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I am studying the existence of harmonic coordinates at infinity on an asymptotically flat manifold. My Reference papers are, The Mass of Asymptotically Flat Manifold, by Bartnik [B] and The Yamabe Problem by Lee and Parker [LP].

Some definitions:

$$W^{k,p}_\delta =\{f\in L^p_{loc}(\mathbb{R}^n) \, \vert \, \sum_{i\leq k} \Vert \nabla^i f\Vert_{p,\delta-i} <+\infty \}$$ where

$$\Vert f\Vert_{p,\delta}^p= \int_{\mathbb{R}^n} \vert f\vert^p \sigma^{-p\delta-n}\ dx,$$ where $\sigma= 1+\vert x\vert$.

Exceptionnal values are the order of decreasing of harmonic function at infinity, namely $\mathbb{Z}\setminus\{-1, \dots,-n-1\}$.

A metric is assumptotically flat if there exists a compact $K$ and a chart $\phi :M\setminus K \rightarrow \mathbb{R}^n\setminus B_1$ such that $\phi_*(g)$ is uniformly equivalent to $\delta$ and $\phi_*(g) -\delta \in W^{1,q}_{-\tau}$ for some $\tau \geq 0$ and $q>n$, the decrease rate of the metric.

Both state that if $\tau$ is positive and $1<p\leq q$ then $\Delta_g:W^{2,p}_\delta \rightarrow L^{p}_{\delta-2}$ is surjective if $\delta>2-n$ and non-exceptional, and injective if $\delta<0$ and non-exceptional. As consequence of the Fredhom theory and explicit knowledge of the Kernel of the flat laplacian. I totally agree with this.

Then in order to prove the existence of harmonic coordinates at infinity , their statements diverge a bit. They both remark, as consequence of the definition of asymptotic flatness, that $$\Delta_g x^i\in L^{p}_{-1-\tau},$$ which I agree too. But Bartnik say it is enough to solve $$\Delta_g v_i =\Delta_g x^i$$ in $W^{1,p}_{1-\tau}$, but we should have $1-\tau >2-n$, i.e. $\tau < n-1$ , which is specify nowhere. And Lee and Parker make almost the same for $n\geq 4$ but replace $1-\tau$ by $1-\tau +\epsilon$ when $n=3$, with $\epsilon$ in order to have $1-\tau +\epsilon>n-2$ which seems more reasonable but I don't see why they don't face this issue in dimension $\geq4$.

I would appreciate any enlightenment or clearer reference about this. Thx

EDIT: I think that the sentence " ${\Delta_g}_{\vert W^{2,p}_{1-\tau}}$ has trivial coKernel so there is $v_i \in W^{2,p}_{1-\tau}$ such that $\Delta_g (v^i -x^i)=0$" page 678 of [B]... is absolutely not correct (or I miss something). The real reason why this equation have solution is that $x^i$ is itself an harmonic function for the flat laplacian which gives that $\Delta_g x^i$ is orthogonal to its CoKernel.

does someone agree?

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    $\begingroup$ You are looking at the wrong end point, I think for the Lee and Parker statement. The replacement by $1-\tau + \epsilon$ when $n = 3$ is because $\tau > (n-2)/2$, the condition they did suppose, allows $\tau < 1$ when $n = 3$, in which case $1-\tau > 0$. // I think for practical purposes you can suppose that $1 - \tau > 2 - n$ throughout, since for the PMT the mass term appears exactly at the level $\tau = n- 2 < n-1$ and similarly for the Yamabe problem (see Theorem 6.5 in Lee and Parker). $\endgroup$
    – Willie Wong
    Commented Jan 30, 2018 at 19:34
  • $\begingroup$ @WillieWong I agree that it should work for the PMT. However, I would like a general staement. Moreover I don't understand why they need $1-\tau$ should be negative since we care about surjectivity and not injectivity. $\endgroup$
    – Paul
    Commented Jan 30, 2018 at 20:59
  • $\begingroup$ @AntonPetrunin I am going to edit my post. $\endgroup$
    – Paul
    Commented Jan 30, 2018 at 21:00
  • $\begingroup$ The requirement of injectivity is presumably useful to obtain quantitative bounds, as well as for proving uniqueness results (for example, I think the uniqueness of the harmonic coordinates for sufficiently large $\tau$ (or sufficiently small $1-\tau$) is used in the proof of Theorem 9.5 in Lee and Parker). $\endgroup$
    – Willie Wong
    Commented Jan 31, 2018 at 16:50
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    $\begingroup$ For existence: using the trivial embeddings $W^{k,p}_\tau \subset W^{k,p}_{\sigma}$ if $\tau < \sigma$, if $1 - \tau < 2-n$ you at least have the existence of harmonic coordinates with "worse decay". In terms of the elliptic theory you can't really do better: just think about solving the Poisson equation $\triangle \phi = f$ on $\mathbb{R}^n$ for $n \geq 3$. Even if $f$ has compact support, the best you can generally expect is that $\phi$ decays like the Newton potential at rate $r^{2-n}$. So the obvious interpretation is really the best you can do. $\endgroup$
    – Willie Wong
    Commented Jan 31, 2018 at 17:03

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