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The following is a well-known result for elliptic operators.

Theorem. Let $P: \Gamma(E)\to \Gamma(F)$ be an elliptic operator of order $m$ between vector bundles $E$ and $F$ over a compact manifold $X$. Then $P$ extends to a Fredholm map $P: W^{k,2}(E) \to W^{k-m,2}(F)$ whose index is independent of $k$ (cf. p193. Lawson-Michelsohn's book 'Spin Geometry')

This kind of extension involves Hilbert spaces and thus is relatively not hard to work with. However, concerning Floer theory (or pseudoholomorphic curvers theory), the linearized maps we concern have to be between non-Hilbert Banach spaces, such as $W^{1,p}(\mathbb R \times S^1, u^*TM)$ or $L^p(S^2, \Lambda^{0,1} \otimes_J u^*TM)$, where $p$ is always assumed to be greater than 2.

In the case $p\neq 2$, it seems much harder to show any Fredholm properties, and the proofs are usually not natural but very technical. You may agree to me if you refer to Audin-Damian's book(section 8.7) or McDuff-Salamon's book(Proposition 3.1.11 which requires Theorem C.2.3) respectively.

So, I would like to ask does there exist any general theorem concerning Fredholm properties, which is similar to the above one? If exist, what could be a general principle to show the Fredholm properties? In the above theorem, can we relax the condition that the base manifold $X$ is compact, as in Floer theory, the base manifold $\mathbb R \times S^1$ is non-compact? Any great reference?

Edit: I recently found a paper (cf. (0.1) in the very beginning) which states that (if I haven't misunderstood): let $E$ and $F$ be two vector bundles over a compact manifold $X$ and let $P: \Gamma(E)\to \Gamma(F)$ is an elliptic differential operator of order $m$, then the paper claims that it is well-known that it has a Fredholm extension $$ P:W^{k+m,p}(E) \to W^{k,p}(F)$$ for every $1<p<\infty$. This should be a really good result but I have trouble in finding a serious reference to confirm this.

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    $\begingroup$ There are some general theorems you can use. Google Sholomo Sternberg and "Fredholm property", and you will find very nice typed-up slides originated by Zworji. In general this is related to a subject called analytical Fredholm theory, but I think for $p\not=2$ case specifically you do not really need so much machinery. The reason we had $2$ at here is (partly) due to Hormander's square root trick. So I think in general you will get similar statements if your manifold is not too bad. $\endgroup$ – Bombyx mori Jun 5 '17 at 1:49
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    $\begingroup$ @Hang Regarding the last part of your question, Is not the the Laplacian $\Delta$ on $X=\mathbb{R}^2\setminus \{0\} \simeq \mathbb{R} \times S^1$ a counter example, since the space of Harmonic functions is an infinite dimensional space? $\endgroup$ – Ali Taghavi Jun 20 '17 at 18:56
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    $\begingroup$ You can compactify the domain of a Floer cylinder, and as far as I know the complications in showing that something is Fredholm in Floer theory do not stem from $p\neq2$ but from the fact that elliptic does not imply Fredholm in the presence of a boundary. This is why you need 'asymptotic boundary conditions' which go into the Fredholm theory. In particular these boundary condition are not necessarily local. I believe there exists some theory for elliptic operators on domains with boundary, but it doesn't seem to be applied in the Floer theory literature (yet). $\endgroup$ – Jan-David Salchow Jul 12 '17 at 6:40
  • $\begingroup$ For Floer theory, you can almost always use Sobolev spaces with higher differentiability (i.e. $W^{k,p}$ with large $k$), and then there is no problem to take $p=2$. $\endgroup$ – John Pardon Oct 18 '17 at 2:33
  • $\begingroup$ (Perhaps you could ask another question if there's some specific part of Floer theory which you don't know how to do with "large $k$" Sobolev spaces). $\endgroup$ – John Pardon Oct 18 '17 at 2:36
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See Chapter 3 of Matthias Schwarz' thesis or Chapter 3 of Donaldson's book on Floer homology. Both do L^p theory on cylindrical end manifolds (following Maz'ya etc.)

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  • $\begingroup$ I take a look on Google books but I did not see any Fredholm theory being involved. Thanks for the reference. $\endgroup$ – Bombyx mori Jun 14 '17 at 19:26
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Too long for a comment:

The paper you linked of Lockhart-McOwen shows you the faults for noncompact manifolds (independent of $p\ne 2$). Look up the Atiyah–Patodi–Singer spectral boundary-value problem. We need asymptotic conditions (or boundary conditions, there is a way to pass between the two) otherwise the kernel of our operator is infinite-dimensional. I've learned the most from reading the analysis chapters in Kronheimer-Mrowka's "Monopoles and 3-manifolds", specifically Chapter 17.

About your question for $p>2$ and $X$ compact, I don't know in general. We can use the Rellich–Kondrachov embedding theorem to get compactness for $W^{k+m,\,p}\to W^{k,\,p}$ with $m\ge0$. It then suffices to prove (what I call) this "Calderon–Zygmund" inequality to get Fredholmness: $||s||_{k+m,\,p} \le C(||Ps||_{k,\,p} + ||s||_{k,\,p})$. To my understanding, this is a rephrasal of finding a parametrix (Green's operator) for $P$ such that $QP-1$ (and $PQ-1$) extends to $W^{k,\,p}$. In other words, we know that an operator is Fredholm when it is invertible up to compact operators, and an elliptic operator $P$ admits a parametrix $Q$ with $QP-1$ and $QP-1$ "smoothing operators". Working over compact manifolds, the hope is that these smoothing operators have a continuous extension to $W^{k,\,p}$. Is that granted by the property "smoothing"? This is probably basic (i.e. the proof using $p=2$ might allow freedom in $p$).

There is a trick in Hormander's "Analysis of linear PDO's III", Corollary B.1.6, which shows how you can take knowledge of your (linear) elliptic operator on $W^{k,2}$ and get the same knowledge over certain Besov spaces $B^{p,k}$ for any $1\le p\le\infty$, but I don't think this is good enough for $W^{k,\,p}$ (that is, I don't think the norms are equivalent (and if it were then we'd get knowledge for $p=1,\infty$ which seems too strong)). But I can very easily be wrong, maybe this gets us what we want... in either case there should be a more direct route.

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  • $\begingroup$ Thanks for the reference, I did not know! $\endgroup$ – Bombyx mori Jul 18 '17 at 4:53

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