I am reading a paper of Yves Benoist (Tores Affines) and I can't figure out how to answer the following question.
Let $\rho :L\to \text{Aff}(\mathbb{R}^2)$ be a Lie group homomorphism, where $L=\mathbb{R}^2$ and $\text{Aff}(\mathbb{R}^2)$ denotes the group of affine homeomorphisms of $\mathbb{R}^2$. Assume that $L$ has an open orbit, that is there exists $\Omega\subset\mathbb{R}^2$ such that for every $l\in L$, $\rho(l)(\Omega)=\Omega$ and also assume that $L$ acts transitively on $\Omega$.
Question: what is the Lie algebra $\mathfrak{k}$ of the image $K=\rho(L)$?
In the paper, Benoist says that, up to conjugation, $\mathfrak{k}$ falls into one of the following commutative sub Lie algebras:
$$\mathfrak{k}_1=\left\lbrace \left(\begin{array}{cc} 0 & 0 \\ 0 & 0\\ \end{array}\right)\left(\begin{array}{c} x \\ y \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$ $$\mathfrak{k}_2=\left\lbrace \left(\begin{array}{cc} 0 & y \\ 0 & 0\\ \end{array}\right)\left(\begin{array}{c} x \\ y \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$ $$\mathfrak{k}_3=\left\lbrace \left(\begin{array}{cc} 0 & 0 \\ 0 & y\\ \end{array}\right)\left(\begin{array}{c} x \\ 0 \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$ $$\mathfrak{k}_4=\left\lbrace \left(\begin{array}{cc} x & y \\ 0 & x\\ \end{array}\right)\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$ $$\mathfrak{k}_5=\left\lbrace \left(\begin{array}{cc} x & 0 \\ 0 & y\\ \end{array}\right)\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$ $$\mathfrak{k}_6=\left\lbrace \left(\begin{array}{cc} x & y \\ -y & x\\ \end{array}\right)\left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right),\quad (x,y)\in\mathbb{R}^2 \right\rbrace$$
I don't know how to solve this, I think that we can use the homomorphism $$ \text{Aff}(\mathbb{R}^2)\hookrightarrow \text{M}_3(\mathbb{R})$$ sending $x\mapsto Ax+b$ to $\left(\begin{array}{cc} A & b \\ 0 & 1\\ \end{array}\right)$ but that's it.
Thanks in advance for your help !