Is there any way to represent every element of the mapping class group of a surface as an arc on that surface?

  • 2
    $\begingroup$ You may be interested in the Alexander method, detailed in chapter 2 of the Primer on Mapping Class Groups. I believe Proposition 2.6 is the right reference. $\endgroup$ – Santana Afton Dec 30 '19 at 21:30

The natural action (of mapping classes act on isotopy classes of arcs) has large stabilisers. So the "correct" answer to your question is "no".

Now, the mapping class group is countable. The set of isotopy classes of arcs is also countable. With a bit of work, you can construct a bijection between them. So in that sense the answer to your question is "yes". (In fact, you will learn a lot from constructing such a bijection.) But making the bijection will involve many choices; so it cannot be used to prove anything...

  • $\begingroup$ I’m confused about your last sentence. There is no canonical Alexander system on a surface, but surely that doesn’t invalidate its usefulness as a tool. And, in fact, there is a nice useful bijection between mapping classes and Alexander systems (fixing a particular reference system). $\endgroup$ – Santana Afton Dec 30 '19 at 21:24
  • $\begingroup$ Is an "Alexander system" a maximal arc system (aka a collection of disjoint arcs cutting the surface into hexagons)? If so, then yes, the mapping class group acts on these with finite stabilisers (and if you allow oriented, labelled arcs then the stabiliser is trivial). It is also true that, up to the action of the mapping class group, there are only finitely many such systems. But this is not what the original post was asking about. There the question concerned single arcs, not arc systems. $\endgroup$ – Sam Nead Dec 30 '19 at 22:06

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.