley:

Theorem(Chevalley): Let $G$ be a connected algebraic group, and let $P$ be a parabolic subgroup of $G$. Then, $P$ is connected and and $N_G(P)=P$.

Proof: See Theorem 1.3.1 of this. $\blacksquare$

Suppose now that $B$ is a Borel subgroup of $G$, a closed subgroup of $\mathrm{GL}_n$. Note then $B$ is a connected solvable smooth subgroup of $\GL_n$ and thus is contained in a maximal connected smooth subgroup $B'$ of $\GL_n$ or, in other words, a Borel $B'$ of $\GL_n$.

Let us note that $B'\cap G$ is smooth and solvable, and contains $B$. Thus, we will know that $B'\cap G$ is $B$ as soon as we know that $B'\cap G$ is connected. But, by Chevalley's theorem it suffices to show that $G/(B'\cap G)$ is proper. But, note that we have a surjection $G/B\to G/(B'\cap G)$ so that $G/(B'\cap G)$ is proper since $G/B$ is.

Here's an alternative, less algebro-geometric, proof using (the second part of) Chevalley's theorem due to my friend A. Bertoloni Meli. Note that $H:=B'\cap G$ while connected and smooth may not be connected. That said, consider that $H^\circ$, the connected component of $H$ is connected, smooth, and solvable. Thus, we certainly know that $H^\circ=B$. Note that $H^\circ$ is a normal subgroup of $H$. Thus, we see that

$$H\subseteq N_G(H^\circ)=N_G(B)=B$$

with the last equality following from Chevalley's theorem. Thus, $H=H^\circ=B$.

If you don't want to assume that $\mathrm{char}(k)=0$, then you can extend to the case when $k$ is perfect I believe if you allow the claim that $B=(G\cap B')_\mathrm{red}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.