Lipschitz metric isometries between Outer Spaces of virtually free groups

We follow the proof of [FM11, Lemma 3.14]. The first claim is that we can choose $g$ such that the image of $C_{g}$ in $F\backslash\mkern-5.0mu\backslash\Gamma$ has no triple points. That is, if we can write $\gamma=\delta_{1}\delta_{2}\delta_{3}$, where the endpoints of each $\delta_{i}$ map to the same point in $F\backslash\mkern-5.0mu\backslash\Gamma$, then we may replace $\gamma$ (and thus $g$) with a shorter loop. If some $\delta_{i}$ maps to a cyclically reduced loop in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, choose that $\delta_{i}$. If all $\delta_{i}$ map to reduced but not cyclically reduced paths in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, then (after shuffling vertex group elements around,) the labellings of $\delta_{i}$ by their images in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$ may be written as

where, since $\gamma$ immerses into $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, we must have $e_{1}\neq e_{2}$, so the loop $\delta_{1}\delta_{2}$ is immersed in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, and we choose it instead. This process shortens $\gamma$ in $F\backslash\mkern-5.0mu\backslash\Gamma$, so eventually we obtain $g$ such that $C_{g}$ has no triple points.

The second claim is that we can choose $g$ such that the image of $C_{g}$ in $F\backslash\mkern-5.0mu\backslash\Gamma$ has no double points which cross in the sense that we have $\gamma=\delta_{1}\delta_{2}\delta_{3}\delta_{4}$, where the initial endpoints of $\delta_{1}$ and $\delta_{3}$ are equal and the initial endpoints of $\delta_{2}$ and $\delta_{4}$ are equal. So suppose we do have such a configuration. Now if some $\delta_{i}\delta_{i+1}$ (with subscripts taken mod $4$) maps to a cyclically reduced path in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, then we can choose that path as our new $g$. Otherwise, as before we have that (after shuffling vertex group elements around) the labellings of $\delta_{i}$ by their images in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$ may be written as

Therefore we conclude that

with subscripts taken mod $4$. Since $\gamma$ immerses into $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, we have $e_{1}\neq e_{3}$ and $e_{2}\neq e_{4}$, so the loop $\delta_{1}\bar{\delta}_{3}$ is immersed in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, and we choose that as our new $g$. Again this process shortens $\gamma$ in $F\backslash\mkern-5.0mu\backslash\Gamma$, so eventually we obtain $g$ such that $C_{g}$ has no crossing double points.

Finally, we claim that we can avoid bad triangles: i.e. a configuration $\gamma=\delta_{1}\delta_{2}\delta_{3}\delta_{4}\delta_{5}\delta_{6}$, where $\delta_{1}$, $\delta_{3}$ and $\delta_{5}$ are closed paths and thus the other three form a “triangle,” which need not be embedded. If any of the paths $\delta_{1}$, $\delta_{2}$ or $\delta_{3}$ is cyclically reduced in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, then we may take that loop as our new $g$. If none of these subpaths are cyclically reduced in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$, then arguing as before, we have that $\delta_{1}\delta_{2}\delta_{3}\bar{\delta}_{2}$ is an immersed closed path in $F\backslash\mkern-5.0mu\backslash\Gamma$ whose image in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$ is cyclically reduced. The only worry is that this path may be longer than $\gamma$ because $\delta_{2}$ is longer than $\delta_{4}\delta_{5}\delta_{6}$. This would imply that the path $\delta_{3}\delta_{4}\delta_{5}\bar{\delta}_{4}$, to which the same argument applies, is shorter than $\gamma$, so we choose it. Again this process shortens $\gamma$ in $F\backslash\mkern-5.0mu\backslash\Gamma$, so eventually we obtain $g$ such that $C_{g}$ has no bad triangles.

We now think of $C_{g}$ as labeled by the images of edges and vertices (with vertex group elements) in $F\backslash\mkern-5.0mu\backslash\Gamma$. Identify vertices with the same vertex label and (oriented) edges with the same edge label. We claim that the resulting graph is a sausage. If there are no identifications or only a single pair of vertex identifications, we are done. By assumption there are no triple points, so each vertex of the resulting graph is at most $4$-valent. Since double points do not cross, if we have two pairs of double points in $C_{g}$, they “nest”.

If there is only a single pair of double points, we are done, as $\gamma$ is a figure-eight. If we suppose there are at least two pairs of double points, then there are two “innermost” pairs which map to vertices in $F\backslash\mkern-5.0mu\backslash\Gamma$. Unlike in the original [FM11], if $v$, $w$ is an innermost pair, the path between $v$ and $w$ may be of the form $eg\bar{e}$, so is not embedded. In this case, choose the middle vertex of $eg\bar{e}$, while in other cases choose a point on the embedded path from $v$ to $w$ arbitrarily. Since there are two innermost pairs this yields two points on $C_{g}$ and therefore two subpaths $\gamma_{1}$ and $\gamma_{2}$ from the first point to the second,, so that $\gamma=\gamma_{1}\bar{\gamma}_{2}$. Since we have no bad triangles and no crossing double points, both $\gamma_{1}$ and $\gamma_{2}$ are embedded in $F\backslash\mkern-5.0mu\backslash\Gamma$. This shows that the graph is an embedded sausage: the intervals $I_{j}$ correspond to subpaths (which may be single vertices) which have more than one preimage in $\gamma$. Since double points do not cross, these intervals appear in the same order in $\gamma_{1}$ and $\gamma_{2}$. ∎

Again the proof is nearly identical to [FM11, Proposition 3.15]. Let $f\colon\Gamma\to\Gamma^{\prime}$ be an optimal map. We begin with the loop $\gamma=\gamma_{1}\bar{\gamma}_{2}$ supplied by 2.1. If the family of intervals $I_{j}$ is empty or contains at most a single interval $I$, then $\gamma$ is a candidate and we are done.

So suppose that the family of intervals $I_{j}$ contains at least two intervals. We will replace $\gamma$ with a candidate. Let $[a,b]$ and $[c,d]$ be the two extremal intervals of $I_{j}$, i.e. $0\leq a\leq b<c\leq d\leq 1$, and if $0<a$ or $d<1$, there is no $I_{j}$ in the interval $(0,a)$ or $(d,1)$ respectively. We replace the path $\gamma_{2}$ with the path that follows $\gamma_{2}$ if $t<b$ or $t>c$ and follows $\gamma_{1}$ on the interval $[b,c]$. Note that $\delta_{2}$ embeds in $F\backslash\mkern-5.0mu\backslash\Gamma$ because $\gamma_{1}(t)=\gamma_{2}(s)$ if and only if $t=s$. Note as well that the $F\backslash\mkern-5.0mu\backslash f$-image of $\delta_{2}$ in $F\backslash\mkern-5.0mu\backslash\Gamma^{\prime}$ is reduced for the same reason and because the $F\backslash\mkern-5.0mu\backslash f$-images of $\gamma_{1}$ and $\gamma_{2}$ are reduced. The loop $\gamma^{\prime}=\gamma\bar{\delta}_{2}$ is a candidate. ∎

As in Culler–Vogtmann [CV86], we may view vertices of the spine as corresponding to trees in $\mathscr{PT}(F)$, all of whose natural edges have the same length. (That is, if a vertex of $\Gamma$ has valence two, so that the incident edges are in the same orbit, we set the lengths of these edges to half the length of edges that are not incident to vertices of valence two).

Under the isometric embedding of A, if $H\leq F$ is a finite-index subgroup, vertices of the spine $K(F)$, thought of as trees in $\mathscr{PT}(F)$ as above, are sent to vertices of the spine $K(H)$. It is clear that this induces a poset map from the vertex set of $K(F)$ to $K(H)$.

In the situation of the statement, the action of $F$ on $\Gamma$ descends to an action of $G$ on $F_{n}\backslash\mkern-5.0mu\backslash\Gamma$ such that the quotient graph of groups in the sense of Bass [Bas93, 3.2] is $F\backslash\mkern-5.0mu\backslash\Gamma$, and conversely every $F_{n}$-tree $\Gamma$ in $K(F_{n})$ such that the quotient $F_{n}\backslash\mkern-5.0mu\backslash\Gamma$ is equipped with a $G$-action via $\alpha$ is actually an $F$-tree in $K(F)$. We will show that a non-surviving edge in a tree $\Gamma$ corresponding to a vertex of $K(F)$ corresponds to an inessential $G$-invariant forest in $F_{n}\backslash\mkern-5.0mu\backslash\Gamma$. Indeed, the image in $F\backslash\mkern-5.0mu\backslash\Gamma$ of a maximal collapsible forest in $\Gamma$ is a (collapsible) forest in the sense of the author in [Lym22, Section 2], and thus a $G$-invariant forest in $F_{n}\backslash\mkern-5.0mu\backslash\Gamma$. An edge is non-surviving if it is contained in every forest in $F\backslash\mkern-5.0mu\backslash\Gamma$, and thus in every $G$-invariant forest in $F_{n}\backslash\mkern-5.0mu\backslash\Gamma$ and conversely. Thus the poset map above sends $L(F)$ isomorphically to $L_{\alpha}$.

If $\varphi\in\operatorname{Out}(F_{n})$ belongs to the normalizer of $\alpha(G)$, then for all $g\in G$ we have $\varphi\alpha(g)=\alpha(g^{\prime})\varphi$ for some $g^{\prime}\in G$ and for $\Gamma\in L_{\alpha}$, we have

so $\Gamma.\varphi$ belongs to the fixed-point set of $\alpha(G)$, and it is clear that all of its edges remain essential, so we conclude $\varphi$ leaves $L_{\alpha}$ invariant.

Let $\operatorname{Aut}^{0}(E)$ be the subgroup of automorphisms leaving $F_{n}$ invariant and let $\operatorname{Out}^{0}(E)$ be its image in $\operatorname{Out}(E)$. Similarly, write $N$ for the normalizer of $\alpha(G)$ in $\operatorname{Out}(F_{n})$ and write $M$ for its full preimage in $\operatorname{Aut}(F_{n})$. Following Krstić [Krs92], we will show that $M$ is isomorphic to $\operatorname{Aut}^{0}(E)$. The final claim in the theorem will then follow from the snake lemma (which applies because the image of each subgroup in the top row is normal in the bottom row) applied to the following commutative diagram

which yields the following exact sequence with indicated connecting homomorphism

By definition, $M$ consists of automorphisms $\Phi\colon F_{n}\to F_{n}$ such that the outer class $\varphi$ satisfies $\varphi\alpha(G)\varphi^{-1}=\alpha(G)$. Since $E$ is the preimage of $\alpha(G)\leq\operatorname{Out}(F_{n})$ in $\operatorname{Aut}(F_{n})$, for $e\in E$, we have that the automorphism of $F_{n}$ defined by $x\mapsto\Phi(e\Phi^{-1}(x)e^{-1})$ is in $E$, say it is equal to $x\mapsto e_{\Phi}xe_{\Phi}^{-1}$. We claim that the map $\Psi_{\Phi}$ defined as $e\mapsto e_{\Phi}$ is an automorphism of $E$. For $f\in F_{n}$, we have $f_{\Phi}$ is the automorphism

so the restriction of $\psi_{\Phi}$ to $F_{n}$ is $\Phi$. An easy calculation expressing $\Phi(ee^{\prime}\Phi^{-1}(x){(ee^{\prime})}^{-1})$ two ways shows that $\psi_{\Phi}$ is a homomorphism. Since the restriction of $\psi_{\Phi}$ to $F_{n}$ is an automorphism, its kernel is a finite normal subgroup of $E$, hence trivial. Its image contains a finite-index subgroup so has finite index. But $E$ has a well-defined, negative Euler characteristic, from which we conclude that $\psi_{\Phi}$ is an automorphism of $E$. Similarly we have $\psi_{\Phi^{\prime}}\psi_{\Phi}=\psi_{\Phi^{\prime}\Phi}$, so the rule $\Phi\mapsto\psi_{\Phi}$ defines a homomorphism $M\to\operatorname{Aut}^{0}(E)$.