Let X be a compact symplectic manifold and $H_t,J_t$ a Floer regular pair of $\mathbb{S}^1$ dependent Hamiltonians and complex structures. The PSS maps are defined by considering $\mathbb{C}$ with a single marked point at the origin $\mathfrak{o}$ and a positive strip like end at $+\infty$. The maps satisfy Floer's equation for the pair $(\mathbb{J},\mathbb{H})$ below.

We consider a surface dependent complex structure $\mathbb{J}$ which is domain-independent (agreeing with some generic J') near the origin $\mathfrak{o}$ and agrees with $J_t$ along the strip-like end.

Similarly, we have a surface dependent $\mathbb{H}$ which is 0 near the origin and which agrees with $H_t$ along the strip-like end.

In proposition 4.1 of their paper on gluing Floer trajectories:


Oh and Zhu make the following claim which "can be derived by a standard argument."

Transversality for maps as above from $\mathbb{C} \to X$ can be achieved using a generic $\mathbb{J},\mathbb{H}$ of the above form.

This proposition doesn't raise any eyebrows because my general impression is that most of the transversality difficulties come from sphere bubbling in these problems. But I'm wondering exactly how it is proven? I could imagine two slightly different types of arguments. I SHOULD NOTE THAT I'M MOSTLY INTERESTED IN THE FACT THAT FOR A FIXED NONDEGENERATE $\mathbb{H}$, we can perturb $\mathbb{J}$ in the class above to achieve transversality.

One approach would be to use the fact that our domain is stable and to draw upon the circle of ideas from Ruan-Tian as explained for example in section 6.7 of MacDuff and Salamon's big textbook on J-holomorphic curves.

The second is to follow the lines of Floer, Hofer and Salamon's paper http://www.math.ethz.ch/~salamon/PREPRINTS/trans.pdf in particular try to prove somewhere injectivity for such solutions as in Theorem 4.3 and then follow the proof of Theorem 5.1.

Can either of these approaches be adapted to the present situation? If not, what is this "standard argument ?" I'm curious just because the Floer datum are somewhat constrained in that one isn't allowing arbitrary domain-dependent perturbations so I'm wondering how that might enter into the proof.


1 Answer 1


I recommend looking at M. Shwarz's thesis, http://www.math.uni-leipzig.de/~schwarz/diss.pdf, Section 4.2. A "generic" almost complex structure $J$ makes all non-constant maps regular. One first sets up a universal moduli space and shows that it is a Banach manifold. There are various ways of setting up the universal moduli spaces, depending on which kind of perturbations you want to allow (on which neighborhood of infinity is the almost complex structure fixed, or how does it decay at infinity) That the universal moduli space is a Banach manifold follows from the implicit function theorem: to show that the linearized operator is surjective, by unique continuation, it suffices to show that an element of the cokernel vanishes in an open neighborhood, since your domain is connected. Then the Sard-Smale theorem shows that regular values of the projection onto the space of almost complex structures are comeager. You have to check that constant maps are regular separately; as you say, this involves a non-degeneracy hypothesis.

  • $\begingroup$ Thank you for your answer and I will study the thesis you recommend. There is still one thing which confuses me. In the Floer-Hofer-Salamon paper they follow your sketch faithfully, but to prove the cokernel vanishes in an open neighborhood they require their theorem 4.3 which is about the density of somewhere injective points. At first glance Schwarz doesn't mention this point. How does he get around it? Does it hold in the case of the PSS map? $\endgroup$
    – user36931
    Feb 18, 2014 at 2:27
  • $\begingroup$ Is the point that one only needs the "somewhere injectivity" to prove that the universal moduli space is a Banach manifold using the argument that you mentioned when there is a continuous family of automorphisms of the domain? $\endgroup$
    – user36931
    Feb 18, 2014 at 8:44
  • $\begingroup$ Yes, exactly. . $\endgroup$ Feb 19, 2014 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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