As the title suggests, I want to understand the strip breaking phenomenon that happens when I Gromov-compactify the moduli space of pseudoholomorphic curves from the holomorphic strip $\Bbb R \times [0,1]$ into a symplectic manifold $(M,\omega,J)$. We require that these curves sends $\Bbb R \times \{0\}$ to a Lagrangian submanifold $L_0$, and similarly $\Bbb R \times \{1\}$ to a Lagrangian submanifold $L_1$. Let $p,q \in L_0\cap L_1$, we require that $u(s,t)\to p,q$ for $|s|\to \infty$.

(From Auroux notes)

As far I understand (See here bottom of page 9) the strip breaking phenomenon appears when the energy of my family of $J$-holomorphic curves $\{u_n\}_n$ is moving towards both ends of my strip. Intuitively I can (somehow) see what's going on, by either reparametrizing the strip to keep one of the two concentrations of energy fixed at the origin and let the other go to the end I somehow see 2 strips.

I would like to formalise this procedure and actually see the strip breaking

Here is my attempt: Let $\{a_n\}_n \subset \Bbb R$ be a sequence converging to $+\infty$, and let $\{ b_n\}_n$ converging to $-\infty$. Assume that the sequence $\{ a_n\}$ is such that it moves the energy concentration going towards the $+ \infty$ end of my strip, back to the origin. Similarly for the sequence $\{b_n\}$ and the other energy concentration.

I think what we want to consider are the two limits \begin{align} &\lim_{n\to +\infty} u_n(s-a_n,t) \\&\lim_{n\to +\infty} u_n(s-b_n,t)\end{align}

Under the assumption (which seems to me quite reasonable) that both limits converge uniformly to functions $w(s,t)$ and $v(s,t)$. I want to see now their behaviour near the ends. To this end we compute
\begin{align} &\lim_{s\to +\infty} w(s,t) \\ &\lim_{s\to +\infty}\lim_{n\to +\infty} u_n(s-a_n,t)\\ \end{align} If I'm allowed to exchange the order of the limits (which seems the case since I asked for uniform convergence) then it's easy to see that $v$ tends to the constant maps $p$ or $q$ near the ends of the strip, similarly for $w$. On the other hand, the typical picture for this situation is the following from Auroux notes

which doesn't exclude the possibility that $r\neq p,q$.

Therefore I must do something wrong in my visualisation process. Can someone help me seeing the strip-breaking process more clearly/formally?


1 Answer 1


The problem is that you do not have uniform convergence. The compactness argument is usually based on the boundedness of the $C^1$-norm (often you do not even have this, see the bubbling described below). The $C^1$-bound allows you to extract from the Arzelà-Ascoli theorem for any compact subset of the domain a subsequence that converges in $C^0$ (on the chosen domain!) If you choose a larger compact subdomain, you have to choose a subsequence of the subsequence, and this way it is not possible to get a subsequence converging uniformly (you only obtain locally uniform convergence).
Also you will need to choose a metric on the strip, and certainly whether you really get uniform boundedness or not, will also depend on the chosen metric.

I can try to explain how bubbling works on holomorphic curves with compact domain (let's say a sphere, which simplifies things a lot since there is only one complex structure, and there is no need to worry about Deligne-Mumford compactification etc.). Also since the domain is compact, we do not need to worry about the choice of the metric. I hope that this allows you to see what you have to think about to understand the holomorphic strip.

Suppose you have a family $u_n\colon S^2 \to M$. There are two cases:

  1. If the family $u_n$ is uniformly bounded in $C^1$,then by Arzelá-Ascoli we have a subsequence that converges in $C^0$ to a map $u_\infty\colon S^2 \to M$ (same domain!), and by elliptic regularity, the subsequence will even converge in $C^\infty$ and $u_\infty$ is $J$-holomorphic. [Remark: Think about this as Morera's theorem from classical complex analysis. $C^0$-convergence guarantees the convergence of the integral, since all integrals are $0$, so must be their limit, thus $u_\infty$ is holomorphic.]

    1. If the family $u_n$ is not uniformly bounded in $C^1$,then (assuming that $M$ is compact, so that the $u_n$ cannot simply "run off"), there will be a sequence $z_n\in S^2$ such that the differentials $\|Du_n(z_n)\| \to \infty$ as $n\to \infty$. (strictly speaking you have to choose a subsequence etc. but I want to keep the notation simple). This of course spoils uniform convergence, but we still can obtain some information by a process called rescaling. We can choose a complex chart around each $z_n \in S^2$. Let's say $\phi_n\colon D^2\subset \mathbb{C} \to S^2$ such that $\phi_n(0) = z_n$. Assume that $\phi(D^2)$ are all disks of radius $1$ in $S^2$. Compose $$\hat u_n := u_n\circ \phi_n \colon D^2 \to M$$ to obtain a family of holomorphic disks where $R_n := \|D\hat u_n(0)\| \to \infty$. The maps $\hat u_n$ do not convergence uniformly either, but we can rescale the domains by multiplying them with $R_n$ and define rescaled holomorphic maps as follows $$\hat v_n\colon R_n\cdot D^2 \to M, \, z \mapsto \hat u_n(\frac{z}{R_n})$$

Observe that

  • The domain $R_n\cdot D^2$ of the maps $\hat v_n$ increases and eventually exhaust $\mathbb{C}$ since $R_n\to \infty$.
  • The differential $D\hat v_n$ has norm $\|D\hat v_n\| = \frac{1}{R_n} \cdot \|D\hat u_n|\|$, and if we have chosen the $z_n$ suitably, the norm of $\|D\hat v_n\|$ will be bounded by $1$ on all of $R_n\cdot D^2$.

Consequence: The $\hat v_n$ do not convergence uniformly but only locally uniformly, that is, fix any radius $R\gg 1$, then from some sufficiently large $n$ on, there will be a subsequence such that we find a subsequence such that $\hat v_n$ converges uniformly (first in $C^0$ and then by elliptic regularity in $C^\infty$) on the disk of radius $R$.

Unfortunately (and contrary to your sketch above), we cannot say what happens at infinity, we can only guarantee existence of a holomorphic plane $u_\infty\colon \mathbb{C} \to M$ that will be the local uniform limit.

What do you do next? Next, you use the removal of singularity theorem. It tells you that any map $u_\infty\colon \mathbb{C} \to M$ with finite energy extends to a holomorphic map $\hat u_\infty\colon \hat {\mathbb{C}} = S^2 \to M$, and you have recovered your first bubble.

The rest of the compactness theorem is unfortunately much more technical: You have shown that there are small neighborhoods of the points $z_n$ in $S^2$ that produce a holomorphic sphere. Next you have to study $S^2$ with small disks around the $z_n$ removed and think if this also converges, possibly repeating the above argument to recover more bubbles, and once you have everything under control, because you know that every bubble has a minimal energy and the sum of all bubbles cannot have more energy than your initial curves, you still have show that all bubbles "connect" to give a "tree".

Conclusion: Usually you do not start out with uniform convergence, but you obtain from the $C^1$-boundedness (possibly after rescaling) a subsequence that converges uniformly on a fixed compact domain. As you increase the domain you have to choose a further subsequence, and your limit curve will not be the uniform limit of any sequence, but only the uniform limit of a sequence after restricting to a compact subset.

  • 3
    $\begingroup$ I don't think this helps the OP, and I think he already understands sphere bubbling. Strip breaking is slightly different than sphere bubbling (even though you can phrase strip breaking as "sphere bubbling at the endpoints" of the strips) -- the issue of the OP is that after rescaling he doesn't see how one end of the strip (in a broken strip) is an intersection point $r$ of the Lagrangians, with $r$ not equal to $p$ or $q$. Anyway, I believe this is sufficiently explained in Floer's original paper: "The Unregularized Gradient Flow of the Symplectic Action" (the section on compactness). $\endgroup$ Mar 15, 2019 at 17:10
  • $\begingroup$ The first key point is "Do we have C¹-bounds or not?" if we do not, we will get bubbling. Maybe you're right that the OP does already assume that we have C¹-bounds (but his formulation is vague, so I'm not sure what he really knows and what not). $\endgroup$ Mar 15, 2019 at 17:53
  • $\begingroup$ I think what I did work out in my answer is how you get convergence from the C¹-bounds. This is done via Arzela-Ascoli, which only works on compact subsets of the domain. Of course you can make the compact subsets as large as you want, but this will never guarantee you a sequence of maps converging uniformly on the whole non-compact domain, and since r lies at infinity, there is not much to say about it. You can deduce that there is an r, because you must have a kind of removable singularity theorem for strips in C⁰, in the same way you have this for punctured spheres or punctured disks. $\endgroup$ Mar 15, 2019 at 17:56
  • $\begingroup$ Dear @KlausNiederkrüger yeah you are right, I was being sloppy with my assumptions. I "think" I understand the traditional bubbling of spheres or disks the point of my question is to formally see the strip breaking. I apologise for my lack of clarity. $\endgroup$
    – Riccardo
    Mar 15, 2019 at 19:01
  • 2
    $\begingroup$ @ChrisGerig that's exactly my problem, I should have been more clear when explaining my issue. I will check Floer's paper! thanks a lot for the reference! $\endgroup$
    – Riccardo
    Mar 15, 2019 at 19:03

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