Timeline for Completeness of asymptotically Euclidean manifolds
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2022 at 3:22 | comment | added | user900940 | Oops, I forgot that Hopf-Rinow is for Riemannian manifolds...my bad. I appreciate it! | |
| Mar 13, 2022 at 2:46 | comment | added | Willie Wong | In the Lorentzian case the proof above doesn't work. Hopf-Rinow is not true in non-Riemannian pseudo-Riemannian manifolds. // However, I am pretty sure that if you assume sufficiently strong decay you can use the Lemme du Bout argument given by Gicquaud's comment. But in this case you probably want sufficiently strong decay of the metric in at least the $C^{1,1}$ sense. | |
| Mar 13, 2022 at 2:14 | comment | added | user900940 | 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. | |
| Mar 11, 2022 at 6:56 | comment | added | Willie Wong | 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. | |
| Mar 10, 2022 at 22:55 | comment | added | user900940 | 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? | |
| Mar 10, 2022 at 18:56 | comment | added | user900940 | Thank you, this makes sense! | |
| Mar 10, 2022 at 18:55 | vote | accept | user900940 | ||
| Mar 10, 2022 at 18:52 | comment | added | Willie Wong | 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).) | |
| Mar 10, 2022 at 18:50 | history | answered | Willie Wong | CC BY-SA 4.0 |