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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
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3 Answers

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
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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 ams.org/journals/tran/1902-003-02/S0002-9947-1902-1500598-9/…. (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
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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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