Comparing layered triangulations of 3-manifolds which fiber over the circle. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:01:20Z http://mathoverflow.net/feeds/question/106426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106426/comparing-layered-triangulations-of-3-manifolds-which-fiber-over-the-circle Comparing layered triangulations of 3-manifolds which fiber over the circle. leone slavich 2012-09-05T12:51:07Z 2012-10-03T22:56:25Z <p>I am sorry but I am reposting this question because I wasn't logged in when I first asked it.</p> <p>Ian Agol has produced a method to build an ideal layered triangulation of a hyperbolic 3-manifold which fibers over the circle with pseudo-Anosov monodromy of the fiber through periodic splitting of train tracks. Such triangulation is transverse-taut and veering.</p> <p>On the converse, he proves that a layered triangulation of a manifold coming from a periodic sequence of Whitehead moves comes from a periodic train track splitting only if it is veering (see arxiv.org/pdf/1008.1606).</p> <p>Both this construction and the notion of layered triangulation require to specify the fiber and the monodromy.</p> <p>From Thurston's work, it is known that a hyperbolic 3-manifold \$M\$ which fibers over the circle fibers in many different ways, and in fact the fibers of the different fibrations are integral points of the cones over certain faces of the unit ball for the thurston norm in \$H_2(M)\$.</p> <p>I was wondering how to compare the different layered triangulations of the same manifold that one can construct from the different fiberings. More precisely:</p> <ol> <li><p>Are two veering layered triangulations built from two different fibrations whose fibers lie in the cone over a common fibering face for the thurston norm combinatorially the same? If this is true, are their taut structures the same? If not, how can we relate the taut structures?</p></li> <li><p>Is there a way to compare triangulations built from fibers in different fibered faces for the thurston norm?</p></li> </ol> <p>Thank you very much for your attention.</p> http://mathoverflow.net/questions/106426/comparing-layered-triangulations-of-3-manifolds-which-fiber-over-the-circle/106462#106462 Answer by Agol for Comparing layered triangulations of 3-manifolds which fiber over the circle. Agol 2012-09-05T22:49:52Z 2012-10-03T22:56:25Z <p><strong>Revision 10/3/12</strong></p> <ol> <li>is true. First of all, by a <a href="https://wikis.uit.tufts.edu/confluence/display/~dmarga01/FLP" rel="nofollow">result of Fried</a>, associated to a fibered face of the Thurston norm on homology, there is a unique (up to isotopy) pseudo-Anosov flow which is transverse to every fibration in (the interior of) the face. Thus, the construction of a veering triangulation associated to each fiber in the face will give a taut ideal triangulation of the same cusped manifold obtained by drilling out the singular fibers. We may replace the original manifold with this cusped manifold, and ask if taut ideal triangulations are the same for every fiber in a fibered face of the cusped manifold?</li> </ol> <p>For any given fibration, some multiple of it will be fully carried by the branched surface associated to the taut ideal veering triangulation (this may be seen by the construction of a veering triangulation as a cycle of Whitehead moves of ideal triangulations of the surface). Since the branched surface is oriented, for each surface carried by the branched surface, one obtains a homologically non-trivial surface (more generally, for non-integral weights, one obtains a measured foliation representing a non-trivial homology class). <img src="https://dl.dropbox.com/u/8592391/branang.jpg" alt="alt text"> One sees that the weight space of the branched surface surjects the relative homology of the cusped manifold (since this is just isomorphic to the simplicial 2-cycles). Moreover, the subspace of cycles which are fully carried with positive weights by the branched surface form an open piecewise linear rational cone in the homology space. One also sees that any such cycle (with positive integral weights) must be fibered. So one obtains an open cover of the rational points in the interior of the fibered face by open rational cones. Since there is always a rational point on the boundary of a rational cone, if one of the boundary points of a cone associated to a veering triangulation lies in the interior of the fibered face, then two of these open sets must intersect, which gives a contradiction unless every boundary point lies in the boundary of the fibered face. Thus, the veering triangulation associated to every fiber in the face will be the same. </p> <p>For 2., I see no relation, unless there happens to be a symmetry which permutes the fibered faces. It might be interesting to find ideal triangulations that admit two distinct taut veering structures (although these won't necessarily be associated to fibered faces). </p>