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Strongly related to the question Prove isometry preserving excision is Killing-like?

There the question was (loosely): if I drill a smooth 4D hole through Minkowski spacetime, is a timelike isometry preserved only if the hold drilled is everywhere parallel to (bounded by) integral curves of the vector field generating the isometry?

@Valter-Moretti re-expressed the proposition rather better and then proved it in exemplary fashion.

I was just writing up some notes on this and then wrote (assuming the hole boundary is in fact smooth),

"it is obvious, however, that if the isometry condition is relaxed, i.e. the metric is allowed to become non-flat, then diffeomorphism between spacelike hypersurfaces may be retained."

And then - in classic fashion - wondered, "Obvious?"

Hence the question: is it "obvious"??

(If so... from what perspective? :) )

UPDATE - SHOWING SOME EFFORT

After further thought, how about this (also referencing the Q&A What's the difference between a manifold and a topological space?).

In answer to the question Are homeomorphic differentiable manifolds actually diffeomorphic?, @Mariano-Suárez-Álvarez said,

Now, if f:M→N is a homeomorphism of smooth manifolds, you can always «adjust» the smooth structures so that f becomes a diffeo: indeed,

and offered the OP there the suggestion

you should be able to prove the following: if M is a smooth manifold, N a topological space and f:M→N a homeomorphism, then there is a structure of smooth manifold on N such that f becomes a diffeomorphism.

Now, since a manifold is a topological space with additional structure, it would seem that for any pair of spacelikes hypersurface, a) they are homeomorphic by construction above, and b) both are already (sub)manifolds and hence toplogical spaces, so it follows immediately (?).

(I do not have the skill to generate the proof referred to though)

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  • $\begingroup$ But, but, if you drill a hole through Minkowski, you'll get Schwarzschild! $\endgroup$ – Dr. Ikjyot Singh Kohli Aug 24 '17 at 15:00
  • $\begingroup$ @Dr.IkjyotSinghKohli No, a) singularity (if considered the excision) $\times \mathbb{R}$ is 2D not 4D, the hypothesis), b) No reference to the Einstein Field Equations so the metric is unconstrained in that way, c) the excision is merely smooth, no symmetry constraint was imposed, and even if EFEs applied, Schwarzschild is a spatially spherically symmetric solution, d) even if EFE and spherically symmetric there are other solutions: e.g. Kerr, Kerr-Newman. $\endgroup$ – Julian Moore Aug 25 '17 at 6:57

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