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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

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

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