Consider a symplectic manifold $(M,\omega)$ with the property that $\pi_2(M) = 0$. Given a time dependent hamiltonian $H_t$ on $M$, and a $\omega$-compatible almost complex structure J on M, we may look at smooth,contractible finite energy solutions to the floer equation. Let

$$\mathscr{M}:=\{ u:\mathbb{R} \times S^1|\text{u is contractible and is solution of the Floer equation with finite energy}\}$$

It can be shown that $\mathscr{M}$ is compact with respect to $C^\infty$ uniform convergence on compact sets topology. Also $\mathbb{R}$ acts on $\mathscr M$ by reparametrization.

On the other hand we may fix a homology class $A \in H_2(M)$ which is indecomposable ( meaning that we cannot get homology classes $B$ and $C$ in $H_2(M)$ such that $A= B +C$ such that $\omega(B) > 0$ and $\omega(C) > 0$) on $M$ and consider the moduli space of J-holomorphic spheres in the class A. Let us denote this moduli space by $\mathscr{M}^\prime$

This space on the other hand is not compact but if we look at the space of unparametrised J-spheres i.e ${\mathscr{M^\prime}} / {PSL(2,\mathbb{C}})$ then the space becomes compact(As $PSL(2,\mathbb{C})$ is the automorphism group for a sphere) . This means that given a sequence $u_k \in \mathscr{M^\prime}$ there is a sub sequence $u_{k^\prime}$ that converges to a unparametrised curve $u \in {\mathscr{M^\prime}} / {PSL(2,\mathbb{C}})$.

Given that we may think of the Cauchy Riemann equation as the Floer equation when the time dependent hamiltonian are constant, why are we able to recover the parametrization of the limiting curve in the Floer case but not able to recover the parametrization in the case of a J-holomorphic sphere? Is there any fundamental reason connected to how the reparameterization groups acts on the moduli spaces that makes this happen?


1 Answer 1


Short answer: the cylinder is non-compact so $C^\infty_{loc}$ convergence is pretty lousy. The non-compactness of the cylinders (= sphere with 2 marked points) encodes the same thing as the non-compactness of $U(1)$ or $PSL(2, \mathbb{C})$ (depending on whether you see the spheres as having the two marked points or not).

Long version of the answer:

First of all, as you correctly point out, in the Floer case, you have the domain $\mathbb{R}$ translation ambiguity in the cylinder[s] you get. In particular, then, if you take some sequence of (parametrized) cylinders from your moduli space, you know they converge in $C^\infty_{loc}$ to a Floer cylinder. That said, the limit you get may be trivial, and if the limiting building consists of multiple broken cylinders, you will need to consider various different parametrizations to capture all the possible pieces of the limit. One sees this behaviour also in considering a sequence of gradient flow lines in the Morse setting.

A silly example of this is is to consider a fixed cylinder $u(s,t)$. Let $u_k(s,t) = u(s-k, t)$. This converges in $C^\infty_{loc}$ to a trivial cylinder.

Interpreted in this way, your moduli space of spheres is also compact. Let's eliminate all the complications from this example and consider degree 1 spheres in $\mathbb{CP}^1$. As an unparametrized sphere, the moduli space is a point. To make these into cylinders, we need to choose two distinguished marked points that correspond to $\pm \infty$ in the cylinder. In this case, the Hamiltonian has a Morse-Bott manifold of orbits, but let's choose two distinguished points in the target $\mathbb{CP}^1$ and declare them to be the orbits we want to converge to.

The only ambiguity that now remains is a $\mathbb{C}^*$ choice in parametrization (you can think of it as placing a third marked point). You are free to put this marked point anywhere, and in this case, we can identify where we put it in the domain with where we put it in the image. If we consider a sequence of these marked points that converge, the corresponding sequence converges. If, instead, we consider a sequence of marked points that go to the marked points that correspond to $\pm \infty$, the resulting Floer cylinders will still converge in $C^\infty_{loc}$, but to constant solutions.


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