Hyperbolic 3-manifolds with a single cusp have canonical ideal triangulations constructed by Epstein and Penner via a certain convex hull construction involving the light-cone in the hyperboloid model of hyperbolic space. (Occasionally, these “triangulations” are really cellulations more complicated cells.) When such a 3-manifold fibers over the circle, there is another type of natural triangulation, called a layered triangulation. Roughly, one starts with a certain ideal triangulation of the fiber surface, looks at the image of this triangulation under the bundle monodromy, interpolates between these by a series of Pachner moves which can then be realized geometrically by layering on tetrahedra.

When the fiber is a once-punctured torus, these two type of triangulations coincide. This was shown by Marc Lackenby using a remarkably soft and elegant argument. It’s natural to wonder whether this phenomena occurs more broadly. For instance Sakuma suggested considering the following:

**Conjecture:** *Canonical triangulations of punctured surface bundles are always layered.*

Saul Schleimer and I have discovered that this is false in general. In particular, the manifold *v1348* from the SnapPea census is fibered by a once-punctured surface of genus 5 yet has a canonical triangulation which is not layered. Precisely, the canonical triangulation does not admit one of Lackenby’s taut structures so that the resulting branched surface carries something with positive weights. Note that *v1349* is in fact the complement of a certain knot in the 3-sphere [CFP].

**Technical details:** The canonical triangulation of *v1348* is in fact just the triangulation encoded in the SnapPea census (and it is a triangulation, not a celluation). It’s easy to check that it fibers using the BNS invariant (cf. [DT]), and compute the genus of the fiber from the Alexander polynomial. One then checks that there are no taut structures of this type using Marc Culler and I’s t3m Python package.

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It’s nice that you’ve found a counterexample – I thought this sort of conjecture was a bit optimistic. Maybe there’s still a layered triangulation which is geometric, i.e. so that all the ideal tetrahedra have positive volume?

One interesting thing to note is that there are canonical cell structures for the complement of the singular locus of a pseudo-Anosov map of a surface. The mapping class is realized by an affine map of the singular Euclidean structure, interpolated by a 1-parameter family of Euclidean structures varying by affine maps with the same (perpendicular) eigenspaces. For a given singular Euclidean structure, there is a canonical cell structure in the complement: just take the Dirichlet decomposition with respect to the singular points. As one varies the Euclidean structure, the Dirichlet domains change finitely many times. This gives rise to a layered triangulation (or rather some sort of layered cell structure – I haven’t thought through the appropriate precise notion). One could guess that this cell structure might be related to the Ford domain of the hyperbolic structure on the complement of the singular locus. I can imagine a possible approach to this, by trying to “regenerate” the hyperbolic structure from the singular solve structure by varying cone angles around the singular points, and try to show that a canonical cell structure persists until the cone angle is decreased to zero at the cusped hyperbolic structure. This is probably too optimistic, but it may be possible to reprove Lackenby’s result this way. Incidentally, Lackenby’s theorem was actually claimed by Jorgensen a long time ago (this was how he originally discovered hyperbolic structures on punctured torus bundles), but wasn’t written up completely until recently (I think Akiyoshi may have given a complete proof). Another proof follows from an approach of Gueritaud to hyperbolize punctured torus bundles.

Comment by Ian Agol — August 20, 2009 @ 8:59 pm |

Maybe there’s still a layered triangulation which is geometric, i.e. so that all the ideal tetrahedra have positive volume?Quite possibly, though I haven’t managed to find any. I randomized the triangulation to generate other geometric triangulations, but these too lacked a taut angle structure carrying a surface with positive weights. Of course, we know there are layered triangulations of this surface bundle, but without knowing the monodromy explicitly it’s hard to generate one of them.

Comment by Nathan Dunfield — August 21, 2009 @ 3:44 am |

Nathan, could you please explain the “branched surface with positive weights” that tells you whether a particular triangulation is layered? For example, m015 (the complement of the (-2,3,7) pretzel knot) is fibered with a fiber of genus 5. But could it be that the 3-tetrahedron triangulation is layered in some funny way?

Comment by Dave Futer — August 31, 2009 @ 9:20 am |

More precisely, a layered triangulation has a natural corresponding taut structure, which is just a choice of dividing the faces of each ideal tetrahedron into two ingoing and two outgoing faces, subject to certain conditions. When we layer on a tetrahedra, the upper faces should be viewed as the outgoing ones, the bottom faces the ingoing ones. In the layered construction, there are many copies of the fiber that can be realized as an embedded subset of the two skeleton of the triangulation. The union of these surfaces are carried, with positive weights, by the branched surface corresponding to the taut structure.

The manifold “m015” is not a counterexample to Sakuma’s conjecture; as you suggest, the 3-tetrahedron canonical triangulation is layered in a “funny way”.

Comment by Nathan Dunfield — August 31, 2009 @ 1:17 pm |

Sakuma gave another question/conjecture: Suppose that M is a hyperbolic manifold that is a surface bundle over the circle, with fibre F having one puncture. Is every edge of the canonical decomposition of M isotopic to a proper arc in F?

Comment by Saul Schleimer — September 4, 2009 @ 1:05 am |

Actually, this weak form of Sakuma’s conjecture is equivalent to the strong form (hence, v1348 is a counterexample to both). Here is a cute fact.

Lemma:Let T be an ideal triangulation of a fibered 3-manifold M with torus boundary. Then T is a layered triangulation if and only if every edge of T is isotopic to a proper arc in the fiber F.The ”only if” direction is obvious. For the ”if” direction, the proof basically involves bootstrapping our way up the dimensions of the skeleta. For simplicity, I am thinking of everything in the infinite cyclic cover of M, where there is a nice well-defined ”projection map” to the fiber F.

Step 1 is to check that if every edge projects to a proper arc in F, then every triangle projects to an embedded triangle (to be precise: the interior is embedded, but two edges might be identified). Given two sides of a triangle, one can obtain the third side by following the first two. Thus, if two sides of a triangle project to disjointly embedded arcs in F, the entire triangle has an embedded projection. But if two sides are

do notproject to disjointly embedded arcs, the projection of the third side would have to self-intersect.Step 2 is to check that in the projection of each tetrahedron, there are only two (opposite) edges that project to intersecting arcs. This follows from Step 1, with a little case-by-case verification.

From Step 2, it follows that then every tetrahedron projects to a quadrilateral. To echo Nathan, two of the faces of the tetrahedron are bottom, incoming faces and two are top, outgoing faces. Once we have this kind of taut structure with respect to the fiber, the triangulation is layered.

Comment by Dave Futer — September 4, 2009 @ 9:02 am |

Dave, thanks for pointing out the equivalence of these two conditions. One way to formalize Step 1 and make Step 2 obvious is to use hyperbolic geometry. In particular, we can assume the proper arcs are geodesics in some fixed hyperbolic structure on

F. Then lifting the map of the projected triangle to gives an ideal triangle. If one edge of the original triangle downstairs crossed another, there would be some translate (by an element of ) of the ideal triangle in which intersected itself; the only configuration which doesn’t lead to a self-crossing is to rotate the triangle by 1/3 of a turn, but that results in an elliptic element in , which is silly.Thus one has a well-defined map from the 2-skeleton of the triangulation of the universal cover of

Mto which takes triangles to geodesic ideal triangles. Step 2 is now clear, since there’s really only one possibility for how four distinct points can be arranged on .Comment by Nathan Dunfield — September 4, 2009 @ 1:35 pm

Thanks Nathan, that does make the argument clearer. Out of curiosity, how do you insert the little bits of TeX into the blog comment?

Comment by Dave Futer — September 4, 2009 @ 4:39 pm

Dave, see here for how to use LaTeX on any WordPress blog. It works well, except for the vertical alignment tends to be messed up, which I find visually distracting.

Comment by Nathan Dunfield — September 4, 2009 @ 4:49 pm