Proof. For part (i), suppose that the axes of \(f\) and \(g\) cross at an angle \(\theta\). Conjugate \(f\) and \(g\) by a transformation \(\phi\in\mathrm{PSL}(2,\mathbb{C})\) so that they act on the unit disc, their axes meet at the origin and the diameter through \(i\) and \(-i\) dissects the angle \(\theta\) (see Figure 2 on the left). If we denote \(F=\phi\circ f \circ \phi^{-1}, G=\phi\circ g \circ \phi^{-1}\), then \(C(F,G)=C(f,g)\) by the invariance of the cross-ratio. Due to the symmetry of this configuration we have that \(u(F)=\overline{s(G)},\;u(F)=-s(F)\) and \(u(G)=-s(G)\). Hence, \[C(F,G)=\frac{u(F)-u(G)}{u(F)-s(G)}\frac{s(F)-s(G)}{s(F)-u(G)}= \frac{-2\relax s(F)}{-2i\relax s(F)} \frac{2\relax s(F)}{2i\relax s(F)}= -\left(\frac{\relax s(F)}{\relax s(F)}\right)^2.\] The law of cosines applied to the Euclidean triangles with vertices \(0, u(F),s(G)\) and \(0,u(F),u(G)\), yield \(\left(2\relax u(F)\right)^2 = 2(1-\cos\theta )\) and \(\left(2\relax u(F)\right)^2=2(1-\cos(\pi-\theta))\), respectively. Therefore, \[C(F,G) = -\frac{1-\cos\theta}{1-\cos(\pi-\theta)}=-\frac{1-\cos\theta}{1+\cos\theta}=-\tan^2\tfrac{1}{2}\theta.\] Moving on to part (ii), assume that the axes of \(f\) and \(g\) are disjoint and a hyperbolic distance \(d\) apart. Since \(C(f^{-1},g)=1/C(f,g)\) and \(C(f,g)\neq 1\), it suffices to show that \(C(f,g)=\tanh^2\tfrac{1}{2}d\) whenever \(0<C(f,g)<1\). Conjugate \(f\) and \(g\) in \(\mathrm{PSL}(2,\mathbb{R})\) so that \(f\) fixes \(-\sigma, \sigma\) and \(g\) fixes \(-\lambda, \lambda\), for some \(0<\sigma<\lambda\), as shown in Figure 2 on the right. Then, \[C(f,g)=\frac{(\lambda-\sigma)^2}{(\lambda+\sigma)^2}.\] Since the imaginary half-axis is perpendicular to both \(\textrm{Ax}(f)\) and \(\textrm{Ax}(g)\), we have that \(d=\rho_{\mathbb{H}}(i\sigma,i\lambda)\). Using formula 3 yields \[\tanh^2\tfrac{1}{2}d=\tanh^2\tfrac{1}{2}\rho_{\mathbb{H}}(i\sigma,i\lambda)=\frac{\lvert i\sigma - i \lambda \rvert^2}{\lvert i\sigma -\overline{i\lambda} \rvert^2} = \frac{(\lambda-\sigma)^2}{(\lambda+\sigma)^2}=C(f,g),\] completing our proof. ◻

Proof. Since \(C(f,g)>1\), the axes \(\mathrm{Ax}(f)\) and \(\mathrm{Ax}(g)\), of \(f\) and \(g\) respectively, are disjoint. Let \(d>0\) be the hyperbolic distance between \(\mathrm{Ax}(f)\) and \(\mathrm{Ax}(g)\). The proof revolves around evaluating the trace of the composition of \(f\) and \(g\), which is given by the following equation found in [@Be1995]: \[\label{eq:32antiparallel32trace32eq} \frac{1}{2}\lvert\text{tr}(f\circ g)\rvert=\lvert\cosh d\sinh\tfrac{1}{2}T_f\sinh\tfrac{1}{2}T_g-\cosh\tfrac{1}{2}T_f\cosh\tfrac{1}{2}T_g\rvert,\tag{7}\] where \(T_f,T_g\) are the translation lengths of \(f\) and \(g\), respectively. Let \(\mathbb{R}_+^2\) be the first quadrant of \(\mathbb{R}^2\), endowed with the Euclidean topology, and consider the function \(h\colon \mathbb{R}_+^2\to\mathbb{R}\), with \[\label{eq:32antiparallel32def32of32h} h(x,y)=\cosh d\sinh x\sinh y-\cosh x\cosh y.\tag{8}\] So, equation 7 can be rewritten as \(\frac{1}{2}\lvert\text{tr}(f\circ g)\rvert = \rvert h\left(\tfrac{1}{2}T_f,\tfrac{1}{2}T_g\right)\rvert\). Also, from 5 we have that \(T_{f^n}=n\;T_f\), meaning that \[\label{n44m-traces} \frac{1}{2}\lvert\text{tr}(f^n\circ g^m)\rvert=\left\lvert h\left(\tfrac{n}{2}T_f,\tfrac{m}{2}T_g\right) \right\rvert, \quad \text{for all}\;n,m\in\mathbb{N}.\tag{9}\] Define the open set \(D=\left\{(x,y)\in\mathbb{R}_+^2\colon -1<h(x,y)<1\right\}\). Since a transformation \(h\) is elliptic if and only if \(\lvert \mathrm{tr}(h)\rvert <2\), we have that the composition \(f^n\circ g^m\) is elliptic, for some \(n,m\in\mathbb{N}\), if and only if \(\left\lvert h\left(\tfrac{n}{2}T_f,\tfrac{m}{2}T_g\right) \right\rvert<1\), or equivalently \(\left(\tfrac{n}{2}T_f,\tfrac{m}{2}T_g\right)\in D\).
So, our goal is to show that if \[\max\left\{\tfrac{1}{2}\lvert \mathrm{tr}(f)\rvert, \tfrac{1}{2}\lvert \mathrm{tr}(g)\rvert\right\}\leqslant\;\frac{5C(f,g)-1}{3C(f,g)+1},\] then there exist integers \(n,m\geqslant 1\), so that \(\left(\tfrac{n}{2}T_f,\tfrac{m}{2}T_g\right)\in D\).
Let \(a,b>0\) be such that \[\begin{align} \label{sinh} &\sinh b =\frac{\sqrt{2}}{\sqrt{\cosh d-1}} &\text{and} & &\sinh a= \frac{\sqrt{2}}{(2\cosh d+1)\sqrt{\cosh d-1}}. \end{align}\tag{10}\] These are well-defined since \(d>0\). Also, 10 are equivalent to \[\begin{align} \label{cosh} &\cosh b =\sqrt{\frac{\cosh d+1}{\cosh d-1}} &\text{and} & &\cosh a= \frac{(2\cosh d-1)}{(2\cosh d+1)}\sqrt{\frac{\cosh d+1}{\cosh d-1}}. \end{align}\tag{11}\] Observe that \(b>a\), since \(\sinh b > \sinh a\). Let \(S\) be the square with vertices \((a,b), (b,b), (b,a), (a,a)\) and \(C\) the open region in \(\mathbb{R}^2_+\) bounded by \(S\). That is, \(\overline{C}=S\cup C\). Our main task is to show that \(\overline{C}\setminus\{(a,b), (b,b), (b,a)\}\subseteq D\), as shown in Figure 4.

We start by showing that the edges of \(S\), without the vertices \((a,b), (b,b), (b,a)\), are contained in \(D\). We shall make heavy use of the fact that the line \(y=x\) is an axis of symmetry for \(h\); i.e. \(h(x,y)=h(y,x)\), for all \(x,y>0\).
Using the formula for \(h\) in 8 and equations 10 , 11 we can evaluate \(h\) on the vertices of \(S\) as follows: \[\begin{align} h(a,a)&=\cosh d \sinh^2a-\cosh^2a=(\cosh d -1)\sinh^2a-1=\frac{2}{(2\cosh d+1)^2}-1, \tag{12}\\ h(b,b)&= \cosh d \sinh^2b - \cosh^2b= \cosh d \frac{2}{\cosh d-1} - \frac{\cosh d +1}{\cosh d-1}=1,\tag{13} \end{align}\] and \[\begin{align} \label{hab} h(b,a)=h(a,b) &= \cosh d \sinh a \sinh b - \cosh a \cosh b=\frac{2\cosh d - (2\cosh d-1)(\cosh d+1)}{(2\cosh d+1)(\cosh d-1)}=\nonumber \\ &=\frac{-2\cosh^2 d+\cosh d+1}{(2\cosh d+1)(\cosh d-1)}=-1. \end{align}\tag{14}\] Observe that 12 implies that \[\label{eq:32antiparallel32haa322} -1<h(a,a)<-\frac{7}{9}.\tag{15}\] Now consider the partial derivative \[\label{increasing} \frac{\partial h}{\partial x}(x,b) = \cosh d \cosh x \sinh b - \sinh x \cosh b, \quad \text{for} \quad x>0.\tag{16}\] Equations 10 and 11 yield \[\frac{1}{\cosh d}\;\frac{\cosh b}{\sinh b}=\frac{\sqrt{\cosh d +1}}{\sqrt{2}\;\cosh d}<1<\frac{\cosh x}{\sinh x},\quad \text{for all}\: x>0.\] Applying this inequality to 16 , we obtain that \(h\) is strictly increasing on the horizontal line \(y=b\). This implies that \[-1=h(a,b)<h(x,b)<h(b,b)=1,\quad \text{for any}\: x\in(a,b),\] meaning that \(h(x,b)\in D\), for all \(x\in(a,b)\). This proves that the horizontal edge of \(S\) joining \((a,b)\) and \((b,b)\) (without these vertices) is contained in \(D\). By the symmetry of \(h\) we also obtain that the vertical edge of \(S\) joining \((b,a)\) and \((b,b)\) (excluding the vertices again) is contained in \(D\).
We now turn to the vertical edge of \(S\) joining \((a,a)\) and \((a,b)\). Consider the partial derivative \[\frac{\partial h}{\partial y}(a,y) = \cosh d \sinh a \cosh y - \cosh a \sinh y, \quad \text{for}\quad y>0.\] Since \(\frac{\partial h}{\partial y}(a,a)= (\cosh d -1) \sinh a \cosh a>0\), by continuity there exists \(y_+\in(a,b)\), close enough to \(a\), so that \(\frac{\partial h}{\partial y}(a,y_+)>0\).
Suppose that \(\frac{\partial h}{\partial y}(a,y)\neq0\) for all \(y\in(a,b)\). Then, again by continuity and the fact that \(\frac{\partial h}{\partial y}(a,y_+)>0\), we have that \(\frac{\partial h}{\partial y}(a,y)>0\), for all \(y\in(a,b)\), which implies that \(h(a,y)\) is increasing in \((a,b)\). But \(h(a,a)>-1=h(a,b)\), from 14 and 15 , which is a contradiction.
We conclude that \(\frac{\partial h}{\partial y}(a,y_0)=0\), for some \(y_0\in(a,b)\). That is \[\begin{align} &\frac{\partial h}{\partial y}(a,y_0)=\cosh d \sinh a \sinh y_0 - \cosh a \sinh y_0=0 & \iff& &\coth y_0 = \frac{\coth a}{\cosh d }. \end{align}\] If \(0<y<y_0\), then \[\begin{align} &\coth y >\coth y_0 = \frac{\coth a}{\cosh d } & &\iff& &\cosh d \cosh y \sinh a > \cosh a \sinh y & &\iff& &\frac{\partial h}{\partial y}(a,y)>0. \end{align}\] Similarly we can show that if \(y>y_0\), then \(\frac{\partial h}{\partial y}(a,y)<0\). Therefore the maximum of \(h(a,y)\), for all \(y>0\), is attained only at \(y_0\). This means that \[\min\{h(a,a), h(a,b)\} < h(a,y) \leqslant h(a,y_0),\quad \text{for all}\;y\in(a,b).\] But, \(h(a,a)>-1=h(a,b)\), again by 14 and 15 , and so \(h(a,y)>-1\), for all \((a,b)\). In order to conclude that \(h(a,y)\in D\), for all \(y\in(a,b)\), it suffices to show that \(h(a,y_0)<0\), which we now prove. \[\begin{align} h(a,y_0)<0 &\iff \cosh d \sinh a \sinh y_0 <\cosh a \cosh y_0 \\ &\iff \cosh d < \coth a \coth y_0= \coth a\;\frac{\coth a}{\cosh d }\\ &\iff \cosh d <\coth a. \end{align}\] But, equations 10 and 11 yield that \[\coth a = \frac{(2\cosh d-1)\sqrt{\cosh d+1}}{\sqrt{2}}=\frac{(\cosh d + \cosh d -1)\sqrt{\cosh d+1}}{\sqrt{2}} >\frac{\cosh d\sqrt{2}}{\sqrt{2}}=\cosh d,\] since \(d>0\). This concludes the proof of the fact that \(h(a,y)\in D\), for all \(y\in(a,b)\), which implies that the vertical edge of \(S\) joining \((a,a)\) and \((a,b)\) (apart from the vertex \((a,b)\)) lies in \(D\). Using the symmetry of \(h\), again, we obtain that the horizontal edge of \(S\) joining \((a,a)\) and \((b,a)\) (apart from the vertex \((b,a)\)) lies in \(D\).
So, as we claimed, all edges of \(S\) lie in \(D\), apart from the vertices \((a,b),(b,b),(b,a)\), which lie on the boundary of \(D\), since \(h(a,b)=h(b,a)=-1\) and \(h(b,b)=1\). With these facts in mind, we move on to the proof of the inclusion \(C\subseteq D\).
First, a simple computation shows that the function \(h(x,x)\) is strictly increasing for all \(x>0\). Therefore, recalling 8 and 13 , we get that \[\label{eq:32anitparallel32y61x} -1=\lim_{x\to0^+}h(x,x)<h(x,x)<h(b,b)=1,\quad \text{for all}\quad 0<x<b.\tag{17}\] This means that \(C\cap\{(x,x)\colon x>0\}\subseteq D\). Now, fix \(x_0>0\) and consider the derivative \[\label{eq:32antiparallel32final32der} \frac{\partial h}{\partial y}(x_0,y)=\cosh d \sinh x_0 \cosh y- \cosh x_0 \sinh y.\tag{18}\] Since the hyperbolic tangent is an increasing function, we obtain that for \(y<x_0\) \[\label{eq:32antiparallel32final32der322} \tanh y < \tanh x_0 < \cosh d \tanh x_0.\tag{19}\] Combining 18 and 19 yields that \(h(x_0,y)\) is strictly increasing for all \(y\in(0,x_0)\), and any \(x_0>0\). So, if \(x_0\in(a,b)\) then \[-1<h(x_0,a)<h(x_0,y)<h(x_0,x_0)<1,\] where the inequality \(-1<h(x_0,a)\) follows from the fact that \((x_0,a)\) lies on an edge of \(S\) and inequality \(h(x_0,x_0)<1\) is 17 applied to \(x_0\). We conclude that the vertical line segment joining \((x_0,a)\) and \((x_0,x_0)\), which is contained in \(C\), lies in \(D\) for all \(x_0\in(a,b)\). Using the symmetry of \(h\) one last time, we obtain that \(C\subseteq D\), as claimed.
To finish the proof of the lemma, note that since the edges of \(S\) have length \(b-a\), if a point \((x,y)\in\mathbb{R}_+^2\) is such that \(\max\{x,y\}<b-a\), then there exist integers \(n,m\geqslant 1\) such that \((nx,my)\in C\subseteq D\). Therefore, recalling equation 9 , we have that if \[\max\left\{\tfrac{1}{2}T_f, \tfrac{1}{2}T_g \right\}< b-a,\] then there exist integers \(n,m\geqslant 1\) such that \(f^n\circ g^m\) is elliptic. Next, assume that \((x,y)\in\mathbb{R}_+^2\) is such that \(\max\{x,y\}=b-a\). If \(x<y\) (or, similarly, \(y<x\)), then there exist \(n,m\geqslant 1\) so that \((nx,my)\in \overline{C}\setminus\{(a,b),(b,b),(b,a)\}\subseteq D\) and we are done once again. On the other hand, if \(x=y=b-a\), then since \(b-a<b\) we can use 17 to deduce that \(-1<h(b-a,b-a)<1\), meaning that \((x,y)\in D\).
We have thus shown that if \[\max\left\{\tfrac{1}{2}T_f, \tfrac{1}{2}T_g \right\}\leqslant b-a,\] then there exist integers \(n,m\geqslant 1\) so that \(f^n\circ g^m\) is elliptic. As such, the proof of this lemma will now be complete upon showing that \[\max\left\{\tfrac{1}{2}T_f, \tfrac{1}{2}T_g \right\}\leqslant b-a \quad \iff \quad \max\left\{ \tfrac{1}{2}\lvert\mathrm{tr}(f)\rvert, \tfrac{1}{2}\lvert \mathrm{tr}(g)\rvert\right\}\leqslant\frac{5\;C(f,g)-1}{3\;C(f,g)+1}.\] Recall that from 4 we have that \(\cosh \left(\tfrac{1}{2}T_f\right) = \tfrac{1}{2} \lvert \mathrm{tr}(f)\rvert\) and so we only need to show that \[\label{eq:32antiparallel32end32goal} \cosh (b-a) = \frac{5C(f,g)-1}{3C(f,g)+1}.\tag{20}\] But, \(\cosh(b-a) = \cosh b\cosh a - \sinh b\sinh a\), and all quantities on the right-hand side are given by equations 10 and 11 . So, \[\begin{align} \cosh(b-a) &= \frac{(2\cosh d -1)}{(2\cosh d+1)}\frac{(\cosh d+1)}{(\cosh d-1)} - \frac{2}{(2\cosh d+1)(\cosh d-1)}\nonumber \\ &=\frac{2\cosh^2 d +\cosh d-3}{(2\cosh d+1)(\cosh d-1)}\nonumber \\ &=\frac{2\cosh d+3}{2\cosh d+1}. \label{coshba} \end{align}\tag{21}\] Now, by Corollary 2 (ii) and the fact that \(C(f,g)>1\), we have that \(\cosh d = \frac{C(f,g)+1}{C(f,g)-1}\) which when substituted to 21 yields 20 . ◻

Proof. Since \(C(f,g)>1\) the axes of \(f,g\) are disjoint and a hyperbolic distance \(d>0\) apart. Also, from 4 and our assumption, we have that \[\tfrac{1}{2}t=\tfrac{1}{2}\lvert \textrm{tr}(f)\rvert =\cosh \tfrac{1}{2} T_f = \tfrac{1}{2}\lvert \textrm{tr}(g) \rvert =\cosh \tfrac{1}{2} T_g.\] This means that \(T\vcentcolon=T_f=T_g\).
As in the proof of Lemma 2, we have that \[\label{eq:32equal32trace32eq1} \frac{1}{2}\lvert \textrm{tr}(f\circ g)\rvert= \lvert h\left(\tfrac{1}{2}T_f,\tfrac{1}{2}T_g\right)\rvert= \lvert h\left(\tfrac{1}{2}T,\tfrac{1}{2}T\right)\rvert,\tag{22}\] where \(h(x,y)=\cosh d\sinh x\sinh y-\cosh x\cosh y\). In 17 we showed that \(-1<h(x,x)<h(b,b)=1\), for all \(x\in(0,b)\), where \(b>0\) satisfies \[\label{eq:32equal32trace32eq2} \cosh b =\sqrt{\frac{\cosh d+1}{\cosh d-1}}.\tag{23}\] So, 22 tells us that if \(\tfrac{1}{2}T\leqslant b\) we have that \(f\circ g\) is either elliptic (when \(T<b\)) or parabolic (when \(T=b\)). Equivalently, by 23 , the inequality \(\tfrac{1}{2}T\leqslant b\) can be written as \[\label{eq:32equal32trace32eq3} \tfrac{1}{2}t=\cosh\tfrac{1}{2}T\leqslant\cosh b = \sqrt{\frac{\cosh d+1}{\cosh d-1}}.\tag{24}\] But, from Corollary 2 (ii), and the fact that \(C(f,g)>1\), we have that \[\cosh d = \frac{C(f,g)+1}{C(f,g)-1},\] which when substituted to 24 yields the desired result. ◻

Proof. By assumption, Lemma 2 is applicable and yields that \(w=f^n\circ g^m\) is elliptic for some integers \(n,m\geqslant 1\). Assume first that \(w\) has infinite order. Since \(C(f,g)>1\), the transformations \(f\) and \(g\) do not commute, and so neither do \(w\) and \(f\). Thus Theorem 5 is applicable and yields that the semigroup generated by \(\{f,w\}\) is dense in \(\mathrm{PSL}(2,\mathbb{R})\).
If, on the other hand, \(w\) has finite order, Lemma 4 implies that the semigroup generated by \(\{f,g\}\) is in fact a group. If this group is discrete, we are done. Otherwise, again because \(C(f,g)>1\), we have that the group generated by \(\{f,g\}\) is non-elementary. So, using Theorem 6 we get that the semigroup generated by \(\{f,g\}\) contains an elliptic of infinite order and we can follow the arguments we presented in the previous paragraph to establish that it is dense in \(\mathrm{PSL}(2,\mathbb{R})\). ◻

Proof. Let \(f,g\) be hyperbolic transformations as in the statement of the theorem. Observe that if we prove the existence of the intervals \(U_f,S_f,U_g,S_g\), then Theorem 4 immediately tells us that \(\{f,g\}\) is dominated. Since \(C(f,g)>0\) the axes of \(f\) and \(g\) are disjoint and a hyperbolic distance \(d>0\) apart. We first consider the case where \(C(f,g)>1\). By assumption we have that \[\label{eq:32newlem32proof32eq1} \min\left\{\tfrac{1}{2}\lvert \mathrm{tr}(f) \rvert, \tfrac{1}{2}\lvert \mathrm{tr}(g) \rvert \right\}> \sqrt{C(f,g)}=\max\left\{\sqrt{C(f,g)},\frac{1}{\sqrt{C(f,g)}}\right\}.\tag{25}\] Equation 4 tells us that \(\tfrac{1}{2}\lvert \textrm{tr}(f)\rvert =\cosh \tfrac{1}{2}T_f\), and similarly for \(g\). Thus, 25 is equivalent to \[\min\left\{\cosh\tfrac{1}{2}T_f, \cosh\tfrac{1}{2}T_g \right\}> \sqrt{C(f,g)},\] which in turn can be rewritten as \[\label{eq:32newlem32proof32eq2} \min\left\{\sinh\tfrac{1}{2}T_f, \sinh\tfrac{1}{2}T_g \right\}> \sqrt{C(f,g)-1}.\tag{26}\] Recall that Lemma 1 tell us that \(C(f,g)=\coth^2\tfrac{1}{2}d\), meaning that \(\frac{1}{\sinh^2\tfrac{1}{2}d}=C(f,g)-1\). We are thus led to \[\label{eq:32newlem32proof32eq3} \min\left\{\sinh\tfrac{1}{2}T_f, \sinh\tfrac{1}{2}T_g \right\}> \frac{1}{\sinh \tfrac{1}{2}d}.\tag{27}\] Now, let \(\ell\) be the unique hyperbolic geodesic that is perpendicular to \(\mathrm{Ax}(f)\) and \(\mathrm{Ax}(g)\), and let \(\sigma\) the reflection in \(\ell\). Also, define the reflections \(\sigma_f=f\circ \sigma\) and \(\sigma_g=\sigma\circ g\) and let \(\ell_f, \ell_g\) be their geodesics of reflection respectively. Observe that \(\ell_f\) is also perpendicular to \(\mathrm{Ax}(f)\) and \(\ell_g\) is perpendicular to \(\mathrm{Ax}(g)\) (see Figure 5).
To fully justify the configuration shown in Figure 5, we first claim that the geodesic \(\ell_f\) does not intersect \(\mathrm{Ax}(g)\) in \(\overline{\mathbb{H}}\) and \(\ell_g\) does not intersect \(\mathrm{Ax}(f)\). Assuming that the geodesic \(\ell_f\) meets \(\mathrm{Ax}(g)\) in \(\overline{\mathbb{H}}\) at an angle \(\phi\in [0,\pi/2)\) we can see that the geodesics \(\ell_f,\mathrm{Ax}(f),\ell\) and \(\mathrm{Ax}(g)\) form a quadrilateral in \(\mathbb{H}\) that has three right angles and the fourth is \(\phi\). Moreover, \(\phi\) cannot be \(\pi/2\) because then \(\ell_f\) would coincide with \(\ell\). Applying [@Be1995] to this quadrilateral implies that \[\sinh\tfrac{1}{2}T_g =\frac{\cos\phi}{\sinh d }\leqslant\frac{1}{\sinh\tfrac{1}{2} d},\] which directly contradicts 27 . We reach the same contradiction if we assume that \(\ell_g\) intersects \(\mathrm{Ax}(f)\) in \(\overline{\mathbb{H}}\).
Next, we show that the geodesics \(\ell_f,\ell_g\) do not intersect in \(\overline{\mathbb{H}}\). Working towards a contradiction we suppose that \(\ell_f\) and \(\ell_g\) meet in \(\overline{\mathbb{H}}\) at an angle \(\theta \in[0,\pi/2]\). Then, because the pairs \(\ell_f,\mathrm{Ax}(g)\) and \(\ell_g,\mathrm{Ax}(f)\) are disjoint, we get that the geodesics \(\ell,\ell_f,\ell_g,\mathrm{Ax}(f), \mathrm{Ax}(g)\) define a pentagon in \(\overline{\mathbb{H}}\) with four right angles and the fifth being \(\theta\). Using [@Be1995] we get \[\label{eq:32newlem32proof32eq4} \sinh\tfrac{1}{2}T_f\sinh\tfrac{1}{2}T_g\cosh d - \cosh\tfrac{1}{2}T_f\cosh\tfrac{1}{2}T_g = \cos\theta\leqslant 1.\tag{28}\] Consider the function \(h\colon \mathbb{R}^2\to \mathbb{R}\), with \(h(x,y)=\sinh x\sinh y\cosh d-\cosh x\cosh y\), and let \(b\in\mathbb{R}\) be such that \[\sinh b=\frac{1}{\sinh \tfrac{1}{2}d}=\sqrt{\frac{2}{\cosh d-1}}.\] The function \(h\) and the number \(b\) also appeared in the proof of Lemma 2, where we showed that \(h(x,b)\) is increasing for all \(x>0\) (see the arguments succeeding 16 ), and \(h(b,b)=1\) (see 13 ). Also, \(h(x,x)=\cosh d \sinh^2x -\cosh^2x\) is an increasing function of \(x\). Combining these facts we obtain that \(h(x,b),h(x,x)>1\), for all \(x>b\). Let us fix \(x_0>b\). Now, again in the proof of Lemma 2 we showed that \(h(x_0,y)\) is increasing for all \(y\in(0,x_0)\) (see the arguments succeeding 18 ). We thus have that \[h(x_0,y)>h(x_0,b)>1,\quad\text{for all}\;y\in (b,x_0).\] Since \(h(x,y)=h(y,x)\), we can conclude that all points \((x,y)\in\mathbb{R}^2\) with \(\min\{x,y\}>b\) satisfy \(h(x,y)>1\). Now note that by 27 we have that \(\tfrac{1}{2}T_f,\tfrac{1}{2}T_g>b\), meaning that \(h\left(\tfrac{1}{2}T_f,\tfrac{1}{2}T_f\right)>1\), contradicting 28 .
To sum up, we just showed that \(\ell_f\) is disjoint with \(\mathrm{Ax}(g)\) and \(\ell_g\), and similarly \(\ell_g\) is disjoint with \(\mathrm{Ax}(f)\) and \(\ell_f\). Hence, if we denote by \(U_f\) the open interval in \(\overline{\mathbb{R}}\) that has the same endpoints as \(\ell_f\) and contains \(u(f)\), and by \(I_g\) the open interval that has the same endpoints as \(\ell_g\) and contains \(s(g)\), we obtain that \(\overline{U_f}\cap \overline{I_g}=\emptyset\). Also, let \(I_f=\sigma(U_f)\) and \(U_g=\sigma(I_g)\). Since \(f=\sigma_f\circ\sigma\) and \(g = \sigma\circ \sigma_g\), we have that \(f(I_f^c)= \overline{U_f}\) and \(g(I_g^c)= \overline{U_g}\). It is now easy to see that we can choose slightly larger intervals \(I_f\subseteq S_f\) and \(I_g\subseteq S_g\) so that \(U_f,S_f,U_g,S_g\) all have disjoint closures, \(f(S_f^c)\subseteq U_f\) and \(g(S_g^c)\subseteq U_g\), as required. This concludes our proof for the case \(C(f,g)>1\).
Recall that because \(f,g\) are hyperbolic, we have that \(C(f,g)\neq1\). So the next case to consider is when \(0<C(f,g)<1\). But, we have that \(C(f^{-1},g)=1/C(f,g)>1\) and so we can apply the first case to the pair \(\{f^{-1},g\}\), in order to obtain the desired intervals. ◻

Proof of Theorem 3. We assume that \(f,g\) are hyperbolic transformations with \(C(f,g)>1\) and \(\lvert \textrm{tr}(f)\rvert=\lvert \textrm{tr}(g)\rvert=t\). Our goal is to show that \(\{f,g\}\) is dominated if and only if \(\tfrac{1}{2}t>\sqrt{C(f,g)}\).
By Lemma 3 we have that if \(\tfrac{1}{2} t\leqslant\sqrt{C(f,g)}\) then \(f\circ g\) is either elliptic or parabolic. Since the semigroup generated by a dominated set only contains hyperbolic transformations, \(\{f,g\}\) is not dominated in this case. On the other hand, if \(\tfrac{1}{2} t> \sqrt{C(f,g)}\), Lemma 5 immediately yields that \(\{f,g\}\) is dominated, since \[\max\left\{\sqrt{C(f,g)},\frac{1}{\sqrt{C(f,g)}}\right\} = \sqrt{C(f,g)}.\qedhere\] ◻

Proof. Since \(C(f,g)<0\), the axes of \(f\) and \(g\) meet at an angle \(\theta\in(0, \pi)\). For convenience, we conjugate \(f\) and \(g\) with a transformation in \(\mathrm{PSL}(2,\mathbb{C})\), so that they act on the unit disc \(\mathbb{D}\), their axes cross at the origin and the Euclidean diameter joining \(i\) and \(-i\) bisects \(\theta\). By the symmetry of this configuration, we also have that the Euclidean diameter joining \(1\) and \(-1\) bisects \(\pi-\theta\). Throughout the proof, we use the notation \([z,w]\) to denote the closed arc of the unit circle running counter-clockwise from \(z\in\partial\mathbb{D}\) to \(w\in\partial\mathbb{D}\). Similarly, \((z,w)\) will denote the respective open arc. By applying a Euclidean rotation about the origin, we can also assume that both \(f\) and \(g\) map the arc \([-1,1]\) inside itself, and without loss of generality assume that the arc \([u(f),u(g)]\) does not contain \(s(f)\) and \(s(g)\). See Figure 6 for the geometry of the configuration of \(f\) and \(g\).
We first consider the case where \(\theta\in(0, \tfrac{\pi}{2}]\). Due to Corollary 2 this is equivalent to \(C(f,g)\in[-1,0)\), meaning that in this case \[\label{eq:32intersecting32eq0} \max\left\{\sqrt{1-C(f,g)},\sqrt{1-\frac{1}{C(f,g)}}\right\}=\sqrt{1-\frac{1}{C(f,g)}}.\tag{29}\] We shall work independently for \(f\) and \(g\). Starting with \(f\), let \(\ell_f\) be the hyperbolic geodesic of \(\mathbb{D}\) landing at \(-i\) that is perpendicular to \(\mathrm{Ax}(f)\), the axis of \(f\). Also, write \(z_f\in\partial\mathbb{D}\) for the other endpoint of \(\ell_f\). Since \(\theta\leqslant\tfrac{\pi}{2}\) and the diameter joining \(-i\) and \(i\) bisects \(\theta\), we have that \(z_f\) lies in \([-1,-i]\). Let \(a_f\in\mathbb{D}\) be the point of intersection of \(\ell_f\) and \(\mathrm{Ax}(f)\). By the symmetry of our configuration, we have that the hyperbolic geodesic \(\widetilde{\ell_f}\) with endpoints \(i\) and \(-z_f\) is also perpendicular to \(\mathrm{Ax}(f)\) and \(-z_f\) lies in \([1,i]\). Also, the point of intersection of \(\widetilde{\ell_f}\) and \(\mathrm{Ax}(f)\) is \(-a_f\). By Remark 7 we have that if \(T_f>d_\mathbb{D}(a_f,-a_f)\), then there exist open arcs \(U_f\subseteq(z_f,-i)\) and \(S_f\subseteq(-z_f,i)\), that are symmetric with respect to \(f\), so that \(f(S_f^c)\subseteq U_f\).
Our goal now is to show that the condition \(T_f>d_\mathbb{D}(a_f,-a_f)\) is equivalent to \(\tfrac{1}{2}\lvert\textrm{tr}(f)\rvert>\sqrt{1-\frac{1}{C(f,g)}}\). Observe that we have \(d_\mathbb{D}(a_f,-a_f)=d_\mathbb{D}(a_f,0)+d_\mathbb{D}(0,-a_f)\) and \(d_\mathbb{D}(a_f,0)=d_\mathbb{D}(0,-a_f)\). So the condition \(T_f>d_\mathbb{D}(a_f,-a_f)\) is equivalent to \(\tfrac{1}{2}T_f>d_\mathbb{D}(a_f,0)\). Consider the hyperbolic triangle with vertices \(a_f,0\) and \(-i\). This triangle has angles \(0,\tfrac{\pi}{2}\) and \(\tfrac{\theta}{2}\), meaning that we can use the Angle of Parallelism [@Be1995] in order to obtain that \[\label{eq:32intersecting32eq1} \cosh d_\mathbb{D}(a_f,0)\;\sin\tfrac{\theta}{2}=1.\tag{30}\] Equation 30 can be rewritten as \[\label{eq:32intersecting32eq2} \cosh d_\mathbb{D}(a_f,0) = \frac{1}{\sqrt{1-\cos^2\tfrac{\theta}{2}}}=\sqrt{1-\frac{1}{C(f,g)}},\tag{31}\] where for the last equality we used Corollary 2 (i). Hence, using 31 and the equation \(\cosh\tfrac{1}{2}T_f=\tfrac{1}{2}\lvert\textrm{tr}(f)\rvert\), i.e. 4 , we have that \[\begin{align} \tfrac{1}{2}T_f>d_\mathbb{D}(a_f,0)& & \iff & &\tfrac{1}{2}\lvert\textrm{tr}(f)\rvert=\cosh\tfrac{1}{2}T_f> \cosh d_\mathbb{D}(a_f,0) = \sqrt{1-\frac{1}{C(f,g)}}. \end{align}\] Following the same arguments for \(g\) in place of \(f\), we can see that if \(\tfrac{1}{2}\lvert\textrm{tr}(g)\rvert>\sqrt{1-\frac{1}{C(f,g)}}\), then there exists a point \(z_g\in[-i,1]\) and arcs \(U_g\subseteq(-i,z_g)\) and \(S_g\subseteq(i,-z_g)\), symmetric with respect to \(g\), and such that \(g(S_g^c)\subseteq U_g\). By construction, we have that the arcs \(U_f,S_f,U_g,S_g\) have pairwise disjoint closures, which concludes the proof in the case where \(\theta\in(0,\tfrac{\pi}{2}]\).
For the case \(\theta\in(\tfrac{\pi}{2},\pi)\), we can apply the above arguments to the pair \(f^{-1},g\), since their axes cross at angle \(\pi-\theta\in(0,\tfrac{\pi}{2})\). Recalling that \(C(f^{-1},g)=1/C(f,g)\) yields the desired conclusion. ◻

Proof. If \(C(f,g)<0\), then the the statement of the theorem is identical to the statement of Lemma 6. If \(C(f,g)>0\), then \[\begin{align} \min\left\{\tfrac{1}{2}\lvert\textrm{tr}(f)\rvert, \tfrac{1}{2}\lvert\textrm{tr}(g)\rvert\right\}&>\max\left\{\sqrt{1+\left\lvert C(f,g)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f,g)\right\rvert}}\right\}\\ &= \max\left\{\sqrt{1+ C(f,g)},\sqrt{1+\frac{1}{ C(f,g)}}\right\}>\max\left\{\sqrt{C(f,g)},\sqrt{\frac{1}{C(f,g)}}\right\}, \end{align}\] and so the conclusion is obtained by applying Lemma 5. ◻

Proof. Suppose that \(f_1,f_2,\dots,f_N\) are hyperbolic transformations, with the properties described in the statement of the theorem. Note that, by assumption, the fixed points of all \(f_i\) are distinct, meaning that \(C(f_i,f_j)\) is well-defined and \(C(f_i,f_j)\neq0\), for all \(i\neq j\). Due to the geometric characterisation of domination in Theorem 4, it suffices to show that under the trace condition in ?? we can find open intervals \(U^1,U^2,\dots, U^k\), with disjoint closures, so that each \(f_k\) maps \(\overline{\cup_{k=1}^NU^k}\) into \(U^k\).
Let \(\varepsilon>1\) be such that for all \(i=1,2,\dots,N\) we have that \[\label{eq:32proof32epsilon} \tfrac{1}{2}\lvert\textrm{tr}(f_i)\rvert > 4\varepsilon\;\max_{i\neq j}\{M_{i,j}\}\cdot \max_{i\neq j}\left\{\frac{\lvert C(f_i,f_j)\rvert +1}{\lvert C(f_i,f_j)-1\rvert}\right\}.\tag{32}\] Fix \(k=1,2,\dots,N\). For any \(i\neq k\), the hypothesis ?? and 32 yield \[\begin{align} \min\{\tfrac{1}{2}\lvert \textrm{tr}(f_k)\rvert,\tfrac{1}{2}\lvert \textrm{tr}(f_i)\rvert\}&> 4\varepsilon\cdot M_{k,i}\cdot \frac{\lvert C(f_k,f_i)\rvert +1}{\lvert C(f_k,f_i)-1\rvert} \nonumber\\ &=4\varepsilon\cdot \max\left\{1+\left\lvert C(f_k,f_i)\right\rvert,1+\frac{1}{\left\lvert C(f_k,f_i)\right\rvert}\right\}\cdot \frac{\lvert C(f_k,f_i)\rvert +1}{\lvert C(f_k,f_i)-1\rvert} \nonumber\\ &>\sqrt{\varepsilon} \cdot \max\left\{\sqrt{1+\left\lvert C(f_k,f_i)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_i)\right\rvert}}\right\} \label{eq:32proof32estimate322} \\ &> \max\left\{\sqrt{1+\left\lvert C(f_k,f_i)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_i)\right\rvert}}\right\}\nonumber , \end{align}\tag{33}\] which means that Theorem 8 is applicable to the pair of hyperbolic transformations \(f_k,f_i\). So we can find intervals \(U_i^k,U_k^i,S_i^k,S_k^i\subseteq\overline{\mathbb{R}}\) with pairwise disjoint closures so that \(U_i^k,S_i^k\) are symmetric with respect to \(\mathrm{Ax}(f_k)\), the axis of \(f_k\), and \(U_k^i,S_k^i\) are symmetric with respect to \(\mathrm{Ax}(f_i)\) . Also the intervals satisfy \(f_k\left(\overline{\mathbb{R}}\setminus S_i^k\right)\subseteq U_i^k\) and \(f_i\left(\overline{\mathbb{R}}\setminus S_k^i\right)\subseteq U_k^i\).
Moreover, in 33 we got that \[\min\{\tfrac{1}{2}\lvert \textrm{tr}(f_k)\rvert,\tfrac{1}{2}\lvert \textrm{tr}(f_i)\rvert\}> \sqrt{\varepsilon}\cdot \max\left\{\sqrt{1+\left\lvert C(f_k,f_i)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_i)\right\rvert}}\right\}.\] Hence, according to Remark 9, we can choose the intervals \(U_i^k,U_k^i,S_i^k,S_k^i\) to have the following properties:
Let \(\ell_i^k,\ell_k^i\) be the hyperbolic geodesics of \(\mathbb{H}\) with the same endpoints as \(U_i^k\) and \(U_k^i\), respectively. Similarly, \(\gamma_i^k,\gamma_k^i\) are the hyperbolic geodesics of \(\mathbb{H}\) with the same endpoints as \(S_i^k\) and \(S_k^i\), respectively. Then the geodesics \(\ell_i^k,\ell_k^i,\gamma_i^k,\gamma_k^i\) are pairwise distinct and satisfy \[\label{eq:32proof32distance32of32intervals} \cosh\tfrac{1}{2}d_\mathbb{H}(\ell_i^k,\gamma_i^k)= \cosh\tfrac{1}{2}d_\mathbb{H}(\ell_k^i,\gamma_k^i)=\max\left\{\sqrt{1+\left\lvert C(f_k,f_i)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_i)\right\rvert}}\right\}\cdot \sqrt{\varepsilon}.\tag{34}\] By repeating this process for all \(i=1,2,\dots,N\), with \(i\neq k\) we obtain two collections of \(N-1\) intervals \(\mathbb{U}^k=\{U_1^k,\dots, U^k_{k-1},U^k_{k+1},\dots, U_N^k\}\) and \(\mathbb{S}^k=\{S_1^k,\dots, S^k_{k-1},S^k_{k+1},\dots, S_N^k\}\). Since \(U^k_i,S^k_i\) are symmetric with respect to the axis of \(f_k\), the collections \(\mathbb{U}^k\) and \(\mathbb{S}^k\) are totally ordered with the set inclusion operation; i.e. for any \(U^k_i,U^k_j\in\mathbb{U}^k\) we have that either \(U^k_i\subseteq U^k_j\) or \(U^k_j\subseteq U^k_i\), and similarly for the intervals in \(\mathbb{S}^k\). Let \(U^k_n\) be a least element of \(\mathbb{U}^k\) (an inner-most interval in the collection) and \(S^k_m\) a least element of \(\mathbb{S}^k\). Here, we say a least element, and not the, since they are not necessarily unique. Furthermore, the indices \(n\) and \(m\) might not coincide. Let us also consider the intervals \(U^k_m\) and \(S^k_n\). Because \(U^k_n\) and \(S^k_m\) are least elements, we have that \(U^k_n\subseteq U^k_m\) and \(S^k_m\subseteq S^k_n\). Now, observe that \[\overline{U^k_n}\cap \overline{S^k_m}\subseteq\overline{U^k_n}\cap \overline{S^k_n}=\emptyset,\] because, by construction, \(U^k_n\) and \(S^k_n\) have disjoint closures.
Now let \(k_0=1,2,\dots,N\) with \(k\neq k_0\). Following the process described above for \(k_0\) in place of \(k\), we obtain collections \(\mathbb{U}^{k_0}, \mathbb{S}^{k_0}\) of intervals that are symmetric with respect to \(\mathrm{Ax}(f_{k_0})\). Let \(U^{k_0}_{n_0}\) and \(S^{k_0}_{m_0}\) be some least elements in \(\mathbb{U}^{k_0}\) and \(\mathbb{S}^{k_0}\), respectively. We want to also show that the intervals \(U^k_n,S^k_m,U^{k_0}_{n_0},S^{k_0}_{m_0}\) have pairwise disjoint closures, where \(U^k_n,S^k_m\) are the intervals that were constructed above. First, observe that as we showed before, we have that \(\overline{U^{k_0}_{n_0}}\cap\overline{S^{k_0}_{m_0}}=\emptyset\). Since \(U^k_n,U^{k_0}_{n_0}\) are least elements in their respective collections, we have that \[\overline{U^{k_0}_{n_0}}\cap\overline{U^k_n}\subseteq\overline{U^{k_0}_k}\cap\overline{U^k_{k_0}}=\emptyset,\] because by construction we have that \(U^k_{k_0},S^k_{k_0},U^{k_0}_k,S^{k_0}_k\) have pairwise disjoint closures. Similar arguments yield the desired conclusion.
The process we just described allows us to construct intervals \(U^1,U^2,\dots,U^N, S^1,S^2,\dots,S^N\) (we suppress the indices and keep the exponents, for visual clarity) that have pairwise disjoint closures, and so that each pair \(U^i,S^i\) is symmetric with respect to the axis \(\mathrm{Ax}(f_i)\) of \(f_i\), for \(i=1,2,\dots,N\). Therefore, our proof will be complete upon showing that \(f_i\left(\overline{\mathbb{R}}\setminus S^i\right)\subseteq U^i\), for all \(i=1,2,\dots,N\).
Focusing on the exponent \(k\) we fixed earlier, and reintroducing the indices in our intervals, that is considering the inner-most intervals \(U^k_n\) and \(S^k_m\) described above, we have to show that \[\label{eq:32end32goal} f_i\left(\overline{\mathbb{R}}\setminus S^k_m\right)\subseteq U^k_n.\tag{35}\] First, observe that if \(n=m\), then we already have the desired conclusion, by construction. So, for the rest of the proof, we assume that \(n\neq m\). We will also use the intervals \(U^k_m\) and \(S^k_ n\). Let \(\ell^k_n,\ell^k_m\) be the geodesics of \(\mathbb{H}\) with the same endpoints as \(U^k_n,U^k_m\), respectively. Similarly, \(\gamma^k_n,\gamma^k_m\) be the geodesics of \(\mathbb{H}\) with the same endpoints as \(S^k_n,S^k_m\), respectively. Because the intervals in our construction satisfied 34 , we have that \[\label{eq:32distance32of32intervals32n} \cosh\tfrac{1}{2}d_\mathbb{H}(\ell_n^k,\gamma_n^k)=\max\left\{\sqrt{1+\left\lvert C(f_k,f_n)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_n)\right\rvert}}\right\}\cdot \sqrt{\varepsilon},\tag{36}\] and \[\label{eq:32distance32of32intervals32m} \cosh\tfrac{1}{2}d_\mathbb{H}(\ell_m^k,\gamma_m^k)=\max\left\{\sqrt{1+\left\lvert C(f_k,f_m)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_m)\right\rvert}}\right\}\cdot \sqrt{\varepsilon}.\tag{37}\] Observe that if \(T_{f_k}\), the translation length of \(f_k\), satisfies \(T_{f_k}>d_\mathbb{H}(\ell_n^k,\gamma_m^k)\), then we have the desired inclusion 35 . Equivalently, it suffices to show that \[\label{eq:32end32goal322} \tfrac{1}{2}\lvert \textrm{tr}(f_k)\rvert =\cosh\tfrac{1}{2}T_{f_k}>\cosh\tfrac{1}{2}d_\mathbb{H}(\ell_n^k,\gamma_m^k).\tag{38}\] For simplicity, let us write \(d_0=d_\mathbb{H}(\ell_n^k,\gamma_m^k)\), \(d_n=d_\mathbb{H}(\ell_n^k,\gamma_n^k)\) and \(d_m=d_\mathbb{H}(\ell_m^k,\gamma_m^k)\). To simplify future estimates we observe that due to 36 and 37 we have that \[\begin{align} \cosh \tfrac{1}{2}d_n\;\cosh \tfrac{1}{2}d_m &= \max\left\{\sqrt{1+\left\lvert C(f_k,f_n)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_n)\right\rvert}}\right\}\cdot \max\left\{\sqrt{1+\left\lvert C(f_k,f_m)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_k,f_m)\right\rvert}}\right\} \cdot \varepsilon\nonumber\\ &\leqslant\left(\max_{i\neq j}\left\{ \max\left\{\sqrt{1+\left\lvert C(f_i,f_j)\right\rvert},\sqrt{1+\frac{1}{\left\lvert C(f_i,f_j)\right\rvert}}\right\} \right\}\right)^2\cdot \varepsilon\nonumber\\ &=\max_{i\neq j} \{M_{i,j}\}\cdot \varepsilon. \label{eq:32proof32estimate} \end{align}\tag{39}\] We consider two cases, in which we will use the trivial inequality \(\cosh(x+y)\leqslant 2\cosh x \;\cosh y\), for all \(x,y\leqslant 0\).
Case 1: \(d_0\leqslant d_n + d_m\) (see left-hand side of Figure 7)
Using 32 and 39 we get that \[\begin{align} \cosh \tfrac{1}{2}d_0 &\leqslant\cosh \left( \tfrac{1}{2}d_n + \tfrac{1}{2}d_m\right)\leqslant 2\;\cosh \tfrac{1}{2}d_n \;\cosh\tfrac{1}{2} d_m\\ &\leqslant 2 \;\max_{i\neq j} \{M_{i,j}\} \cdot \varepsilon\leqslant 2\;\max_{i\neq j}\{M_{i,j}\}\cdot \max_{i\neq j}\left\{\frac{\lvert C(f_i,f_j)\rvert +1}{\lvert C(f_i,f_j)-1\rvert}\right\} \cdot \varepsilon\\ &< 4\;\max_{i\neq j}\{M_{i,j}\}\cdot \max_{i\neq j}\left\{\frac{\lvert C(f_i,f_j)\rvert +1}{\lvert C(f_i,f_j)-1\rvert}\right\}\cdot\varepsilon< \tfrac{1}{2}\lvert \textrm{tr}(f_k)\rvert , \end{align}\] as required for 38 .
Case 2: \(d_0>d_n+d_m\) (see right-hand side of Figure 7)
In this case we have that \(d_0=d_n+d_m+d_\mathbb{H}(\gamma_n^k,\ell_m^k)\), or equivalently \[\label{eq:32case32232eq1} \cosh \tfrac{1}{2}d_0 = \cosh\left( \tfrac{1}{2}d_n+\tfrac{1}{2}d_m + \tfrac{1}{2}d_\mathbb{H}(\gamma_n^k,\ell_m^k)\right) \leqslant 4\;\cosh \tfrac{1}{2}d_n\;\cosh\tfrac{1}{2}d_m \;\cosh \tfrac{1}{2}d_\mathbb{H}(\gamma_n^k,\ell_m^k).\tag{40}\] Let \(H_n^k\subseteq\mathbb{H}\) be the open hyperbolic half-plane that is bounded by the geodesic \(\gamma_n^k\) and contains \(\ell_n^k\). Similarly, \(H_m^k\) denotes the open hyperbolic half-plane bounded by \(\ell_m^k\), that contains \(\gamma_m^k\). Note that the half-planes \(H_n^k, H_m^k\) have disjoint closures, and \[\label{eq:32case32232eq2} \inf\left\{d_\mathbb{H}(z,w)\colon z\in H_n^k\;\text{and} \;w\in H_m^k\right\}=d_\mathbb{H}(\gamma_n^k,\ell_m^k).\tag{41}\] Moreover, by construction of the intervals \(U_n^k, S_n^k,U_m^k, S_m^k\), we have that \(\mathrm{Ax}(f_n)\subseteq H_n^k\), and similarly \(\mathrm{Ax}(f_m)\subseteq H_m^k\). Therefore the axes of \(f_n\) and \(f_m\) are disjoint, and \[\label{eq:32case32232eq3} d_\mathbb{H}\left( \mathrm{Ax}(f_n), \mathrm{Ax}(f_m)\right)<\inf\left\{d_\mathbb{H}(z,w)\colon z\in H_n^k\;\text{and} \;w\in H_m^k\right\}.\tag{42}\] So, due to 41 and 42 , inequality 40 yields \[\begin{align} \cosh \tfrac{1}{2}d_0 &< 4\;\cosh \tfrac{1}{2}d_n\;\cosh\tfrac{1}{2}d_m \;\cosh\tfrac{1}{2}d_\mathbb{H}\left( \mathrm{Ax}(f_n), \mathrm{Ax}(f_m)\right)\nonumber \\ &\leqslant 4\;\cosh \tfrac{1}{2}d_n\;\cosh\tfrac{1}{2}d_m \;\cosh d_\mathbb{H}\left( \mathrm{Ax}(f_n), \mathrm{Ax}(f_m)\right). \label{eq:32case32232eq4} \end{align}\tag{43}\] But, according to Corollary 2, we have that \(\cosh d_\mathbb{H}\left( \mathrm{Ax}(f_n), \mathrm{Ax}(f_m)\right) = \frac{C(f_n,f_m)+1}{\lvert C(f_n,f_m)-1\rvert}\). Substituting this to 43 and using 32 , 39 yield \[\begin{align} \cosh \tfrac{1}{2}d_0 \leqslant 4\;\max_{i\neq j}\{M_{i,j}\}\cdot \varepsilon\cdot \frac{C(f_n,f_m)+1}{\lvert C(f_n,f_m)-1\rvert}<\tfrac{1}{2}\lvert\textrm{tr}(f_k)\rvert, \end{align}\] and we are once more led to 38 . ◻