This question has been asked before here and there but has not received answers which make clear my difficulties understanding this argument. I am quite rusty in both group theory and topology, and I suppose this is why it's getting the better of me. The proposition (and the arguments I fail to comprehend) comes from Boothby's "An Introduction to Differentiable Manifolds and Riemannian Geometry" on page 96.

It says the following. Suppose $M$ is a Lie group and $G$ a discrete subgroup of $M$. Then for $g\in G$ the left translation $g:M\ni x\mapsto gx\in M$ is properly discontinuous.

A group action $g$ is properly discontinuous if (i) and (ii) hold:

(i) There is a neighbourhood $U\subset M$ of every $x\in M$ making the set $\{g\in G : gU\cap U \neq \emptyset \}$ finite, (ii) If $Gx\neq Gy$ then there are respective neighbourhoods $U,V\subset M$ of $x,y$ such that their intersection $gU\cap V = \emptyset$ for all $g\in G$.

Specifically I am having trouble showing (ii). I could give more detail as to up to which point I am stuck, but perhaps it is better to simply leave it as is since the argument as a whole is quite short. I should mention that it makes use of the fact found in there and on request I can outline what parts I have understood so far.



Fact: Let $G$ be a topological group. For any neighbourhood $U$ of $1$, there exists a (possibly smaller) open neighbourhood $V\ni 1$ such that $V=V^{-1}$ and $V^2\subseteq U$.

Proof: Let $V=U_1\cap U_2\cap U_1^{-1}\cap U_2^{-1}$, where $U_1\times U_2\subseteq m^{-1}(U)$ is a neighbourhood of $(1,1)$, $m\colon G\times G\to G$ is the multiplication in $G$.

Now for your question, note that WLOG we may assume $x=1$. Take the open neighbourhood $U=M-Gy$ (note $Gy$ is closed since $M$ is Hausdorff) of $1$ and obtain $V\ni 1$ with $V^2\subseteq U$ and $V=V^{-1}$. We claim the neighbourhoods $V$ of $x$ and $yV$ of $y$ works. Indeed, if $z\in gV\cap yV$, then $g^{-1}z\in V$ and $y^{-1}z\in V$, so $g^{-1}y=(g^{-1}z)(y^{-1}z)^{-1}\in VV^{-1}=V^2$, contradicting the construction $V^2\subseteq M-Gy$.

  • $\begingroup$ Crystal clear, thank you very much! $\endgroup$ – plebmatician Apr 14 at 9:20

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