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Dec 11, 2023 at 5:47 history edited Laithy CC BY-SA 4.0
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Dec 10, 2023 at 4:20 comment added Laithy The idea is that if you a priori know that your vector field $X$ goes to $0$ as $r$ goes to $\infty$, then $X$ is conformal Killing implies that it's the $0$ vector field. Check the footnote in pg 324.
Dec 9, 2023 at 14:42 comment added Tim Carson I'm confused by the argument for the case when $g$ is asymptotically flat. It appears in the cited paper as: "On an asymptotically flat manifold $Z^a$ must approach one must approach asymptotically one of the familiar translation or rotation Killing vectors of Euclidean space." How do we know that the limit is a non-zero Killing vector field? Indeed, isn't that what we're trying to prove?
Dec 9, 2023 at 0:20 history asked Laithy CC BY-SA 4.0