Nielsen-Thurston classification via the curve complex? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:39:59Z http://mathoverflow.net/feeds/question/91323 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex Nielsen-Thurston classification via the curve complex? Dave Futer 2012-03-15T19:13:35Z 2012-07-06T13:18:15Z <p>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. </p> <p>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?</p> http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex/91325#91325 Answer by Igor Rivin for Nielsen-Thurston classification via the curve complex? Igor Rivin 2012-03-15T19:31:13Z 2012-03-15T20:35:19Z <p>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)</p> http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex/91333#91333 Answer by Lee Mosher for Nielsen-Thurston classification via the curve complex? Lee Mosher 2012-03-15T22:22:44Z 2012-03-16T04:44:18Z <p>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.</p> <p>-----edited after Dave Futar's comment-----</p> <p>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.</p> http://mathoverflow.net/questions/91323/nielsen-thurston-classification-via-the-curve-complex/101489#101489 Answer by Dan Margalit for Nielsen-Thurston classification via the curve complex? Dan Margalit 2012-07-06T13:18:15Z 2012-07-06T13:18:15Z <p>Dave, you might find this interesting:</p> <p>www.mth.kcl.ac.uk/staff/wj_harvey/harcsh.pdf</p> <p>(although this doesn't give the Nielsen-Thurston classification).</p>