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An isometry of hyperbolic space with positive attained translation length has an invariant complete geodesic joining its two ideal fixed points. This is its axis. It translates along the axis by its translation length; in three dimensions it may additionally rotate around it. In upper half-space, $(z,t)\mapsto (kz,|k|t)$ with $|k|\ne 1$ has the vertical axis through zero and length $|\log |k||$. Every connecting path has length at least the change in $\log t$, and a vertical segment attains the bound.

Represent the Möbius transformation by $A=\left(\begin{array}{cc}3& 2\\ 7& 5\end{array}\right)$, which has determinant one and trace eight. Its distinct eigenvalues are ${\lambda }_{\pm }=4\pm \sqrt{15}$ and their product is one. A change of basis to eigenvectors conjugates its projective action to a dilation by ${\lambda }_{+}/{\lambda }_{-}=(4+\sqrt{15}{)}^2$ or its reciprocal. Translation length is invariant under conjugacy, so part (i)'s dilation calculation givesThe fixed boundary points solve $7z^2+2z-2=0$, namely $(-1\pm \sqrt{15})/7$. The axis of a hyperbolic-space isometry is the semicircular geodesic with those endpoints in the vertical real plane.

A model of three-dimensional hyperbolic space is ${\mathbb{H}}^3=\{(z,t):z\in \mathbb{C},t>0\}$ with metric $d{s}^2=(|dz{|}^2+d{t}^2)/t^2$, the complete simply connected metric of curvature $-1$. The translation length of an isometry $g$ is $\ell (g)={\inf }_{P}d(P,gP)$. A hyperbolic isometry has positive translation length attained on its invariant complete geodesic, the axis of a hyperbolic-space isometry. It joins the two fixed ideal points and is translated by distance $\ell (g)$; a three-dimensional axial isometry can also rotate around it.

For the boundary map $z\mapsto kz$, the hyperbolic extension of a Möbius transformation is $(z,t)\mapsto (kz,|k|t)$. Every path from $P$ to $gP$ has length at least the absolute change of $\log t$, namely $|\log |k||$, because the metric length dominates $|dt|/t$. On the vertical line $z=0$ a vertical segment attains that bound. HenceIf $|k|\ne 1$ the vertical line is the axis. If $|k|=1$ the map is a rotation with zero translation length, including the identity when $k=1$; it is not a positive-length hyperbolic isometry.

Suppose that $a$ acted as a hyperbolic isometry of a tree, with translation length $\tau >0$. Nonzero powers have the same axis andTranslation length is invariant under conjugacy, whereas the defining relation in the Baumslag-Solitar group says that $a^m$ and $a^n$ are conjugate. Hence $m\tau =n\tau$, contradicting $m\ne n$. Therefore $a$ acts elliptically in every combinatorial tree action.