Proof. Omitting for convenience the arguments $\lambda ,b$ from the partition functions, we have

$$ \begin{align*} Z_{T,\mathrm{\texttt{+}} r_2}=Z_{T_2,\mathrm{\texttt{+}} r_2} ( Z_{T_1,\mathrm{\texttt{+}} r_1}+b Z_{T_1,\mathrm{\texttt{-}} r_1})^{k}, \quad Z_{T,\mathrm{\texttt{-}} r_2}=Z_{T_2,\mathrm{\texttt{-}} r_2} (b Z_{T_1,\mathrm{\texttt{+}} r_1}+ Z_{T_1,\mathrm{\texttt{-}} r_1})^{k}. \end{align*} $$

Dividing these yields the result (note, $Z_{T_2,\mathrm {\texttt {-}} r_2}\neq 0$ and $\xi _1=\tfrac {Z_{T_1,\mathrm {\texttt {+}} r_1}}{Z_{T_1,\mathrm {\texttt {-}} r_1}}\in \mathbb {S}$, so $Z_{T,\mathrm {\texttt {-}} r_2}\neq 0$); the fact that $\xi \in \mathbb {S}$ follows from footnote 6.

Proof. For $b\in (\frac {k-1}{k+1}, 1)$, the range of $\Lambda _k(b)$ follows from [Reference Peters and Regts42, Theorem 14]. For $b\in (0,\frac {k-1}{k+1}]$, the range of $\Lambda _k(b)$ follows from item (i) in Lemma 10. The characterisation of the Julia set for $\lambda \in \Lambda _k(b)$ is shown in [Reference Peters and Regts42, Proof of Proposition 17].

Proof. We refer to [Reference Peters and Regts42] for proofs of items (i)–(iv). Specifically, item (i) follows from [Reference Peters and Regts42, Lemma 8 & Equation (3.1)], item (ii) from [Reference Peters and Regts42, Lemma 13, Theorem 14, Proof of Proposition 17], item (iii) from [Reference Peters and Regts42, Proof of Theorem 14] and item (iv) from [Reference Peters and Regts42, Theorem 14, Proof of Theorem 5(i)].

We will prove item (v). By taking the derivative of both sides of the equality $f_{\lambda ,k}(R_k(\lambda )) = R_k(\lambda )$ with respect to $\lambda $ and rewriting, we obtain

$$ \begin{align*} R_k'(\lambda) = \frac{R_k(\lambda)}{\lambda\Big(1 - f_{\lambda,k}'\big(R_k(\lambda)\big)\Big)}. \end{align*} $$

Using equation (2) for $z = R_m(1) = 1$, we obtain that $R_{i+1}'(1)> R_{i}'(1)$ for $i \in \{1, \dots , d-2\}$, and thus for $\lambda \in \mathbb {S}$ in the upper half-plane near $1$ we find that $\text {Arg}{(R_{i+1}(\lambda ))}> \text {Arg}{(R_{i}(\lambda ))}$.

The derivative at a fixed point of a map of the unit circle to itself is real (see also [Reference Peters and Regts42, Lemma 11]). Furthermore, if such a map is orientation-preserving, the derivative at a fixed point is positive. Because the map $f_{\lambda ,i}$ is orientation-preserving with attracting fixed point $R_i(\lambda )$, we find that $f_{\lambda ,i}'(R_i(\lambda )) \in (0,1)$ for $\lambda \in \operatorname {\mathrm {Arc}}{(\overline {\lambda _i}, \lambda _i)}$. From this, we deduce that we can write

(4)$$ \begin{align} \left|R_k'(\lambda)\right| = \frac{1}{1 - f_{\lambda,k}'\left(R_k(\lambda)\right)}. \end{align} $$

From equation (2) it can be seen that $\left |f_{i}'(z)\right |$ is increasing both with respect to $\text {Arg}{(z)}$ when $\mathrm {Im}(z)>0$ and with respect to the index i and thus, as long as $R_{i}(\lambda )$ and $R_{i+1}(\lambda )$ are both defined and $\text {Arg}(R_{i+1}(\lambda ))> \text {Arg}(R_{i}(\lambda ))$, we deduce that $|R_{i+1}'(\lambda )|> |R_{i}'(\lambda )|$. Since $\text {Arg}(R_{i+1}(\lambda ))> \text {Arg}(R_{i}(\lambda ))$ for $\lambda $ in the upper half-plane close to $1$, we conclude that there cannot be any $\lambda $ in the upper half-plane such that $\text {Arg}(R_{i+1}(\lambda )) \leq \text {Arg}(R_{i}(\lambda ))$.

Now suppose that there is some index i such that $\text {Arg}{(\lambda _i)} \leq \text {Arg}{(\lambda _{i+1})}$. Note that $R_i(\lambda _i)$ is a parabolic fixed point of $f_{\lambda _i,i}$, which means that $f_{\lambda _i,i}'(R_i(\lambda _i)) = 1$. Because we assumed that $\text {Arg}{(\lambda _i)} \leq \text {Arg}{(\lambda _{i+1})}$, we see from item (ii) that $R_{i+1}(\lambda _i)$ must be well-defined. We already deduced that $\text {Arg}{(R_{i+1}(\lambda _i))}> \text {Arg}{(R_{i}(\lambda _i))}$ and thus $f_{\lambda _i,i+1}'(R_{i+1}(\lambda _i))> f_{\lambda _i,i}'(R_i(\lambda _i)) = 1$, which contradicts the fact that $R_{i+1}(\lambda _i)$ is an attracting fixed point of $f_{\lambda _i,i+1}$. This concludes the proof of the first two claims of item (v).

Finally, we show the final claim of item (v). For indices $0\leq i\leq j\leq m$, it will be convenient to denote by $A_{i,j,\lambda }$ the arc $\operatorname {\mathrm {Arc}}{[R_{i}(\lambda ),R_{j}(\lambda )]}$, under the convention that $R_0(\lambda )=1$. For $\lambda \in \operatorname {\mathrm {Arc}}{(1,\lambda _{m}]}$, our goal is hence to show that $\ell \left (A_{i,i+1,\lambda }\right ) \leq \ell \left (A_{i+1,i+2,\lambda }\right )$ for all $i\in \{1, \dots , m-2\}$.

For any $\lambda \in \operatorname {\mathrm {Arc}}{(1,\lambda _{k}]}$, we observe for $i = 1, \dots , k$ that

$$ \begin{align*} \ell(A_{0,i,\lambda}) = \int_{\operatorname{\mathrm{Arc}}{[1,\lambda]}} |R_i'(z)| \,|dz|, \end{align*} $$

and thus for $i \in \{1, \dots , k-1\}$ we have $\ell (A_{i,i+1,\lambda }) = \int _{\operatorname {\mathrm {Arc}}{[1,\lambda ]}} |R_{i+1}'(z)| - |R_{i}'(z)| \,|dz|$.

We first show item (v) for $\lambda $ near $1$. As we let $\lambda $ approach $1$ along the circle, we obtain that

$$ \begin{align*} \lim_{\lambda \to 1} \frac{\ell(A_{i,i+1,\lambda}) }{\ell{\left(\operatorname{\mathrm{Arc}}{[1,\lambda]}\right)}} &= |R_{i+1}'(1)| - |R_{i}'(1)| = \frac{1}{1 - f_{\lambda,i+1}'(1)} - \frac{1}{1 - f_{\lambda,i}'(1)} \\ &=\frac{(1+b)/(1-b)}{\big(i - (1+b)/(1-b)\big)\big(i-2b/(1-b)\big)}. \end{align*} $$

The second equality can be obtained by using (4) and the third by using (2) and simplifying. If we denote this expression by $g(i)$, then it is not hard to see that $g(i+1)> g(i)$ as long as $i + 1 < 2b/(1-b)$. Because $b\in (\tfrac {d-2}{d},1)$, we have $2b/(1-b)>d-2$. So, indeed, $g(i+1)> g(i)$ for $i \in \{1, \dots , d-3\}$, which contains $\{1, \dots , k-2\}$. This shows that the inequality in (3) is true for $\lambda $ near $1$.

Now suppose that there is $\lambda \in \mathbb {S}$ and index i for which the inequality in (3) does not hold. Then, by continuity, because the inequality does hold near $1$, there is a $\lambda \in \mathbb {S}$ for which the inequality is an equality; that is,

(5)$$ \begin{align} \ell\left(A_{i,i+1,\lambda}\right) = \ell\left(A_{i+1,i+2, \lambda}\right). \end{align} $$

For convenience, we will henceforth drop the subscript $\lambda $ from the notation for the arcs $A_{i,j,\lambda }$ and simply write $A_{i,j}$. The maps $f_{\lambda ,j}$ are orientation-preserving for any j and thus

$$ \begin{align*} \ell{\left(\operatorname{\mathrm{Arc}}[\lambda,R_j(\lambda)]\right)} = \ell{\left(f_{\lambda,j}(A_{0,j})\right)} = \int_{A_{0,j}} |f_j'(z)|\, |dz|. \end{align*} $$

Using this equality we can write

$$ \begin{align*} \ell{\left(A_{j,j+1}\right)} = \int_{A_{0,j+1}} |f_{j+1}'(z)|\, |dz| - \int_{A_{0,j}}|f_j'(z)|\, |dz|. \end{align*} $$

We use this equality for $j= i$ and $j= i+1$ and rearrange (5) to obtain

$$ \begin{align*} 2\int_{A_{0,i+1}} |f_{i+1}'(z)|\, |dz| = \int_{A_{0,i+2}} |f_{i+2}'(z)|\, |dz| + \int_{A_{0,i}} |f_i'(z)|\, |dz|. \end{align*} $$

We use (2) to rewrite the left-hand side of this equation as

$$ \begin{align*} 2(i+1) \int_{A_{0,i}} |f_{1}'(z)|\, |dz| + 2 \int_{A_{i,i+1}} |f_{i+1}'(z)|\, |dz|, \end{align*} $$

and we rewrite the right-hand side as

$$ \begin{align*} (i+2)\int_{A_{0,i}} |f_{1}'(z)|\, |dz| + \int_{A_{i,i+2}} |f_{i+2}'(z)|\, |dz| + i\cdot \int_{A_{0,i}} |f_{1}'(z)|\, |dz|. \end{align*} $$

These two being equal implies that

$$ \begin{align*} 2 \int_{A_{i,i+1}} |f_{i+1}'(z)|\, |dz| = \int_{A_{i,i+2}} |f_{i+2}'(z)|\, |dz|. \end{align*} $$

We will show that this yields a contradiction. We rewrite the right-hand side as

$$ \begin{align*} \int_{A_{i,i+2}} |f_{i+2}'(z)|\, |dz| = \int_{A_{i,i+1}} |f_{i+2}'(z)|\, |dz| + \int_{A_{i+1,i+2}} |f_{i+2}'(z)|\, |dz|, \end{align*} $$

and we will show that both summands are greater than $\int _{A_{i,i+1}} |f_{i+1}'(z)|\, |dz|$, which will yield the contradiction. The inequality for the first summand follows easily from the fact that $|f_{i+2}'(z)|> |f_{i+1}'(z)|$ for all $z \in \mathbb {S}$; cf. (2). The second inequality uses the fact that $|f_{i+2}'(z)|$ increases as $\text {Arg}(z)$ increases (when $\operatorname {\mathrm {Im}}{z}> 0$) and thus

$$ \begin{align*} \int_{A_{i+1,i+2}} |f_{i+2}'(z)|\, |dz| &> |f_{i+2}'(R_{i+1}(\lambda))| \cdot \ell{\left(A_{i+1,i+2}\right)} > |f_{i+1}'(R_{i+1}(\lambda))| \cdot \ell{\left(A_{i,i+1}\right)} \\ &> \int_{A_{i,i+1}} |f_{i+1}'(z)|\, |dz|. \end{align*} $$

The second inequality of this derivation uses the assumed equality in (5). This yields the desired contradiction.

Proof. Write $\lambda = x + i y$ with $x,y \in \mathbb {R}$ such that $x^2 + y^2 = 1$. A short calculation gives that

$$ \begin{align*} \operatorname{\mathrm{tr}}^2(f_{\lambda,1}) = \frac{2 \left(x + 1\right)}{1 - b^2}. \end{align*} $$

The value of $tr^2(f_{\lambda ,1})$ strictly increases from $0$ to $4/(1-b^2)$ as x increases from $-1$ to $1$. It follows that there is a unique value $x \in (-1,1)$ such that $\operatorname {\mathrm {tr}}^2(f_{\lambda ,1}) = 4$. This value must coincide with $\textrm {Re}(\lambda _1)$, where $\lambda _1=\lambda _1(b) \in \mathbb {S}$ is as in Lemma 8, completing the proof.

Proof. Let $\xi = x + i y$ with $x,y \in \mathbb {Q}$ such that $x^2 + y^2 = 1$. Because $f_{\xi ,1}$ is elliptic, it is conjugate to a rotation $z \mapsto e^{i \theta } z$ with

(6)$$ \begin{align} 2 \cdot \left(\cos (\theta )+1\right) = \frac{2 \left(x + 1\right)}{1 - b^2}. \end{align} $$

Let $t = 2 \cdot \left (\cos (\theta )+1\right )$. Suppose $\theta $ is an angle corresponding to a rational rotation; that is, if we let $z = e^{i \theta }$, then there is a natural number n such that $z^n = 1$. It follows that then $\overline {z}^n = 1$ and thus both z and $\overline {z}$ are also algebraic integers. Therefore, $z + \overline {z} = 2\cos (\theta )$ is an algebraic integer. It follows that t is an algebraic integer, while the right-hand side of (6) shows that t must also be rational. Because the only rational algebraic integers are integers, we can conclude that t must be an integer and thus $t \in \{0,1,2,3\}$. If $t=0$, we see that $\xi = x = -1$, which we excluded, so only three possible values of t remain. Let $X = t(1+b)/(1-b)$ and $Y = 2ty/(1-b)^2$; then $(X,Y)$ is a rational point on the elliptic curve $E_t$ given by the following equation:

$$ \begin{align*} E_t: Y^2 = X^3 - (t-2)t\cdot X^2 + t^2 \cdot X. \end{align*} $$

The set of rational points of an elliptic curve together with an additional point has a group structure that is isomorphic to $\mathbb {Z}^r \times \mathbb {Z}/N\mathbb {Z}$. The number $r \geq 0$ is called the rank of the curve and the subgroup isomorphic to $\mathbb {Z}/N\mathbb {Z}$ is called the torsion subgroup. The rank and the torsion subgroup of a particular curve can be found using a computer algebra system. Using Sage, if the variable $\texttt{t}$ is declared to be either $1,2$ or $3$, the curve $E_t$ can be defined with the code $\texttt{Et = EllipticCurve([0, -(t-2)*t, 0, t**2, 0])}$. The rank and the torsion subgroup can subsequently be found with the commands $\texttt{Et.rank()}$ and $\texttt{Et.torsion\_subgroup()}$. We find that $E_t$ has rank $0$ for all $t \in \{1,2,3\}$. For $t \in \{1,3\}$ the torsion subgroup is isomorphic to $\mathbb {Z}/2\mathbb {Z}$, and for $t=2$ it is isomorphic to $\mathbb {Z}/4\mathbb {Z}$. This means that there is one rational point on $E_1$ and $E_3$, which we can see is the point $(0,0)$, and there are three rational points on $E_2$, namely, $\{(0,0), (2, \pm 4)\}$. These points do not correspond to values of b within the interval $(0,1)$, which means that $\theta $ cannot correspond to a rational rotation.

Proof. For $k = 1$ it follows from Corollary 13 that $f_{\xi ,1}$ is conjugate to a rotation. If $f_{\xi ,1}$ is conjugate to an irrational rotation, then the orbit of any initial point $z_0$ will get arbitrarily close to $-1$ for which $|f_{1}'(-1)| = \frac {1+b}{1-b}> 1$. Otherwise, if $f_{\xi ,1}$ is conjugate to a rational rotation, there is an integer $N> 1$ such that $f_{\xi ,1}^N(z) = z$ for all z; consider the smallest such integer N. Let $\theta \in (0,\pi ]$ be the angle such that $f_{\xi ,1}$ is conjugate to the rotation $z \mapsto e^{i\theta }\cdot z$. Equation (6) then states that

$$ \begin{align*} 2 \cdot \left(\cos (\theta )+1\right) = \frac{2 \left(\operatorname{\mathrm{Re}}(\xi) + 1\right)}{1 - b^2}. \end{align*} $$

If $N=2$, then $\theta = \pi $ and thus $\operatorname {\mathrm {Re}}(\xi ) = -1$, contradicting $\xi \neq -1$. Hence, $N>2$. From $f_{\xi ,1}^N(z) = z$, we obtain

(7)$$ \begin{align} \prod_{n=0}^{N-1} f_{\xi,1}'(z_n) = (f_{\xi,1}^N)'(z_0) = 1. \end{align} $$

From (2), there are precisely two values of $w\in \mathbb {S}$ such that $|f_{\xi ,1}'(w)| = 1$. Because $N> 2$ and N is the smallest integer such that $f_{\xi ,1}^N(z_0) = z_0$, we conclude there is at least one term, say with index m, of the product in (7) for which $|f_{\xi ,1}'(z_m)|> 1$.

Consider now the case $k \geq 2$ and denote $f = f_{\xi ,k}$. By Lemma 8, for $\xi \in \Lambda _{k}(b)$ the Julia set of f is the circle $\mathbb {S}$. In [Reference Peters and Regts42, Proof of Proposition 17], it is shown that the two Fatou components of f, denoted by $\mathbb {D}$ and $\overline {\mathbb {D}}^c$, are attracting basins and contain the critical points $-b$ and $-1/b$. From [Reference Milnor38, Theorem 19.1], we therefore conclude that the map f is hyperbolic; that is, there exists a conformal metric $\mu $ on a neighbourhood U of $\mathbb {S}$ such that $||D f_z||_\mu \geq \kappa> 1$ for a constant $\kappa $ and all $z \in \mathbb {S}$. We will briefly expand on the meaning of this notation.

For any $z \in U$ the metric $\mu $ induces a norm $||\cdot ||_\mu $ on the tangent space of U at z denoted by $TU_z$. For $z\in \mathbb {S}$ the map f induces a linear map, the derivative of f at z, $Df_z: TU_z \to TU_{f(z)}$. For nonzero $v\in Df_z$, the ratio $||D f_z(v)||_\mu /||v||_\mu $ is independent of the choice v and is denoted by $||D f_z||_\mu $, the norm of $Df_z$.

Because $\mathbb {S}$ is compact and the metric $\mu $ is conformal, there is a constant $c> 0$ such that $|g'(z)|> c \cdot ||D g_z||_\mu $ for all $z \in \mathbb {S}$ and maps $g: \mathbb {S} \to \mathbb {S}$. It follows that for all $N> 0$,

$$ \begin{align*} \prod_{n=0}^{N-1} |f'(z_n)| = |(f^N)'(z_0)|> c \cdot ||D f^N_{z_0}||_\mu \geq c \cdot \kappa^N. \end{align*} $$

There is an $N> 0$ such that the right-hand side of this equation is greater than $1$. The product on the left-hand side of the equation shows that for such an N there must be at least one index $m \in \{0,\dots ,N-1\}$ such that $|f'(z_m)|> 1$.

Proof. Consider the orbit

$$ \begin{align*} \mathcal{S} = \left\{f_{\lambda,1}^n(1): n \geq 1 \right\}. \end{align*} $$

Note that the elements of $\mathcal {S}$ are fields of paths. We have seen in Subsection 3.2 that either $f_{\lambda ,1}$ is conjugate to an irrational rotation or the orbit of $1$ tends towards an attracting or a parabolic fixed point. In either case there is $\sigma _0 \in \mathbb {S}$ such that $\sigma _0 \not \in \mathcal {S}$ and the elements of $\mathcal {S}$ accumulate on $\sigma _0$. It follows from Lemma 15 that there is a positive integer N such that $\sigma := f_{\xi , k}^N(\sigma _0)$ has the property $|f_{k}'(\sigma )|> 1$. Now define

$$ \begin{align*} \mathcal{R} = \left\{f_{\xi, k}^N(s): s \in \mathcal{S} \right\}. \end{align*} $$

By assumption, $\xi $ can be implemented by a rooted tree in $\mathcal {T}_{d+1}$ with root degree $m\leq d-k$, so by inductively applying Lemma 4, the elements of $\mathcal {R}$ are fields of trees in $\mathcal {T}_{d+1}$ whose root degree is $m+k\leq d$. There is a sequence $\{\zeta _n\}_{n \geq 1} \subseteq \mathcal {R}$ accumulating on $\sigma $ without being equal to $\sigma $, which is what we wanted to show.

Proof. We start with the proof of part (b). Let $\lambda '\in \mathbb {S}\setminus I(\theta _b)$. By [Reference Peters and Regts42, Corollary 4 and Theorem 5], it follows that there exists $\lambda \in \mathbb {S}$ arbitrarily close to $\lambda '$ for which there exists a tree $T\in \mathcal {T}_{d+1}$ such that $Z_T(\lambda ,b)=0$. Choose such a tree T with the minimum number of vertices and let v be a leaf of T, from now on referred to as the root of T. Denote $T'=T-v$ and let u be the unique neighbour of v in T. Then $Z_{T,\mathrm {\texttt {-}} v}(\lambda ,b)\neq 0$. Indeed, if $Z_{T,\mathrm {\texttt {-}} v}(\lambda ,b)= 0$, then $Z_{T,\mathrm {\texttt {+}} v}(\lambda ,b)= 0$. Since

$$ \begin{align*}\left (\begin{array}{lr} \lambda &\lambda b \\ b&1 \end{array}\right) \left (\begin{array}{l} Z_{T',\mathrm{\texttt{+}} u}(\lambda,b)\\Z_{T',\mathrm{\texttt{-}} u}(\lambda,b)\end{array}\right) = \left (\begin{array}{l} Z_{T,\mathrm{\texttt{+}} v}(\lambda,b)\\Z_{T,\mathrm{\texttt{-}} v}(\lambda,b)\end{array}\right) \end{align*} $$

and since the matrix is invertible (as $|b|\neq 1$), this would imply $Z_{T',+u}(\lambda ,b)=Z_{T',-u}(\lambda ,b)=0$ and hence $Z_{T'}(\lambda ,b)=0$, contradicting the minimality of T. Therefore, $R_{T,v}=-1$.

Since the map $ z \mapsto \xi (z) :=\frac {Z_{T,\mathrm {\texttt {+}} v}(z,b)}{Z_{T,\mathrm {\texttt {-}} v}(z,b)}$ is holomorphic near $z = \lambda $, it follows that there exists $\lambda ''\in \mathbb {S}_{\mathbb {Q}}$ arbitrarily close to $\lambda '$ such that $ \xi =\xi (\lambda '')\in \operatorname {\mathrm {Arc}}(\lambda _1(b),\overline {\lambda _1(b)})\setminus \{-1\}$. Therefore, by Lemma 14 and Theorem 12, the orbit $\{f^n_{\xi ,1}(1)\}$ is dense in $\mathbb {S}$. So from Lemma 4, by using paths with the tree T attached to all but one of its vertices at the root v of T, we obtain a collection of trees contained in $\mathcal {T}_{d+1}$ whose fields are dense in $\mathbb {S}$.

We next prove part (a) for all $d\geq 2$ and $b \in (\frac {d-2}{d},\frac {d-1}{d+1}]$. The case $b \in (0,\frac {d-2}{d}]$ follows from this by invoking smaller values for d. Let $\lambda \in \mathbb {S}_{\mathbb {Q}}\setminus \{\pm 1\}$.

Let $k_0 = d$ and $m_0 = 0$, and define the sequences $k_n$ and $m_n$ by $k_{n+1} = \left \lfloor \frac {k_n}{2}\right \rfloor $ and $m_{n+1} = m_n + k_{n+1}$. Inductively we show that $m_n \leq d - k_n$: we have $m_0 = d - k_0$ and then

$$ \begin{align*} m_{n+1} = m_n + k_{n+1} \leq d- k_n + k_{n+1} = d - \big(k_n - \left\lfloor\tfrac{k_n}{2}\right\rfloor\big) \leq d - \left\lfloor\tfrac{k_n}{2}\right\rfloor = d - k_{n+1}. \end{align*} $$

Clearly, there is an integer N such that $k_{N+1} = 1$. We claim that for every $n \in \{0, \dots , N\}$ there is a rooted tree in $\mathcal {T}_{d+1}$ with root degree $m_n$ that implements a field $\xi _n$ so that $|f'(\xi _n)|>1$ and at least one of the following holds:

1. 1. There is a tree in $\mathcal {T}_{d+1}$ with root degree at most $m_n$ that implements a field inside $\operatorname {\mathrm {Arc}}{(\lambda _{k_{n+1}},\overline {\lambda _{k_{n+1}}})}\setminus \{-1\}$, or else

2. 2. the set of fields implemented by trees in $\mathcal {T}_{d+1}$ is dense in $\mathbb {S}$.

To show this for $n = 0$, we consider the tree consisting of a single vertex. This tree implements the field $\lambda $ and its root degree is $0$. By equation (2) of Lemma 10 and since $b\leq \frac {d-1}{d+1}$, we have that $|f_{d}'(z)|> 1$ for all $z \in \mathbb {S} \setminus \{1\}$ and, in particular, we have $|f_{d}'(\lambda )|> 1$. If $\lambda \in \operatorname {\mathrm {Arc}}[\overline {\lambda _{k_1}},\lambda _{k_1}]$, then we apply Lemma 17 to obtain Item (2). If $\lambda \in \operatorname {\mathrm {Arc}}{(\lambda _{k_{1}},\overline {\lambda _{k_{1}}})}$, then the tree consisting of a single vertex satisfies the conditions of Item (1).

Now suppose that we have shown the claim for $n-1$ for some $n\geq 1$ and assume that we are in the case of Item (1) (otherwise, we are done); that is, there is a tree in $\mathcal {T}_{d+1}$ with root degree $m_{n-1}$ that implements a field $\xi $ inside $\operatorname {\mathrm {Arc}}{(\lambda _{k_{n}},\overline {\lambda _{k_{n}}})}\setminus \{-1\}$. We can apply Lemma 16 to obtain a tree T in $\mathcal {T}_{d+1}$ with root degree at most $k_n + m_{n-1} = m_n$ that implements a field $\zeta \neq -1$ such that $|f_{k_{n}}'(\zeta )|> 1$. If $\zeta \in \operatorname {\mathrm {Arc}}[\overline {\lambda _{k_{n+1}}},\lambda _{k_{n+1}}]$, we can apply Lemma 17 to obtain Item (2). We can apply this lemma because $m_n \leq d - k_n$. If $\zeta \in \operatorname {\mathrm {Arc}}{(\lambda _{k_{n+1}},\overline {\lambda _{k_{n+1}}})}$, then T itself satisfies the conditions of Item (1), which proves the claim.

To finish the proof, it remains to consider the case of Item (1), where we can find a tree in $\mathcal {T}_{d+1}$ with root degree at most $m_N < d - 1$ which implements a field $\xi $ inside $\operatorname {\mathrm {Arc}}{(\lambda _{1},\overline {\lambda _{1}})}\setminus \{-1\}$. We have shown in Lemma 14 that $f_{\xi ,1}$ is conjugate to an irrational rotation and thus the orbit $\{f_{\xi ,1}^n(1)\}_{n \geq 1}$ is dense in $\mathbb {S}$. The elements of this orbit correspond to rooted trees in $\mathcal {T}_{d+1}$ and, hence, we can conclude Item (2) in this case as well.

Let $\mathcal {R}$ denote the set of all fields implemented by rooted trees in $\mathcal {T}_{d+1}$. Let $\zeta \in \mathcal {R}$ and T be a tree in $\mathcal {T}_{d+1}$ that implements $\zeta $. We construct the tree $\tilde {T}$ with root r obtained by attaching r to the root of T with an edge. Then, the root of $\tilde {T}$ has degree 1 and the field implemented by $\tilde {T}$ is $f_{\lambda ,1}(\zeta )$. So the set of fields implemented by rooted trees in $\mathcal {T}_{d+1}$ whose root degrees are $1$ contains $f_{\lambda ,1}(\mathcal {R})$. Since $f_{\lambda ,1}(\mathbb {S}) = \mathbb {S}$ and $\mathcal {R}$ is dense in $\mathbb {S}$, we conclude that $f_{\lambda ,1}(\mathcal {R})$ is dense in $\mathbb {S}$ as well.

Proof. For $m \in \{1,\dots , k\}$, define the closed interval $I_m = f_m\left ([0,1]\right ) = [f_m(0),f_m(1)]$ and note that $f_m: [0,1] \to I_m$ is bijective with a differentiable inverse. We define a sequence of intervals in the following way. Let $J_0 = J$, and as long as there exists an index m such that $J_n \subseteq I_m$, we define $J_{n+1} = f_m^{-1}(J)$. We will show that this cannot be done indefinitely; that is, there will be some interval $J_n$ such that $J_n \not \subseteq I_m$ for all m.

For an interval $I \subseteq [0,1]$, let $\ell (I)$ denote the length of the interval and denote $\ell (J)$ by $\epsilon $. For each index m choose a partition $I_m = I_{m,L} \cup I_{m,M} \cup I_{m,R}$, where $I_{m,L}, I_{m,M}, I_{m,R}$ are of the form $[f_m(0),a), [a,b], (b, f_m(1)]$, respectively, for a choice of $a,b \in \mathrm {int}(I_m)$ such that $a<b$ and both $\ell (I_{m,L})$ and $\ell (I_{m,R})$ are less than $\epsilon /4$. We can choose $C> 1$ such that ${f_m^{-1}}'(x)> C$ for all indices m and $x \in I_{m,M}$. We will show inductively that for all $n \geq 0$ for which $J_n$ is defined it is the case that $\ell (J_{n}) \geq \epsilon \cdot (1+C^{n})/2$. For $n = 0$ the statement is true. Suppose that the statement is true for $n \geq 0$ for which $J_{n+1}$ is defined. By definition, there is an index m such that $J_n \subseteq I_m$ and $J_{n+1} = f_m^{-1}(J_n)$. We find

$$ \begin{align*} \ell(J_{n+1}) &= \ell(f_m^{-1}(J_n \cap I_{m,M})) + \ell(f_m^{-1}(J_n \cap I_{m,L})) + \ell(f_m^{-1}(J_n \cap I_{m,R}))\\ &\geq C \cdot \ell(J_n \cap I_{m,M}) + \ell(J_n \cap I_{m,L}) + \ell(J_n \cap I_{m,R})\\ &= C \left(\ell(J_n) - \ell(J_n \cap I_{m,L}) - \ell(J_n \cap I_{m,R})\right) + \ell(J_n \cap I_{m,L}) + \ell(J_n \cap I_{m,R}),\\[-15pt] \end{align*} $$

where we have used that ${f_m^{-1}}'(x) \geq 1$ for $x \in I_m$. Because $\ell (J_n \cap I_{m,L}) + \ell (J_n \cap I_{m,R}) \leq \epsilon /2$, this is again at least equal to

$$ \begin{align*} C (\ell(J_n) - \epsilon/2) + \epsilon/2 \geq C (\epsilon \cdot (1+C^{n})/2 - \epsilon/2) + \epsilon/2 = \epsilon \cdot (1+C^{n+1})/2.\\[-15pt] \end{align*} $$

It follows that there is an index n such that $J_n$ is not totally contained inside $I_m$ for any index m. This means that there is an m such that $J_n$ contains at least one of the endpoints of $I_m$. Without loss of generality, we can assume that $J_n$ contains the left endpoint of $I_m$. It follows that there is an $a> 0$ such that $f_m([0,a]) \subset J_n$ and thus there is a sequence $m_1, \dots , m_n$ such that $(f_{m_1} \circ \cdots \circ f_{m_n} \circ f_m)([0,a]) \subset J$. We complete the proof by showing that for at least one of the maps $f_i$ there is an index $N_a$ for any $a> 0$ such that $f_i^{N_a}([0,1]) \subset [0,a]$.

Observe that there must be at least one map $f_i$ such that $f_i(0) = 0$. We obtain an inclusion of intervals $[0,1] \supset f_i([0,1]) \supset f_i^2([0,1]) \supset \cdots $, where $f_i^N([0,1]) = [0, f_i^N(1)]$. This shows that the sequence $\{f_i^N(1)\}_{N \geq 0}$ is decreasing and thus has a limit L. If $L \neq 0$, we would have $f_i([0,L]) = [0,L]$, which contradicts the fact that $f_i'(x) < 1$ for all $x \in (0,L)$, so $L = 0$. This concludes the proof.

Proof. We can assume that $\xi $ lies in the upper half-plane. Since all maps in this argument use the parameter $\xi $, we will write $f_{m}$ instead of $f_{\xi ,m}$ for all m. Define the arc $A = \operatorname {\mathrm {Arc}}{[R_{k}(\xi ), R_{k+1}(\xi )]}$. Using equation (2), we find that for every m,

$$ \begin{align*} |f_m'(R_{k}(\xi))| = \frac{m}{2k + 1} \cdot |f_{2k+1}'(R_{k}(\xi))| \geq \frac{m}{2k + 1}. \end{align*} $$

By using the fact that $R_{k+1}(\xi )$ is either a parabolic or an attracting fixed point of $f_{k+1}$, we deduce that for all $z \in A$,

$$ \begin{align*} |f_{k}'(z)| < |f_{k+1}'(z)| \leq |f_{k+1}'(R_{k+1}(\xi))| \leq 1, \end{align*} $$

where the second inequality is strict when $z \neq R_{k+1}(\xi )$. It follows that for all $z \in A$ not equal to $R_{k+1}(\xi )$ we have $k/(2k+1) \leq |f_{k}'(z)| < 1$ and $(k+1)/(2k+1) \leq |f_{k+1}'(z)| < 1$. Therefore,

$$ \begin{align*} \ell(A)> \ell(f_{k}(A)) > \frac{k}{2k+1}\ell(A) \quad\text{ and }\quad \ell(A) > \ell(f_{k+1}(A)) > \frac{k+1}{2k+1}\ell(A). \end{align*} $$

From this we deduce that $\ell (f_{k}(A)) + \ell (f_{k+1}(A))> \ell (A)$. Thus, because $f_{k}(A)$ is of the form $\operatorname {\mathrm {Arc}}{[R_{k}(\xi ), a]}$ and $f_{k+1}(A)$ is of the form $\operatorname {\mathrm {Arc}}{[b, R_{k+1}(\xi )]}$ for some $a,b \in A$, we conclude that $f_{k}(A) \cup f_{k+1}(A) = A$.

It follows from item (iv) of Lemma 10 that there is some M such that $f_{k+1}^M(1) \in A$. Let $J \subseteq A$ be any open arc. According to Corollary 20, there is a sequence of indices $m_1,\dots , m_N\in \{k,k+1\}$ such that $\left (f_{m_1} \circ \cdots \circ f_{m_N} \circ f_{k+1}^M\right ){\left (1\right )} \in J$. The fact that J was chosen as an arbitrary open arc in A concludes the proof.

Proof. We can assume that $\xi _1$ and $\xi _2$ lie in the upper half-plane and that $\text {Arg}(\xi _1) < \text {Arg}(\xi _2)$. Let $A = \operatorname {\mathrm {Arc}}{[R_k(\xi _1),R_k(\xi _2)]}$; then for all $z \in A$ we have

$$ \begin{align*} \frac{1}{2} \leq \frac{1}{2}\cdot \left|f_{2k}'(R_k(\xi_1))\right| = \left|f_{k}'(R_k(\xi_1))\right| \leq \left|f_{k}'(z)\right| \leq \left|f_{k}'(R_k(\xi_2))\right| \leq 1, \end{align*} $$

where the second to last inequality is strict when $z \neq R_{k}(\xi _2)$. Therefore, for $i \in \{1,2\}$ we have

$$ \begin{align*} \ell(A)> \ell(f_{\xi_i,k}(A)) > \frac{1}{2} \cdot\ell(A), \end{align*} $$

and from this we deduce that $\ell (f_{\xi _1,k}(A)) + \ell (f_{\xi _2,k}(A))> \ell (A)$. The rest of the proof proceeds exactly as the proof of Lemma 21.

Proof. W.l.o.g., we may assume that $\xi $ lies in the upper half plane. We write $f_{m} = f_{\xi ,m}$ for all indices m. Define the arcs $A_1 = \operatorname {\mathrm {Arc}}{[R_{k-2}(\xi ),R_{k-1}(\xi )]}$ and $A_2 = \operatorname {\mathrm {Arc}}{[R_{k-1}(\xi ),R_{k}(\xi )]}$. Analogous to the proofs of Lemmas 21 and 22, we can use Corollary 20 to show that the orbit of $1$ under the action of the semigroup generated by $f_{k-2}, f_{k-1}$ and $f_{k}$ is dense in $A_1 \cup A_2$ if

(8)$$ \begin{align} f_{k-2}(A_1 \cup A_2) \cup f_{k-1}(A_1 \cup A_2) \cup f_{k}(A_1 \cup A_2) = A_1 \cup A_2. \end{align} $$

We will assume that this is not the case and show that this leads to statement (ii). First, we will show that the left-hand side of equation (8) does cover $A_1$. For any arc A in the upper half-plane such that $\text {Arg}{(x)} \geq \text {Arg}{(\xi )}$ for all $x \in A$ and index m, we have

(9)$$ \begin{align} \ell{\left(f_m(A)\right)}> |f_m'(\xi)| \cdot \ell{\left(A\right)} = \frac{m\cdot |f_p'(\xi)|}{p} \cdot \ell{\left(A\right)} \geq \frac{m}{p} \cdot \ell{\left(A\right)}. \end{align} $$

We use this and the fact that $\ell {\left (A_2\right )} \geq \ell {\left (A_1\right )}$, which follows from item (v) of Lemma 10, to conclude the following:

$$ \begin{align*} \ell{\left(f_{k-2}(A_1 \cup A_2)\right)} + \ell{\left(f_{k-1}(A_1)\right)} &\geq \frac{k-2}{p} \ell{\left(A_1 \cup A_2\right)} + \frac{k-1}{p} \ell{\left(A_1\right)} \\ &\geq \frac{2(k-2)}{p} \ell{\left(A_1\right)} + \frac{k-1}{p} \ell{\left(A_1\right)} \\ & = \frac{3k - 5}{p} \ell{\left(A_1\right)} \geq \ell{\left(A_1\right)}. \end{align*} $$

Because $f_{k-2}(A_1 \cup A_2)$ is of the form $\operatorname {\mathrm {Arc}}{[R_{k-2}(\xi ),a]}$ and $f_{k-1}(A_1)$ is of the form $\operatorname {\mathrm {Arc}}{[b,R_{k-1}(\xi )]}$, we have that $A_1$ is covered by $f_{k-2}(A_1 \cup A_2) \cup f_{k-1}(A_1)$. Our assumption can be formulated as

$$ \begin{align*} \ell{\left(A_2\right)} \geq \ell{\left(f_{k-1}(A_2)\right)} + \ell{\left(f_{k}(A_1 \cup A_2)\right)}. \end{align*} $$

Note that

$$ \begin{align*} \ell{\left(f_{k-1}(A_2)\right)} + \ell{\left(f_{k}(A_1 \cup A_2)\right)} \geq \frac{k-1}{p} \ell{\left(A_2\right)} + \frac{k}{p} \left(\ell{\left(A_1\right)} + \ell{\left(A_2\right)}\right). \end{align*} $$

Combining the previous two inequalities, we get

(10)$$ \begin{align} \ell{\left(A_1\right)} \leq \frac{p-2k+1}{k} \cdot \ell{\left(A_2\right)}. \end{align} $$

Let $A_0 = \operatorname {\mathrm {Arc}}{[1,R_{k-2}(\xi )]}$. By using the fact that $R_m(\xi )$ is a fixed point of $f_m$ and $f_m(1) = \xi $ for every m, we see that $f_{k-2}(A_0) = \operatorname {\mathrm {Arc}}{[\xi , R_{k-2}(\xi )]}$, $f_{k-1}(A_0 \cup A_1) = \operatorname {\mathrm {Arc}}{[\xi , R_{k-1}(\xi )]}$ and $f_{k}(A_0 \cup A_1 \cup A_2) = \operatorname {\mathrm {Arc}}{[\xi , R_{k}(\xi )]}$. It follows from the relation between the derivative of different maps given in item (i) of Lemma 10 that for any arc A on which $f_{m_1}$ and $f_{m_2}$ are injective we have

$$ \begin{align*} \ell{\left(f_{m_1}(A)\right)} = m_1 \cdot \ell{\left(f_{1}(A)\right)} = \frac{m_1}{m_2} \cdot \ell{\left(f_{m_2}(A)\right)}. \end{align*} $$

These observations can be used to write $\ell {\left (A_1\right )}$ and $\ell {\left (A_2\right )}$ as follows:

$$ \begin{align*} \ell{\left(A_1\right)} &= \ell{\left(f_{k-1}(A_0 \cup A_1)\right)} - \ell{\left(f_{k-2}(A_0)\right)}\\ &= \frac{k-2}{k-1}\ell{\left(f_{k-1}(A_0 \cup A_1)\right)} + \frac{1}{k-1}\ell{\left(f_{k-1}(A_0 \cup A_1)\right)} - \frac{k-2}{k-1}\ell{\left(f_{k-1}(A_0)\right)}\\ &= \frac{k-2}{k-1}\ell{\left(f_{k-1}(A_1)\right)} + \frac{1}{k-1}\ell{\left(f_{k-1}(A_0 \cup A_1)\right)} \end{align*} $$

and

$$ \begin{align*} \ell{\left(A_2\right)} = \ell{\left(f_{k}(A_0 \cup A_1 \cup A_2)\right)} - \ell{\left(f_{k-1}(A_0 \cup A_1)\right)} = \ell{\left(f_{k}(A_2)\right)} + \frac{1}{k-1} \cdot \ell{\left(f_{k-1}(A_0 \cup A_1)\right)}. \end{align*} $$

By considering our way of writing $\ell {\left (A_1\right )}$ and the inequalities given in (9) and (10), we obtain the following inequalities:

$$ \begin{align*} \frac{1}{k-1}\cdot\ell{\left(f_{k-1}(A_0 \cup A_1)\right)} &= \ell{\left(A_1\right)} - \frac{k-2}{k-1}\cdot\ell{\left(f_{k-1}(A_1)\right)}< \ell{\left(A_1\right)} - \frac{k-2}{k-1} \cdot \frac{k-1}{p} \ell{\left(A_1\right)} \\ &= \frac{p-k+2}{p} \cdot \ell{\left(A_1\right)}< \frac{p-k+2}{p} \cdot \frac{p-2k+1}{k} \cdot \ell{\left(A_2\right)}. \end{align*} $$

It follows from the fact that $\text {Arg}{(R_{k}(\xi ))} \geq \text {Arg}{(x)}$ for all $x \in A_2$ that $\ell {\left (f_{k}(A_2)\right )} < |f_k'(R_{k}(\xi ))| \cdot \ell {\left (A_2\right )}$. By using this inequality and the previous inequality, we obtain

$$ \begin{align*} \ell{\left(A_2\right)} &= \ell{\left(f_{k}(A_2)\right)} + \frac{1}{k-1} \cdot \ell{\left(f_{k-1}(A_0 \cup A_1)\right)} \\ &< |f_k'(R_{k}(\xi))| \cdot \ell{\left(A_2\right)} + \frac{p-k+2}{p} \cdot \frac{p-2k+1}{k} \cdot \ell{\left(A_2\right)}. \end{align*} $$

We can cancel $\ell {\left (A_2\right )}$ and rewrite to obtain

$$ \begin{align*} |f_k'(R_{k}(\xi))|> 1-\frac{p-k+2}{p} \cdot \frac{p-2k+1}{k}, \end{align*} $$

which is what we set out to prove.

Proof. We will again assume that $\xi $ lies in the upper half-plane. We apply Lemma 23 with $k = m-1$ and $p = 2m$ for item (a) and $p = 2m + 1$ for item (b). If the first statement of that lemma holds, we see that orbit of $1$ under the action of $f_{m-3},f_{m-2}$ and $f_{m-1}$ generates an arc, in which case we are done. If we assume that the second statement holds, we obtain

$$ \begin{align*} f_{m-1}'(R_{m-1}(\xi))> 1-\frac{2m-(m-1)+2}{2m} \cdot \frac{2m-2(m-1)+1}{m-1} > \frac{1}{2} \end{align*} $$

in the case where $p = 2m$ and

$$ \begin{align*} f_{m-1}'(R_{m-1}(\xi))> 1-\frac{2m+1-(m-1)+2}{2m+1} \cdot \frac{2m+1-2(m-1)+1}{m-1} > \frac{1}{2} \end{align*} $$

in the case where $p = 2m+1$. It follows that for $x \in \operatorname {\mathrm {Arc}}{[R_{k-1}(\xi ),R_{k}(\xi )]}$, we obtain $1>|f_{m}'(x)| > |f_{m-1}'(x)| > 1/2$. Therefore, with $A = \operatorname {\mathrm {Arc}}{[R_{k-1}(\xi ),R_{k}(\xi )]}$, we get

$$ \begin{align*} \ell{\left(f_m\left(A\right)\right)} + \ell{\left(f_{m-1}\left(A\right)\right)} \geq \ell{\left(A\right)}. \end{align*} $$

This, together with Corollary 20, implies that the orbit of $1$ under the action of the semigroup generated by $f_{m-1}$ and $f_{m}$ is dense in A.

Proof. We will prove this by induction on k. For $k=0$, the field $\xi $ has to lie in $\operatorname {\mathrm {Arc}}(\lambda _1,\overline {\lambda _1})\setminus \{-1\}$ and the root degree of the tree in $\mathcal {T}_{d+1}$ implementing $\xi $ is at most $d-1$. From Corollary 13 and Lemma 14, we have that $f_{\xi ,1}$ is conjugate to an irrational rotation and thus the orbit of any initial point $z_0\in \mathbb {S}$ is dense in $\mathbb {S}$. By Lemma 4, every element of the set $\{f_{\xi ,1}^n(\lambda )\}_{n \geq 1}$ is the field implemented by a tree in $\mathcal {T}_{d+1}$ and hence we obtain the theorem for $k=0$.

Now suppose that $k \geq 1$ and that we have proved the statement for $k-1$. If $b < (2^{k-1}-1)/(2^{k-1}+1)$, then we must have $k>1$, and by Lemma 8 we can immediately apply the induction hypothesis with $\xi =\lambda $ and tree consisting of a single vertex. So, assume that $b \geq (2^{k-1}-1)/(2^{k-1}+1)$ and observe that the parameter $\lambda _{2^{k-1}}\in \mathbb {S}$ from Lemma 10 exists. It follows from Lemma 16 that there is $\sigma \in \mathbb {S}$ with $|f_{2^k}'(\sigma )|> 1$ and a set $\mathcal {R}= \{\zeta _n\}_{n \geq 1}$ accumulating on $\sigma $ such that each $\zeta _n$ is implemented by a tree in $\mathcal {T}_{d+1}$ whose root degree is at most $d - (2^{k+1} - 1) + 2^k = d-(2^{k} - 1)$. If $\mathcal {R}$ has a nonempty intersection with $\operatorname {\mathrm {Arc}}{(\lambda _{2^{k-1}},\overline {\lambda _{2^{k-1}}})}\setminus \{-1\}$, we can apply the induction hypothesis to the tree corresponding to the field in this intersection. Therefore, we assume that the elements of $\mathcal {R}$ accumulate on $\sigma $ from inside $\operatorname {\mathrm {Arc}}{[\overline {\lambda _{2^{k-1}}},\lambda _{2^{k-1}}]}$. It follows that we can find two distinct elements $r_1,r_2 \in \mathcal {R}$ such that they both lie in either $\operatorname {\mathrm {Arc}}{(\overline {\lambda _{2^{k-1}}},1)}$ or $\operatorname {\mathrm {Arc}}{(1,\lambda _{2^{k-1}})}$ and such that $|f_{2^k}'(r_i)|>1$ for $i =1,2$. By Remark 11, we have $|f_{2^k}'(R_{2^{k-1}}(r_i))|>|f_{2^k}'(r_i)|>1$ and thus we can apply Lemma 22 to conclude that the following set is dense in an arc A of the circle:

$$ \begin{align*} \mathcal{A} = \left\{ (f_{r_{i_1},2^{k-1}} \circ \cdots \circ f_{r_{i_n},2^{k-1}})(1) : n\in \mathbb{Z}_{\geq 1} \text{ and } i_1, \dots, i_n \in \{1,2\} \right\}. \end{align*} $$

Since $r_1,r_2$ are implemented by trees in $\mathcal {T}_{d+1}$ whose root degrees are at most $d - (2^{k+1} - 1) + 2^k = d-(2^{k} - 1)$, by Lemma 4, every element of $\mathcal {A}$ is implemented by a tree in $\mathcal {T}_{d+1}$ whose root degree is bounded by $d-(2^{k} - 1)+ 2^{k-1} = d - (2^{k-1} -1 ) \leq d$. We have seen that (2) implies that for $b \leq (d-1)/(d+1)$ it holds that $|f_d'(z)|> 1$ for all $z \in \mathbb {S}\setminus \{1\}$. This implies that there is some $N \in \mathbb {Z}_{\geq 1}$ such that $f_d^N(A) = \mathbb {S}$. It follows that the set $\{f_{\lambda ,d}^N(a): a \in \mathcal {A}\}$ is dense in $\mathbb {S}$, finishing the proof, since every element of this set corresponds to the field of a tree in $\mathcal {T}_{d+1}$ (again using Lemma 4).