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

In the Introduction to Symplectic Field Theory paper, Eliashberg, Givental and Hofer introduce a differential graded algebra generated by periodic orbits of the Reeb vector field on a contact manifold $M$. Instead of taking all periodic orbits, they throw out the "bad orbits". Given a simple periodic orbit $\gamma : S^1 \to M$, denote by $\gamma^k$ the periodic orbit that is a $k$-fold cover of $\gamma$. They define the even covers of an embedded orbit $\gamma$ to be bad if the Conley-Zehnder indices of $\gamma^{2k}$ have different parity from the Conley-Zehnder indices of $\gamma^{2k-1}$. (Note that this is well-defined, since by properties of the Conley-Zehnder index, the even covers all have the same parity, and the odd ones all have the same. Furthermore, the parity of the Conley-Zehnder index is independent of the choice of trivialization of the contact structure along the orbit.)

The question then has three parts:

  1. Why must they throw out the "bad orbits"? Is there a possible theory that includes them?

  2. Why can they throw out the "bad orbits" and still get a differential on the chain complex? doesn't the compactness theorem for pseudholomorphic curves allow for breaking at a "bad orbit"? By analogy, one cannot arbitrarily exclude certain critical points from a Morse complex and still obtain a chain complex.

  3. Naively, one would hope that if $a$ is a good orbit, $b$ is a bad orbit, and the moduli space $\mathcal M(a,b)$ is 1-dimensional, then the signed count $\# \mathcal{M}(a,b)/\mathbb{R}$ is zero. Is this true? Is there a way in which it can be arranged to be true?

share|improve this question
2  
Some context, or an explanation of your notation, can do wonders in attracting useful answers! –  Mariano Suárez-Alvarez Mar 29 '11 at 5:36
1  
(As can proper capitalization and grammar, of course) –  Mariano Suárez-Alvarez Mar 29 '11 at 5:37
2  
Please see mathoverflow.net/howtoask –  David Roberts Mar 29 '11 at 6:06
    
If you (or someone else willing to put in the time) can turn this into a meaningful question, please use the "edit" link to fix it, and then flag for moderator attention. –  S. Carnahan Mar 29 '11 at 9:33
1  
@Scott, I can't find any "edit" link, on this question or on any other one for that matter. (I can find it on math.SE, where it is between "flag" and "link", whereas here I have nothing between "flag" and "cite", so I suspect it's been disabled on MO?) I would be happy to edit this question (and provide an answer). The author is presumably asking about why the so-called "bad orbits" are thrown out of the symplectic field theory chain complex. In dimension 3, these are hyperbolic orbits whose stable/unstable manifolds are not orientable. –  Sam Lisi Mar 29 '11 at 13:19

2 Answers 2

I think Eigenbunny's explanation cuts to the heart of the matter. Here are some more specific details in the case of symplectic field theory.

First of all, to get some intuition for what bad orbits look like, in the case of a 3 dimensional contact manifold, the bad orbits are precisely the hyperbolic orbits that have non-orientable stable and unstable manifolds. For simplicity, I will talk about cylinders, but the general case is very similar.

(1) The bad orbits must be thrown out because of an orientation problem. In SFT, the simple orbits are decorated with a choice of marked points on each one, called markers. Also, the domain curves are decorated with asymptotic markers. These are a choice of marked point in the circle at infinity for each of the cylindrical ends of the domain. Now consider a holomorphic half-cylinder asymptotic to the $m$-fold cover of a simple orbit $\gamma$. By the convergence result, the image of the asymptotic marker makes sense, and is a point on $\gamma$. We want the image of the asymptotic marker to match the marker on the orbit. Given an unmarked curve, we have $m$ possible locations for the asymptotic marker on the domain.
To understand how to glue, we want to think of how a level 2 building can arise as the limit of cylinders. Suppose we break at the $m$-fold cover of $\gamma$. The two ends asymptotic to this orbit don't obtain an asymptotic marker -- only a marker relative to each other. We can think of this as having $m$ possible choices of asymptotic marker on each end, and then identifying the ones obtained by a simultaneous rotation, giving us $m$ possible choices in the end. Now, for us to count these curves, we need the rotation of the markers to consistent with the orientations. In general it is, but for bad orbits, this is orientation reversing.

This business with markers is one of the places where we see the equivariant nature of SFT: we break the $S^1$ symmetry with these markers, and then quotient out by all possible choices to remove the dependence.

For more details, Bourgeois and Mohnke's Coherent orientations in symplectic field theory does a good job explaining this in detail. Also, see the Introduction to SFT paper by Eliashberg, Givental & Hofer (Section 1.8.4 and Remarks 1.9.2, 1.9.6)

(2) This is also because of the $\mathbb Z_m$ action. Consider the simplest example, where we look at the boundary of a 1-dimensional moduli space of cylinders connecting orbits a and a'. (By this, I mean 1 dimensional after the quotient by the translation.) Suppose one boundary component consists of a level 2 building with two cylinders, a to b and b to a', and b is a bad orbit. Suppose b has multiplicity $m=2k$. Then, the boundary building has $m$ different possible decorations by asymptotic markers. If you chase through the orientations, the end result is that these terms all cancel out, thus saving us from having problems with $d^2$.

(3) The answer to the most optimistic version of this question is no: here is no reason for the moduli space to be empty in general. In order to get a signed count of zero, one needs to make sense of orientations. The way orientations are done in SFT, these moduli spaces asymptotic to bad orbits don't even get one, so the signed count doesn't make sense. Arguably, this isn't the only way we might define orientations (to get the same theory). A related situation in which something like this is true is in the isomorphism between linearized contact homology and $S^1$-equivariant symplectic homology (actually the $SH^+$ part), due to Bourgeois and Oancea. Their non-equivariant contact homology involves decorating each Reeb orbit with a Morse function and imposing markers. The ``bad'' orbits end up dying (over $\mathbb Q$) because the differential of the maximum, instead of being $0$, is then twice the minimum.

share|improve this answer

That's because symplectic field theory is secretly an $S^1$-equivariant theory. As a finite-dimensional model, suppose that you have a manifold $M$ with an $S^1$-action and an invariant function $h: M \rightarrow \mathbb{R}$. Take an orbit of the $S^1$-action (not a fixed point) which consists of critical points of $h$, and such that the Hessian is transversally nondegenerate. If the negative eigenspaces of the Hessian form a nontrivial vector bundle over our orbit, the local contribution to the Morse homology is a twisted homology of $S^1$, which vanishes with rational coefficients. This vanishing also holds for equivariant homology. Note that this phenomenon can never happen for free orbits, since the $S^1$-action itself provides a trivialization of the bundle of negative eigenspaces, but it does happen for orbits with finite even stabilizer.

The analogy is of course that $h$ corresponds to the action functional on free loop space. Free orbits of critical points correspond to simple periodic (Reeb) orbits, and ones with finite stabilizers to multiple covers of simple orbits.

share|improve this answer

Your Answer

 
discard

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

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