Let $H$ be a Heisenberg group, i.e. $$ H=\left\langle a,b,c |[a,b]=c,[a,c]=[b,c]=1\right\rangle. $$ $H$ acts on the circle by homeomorphism which preserves the orientation. If the rotation number of $c$ is zero, it implies there is a fixed point of $c$, then does it imply the set of fixed points of $c$ is finite? Thanks in advance.
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2$\begingroup$ As you did not specify the action, there is a simple counterexample: assume that the action factors through the rotation group $SO(2)$. This is abelian, so $c$ necessarily acts trivially and hence has an infinite fixed set. $\endgroup$– Matthias WendtCommented Jun 15, 2014 at 11:57
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$\begingroup$ @MatthiasWendt We suppose the action of $c$ is non trivial. $\endgroup$– user50402Commented Jun 15, 2014 at 13:40
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1$\begingroup$ You might want to have a look at arXiv:0910.0218v4. There are several interesting statements on the group structure of the homeomorphism group. In particular, the lemma of Beklaryan and Margulis (Lemma 1.8 in the arXiv-paper mentioned) implies that the rotation number of $c$ is always $0$. Other results in that paper say that a certain amount of differentiability of the action implies that it factors through rotations. I think (but I'm not sure) that Theorem 1.1 in that paper implies that there is no faithful action of $H$ via homeomorphisms. $\endgroup$– Matthias WendtCommented Jun 15, 2014 at 15:26
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$\begingroup$ @MatthiasWendt In fact, I want to prove such $c$ is trivial, so the first step is to prove the fixed points of $c$ is finite under the hypothesis, then there would be a contradictory. Thanks for the reference, but it doesn't show the set of fixed point of $c$ is finite. $\endgroup$– user50402Commented Jun 15, 2014 at 16:19
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In fact, there are many faithful (one-to-one) actions of $H$ by homeomorphisms of the circle. For a concrete example, see http://www.math.uchicago.edu/~farb/papers/nilpotent.pdf They give an action by homeos (in fact, $C^1$ diffeos) of $[0, 1]$; by gluing the ends together, you get an action on $S^1$. In this action, $c$ has infinitely many fixed points (but certainly doesn't fix every point).
One thing you can say, in general, is that due to the amenability of $H$, any action $\phi\colon H \to Homeo(S^1)$ will have an invariant probability measure $\mu$. It follows that the rotation number $\rho$, restricted to $\phi(H)$, is a homomorphism. In particular, $\rho(\phi(c)) = 0$, so $c$ must have at least one fixed point. In fact, you can then see that the support of $\mu$ has to be contained in the set of fixed points of $\phi(c)$...
You can ask, what could the rotation numbers of $\phi(a)$ and $\phi(b)$ be, assuming $\phi$ is 1-1? And the answer to that is, anything. See Rotation numbers for amenable group actions on the circle.
See also the paper http://arxiv.org/abs/1403.7781, where I discuss in detail the structure that nilpotent actions on one-manifolds ($\mathbb{R}$ or $S^1$) can have.
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$\begingroup$ You are right, what I claimed was not correct. I deleted the post. $\endgroup$ Commented Jun 17, 2014 at 7:57
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$\begingroup$ Still, there is one thing not clear to me. If the probability measure has no atoms, then it can be straightened out and the action factors through $SO(2)$. (Assuming this tacitly was my mistake.) If there are atoms, there are only finitely many of those. Do they coincide with the fixed point set? In particular, is the answer to the question positive or negative? $\endgroup$ Commented Jun 17, 2014 at 8:42
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$\begingroup$ The answer to the question is negative. It's not difficult to make actions of $H$ on $S^1$ with a single point $x$ fixed by the whole group, so $x$ will be the unique atom for any invariant probability measure, but $c$ may have infinitely many fixed points. $\endgroup$ Commented Jun 17, 2014 at 9:10
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$\begingroup$ However, there are two mistakes in what you say. There can be a (countably) infinite number of atoms. Picture a set of points $x_n \in S^1$ which are discrete except that they accumulate on some point $x$. Put some mass $m_n$ on each $x_n$ such that $\sum m_n = 1$. Also, it s not enough for the measure to be non-atomic to straighten it out to Lebesgue measure. You need that it assign positive measure to open sets (see Ghys). For instance, you could have a measure whose support is a Cantor set $C \subset S^1$, with the measure of each $x \in C$ equal to 0. $\endgroup$ Commented Jun 17, 2014 at 9:14
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$\begingroup$ Great, thanks for the clarification. $\endgroup$ Commented Jun 17, 2014 at 9:22