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This item is a bundle of definitions (with a few small auxiliary facts proved along the way). They share the following vocabulary.

Shared vocabulary

The circle. $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ is the additive quotient of $\mathbb{R}$ by the subgroup of integer multiples of $1$. For $x \in \mathbb{R}$, $[x] \in \mathbb{T}$ denotes its class. Every point of $\mathbb{T}$ has exactly one representative in the half-open interval $[0,1)$; the "representative map" $\rho : \mathbb{T} \to [0,1)$ sends a class to that representative, and its inverse is $x \mapsto [x]$.

Permutations of the circle. $\mathrm{Sym}(\mathbb{T})$ is the group of all bijections $\mathbb{T} \to \mathbb{T}$ — no continuity or order condition is built in. The product is composition with the right factor applied first: $(fg)(p) = f(g(p))$.

Dyadic rationals. A real number $x$ is dyadic if there are an integer $m$ and a natural number $k \ge 0$ with

$$x = \frac{m}{2^k}.$$

($k = 0$ is allowed, so every integer is dyadic.)

The unit interval. $I = [0,1] \subset \mathbb{R}$ (closed), with its usual order. An order isomorphism of $I$ is an order-preserving bijection $I \to I$ whose inverse is also order-preserving.

Powers of 2. Wherever $2^n$ appears with $n$ an integer, it is the real number $2^n$ with $n$ possibly negative (so $2^{-3} = 1/8$).

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1. Circle maps of "Thompson type"

A bijection $f \in \mathrm{Sym}(\mathbb{T})$ is of Thompson type if there exists an order isomorphism $L : \mathbb{R} \to \mathbb{R}$ (an order-preserving bijection with order-preserving inverse) satisfying all four of the following:

1. Periodicity: $L(x+1) = L(x) + 1$ for every real $x$.

2. $L$ lifts $f$: $f([x]) = [L(x)]$ for every real $x$.

3. $L$ preserves dyadics: for every real $x$, if $x$ is dyadic then $L(x)$ is dyadic. (Only this direction is required.)

4. Piecewise dyadic-affine with finitely many dyadic breakpoints mod 1: there exists a finite set $B \subset \mathbb{R}$, all of whose elements are dyadic, such that for all reals $x < y$:

   if the open interval $(x,y)$ contains no point of the form $b + k$ with $b \in B$ and $k \in \mathbb{Z}$, then there exist an integer $n$ and a real number $c$ such that

$$L(z) = 2^n z + c \qquad \text{for every } z \in [x,y].$$

Remarks on the fine print:

• In (4) the hypothesis is only about the open interval $(x,y)$; the endpoints $x,y$ may themselves be of the form $b+k$, and the affine formula is then asserted on the closed interval $[x,y]$.
• The set $B$ is any finite set of dyadic reals; it is not required to lie in $[0,1)$, and it may be empty. If $B$ is empty, the hypothesis of (4) holds for every $x<y$.
• The integer $n$ and the real $c$ may depend on $x$ and $y$. The constant $c$ is not itself required to be dyadic.
• Nothing is said about the inverse of $f$ or of $L$ beyond $L$ being an order isomorphism.

2. The group $T$

$T$ is the subgroup of $\mathrm{Sym}(\mathbb{T})$ generated by all bijections of Thompson type (Section 1), i.e. the smallest subgroup of $\mathrm{Sym}(\mathbb{T})$ containing every such bijection. Its elements are exactly the finite products of Thompson-type bijections and their inverses (the empty product being the identity).

3. Two auxiliary facts about order isomorphisms of $I$ (proved, not assumed)

• For every order isomorphism $g$ of $I=[0,1]$: $g(1) = 1$.
• For every order isomorphism $g$ of $I$ and every $x \in I$: $g(x) < 1$ if and only if $x < 1$.

4. From interval maps to circle maps

Restriction to $[0,1)$. Given an order isomorphism $g$ of $I$, define the bijection $\hat g$ of the half-open interval $[0,1)$ by $\hat g(x) = g(x)$. (It lands in $[0,1)$ by the second fact of Section 3; its inverse is $x \mapsto g^{-1}(x)$.)

Transport to the circle. For an order isomorphism $g$ of $I$, $\Phi(g) \in \mathrm{Sym}(\mathbb{T})$ is the bijection

$$\Phi(g) = \rho^{-1} \circ \hat g \circ \rho, \qquad\text{i.e.}\qquad \Phi(g)([x]) = [\,g(x)\,] \text{ for } x \in [0,1).$$

In words: take the representative in $[0,1)$, apply $g$, and take the class again.

5. The map $C$ on the circle

Define $\gamma : \mathbb{R} \to \mathbb{R}$ by

$$\gamma(x) = \begin{cases} x/2 + 3/4, & x < 1/2,\\ 2x - 1, & 1/2 \le x < 3/4,\\ x - 1/4, & x \ge 3/4,\end{cases}$$

and $\delta : \mathbb{R} \to \mathbb{R}$ by

$$\delta(y) = \begin{cases} (y+1)/2, & y < 1/2,\\ y + 1/4, & 1/2 \le y < 3/4,\\ 2y - 3/2, & y \ge 3/4.\end{cases}$$

Only their values on $[0,1)$ matter below. Proved auxiliary facts: $\gamma$ and $\delta$ each map $[0,1)$ into $[0,1)$, and on $[0,1)$ they are inverse to each other ($\delta(\gamma(x)) = x$ and $\gamma(\delta(x)) = x$ for $x \in [0,1)$).

Concretely on $[0,1)$: $\gamma$ sends $[0,1/2)$ to $[3/4,1)$ by $x \mapsto x/2+3/4$; $[1/2,3/4)$ to $[0,1/2)$ by $x \mapsto 2x-1$; and $[3/4,1)$ to $[1/2,3/4)$ by $x \mapsto x-1/4$.

$\hat\gamma$ is the resulting bijection of $[0,1)$ (inverse $\delta$), and

$$C = \rho^{-1} \circ \hat\gamma \circ \rho \in \mathrm{Sym}(\mathbb{T}), \qquad C([x]) = [\gamma(x)] \text{ for } x \in [0,1).$$

6. The interval maps $A_I$ and $B_I$ (from the imported vocabulary)

$A_I$ is the order isomorphism of $I = [0,1]$ given by

$$A_I(x) = \begin{cases} x/2, & 0 \le x \le 1/2,\\ x - 1/4, & 1/2 \le x \le 3/4,\\ 2x - 1, & 3/4 \le x \le 1,\end{cases}$$

and $B_I$ is the order isomorphism of $I$ given by

$$B_I(x) = \begin{cases} x, & 0 \le x \le 1/2,\\ x/2 + 1/4, & 1/2 \le x \le 3/4,\\ x - 1/8, & 3/4 \le x \le 7/8,\\ 2x - 1, & 7/8 \le x \le 1.\end{cases}$$

(The formulas agree at the shared breakpoints. Each is the restriction to $[0,1]$ of an order isomorphism of $\mathbb{R}$ that is the identity outside $(0,1)$.)

7. The generator assignment $\sigma$

A three-letter alphabet $\{A, B, C\}$ is introduced (three distinct symbols). The function $\sigma$ from this alphabet to $\mathrm{Sym}(\mathbb{T})$ is

$$\sigma(A) = \Phi(A_I), \qquad \sigma(B) = \Phi(B_I), \qquad \sigma(C) = C.$$

So $\sigma(A)([x]) = [A_I(x)]$ and $\sigma(B)([x]) = [B_I(x)]$ for $x \in [0,1)$. This is only a function on the three letters; no group homomorphism out of any group is defined or claimed here.

8. The presented group $T_1$

Write $a, b, c$ for the free generators of the free group $F(a,b,c)$ on the three letters. The relator set consists of the following six elements of $F(a,b,c)$ (products read left to right; $a^{-2}$ means $(a^{-1})^2$):

1. $(ab^{-1})\,(a^{-1}ba)\,(ab^{-1})^{-1}\,(a^{-1}ba)^{-1}$
2. $(ab^{-1})\,(a^{-2}ba^{2})\,(ab^{-1})^{-1}\,(a^{-2}ba^{2})^{-1}$
3. $c^{-1}\,b\,(a^{-1}cb)$
4. $\big((a^{-1}cb)(a^{-1}ba)\big)^{-1}\,b\,(a^{-2}cb^{2})$
5. $(ca)^{-1}\,(a^{-1}cb)^{2}$
6. $c^{3}$

$T_1$ is the group $\langle a, b, c \mid \text{relators 1–6}\rangle$: the quotient of $F(a,b,c)$ by the normal subgroup generated by the six relators (so each relator becomes the identity). Below, $a, b, c$ also denote their images in $T_1$.

9. The elements $X_n$ and $C_n$ of $T_1$

For $n \in \mathbb{N} = \{0,1,2,\dots\}$:

$$X_0 = a, \qquad X_{n+1} = (a^{n})^{-1}\, b\, a^{n} \quad (n \ge 0).$$

So $X_1 = b$, $X_2 = a^{-1}ba$, $X_3 = a^{-2}ba^{2}$, and so on. Note the index shift: $X_{n+1}$ conjugates by $a^{n}$, not $a^{n+1}$.

$$C_0 = 1, \qquad C_{n+1} = (a^{n})^{-1}\, c\, b^{n} \quad (n \ge 0).$$

So $C_1 = c$, $C_2 = a^{-1}cb$, $C_3 = a^{-2}cb^{2}$, and so on. (The right-hand factor is a power of $b$, not of $a$.)

In this notation relators 1–6 read: $[ab^{-1}, X_2]$, $[ab^{-1}, X_3]$ (commutator $[u,v] = uvu^{-1}v^{-1}$), $C_1^{-1} X_1 C_2$, $(C_2X_2)^{-1}X_1C_3$, $(C_1a)^{-1}C_2^{2}$, $C_1^{3}$.

10. Positive elements of $T_1$

An element $g \in T_1$ is positive if it lies in the submonoid of $T_1$ generated by $\{X_n : n \in \mathbb{N}\}$ — that is, $g$ is a finite product $X_{n_1}X_{n_2}\cdots X_{n_r}$ with $r \ge 0$ and each $n_i \in \mathbb{N}$ (no inverses; the empty product $r=0$ gives $g = 1$, so the identity is positive). Note $X_0 = a$ is among the generators. Positivity is a property of the group element, not of a chosen word: $g$ is positive if some such product equals $g$ in $T_1$.