Maximal Lie subalgebra of $sl_n$

Question

How to find the maximal solvable Lie subalgebra of $\mathfrak sl(n,\mathbb R)$?

Maybe the invariance lemma is the key!

Answer 1

Note that a maximal solvable subalgebra need not be unique. For simplicity take $n=2$. So in $\mathfrak{sl}(2,\mathbb{R})$ we have several maximal solvable subalgebras, e.g., $\mathfrak{a}=\langle h,x\rangle$, or $\mathfrak{b}=\langle h,y\rangle$. Here $(x,y,z)$ is a basis with brackets $[x,y]=h,\,[[h,x]=2x,\,[h,y]=2y$. In general, a maximal solvable subalgebra is called a Borel subalgebra. Over the complex numbers they a re all conjugated. The so-called standard Borel subalgebra can be constructed via the weight space decomposition:

Proposition Let $L$ be a semisimple Lie algebra, and $H$ be a Cartan subalgebra with root system $\Phi$ and base $\Delta$. Then the standard Borel subalgebra is given by $$ B:=B(\Delta)=H\oplus \bigoplus_{\alpha\in \Phi^+} L_{\alpha}. $$ In terms of matrices, the standard Borel subalgebra of $\mathfrak{sl}(n)$ consists of the solvable Lie subalgebra of all upper-triangular matrices in $\mathfrak{sl}(n)$.