I need to prove the following statement :

Assume that $\mathrm{SL}_{2}(\mathbb{R})$ acts on $(X,d)$ by isometries. Assume that $x\in X$ and $a_{n}\in A=\{\begin{pmatrix} e^{t/2}&0 \\ 0&e^{-t/2} \end{pmatrix}, t \in \mathbb{R}\backslash\{0\}\}$ such that $a_{n}\to\infty$ and $a_{n}\cdot x\to x$. Then $x$ is a fixed point.

So I need to show that $\forall g \in SL_{2}(\mathbb{R}) : g.x=x$.

We know that since $\mathrm{SL}_{2}(\mathbb{R})$ acts on $(X,d)$ by isometries : $\forall g \in SL_{2}(\mathbb{R}), \forall x,y \in X : d(g.x,g.y)=d(x,y)$

Since $a_{n}\cdot x\to x$ : $d(a_{n}.x,x)\longrightarrow 0$ as $n\rightarrow \infty$ and since $a_{n}\to \infty$ we have a sequence $(t_{n})$ in $\mathbb{R}\backslash\{0\}$ with $t_{n}\longrightarrow \infty$

KAK decomposition : For any $g\in SL_{2}(R)$ there exists $k,l \in SO(2)$ and $a \in A$ so that $g=kal$.

I thought that maybe I could use the KAK decomposition to reduce my problem to the case were $a.x=x$ ($*$) for $a \in A$ but I am not sure if I can do this, and if I can I don't know how to prove that ($*$) would hold.

I really don't know where to start and I would appreciate any kind of hints !

Thank you beforehand for your help

  • $\begingroup$ This is false for a trivial reason. Let $X=\mathbb R^2$ and consider the sequence of matrices $$a_n=\begin{bmatrix} \exp(\frac1n) & 0 \\ 0 & \exp(-\frac1n)\end{bmatrix}.$$ Any vector $x\in\mathbb R^2$ satisfies $a_nx\to x$. You probably want the parameters in the exponentials to diverge to $\infty$. In that case, the only vector $x\in\mathbb R^2$ satisfying the condition is $x=(0,0)$. Can this be a useful piece of information for you? $\endgroup$ – Giuseppe Negro May 28 '18 at 11:45
  • $\begingroup$ I indeed want the exponentials to diverge($a_{n}\rightarrow \infty$). I am sorry but I don't see how this is a counter example to my statement. The only vector satisfying this condition is $(0,0)$ but $(0,0)$ is a fixed point in $\mathbb{R}^{2}$, which is precisely the statement I want to prove. Sorry if I misunderstood your comment ! $\endgroup$ – Em_ May 28 '18 at 12:06
  • $\begingroup$ No, it's OK, we're saying the same thing. I was wondering if it is sufficient to prove the claim on $\mathbb R^2$. Intuitively I think that it should be the case. $\endgroup$ – Giuseppe Negro May 28 '18 at 12:10
  • $\begingroup$ What do you know about the action of $SL(2)$ on $X$? What do you mean by "action", actually? Is the map $a\in SL(2)\to \mathrm{Isom}(X)$ a continuous group homomorphism? $\endgroup$ – Giuseppe Negro May 28 '18 at 12:52
  • $\begingroup$ By action I mean group action of SL2(R) on the metric space X (so there exists a neutral element (here the identity matrix) and there is compatibility ($gh.x=g.(h.x)$)). I also know that the action is isometric. And I don't think this map is a group homomorphism. $\endgroup$ – Em_ May 28 '18 at 13:04

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