The automorphism group of the real line with standard topology

Question

How much is known about the automorphism group of the real line with the standard topology? I have been unable to find a reference for this question. Any information about $\mathrm{Aut}(\mathbb R)$ or its subgroups would be appreciated. The only subgroups I'm aware of are the isometry group $\mathrm{Iso}(\mathbb R)$ and the subgroups of $\mathrm{Iso}(\mathbb R)$.

Edits 4/25/14:

Just so it's clear, by "the automorphism group $\mathrm{Aut}(\mathbb R)$" I mean the set of all homeomorphisms $f : \mathbb R \to \mathbb R$ under function composition.

So my main question is: What are some interesting subgroups of $\mathrm{Aut}(\mathbb R)$ which are not contained in $\mathrm{Iso}(\mathbb R)$?

Answer 1

As I said in my comments, there are many groups which embed (as subgroup) in the group $G=Homeo_+(R)$, the group of orientation-preserving homeomorphisms of $R$ (which is index $2$ subgroup in $Homeo(R)$). Whether these groups qualify as interesting is the matter of taste, since some people find only finite groups to be interesting (and among finite groups only the trivial group embeds in $G$).

To begin with, a countable group is left orderable if and only if it embeds in $G$. (See here for further details on orderability of groups.) It is known that all locally indicable groups, all braid groups, all (finitely generated) right angled artin groups, all residually torsion-free nilpotent groups, embed in $G$. For something more concrete: all finitely generated free groups, all surface group embed in $G$, the fundamental group of each compact 3-dimensional hyperbolic manifold contains a finite index subgroup which embed in $G$, same for the fundamental groups of all knot complements. See this very recent preprint for proofs and references.

Addendum: Topology of $G$, which is a topological group when equipped with topology of uniform convergence on compacts.

I added it here in order to justify the guess made by Oliver Begassat:

Lemma. The group $G$ is contractible.

Proof. Let $G_0<G$ denote the subgroup of homeomorphisms fixing the origin. Then the formula $$ F(x,t)=f(x) - tf(0) $$ defines a homotopy of every $f\in G$ to $f_1=F(x,1)\in G_0$, thereby establishing that the inclusion $G_0\to G$ is a homotopy equivalence.

For $f\in G_0$ define the homotopy $$ H(x,t)= (1-t)f(x) + tx. $$ It s elementary to see that for every fixed $t$, the map $H(\cdot, t)$ is continuous, strictly increasing and surjective, as a map ${\mathbb R}\to {\mathbb R}$. For $t=1$, $H(x,t)=x$. Therefore, we obtain that $G_0$ is contractible. qed

As for Ittay's concerns about torsion: Every nontrivial element of the group $G$ has infinite order.

Answer 2

One very interesting subgroup is the set $$ \{ \phi \in Aut(\mathbb{R}) : \phi(x+1)=\phi(x)+1 \} $$

These are precisely those homeomorphisms which are the lifts of circle homeomorphisms. Circle homeomorphisms and their properties have been studied for some time in dynamical system theory.

Poincare started the study of circle maps when proved that the rotation number $$\alpha = \lim_{n\to\infty} \frac{\phi^n(x)-x}{n} $$ exists and is independent of the point $x$.

Answer 3

There is an apparently open problem about that group, posed by J. Schreier as Problem 111 in The Scottish Book:

Does there exist a noncountable group with the property that every countable sequence of elements of this group is contained in a subgroup which has a finite number of generators? In particular, do the groups $S_\infty$ and the group of all homeomorphisms of the interval have this property? [Emphasis added.]

The other parts of Problem 111 have been answered affirmatively (with the "finite number of generators" equal to $2$), but the question about $\operatorname{Aut}(\mathbb R)$ is still open, as far as I know. That is, is every countable subset of $\operatorname{Aut}(\mathbb R)$ contained in a finitely generated subgroup of $\operatorname{Aut}(\mathbb R)$?

Answer 4

Unsurprisingly, the set of homeomorphisms of $\Bbb R$ has numerous and interesting subgroups. And you do not have to look far to find some. The easiest way to find some is to consider what other constructs do we consider $\Bbb R$ to be? A small list is:

• A group (under +)
• A vector space (over itself or over $\Bbb Q$)
• A metric space
• A uniform space
• A linearly-ordered set
• A differentiable manifold
• A measurable space (with either the Borel or Lebesgue $\sigma$-algebra)

Investigating $\Bbb R$'s Euclidean topology paired with any of these other concepts will lead you to investigate continuous functions which also preserve the extra structure. For example, when you consider $\Bbb R$ as a group you investigate the continuous maps which are also homomorphisms. These maps turn out to be $(x\mapsto \alpha x)$ for some $\alpha\in\Bbb R$. These are also the same continuous maps which preserve $\Bbb R$'s vector space structure (either over itself or over $\Bbb Q$).

You can also consider $\Bbb R$ as a metric space. Every metric induces a topology on its underlying set and is called the metric topology. $\Bbb R$'s Euclidean topology is the metric topology induced by $d(x,y)=|x-y|$. However, you can have different metrics induce the same topology. Thus you can consider subgroups of the homeomorphisms of $\Bbb R$ which also preserve some specific metric. Metric spaces give rise to metric maps (which are continuous maps when considering only the topology that the metric induces). Can you find a bijective metric map which is not an isometry?

Uniform spaces are a midway point between topologies and metric spaces. They are the proper setting to study uniform continuity and Cauchy sequences. Just like a metric induces a topology, so does a uniformity (the generalization of a metric) and is called the uniform topology. Metrics also induce uniformities. Just like with metrics, you can have distinct uniformities induce the same topology. Thus you can consider a subgroup of the homeomorphisms of $\Bbb R$ which preserves some uniformity. Generalizing from metrics to uniformities is not fruitless as it may seem, as you can have uniformities which are not induced by any metric. Thus you get subgroups which you may not witness by only considering metrics.

Considering $\Bbb R$ as a linearly-ordered set does not lead to many interesting subgroups, as $\Bbb R$'s topology coincides with its order topology. Every homoemorphism of $\Bbb R$ is monotone, and every continuous, (strictly) monotone function is a homeomorphism.

Considering $\Bbb R$ as a differentiable manifold leads us to investigate diffeomorphisms of $\Bbb R$. Of course, every diffeomorphism of $\Bbb R$ is continuous, but the converse does not hold. An example is $x\mapsto x^3$ because its inverse map is not differentiable at the origin. Thus you can consider the subgroup of diffeomorphisms. This investigation leads you to an increasing order of subgroups: $C^1\subseteq C^2\subseteq C^3\subseteq \cdots\subseteq C^\infty\subseteq C^\omega$ which represents the number of times you can differentiate the function (the last one represents analytic functions).

Considering $\Bbb R$ as a measure space with either its Borel or Lebesgue algebra leads to interesting subgroups. You can consider homeomorphisms which preserve measure (which turn out to be the isometries). You can consider homeomorphisms which preserve measure for 'almost all' subsets of $\Bbb R$---in the sense that every subset except one of finite measure has its measure preserved. You can consider the subgroup of homeomorphisms which are almost-everywhere differentiable.

Other miscellany subgroups follow here: monotone, continuous functions which differ from the identity only on a bounded set; injective polynomials; the subset of homeomorphisms which fix a certain subset; the strictly increasing homeomorphisms; and the homeomorphisms with at most polynomial growth.

You can always consider the subgroup generated by any combination of these subgroups. And of course, there are many more examples.