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This question is related to my old question Bubbling off of a pseudo holomorphic sphere on surface with cylindrical ends about the bubbling off argument in Seidel's paper The symplectic Floer homology of a Dehn twist in the case of a surface.

I was trying to find a reason for the following seemingly obvious passage:

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 $\Sigma'=\Sigma \setminus [-1,1]\times S^1$

At this point we have two cases:

  1. The distance of $u^i(z_i)$ from $\Sigma'\subset \Sigma^{R_i}$ goes to infinity.
  2. Such distance is bounded.

My previous question was about case 2, now I'm actually curious about the fine details of case 1. In that case the author concludes that we would have a finite energy "bubble" entirely contained in the cylinder $\Bbb R \times S^1$ and then he proceed in ruling that out.

My question now is the following: how can we ensure that in case 1, the bubble is entirely contained in the cylinder?

The usual rescaling argument implies that the image of the rescaled maps $$ u^i(C_i^{-1}z+z_i) : D_{C_i}(0)\to \Sigma^{R_i}$$ coincides with the image of the restriction of the maps to the unit disk $$B_i:=u^i(D_1(z_i))$$ and therefore the image of my limit bubble $\tilde{u}$ will be the limit of these $B_i$ in the split manifold $\Sigma'\cup \Bbb R_{\pm}\times S^1 \cup \Bbb R \times S^1$.

I honestly don't see how to show that the limit of these $B_i$ will be entirely contained in $\Bbb R \times S^1$.

This is what I tried: let $$ d_i := d(u^i(z_i)), \Sigma')\to \infty$$ If $$\lim_{i\to \infty} \frac{d_i}{C_i}>0$$ then the usual rescaling/bubbling reasoning works fine and gives a bubble completely contained in the neck. If $$\lim_{i\to \infty} \frac{d_i}{C_i}=0$$ I run into some troubles in showing that the usual rescaling works. In fact, I have to restrict the functions $u^i$ to a domain contained in the disks $D_{\frac{d_i}{C_i}}(z_i)$ to ensure that the maps land in the neck. But those disks are shrinking! can I still apply the usual bubbling argument even if the domain of my family of functions is degenerating to a point? I would have the following rescaled/restricted functions $$ \tilde{u}^i(\frac{z}{C_i}+ z_i) \colon D_{d_i}(0)\to [-R_i,R_i]\times S^1$$ and they have uniformly bounded $L^{\infty}$-norm and their domain is approaching $\Bbb C$. So it seems to me we have a "bubble". My only problem is that all the references start by working with functions with fixed domain, and not one that is degenerating to a point

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  • $\begingroup$ I haven't looked at this in over a year, but typical arguments argue by contradiction and compare the resulting energies/topologies (of the curves). $\endgroup$
    – Chris Gerig
    Commented Apr 7, 2023 at 0:20
  • $\begingroup$ @ChrisGerig something like: we look at the limit in the split manifold, if the image of the bubbled off map $\tilde{u}$ is not entirely contained in the neck then it must have infinite energy? $\endgroup$
    – Riccardo
    Commented Apr 7, 2023 at 2:39
  • $\begingroup$ Recall that this rescaling argument to obtain a plane (as opposed to a halfplane) only works if you know already that the boundary of the strip does not enter deep into the neck. It is then easy to see that larger and larger subdisks of $D_{r_i}(0)⊂D_{C_i}$ map into $\mathbb{R}\times S^1$ (since the gradient of the $u_i$ is uniformly bounded and case 1) and thus the maps restricted to these subdisks conv. to a holomorphic plane u in $\mathbb{R}\times S^1$ (with bounded gradient). To exclude this, one needs however to show that the finite energy of the $u^i$ implies finite Hofer energy of $u$. $\endgroup$
    – user_1789
    Commented Apr 7, 2023 at 13:44
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    $\begingroup$ It is true in Seidel's situation, an argument is outlined in the comments to your previous question. But again, even if one has this, one still needs to pass from "bounded energy of $u^i$" to "finite Hofer energy of $u$" to derive a contradiction. $\endgroup$
    – user_1789
    Commented Apr 7, 2023 at 14:20
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    $\begingroup$ In the given situation yes: Note that the computations give bounds on the $\lambda$-energy (and the Hofer energy) of the curves in terms of the $\omega$-energy (which in your case is unif. bounded) of the part of the curves contained in the neck (in it, but near its boundary). But since the boundary does not enter arbitrary far into the neck, we can assume without loss of generality (by shrinking the neck by a fixed amount), that the boundary is mapped to $\Sigma'$. Then the energy estimates in the neck work as in BEHWZ, since one does not see there anything from the boundary of the $u^i$. $\endgroup$
    – user_1789
    Commented Apr 8, 2023 at 13:28

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