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This is somewhat related to this previous quesiton. Suppose I give you a Heegard splitting of $M^3$ of genus $g$ with a gluing map $\phi.$ Is there some condition on $\phi$ which would guarantee that $M^3$ was Haken?

EDIT of course, there are conditions which tell you that $M^3$ has nontrivial rational homology, but I am looking for something more...

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Yes, there is a well-known condition of Casson & Gordon, that a Heegaard splitting which is weakly reducible, but not reducible, gives a manifold which is Haken. In terms of the curve complex, this states that the Heegaard splitting has distance precisely $=1$. Distance 0 means reducible, distance 1 means weakly reducible, where we are considering the distances between the sets of meridians for the two handlebodies in the curve complex of the Heegaard surface.

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    $\begingroup$ Well-known to some :) I am actually interested in things with high curve complex distance, which means that (a) the criterion you mention does not apply but also (b) by a result of one Kevin Hartshorn the genus of an incompressible surface is big (if there is one at all, that is). $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2014 at 14:17
  • $\begingroup$ @IgorRivin: Ok, that clarifies things. Hyam Rubinstein gave some conditions a while back to imply that a genus 2 Heegaard splitting will have an incompressible surface in terms of the curve complex. If you like, I could try to recall that and spell it out. In some sense, the construction is a generalization of Hatcher-Thurston for incompressible surfaces in 2-bridge links. $\endgroup$
    – Ian Agol
    Commented Mar 25, 2014 at 15:12
  • $\begingroup$ Ian, yes, that would be interesting, if you could dredge it up.. Thanks! $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2014 at 15:49

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