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By the Arnold Conjecture, I mean the following statement:

Let $M$ be a closed symplectic manifold, and $\phi:M\to M$ a Hamiltonian symplectomorphism with nondegenerate fixed points. Then $ \# \operatorname{Fix} (\phi) \geq \dim H_\bullet (M;\mathbb Q)$.

The proof is via defining the "Floer homology" $HF(M;H)$ of $M$ with respect to any $1$-periodic Hamiltonian $H:M\times\mathbb S^1\to\mathbb R$, which is then shown to be independent of $H$ (and thus can be written as $HF(M)$). There are some technicalities involved in this definition (Novikov rings, transversality, bubbling, etc.), which for the purposes of this question I would like to ignore.

The Arnold Conjecture is proven by exhibiting an isomorphism $HF(M)=H(M;\mathbb Q)$. My question is about whether this can be proven without "$\mathbb S^1$-localization". I know of the following two proofs:

  1. Floer's original proof is as follows. We use the Hamiltonian $H(x,t)=\epsilon\cdot f(x)$ for some Morse--Smale function $f:M\to\mathbb R$. Then the generators of the Floer chain complex correspond to the critical points of $f$, so it remains to identify the differential with the Morse--Smale differential. Floer exploits an $\mathbb S^1$-symmetric in the problem to show that any part of the Floer differential which does not come from a Morse--Smale flow line must carry a free $\mathbb S^1$-action. Since we only count zero-dimensional moduli spaces for the differential, this is enough since the only $0$-manifold with a free action of $\mathbb S^1$ is the empty set.

  2. The PSS isomorphism (Piunikhin-Salamon-Schwarz). I won't write the details on this, but it is another approach to the isomorphism, which also uses the freeness of the action of $\mathbb S^1$ on "bad" moduli spaces to show they do not contribute to certain maps (EDIT: this is wrong, the PSS approach yields an isomorphism without appealing to $S^1$-localization; see Thomas Kragh's answer below). See, for example, these notes by Salamon: http://www.math.ethz.ch/~salamon/PREPRINTS/floer.pdf

Both of these proofs use $\mathbb S^1$-localization, meaning some statement to the effect that a free $\mathbb S^1$ action on a zero-dimensional moduli space implies the fundamental class is zero.

Are there approaches which genuinely avoid $\mathbb S^1$-localization?

In fact, I have heard of two such approaches, but so far have not understood either of them:

  1. Katrin Wehrheim in research statement http://math.mit.edu/~katrin/slides/research.pdf (see section 5.6) says she and Albers and Fish have such an approach, summarized in some slides http://www-math.mit.edu/∼katrin/slides/arnold.pdf, but the paper ("Polyfold Proof of the Weak Arnold Conjecture") remains unreleased as far as I can tell (probably the slides contain something of an outline of their strategy, but I have yet to read them carefully).

  2. Fukaya--Oh--Ohta--Ono in Remark 31.18 of their recent paper http://arxiv.org/abs/1209.4410 say there is such an approach, but it is based on the machinery of their books "Lagrangian intersection Floer theory: anomaly and obstruction I - II" which are beyond my current knowledge.

Is there a down-to-earth explanation of how these approaches to the isomorphism $HF(M)=H(M;\mathbb Q)$ manage to avoid $\mathbb S^1$-localization arguments (which seem crucial for both Floer's approach and the PSS approach)? Do the approaches of Albers--Fish--Wehrheim and Fukaya--Oh--Ohta--Ono genuinely get around the need for $\mathbb S^1$-localization, or is the same difficulty hiding somewhere else?

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As a fan of $S^1$-localization, I'm curious why you want to avoid it -- it seems like the easy part! Or is it that you don't, and are just curious whether these approaches actually manage to do so? –  Allen Knutson Jun 25 '13 at 13:55
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For proving the Arnold conjecture in full generality, one runs into situations where one cannot achieve transversality by using a generic $J$. This makes the $\mathbb S^1$ localization step a lot harder, basically. –  John Pardon Jun 25 '13 at 15:59

2 Answers 2

up vote 6 down vote accepted

I was thinking about this on and of for a long time. However, after discussing it with several people (mostly Mohammed Abouzaid) I have arrived at the following explanation:

The PSS approach works without having to consider S^1 actions.

First of all it is very very important in this discussion not to ignore Novikov rings. Indeed, an alternative to Novikov rings is to define the action integral on a covering space of the free loop space. This means that the constant curves are represented several times (on different layers of the covering space - the layers are encoded in the Novikov coefficients). Let $R$ be the Novikov ring.

Now look at the Hamiltonian $\epsilon f$. Then all the critical points of $f$ determines infinitely many critical points of the action integral, and the fact that $FH_* (M)$ is $H_*(M;R)$ is not at all clear. Indeed the differential in the Morse complex for the action counts the same differentials as those for $f$ (on each layer), but there might exist differentials going from one layer to a lower layer, which cancels parts of the Homology.

Alternatively one can think of an action filtration on the action such that page two of the associated spectral sequence is a copy of $H_*(M;\mathbb{Z})$ per layer, but then might have differentials on higher pages canceling parts of this picture, and thus we would not have a PSS isomorphism.

Now, the PSS isomorphism counts the moduli space of discs with a tail starting with a critical value of $f$ going to a critical value of the action (after some perturbation perhaps). The fact that this respects the Novikov coefficients is equivalent in the covering space picture to the fact that even though the action is only defined on the covering space its gradient descents to the actual loop space. This also explains why the differential in $FC_* (M)$ respects the Novikiv coefficients.

The fact that this is a chain map can be proved without using any $S^1$ action hand waving. Indeed, it can be proved using standard 2 dimensional moduli space hand waving (some one correct me if I am wrong here) similar to that of $d^2=0$.

The fact that this is a chain map is enough to see that it is an isomorphisms. Indeed, on the level of chain complexes it is now an isomorphism. Indeed, by considering the action this is an upper triangular matrix with 1's on the diagonal (albeit infinite).

Note, that in this argument all the subtlety of the isomorphism is encoded in the fact that the map $MC_* (M,R) \to FC_* (M)$ is in fact a well-defined chain map which is the "identity" on the generators plus lower action critical terms.

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The very notes of Salomon you are linking describe the approach to prove of weak Arnold conjecture (statement you made) without using localization. First one establishes existence of continuation quasi-isomorphisms $CF (H_0) \to CF(H_1)$ for a pair of regular $H_0, H_1$. This is a lot like PSS map. Then for $H_1$, $C^2$ small, and time independent all periodic orbits correspond to critical points of $H_2$, and all Floer trajectories, correspond to negative gradient trajectories for $H_2$ (depending on conventions for action, etc). That does it. I believe full details are in the book Hofer-Zehnder on Hamiltonian dynamics.

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This is essentially the first argument I describe in the question statement, and as far as I know, it needs $S^1$-localization to be complete. The problem is with your statement "all Floer trajectories correspond to negative gradient trajectories". This is false: there can be extra trajectories obtained by gluing a J-holomorphic sphere to a Morse-type (gradient) trajectory. It turns out that these extra trajectories contribute zero to the differential, but I don't know how to prove this without using $S^1$-localization. –  John Pardon Nov 7 '13 at 17:52
    
Sorry. I was under impression that you are working with nice symplectic manifolds, e.g. semi-positive. In this case the moduli spaces can be regularized so that there are no nodal elements of the type you describe in Fredholm index 1 and 2 moduli spaces (substract 1 to get the dimension of the moduli space). It has been a while since I've done this though so I may be missing something. Like I said the Hofer-Zehnder book is a very detailed reference from what I remember. –  Yasha Nov 6 at 17:17

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