I want to show that a set $B\subset L$ is a maximal solvable subalgebra.
With $L = \mathscr{o}(8,F)$, $F$ and algebraically closed field, and $\operatorname{char}(F)=0$ and
$$B= \left\{\begin{pmatrix}p&q\\0&s \end{pmatrix}\mid p \textrm{ upper triangular, }q,s\in\mathscr{gl}(4,F)\textrm{ and }p^t=-s, q^t=-q\right\}.$$
$B$ is a subalgebra by construction. So my problem now is how I can show that it is maximal solvable.
I tried the approach by '$L$ solvable $\Leftrightarrow$ $[L,L]$ nilpotent'. I am not sure how good this idea is, but here what I have:
$[B,B]= span\{[x,y]\mid x,y\in B\}$. So $x$ and $y$ are matrices of the form above.
That means, that for the '$p$-part' we see already that it is nilpotent. Since with each multiplication there is one more zero-diagonal.
But how can I show now, that we will also get rid of the $q$ and $s$ part?
I tried to look at $[x,y] = xy-yx = \begin{pmatrix}\bigtriangledown&*\\0&ss' \end{pmatrix}-\begin{pmatrix}\bigtriangledown&*\\ 0&s's \end{pmatrix}$
(with $\bigtriangledown$ any upper triangle matrix minus one diagonal, $s$ the part in the $x$ matrix and $s'$ in the $y$ matrix). (Sorry for the chaotic notation. I don't realy know how to write it easier..)
Is this a beginning where I should go on with? Or does someone have a hint how to approach this problem?
I don't really know, how to go on from here on. So I'd be very happy for any hint :)
Best, Luca
As far as I can think, use the Thm c at page 84, however i have no idea of how to compute this $B(triangular)=H+\cup L_{\alpha}$ which are the positive.