Simplicial Trees and Group Actions

If the graph is a simplicial tree, then [L1] gives connectedness, so every two vertices are joined by some reduced path. If there were two distinct reduced paths between the same vertices, follow one and return along the reverse of the other; after cancelling any common initial segment, one obtains a nontrivial reduced closed path, contradicting [L1].

Conversely, assume the graph is nonempty and every two vertices are joined by a unique reduced path. Then it is connected. A nontrivial reduced closed path based at $v_0$ with first edge $e_1$ would give two reduced paths from $v_0$ to $t(e_1)$: the one-edge path $e_1$ and the reverse of the remaining part of the cycle. That contradicts uniqueness. Hence no nontrivial reduced closed path exists, so the graph is a simplicial tree by [L1].

By [L2], the reduced path from $v$ to $w$ exists and is unique. Its length is an integer, and [L1] defines $d_T(v,w)$ to be exactly that integer. So $d_T$ is integer-valued.

Let $v=v_0,e_1,\dots ,e_n,v_n=w$ be the unique reduced path. Every subpath is again reduced, so [L1] gives $d_T(v_i,v_j)=j-i$ for $i\le j$. Hence the path realizes distance on each of its segments, which is the geodesic property claimed in the statement.

By the no-loop/no-parallel-edge hypothesis and [L1], a reduced simplicial path in $X$ is exactly a path in the underlying finite simple graph $\mathrm{\mid }X\mathrm{\mid }$, and connectedness means the same thing in both models. Therefore the uniqueness criterion in [L3] translates verbatim into the uniqueness criterion in [L2].

Applying [L2] and [L3] to the translation from step 1.1 shows that $X$ is a simplicial tree exactly when $\mathrm{\mid }X\mathrm{\mid }$ is a published finite tree.

Every automorphism of $T$ permutes geometric edges, so it sends the midpoint of an edge to the midpoint of its image edge. Hence the original action extends uniquely to the subdivided graph $T^{\mathrm{\prime }}$.

The graph $T^{\mathrm{\prime }}$ is connected because each original edge has only been split in two. A reduced closed path in $T^{\mathrm{\prime }}$ would project, after deleting midpoint backtracks, to a nontrivial reduced closed path in $T$, contradicting [L2]. So $T^{\mathrm{\prime }}$ is again a simplicial tree.

In $T^{\mathrm{\prime }}$, every half-edge joins an original vertex to a midpoint vertex. Its reverse joins a midpoint vertex to an original vertex, so an automorphism cannot send a half-edge to its reverse without swapping the two vertex types. That cannot happen in the extended action. Therefore the action on $T^{\mathrm{\prime }}$ is without inversions in the sense of [L1].

If $g\in G_e$, then $g\cdot e=e$, so applying origin and terminus to this equality gives $g\cdot v=v$ and $g\cdot w=w$. Thus $G_e\subseteq G_v\cap G_w$. Since each stabilizer is an intersection of solution sets to $g\cdot x=x$, it is closed under products and inverses, hence is a subgroup by [L2].

If $e^{\mathrm{\prime }}=h\cdot e$, then $G_{e^{\mathrm{\prime }}}=h{G}_e{h}^{-1}$ and similarly $G_{h\cdot v}=h{G}_v{h}^{-1}$ and $G_{h\cdot w}=h{G}_w{h}^{-1}$. Therefore passing to another representative of the quotient edge conjugates the two inclusion maps by the same $h$, so the quotient incidence data from [L1] is well defined up to that canonical conjugacy.

Let $u$ and $v$ be fixed vertices of $g$. By [L2] there is a unique reduced path $P$ from $u$ to $v$. Applying $g$ to $P$ gives another reduced path from $u$ to $v$, so uniqueness in [L2] forces $g(P)=P$.

Every vertex on $P$ is therefore fixed by $g$, because $g$ preserves the ordered path and fixes its endpoints. Hence the fixed vertices are closed under the unique geodesic between any two of them, which is exactly the subtree condition described in [L1].

By [L4], every displacement $d_T(v,gv)$ is a natural number, so choose $v$ with minimal displacement $m=d_T(v,gv)$. Then $m=\mathrm{\ell }(g)$ by [L2]. If $m=0$, the vertex $v$ is fixed and [L3] describes the fixed subtree.

Assume $m>0$, and let $P$ be the unique reduced path from $v$ to $gv$. Its translate $gP$ is the unique reduced path from $gv$ to $g^{2}v$. If $P$ and $gP$ met in more than the vertex $gv$, then some interior point of the overlap would have displacement strictly smaller than $m$, contradicting step 1.1; if they shared an edge with opposite orientations, that edge would be inverted, contradicting [L1]. Hence consecutive translates $g^{n}P$ and $g^{n+1}P$ meet only at one endpoint. [L1, L4, step 1.1, assume-case[hyperbolic], algebra]

Therefore $A_g:={\bigcup }_{n\in \mathbb{Z}}{g}^{n}P$ is a bi-infinite reduced path. It is preserved by $g$, and $g$ sends each segment $g^{n}P$ onto $g^{n+1}P$, so every vertex on $A_g$ moves distance exactly $m$ along that line. Thus $g$ acts on $A_g$ as translation by $\mathrm{\ell }(g)=m$.

Let $x$ be a vertex not on $A_g$, and let $y$ be the first vertex of $A_g$ on the unique reduced path from $x$ to $A_g$. Then the geodesic from $x$ to $gx$ runs from $x$ to $y$, then along $A_g$ from $y$ to $gy$ by length $m$, and then from $gy$ to $gx$, so $d_T(x,gx)=2d_T(x,A_g)+m>m$. Hence the vertices of minimal displacement are exactly those on $A_g$, which makes $A_g$ unique. This is the hyperbolic case, and it excludes fixed vertices.

If $g$ is elliptic, [L1] gives a fixed vertex, so every power $g^n$ fixes that same vertex and has translation length $0$. Conjugating by $h$ carries a fixed vertex of $g$ to a fixed vertex of $hg{h}^{-1}$, so conjugation preserves the elliptic case and its translation length.

If $g$ is hyperbolic with axis $A_g$ and translation length $\mathrm{\ell }(g)$, then $hg{h}^{-1}$ preserves the line $h{A}_g$ and translates it by the same distance, so $\mathrm{\ell }(hg{h}^{-1})=\mathrm{\ell }(g)$. Also $g^n$ preserves $A_g$ and shifts every vertex on it by $\mathrm{\mid }n\mathrm{\mid }\mathrm{\ell }(g)$, so $\mathrm{\ell }(g^n)=\mathrm{\mid }n\mathrm{\mid }\mathrm{\ell }(g)$ for $n\mathrm{\ne }0$.

The case $n=0$ gives $\mathrm{\ell }(1)=0$, already covered by step 1.1, so the displayed formulas hold for every integer $n$.

Replace $T$ by its barycentric subdivision $T^{\mathrm{\prime }}$ using [L1]. Choose a vertex $v$ of $T^{\mathrm{\prime }}$. Because $G$ is finite, the orbit $G\cdot v$ is finite, and the union of the geodesics joining pairs of orbit vertices is therefore a finite $G$-invariant subtree $U\subseteq T^{\mathrm{\prime }}$.

Let $D$ be the diameter of $U$, choose vertices $x,y\in U$ with $d(x,y)=D$, and let $m$ be the midpoint of the unique geodesic from $x$ to $y$. In a finite tree every diameter geodesic has the same midpoint: if two diameters had different midpoints, the unique path joining those midpoints would extend one of them past length $D$, contradicting maximality. So $m$ depends only on $U$, not on the chosen diameter.

Any automorphism of $U$ sends diameter geodesics to diameter geodesics, so it fixes the intrinsic midpoint $m$ from step 2.1. If $D$ is even, $m$ is a vertex of $U$ and is fixed. If $D$ is odd, $m$ is the midpoint of a unique geometric edge of $U$, so every element of $G$ preserves that edge setwise. The action on $T^{\mathrm{\prime }}$ is without inversions by [L1], hence no element can swap the two endpoints; both endpoints are fixed. In either case the induced action on $T^{\mathrm{\prime }}$ fixes a vertex.

Let $g\in G$ have finite order. Then the cyclic subgroup $⟨g⟩$ is finite, so [L3] gives a fixed vertex for its action on the barycentric subdivision of $T$.

Let $p$ be the fixed vertex from step 1.1 in the barycentric subdivision. If $p$ is an original vertex of $T$, then [L1] forces $g=e$. If $p$ is the midpoint of an original geometric edge, then $g$ preserves that edge setwise. Because the original action is without inversions by [L2], $g$ cannot swap its two orientations, so it fixes both endpoints of that edge. Now [L1] again gives $g=e$. Hence no nonidentity torsion element exists.

Let $A_g$ and $A_h$ be the disjoint axes from [L1], and let $[p,q]$ be the unique geodesic joining them with $p\in A_g$ and $q\in A_h$. Removing $p$ from the tree leaves two components meeting the two rays of $A_g$, call them $X_g^{+}$ and $X_g^{-}$. Define $X_h^{+}$ and $X_h^{-}$ similarly at $q$. These four half-trees are pairwise disjoint.

Choose $m,n>0$ so large that $m\mathrm{\ell }(g)>d(p,q)$ and $n\mathrm{\ell }(h)>d(p,q)$. Then [L1] implies $g^m(T\setminus X_g^{-})\subseteq X_g^{+}$ and $g^{-m}(T\setminus X_g^{+})\subseteq X_g^{-}$, and similarly $h^n(T\setminus X_h^{-})\subseteq X_h^{+}$ and $h^{-n}(T\setminus X_h^{+})\subseteq X_h^{-}$. Indeed, a sufficiently long translation along an axis carries everything on one side of the basepoint across the bridge segment $[p,q]$ into the corresponding attracting half-tree.

Pick a vertex $x$ outside the four half-trees. Any reduced word in $g^{\pm m}$ and $h^{\pm n}$ sends $x$ into the domain attached to its first letter by step 2.1, and successive letters cannot bring it back because the four domains are disjoint. So no nonempty reduced word acts trivially. Therefore $g^m$ and $h^n$ freely generate a free subgroup of rank $2$.

The bi-infinite line is a simplicial tree with infinitely many vertices, so it lies outside the finite scope named in [L1].

Therefore [L1] does not identify every simplicial tree with a published finite tree, and the statement is false.

Let $T$ be a single geometric edge with endpoints $u,v$, and let the nontrivial element of $C_2$ swap $u$ and $v$. It sends one oriented edge $e:u\to v$ to its reverse $\bar{e}:v\to u$, so the action has an inversion in the sense of [L1].

Since [L2] would be unnecessary if every action were already inversion-free, the example in step 1.1 disproves the statement.

Translation by $n\mapsto n+1$ on the bi-infinite line is hyperbolic: it preserves the whole line and moves every vertex a positive distance. This is the second case of [L1].

A hyperbolic automorphism has no fixed vertex, so the translation from step 1.1 is a counterexample. Hence the statement is false.

Reflect the bi-infinite line by $r(n)=-n$. The vertex $0$ is fixed, so [L2] gives $\mathrm{\ell }(r)=0$ via [L1].

At the basepoint $v=1$ one has $d(1,r(1))=d(1,-1)=2\mathrm{\ne }0$. So $\mathrm{\ell }(r)$ is not the displacement from an arbitrary basepoint, and the statement is false.

Let $g$ be translation by $n\mapsto n+3$ on the bi-infinite line. By [L2] this is a hyperbolic action without inversions on a tree.

In the quotient graph from [L1], the vertex orbits are the residue classes of $n$ modulo $3$, and the edge orbits join them in a $3$-cycle. That quotient is not a tree, so the statement is false.