Why do strongly irreducible Heegaard surfaces look like fibers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:02:11Z http://mathoverflow.net/feeds/question/39121 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39121/why-do-strongly-irreducible-heegaard-surfaces-look-like-fibers Why do strongly irreducible Heegaard surfaces look like fibers? bb 2010-09-17T17:13:55Z 2010-09-17T20:47:49Z <p>I remember hearing somewhere that strongly irreducible Heegaard surfaces in hyperbolic 3-manifolds "look like" fibers.</p> <p>I know that Otal's result about short geodesics in hyperbolic mapping tori being unlinked with respect to fibers has an analogue in the setting of strongly irreducible Heegaard surfaces.</p> <p>Can someone tell me some other similarities between strongly irreducible Heegaard surfaces and fibers in hyperbolic 3-manifolds? Or why someone would say that strongly irreducible Heegaard surfaces look like fibers? Any references would be greatly appreciated.</p> http://mathoverflow.net/questions/39121/why-do-strongly-irreducible-heegaard-surfaces-look-like-fibers/39140#39140 Answer by Sam Nead for Why do strongly irreducible Heegaard surfaces look like fibers? Sam Nead 2010-09-17T20:17:07Z 2010-09-17T20:47:49Z <p>Something more basic is true. Strongly irreducible Heegaard splittings act a lot like incompressible surfaces (of which fibers are a special case). Here are two results as evidence. </p> <p>First, suppose that the three-manifold is equipped with a triangulation. Then Haken showed that incompressible surfaces can be normalized. That is, isotoped to intersect every tetrahedron in a collection of standard disks; normal triangles and normal quads. Rubinstein (and Stocking) showed that strongly irreducible splittings can be "almost normalized"; in every tetrahedron it has normal disks and in at most one tetrahedron there is an almost normal annulus or octagon. Michelle Stocking's thesis is a standard reference for this material. Many people (Rubinstein, Hass, Bachman, Scott, ...) will say in public that the above is a PL version of an analytic truth: incompressible surfaces can be isotoped to be minimal surfaces of index zero while strongly irreducible surfaces can be made minimal of index one. </p> <p>Second, suppose that \$S, T\$ are surfaces, with \$T\$ incompressible. If \$S\$ is also incompressible then it is an exercise in innermost disks to show that, after isotoping \$S\$ to meet \$T\$ minimally, all curves of intersection are essential on both surfaces. If \$S\$ is instead strongly irreducible then a one-parameter sweep-out argument, followed by an innermost disk argument shows that there is some position of \$S\$ where all curves of intersection are essential on both surfaces. </p> <p>Ok, third (I couldn't resist), suppose that the ambient three-manifold is hyperbolic. Minsky's approach to the ending lamination conjecture says that the geometry "around" the incompressible surface or a strongly irreducible splitting can be modelled using "blocks" based on the four-holed sphere or once-holed torus, where the blocks are glued to each other vertically using pairs of pants and horizontally using solid tori (Margulis tubes). </p>