Question on the proof of $\mathrm{scl}(a) = |\mathrm{rot}(a)|/2$ in $\mathrm{Homeo}^+(\Bbb R)^{\Bbb Z}$

Question

I'm currently reading a part of Calegari's scl and encounter a proof I can't understand well.

Theorem 2.43. Let $\mathrm{Homeo}^+(\Bbb R)^{\Bbb Z}$ denote the full preimage of $\mathrm{Homeo}^+(S^1)$ in $\mathrm{Homeo}^+(\Bbb R)$. Then $\mathrm{scl}(a) = |\mathrm{rot}(a)|/2$ in $\mathrm{Homeo}^+(\Bbb R)^{\Bbb Z}$.

$\mathrm{scl}$ here means stable commutator length and $\mathrm{rot}$ here means Poincare rotation number (rotation number of $g\in\mathrm{Homeo}^+(\Bbb R)^{\Bbb Z}$ is defined by $\lim_{n\to\infty}{g^n(0)\over n}$.)
Proof. Let $b$ be an element which translates some elements in the positive direction and some elements in the negative direction. Then for any $p\in\Bbb R$ and any small $\epsilon>0$, some conjugate of $b$ takes $p$ to $p+1-\epsilon$. Similarly, some conjugate of $b^{-1}$ takes $b(p)$ to $b(p)+1-\epsilon$. It follows that for any $p\in\Bbb R$ and any small $\epsilon>0$ there is a commutator which takes $p$ to $p+2-2\epsilon$.
Given $a$ with $|\mathrm{rot}(a)| = r$, the power $a^n$ moves every point a distance less than $nr+1$. It turns out that for any $p\in\Bbb R$ and any $|s|<2$ one can find a commutator $g$ such that $g(p)-p =s$. Therefore $a^n$ can be written as a product of at most $|(nr+1)/2|+1$ commutators with an element $a'$ which fixes some point. The dynamics of $a'$ on every complementary intervals to $\text{fix}(a')$ is topologically conjuguate to a translation of $\Bbb R$, which is the commutator of two dilatations. ...

There are four sentences I can't understand. For the third one, I can't see the role of $a'$ with some fixed point. And for the last one, the commutator of two dilations is just an identity isn't it? I can't see how translation can be produced.

Answer 1

Here $G = \mathrm{Homeo}^+(\mathbb R)^{\mathbb Z}$ denotes the set of orientation-preserving homeomorphisms $f$ of $\mathbb R$ satisfying $f(x+n) = f(x) + n$ for all $x \in \mathbb R$ and $n \in \mathbb Z$ (in other words, such that $f(x) - x$ is $1$-periodic). Also $b$ is any fixed element of $G$ that moves some elements in the positive direction and some elements in negative direction, e.g., we could take $b(x) = x + \sin(2 \pi x) / 1000$.

1. Let $\tau_h(x) = x+h$ and observe that $\tau_h\circ b \circ \tau_h^{-1}(x) = b(x-h) + h$, so we can easily translate $b$ around. Therefore we may assume $p = \epsilon/2$. Then $b(p) - p$ is also small and positive. Let $f$ be an orientation-preserving homeomorphism of $[0,1]$ such that $f(p) = p$ and $f(b(p)) = 1-\epsilon/2$ (e.g., a suitable piecewise-linear function), and extend $f$ to an element of $G$ in the unique way. Then $f \circ b \circ f^{-1}$ maps $p=\epsilon/2$ to $1-\epsilon/2 = p + 1 - \epsilon$.

2. The point is that a commutators are the same thing as quotients of conjugates. Indeed $[g, h] = g^{-1} g^h$ and $g^{-h} g^{h'} = [g^h, h^{-1}h']$. We can find a conjugate $b^{g_1}$ that maps $p$ to $p+1-\epsilon$ and a conjugate $b^{-g_2}$ of $b^{-1}$ that maps $p+1-\epsilon$ to $p+2-2\epsilon$, so their composition $b^{-g_2} b^{g_1} = [b^{g_2}, g_2^{-1} g_1]$ maps $p$ to $p+2-2\epsilon$. An easy modification of this argument achieves the same with $p+s$ for any required $s \in (-2,2)$.

3. $a^n$ moves $p$ to $p+\delta$ where $|\delta| \le nr+1$ (as stated). By 2 we can also find a product $c_1 \cdots c_k$ of $k \le \lfloor (nr+1)/2 \rfloor + 1$ commutators that does the same. Then $a^n = c_1 \cdots c_k a'$ where $a'(p) = p$.

4. Finally we want to determine the commutator length of an element $a'$ that fixes some point $p$. Without loss of generality (conjugating by $\tau_p$) we may assume $p = 0$. Hence $a'$ defines an orientation-preserving homeomorphism of $[0,1]$. Consider the fixed points of $a'$ in $[0,1]$. Between any two fixed points, between $u$ and $v$ say, we have $a(x) > x$ identically or $a(x) < x$. Assume $a(x) > x$ identically on $(u,v)$. We can define an orientation-preserving homeomorphism $f : (u,v) \to \mathbb R$ so that $f(a(x)) = f(x) + 1$ for all $x \in (u,v)$. Indeed, start by choosing any $x_0 \in (u,v)$, let $x_1 = a(x_0) > x_0$. Define $f$ on $[x_0, x_1]$ to be any orientation-preserving homeomorphism $[x_0,x_1] \to [0,1]$ (e.g., linear). Now extend $f$ to the rest of $(u,v)$ uniquely by the rule $f(a^n(x)) = f(x) + n$. This explains the sentence "the dynamics of $a'$ on complementary intervals of $\mathrm{fix}(a')$ is topologically conjugate to a translation of $\mathbb R$". Now consider the dilations $d_1(x) = 2x$ (centered at $0$) and $d_2(x) = 2x - 4$ (centered at $4$). We have $[d_2,d_1](x) = d_2^{-1}d_1^{-1}d_2d_1(x) = x+1 = \tau_1(x)$, so indeed a translation is the commutator of two dilations.