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Let $G$ be a finitely generated group. By a theorem of Culler and Morgan, the set of non-abelian (not necessarily simplicial) minimal $\mathbb{R}$-trees with isometric $G$-action injects into the infinite-dimensional vector space $\mathbb{R}^G$ by assigning to each non-abelian tree $T$ its translation length function $l_T\colon G\to\mathbb{R},\ g\mapsto l_T(g)=\inf_{x\in T}d(x,gx)$.

Is there a way to read off of $l_T\in\mathbb{R}^G$ if $T$ is simplicial? For example, if the entries of $l_T$ are bounded below by some positive constant $c>0$, can $T$ still be non-simplicial?

Here, by a simplicial tree I mean an $\mathbb{R}$-tree whose set of vertices (i.e. points that when removed disconnect the tree into more than two components) is discrete and closed.

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    $\begingroup$ First, if you want an injection, you certainly should restrict to minimal actions. Also, I don't know what an abelian tree is. Why do you expect simplicial to be related to boundedness? Shouldn't it be related to integrality? Finally, $l_T$ is not a length function (length is usually supposed to be subadditive, but for your $l=l_T$ you can have $l(g)=l(h)=0\neq l(gh)$). $\endgroup$ – YCor Oct 15 '13 at 19:39
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    $\begingroup$ If by simplicial you mean a simplicial tree with real edge lengths, then I think by Lyndon-Chiswell theory you get a simplicial action with the same length function iff the image of the translation length is discrete. I'm assuming the action is minimal so that the translation length determines the action. You might also want to used the based version. $\endgroup$ – Benjamin Steinberg Oct 15 '13 at 19:40
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    $\begingroup$ @Sebastian: you should also provide a precise definition to simplicial. If you start with an action on a simplicial tree (with edges of length 1) and multiply the distance on the tree by $\sqrt{2}$, do you call the resulting $\mathbb{R}$-tree simplicial? etc. Please give a definition. $\endgroup$ – YCor Oct 15 '13 at 19:59
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    $\begingroup$ Sebastian: Take a look at Chiswell's book "Introduction to $\Lambda$-trees", chapter 4, where he described construction of the tree from the length function $\ell$ on $G$. I think, if you follow the arguments, discreteness of the set of values of $\ell$ will imply discreteness of the tree. $\endgroup$ – Misha Oct 15 '13 at 21:06
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    $\begingroup$ Here is the alternative approach: Given a discrete length function (I think, it suffices to assume it us bounded below) one defines a discrete pretree on which the group still acts. This discrete pretree extends canonically to a discrete tree, so the group still acts. This was worked out by Bowditch and Guirardel long ago. $\endgroup$ – Misha Oct 16 '13 at 18:42
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A partial answer: suppose in addition that there are finitely many orbits of branch points. (Branch point = vertex in the OQ.) This condition has been considered by Jiang and Guirardel. For example Jiang showed that this condition is satisfied if the given minimal action (of f.g. $G$ on an $\mathbb{R}$-tree) is free.

Note $\ell(g)\leq L_x(g)=d(x,gx)$ for all $x\in X$, so if the non-zero values of $\ell$ are bounded away from zero by $c$ then $d(x,gx)>c$ for all $x\in X$ and $g\in G$, that is, orbits of each point are discrete. Therefore the set of all branch points, as a finite union of closed and discrete subsets is closed and discrete.

In the opposite direction, if $x$ is a branch point and $c_x>0$ such that every other branch point is further from $x$ than $c_x$, then $L_x(g)> c_x$ for all $g$ that doesn't fix $x$. Since $L_{\gamma x}(g)=L_x(\gamma^{-1}g\gamma)$, the number $c_x$ depends only on the orbit, not on $x$ itself. Choosing $c_x$ for each orbit and taking the minimum $c$ we get $L_x(g)=d(x,gx)>c$ for all branch points $x$.

However, making the step from $L_x(g)$ bounded away from zero to $\ell(g)$ bounded away from zero requires more input.

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  • $\begingroup$ Thank you, this is enlightening. Unfortunately, having only finitely many orbits of branch points seems to be a condition that is, in general, hard to check. As pointed out in my comment above, the image of $ l $ being discrete seems to suffice. It would be interesting to know if the values of $ l $ being bounded below also does. $\endgroup$ – Sebastian Oct 16 '13 at 19:43
  • $\begingroup$ I've deleted an observation to the effect that an action may be simplicial even if the image of $\ell$ is not discrete. Unfortunately, the example originally given doesn't support this. $\endgroup$ – shane.orourke Oct 18 '13 at 9:25
  • $\begingroup$ @shane.orourke if the action is simplicial (and minimal), then since $\Gamma$ is f.g., Bass-Serre theory tells you that there are finitely many edge orbits, and hence finitely many edge lengths. So the set of translations lengths (i.e. the image of $\ell$) is contained in the semigroup of $\mathbf{R}$ generated by these finitely many positive elements, and hence is discrete. $\endgroup$ – YCor Oct 18 '13 at 18:49
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This question has been answered by Misha, in the comments, as follows.

Given a discrete length function (I think, it suffices to assume it us bounded below) one defines a discrete pretree on which the group still acts. This discrete pretree extends canonically to a discrete tree, so the group still acts. This was worked out by Bowditch and Guirardel long ago

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  • $\begingroup$ In an answer it would have been nice to be a little less sloppy... $\endgroup$ – YCor Dec 17 '13 at 11:16
  • $\begingroup$ @YvesCornulier - I don't have time to be more careful. But the answer is CW, so feel free to improve it! $\endgroup$ – HJRW Dec 17 '13 at 11:18
  • $\begingroup$ I'll try if I find some time but I'd first need to understand the picture correctly. $\endgroup$ – YCor Dec 17 '13 at 14:17
  • $\begingroup$ I am still worried about this answer. If the branch points of $T$ are a discrete pretree then $T$ must be discrete. The pretree structure on the branch points of $T$ is described in "Parry - Axioms for Translation Length Functions"; it is discrete if $max\left\{ 0 , l(xy) - l(x) - l(y)\right\}$ is discrete, where $x,y$ range over all hyperbolic elements of $G$. Each summand in $l(xy) - l(x) - l(y)$ is discrete if $l$ is discrete. But linear combinations of discrete functions need not be discrete; think of $l(xy)$ taking values which are multiples of $\sqrt{2}$ and $l(x)$ taking integer values. $\endgroup$ – Sebastian Dec 30 '13 at 17:07

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