TITLE: In the definition of the Heegard Floer surgery exact triangle, what exactly is the correspondence between Whitney triangles and periodic domains?
QUESTION [7 upvotes]: I'm reading Osváth-Szabó's notes on Heegard Floer homology, in particular about the surgery exact triangle.
On page 14 (numbered 42 on the document), they describe an isomorphism between the space of homology classes of Whitney triangles $\pi_2(\mathbf{x},\mathbf{y},\mathbf{z})$ in $\mbox{Sym}^g(\Sigma)$ and $\mathbb{Z} \times \mathcal{P} $, where $\mathcal{P}$ denotes the group of periodic domains in $\Sigma$.
I'm not sure if I understand this isomorphism correctly, or if there are some typos, or both. Here is how I understand the isomorphism works:
Given two elements $\psi, \psi_0 \in \pi_2(\mathbf{x},\mathbf{y},\mathbf{z})$, we can associate domains $\mathcal{D}(\psi), \mathcal{D}(\psi_0)$, by taking $$\mathcal{D}(\psi) = \sum n_{z_i}(\psi) D_i,$$
where $D_i$ are the components of $\Sigma - \{\mathbf{\alpha} \cup \mathbf{\beta}\cup\mathbf{\gamma}\}$, and $z_i \in D_i$; and $n_{z_i}(\psi)$ is the algebraic intersection of $\psi$ with $z_i \times \mbox{sym}^{g-1}(\Sigma)$.
It follows that if $n_z(\psi) = n_z(\psi_0)$, the domain $E = \mathcal{D}(\psi) - \mathcal{D}(\psi_0)$ is periodic: i.e. it satisfies $n_z(E) = 0$ (here, $n_z(E)$ denotes the coefficient of the component of $\Sigma - \{\mathbf{\alpha} \cup \mathbf{\beta}\cup\mathbf{\gamma}\}$ containing $z$ in $E$).
So all we have to do to define the isomorphism is pick a $\psi_0$ such that $n_z(\psi_0) = 1$: then we can subtract off $n \mathcal{D}(\psi_0)$ from $\mathcal{D}(\psi)$ to get a periodic domain. Given a fixed $\psi_0$, the isomorphism is then given by
$$\psi \mapsto (n, \mathcal{D}(\psi) - n\mathcal{D}(\psi_0)).$$
The inverse of the isomorphism should then be given by sending $(n, E)$ to something like $n \psi_0 + \phi$, where $\mathcal{D}(\phi) = E$; but this seems dependent on choice and I'm not entirely sure how to make sense of $n$ times a holomorphic triangle.
Is this right? How do I fix the last part?
Thanks very much.

REPLY [4 votes]: If I had to guess, I'd go for the typo and a small misunderstanding.
I think that the isomorphism should indeed depend on the choice of a triangle $\psi_0$ with $n_z(\psi_0) = 0$ (careful here: this has to be 0 and not 1, as you wrote above), but that it should be defined as
$$\psi\mapsto (n_z(\psi), \mathcal{D}(\psi)-\mathcal{D}(\psi_0)-n_z(\psi)\cdot\Sigma)$$
whose inverse is given by (notice that now you only sum a triangle and some periodic domain, plus multiples of the Heegaard surface)
$$(n,P)\mapsto \psi_0 + P + n\Sigma$$
The isomorphism I gave above agrees with the isomorphism in the notes only for domains with $n_z = 0$ (which are the only ones you consider in the hat version).

I don't think that you need a canonical choice for $\psi_0$ (hence for the isomorphism), nor that it's possible to have one.
One final remark: here you're not really summing holomorphic triangles, but rather domains, so you shouldn't worry about the "holomorphic meaning" of the sum of domains.