I am attempting to get to grips with Thurston's hyperbolization theorem for Haken $3$--manifolds. In particular I was looking at the section related to gluing up along hierarchy surfaces in Otal, Jean-Pierre (1998), "Thurston's hyperbolization of Haken manifolds"

From what I understand we take hierarchy for a Haken $3$--manifold $M$ with corners (a decomposition of the manifold with corners along incompressible surfaces into a collection of $3$--balls), use something like Andreev's theorem to put a hyperbolic structure on the $3$--balls, and use an inductive argument to re-glue the manifold along the hierarchy surfaces in such a way that the original manifold carries a complete hyperbolic metric. I am speaking very loosely here since I can't say I understand all the details.

My question is as follows. Is it possible to re-formulate the this construction in terms of manifolds with boundary pattern (in the spirit of Johannson), and if so, if the boundary pattern on the collection of $3$--balls corresponds to the cuts along hierarchy surfaces, does the boundary pattern carry all the data needed for re-gluing?


1 Answer 1


The sorts of hierarchies that Johannson makes use of are called "simple hierarchies", and go back to Waldhausen (used by him in the solution of the word problem and homotopy rigidity of Haken 3-manifolds). The boundary patterns associated to these hierarchies don't contain all of the information one needs to glue up the hierarchy. As an example, consider the first cut of a Haken manifold, say along a separating incompressible surface. The boundary pattern here is trivial, consisting of the entire boundary, which is two copies of the first surface. Then the second cut will be along a (boundary)-incompressible surface (possibly disconnected), which we may assume meets both components of the boundary, and the resulting boundary pattern will be the second surface and the first surface cut up along the boundary of the second surface. To reglue the manifold, we may glue the 2nd boundary pattern by any automorphism of this surface (with boundary), and then glue the 1st stage by any of the automorphisms of the 1st surface. Notice that the boundary pattern associated to one copy of the 1st surface does not ``see" the boundary pattern associated to the other side, which is partly why the pattern is not enough to recover the gluing. For a discussion of simple hierarchies, see the paper of Aitchison-Rubinstein and references therein.

One remark (since you mention Andreev's theorem): Thurston told me that he originally came up with a new proof of Andreev's theorem for hyperbolic reflection polyhedra by the sort of inductive argument used in the proof of geometrization of Haken 3-manifolds. It turns out that reflection orbifolds are either tetrahedral or Haken. The theory here is much simpler, because compactification of the hyperbolic structure on non-compact reflection groups is simpler (the bending laminations of the boundary of the convex hull are finitely many arcs). After succeeding in this case, he worked on the general case of Haken manifolds, and in fact incorporated the reflection group techniques in his "orbifold trick" to make the gluing problem have no boundary.

  • $\begingroup$ Thanks Ian, yes this makes sense. Just wondering whether the obstruction you mention is the only known obstruction. For instance if I have a decomposition along hierarchy surfaces that are non-separating (I can do this for example by assumptions on the first homology of the manifold) and postpone cuts by compressing disks till the end, then I think I have a construction where each hierarchy surface is non-separating. I tried this out on closed Haken $3$--manifolds with infinite homology and it seems to work. Thus I end up with a single $3$--ball and all cuts would be visible...(continued) $\endgroup$
    – Don Shanil
    Jul 26, 2013 at 4:18
  • $\begingroup$ (continued)...I guess the question is whether in this case the boundary pattern carries enough information for re-gluing. Thanks again....was just curious whether this could be done...you have more than adequately answered the original question btw:) $\endgroup$
    – Don Shanil
    Jul 26, 2013 at 4:22

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