# If the orbit space is Hausdorff, must the action be proper?

Let $G$ be a Lie group acting continuously on a manifold $M$. It is well known that if the action is proper, then the quotient space $M/G$ is Hausdorff.

Does the converse hold? If $M/G$ is Hausdorff, must the action be proper?

No, let $G$ be a non locally compact Lie group (infinite dimensional) the left action of $G$ on $G$ is not proper, for example, take any neighborhood $U$ of the identity $1$, the set $\{g\in G, g(U)\cap U\neq \phi\}$ is infinite. but the quotient of $G$ by $G$ is a point.

An interesting question is to know if your question is true when $G$ is a discrete group acting on a manifold $M$.

• Wait, how do you define a proper action? For me, it means that for every compact subset K, the set of elements g in G such that g(K) intersects K is compact. Your suggestion does not seem to be a counterexample for this definition. – Ilia Smilga Feb 22 '17 at 23:54

This is a silly counterexample but it's a counterexample. Let $M=\{0\}$ and let $G=\mathbb{R}$ act on $M$ by multiplication. Then $M/G$ is still only one point but the action is not proper, since the preimage of $(0,0) \in M\times M$ is $G\times M$.

This is a instant of a more general phenomena, namely that if $M$ is compact and $G$ is not then the action is not proper. If it happens (as in this case) that the action is transitive your quotient will be one point and thus Hausdorff.

This should be seen as a complement to Tsemo Aristide's answer.