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Is there an algorithm to detect the Non-Haken Manifold? Or, is there a sufficient condition for a manifold to be a non-Haken manifold? (off course, I hope that condition is not the ones in its definition.)

Note: a Non-Haken manifold means that either it is reducible or it contains no 2-sided properly embedded incompressible surface.

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    $\begingroup$ I don't understand the question: since there is an algorithm to find an incompressible surface (e.g., Haken's algorithm), that algorithm will return with failure when the manifold is not Haken (or it will say that the surfaces it found were all spheres). $\endgroup$
    – Igor Rivin
    Commented Jul 11, 2013 at 3:27
  • $\begingroup$ yanqing: Your definition of a non-Haken manifold is incorrect. A manifold is Haken if it is irreducible and sufficiently large. Negation would be: $M$ is Either reducible or is not sufficiently large. $\endgroup$
    – Misha
    Commented Jul 11, 2013 at 6:14
  • $\begingroup$ @Misha: Yeah, you are right. $\endgroup$
    – yanqing
    Commented Jul 11, 2013 at 6:33
  • $\begingroup$ @Igor:I have constructed a closed manifold from Heegaard splitting. But I do not know whether it is a Haken or not. Since it is hard to determine it from the definition, I want to know some other sufficient conditions. $\endgroup$
    – yanqing
    Commented Jul 11, 2013 at 6:37

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Yes, there is an algorithm due to Jaco-Oertel to detect if an irreducible manifold is Haken, and therefore to detect a non-Haken manifold (if it is given to be irreducible). The phrasing is a bit confusing (are you assuming the given manifold is irreducible?), but given Rubinstein's solution to sphere recognition, there is also an algorithm to detect if a manifold is irreducible. See Jaco and Tollefson's exposition for improvements to these algorithms, and the papers of Ben Burton (arXiv, MathSciNet) for implementations.

Addendum: Sufficient conditions for being Haken are easier than for being non-Haken. For example, if a manifold has $b_1>0$, or has positive-dimensional $SL_2\mathbb{C}$ character variety, then it is Haken. In fact, verifying a manifold is Haken is likely NP. However, showing something is non-Haken is more difficult. There are also techniques to analyze when a group has no fixed-point free action on a tree. Some version of this was implemented by Fenley (Laminar free hyperbolic 3-manifolds, Comment. Math. Helv. 82 (2007), no. 2, 247–321) to find laminar-free 3-manifolds, and in principle could be used to prove that a 3-manifold is non-Haken. However, it seems that this requires an exponential search.

Work of Dani Wise implies that a compact 3-manifold is Haken or reducible if and only if it surjects an infinite virtually-free group.

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  • $\begingroup$ As Ian mentions, not only is there an algorithm, there's an effective algorithm that Ben Burton has implemented in the software package "Regina". It's not lightning fast but it works. $\endgroup$ Commented Jul 11, 2013 at 6:30
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    $\begingroup$ @Ryan: it's actually gotten quite good - Burton-Coward-Tillmann verified quickly that the Seifert-Weber space is non-Haken, and all the manifolds in the Snappea census. arxiv.org/abs/1212.1531 $\endgroup$
    – Ian Agol
    Commented Jul 11, 2013 at 16:58
  • $\begingroup$ (see the abstract of Stephan Tillmann here for the claim about census manifolds: math.univ-toulouse.fr/top-geom-conf-2013/documents/booklet.pdf) $\endgroup$
    – Ian Agol
    Commented Jul 11, 2013 at 17:06

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