Some Questions — Question 3

Definitions and mathematical statement

Let $T$ be a rooted locally finite tree without terminal vertices and with infinitely many boundary rays. Its boundary $\partial T$ consists of infinite rays starting at the root, with the compact topology in which rays are close if they share a long initial segment. Give $\operatorname{Aut}(T)$ the topology of agreement on finite rooted subtrees. Let $G\leq\operatorname{Aut}(T)$ be a closed subgroup whose action on $\partial T$ is transitive.

For an integer $m\geq0$, the level stabilizer is the subgroup fixing every vertex at distance $m$ from the root: \[ \operatorname{St}_G(m)=\{g\in G:gv=v\text{ for every level-}m\text{ vertex }v\}. \] Must this subgroup contain an element $g$ that fixes no boundary ray? In quantified form the question is whether \[ \forall m\geq0\ \exists g\in\operatorname{St}_G(m)\quad \operatorname{Fix}_{\partial T}(g)=\varnothing. \] Fixed-point-freeness refers to the boundary, not to vertices of the tree: every automorphism fixes the root, and a level stabilizer necessarily fixes the first $m$ levels. The same selected element must move every infinite ray, rather than choosing a different element for each ray. Closedness is part of the hypothesis, making $G$ a compact group. Transitivity likewise means actual transitivity on complete rays, not merely that an individual orbit is dense.

Short English statement

Must every level stabilizer contain a single automorphism that moves every boundary ray?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 1200003 (AMR-011-0003), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Abert - Some questions (pdf) (2010), Question 3. Original source indexed by AMR-011-0003. https://www.renyi.hu/~abert/questions.pdf.