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I'm trying to learn about Embedded Contact Homology. To understand the proof of $d^2=0$, I started by watching Hutchings' lectures on Obstruction Bundle Gluing on YouTube (1, 2, 3) and have now started reading the relevant two-part paper by Hutchings-Taubes to learn more about the details of the gluing argument involved ("Gluing pseudoholomorphic curves along branched covered cylinders" 1, 2). I am stuck at the very beginning of Part I, specifically I'm unable to understand most of the proof of Lemma 1.11 which is about trajectories which are close to breaking (in what follows, I use the notation from the paper).

Can someone explain (with more details) why one of (i) or (ii) must occur (it's not completely clear to me that this formally follows from the SFT compactness theorem)? I'm guessing that part of the reason for this is that the curves $u_{\pm}$ are transversally cut out and isolated in $\mathcal M^J(\alpha_+,\beta_+)/\mathbb R$ and $\mathcal M^J(\beta_-,\alpha_-)/\mathbb R$, and so taking $\delta$ sufficiently small would imply that any index $1$ component in the limit is close to one of $u_{\pm}$ and thus, is one of $u_{\pm}$.

Also, why must all the other components in all other levels be branched covers of trivial cylinders (I realize they must be curves whose Fredholm indices add up to $0$, but I'm not able to rule out multiply covered curves which are not trivial cylinders from the limit - and these could have index $<0$). Finally, why is it that in case (ii) there are no other levels?

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I was there during this IHES conference to scribe the lectures and also give a discussion session on it. So in case it helps, my notes from both of these are available here.

Your thought on the reason for (i) and (ii) occurring is correct, in lieu of the fact the sum of the Fredholm indices must equal 2. SFT compactness is only used to get a broken curve from a sequence of curves in $\mathcal{M}$. But we know more about our curves, since they live in $\mathcal{M}\cap\mathcal{C}_\delta$ for $\delta$ very small.

All other components in other levels are (as maps) going to be covers of trivial cylinders because of the topological type of the image of the curves that live in $\mathcal{C}_\delta$. The image can be decomposed as $C_-\cup C_0\cup C_+$, such that for $\delta$ very small we have $C_\pm$ being translates of something that is "very close" to $u_\pm$ (and that $C_0$ is "very close" to being a cylinder). For example, in (i) we can't have new genus magically appearing "on top of" $u_+$ (what's really happening is that the "cylindrical ends" of $u_+$ are being stretched, and we know that trivial cylinders have energy zero). In the limit, we therefore might get a component which is a multiply covered map over a cylinder, but by Lemma 1.7 it cannot have negative index.

Finally, in (ii) there are no other levels because we already know that one level contains the "nontrivial positive index curve". So any other level would have index 0 and live "at the top or bottom" of the broken curve, which as was explained in the proof, can't happen. (Note that in (i) there could possibly be index 0 levels betwixt the two nontrivial levels.)

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    $\begingroup$ Thanks for the reference to your notes, I'll take a look at them. I'm not completely convinced by your argument (to be clear, I already intuitively believe the statement, but am trying to write an honest proof). Could you please expand on the following points: (1) Why is $C_0$ close to being a cover of a cylinder (all we seem to know is that its image is close to a cylinder, and don't know it as a map) (2) As I see it, every other level being of index zero should be because all the indices are non-negative. But why can't multiple cover components of negative index appear in the limit? $\endgroup$ – Mohan Swaminathan Jun 4 '18 at 1:19
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    $\begingroup$ Sorry I meant to say image not domain (as in the definition of $\mathcal{C}_\delta$), and I blurred the distinction between a map and its image. The edits should make this more clear now. $\endgroup$ – Chris Gerig Jun 4 '18 at 1:52
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SFT compactness is, as far as I know, in this situation not directly applicable, since the genera of the curves $u_n$ may be unbounded (maybe I am missing something which ensures genus bounds for them in this situation, or one can use the additional assumptions to make SFT compactness applicable). In any case, there is in dimension 4 a compactness result for $J$-holomorphic currents by Taubes, which is often used for ECH, and could presumably be used here.

Assuming for the moment genus bounds on the $u_n$, SFT compactness shows that a subsequence of $u_n$ converges to a holomorphic building $v$, that is "a curve" with potentially several levels $v=(v_1=:v_+, ...,v_k=:v_-)$.

If there is no splitting ($k=1,v_+=v_-$), then SFT compactness ensures that a subsequence of curves converges (up to shifts in the $\mathbb{R}$-direction) to a curve with the same asymptotes, that is to some $[v]\in \mathcal{M}^J(\alpha_+,\alpha_-)/\mathbb{R}$.

If there is a splitting ($k>1$), note first that the assumptions relating to "parametrization by $C_\pm$" ensure that the parts of $u_n$, whose image is parametrized by $C_\pm$, are both contained in only one level (that is, in this region there is no further splitting possible). If we use then in addition the assumption on $C_0$, we can also conclude that the lowest level $v_-$ has the same asymptotes as $u_-$ and the top level $v_+$ has the same asymptotes as $u_+$ and in the middle we have a piece (consisting potentially of none, one or several levels), whose top asymptotes coincide(!) with the bottom asymptotes (since these coincide with the asymptotes of $u_{\mp}$ (or also of $v_\mp$) at $\pm\infty$ ).

Here we used throughout, that $\delta$ can be chosen so small, such that the $\delta$-neighborhood $N_\delta$ of the asymptotic orbits deformation retracts to this collection of the asymptotic orbits. In other words the asymptotes (up to time shifts) are determined by these $C^0$-data.

It follows that the $d\alpha$ energy of the whole middle piece is 0 (the $d\alpha$-energy can be computed as difference of the periods of the asymptotic orbits, and depends only on the relative homology class). Any proper nonconstant curve of finite Hofer energy with $d\alpha$ energy zero is necessarily a branched cover over a collection of cylinders. Thus the whole middle piece is a branched cover over a collection of cylinders (potentially in different middle levels).

It remains to identify the top and the bottom component as $u_\pm$ (we know already that the images have the same asymptotes). Since the asymptotes are given and the $d\alpha$-energy is computed from the asymptotes, it follows that the sum of the $d\alpha$-energies of the simple curves underlying $v_\pm$ coincides with the sum of $d\alpha$ energies of $u_\pm$. Since moreover the $d\alpha$ energy converges (and the $d\alpha$ energy is additive over the levels), the $d\alpha$ energy of $v$ is equal to the $d\alpha$-energy of the $u_n$'s, which is in turn the same as the sum of the $d\alpha$-energies of $u_\pm$.

Therefore all of the components of $v_\pm$, which contribute $d\alpha$-energy, are simple, i.e. they are not nontrivial branched covers of other curves. Thus at most trivial cylinders in the image of $v_\pm$ are (branched) covered. On the other hand, condition 1d ensures now exactly that any (possibly branched) cover over the trivial cylinders with the given asymptotes $(a_i,b_i ...)$ are necessarily unbranched, and thus these components are determined as maps by the asymptotes; hence the components in $v_\pm$ which are (apriori branched) covers of trivial cylinders coincide with the unbranched covers occuring in $u_\pm$.

The components of $v_\pm$ contributing nonzero $d\alpha$-energy are therefore parametrized by surfaces of fixed topological type (the same as for $u_\pm$), and we can (as you mentioned) choose $\delta$ small enough, to ensure that also these components coincide (as parametrized curves up to translation) with the corresponding components of $u_\pm$, as the $[u_\pm]$ are isolated in their moduli space.

Edit: More on why the ends of $u_+$ coincide with the ends of $v_+$. First I should be more precise above (in paragraph 4) to say, that at that point one only knows, that the ends of $u_+$ coincide (with multiplicity) with the the ends of the simple curve underlying $v_+$.

This more precise statement should hold for the following reasons: First, all negative ends of $u_+$ can be distinguished on a $\delta$-level in the target (as $u_+$ is simple and by "unique continuation as almost complex submanifold" or perhaps more directly studying the asymptotes in coordinates). Then there is for each end $e_u$ of $u_+$ negatively asymptotic to some Reeb orbit (almost parametrized by some circle $c$ in $e_u$), a circle $c'$ in the curve $v_+$ (parametrized via $\psi_+$ by $c \subset e_u$), which lies in the same component of the $\delta$-neighborhood $N_\delta$ and is homotopic in $N_\delta$ to $c$. Then the "parametrized component" of $c'$ in $N_\delta \cap v_+$ does not leave $N_\delta$ again by the assumption on $C_0$. Thus $c'$ determines an end $e_v$ of the simple curve underlying $v_+$, which wraps around the Reeb orbit like $c'$ and thus with the same multiplicity as $e_u$. This procedure works for all ends of $u_+$ and all ends of the simple curve underlying $v_+$ are among these.

Then one proceeds with the discussion as above, using only what we know about the asymptotics of the simple curves underlying $v_\pm$.

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  • $\begingroup$ Could you say a little more about why $v_+$ has the same negative asymptotes as $u_+$ (with multiplicities)? I don't think I understand that part of the argument. $\endgroup$ – Mohan Swaminathan Jun 4 '18 at 13:58
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    $\begingroup$ We typically first make use of Gromov compactness for (a sequence of) currents in order to get a genus bound, and then (passing to a subsequence) use SFT compactness for curves (this is explained for example in the proof of Lemma 7.19 of the same paper). Better yet, in our scenario we're not looking at a random sequence of curves in $\mathcal{M}$, they're a sequence of curves in $\mathcal{M}\cap\mathcal{C}_ \delta$ already close to breaking into $(u_+,u_-)$, so I think that if there were not a genus bound (for any $\delta$) we'd take $\delta\to0$ to get a contradiction. $\endgroup$ – Chris Gerig Jun 4 '18 at 17:07
  • $\begingroup$ I added some more details on why the negative ends of $u_+$ should coincide with those of the simple curve underlying $v_+$. $\endgroup$ – user_1789 Jun 4 '18 at 17:16

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