12
$\begingroup$

My question is as follows: "Is a generic closed orientable hyperbolic 3-manifold Haken?" Of course the word 'generic' can be interpreted in many ways, and the answer might depend on the way how one interprets 'generic'. As an example, let's consider the related question "Is a generic closed (aspherical) 3-manifold hyperbolic?" In this case there are various notions of 'generic', in each case the answer is a resounding yes. For example, almost all Dehn fillings on a hyperbolic knot result again in a hyperbolic 3-manifold. Furthermore Maher (Random Heegaard splittings) showed in a precise sense that a 'random gluing of two handlebodies of genus >1' gives rise to a hyperbolic 3-manifold. Similarly, in a precise sense, a 'random' fibered 3-manifold is hyperbolic.

My hunch would have been that a generic hyperbolic 3-manifold is non-Haken (why should it have an incompressible surface?), but I just had lunch with another 3-manifold topologist and his guess was that a generic hyperbolic 3-manifold should be Haken.

$\endgroup$
  • 4
    $\begingroup$ I don't agree that a "generic closed 3-manifold is hyperbolic! for all notions of "generic". For instance, if you glue tetrahedra together randomly so that the result is a 3-manifold, then it will not be prime. $\endgroup$ – Dylan Thurston Aug 24 '14 at 14:02
  • $\begingroup$ thanks for the comment. I should have been more careful, all notions of 'generic' I could think of. I presume there is no rigorous 'citable' proof of that statement. If yes, then I would be very interested. $\endgroup$ – Stefan Friedl Aug 24 '14 at 18:14
  • 4
    $\begingroup$ @DylanThurston: Dylan, how do you prove this? $\endgroup$ – Misha Aug 24 '14 at 19:33
  • 3
    $\begingroup$ Let $(M_{n})$ be a sequence of random splittings corresponding to a random walk on the mapping class group. Then for fixed g, the probability that $M_{n}$ contains an incompressible surface of genus at most $g$ tends to $0$ as $n \rightarrow \infty$. This follows from Maher's "Random Heegaard Splittings" paper mentioned in the post; he shows that the Heegaard distance of $M_{n}$ will grow linearly with overwhelming probability. Then by a theorem of Hartshorn, the minimal genus of an incompressible surface in $M_{n}$ also grows linearly. I think Maher mentions this in the intro of that paper. $\endgroup$ – Tarik Aougab Aug 25 '14 at 0:06
  • 1
    $\begingroup$ Some references are given in Remark 1.12 of books.google.co.kr/… $\endgroup$ – ThiKu Aug 25 '14 at 2:41
3
$\begingroup$

Firstly, the random Heegard splitting model and the random fibering model are discussed at length in my preprint. (I believe the OP is familiar with it). I believe the question is asked there, though not answered - it is pointed out that the smaller genus of an incompressible surface will grow, but that seems to be neither here nor there. A related question is whether a random fibered manifold has an incompressible surface which is not the fiber of the fibration you started with. Here, the answer is YES, for genus 1 (Floyd-Hatcher), and probably YES for higher genus, which would tend to suggest that a random Dunfield-Thurston manifold might be Haken, but nobody knows for sure.

As for Dylan's comment, it is certainly plausible. Any given configuration of tetrahedra (in particular one with a normal sphere which is not the link of a vertex), in fact, you might as well assume that it is topologically a thickened sphere, will appear with probability one, and it is highly unlikely that it will bound balls on both sides.

$\endgroup$
  • 1
    $\begingroup$ are we sure that any configuration of tetrahedra appears with probability one? This does not hold for graphs (some graphs almost never appear as subgraphs of a random 4-valent graph). $\endgroup$ – Bruno Martelli Aug 25 '14 at 6:30
  • $\begingroup$ @BrunoMartelli Well, "sure" is a strong word, but I don't see why not (this seems like a $0-1$ law sort of thing, and I don't see why it would be $0$... $\endgroup$ – Igor Rivin Aug 25 '14 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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