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Hello.

I am a beginning graduate student who wants to study Heegaard Floer Homologies.

I am now reading the paper

https://arxiv.org/abs/1301.4919 Errata to 'A cylindrical reformulation of Heegaard Floer homology'

and I am now stuck on the proof of Lemma 4.1'(p.5).

The statement is about that there is a certain map with some conditions representing a positive homology class in $\pi_2 ({\overrightarrow{x}, \overrightarrow{y}})$ in his cylindrical setting.

After gluing domains away from the $\overrightarrow{x} \cup \overrightarrow{y}$, he made special arrangement for gluing near intersection points.

The point I could not understand is the following.

For given point $p \in \overrightarrow{x} \cup \overrightarrow{y}$, in a neighborhood of that point there are 4 domains and their coefficients(intersection number of homology class and base points $\times$ a disc).

And the arrangement of these coefficients are given by

$\{ n, n+k, n+l , n+k+l\}$ or $\{n+1, n+k, n+l, n+k+l\}$ or other variants that 1 goes to other parts.

I could not understand why coefficients are given like this.

I think I miss something easy, but I could not see the point.

Please give me any hint or answer to understand this.

Thank you.

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    $\begingroup$ Why not just e-mail, Robert? $\endgroup$
    – Aleksey
    Apr 14, 2013 at 21:27

1 Answer 1

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The coefficients of the regions that meet at $p$ reflect the fact that the domain $\mathcal{D}$ has (or doesn't have) a corner at $p$.

If $p$ is not a corner (i.e. if either $p\not\in\mathbf{x}\cup\mathbf{y}$ or $p\in\mathbf{x}\cap\mathbf{y}$), then the domain is the projection of an embedded surface $S$ such that the preimages of $p$ lie either on the boundary or in the interior of $S$. If you think about it, there are five different cases, and each one of them contributes as $+(1,1,1,1)$ or $+(1,1,0,0)$ (or cyclic permutations thereof).

Analogously, if $p$ is a corner (i.e. $p\in\mathbf{x}\triangle \mathbf{y}$), one of the preimages of $p$ is a corner of $S$, and this gives you a contribution of $+(1,0,0,0)$ or $+(1,1,1,0)$ (or cyclic permutations thereof).

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    $\begingroup$ I need to think about it more carefully, but it seems your answer solves my question. Thanks! $\endgroup$
    – HFH Fan
    Apr 14, 2013 at 22:25

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