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Let $M = [1,\infty)\times S^2$. Let $\Omega$ be any smooth positive function on $S^2$. Is the metric $dr^2+ \Omega^2 r^2 g_\text{round}$ asymptotically flat (where $g_\text{round}$ is the round metric on $S^2$)? If not, what assumptions do we need on $\Omega$ for the metric to be asymptotically flat?

Any references are really appreciated.

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    $\begingroup$ You should specify what you mean by asymptotically flat, but if you mean that there is a chart at infinity so that $|\nabla^k(g-g_\textrm{flat})| = o(r^{-k})$ then the only example should be if $\Omega^2 g_\textrm{round}$ is has curvature $K=1$. To see this, you can compute the sectional curvature of your metric and see that it is $o(r^{-2})$ if and only if $K_{\Omega^2 g_\textrm{round}} = 1$. On the other hand, from asymptotic flatness, curvature must be $o(r^{-2})$. $\endgroup$ Dec 5, 2021 at 17:47
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    $\begingroup$ It looks like the Ricci of the metric is $Ric = \frac{1}{r^2} (K_{\Omega^2 g_{\text{round}}}-1) g$ on vector fields tangent to the sphere and $Ric(\partial_r, \cdot) =0$. So the Ricci curvature does decay even if $K_{\Omega^2g_{\text{round}}} \neq 1$. $\endgroup$
    – Laithy
    Dec 6, 2021 at 0:52
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    $\begingroup$ Agreed that any metric has decaying curvature, but the crucial point is the rate of decay. Most metrics don't have fast enough decay to be asymptotically flat. $\endgroup$ Dec 6, 2021 at 1:01
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    $\begingroup$ Oh I see. Suppose I define asymptotically flat as follows: there exists a coordinate system such that the metric satisfies $g_{ij} = \delta_{ij} + O_2(r^{-\tau})$ (where $\tau>0$). Then in particular, that means that $|Ric| = O(r^{-\tau-2})$. And so by observing that the Ricci curvature of our metric doesn't satisfy that, we conclude that $g$ is not asymptotically flat. Is that a correct argument or is that nonsense? $\endgroup$
    – Laithy
    Dec 6, 2021 at 1:11
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    $\begingroup$ exactly, that's correct. $\endgroup$ Dec 6, 2021 at 4:04

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