3
$\begingroup$

(This is a reformulation of an earlier unanswered question. I would like to thank Ian Agol for pointing out to me Walter Parry's characterization of hyperbolic translation length functions.)

Let $G$ be a finitely generated group and $T_1$ and $T_2$ two minimal (and hence cocompact) metric simplicial trees that $G$ acts on isometrically by simplicial automorphisms without inversions. Consider the (unbased) hyperbolic translation length functions $$l_{T_i}\colon G\to [0,\infty),\ g\mapsto\inf_{x\in T_i}d(x,gx),\ i\in\left\{1,2\right\}.$$

Assume that $T_1$ and $T_2$ have the same covolume, say covolume 1. Moreover, suppose that for all $g\in G$ we have $l_{T_1}(g)\leq l_{T_2}(g)$ and $l_{T_1}(g)=0\Leftrightarrow l_{T_2}(g)=0$.

Does this imply that $l_{T_1}$ and $l_{T_2}$ are equal?

I have the feeling that for every $g$ such that $l_{T_1}(g)$ is strictly smaller than $l_{T_2}(g)$ there must be an $h\in G$ such that $l_{T_1}(h)$ is strictly bigger than $l_{T_2}(h)$ in order to preserve the covolume. Indeed, I can prove the statement for free $F_n$-trees (elements of Culler-Vogtmann Outer space), but my arguments do not generalize immediately.

$\endgroup$
  • $\begingroup$ In the earlier version, didn't you insist that your trees were non-abelian? $\endgroup$ – HJRW Oct 24 '13 at 18:32
  • $\begingroup$ Anyway, the answer is 'no', even in the non-abelian case. You can construct counterexamples by pulling elements of a vertex-stabilizer across an edge to increase the size of an edge stabilizer. I'll write a detailed answer when I have some time (maybe later today?). $\endgroup$ – HJRW Oct 24 '13 at 18:55
  • $\begingroup$ Sorry, I need to correct my previous comment. I have some evidence if $T_1$ and $T_2$ are irreducible. If we also assume that $H\leq G$ fixes a point in $T_1$ if and only if it fixes a point in $T_2$ then $G\backslash T_1$ and $G\backslash T_2$ are simplicially isometric. $\endgroup$ – Sebastian Oct 24 '13 at 21:18
  • $\begingroup$ What do you call covolume of $T_i$? It is something intrinsic, or does it involve the $G$-action? $\endgroup$ – YCor Oct 24 '13 at 22:46
  • $\begingroup$ HW answered my question in his answer. Also, could you give a link to the "earlier unanswered question"? I'm looking for Parry's characterization you mention. $\endgroup$ – YCor Oct 25 '13 at 10:32
1
$\begingroup$

Suppose $S$ is a metric $G$-tree of covolume $1$ (ie the sum of the edge lengths in $G\backslash S$ is equal to 1). Now suppose that $T$ is obtained from folding together two adjacent edges in the same $G$-orbit, as follows:

Let $e$ be an edge of $S$ adjacent to a vertex $v$, and suppose that $g\in G$ stabilizes $v$ but not $e$. Then $T$ is obtained from $S$ by ($G$-equivariantly) folding $e$ and $ge$.

This operation doesn't change the underlying graph of the quotient, so $T$ still has covolume equal to 1.

Consider a hyperbolic element $\gamma\in G$. The axis of $\gamma$ in $T$ is contained in the image of the axis of $\gamma$ in $S$, so $l_T(\gamma)\leq l_S(\gamma)$. However, if the axis of $\gamma$ in $S$ traversed a pair of edges in $S$ that are folded in $T$, then we will have a strict inequality

$l_T(\gamma)< l_S(\gamma)$ .

For a specific example, let $G$ be the free group $\langle a,b\rangle$, and let $S$ be the Bass--Serre tree of the HNN extension

$G\cong \langle a\rangle*_1$

where $b$ is the stable letter. Now let $T$ be the Bass--Serre tree of the HNN extension

$G\cong \langle a,a^b\rangle*_{a\sim a^b}$

where, again, $b$ is the stable letter. Note that the commutator $[a,b]$ has strictly smaller translation length in $T$ than in $S$---indeed, it's hyperbolic in $S$ but elliptic in $T$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Folding $G$-equivalent edges may create new elliptic elements, that is, elements that fix points and hence have translation length 0. For example, this will be the case if the axis of a hyperbolic group element $h\in G$ runs in a zig-zag path through $G$-equivalent corners. But maybe we can get away with slightly folding so that no two edges are folded completely. In your example, if by $a^b$ you mean $bab^{-1}$, the element $a^b$ stabilizes a vertex of $S$, namely the vertex corresponding to the coset $[b]=[ba]$, and hence has translation length 0. It also has translation length 0 in $T$. $\endgroup$ – Sebastian Oct 25 '13 at 8:45
  • $\begingroup$ @Sebastian - Sorry, you're correct that I gave the wrong hyperbolic element - the commutator $[a,b]$ is an example of an element that's hyperbolic in $S$ but elliptic in $T$. $\endgroup$ – HJRW Oct 25 '13 at 9:31
  • $\begingroup$ Thank you, this is indeed an interesting example. But I excluded this situation from the very beginning: "Moreover, suppose that ... $l_{T_1}(g)=0\Leftrightarrow l_{T_2}(g)=0$." This probably also explains my comment above regarding the folding construction. $\endgroup$ – Sebastian Oct 25 '13 at 10:19
  • $\begingroup$ @Sebastian - Ah, sorry, I missed that requirement. I suspect we can show that $T_i$ satisfying the hyperbolic-length inequality are necessarily obtained by this sort of folding, and so deduce that the trees really are equivariantly homeomorphic. $\endgroup$ – HJRW Oct 26 '13 at 3:14
  • $\begingroup$ @Sebastian - sorry, I made a slight mistake. You can't deduce that the $T_i$ are equivariantly homeomorphic - many different abelian trees have the same hyperbolic length functions, for instance. You may have to deal with that case separately. $\endgroup$ – HJRW Oct 26 '13 at 13:21
1
$\begingroup$

Here is a counterexample (unfortunately):

We give the two graphs of groups the same metrics. Equivariantly folding the two branching edges in the Bass-Serre covering $T_1$ gives rise to an equivariant 1-Lipschitz map from $T_1$ to $T_2$ and hence we have $l_{T_2}(g)\leq l_{T_1}(g)$ for all $g\in G$. Now observe that the graph of groups of $T_2$ can be obtained from the graph of groups of $T_1$ by an elementary collapse followed by an elementary expansion. Elementary deformations do not create new elliptic subgroups, whence $l_{T_1}(g)=0\Leftrightarrow l_{T_2}(g)=0$.

However, if we give each edge length $1/3$, the hyperbolic group element corresponding to the upper loop followed by a backtrack along the connecting bridge has translation length $1$ in $T_1$ and length $1/3$ in $T_2$.

The two trees are in fact non-abelian (they are irreducible), as the fundamental groups of their graphs of groups contain a free group of rank 2 acting freely. Counterexample

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ So in fact the counterexample is obtained by pulling an element over an edge stabilizer! $\endgroup$ – HJRW Oct 26 '13 at 18:26
  • 1
    $\begingroup$ Yes, exactly. Thank you for your help! At first I expected that it would always be possible to find a hyperbolic element that becomes elliptic under this operation (as in the HNN example), but this is not the case here. $\endgroup$ – Sebastian Oct 26 '13 at 18:33
0
$\begingroup$

I guess this is proved in F.Paulin, The Gromov topology on $\mathbb R$-trees, Topology Appl. 32 (1989).

Abstract:

We are interested in isometric actions of a fixed finitely generated group on R-trees. Using metric methods inspired by Gromov’s work, we define a more geometric topology on sets of such objects. We prove it to be the same as the Morgan-Shalen topology, defined by the translation lengths of the group elements, in the case of minimal irreducible actions.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You should say more, e.g. sketch out how it's proven or link the article and state which proposition it is. Short answers like this sometimes get flagged as low quality, because a reader can't be sure if you're correct or not without digging for a 20 year old paper $\endgroup$ – David White Oct 25 '13 at 13:50
  • $\begingroup$ added link and abstract $\endgroup$ – Carlo Beenakker Oct 25 '13 at 14:38
  • $\begingroup$ I don't think that the statement follows from Paulin's work. If it does, please explain why. Paulin did not consider trees of equal covolume and the answer is definitely 'no' if we drop this assumption. Paulin considered trees that are close, but any slight downscaling of the metric will produce a tree which is close. $\endgroup$ – Sebastian Oct 25 '13 at 15:50
  • 2
    $\begingroup$ I'm sorry about that (as you see I'm new in MO) when I said "I guess" I ment "I guess" and not "I am sure". I just wanted to post a maybe useful reference. Since I'm new and have no reputation, I cannot comment, I'm obliged to "answer" if I want to say something... $\endgroup$ – user126154 Oct 25 '13 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.