Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a topological group, acts on a topological space $X$, such that the map $f: G \times X \rightarrow X:(g,x)\mapsto g*x$ is continuous.

We say that this action is $properly\;discontinuous$ if for every element $x\in X$, there exist an open neighbourhood $U_{x}$ for $x$ such that $gU_{x}\cap U_{x}\neq \phi$ and $g\in G$ implies that $g=1_{G}$

I am trying to show that if the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorphic to $\mathbb{Z}$.

best regards

share|cite|improve this question
1  
Hint: find the free generator of $G$. – Berci Oct 20 '12 at 0:28
    
Small nitpicking: $G$ has to be nontrivial. – Najib Idrissi May 24 at 14:06

The orbit $\mathcal O=G0$ of $0\in\mathbb R$ is discrete. It follows that it is countable and, morerover, there is either a strictly increasing or decreasing bijection $f:\mathbb Z\to\mathcal O$ or $f:\mathbb N_0\to\mathcal O$. Let me suppose it is increasing; if not, we do something similar.

If $g\in G$, there is a $n_g\in\mathbb Z$ such that $g(f(0))=f(n_g)$. I claim that $$g(f(a))=f(a+n_g)\qquad\text{for all $a$ in $\mathbb N_0$ or $\mathbb Z$,}$$ depending on which case we are in. Indeed, there is an $m$ such that $g(f(1))=f(m)$. If we had $m>1+n_g$, then the image of the interval $[f(0),f(1)]$ under $g$ would contain the interval $[f(n_g),f(n_g+2)]$, and there would exist a point in $(f(0),f(1))$ whose image under $g$ is $f(n_g+1)$. This is absurd. Doing this in general gives my claim.

Now I can define a map $\phi:g\in G\mapsto n_g\in\mathbb Z$. It is a morphism of groups, and it must be injective.

share|cite|improve this answer

It suffices to show the orbit $G(0)$ of $0$ is a discrete set, and any $x\in\mathbb{R}$ is "sandwiched" by two representatives of $G(0)$. Then the space $X/G$ is homeomorphic to $\mathbb{S}^1$ (the action is continuous), and so $G\approx\pi_1(X/G)=\pi_1(S^1)=\mathbb{Z}$.

The discreteness is natural. Suppose $G(0)$ is bounded above. Let $s\equiv\sup G(0)$; since $G\neq\{e\}$ and acts properly discontinuous, there is some $s'=gs\neq s$ with $g\neq e$. If $s'<s$, then $g(s,\infty)=(s',\infty)$ contains no representative of $G(0)$, contradicting with the definition of $s$. If $s'>s$, then for an interval $I$ containing $s$ such that $gI\cap I=\emptyset$ (such interval exists due to $G$ acting properly discontinuous), all elements of $gI$ are larger than $s$ and so $gI$ contains no representative of $G(0)$. This is however a contradiction since any interval $I$ containing $s=\sup G(0)$ (and so $gI$) must contain some representative of $G(0)$. An analogous argument using $i\equiv\inf G(0)$ shows $G(0)$ is also not bounded below.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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