Read-back

The statement is a conjunction of two independent claims, (I) and (II), about a group $H_{\mathbb Q}$ of homeomorphisms of the projective line, and (in (II)) Thompson's group $F$ acting on $[0,1]$. It has no hypotheses and no free variables. All the objects involved are defined first.

The objects

The projective line. $\widehat{\mathbb R} = \mathbb R \cup \{\infty\}$ is the one-point compactification of $\mathbb R$: a set $U$ is open iff $U \cap \mathbb R$ is open in $\mathbb R$ and, if $\infty \in U$, the set $\mathbb R \setminus U$ is compact. So the neighbourhoods of $\infty$ are the sets containing $\{x \in \mathbb R : |x| > R\} \cup \{\infty\}$ for some $R$. A real number $x$ is regarded as a point of $\widehat{\mathbb R}$ by the obvious inclusion. The rational points are

$$\mathbb Q \cup \{\infty\} \subseteq \widehat{\mathbb R}.$$

Möbius maps. For a subring $A \subseteq \mathbb R$, $\mathrm{SL}_2(A)$ is the group of $2\times 2$ matrices with entries in $A$ and determinant $1$. Such a matrix $g = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$ acts on $\widehat{\mathbb R}$ (viewed as a real matrix) by

$$g\cdot x = \begin{cases} \dfrac{ax+b}{cx+d} & x\in\mathbb R,\ cx+d \ne 0,\\[4pt] \infty & x \in \mathbb R,\ cx+d = 0,\end{cases}\qquad g\cdot\infty = \begin{cases} a/c & c \ne 0,\\ \infty & c = 0.\end{cases}$$

Only two subrings are used: $A = \mathbb R$ (matrices in $\mathrm{SL}_2(\mathbb R)$) and $A = \mathbb Z$, the smallest subring of $\mathbb R$ (matrices in $\mathrm{SL}_2(\mathbb Z)$).

Piecewise-Möbius condition. For a subring $A$, a set $E \subseteq \widehat{\mathbb R}$ and a homeomorphism $f$ of $\widehat{\mathbb R}$, say $f$ is piecewise-$A$ with breakpoints in $E$ if there is a finite set $B \subseteq E$ such that for every point $x \in \widehat{\mathbb R}\setminus B$ (including $x = \infty$ when $\infty \notin B$) there is a matrix $g\in\mathrm{SL}_2(A)$ and a neighbourhood $V$ of $x$ in $\widehat{\mathbb R}$ with $f(y) = g\cdot y$ for all $y \in V$. Nothing is required at points of $B$; $B$ may be empty.

The groups. The homeomorphisms of $\widehat{\mathbb R}$ form a group under composition, $(f g)(x) = f(g(x))$.

• $G_{pp}$ is the subgroup generated by all homeomorphisms $f$ of $\widehat{\mathbb R}$ that are piecewise-$\mathbb R$ with breakpoints anywhere in $\widehat{\mathbb R}$ (i.e. finitely many breakpoints, local pieces from $\mathrm{SL}_2(\mathbb R)$).
• $G_{\mathbb Q}$ is the subgroup generated by all homeomorphisms $f$ of $\widehat{\mathbb R}$ such that both (a) $f \in G_{pp}$, and (b) $f$ is piecewise-$\mathbb Z$ with breakpoints in $\mathbb Q\cup\{\infty\}$ (finitely many breakpoints, all rational or $\infty$; local pieces from $\mathrm{SL}_2(\mathbb Z)$).
• $H_{\mathbb Q} = \{h \in G_{\mathbb Q} : h(\infty) = \infty\}$, the stabiliser of $\infty$ in $G_{\mathbb Q}$, with the group law of composition.

Thompson's group $F$ on $[0,1]$. A real number is dyadic if it equals $m/2^k$ for some $m\in\mathbb Z$, $k \in \mathbb N = \{0,1,2,\dots\}$. Consider the order-preserving bijections $f\colon[0,1]\to[0,1]$ (order isomorphisms; these form a group under composition, $(fg)(z) = f(g(z))$). Call $f$ a Thompson map if there is a finite set $B$ of dyadic reals (not required to lie in $[0,1]$) such that: for all $x<y$ in $[0,1]$ for which the open interval $(x,y)$ contains no point of $B$, there are $n \in\mathbb Z$ and $c\in\mathbb R$ with

$$f(z) = 2^n z + c \quad\text{for all } z \in [x,y].$$

(The constant $c$ is not required to be dyadic.) $F$ is the subgroup generated by all Thompson maps.

Extension by the identity. For an order isomorphism $f$ of $[0,1]$, $\bar f\colon\mathbb R\to\mathbb R$ is $\bar f(x) = f(x)$ for $x\in[0,1]$ and $\bar f(x) = x$ otherwise. For $t \in (0,1)$ this is just $\bar f(t) = f(t)$, and $f(t) \in (0,1)$, since an order isomorphism of $[0,1]$ fixes $0$ and $1$.

The statement

(I) Smoothness. For every $h \in H_{\mathbb Q}$ there is a function $u\colon\mathbb R\to\mathbb R$ of class $C^1$ (differentiable at every real point, with continuous derivative $u'\colon\mathbb R\to\mathbb R$) such that

$$h(x) = u(x) \qquad\text{for every } x \in \mathbb R,$$

the equality taking place in $\widehat{\mathbb R}$. In particular $h$ sends every real number to a real number (never to $\infty$), and the restriction $h|_{\mathbb R}\colon \mathbb R\to\mathbb R$ is $C^1$ on all of $\mathbb R$. The function $u$ may depend on $h$.

(II) Conjugacy to $F$. There exist

• a function $c\colon\mathbb R\to\mathbb R$ that is strictly increasing on the open interval $(0,1)$ and satisfies $c\big((0,1)\big) = \mathbb R$ (so $c$ restricted to $(0,1)$ is a strictly increasing bijection $(0,1)\to\mathbb R$; the values of $c$ outside $(0,1)$ are unconstrained and play no role), and then
• a group isomorphism $\varphi\colon H_{\mathbb Q}\to F$ (a bijection with $\varphi(h_1 h_2) = \varphi(h_1)\varphi(h_2)$, both group laws being composition),

such that for every $h\in H_{\mathbb Q}$ and every $t \in (0,1)$,

$$h\big(c(t)\big) = c\big(\overline{\varphi(h)}\,(t)\big) = c\big(\varphi(h)(t)\big),$$

the equality again taking place in $\widehat{\mathbb R}$ with both sides real. Equivalently, on $(0,1)$,

$$h\circ c = c\circ \varphi(h), \qquad\text{i.e.}\qquad \varphi(h) = c^{-1}\circ h|_{\mathbb R}\circ c \ \text{ on } (0,1),$$

where $c^{-1}$ is the inverse of the bijection $c|_{(0,1)}\colon(0,1)\to\mathbb R$.

The order of quantifiers is: $c$ is chosen first, then $\varphi$, and the single pair $(c,\varphi)$ works for every $h$ and every $t$. Claims (I) and (II) are asserted jointly (both must hold) but do not share any witnesses.