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I am reading the introductory paper "Heegaard diagrams and holomorphic disks" by Ozsváth and Szabó (https://arxiv.org/abs/math/0403029v1, Section 2.2), and I do not understand one of the examples. They compute the homology of the following Heegaard diagram of $S^3$:

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In order to count holomorphic disks connecting $x_3 \times y_3$ to $x_2 \times y_3$, they consider the uniformization of $\Delta$ as a standard annulus with four marked points on its boundary, corresponding to $x_2, x_3, y_2, y_3$. They call $a$ the angle of the arc in the boundary connecting $x_3$ to $x_2$ which comes from $\alpha_1$, and $b$ the angle of the arc in the boundary connecting $y_3$ to $y_2$ which comes from $\alpha_2$. Then, they consider the one-parameter family of conformal annuli with four marked points obtained from $\Delta \cup \Gamma$ by cutting a slit along $\alpha_2$ starting from $y_3$, where the marked points correspond to $x_2,x_3$ and $y_3$ counted twice. At this point, they state:

A four-times marked annulus which admits an involution (interchanging the two $\alpha$-arcs on the boundary) gives rise to a holomorphic disk connecting $x_3 \times y_3$ to $x_2 \times y_3$. By analyzing the conformal angles of the $\alpha$ arcs in this one-parameter family, one can prove that the mod $2$ count of the holomorphic is $1$ iff $a < b$.

I have many questions. First, I do not really get what $a$ and $b$ are. I cannot fully understand why the involution gives rise to a holomorphic disk, although I am quite convinced. The thing I do not get at all is the analysis of the conformal angles in the end. The idea seems to be that if the arc of $\alpha_2$ from $y_2$ to $y_3$ is too long, there cannot be a holomorphic disk, but I do not really get the reason.

Any help will be greatly appreciated!


1 Answer 1


There are several things going on here, explained rather elliptically in the paper. Let me expand.

First, there's the question of which holomorphic annuli double-cover the disk. More precisely, suppose you are given a holomorphic annulus with each boundary colored in two pieces, red and blue. (Suppose the boundary is sufficiently nice, e.g. locally connected, so that various notions of "boundary" coincide.) When does that annulus have a double-cover to the disk, again with the boundary colored in two colors? The answer is as follows.

Uniformize the annulus to a Euclidean annulus of circumference $2\pi$ with bi-colored boundary. A certain proportion of the upper boundary will be red; call that angle $a$ with $0 < a < 2\pi$. Similarly call the angle of the red portion of the lower boundary $b$. Then the annulus has a double-cover of the disk if and only if $a = b$. I won't go through the proof in detail, but in the case $a=b$ you can get the double cover as follows:

  • Connect the left end of the upper red segment to the right end of the lower red segment, and conversely connect the right end of the upper red segment to the left end of the lower red segment. These intersect in a point $p$.
  • Similarly connect the left/right ends of the upper blue segment to the right/left ends of the lower blue segment. These intersect in a point $q$.
  • Both $p$ and $q$ are exactly halfway up the Euclidean annulus (since we assume $a=b$).
  • The $180^\circ$ rotation around $p$ and $q$ gives an involution of the annulus; the quotient is the disk, and the quotient map is the desired holomorphic map to the disk.

OK, so now we know what kind of annulus we are looking for (and the relevance of comparing $a$ to $b$). Now let's look at the specific annuli in the problem. There is a 1-parameter space of potential conformal annuli here, depending on how long of a slit to cut from $y_3$ to $y_2$ along the boundary between $\Gamma$ and $\Delta$. The question is whether any of the annuli in this family have the balance mentioned above of the amount of $\alpha$ on the two components.

At one end of the parameter space, there is the annulus $A_0 = \Gamma \cup \Delta$ with no slit. This has no $\alpha$-arc appearing on the inner ($y_2$ and $y_3$ boundary), so for that annulus we have $b = 0$.

Near the other end of the parameter space, the slit reaches almost all the way from $y_3$ to $y_2$. Then the conformal structure on the resulting annulus is very close to the conformal structure on $\Delta$ itself: the extra near-bigon $\Gamma$ has almost no effect on the conformal structure.

Thus, if we look at the curve in the space of $a$ and $b$ values, we go from some point with $b=0$ (so necessarily $a>b$) to whatever values you get for $\Delta$. You will have an odd number of crossings iff we have $a < b$ for $\Delta$.

A couple notes.

  • In case it's not obvious, one point of this example is that the holomorphic disk counts are not combinatorial and do depend on the complex structure. You can make other examples where the two choices of which disk is represented are combinatorially equivalent, I think maybe there's an example later in this same paper.
  • I generally find it much easier to think about everything in Lipshitz's cylindrical reformulation, and probably implicitly use that above. But the foundational papers are written in this symmetric product language.
  • I'd love it if this argument appeared more explicitly somewhere in the literature; maybe it's somewhere I don't know about.
  • You can generalize this criterion to cases where there are more than four total marked points on the boundary. Then the condition is that the total angle of $\alpha$-arcs on one component equals the total angle of $\alpha$-arcs on the other. Again I don't know if this appears in the literature.
  • I omitted several things, for instance justifying the treatment of the boundary (using stronger versions of the Riemann mapping theorem/uniformization of annuli), why the extra bigon has no effect on the conformal structure of the annulus (related to the Gromov compactness used everywhere in the theory), and discussion of perturbing the almost-complex structure to make it generic enough to be sure the moduli spaces have the expected dimension doesn't ruin the argument.

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