Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(W,\omega)$ be a closed symplectially aspherical symplectic manifold, and fix a Hamiltonian $H\in C^{\infty}(W\times S^{1};\mathbb{R})$ and a compatible almost complex structure $J$ on $W$. Given two 1-periodic orbits $x$ and $y$ of $H$, let $\mathcal{M}(x,y;H,J)$ denote the set of maps $w\in C^{\infty}(\mathbb{R}\times S^{1};W)$ that satisfy

$\mathcal{F}_{H,J}(w):=\partial_{s}w+J(w)\partial_{t}w+\nabla H_{t}(w)=0$

where $\nabla H_{t}$ is the gradient of $H$ with respect to the metric $\omega(\cdot,J\cdot)$. Let $\mathcal{M}(H,J)$ denote the union of the spaces $\mathcal{M}(x,y;H,J)$ as $x$ and $y$ run through the 1-periodic orbits of $H$. Well known elliptic regularity results tellus that the space $\mathcal{M}(H,J)$ is bounded in the $C_{loc}^{\infty}$ topology on $C^{\infty}(\mathbb{R}\times S^{1};W)$.

It is well known that if:

(1) all 1-periodic orbits of $H$ are non-degenerate, and

(2) the pair $(H,J)$ is regular (meaning that $d\mathcal{F}_{H,J}(w)$ is surjective for all $w\in\mathcal{M}(H,J)$),

then for any 1-periodic orbits $x,y$ the space $\mathcal{M}(x,y;H,J)$ is a finite dimensional manifold.

The spaces $\mathcal{M}(x,y;H,J)$ carry a free $\mathbb{R}$-action via translation, and hence they obviously can never be compact. Instead one can look at the quotient spaces $\mathcal{M}(x,y;H,J)/\mathbb{R}$ obtained by dividing out by this action. In general the spaces $\mathcal{M}(x,y;H,J)/\mathbb{R}$ are still not compact, but it is possible to obtain a very precise description of the closure $\overline{\mathcal{M}(x,y;H,J)/\mathbb{R}}$ (in the topology induced from the $C_{loc}^{\infty}$ topology on $C^{\infty}(\mathbb{R}\times S^{1};W)$).

For instance, if $dim(\mathcal{M}(x,y;H,J))=1$ then $\overline{\mathcal{M}(x,y;H,J)/\mathbb{R}}=\mathcal{M}(x,y;H,J)/\mathbb{R}$ (and hence $\mathcal{M}(x,y;H,J)/\mathbb{R}$ is compact), and if $dim(\mathcal{M}(x,y;H,J))=2$ then the boundary $\partial(\overline{\mathcal{M}(x,y;H,J)/\mathbb{R})}$ is precisely equal to the union $\cup_{z}\overline{\mathcal{M}(x,z;H,J)/\mathbb{R}}\times\overline{\mathcal{M}(z,y;H,J)/\mathbb{R}}$ where $z$ runs over all the 1-periodic orbits such that $dim(\mathcal{M}(x,z;H,J))=1$. This latter fact is precisely why the boundary operator in Floer homology really is a boundary operator.

Suppose now that we are given two pairs $(H^{a},J^{a})$ and $(H^{b},J^{b})$ of Hamiltonians that satisfy (1) and (2) above, and suppose we are given a family $(H_{s},J_{s})$ ($s\in\mathbb{R}$) such that $(H_{s},J_{s})\equiv(H^{a},J^{a})$ for $s$ near $-\infty$ and $(H_{s},J_{s})\equiv(H^{b},J^{b})$ for $s$ near $+\infty$. Then if $x$ is a periodic orbit of $H^{a}$ and $y$ is a periodic orbit of $H^{b}$, we can again look at the spaces $\mathcal{M}(x,y;H_{s},J_{s})$, which are defined in the same way, only now $H$ and $J$ depend on $s$. If $(H_{s},J_{s})$ satisfies the property:

(3) The operator $d\mathcal{F}_{H_{s},J_{s}}(w)$ is surjective for all $w\in\mathcal{M}(H_{s},J_{s})$,

then the spaces $\mathcal{M}(x,y;H_{s},J_{s})$ are again finite dimensional manifolds, and as before we have a good description of the closures $\overline{\mathcal{M}(x,y;H_{s},J_{s})}$. This fact is used to show that $(H_{s},J_{s})$ induces a chain map $FH_{*}(H^{a})\rightarrow FH_{*}(H^{b})$ on the Floer homology groups.

Anyway, my question is this:

Suppose we drop assumptions (1), (2) and (3) above. How bad can the spaces $\mathcal{M}(x,y;H_{s},J_{s})$ be? In order to make this a proper question, I will focus on a special case (although I am also interested in answers that address the general case). Consider the special case where $H^{a}=H^{b}=H$ and $x=y$. Thus $x$ is a (possibly degenerate) orbit of $H$, and $(H_{s},J_{s})$ is any family (not necessarily regular) such that $(H_{s},J_{s})\equiv(H,J)$ for $\left|s\right|$ large. Do we have any information about what the closure of $\mathcal{M}(x,x,H_{s},J_{s})$ is like? For instance, is it possible that there exists $(w_{n})_{n=1,2,3,\dots}$ in $\mathcal{M}(x,x;H_{s},J_{s})$ and $(s_{n})_{n=1,2,3,\dots}$ a bounded sequence of real number such that $w_{n}(s_{n})\rightarrow z$ for a different 1-periodic orbit $z\ne x$ of $H$?

share|improve this question
2  
Perhaps the most basic question in this context is whether $L^2$ solutions to Floer's equation converge to Hamiltonian orbits. I don't know of any reference for this, though perhaps it's in the minimal surfaces literature. The analogous property of instantons is proved by Morgan, Mrowka and Ruberman in "The $L^2$-moduli space and a vanishing theorem for Donaldson polynomial invariants", Monographs in Geometry and Topology, II. International Press, Cambridge, MA, 1994; MR1287851. –  Tim Perutz Apr 24 '11 at 21:34
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.