Given an orientation-preserving homeomorphism $f: S^1 \to S^1$, one can define its rotation number $\rho(f) \in \mathbb{R}/\mathbb{Z}$, as $\rho(f) = (\lim_{n \to \infty} \tilde{f}^n(0)/n) + \mathbb{Z}$, where $\tilde{f}: \mathbb{R} \to \mathbb{R}$ is any lift of $f$. Intuitively, it measures the rate of circulation around the circle.

Now $\rho: Homeo_+(S^1) \to \mathbb{R}/\mathbb{Z}$ is not a homomorphism, but its restriction to any amenable subgroup of $Homeo_+(S^1)$ is. Thus if $G$ is amenable, and $\phi: G \to Homeo_+(S^1)$ is an action, then the composition $\rho \circ \phi: G \to \mathbb{R}/\mathbb{Z}$ is a homomorphism.

I am wondering, what homomorphisms $G \to \mathbb{R}/\mathbb{Z}$ arise as $\rho \circ \phi$ if $\phi$ is required to be 1-1?

(The requirement that $\phi$ be 1-1 is important, since otherwise we can realize any $\psi: G \to \mathbb{R}/\mathbb{Z}$ simply by making $G$ act on the circle via $\psi$.)

This question might be hard, since I don't even know which amenable groups act faithfully on the circle. But I'd even be curious about the case where $G$ is a finitely-generated, torsion-free nilpotent group. These guys do act on the circle faithfully, but the standard construction gives you something trivial in terms of rotation number.

I doubt it helps, but by Ghys and Matsumoto, for an amenable group $G$, two actions $\phi_1, \phi_2: G \to Homeo_+(S^1)$ are semi-conjugate if and only if $\rho(\phi_1(g)) = \rho(\phi_2(g))$ for every $g \in G$ (Matsumoto, "Numerical invariants for semiconjugacy of homeomorphisms of the circle").

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    $\begingroup$ Could you clarify what is $G$? Is it an arbitrary discrete group? amenable? If so, your question includes as a subquestion: which groups admit a faithful action on the circle. There's a way to answer this in terms of left invariant cyclic orders, but in practice I'm not sure it is so helpful to recognize whether a given group has a faithful action on the circle. $\endgroup$
    – YCor
    Oct 8, 2013 at 13:01
  • $\begingroup$ Let us say for now, $G$ is a finitely-generated torsion-free nilpotent group. They act on the circle; see Farb, Franks: "Groups of homeomorphisms on one-manifolds, III: Nilpotent subgroups". This, really, is my main question, and I apologize if it's very easy. I'd also be interested in the question for more general discrete amenable groups. $\endgroup$ Oct 8, 2013 at 13:30
  • $\begingroup$ It looks like a nice question. Do you know the answer for a f.g. abelian group? $\endgroup$
    – YCor
    Oct 8, 2013 at 14:12
  • $\begingroup$ Good question. If $G$ is a f.g. abelian group, then a homomorphism $\psi: G \to \mathbb{R}/\mathbb{Z}$ can be realized in the above way iff $\ker(\psi)$ is torsion-free. We are looking for a faithful action $\phi$ of $G$ on the circle whose image in rotation number has a certain form, which by the result of Matsumoto I mentioned identifies $\phi$ up to semiconjugacy. It turns out, then, that if $\ker(\phi)$ had torsion, there would be a finite-order (orientation-preserving) homeomorphism of an interval, which is impossible. $\endgroup$ Oct 9, 2013 at 12:00
  • $\begingroup$ If $\ker(\psi)$ is torsion-free, we can build a faithful action by taking an orbit of $im(\psi)$, "blowing up" the points in this orbit into (countably many) intervals, and making $\ker(\psi)$ act non-trivially on these intervals. Like I said, f.g. torsion-free nilpotent groups, and in particular abelian groups, act faithfully on the interval [0, 1]. Elements of $G$ not in $\ker(\psi)$ send one interval to another, in whatever way is dictated by their action under $\psi$ on the circle. If you do careful bookkeeping, all the necessary relations of $G$ will hold. $\endgroup$ Oct 9, 2013 at 12:01

2 Answers 2


I think the paper http://arxiv.org/abs/0910.0218 contains nice information related to the questions.


The answer to my question, if $G$ is assumed to be a finitely generated torsion-free nilpotent group, is that any homomorphism $\psi: G \to \mathbb{R}/\mathbb{Z}$ can be realized as $\rho \circ \phi$ for $\phi: G \to Homeo_+(S^1)$ 1-1. I'm sure this was already known, and the paper referenced by Dan Sălăjan definitely contains the right way to think about it.

Suppose we have $\psi: G \to \mathbb{R}/\mathbb{Z}$. Take the orbit of some $x \in S^1$ under $im(\psi)$, and blow up the points in this orbit so that they are intervals. We could call the circle with blown-up intervals $\tilde{S^1}$. Obviously, we can define an action of $im(\psi)$ on $\tilde{S^1}$, for instance by sending one interval to another affinely.

Now $G$ is a f.g. torsion-free nilpotent group, so it acts faithfully on the interval $[0, 1]$, and thus so does $\ker(\psi)$. If we take a copy of this action of $\ker(\psi)$ on each blown-up interval, together with the action of $im(\psi)$ that sends one interval to another, then altogether we have an action of the wreath product $\ker(\psi) \wr im(\psi)$ on $\tilde{S^1}$.

By the Kaloujnine-Krasner Theorem, $\ker(\psi) \wr im(\psi)$ contains a copy of every extension of $\ker(\psi)$ by $im(\psi)$, including $G$ itself. Indeed, under this embedding $\Phi\colon G \hookrightarrow \ker(\psi) \wr im(\psi)$, the projection of $\Phi(g)$ onto the $im(\psi)$ factor is precisely $\psi(g)$. (For more information on the Kaloujnine-Krasner Theorem, see e.g. Lectures on Finitely Generated Solvable Groups, in SpringerBriefs in Mathematics.) So this allows us to define our embedding $\phi\colon G \to Homeo_+(S^1)$ having the desired property.


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