4
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – David May 9 '16 at 4:03
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey May 9 '16 at 4:05
  • $\begingroup$ Perfect! Thank you so much. $\endgroup$ – David May 9 '16 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.