# Irreducibility of a Heegaard splitting from its Heegaard diagram

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?

• 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. – Mike Miller Eismeier Jul 1 '17 at 16:49
• 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. – Mustafa Jul 1 '17 at 18:58
• I guess my ignorance is showing. What does it mean for you that a Heegaard diagram be reducible? – Mike Miller Eismeier Jul 1 '17 at 19:17
• 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) – Mustafa Jul 1 '17 at 19:33

## 1 Answer

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.

• 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. – Mustafa Jul 3 '17 at 14:20
• @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. – Ian Agol Jul 3 '17 at 15:26