5
$\begingroup$

Assume that a Heegaard diagram $(\Sigma_g,\{\alpha_1,\ldots,\alpha_g\},\{\beta_1,\ldots,\beta_g\})$, defining a Heegaard splitting of a closed 3-manifold, is given. So, by attaching 3-dimensional 2-handles to $\Sigma_g$ along $\alpha_i$ and $\beta_i$, one gets a closed 3-manifold with the Heegaard surface $\Sigma_g$.

I wonder if there are any answers to following questions:

Are there sufficient conditions on the diagram for the Heegaard splitting to be irreducible? Furthermore, is there an algorithm to detect whether the splitting is irreducible or not?

$\endgroup$
4
  • $\begingroup$ It seems to me this is equivalent to saying that the complement of the union of the alpha and beta curves is simply connected. Every time you take a connected sum with an irreducible piece, you puncture this complementary region one more time. I guess I don't know how Heegaard diagrams are usually encoded in a computer, but I imagine you should have easy access to the homology of the complements of the curves. $\endgroup$
    – mme
    Jul 1, 2017 at 16:49
  • $\begingroup$ Hey, @mike. It is obvious to me that if the complement of the curves is not simply connected (SC), then the splitting is reducible. Because a non-trivial simple closed curve in the complement of the curves would bound a disk on both sides of $\Sigma_g$. However, I can't see why the converse is necessarily true. I think even the diagram is complicated (in the sense that the complement of the curves is SC), one might get a reducing pair of disks intersecting alpha and beta curves. I will try to construct an example. $\endgroup$
    – Mustafa
    Jul 1, 2017 at 18:58
  • $\begingroup$ I guess my ignorance is showing. What does it mean for you that a Heegaard diagram be reducible? $\endgroup$
    – mme
    Jul 1, 2017 at 19:17
  • $\begingroup$ I don't have a definition of "reducible Heegaard diagrams". I just wonder under what assumptions the splitting suggested by the diagram would be reducible. My definition for reducible Heegaard splittings is standard. A Heegaard splitting $(H_1,H_2)$ of a 3-manifold $M$ is reducible if there are disks $D_i\subset H_i$ such that $\partial D_1=\partial D_2$. (@mike) $\endgroup$
    – Mustafa
    Jul 1, 2017 at 19:33

1 Answer 1

6
$\begingroup$

There is a criterion which is usually used to show that a Heegaard splitting is irreducible, which in fact shows that the Heegaard splitting is strongly irreducible (meaning that any pair of meridian disks for $H_1$ and $H_2$ have intersecting boundaries). The original criterion was given by Casson and Gordon in an unpublished paper (see the appendix of Schultens-Moriah).

The paper of Lustig-Moriah discusses the Casson-Gordon "rectangle condition" further.

Some other sufficient conditions are given by Jung Hoon Lee.

Finally, it is now known how to classify irreducible Heegaard splittings of non-Haken hyperbolic 3-manifolds by Colding-Gabai-Ketover. They show how to give a list (without repetition) of Heegaard splittings in these 3-manifolds. The proof shows how to recognize whether a Heegaard splitting is irreducible, and hence answers your question in this special case.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks, @ian. This was helpful. I wonder if the rectangle condition can be weakened to be sufficient for irreducibility rather than strong irreducibility. Apparently, Jung Hoon Lee tried to do that in the first place, but a mistake was found in his/her argument. $\endgroup$
    – Mustafa
    Jul 3, 2017 at 14:20
  • 1
    $\begingroup$ @Mustafa okay, I hadn't noticed that there was an error in Lees argument. For non Haken manifolds, irreducibility and strong irreducibility are equivalent. But I don't know when there is a heegaard diagram satisfying the rectangle condition. One can detelescope a weakly reducible splitting to get a strongly irreducible generalized splitting. But it might be subtle to determine when this implies the original splitting is irreducible, since it might not be unique. $\endgroup$
    – Ian Agol
    Jul 3, 2017 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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