Suppose that $a$ acted as a <hyperbolic isometry of a tree>, with <translation length> $\tau>0$. Nonzero powers have the same axis and
$$
\ell(a^m)=m\tau,
\qquad
\ell(a^n)=n\tau.
$$
Translation 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.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 2 / c / Solution

{title2=$\ell(g)$}
{wiki}

The translation length of an isometry $g$ of a metric space is $\ell(g)=\inf_x d(x,gx)$. For a hyperbolic tree isometry, it is attained exactly on the <axis of a tree isometry>.

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 Incoming links: Translation length