MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am curious to see if anyone knows a proof of the Nielsen-Thurston classification of mapping classes that does not depend on results in Teichmuller theory.

From a naive point of view, translation distances in the curve complex should serve the same purpose as translation distances in Teichmuller space. For instance, if a mapping class fixes a simplex setwise, then it's reducible (actually, reducibles look much simpler from this point of view). Similarly, a mapping class is pseudo-Anosov iff its stable translation length in the curve complex is strictly positive -- but can one prove this without quoting results of Masur-Minsky that need Teichmuller theory anyhow?

share|cite|improve this question
up vote 8 down vote accepted

Masur and Minsky, in their paper "Quasiconvexity in the curve complex", give a purely combinatorial proof that certain "nested" train track sequences project to quasigeodesics in the curve complex. It follows from this, without too much more trouble, that a pseudo-Anosov mapping class has positive stable translation length. Teichmuller theory is not used in this proof.

-----edited after Dave Futar's comment-----

Here's an argument for the converse, following Dave's suggestion to use Klarreich's boundary. Suppose that $\phi \in MCG(S)$ has positive translation length. So $\phi$ has a unique attracting/repelling pair of fixed points on the Gromov boundary of the curve complex. By Klarreich's theorem, this is a pair of measureable but measureless singular foliations $F_1,F_2$, each of which is arational. They are unequal in the Gromov boundary, so by Klarreich's theorem they are inequivalent (with respect to isotopy and Whitehead moves), and so they can be realized on the surface as a transverse pair of foliations. And there is an actual homeomorphism $\Phi$ in the mapping class $\phi$ that preserves this pair of foliations. What's left is to construct transverse measures on these two foliations with the appropriate stretching properties. This situation occurs as a piece of the proof of Thurston's classification theorem in Fathi-Laudenbach-Poenaru, and the conclusion is that either $\Phi$ is a finite order mapping class or the desired transverse measures exist and $\Phi$ is a pseudo-Anosov homeomorphism. Finite order is ruled out by the assumption of positive translation length in the curve complex.

share|cite|improve this answer
Thanks, Lee. What about the converse implication: positive stable translation length implies pseudo-Anosov? Does that require knowing e.g. Erica Klarreich's work? – Dave Futer Mar 15 '12 at 23:42

I don't know about the curve complex, but Bestvina-Handel (Train tracks for surface homeos, 1995) give a completely combinatorial/algorithmic proof, which, at a quick glance, never mentions teichmuller space)

share|cite|improve this answer
Thanks, Igor! . – Dave Futer Mar 16 '12 at 0:03
Is there a reference for seeing---via Teichmuller theory---that the curve complex has infinite diameter? I mean, as opposed to nested train-tracks a la Masur/Minsky or Bowditch. – J. Martel Mar 16 '12 at 3:16
@jmart: Masur & Minsky give a short proof of infinite diameter that they attribute to Luo. See the parenthetical remark after the statement of Proposition 4.6 in "Geometry of the Curve Complex I". – Lee Mosher Mar 16 '12 at 4:17

Dave, you might find this interesting:

(although this doesn't give the Nielsen-Thurston classification).

share|cite|improve this answer
Dan, I believe that paper is flawed. For example, the 2nd paragraph of the proof of Theorem 2.3 is just wrong. – Lee Mosher Jul 6 '12 at 13:46
Thanks, Lee. I thought maybe I should have looked at it before posting. – Dan Margalit Jul 10 '12 at 15:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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