Let $L$ be a Lie Algebra and let $E(L)$ denote the subgroup of the inner automorphisms, generated by all $\exp(\operatorname{ad}(z))$ for $z\in L$ being strongly ad-nilpotent. Let $\operatorname{char}\mathbb{F}=0$.
I am trying to simplify the general proof for the conjugacy of Borel subalgebras of an arbitrary Lie Algebra $L$ under $E(L)$ given in the book for the special case $L=\mathfrak{sl}(2,\mathbb{F})$.
Let $B=\mathbb{F}h+\mathbb{F}x$ where
$$h=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\,\,\,\, x=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ from the usual basis of $\mathfrak{sl}(2,\mathbb{F})$ be a standard Borel subalgebra and $A$ any other Borel subalgebra of $\mathfrak{sl}(2,\mathbb{F})$. Now I am trying to show that $B$ and $A$ are conjugate.
Following the general proof in the book I considered the cases $\dim(B\cap A)=0,1,2$. I'm quite sure about the cases 0 and 2, but with $\dim(B\cap A)=1$ I always end up with a contradiction and I can´t tell where I am mistaken.
I tried to follow the argumentation of the general proof. There are 2 cases checked:
The set $N$ of nilpotent elements of $B\cap A$ is nonzero.
$N$ is a subspace of $B\cap A$. In this case that means $N=B\cap A$. The normalizer $K$ of $N$ has to be a proper subalgebra, as $\mathfrak{sl}(2,\mathbb{F})$ is simple, so $K$ is at most of dimension 2. $B\cap A$ is properly contained in both $B\cap K$ and $A\cap K$, so $\dim(B\cap K)=\dim(A\cap K)=2$ and $B=K=A$. This contradicts $\dim(B\cap A)=1$.$B\cap A$ has no nonzero nilpotent elements.
It is obvious from preceding chapters, that $B\cap A$ is toral, in this case maximal toral. From a previous Theorem $B\cap A$ is conjugate under $E(L)$ to $\mathbb{F}h$, so like in the general proof I took $B\cap A=\mathbb{F}h$. But that means as in the general proof, that $A$ has to include $\mathbb{F}x$ and $B=A$. This again contradicts $\dim(A\cap B)=1$.
The general proof gives the following arguments for $\mathbb{F}x \subset A$:
First of all the standard Borel subalgebra is fixed as: $B=H+\coprod_{{\alpha\succ0}}L_{\alpha}$, $H$ maximal toral subalgebra, $\alpha$ the positiv roots relative to a fixed base of the root system generatet by $H$ and $L_{\alpha}$ as in the Cartan decomposition. Then it is shown, that in case 2, $B\cap A$ can assumed to be a subset of $H$. The general proof then looks at the cases, that $B\cap A=H$ and that $B\cap A$ ist properly included in $H$. As in this case $B\cap A$ and $H=\mathbb{F} h$ are both onedimensional I only looked at the first case.
The general proof says that, as $A$ properly includes $H$, it has to include at least one $L_{\alpha}$. The only $L_{\alpha}$ in this case is $\mathbb{F}x$, so it has to be in $A$.
Thank you for helping me.