Timeline for Bubbling off of a pseudo holomorphic sphere on surface with cylindrical ends
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2020 at 14:52 | vote | accept | Riccardo | ||
| Dec 2, 2019 at 23:29 | comment | added | user_1789 | The image of ${๐ _๐}×[0,1]$ is a curve of length at least ๐ by the choice of โ and since the endpoints of $u^i(\{๐ _๐\}×[0,1])$ are in $N_i^R$. Thus $๐\leq \int_0^1|๐๐ข^i(๐ _๐,๐ก)|๐๐กโค\sqrt{\int_0^1|๐๐ข^i(๐ _๐,๐ก)|^2๐๐ก} =:\sqrt{e_i(๐ _๐)}$. Let $J_i\subset \mathbb{R}$ be the subset of those $๐ $ for which this is applicable, then $|J_i|\varepsilon^2\leq \int_{J_i} e_i(s)ds\leq E(u^i)$. $J_i$ contains arbitrary large intervals (as $i\rightarrow \infty$) since the gradient of $u^i$ is bounded and since the $s_i$ go arbitrary deep into $N_i^R$. | |
| Dec 2, 2019 at 22:44 | comment | added | user_1789 | a) One has to take care, that one really gets a plane (and not a halfplane). For that I would return to the Floer picture instead of the holomorphic strip. b) The final contradiction seems to come from the nonexistence of (sometimes finite area) holomorphic planes in some Riemann surfaces, not the monotonicity lemma. c) If one still wants to apply the (nonrelative) monotonicity lemma then this is unproblematic if one takes care of the above mentioned boundary conditions, since the surfaces appearing are uniformly geometrically bounded (but the boundary conditions apriori not). | |
| Dec 2, 2019 at 17:15 | history | edited | Riccardo | CC BY-SA 4.0 |
added 1401 characters in body
|
| Dec 2, 2019 at 16:04 | comment | added | Riccardo | @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? | |
| Nov 29, 2019 at 7:50 | comment | added | user_1789 | 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. | |
| Nov 29, 2019 at 7:42 | comment | added | user_1789 | 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 $u^i$ | |
| Nov 29, 2019 at 5:11 | comment | added | Riccardo | @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. | |
| Nov 28, 2019 at 8:56 | comment | added | user_1789 | 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. | |
| Nov 27, 2019 at 5:04 | vote | accept | Riccardo | ||
| Dec 2, 2019 at 14:37 | |||||
| Nov 27, 2019 at 2:36 | answer | added | Chris Gerig | timeline score: 4 | |
| Nov 27, 2019 at 2:21 | history | asked | Riccardo | CC BY-SA 4.0 |