2
$\begingroup$

Let $(M,d\lambda)$ be a compact exact symplectic manifold and $\overline{M}$ its symplectic completion. For simplicity we can think of $\overline{M}$ has a cotangent bundle and $\partial M$ the sphere bundle $SM$. Let $F$ be a defining function for $\partial M$, Rabinowitz Floer homology is the Floer homology of the Rabinowitz action functional, define on $C^1(S^1,\overline{M})\times \mathbb{R}$ :

$$\mathcal{A}_F(v,\eta):=-\int_{S^1}v^*\lambda-\eta\int_0^1F(v(t))dt$$

For a generic choice of $F$ the functional is Morse-Bott. Now one can also consider the perturbed Rabinowitz action functional for a smooth time-dependent Hamiltonian $H:S^1\times \overline{M}\rightarrow \mathbb{R}$:

$$\mathcal{A}_{F,H}(v,\eta):=-\int_{S^1}v^*\lambda-\eta\int_0^1F(v(t))dt-\int_0^1H(v(t),t)dt$$

Now for a generic choice of $H$, the functional is Morse. Furthermore, it turns out that the homologies one can extract from $\mathcal{A}_{F}$ and $\mathcal{A}_{F,H}$, for generic functions, are isomorphic.

Now my question is related with product structures. In Floer homology, for the Hamiltonian action functional, one has a product structure, given by the pair-of-pants product. From my knowledge of looking around in the literature such a product does not seem to happen in the case of the Rabinowitz Floer homology, I'm trying to understand why ? I can't seem to understand what would be the problem of defining a product here such as the pair-of-pants product. Is the extra term $\eta\in \mathbb{R}$ causing the problems here? I know it causes some problems when trying to prove compactness properties for the action functional so maybe it also causes some problems here.

From my understanding, it has be proven that Rabinowitz Floer homology is isomorphic to V-shaped symplectic homology, https://arxiv.org/pdf/0903.0768.pdf, and this carries a product structure, so looking under this isomorphism wouldn't we be able to obtain a product structure such as a pair-of-pants product for the Rabinowitz Floer homology?

Any insight is appreciated, thanks in advance.

$\endgroup$

0

You must log in to answer this question.