2
$\begingroup$

$\DeclareMathOperator\MCG{MCG}$Let $\Gamma=\MCG(S_{g, 0,0 })$ be mapping class group of closed hyperbolic surfaces. Let $V=C^0(S)$ be the set of vertices of the simplicial curve complex. We are studying the well known group action $\Gamma \times V \to V$, yet we find ourselves utterly incapable of computing and translating elements $([\phi], [\alpha]) \mapsto [\phi(\alpha)]$ where $\alpha$ represents a "simple closed curve" in $V$. In otherwords it appears that the action of the MCG on the curve complex is incomputable for closed surfaces.

I am aware of Mark C. Bell's curver program. It's documentation indicates that it uses results of Saul Schleimer relating paths in a "flip graph" with words in the pointed mapping class groups. I am not aware of any flip graph analogy for the closed mapping class groups.

My question: What is best computational approach to being able to solve the following problem for the group action $\Gamma \times V \to V$:

``Given $[\phi] \in \Gamma, [\alpha] \in V, [\beta] \in V$, determine whether $[\phi(\alpha)]=[\beta]$ is True or False in $V$."

Evidently the fact that we have no linear representations of $\Gamma$ is one computational obstruction, for otherwise all the quantities could essentially be represented as matrices.

In my research I have interest in constructing finite subsets $I$ for which a chain sum $\sum_{[\phi] \in I} \sum_{[\alpha] \in B}[\phi(\alpha)]=0$ mod 2. Here $B$ is a finite subset of $V$ based on Nathan Broaddus' homology spheres. (e.g. $B$ is the vertex set if the two dimensional spheres in genus $g=2$ discovered by Broaddus).

(**) I'll accept Sam Nead's answer since it suggests that Birman's exact sequence tells us that forgetful map from pointed mapping classes to unpointed mapping classes is onto, with kernel equal to the image of $\pi_1(S, pt)$ under the Push map. So it's obvious from Birman's sequence that Bell-Webb's curver program can be used to compute pure mapping class actions on the curve complex. Although this raises the question of whether the membership problem for image of $\pi_1(S, pt)$ by the push map is computable in the pointed groups. Assuming it is, then to solve the subset-sum problem I'll start with the pointed groups in curver.

$\endgroup$
4
  • 1
    $\begingroup$ What goes wrong in the following? Take a triangulation of the surface, and (simplicial) representatives of all curves involved; an element $\phi$ of the MCG is given by a self homeomorphism (or just self homotopy equivalence), so you can compute a representative of $[\phi.\alpha]$ (a simplicial curve on the triangulated surface,) say by applying simplicial approximation to $\phi$. To check if it's isotopic to a curve $\beta$ you have to check whether they are conjugate in the fundamental group. The conjugacy problem in these groups is solvable. $\endgroup$ Commented Dec 21, 2021 at 17:06
  • $\begingroup$ To your point about conjugacy, yes, i wish i could compute the action of MCG directly on geodesics in the hyperbolic universal cover. I would find that basically computable in symbolic python or wolfram. W.r.t. simplicial maps, maybe it's possible, and maybe thats what Mark Bell's curver does for pointed groups. I've never seen that in a paper on MCG. $\endgroup$
    – JHM
    Commented Dec 21, 2021 at 17:50
  • 1
    $\begingroup$ I'm pretty sure this does give a method for effective computation and answers the question in the title. More efficient methods exist, for example you don't really need a simplicial approximation to $\phi$, but (depending on how elements of the MCG are given) should be able to apply it to the homotopy class of a given curve directly... I don't know enough about the known algorithms to suggest the "best" or even a good computational approach. But in any case it is computable and can be implemented on a computer with a reasonable amount of work. $\endgroup$ Commented Dec 21, 2021 at 21:15
  • $\begingroup$ Yes, there is a connection to the Birman exact sequence. The algorithm I sketch (in my answer) also solves the membership problem (of being an inner automorphism inside of the automorphism group) you mention. $\endgroup$
    – Sam Nead
    Commented Dec 22, 2021 at 19:53

1 Answer 1

4
$\begingroup$

You write "it appears that the action of the MCG on the curve complex is incomputable for closed surfaces."

This is not correct. Geva Yashfe points out one approach in the comments. Here is another, with references given below.

Fix $S$ a connected, closed, oriented surface of genus $g$. Fix a one-vertex triangulation of $S$. We represent simple closed curves via "edge coordinates" (or we could use normal coordinates). Note that (as $S$ is closed) two curves can have different edge coordinates yet be isotopic.

Thus your problem reduces to determining isotopy equivalence among such representatives. (The presence of a mapping class is dealt with by understanding (say) how Dehn twists act on edge coordinates.) It is a theorem that two curves are isotopic if and only if they cobound an annulus, perhaps after performing a sequence of "bigon" moves. Detecting such annuli or bigons, and performing bigon moves, can be done computationally.

A naive implementation will be polynomial time in the edge weights. Thinking more deeply gives an algorithm that is polynomial time in the logarithm of the edge weights. This brings us to the beginnings of modern research in this area, and so answers your question (in italics).

You can find versions and discussions of this in papers such as

  • Schaefer, Sedgwick, and Štefankovič [2002]
  • Agol, Hass, and Thurston [2005]
  • Erickson and Nayyeri [2012]

as well as work of Bell (partly with Webb).

$\endgroup$
4
  • 1
    $\begingroup$ Your answer suggests that Bell's curver program can be used to compute the action of pure mapping classes on the curve complex. I think I'll ask Mark directly. $\endgroup$
    – JHM
    Commented Dec 22, 2021 at 13:35
  • $\begingroup$ Good idea. If I have not answered your question, please let me know. If I have, please consider accepting my answer (and thus “closing” the question). $\endgroup$
    – Sam Nead
    Commented Dec 22, 2021 at 14:15
  • 1
    $\begingroup$ Thank you yes, I think you made the point. It's not a great question because after your answer it's obvious from Birman exact sequence. And that means all the hard work's been done in curver, and i just need work with curver code. $\endgroup$
    – JHM
    Commented Dec 22, 2021 at 14:55
  • 1
    $\begingroup$ Following the answer given by Sam, I have computed exactly what i've been looking for with Bell's curver program. The action of $MCG$ on curve complex is essentially equivalent to conjugacy action of $MCG$ on Dehn twists. And Bell's curver program is effective enough at this. $\endgroup$
    – JHM
    Commented Dec 27, 2021 at 20:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .