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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?

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    $\begingroup$ Note that Haken's theorem need the 3-manifold to be irreducible. $\endgroup$
    – Bruno Martelli
    Commented Aug 14, 2012 at 15:30

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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.

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  • $\begingroup$ Is there any nontrivial example where the incompressible surfaces do isotope into the two-skeleton? Is there a classification of such manifolds? $\endgroup$
    – Igor Rivin
    Commented Aug 13, 2012 at 23:51
  • $\begingroup$ 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. $\endgroup$
    – Douglas Zare
    Commented Aug 14, 2012 at 0:01
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    $\begingroup$ @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. $\endgroup$
    – Lee Mosher
    Commented Aug 14, 2012 at 0:32
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    $\begingroup$ @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... $\endgroup$
    – Douglas Zare
    Commented Aug 14, 2012 at 0:46
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    $\begingroup$ 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 $S\times 1$. $\endgroup$
    – Bruno Martelli
    Commented Aug 14, 2012 at 15:37

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