I am interested in studying moduli of complex surfaces which arise in computing the differential on the Heegaard Floer homology chain complex. In particular, I am interested in the generic case, when the holomorphic discs in $\operatorname{Sym}^g\Sigma$ are as "bad" as possible. I have searched online for some examples of "generic"/"complicated" Whitney disks in the symmetric product, but most papers I see deal with special cases (namely where one can see by inspection and the Riemann mapping theorem that certain homotopy classes of Whitney disks are uniquely representable by holomorphic disks).

To elaborate on what I mean by "complicated" holomorphic disks, let's recall a certain (now standard) perspective on the Whitney disks. If we have a holomorphic disk $\phi:\mathbb D^2\to\operatorname{Sym}^g\Sigma$, then we can consider the fiber product: $$\begin{matrix} S&\xrightarrow{\tilde\phi}&\Sigma\times\operatorname{Sym}^{g-1}\Sigma\cr \downarrow& &\downarrow\cr \mathbb D^2&\xrightarrow\phi &\operatorname{Sym}^g\Sigma \end{matrix}$$ Then $S\to\mathbb D^2$ is a $g$-fold ramified cover (and the fiber over a point $p\in\mathbb D^2$ is "the $g$ points in $\Sigma$ given by $\phi(p)$"). Thus another way of viewing holomorphic disks in $\operatorname{Sym}^g\Sigma$ is as $g$-fold ramified maps $S\to\mathbb D^2$ along with a map $S\to\Sigma$. Thus, even though we consider only disks mapping to $\operatorname{Sym}^g\Sigma$, moduli spaces of more complicated Riemann surfaces naturally come in to play in Heegaard Floer homology as moduli of the ramified cover $S$.

I am interested in the following "complicated" behaviour of $S$ and of the map $\tilde\phi:S\to\Sigma$:

  1. Can the map $S\to\Sigma$ fail to be an immersion? (I believe the answer is yes; in fact I think I know where to look for more information on this, I just haven't followed up on it yet).

  2. In general, what do the "slits" or "cuts" along the $\alpha$ and $\beta$ curves look like in $S$, and how does $S$ degenerate as these slits vary in length? Do the slits ever interact with each other (e.g. by colliding), or do they all give "independent" degenerations of $S$? (this is a bit vague, but if someone has an enlightening example, I'd really like to see it)

  3. Can $S$ have positive genus? This is the question I'm really most interested in. I have on good authority that the answer is yes, so actually what I really want is an example where $S$ has positive genus and contributes to the differential (or at least has Maslov index one).

  • $\begingroup$ As for 3, doesn't Lemma 9.4 in Ozsváth and Szabó's "Holomorphic disks and topological invariants for closed three-manifolds" exhibit an instance where $S$ is an annulus? $\endgroup$ Mar 11, 2012 at 9:17
  • $\begingroup$ I do know of examples where $S$ is an annulus (I like the pictures in math.ucla.edu/~lwatson/pages/pdf/resume-CIRGET-liam.pdf), and can easily imagine how to get a sphere minus $k$ disks. But these all have genus zero. $\endgroup$ Mar 11, 2012 at 20:15
  • $\begingroup$ Sorry - messed up between genus and Euler characteristic of the associated closed surface.. My comment was fairly useless. $\endgroup$ Mar 11, 2012 at 20:27
  • $\begingroup$ Have you looked at Robert Lipshitz's thesis? He takes something similar to this ramified cover approach. If I recall correctly, I heard him give a talk in which he mentioned in passing that one must consider curves with genus, though I don't remember what justification he gave. (You might also consider asking him directly. I'm sure he knows the answer to these questions. If you get an answer this way, please do come back and write up a summary for mathoverflow.) $\endgroup$
    – Sam Lisi
    Mar 27, 2012 at 0:06

1 Answer 1

  1. Yes, $S \to \Sigma$ can fail to be an immersion. The failure is a branch point of the map $S \to \Sigma$, since it is (close to) a holomorphic map. This is fairly common as soon as the multiplicity of some region in $\Sigma$ gets higher than $1$. Of course, in the simple examples you can actually compute, this tends not to happen.

  2. A "slit" along the $\alpha$ or $\beta$ curves looks like a perfectly ordinary piece of the boundary upstairs in $S$, including at the end of the slit. It's best to think about the end of the slit as a boundary branch point, looking locally like the map $z \to z^2$ restricted to the upper half-plane in the domain. Slits cannot collide, due to boundary monotonicity: above each point on $\partial \mathbb{D}^2$, the $g$ different points map to distinct $\alpha$-curves. I don't know what you mean by "independent" degenerations, sorry....

  3. Yes, higher genus images can happen, and it's not too hard to construct examples that are forced by gluing, although again in most cases where you're able to compute Heegaard Floer homology by directly counting curves it does not. The index formula in Corollary 4.3 of Lipshitz's paper "A cylindrical reformulation of Heegaard Floer homology" will let you easily construct examples of high genus surface with index 1, and a little more playing around should let you see that some of these must actually have representatives.

  • $\begingroup$ Thanks! I understand your answer to #3, and am satisfied with this construction of higher genus S. Given the vague phrasing of #2, your answer gives me the information I wanted (particularly that the slits do not collide). $\endgroup$ Apr 2, 2012 at 2:24
  • $\begingroup$ I'm still confused about #1, though. More specifically, I thought that following the proof of Lemma 2.17 in Oszvath--Szabo "Holomorphic disks and topological invariants for closed three-manifolds" (or Lemma 4.1 in Lipschitz "A cylindrical reformulation of Heegaard Floer homology") gives for any domain D a surface S --> Sigma which is an immersion on its interior. And wouldn't every holomorphic disk in that homotopy class determined by D (say g>2) give the same map S --> Sigma except possibly differing by the lengths of the slits? $\endgroup$ Apr 2, 2012 at 2:29
  • 2
    $\begingroup$ A map with interior branch points can be homotopic to one without them. Specifically, in the construction you refer to, they construct maps with corners of greater than 180 degrees. These can turn into slits, i.e., boundary branch points. Two boundary branch points in turn can trade off for an interior branch point. I now see my answer to #2 above was incomplete. When the boundary of $S$ maps with two U-turns (boundary branch points) in a row along a single boundary component, the two branch points can collide to form an interior branch point. $\endgroup$ Apr 3, 2012 at 9:15

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