# What are the possible Lie algebras of $K=\rho(\mathbb{R}^2)$, where $\rho :\mathbb{R}^2\to \text{Aff}(\mathbb{R}^2)$?

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.

• We can pass to the level of Lie algebras and compute these subalgebras of $\mathfrak{aff}(\Bbb R^2)=\Bbb R^2\rtimes \mathfrak{gl}_2(\Bbb R)$ explicitly. – Dietrich Burde Apr 18 at 11:28
• More details are given here, page $48$. A further reference is T. Nagano, K. Yagi, The affine structures on the real two-torus, Osaka J. Math. 11 (1974), 181-210. – Dietrich Burde Apr 18 at 11:45
• I mean, by differentiation, to pass from $\rho\colon L\rightarrow Aff(\Bbb R^n)$ to $d\rho\colon Lie(L)\rightarrow \mathfrak{aff}(\Bbb R^n)$. – Dietrich Burde Apr 18 at 13:11
• Ok but what should I do then? I'm thinking about taking $d\rho(1,0)=(A,b)$ and $d\rho(0,1)=(A^\prime,b^\prime)$ and using the fact that $Lie(\mathbb{R}^2)$ is abelian to get a relation between the parameters but I can't go on next. – Adam Chalumeau Apr 18 at 14:32