# When does the Borel subgroups of affine (linear) algebraic groups come from Borel subgroups of general linear group?

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 ?

$$\newcommand{\GL}{\mathrm{GL}}$$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 $$B=(G\cap B')_\mathrm{red}$$.