7
$\begingroup$

Morgan and Shalen "Free action of surface groups on R-trees" 1989 shows that surface groups (genus at least 2) act freely on some real trees (R-trees). Their proof seems to be non-constructive, requiring the Baire category theorem. But the object they are constructing sounds quite simple. Their Proposition 17 says:

On every closed non-exceptional hyperbolic surface, there exists a measured geodesic lamination whose leaves and complementary regions are all simply connected.

Does anyone have a picture of this lamination on the genus 2 surface? And, using that, a picture of the R-tree that the genus 2 surface group acts freely on?

Improve this question
$\endgroup$
  • 3
    $\begingroup$ Many examples are constructed using invariant train tracks of pseudo-Anosov homeomorphisms. Do you know about those? $\endgroup$ – Lee Mosher Nov 5 '19 at 23:06
  • $\begingroup$ I'm passingly familiar with train tracks and not at all familiar with pseudo-Anosov homeomorphisms. Thanks for giving me some directions to read up on. $\endgroup$ – user32157 Nov 6 '19 at 13:57
  • $\begingroup$ I wonder if Thurston had a visualization in one of his books of notes where he introduced measured lamination. If not, I doubt a useful visualization exists. $\endgroup$ – user6976 Nov 6 '19 at 19:40
10
$\begingroup$

Here is a picture of a train track, which approximates a lamination:

enter image description here

Another picture:

enter image description here

The leaf space of this lamination (obtained by collapsing each leaf and complementary polygon to a point) is an $\mathbb{R}$-tree. As you contract the leaves to points, you might get a picture looking somewhat like this:

enter image description here

If you fill in with the boundary circle, you get a dendrite which might help you visualize an $\mathbb{R}$-tree:

enter image description here

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks for the reply and these are definitely nice pictures. Is the lamination in the second picture satisfying the desired properties, though? It looks like some complementary regions might be non-simply connected though I can't quite visualize what is what. And similarly, is that the R-tree with a surface group action, or simply an R-tree? If so, what does the group action look like (what do the generator do)? $\endgroup$ – user32157 Nov 6 '19 at 13:54
  • 1
    $\begingroup$ The second picture doesn’t show the leaves on the back side of the surface, which you should think of as spiraling and branching in a way that doesn’t match the front, so that the complementary region consists of 4 ideal triangles. $\endgroup$ – Ian Agol Nov 6 '19 at 14:29
  • 1
    $\begingroup$ As for the picture of the dendrite (which is a compactification of the R-tree), this is one associated to a quadratic map, not a surface group; however the surface one will look similar. In principal, the complement of this dendrite on the Riemann sphere will be a disk conformally equivalent to the hyperbolic plane. Each ideal triangle will have its 3 ideal points converging to a trivalent vertex of the dendrite. So the surface permutes the trivalent points the same way that it permutes the ideal triangles. $\endgroup$ – Ian Agol Nov 6 '19 at 14:47
  • 1
    $\begingroup$ Curt McMullen has computed such a dendrite, but it’s hausforff dimension is 2 so it looks thick. math.harvard.edu/~ctm/gallery/index.html see the Griffin and snowflake. $\endgroup$ – Ian Agol Nov 6 '19 at 14:56
4
$\begingroup$

Let $G$ be the fundamental group of the surface of genus 2. The group $Out(G)$ is the mapping class group $Mod_2$. Take a pseudo-Anosov $\phi$ from $Mod_2$. Then $G$ acts on the asymptotic cone $T$ of $G$ by $$g\cdot (g_1,...,g_n,...)=(\phi(g)\cdot g_1,...,\phi^n(g)\cdot g_n,....).$$That action is free by isometries and $T$ is an $\mathbb{R}$-tree. One of course needs that $(\phi(g),...,\phi^n(g),...)$ is not equal to $(1,...,1,...)$ in the asymptotic cone but that can be achieved by choosing scaling constants of the cone.

Improve this answer
$\endgroup$
  • $\begingroup$ Is it clear that the action is free? The construction makes sense for more general hyperbolic groups, and the action is usually not free (but with virtually cyclic arc-stabilisers). Is there something specific about surface groups? $\endgroup$ – AGenevois Nov 6 '19 at 7:34
  • $\begingroup$ It does not work for all hyperbolic groups. One needs "pseudo-Anosov" automorphism. For example it rules out all groups with property T. $\endgroup$ – user6976 Nov 6 '19 at 7:39
  • $\begingroup$ Sure, one needs $\mathrm{Out}(G)$ to be infinite in order to get an action without a global fixed point. But free products of surface groups are the only hyperbolic groups acting freely on real trees. So I am wondering why the construction leads to a free action for surface groups but not (for instance) for an amalgamated product of two surface groups over $\mathbb{Z}$. $\endgroup$ – AGenevois Nov 6 '19 at 7:50
  • $\begingroup$ A power of a pA takes every element of Mod far from itself (modulo conjugation). $\endgroup$ – user6976 Nov 6 '19 at 9:31
  • $\begingroup$ One can view elements of Mod as multicurves as in Mazur-Minsky. $\endgroup$ – user6976 Nov 6 '19 at 9:33

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.