2
$\begingroup$

This is related to a problem that I posed about a year ago. I was given several references by a number of experts who were kind enough to entertain my rather arcane question. Those references were helpful in that they answered my question using modern methods, but my quest to re-derive the solution using only the methods available to relativists in the 1970’s resulted in a dead end. Basically, I wanted to know why a certain gluing procedure, put forth by Robert Wald in his book General Relativity, but which was originated by Yvonne Choquet-Bruhat and James W. York, Jr. in their paper The Cauchy Problem (published in the book General Relativity and Gravitation - One Hundred Years After the Birth of Albert Einstein. Volume 1) produces a Hausdorff space-time.

The theorem that I am interested in is the following, taken from the Bruhat-York paper mentioned above.

Theorem. Every initial data set $ (\Sigma,\gamma,K) $, where $ \gamma \in {H^{\text{loc}}_{s}}(\Sigma) $ and $ K \in {H^{\text{loc}}_{s - 1}}(\Sigma) $, admits a vacuum development $ (V,g) $, where $ g \in {H^{\text{loc}}_{s}}(V) $.

Note: A tensor field on a manifold $ V $ is said to be in $ {H^{\text{loc}}_{s}}(V) $ if its norm is square-integrable, together with its generalized derivatives of order $ \leq s $ on every compact subset of $ V $.

The proof by the two authors is as follows.

Proof: Let $ (\Sigma_{a},\varphi_{a}) $ and $ (\Sigma_{b},\varphi_{b}) $ be two coordinate patches on $ \Sigma $ compatible with its smooth structure, where $ \varphi_{a}: \Sigma_{a} \to \mathbb{R}^{3} $ and $ \varphi_{b}: \Sigma_{b} \to \mathbb{R}^{3} $. Let $ (U_{a},g_{a}) $ and $ (U_{b},g_{b}) $ be $ {H^{\text{loc}}_{s}} $-vacuum developments of $ (\Sigma_{a},\gamma,K) $ and $ (\Sigma_{b},\gamma,K) $ respectively. By local uniqueness, there exists, if $ U_{a} $ and $ U_{b} $ are conveniently reduced, an admissible isometry $ \psi_{ab} $ between the largest development $ u_{ab} \subseteq U_{a} $ of $ {\varphi_{a}}[\Sigma_{a} \cap \Sigma_{b}] $ with the metric $ g_{a} $, and $ u_{ba} \subseteq U_{b} $ of $ {\varphi_{b}}[\Sigma_{a} \cap \Sigma_{b}] $ with the metric $ g_{b} $. We consider now the quotient of $ (U,g) $, the disjoint union of the $ (U_{a},g_{a}) $’s, by the equivalence relation $ x_{a} \sim x_{b} $ if $ x_{a} \in u_{ab} $ and $ x_{b} = {\psi_{ab}}(x_{a}) $. One can show that $ U $ is a smooth manifold$ ^{\dagger} $ and that $ (U,g) $ is a vacuum development of $ (\Sigma,\gamma,K) $. $ \quad \blacksquare $

$ ^{\dagger} $ A smooth identification between open sets $ U_{1} $ and $ U_{2} $ of two smooth manifolds $ V_{1} $ and $ V_{2} $ respectively leads to a (Hausdorff) smooth manifold if no $ x_{1} \in \partial U_{1} $ is an accumulation of points whose images in $ V_{2} $ also have an accumulation point $ x_{2} \in \partial U_{2} $.

I understand the importance of $ ^{\dagger} $, but I am utterly unable to prove that in the proof above, no $ x_{a} \in \partial u_{ab} $ is an accumulation of points whose images in $ U_{b} $ also have an accumulation point $ x_{b} \in \partial u_{ba} $. I also have no idea why, in the proof, there is a need to conveniently reduce $ U_{a} $ and $ U_{b} $.

I am starting to believe that the proof is woefully incomplete or even wrong, so I would appreciate it if someone could offer any suggestions. Thank you!

$\endgroup$
  • $\begingroup$ This might be naive, but this is what I'd do (and after reading the linked question it's what I think Willie suggested)... Restrict the initial data for the local developments in such a way that the resulting "restricted isometries" domain, e.g. $v_{ab}$, is such that $\overline{v_{ab}}\subset u_{ab}$ and the "restricted isometry" is the restriction of $\psi$ to $v_{ab}$. This should all be doable using the uniqueness properties of the local developments. In this case you get the needed condition for Hausdorffness because of the existence of an extension for the "restricted isometry". $\endgroup$ – Ben Whale Aug 10 '14 at 14:47
  • $\begingroup$ Also it's something that Ward would certainly have thought of. (My comment above went over the word limit). $\endgroup$ – Ben Whale Aug 10 '14 at 14:47
  • $\begingroup$ @BenWhale: Hi Ben. I don’t quite understand your comment on restricted isometries. Could you explain why the Hausdorff property is obtained? $\endgroup$ – Leonard Aug 31 '14 at 1:44
  • 2
    $\begingroup$ Hausdorff is actually the hardest part of the proof, in some sense. You may be interested in Jan Sbierski's preprint where in section 3.2 he proves the Hausdorff-ness using a line of argument not too far removed from the one you are thinking about. I am inclined to agree that the original papers may be incomplete when it comes to this point. // Edit Ah: I see I have already recommended this paper to you before. Sorry for the noise. $\endgroup$ – Willie Wong Oct 8 '14 at 12:03
  • 1
    $\begingroup$ @WillieWong: No problem! You’ve been most helpful in my endeavor. I only wanted to receive some expert opinion as to the completeness of the original proofs. Hans Ringström has told me personally that he thinks there are problems with removing non-Hausdorff pairs of points. One problem is that removing such points might split an inextendible causal curve into two parts so that one part of the curve no longer intersects the intended Cauchy hyper-surface. $\endgroup$ – Leonard Oct 9 '14 at 3:58

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.