Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Haken proved that an incompressible surface in a triangulated irreducible 3-manifold is isotopic to a surface which is normal with respect to the triangulation (Theorie der Normalflächen. Acta Math. 105 1961 245–375).

While normal surfaces are tremendously useful, I want my surfaces to be isotopic into the two-skeleton and I am unsure when I can conclude that they indeed are isotopic into the two-skeleton. I suspect that there are triangulated 3-manifolds out there containing incompressible surfaces which are not isotopic into the two-skeleton, but I am hoping that if the 3-manifold has a metric and the tetrahedra are "small enough" compared to the injectivity radius of the 3-manifold, then incompressible surfaces can be isotoped into the two-skeleton.

Can someone point me to a reference or explain whether or not there are reasonable properties of a triangulation which imply that incompressible surfaces are isotopic into the two-skeleton?

share|improve this question
2  
Note that Haken's theorem need the 3-manifold to be irreducible. –  Bruno Martelli Aug 14 '12 at 15:30

1 Answer 1

up vote 10 down vote accepted

Unfortunately, you can't always expect every incompressible surface to be isotoped into the $2$-skeleton. There are only finitely many subsets of the $2$-skeleton, but there can be infinitely many nonparallel incompressible surfaces. This is simplest in the $3$-torus, but there are many other examples. Jaco's stair construction gives incompressible surfaces of arbitrarily high genus in a surface (of positive genus) times a circle.

share|improve this answer
    
Is there any nontrivial example where the incompressible surfaces do isotope into the two-skeleton? Is there a classification of such manifolds? –  Igor Rivin Aug 13 '12 at 23:51
    
I guess the non-Haken manifolds (perhaps irreducible ones) are what you are calling trivial. I'm not sure when there are only finitely many isotopy classes of $\pi_1$-injective surfaces. Any nice enough triangulation of such a manifold should work. –  Douglas Zare Aug 14 '12 at 0:01
1  
@bb: Your original question posited a fixed underlying triangulation. If instead you fix the surface $S$ and allow yourself to pick and choose amongst triangulations, then certainly you can produce a triangulation into which $S$ may be isotoped. Nonetheless, as Zare's answer shows, for a typical 3-manifold there will not exist any one triangulation into which all incompressible surfaces can be isotoped. –  Lee Mosher Aug 14 '12 at 0:32
1  
@bb: It sounds like you are trying to construct a triangulation from a surface so that the surface is in the $2$-skeleton. If so, just triangulate the $3$-manifold with boundary you get by cutting along $S$, and add a few tetrahedra to make the two triangulations of $S$ consistent. Your construction seems to make some strong assumptions about how $S$ intersects the particular triangulation you have at the start. You can see that this construction has to make some assumptions, since there are only finitely many subsets of the $2$-skeleton of any fixed iteration of barycentric subdivision... –  Douglas Zare Aug 14 '12 at 0:46
1  
Do you have a triangulation of $S\times (0,1)$ with infinitely many tetrahedra, or are you just taking the interior of a (finite) triangulation of $S\times [0,1]$? In the second case you are done: the boundary of the regular neighborhood of $S \times 0$ in the triangulation is always isotopic to $S$, and if you start with a 2nd barycentric subdivision you are sure that it does not intersect $S\times 1$. –  Bruno Martelli Aug 14 '12 at 15:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.