# Example of a free group action that is not proper.

I have been trying to think about Lie group actions on smooth manifolds and what the quotient spaces look like. I have a proof that compact Lie groups produce proper actions on manifolds, as well as some examples of actions that are not proper, but none of them are free. What I am seeking is an example of a Lie group action on a smooth manifold that is free but not proper.

I will need a Lie group that is not compact. My first thought was to use $\mathbb{R}$ (or perhaps, $GL(n,\mathbb{F})$), but I haven't had any luck.

Help would be appreciated. If you could give an idea of what the orbit space is, that would be extra useful, but it is not necessary.

## 1 Answer

A nice instructive example is to take the group $\mathbb{Z}$ and let it act on $S^1$ by an irrational rotation. That is, let $\alpha\in[0,1]$ be some irrational number and define $n\cdot e^{i\theta}=e^{i(\theta+2\pi n\alpha)}$ for $n\in\mathbb{Z}$. The action is free since $\alpha$ is irrational, so $e^{2\pi i n\alpha}\neq 1$ for any nonzero integer $n$. It is not proper because $S^1$ is compact but $\mathbb{Z}$ is not.

Note that since the fractional parts of integer multiples of $\theta$ are dense in $[0,1]$, every orbit of this action is dense in $S^1$. So the orbit space is quite horrible: it has the indiscrete topology!

• Thank you very much! I actually discovered this action, but did not understand that it was not proper. It seems like you are using some kind of theorem, so I would appreciate clarification if you're willing: why is any action by a non compact group on a compact set is not proper? – David May 9 '16 at 4:03
• If $G$ is a noncompact group acting on a nonempty compact space $X$, then the map $f:G\times X\to X\times X$ sending $(g,x)$ to $(gx,x)$ cannot be proper because $X\times X$ is compact but $f^{-1}(X\times X)=G\times X$ is not compact. – Eric Wofsey May 9 '16 at 4:05
• Perfect! Thank you so much. – David May 9 '16 at 4:53