Is the action of $\mathbb Z$ on $\mathbb R$ by translation the only such action?

Question

It is well known that $\mathbb Z$ acts on $\mathbb R$ by translation. That is by $n\cdot r=n+r$. The quotient space of this action is $S^1$.

Could someone give me an example where $\mathbb Z$ acts on $\mathbb R$ in some other (non trivial) way to give (possibly) some other quotient? I am only interested in continuous actions.

Specifically, is there an action such that the quotient is a closed interval?

Thank you.

Answer 1

Recall that a continuous action of $\Bbb{Z}$ on $\Bbb{R}$ consists of a group morphism $\Bbb{Z} \to \operatorname{Aut}(\Bbb{R})$, where $\operatorname{Aut}(\Bbb{R})$ denotes the group of homeomorphisms of $\Bbb{R}$ into itself.

Now, $\Bbb{Z}$ is cyclic, hence any morphism $\mu:\Bbb{Z} \to \operatorname{Aut}(\Bbb{R})$ is uniquely determined by $\mu(1)$. So, if you fix any homeomorphism $f \in \operatorname{Aut}(\Bbb{R})$, you will get the action $$n \cdot x = f^n(x)$$ where $f^n$ is the $n$-iterated map $f(f( \dots))$ (for $n < 0$ it's $(f^{-1})^{-n}$).

Now, let's see some examples:

$f(x)=x+1$ leads to the action in your question.

$f(x)=x$ leads to the trivial action.

$f(x)=-x$ leads to the action $n \cdot r = (-1)^nr$, whose quotient is $\Bbb{R}_{\ge 0}$.

Answer 2

Because $\Bbb Z$ is cyclic, you're equivalently asking "What are the conjugacy classes of elements of $\text{Homeo}(\Bbb R)$?" (That is: what are the topologically distinct discrete dynamical systems on $\Bbb R$?")

There are quite a lot. Here's a silly example not given in the other answer: Pick a homeomorphism $\varphi: \Bbb R \to (-1,1)$. Define a homeomorphism $\psi: \Bbb R \to \Bbb R$ by: $\psi(x) = x$ for $|x| \geq 1$; and $\psi(x) = \varphi(\varphi^{-1}(x)+1)$ when $|x| < 1$. Then the quotient of $\Bbb R$ under this action is not even $T_1$, since $1$ is arbitrarily close to the orbit of 0, but is not identified with it.

You could also set things up so that eg you have a Cantor set of fixed points.

One nice survey of results on these conjugacy classes is here.

Answer 3

You want continuous actiones, i.e., for each $n\in\Bbb Z$, the map $f_n\colon r\mapsto n\cdot r$ should be continuous. Note that the action is completely detemined by $f_1$ and that $f_1$ must have an inverse $f_{-1}$ that is also continuous. In other words: $f_1$ is a homeomorphism. As such, it may either be orientation preserving (i.e., strictly increasing) or not (i.e. strictly decreasing).

Consider first the former case and let $F\subseteq \Bbb R$ be the set of fixedpoints of $f_1$. As $F$ is a closed subset of $\Bbb R$, its complement is open, hence is the disjoint union of countably many open intervals $(a,b)$ (with infinite ends allowed). Each such $(a,b)$ is not only homeomorphic to $\Bbb R$, but in fact is so in a way tkat turns the action of $\Bbb Z$ on $(a,b)$ into the standard action of $\Bbb Z$ on $\Bbb R$. To see this pick $x_0\in (a,b)$, let $x_1=f_1(x_0)$. We start defining our homeomorphism $\phi\colon (a,b)\to\Bbb R$ by setting $f(x)=\frac{x-x_0}{x_1-x_0}$ for $x\in[x_0,x_1]$. After that, for any $x\in (a,b)$ we find $n\in \Bbb Z$ with $f_1^{\circ n}(x)\in[x_0,x_1)$ and can let $\phi(x)=\phi(f_1^{\circ n}(x))-n$.

We see that each open inter-fixpoint interval $(a,b)$ between contributes a copy of $S^1$ to the quotient space. The fixpoints $x\in F$ themselves "survive" the transition to the quotient space. It seems that we have a disjoint union of $F$ (with subspace topology) and a couple of copies of $S^1$, but we have to be a bit careful with the quotient topology: If $a\in F$ is a boundary point of $(a,b)$ (and/or $(c,a)$) then any open neighbourhood of it in $\Bbb R$ contains enough of the adjacent open interval(s) to cover a "full round" of the corresponding $S^1$. In other words, in the quotient space, every open neighbourhood of $a$ contains the one or two adjacent $S^1$'s.

Now back to the beginning - what happens if $f_1$ is strictly decreasing? In that case the action of $2\Bbb Z$ is of the kind described above, and the full action of $\Bbb Z$ intruduces some pairwise identification: The unique fixed point $z$ of $f_1$ is also a fixed point of $f_2$ (so not within one of the $S^1$). The copies of $S^1$ are identified in pairs (which still makes them $S^1$'s), components of $F$ not containing $z$ are also identified in pairs, only the component of $F$ containing $z$ is glued similarly to $[-a,a]\to [0,a]$, $x\mapsto |x|$; however, if $z$ is an isolated point of $F$, then as above the two adjacent $S^1$ are identified with the additional strange effect that the only open neighbourhoods of $z$ in this $S^1$ contains all of $S^1$