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

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Solution

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

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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-133/2/a/solution