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Say that you place an asymptotically Euclidean metric on $\mathbb{R}^3,$ e.g. $\mathbb{R}^3$ is endowed Riemannian metric $g$ such that $\text{supp}(g^{ij}-\delta^{ij})\subseteq\{|x|\leq R\}$ for some large $R>0.$ Does it follow that $(\mathbb{R}^3,g)$ is geodesically complete? This seems to be applied in PDE literature quite often when working with the flow generated by the principle symbol of the Laplace-Beltrami operator, but I am unaware of a reference. It seems intuitively true: $\mathbb{R}^3=\{|x|\leq R\}\cup \{|x|>R\}.$ The first set in the union is compact, and the other is where our metric is just the standard metric.

If this is a simple question, please let me know in the comments and I will delete.

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  • $\begingroup$ My thoughts are: if a vector field is bounded wrt a metric which is geodesically complete, it is complete. You can to apply this to your situation by looking at the vector field generating the geodesic flow ( on the tangent bundle of R^n) $\endgroup$
    – Thomas Rot
    Commented Mar 10, 2022 at 17:57
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    $\begingroup$ Yes. The idea is that the unit tangent bundle is locally compact so if you look at the geodesic flow, if you have an incomplete vector field it must leave any compact (in French this is known as the "lemme du bout" but I could not find the English translation). This can only happen if the geodesic leaves any compact of $\mathbb{R}^3$ and this in turns implies that the length of the geodesic must be infinite, which contradicts incompleteness... $\endgroup$
    – Romain Gicquaud
    Commented Mar 10, 2022 at 18:03
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    $\begingroup$ @RomainGicquaud I don't think it has a standard name in English. Hartman calls (the contrapositive) of the Lemme du Bout as the "extension theorem". Jack Hale's book calls it the "continuation theorem". (This is more an ODE theorem than a geometry one.) $\endgroup$
    – Willie Wong
    Commented Mar 10, 2022 at 19:04

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Yes, and it follows from Hopf-Rinow (let's assume $g$ is at least $C^2$ so that there's no ambiguity about the geodesic flow).

Since $g$ and $\delta$ differs only on a compact set, you have the the lengths defined by $g$ and $\delta$ are globally comparable, that is, there exists $\lambda, \Lambda > 0$ such that for every $x\in \mathbb{R}^3$ and $\xi\in T_x\mathbb{R}^3$ you have

$$ \lambda \delta(\xi,\xi) \leq g(\xi,\xi) \leq \Lambda \delta(\xi,\xi) $$

and so the metric topologies are equivalent. Thus

A set $K$ is closed and bounded w.r.t. $g$

IFF

$K$ is closed and bounded w.r.t $\delta$

IFF

$K$ is compact.

And by Hopf-Rinow you have $g$ is complete.

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  • $\begingroup$ Note that asymptotically Euclidean is not necessary, all that's required is the uniform comparability between the two metrics. (If the $g$ is not bounded w.r.t. to $\delta$, then it is not a continuous Riemannian metric; if $g$ is not coercive on $\delta$, then you have counterexamples (think $g|_x = e^{-|x|^2} \delta|_x$ which is incomplete).) $\endgroup$
    – Willie Wong
    Commented Mar 10, 2022 at 18:52
  • $\begingroup$ Thank you, this makes sense! $\endgroup$
    – user900940
    Commented Mar 10, 2022 at 18:56
  • $\begingroup$ Out of curiosity, do you know how the argument can be modified when we instead modify the assumption on the metric to be that $g$ tends towards flat at some rate dependent on the radius (and added decay for derivatives, say $|\partial_\alpha(g-\delta)|=\mathcal{O}(r^{-|\alpha|-\epsilon})$)? This is closer to the type of assumption that I see on the metric. It seems like the two metrics are uniformly comparable by continuity, is that correct? $\endgroup$
    – user900940
    Commented Mar 10, 2022 at 22:55
  • $\begingroup$ Yes, and you don't even necessarily need the higher derivative bounds. The bounds $|g-\delta| = O(r^{-\epsilon})$ is enough to show the uniform comparability. $\endgroup$
    – Willie Wong
    Commented Mar 11, 2022 at 6:56
  • $\begingroup$ I've been thinking about this a bit more, and I've been trying to figure out what goes wrong in the case of asymptotic flatness. If we assume the same conditions as above but $\delta$ is the Minkowski metric (say that $g$ has time-independent coefficients, as well), then it seems like one should get geodesic completeness. $\endgroup$
    – user900940
    Commented Mar 13, 2022 at 2:14

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