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I was wondering about the following question: if you have a faithful action of a group G on the real line R by orientation-preserving homeomorphisms, it is easy to construct a new action such that a point p in R has the trivial stabilizer in G. Is it possible to make (possibly by a completely new action) the map G -> R defined by g -> g(p) is a group homomorphism always (when R is regareded as a group with addition)? If not, when could it be done?

Any comment and/or advice would be greatly appreciated.

Edit: As many people pointed out already, if such a homomorphism G -> R exists, then G must be abelian - I was over-simplifying the question. What I really want is when G is a countable group acting faithfully on R and S is a finite subset of G, can I construct a new action so that a point p with trivial stabilzer satisfies that g(p) + h(p) = g(h(p)) for all g, h in S?

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Just to be sure I understand correctly, the question asks for a non-trivial homomorphism $G \rightarrow R$ for some $p$, right? – Dan Sălăjan May 23 '13 at 9:58
@Harry: it's unclear from your question if you want an action such that some or every point $p$ has a trivial stabilizer. This dramatically changes the answer! – YCor May 23 '13 at 12:11
@ Dan: Yes. @Yves: I assumed a single point has a trivial stabilizer. I just added what I actually wondered. – Harry Baik May 23 '13 at 21:47

This is true (edit: I mean here: the existence of an action with one point with trivial stabilizer) when $G$ is countable, or more generally when it acts faithfully order-preserving on a countable totally ordered set:

1) If an arbitrary group $G$ acts order-preserving faithfully on a totally ordered set $(D,\le)$, then there exists a left-invariant total ordering $\le'$ on $G$ (so that the action of $G$ on itself by left-translation is free and order-preserving)

Indeed, consider a well-ordering $\preceq$ on $D$ (unrelated to the total order), and define $g\le' h$ if $g=h$ or $g\neq h$ and the $\preceq$-minimal element $y$ of $\{x\in D:g(x)\neq h(x)\}$ satisfies $g(y)\le h(y)$.

2) If an arbitrary group acts order-preserving on a countable totally ordered set $(D,\le)$, then this action can be extended to an action on the reals.

Indeed, first let $G$ act on $D\times ]0,1[$ with the lexicographic ordering (so that each $\{d\}\times ]0,1[$ is convex) by $g(d,t)=(gd,t)$, and extend the action to the Dedekind cut completion. The latter is isomorphic as a totally order set to the real line.

3) Using 1) and 2), if a countable group acts faithfully order-preserving on a totally ordered set, then it also admits a faithful action on the real line with a free orbit (i.e., so that at least one point has a trivial stabilizer).

On the other hand, I'm not sure that the group $\text{Homeo}^+(\mathbf{R})$ admits an order-preserving action on $\mathbf{R}$ with a free orbit.

(edit: obviously, as pointed out by other people, if $G$ is not abelian it's hopeless to expect to realize this action by a homomorphisms into the reals)

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A free orbit doesn t guarantee a homomorphism as desired. – Dan Sălăjan May 23 '13 at 10:40
Thanks a lot, Yves. I realized that the question was silly when G is not necessarily abelian by the obvious reason as you mentioned. I added a question at the end of the post - the homomorphism-like property holds only for some specified finite set of elements. – Harry Baik May 23 '13 at 21:50

Just to expand on Will's answer, prompted by Yves' comment to it.

If $G$ has an orientation-preserving action on $\mathbb{R}$ with trivial stabilisers, then the action is rigid in the sense that no compact interval $[x,y]$ is mapped properly into itself by any $g$ --- otherwise $g^n x$ would converge to a point fixed by $g$. It follows that each $g\neq 1$ is either positive ($gx>x$ for all $x$) or negative ($gx<x$ for all $x$). This defines an order on $G$ which is both left- and right-invariant. (Rigidity is necessary to guarantee right-invariance.)

I claim that this order is archimedean, meaning that for $g,h\in G$ with $h\neq 1$ there exists $n\in\mathbb{Z}$ with $g<h^n$. Otherwise, $x\leq h^n x<gx$ for all $n\geq 0$ which would imply a fixed point for $h$.

So $G$ is an archimedean ordered group. Thus by an old result of H\"older, $G$ is abelian, and in fact can be naturally embedded in $\mathbb{R}$ along the lines of Will Sawin's answer. And it is not hard to show that (additive) subgroups of $\mathbb{R}$ are indeed either discrete or dense.

EDIT (following comments to the original question)

To cover all bases, suppose that $G$ acts on $\mathbb{R}$ and that the stabiliser of $p$ is trivial, but the same is not necessarily true of all points. Then focusing on the action of $G$ on the orbit $Gp$, we still obtain a two-sided order on $G$. Now if $\phi:g\mapsto g\cdot p$ is to be a homomorphism (for a possibly new action) then since the target is abelian the commutator subgroup $[G,G]$ must be contained in the kernel of the (new) action. So for a faithful action, $G$ must be abelian. But not all (even finitely generated) orderable groups are abelian; for example the soluble Baumslag-Solitar group $BS(1,n)$ acts on the line with (at least some) trivial stabilisers: the maps $x\mapsto x+1$ and $x\mapsto nx$ generate a group isomorphic to $BS(1,n)$. But these groups are of course not abelian for $n\geq 2$.

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This nice theorem of Holder is from 1901: O. Holder. Die axiome der quantitat und die lehre von mass. Berichte uber die Ver- handlungen der Koniglichen, Sachsischen Gesellschaft der Wissenschaften zu Leipzig,Mathematische-Physysische Classe , 53:1-64, 1901. (for history lovers :) – Dan Sălăjan May 23 '13 at 11:50
Thanks Dan. The result is also in a 1902 Transactions paper by Edward Huntington: A Complete Set of Postulates for the Theory of Absolute Continuous Magnitude. (It's assertion 22.) It appears to be freely available at…. (I'm afraid my own subscription to the journal Dan mentions doesn't go back that far ;) – shane.orourke May 23 '13 at 17:51
Thank you for the detailed explanation, Shane! Also thanks for the history, Dan :) – Harry Baik May 24 '13 at 1:10

It can be done when every point has a trivial stabilizer. Divide into two cases - $\{g(p) | g \in G\}$ dense and discrete.

In the first case $G$ is a totally ordered group whose order completion is homeomorphic to $\mathbb R$, since, as a totally ordered set, it is isomorphic to $\{g(p) | g \in G\}$. It's ordered completion must clearly be the group $\mathbb R$, the unique complete Archimedean totally ordered group. But this exactly defines an order-preserving isomorphism, thus a homeomorphism between $\mathbb R$ the group and $\mathbb R$ the target of the action of $G$. Thus the action of $G$ on $\mathbb R$ is homoemorphic to its action on $\mathbb R$ the group, which is a homomorphism.

In the second case $G = \mathbb Z$. Fix any homoemorphism $\mathbb R/G = \mathbb R/\mathbb Z$, and lift to an isomorphism of the original $\mathbb R$ to an $\mathbb R$ where the action of $G$ is a homomorphism.

If one point has a nontrivial stabilizer, this provides an upper or lower bond on $g(p)$, so it can never be the image of a homomorphism.

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I don't think that $\{g(p)| g\in G\}$ is always dense or discrete. – YCor May 23 '13 at 8:15
This was now fixed by shane's answer, but the converse requires clarification since the question allows to modify the action. – Dan Sălăjan May 23 '13 at 10:54

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