4
$\begingroup$

I need some clarification about the reason why we have a sphere bubbling off in the situation described by Seidel in The Symplectic Floer Homology of a Dehn Twist.

I’ll try to summarize to the best of my abilities the situation I’m interested in:

We are stretching the neck along a circle in $\Sigma$. Let us fix an identification of the tubular neighborhood of such circle with $[-1,1]\times S^1$. Let $R>0$, by $\Sigma^R$ I’ll denote the surface diffeomorphic to $\Sigma$ but with neck $[-R,R]\times S^1$.

Assume that for $R_i\to \infty$, we have a sequence of $J^{R_i}$-holomorphic strips $\{u^i\}_i$, with $u^i\in \mathcal{M}^{R_i}(x_-,x_+)$ (for a definition see bottom of page 832 of the paper) with uniform bounded energy.

Notice that we can’t have uniform bound on $|du^i|_{L^{\infty}}$, in particular we can find a sequence of points $z_i \in [-R_i,R_i]\times S^1$ such that $|du^i|$ has a maximum there, with value $C_i$, and $C_i\to \infty$. Let $\Sigma’=\Sigma \setminus [-1,1]\times S^1$, (the definition of $\Sigma’$ is independent of the length of the neck!) now assume that, for $i$ big enough, $$d(u^i(z_i),\Sigma’)\leq k < \infty$$

(I.e. no matter the length of the neck that is stretching to infinity, my Maximums stays within finite distance to some edge of the neck)

From there the author immediately concludes that after some reparametrization, the $u^i$’s converged to a $J$-holomorphic sphere in the surface obtained from $\Sigma’$ by attaching a semi-infinite cylinder to each boundary component. I can show that, after reparametrization my curves converges to a non-constant $J$-hol map defined on $\Bbb C$, but I don’t see why this map should extend to the sphere. My understanding is that in order to use the removable of singularity theorem we need a compact image, but $\Sigma’$ With the cylindrical ends is non compact.

Most likely there is some clever way to reparametrize this curves (i used the standard conformal map $z\mapsto z/C_i+z_i$ that centers $u^i$ at $0$ and normalize the norm of its differential at $0$), or to infer something about the image of the limit, but I’m unable to see it.

Any hint is really appreciated, since this is bugging me. Thanks in advance!

Update

I agree with the given answer that monotonicity lemma could be a way to establish compactness of the image of the limiting function. What I don't understand are the following point:

1) Assuming that for small enough balls we can ensure that the topological boundary of our curve is outside such ball (as required by the monotonicity lemma) how do we deal with the fact that the target manifold is not compact? That's the key assumption in the lemma. As I wrote in the comments I though that we could just restrict our attention to some compact sub manifolds of $\Sigma' \cup \partial \Sigma' \times [0,\infty)$ but the lower bound provided by the lemma would depend on such a choice. Hence it's not clear to me how to obtain a uniform lower bound on the energy of the curve passing through a given ball.

2)If we work with a vertical almost complex structure on the neck, then I think the lower bound provided by the monotonicity lemma is the same in every sector $[a-n,a+n]\times \partial \Sigma'$ (since they all are isometric and with the same a.c.s.) In that case I can see how to get a uniform lower bound. But what worries me is that I don't see any problem in applying this reasoning to case #1, but as far as I understood, in that case we can't rule out non-constant $j$-hol planes that easily and we must use a different strategy (namely integrability of the a.c.s.)

$\endgroup$
  • $\begingroup$ Note that the energy bound alone does not imply the unboundedness of $|du^i|$, the specific form of $h$ on $N^R_i$ is also used. Furthermore, the holomorphic strips have free (even though twisted periodic) boundary values, thus it seems not so easy to apply the monotonicity lemma directly. $\endgroup$ – user_1789 Nov 28 at 8:56
  • $\begingroup$ @user_1789 do you mind to elaborate a little further your point of the monotonicity lemma? I'm not an expert but I've not seen any problem in the application of it to the limiting curve $u$. For what concerns the unboundedness of the differential, I agree with you, even though I'm curious about what's the proof you've in mind. $\endgroup$ – Riccardo Nov 29 at 5:11
  • $\begingroup$ The monotonicity lemma on a given ball applies to nonconstant holomorphic curves passing through the center of the ball, whose (topological) boundary lies outside of the ball. The condition on the boundary is important, without it there are easy "counterexamples". If one has curves with boundary or noncompact curves, one needs to ensure this condition is met. This seems to me not that obvious for the strips $u^i$ $\endgroup$ – user_1789 Nov 29 at 7:42
  • $\begingroup$ It is clear that the gradient is unbounded if only interior points $z_i=(s_i,t_i)$ escape arbitrarily far into $N_i^R$ (since then the image of $\{s_i\}\times [0,1]$ is arbitrary long). If on the other hand a boundary point (s,0) goes there, then so does (s,1), and the segment between them has length at least epsilon. If the gradient is bounded and the boundary points go arbitrary deep into $N_i^R$, there is a arbitrary large interval of such $s\in \mathbb{R}$ and this yields a arbitrary large lower bound on the energy. $\endgroup$ – user_1789 Nov 29 at 7:50
  • $\begingroup$ @user_1789 I apologise, but I can't really follow your reasoning. You assume that boundary points $(s_i,0)$ go arbitrarily deep into the neck, by periodicity this means that $(s_i,1)$ go deep into the neck as well. The segment between them (in the domain) has length $1$ (don't understand the $\epsilon$). How do you conclude from there that the energy is exploding? $\endgroup$ – Riccardo Dec 2 at 16:04
4
$\begingroup$

It turns out that the Removable Singularities Theorem and the Monotonicity Lemma do not require compactness, but that the target manifold should have bounded geometry, such as in our case of inserting a cylindrical piece to the compact surface. See for example, Lemma 5.11 in "Compactness results in symplectic field theory" by Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder and Proposition 2.71 of Abbas' book "An introduction to compactness results in symplectic field theory". Seidel brought to my attention Theorem 4.5.1 in Sikorav's paper "Some properties of holomorphic curves in almost complex manifolds", which only needs the (possibly noncompact) target to have bounded geometry and does not need the disc to have bounded image, for removing singularities.

By assumption our points $u^i(z_i)$ don't run away along the semi-infinite ends attached to $\Sigma'$, i.e. our bubble points are stuck in $\Sigma'$ union a collar $\partial\Sigma'\times[0,k]$ and cannot escape along $\partial\Sigma'\times[0,\infty)$, otherwise we're in Case#1 (in the paper) which has the points running away along the neck of $\Sigma^{R_i}$. We now zoom in our around these bubble points (hence away from the boundary of the holomorphic strips), forming a rescaled sequence of holomorphic discs. Note that in both the case at hand (Case#2 in the paper) and Case#1 in the paper, our sequence of rescaled maps have domains being open disks centered around the bubble points. It can happen in Case#1 that these disks just shift along with the bubble points down the infinite neck region, which doesn't happen in Case#2.

Aside: I hoped that the monotonicity lemma would prevent the sequence of rescaled holomorphic discs from "thinning" out and having unbounded image, that is, our maps must use up at least a certain amount of energy for every ball whose center it passes through. We couldn't a priori apply the usual monotonicity lemma (for compact target manifolds) to each compact set in a compact exhaustion of the noncompact surface, because the resulting (lower) energy bound may vary (hence decrease) with each set. But due to the cylindrical end there is a uniform lower bound. In any case, the a priori issue is that the (image of the) boundary of the discs might squeeze into the balls with which we are trying to compute the area bounds.

$\endgroup$
  • $\begingroup$ Thanks for pointing out this version of the monotonicity lemma, I was unaware of it. I'm kind of worried about it's assumptions though. how do we ensure that the sectors in the boundary of the disk are mapped over legendrian sub-manifolds in the neck? $\endgroup$ – Riccardo Dec 2 at 21:29
  • $\begingroup$ yes your clarifications were very helpful. I think I concentrate all my remaining doubts in the following: what prevents from using monotonicity (the very same argument) in case #1? I would proceed as follows. let $\tilde{u}$ my limiting curve from $\Bbb C$ into $S^1\times (-\infty, \infty)$, it has finite energy, i take any path $\gamma$ from $0$ to $\pm \infty$ in the domain a collection of balls of radius $r$ whose centre lies on $u(\gamma)$ and they don't intersect (like a string of pearls). by the monotonicity I know I can have at most $K<\infty$ of them, $\endgroup$ – Riccardo Dec 3 at 0:17
  • $\begingroup$ hence my curve is contained in the region $S^1\times [-2Kr,2Kr]\times S^1$ (translated by $\tilde{u}(0)$) which is compact. $\endgroup$ – Riccardo Dec 3 at 0:17
  • $\begingroup$ @ChrisGerig How do you ensure that $\tilde{u}(I_1)\cap B_r(\tilde{u}(z_0))=\emptyset$ in the notation of Abbas? $\endgroup$ – user_1789 Dec 3 at 0:18
  • $\begingroup$ @user_1789 I was wondering if we could just take a bigger disk. As far as I understood, the proof should be the same, I just get a bigger constant. The only thing that is unclear to me is that, while it's clear that we can assume that $\text{Im}(\tilde{u}(D))\not\subset B_r$, I don't know if we can assume that for big enough disk all the boundary is going to be outside it. Seems like the case but I can't find a precise way to show it $\endgroup$ – Riccardo Dec 3 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.