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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:

  1. 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 $\dim(B\cap A)=1$.

  2. $B\cap A$ 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.

Idun E.
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  • This question has a lot of text that's bunched up together and hard to read. If you'd like people to read and answer it, you might consider writing less and spacing out what you write more. – Qiaochu Yuan Jun 01 '15 at 03:28
  • I shortened the question and hopefully it is better to read now. – Idun E. Jun 01 '15 at 11:24
  • There's clearly a problem in your final case 2, since you could have $B$ the upper triangular Borel and $A$ the lower triangular. Can you explain why the general proof then forces $x \in A$? – Matthew Towers Jun 02 '15 at 08:16
  • At first, thank you for editing. It is the second question I posted and I am not yet used to it. – Idun E. Jun 02 '15 at 15:24
  • I wrote the explaination, why $x\in A$ into the question, as it is to long for a comment. – Idun E. Jun 02 '15 at 15:55
  • Ok, I located a mistake: $A$ doesn´t include $\mathbb{F}x$ but $\mathbb{F} y$. I read $0\prec \alpha$ instand of $0\succ \alpha$. So $A$ hast to be $A=\mathbb{F}h+\mathbb{F}y$. That leads me to the question, if those are the only Borel subalgebras of $\mathfrak{sl}(2, \mathbb{F})$? – Idun E. Jun 03 '15 at 07:50
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    They're not the only Borels (e.g. take one of them and conjugate it by some invertible matrix and you usually get something different), but remember that in the proof of your last case you conjugate $B \cap A$ to $\mathbb{F}h$ so you are no longer dealing with arbitrary $B$. Possibly the only Borels of sl2 containing this subspace are the upper and lower triangular ones. – Matthew Towers Jun 03 '15 at 15:34
  • So I should take $B=H+L_{\alpha}$ to be an arbitrary standard Borel subalgebra instead of $\mathbb{F}h+\mathbb{F}x$. Doing the same I as above I get $A=H+L_{-\alpha}$, what is conjugate to $B$. That $B$ is conjugate to any other Borel subalgebra of $\mathfrak{sl}(2, \mathbb{F})$ then yields from the fact, that I \textbf{conjugate} $B\cap A$ to $H$, right? – Idun E. Jun 03 '15 at 17:31
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    $H+L_\alpha$ won't be big enough to be a Borel for larger Lie algebras, but yes I think this is how it works for $\mathfrak{sl}(2,\mathbb{F})$. – Matthew Towers Jun 04 '15 at 12:56

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