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](https://web.math.princeton.edu/~szabo/clay.pdf)). 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. 



Untwisting Heegaard diagrams
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:


Maher: <http://dx.doi.org/10.1112/jtopol/jtq031>


Lustig-Moriah: <http://arxiv.org/abs/1002.4292>


**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


