# $\mathrm{Homeo}_{\mathbb{Z}} (\mathbb{R})$ is a perfect group

I would like to show that $$\mathrm{Homeo}_{\Bbb Z} (\Bbb R)= \{ f : \Bbb R \to \Bbb R \in \mathrm{Homeo}(\mathbb{R}) \mid \forall x, f(x+1)=f(x)+1 \}$$ is a perfect group (ie. equal to its derived subgroup). I already know that the group $\mathrm{Homeo}^+(\mathbb{S}^1)$ of increasing homeomorphisms of the circle is perfect.

If $\tilde{f}_1,\dots,\tilde{f}_r : \mathbb{R} \to \mathbb{R}$ lift $f_1, \dots, f_r : \mathbb{R} / \mathbb{Z} \to \mathbb{R} / \mathbb{Z}$ respectively, then $w(\tilde{f}_1,\dots,\tilde{f}_r)$ lifts $w(f_1,\dots,f_r)$ for any word $w$. Therefore, because any element of $\mathrm{Homeo}_{\Bbb Z} (\Bbb R)$ is a lift of an element of $\mathrm{Homeo}^+(\mathbb{S}^1)$ and that two lifts disagree by a power of $T : x \mapsto x+1$, we deduce that any element of $\mathrm{Homeo}_{\Bbb Z}(\mathbb{R})$ can be written as a power of $T$ followed by a product of commutators.

To conclude, it is then sufficient to show that $T$ is a commutator in $\mathrm{Homeo}_{\Bbb Z}(\mathbb{R})$. Although it is easy in $\mathrm{Homeo}(\mathbb{R})$ (a translation is a commutator of two reflections), it seems to be more difficult in $\mathrm{Homeo}_{\Bbb Z}(\mathbb{R})$. Do you have an idea?

• If you use the Euler class of the standard (genus $g$) surface group action on the circle, you see that the translation by $2g-2$ is a product of commutators of elements of $Homeo_Z(R)$. Thus, (using $g=2$), translation by 2 is in the commutator subgroup. Working a bit harder one constructs surface group actions (nondiscrete ones) with Euler class 1 too. Nov 27, 2013 at 11:32
• Unfortunately, I never heard about what you are dealing with... Nov 27, 2013 at 12:24
• I think, Ghys has several papers which you would enjoiy reading, which discuss relation of the rotation number and the Euler class. I will dig out some references... Nov 27, 2013 at 13:09

Here is a brief explanation, I will write more details later on, when I have time. Consider a closed oriented hyperbolic surface $S$ of genus $g\ge 2$. The hyperbolic structure on $S$ determines a properly discontinuous isometric action of the group $\Gamma=\pi_1(S)$ on the unit disk $D$ (Poincaré model of hyperbolic plane). In particular, we get an action of the surface group on the unit circle $S^1$ bounding $D$. Consider the associated $S^1$-bundle $\xi: E\to S$. The Euler number $e(\xi)$ of this bundle equals $\chi(S)$ (since this circle bundle is isomorphic to the unit tangent bundle of $S$). Let $a_1,b_2,\dots,a_g, b_g$ be the standard generators of $\Gamma$, $$\prod_i [a_i, b_i]=1.$$
Now, let's lift the action of $\Gamma$ from the circle to the real line. The obstruction to the lifting is precisely the Euler class of $\xi$ and one verifies that if $\tilde a_i, \tilde b_i$ are lifts of the generators of $\Gamma$ to homeomorphisms of ${\mathbb R}$ (it does not matter which lifts you take), then $$\prod_i [\tilde a_i, \tilde b_i] =\tau, \tau(x)=x+(2-2g),$$ translation of the real line. In particular, all even translations of the real line are in the commutator subgroup of $\mathrm{Homeo}_Z({\mathbb R})$.
One can also get isometric actions of $\Gamma$ on the hyperbolic plane (no longer properly discontinuous) for which the Euler number can take any integer value between $\chi(S)$ and $-\chi(S)$. (Ghys likely discusses these as well in one of his papers.) Thus, by repeating the above argument, one shows that the entire group of integers lies in the commutator subgroup.