The abelian group of homeomorphism

Question

Let $G$ be a subgroup of the group of homeomorphisms on the circle, and we suppose $G$ is abelian, if every element of $G$ has a fixed point on the circle, does it imply that $G$ has a common fixed point?

Answer 1

Yes, $G$ must have a common fixed point.

Let me give the proof first under the assumption that $G$ preserves orientation.

First a special case: assume $G$ is generated by two elements $f,g$. Arguing by contradiction, suppose $Fix(f) \cap Fix(g) = \emptyset$. Then there exists a component $J$ of $S^1 - Fix(f)$ which contains at least one element of $Fix(g)$. The subset $J$ is homeomorphic to the real line, and $f$ restricts to an orientation preserving, fixed point free homeomorphism of $J$, and so there is a homeomorphism $J \mapsto \mathbb{R}$ that conjugates $f \, | \, J$ to the translation map $x \mapsto x+1$. On the other hand, since $f$ and $g$ commute, $f$ must take the set $J \cap Fix(g)$ to itself. This is impossible because $J \cap Fix(g)$ is a compact subset of $J$ and the translation map $x \mapsto x+1$ does not preserve any compact subset.

One can use transfinite induction on a well-ordered generating set to get the general case that $G$ preserves orientation: the above argument can be used for the finite induction step; and for the limiting induction step one can use that the intersection of compact nested sets is nonempty.

I also had a proof for the case that $G$ has an orientation reversing element, but that proof had an error and I have removed it. I'll try to fix that up later.