In Gallot, Hulin, Lafontaine's Riemannian Geometry:
Definition Let $G$ be a discrete group, acting continuously on the left on a locally compact topological space $E$. One says that $G$ acts properly on $E$ if, for $x,y\in E$ where $y$ does not belong to the orbit of $x$ under $G$, there exist two neighborhoods $V$ and $W$ of $x$ and $y$ respectively such that, for any $g\in G$, $gV\cap W=\emptyset$ holds. The group $G$ acts freely if, for $x\in M$ and $g\in G$ with $g\neq\operatorname{id}$, then $gx\neq x$.
Now in the proof of the following theorem
Theorem Let $G$ be a discrete group of diffeomorphisms of a manifold $M$. If $G$ acts properly and freely on $M$, there exists on the topological quotient space $M/G$ a unique structure of smooth manifold such that the canonical projection $\pi\colon M\to M/G$ is a covering map.
it's claimed that
Moreover, since $G$ acts freely, every $x$ in $M$ is contained in an open set $U$ such that all the open sets $gU$ are pairwise disjoint when $g$ runs through $G$.
I cannot see the reason for this, according to the prior definitions. As far as I've searched on the Internet or books, usually the condition is assumed in the definition of a proper action, so I don't know whether this is implied in the preceding definitions. I wonder how to prove this.
Any idea? Thanks!
Best Answer
The definition of proper you quote is not the standard definition, and it is not strong enough to imply this result. Consider the following example: Let $G = S^1$ act on $M = \mathbb{R}^2\smallsetminus\{(0,0)\}$ in the usual way, i.e. by rotations about the origin. This action is clearly free. However, equip $G$ with the discrete topology. Note that this makes the topology on $G$ finer, so the action map $G\times M\to M$ is still continuous (the preimages of open sets were open using the usually topology on $G$, so by making the topology strictly finer, they are still open).
Now, I claim that this action satisfies the definition of proper you state in the first quote. Pick $x$, and then the orbit of $x$ is the circle about the origin of radius $\|x\|$. Pick $y$ not in the orbit of $x$ (so $\|x\|\neq\|y\|$), let $\epsilon = |\|x\| - \|y \||$, and let $V$ and $W$ be the $\epsilon/2$ balls about $x$ and $y$, respectively. It is easy to see that $gV\cap W = \emptyset$ for each $g \in G$, as $gV$ is contained in the "$\epsilon/2$-annulus" about the orbit of $x$, while $W$ doesn't intersect this annulus.
However, obviously, there is no open neighborhood $U$ of any $x$ such that the $gU$ are pairwise disjoint. In fact, for each $x\in M$, any open neighborhood $U$ of $x$ intersects the orbit $Gx$ at infinitely many points.
I'll also mention that the quoted definition of proper doesn't imply many of the other important consequences of proper actions. Proper actions of discrete groups (by the standard definition) have finite stabilizers. But consider the example of $\mathbb{Z}$ acting trivially on a point; it satisfies the quoted definition vacuously (only one orbit), but the stabilizer of the point is $\mathbb{Z}$. Similarly, proper actions by the usual definition have Hausdorff quotients (distinct orbits can be separated by open sets)--in my opinion, this is the the main reason for restricting to proper actions (at least for many purposes). However, the example here satisfies the quoted definition, but the orbits of $(1,0)$ and $(0,1)$ contain arbitrarily close points, as explained at that page.
I'm not familiar with the book you are using, but it sounds like they intended the standard definition: the map $G\times M \to M\times M$ given by $(g,x)\mapsto (x,gx)$ is a proper map, i.e. a closed map such that the preimage of each compact set is compact. tom Dieck's Transformation Groups gives a pretty nice treatment of this definition and its consequences in a general setting starting on page 27.