Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Most Heegaard diagrams contain many rectangles, for instance from loops that circle around one of the handle disks. You can always `twist' a Heegaard diagram to get more and more rectangles (as in page 5 of this paper). Can one reverse this process to eliminate rectangles?

My real question is, is there a systematic way of constructing Heegaard diagrams without rectangles?

I am interested in Heegaard diagrams without prismatic circuits of length $\leq 4$; in particular, such diagrams can have no rectangles. I'm interested to know how common such Heegaard diagrams are, so I'm asking this question as a first step.

share|cite|improve this question
I don't think this is quite what you're after, but perhaps the Casson-Gordon Rectangle Condition is a helpful concept. See (e.g.) this paper by Kobayashi:… –  Scott Taylor Apr 19 '13 at 1:55
@Scott Taylor: That paper is very interesting; thanks! I am really just interested in knowing more about combinatorial properties of Heegaard diagrams, so that paper is great. –  Brian Rushton Apr 19 '13 at 1:59

1 Answer 1

up vote 8 down vote accepted

I'm sorry to say that this condition is quite rare in practice. If the splitting has high enough distance in the curve complex, any pair of curves from the two disk sets will intersect a lot, resulting in many rectangles. Furthermore, high-distance splittings are "generic."

The word "generic" can be made precise in two ways. By the work of Joseph Maher, random walks in the mapping class group result in a high-distance splitting with probability approaching 1. In a different direction, the work of Lustig and Moriah implies that high-distance splittings are "generic" in a measure-theoretic sense.

Here are the references:



Update: Actually, I am becoming convinced that only very low distance splittings can have rectangles.

Lemma: Let $S$ be a Heegaard splitting surface of genus $g \geq 5$, with Hempel distance $\geq 2$. Then any Heegaard diagram for $S$ contains rectangles.

In other words, for genus $g \geq 5$, any Heegaard diagram without rectangles must come from a weakly reducible splitting. I strongly suspect this is true in every genus.

Proof: Let $\alpha_1, \ldots, \alpha_g$ and $\beta_1, \ldots, \beta_g$ be the curves of any Heegaard diagram for this splitting. By hypothesis, every $\alpha_i$ intersects every $\beta_j$. Now, cut $S$ along all the $\alpha_i$. We get a sphere with $2g$ holes. The maximal number of disjoint, non-parallel arcs in this surface is $6g-6$. On the other hand, since every $\alpha_i$ intersects every $\beta_j$, there are at least $g^2$ remnants of the $\beta$ curves in this sphere. Since $g^2 > 6g-6$, when $g \geq 5$, some of these arcs must run in parallel. QED

share|cite|improve this answer
Which is not to say that low-distance splittings are not plentiful or interesting. (Perhaps with the definition of "interesting" as "not generic" :) ) –  Scott Taylor Apr 24 '13 at 3:24
Scott: I take your point, and mostly agree with it. In fact, I would argue that the most interesting splittings are of distance 2. (Either less than 2 or more than 2 provides a lot of additional information.) But by the above update, splittings with rectangles must be weakly reducible -- at least in high genus, and probably in every genus. –  Dave Futer Apr 24 '13 at 13:00
Dave: I agree! Although the distance 1 case in some situations isn't as tractable as I might like. For instance after untelescoping the thick surfaces might all have distance 2, so although you have essential surfaces the Heegaard splitting structure is still opaque. –  Scott Taylor Apr 24 '13 at 16:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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