Simplicial Trees and Group Actions - Examples

Every two integers are joined in $L$ by the obvious consecutive edge path, and no reduced closed path exists, so $L$ is a simplicial tree. The map $\tau (n)=n+1$ preserves adjacency and has no fixed vertex.

Every vertex moves distance $1$, so [L1] gives $\mathrm{\ell }(\tau )=1$. Therefore [L2] places $\tau$ in the hyperbolic case, with axis equal to the whole line $L$.

By [L1], the underlying simple Cayley graph is a tree. Each vertex has one edge labelled by each basis element and by its inverse, so its valence is $2r$.

Replacing every geometric edge by the two corresponding orientations does not create a cycle; it only records both directions explicitly. Thus the same graph becomes a simplicial tree, in agreement with the finite bridge principle [L2].

Before subdivision, the nontrivial element sends the oriented edge $u\to v$ to the reverse edge $v\to u$, so there is an inversion.

After subdivision, the midpoint is a vertex fixed by the action, and each half-edge is sent to the other half-edge with the same orientation type rather than to its reverse. This is exactly the mechanism asserted in [L1].

The reflection preserves adjacency and the unique diameter path of $P_5$, so it is an automorphism of the tree.

Its midpoint is already the vertex $0$, so the fixed vertex promised by [L1] occurs without needing any further subdivision in this example.

The reflection $r$ fixes the vertex $0$, so [L1] places it in the elliptic case.

The translation ${\tau }_2$ has no fixed vertex and preserves the whole line while moving every vertex two steps, so [L1] places it in the hyperbolic case.

By [L2], the bi-infinite line is a simplicial tree. Let $g$ be translation by three steps, so $⟨g⟩$ acts without inversions on that tree.

In the quotient graph from [L1], the three vertex orbits are the residue classes modulo $3$, and the edge orbits join them cyclically. The quotient is therefore a $3$-cycle, not a tree. This refutes the statement.