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

Let $x,y\in \mathrm{Fix}(\varphi )$. A tree has a unique geodesic segment $[x,y]$. The isometry $\varphi$ sends this segment to the segment $[\varphi x,\varphi y]=[x,y]$. An isometry of a segment that fixes both endpoints fixes every point of it, soThus the fixed-point set of an elliptic isometry of a tree is a convex subtree, in particular it is path-connected.

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Let $A=\mathrm{Axis}(\varphi )$ and let $p$ be the nearest-point projection of $x$ to $A$. If the translation length is $\tau >0$, the unique path from $x$ to $\varphi x$ is the concatenationThe first and last pieces both have length $d(x,A)$, while the middle one has length $\tau$. ThereforeThe displacement is minimized exactly when $x\in A$, proving that the axis is the minimum set of the displacement function.

On $A$, the power ${\varphi }^r$ translates through $r\tau$ when $r>0$ and in the opposite direction through $|r|\tau$ when $r<0$. The same formula applied to ${\varphi }^r$ shows that its minimum set is $A$. Hence

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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.

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Suppose $G=A\ast B$ were a nontrivial free product. Its Bass-Serre tree action has trivial edge stabilizers and no global fixed vertex. Part c makes $a$ elliptic. Since $a$ has infinite order, the fixed set of every nonzero power $a^r$ is a single vertex: it is nonempty, while fixing two vertices would fix the intervening edge and put the infinite-order element $a^r$ in a trivial edge stabilizer.

Let this vertex be $v$. The relation $b{a}^m{b}^{-1}=a^n$ givesand both sides are the singleton $\{v\}$. Thus $b$ also fixes $v$. Since $a$ and $b$ generate the Baumslag-Solitar group, the entire group fixes $v$, contradicting the Bass-Serre action of a nontrivial free product. Hence no such decomposition exists.

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