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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 !

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  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Apr 18 at 11:28
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Apr 18 at 11:45
  • $\begingroup$ What do you mean by passing to the level of Lie algebras (to be fair I haven't had a good course on Lie Algebras)? I know the paper of T. Nagano and K. Yagi but it seems hard. I'll try to read the recent one of Baues. Thanks so much for your help ! $\endgroup$ – Adam Chalumeau Apr 18 at 12:52
  • $\begingroup$ I mean, by differentiation, to pass from $\rho\colon L\rightarrow Aff(\Bbb R^n)$ to $d\rho\colon Lie(L)\rightarrow \mathfrak{aff}(\Bbb R^n)$. $\endgroup$ – Dietrich Burde Apr 18 at 13:11
  • $\begingroup$ 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. $\endgroup$ – Adam Chalumeau Apr 18 at 14:32

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