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I am reading Abouzaid's paper "A geometric criterion for generating the Fukaya category" (https://arxiv.org/abs/1001.4593), and it is claimed there, without proof, in section C.4 in the appendix (pp.34) that the compactificaiton of the moduli space of holomorphic annuli is given by two discs meeting in an interior node at $R\to\infty$ and by two discs meeting at two boundary nodes at $R\to0$, where $e^R$ is the conformal parameter (i.e. annuli is given by $\{1\le|z|\le e^R\}$. I tried searching the literature for a proof, but other than finding the same claim in other papers I have found none.

1) Could someone please suggest a proof, or refer me to one?

2) Is there an explicit way to "calculate" the DM-compactification of a space of Riemann surfaces with boundary (and maybe marked points).?

3) While I can imagine the limit as $R\to\infty$ as a thin neck being developed, I am especially curious about the limit as $R\to 0$ as it seems somewhat arbitrary, why do only 2 chords shrink to a point? why not 1 or 3?

Thank you!

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  • $\begingroup$ The right person to ask is Aleksey Zinger $\endgroup$ – user21574 May 17 '17 at 19:26
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    $\begingroup$ Let me note the sentence just after Equation (C.8) requires that there be two marked points at $\{1,-r\}$. These are "diametrically opposite," which is why the limit breaks up into two discs. If there was only one component, then, after gluing the node, you would find that the two marked points don't satisfy the required constraint. If there were three or more components, then one of them would not carry any marked point, and hence would be unstable. Less importantly, since you link to the arXiv version, you should probably say that it's on p. 38. $\endgroup$ – Mohammed Abouzaid May 18 '17 at 12:57
  • $\begingroup$ Thank you, Mohammed, it all makes sense now. also thanks for the reference to Liu's thesis, exactly the kind of exposition I needed. $\endgroup$ – Yaniv Ganor May 18 '17 at 17:34
  • $\begingroup$ @MohammedAbouzaid Sorry for commenting such an old post but I can't figure out what do you mean by "if there were only one component [...] The two marked points don't satisfy the required constraint". How do you encode in the moduli space that the two marked points must belong to different boundary components? $\endgroup$ – Luigi M Dec 1 '20 at 15:54
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    $\begingroup$ I think what Mohammed meant is that if one glues the node on a singly-pinched annulus (with two marked points) the two marked points will never be diametrically opposite, which now I don't actually understand why, I have to admit. Mohammed, could you please elaborate on that? @MohammedAbouzaid $\endgroup$ – Yaniv Ganor Dec 2 '20 at 12:14
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The study of the Deligne-Mumford compactification of Riemann surfaces with boundary can be reduced to the study of the Deligne-Mumford compactification of closed Riemann surfaces together with the study of separating antiholomorphic involutions on them.

More precisely, the data of a compact Riemann surface A with (nonempty) boundary can be recovered from its double S=double(A) (a closed Riemann surface), the natural antiholomorphic involution $\iota$ on $S$ and the choice of one connected component C of $S\setminus Fix(\iota)$.

Thus Isomorphism classes of compact Riemann surfaces A of a given type are naturally identified with Isomorphism classes of triples $(S,\iota,C)$ (for a separating antiholomorphic involution $\iota$) such that C is of the same topological type as A.

In your specific case, the doubles are tori. As Riemann surfaces these doubles can only degenerate in one way, but there are two distinct isomorphism types of involutions on the limit: In one case the fixed point set of the involution has a component consisting of the node and a circle disjoint from the node; in the other case the fixed point set consists of two circles, both passing through the node, but otherwise disjoint.

The first case gives rise to a disk with an interior puncture, the second to a disk, meeting with two punctures on the boundary, which form together a boundary node.

If the conformal modulus of the annulus A goes to zero (i.e. $R\rightarrow 0$, the latter case), then the geodesic contracting is simply obtained by doubling a segment joining the two boundary components of $A$.

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