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  1. What can we say about the transition map if the manifold is a Hausdorff space?

  2. Why do we need the manifolds to be Hausdorff and paracompact in General Relativity?

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take a look at this math.se question: math.stackexchange.com/q/98105 –  user37496 Apr 24 at 16:57

3 Answers 3

You can't do calculus on a manifold that is not Hausdorff and paracompact. If you can't do calculus, doing physics with field equations is pretty pointless.

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Why can't you do calculus if the manifold is not Hausdorff? The manifold will still be locally Hausdorff. –  Pratyush Sarkar Oct 9 at 21:52

Regarding your first question, please state it more clearly.

Without Hausdorff property, uniqueness property of limits of sequences fails and it implies many bad consequences for several results concerning abstract uniqueness, e.g. of solutions of differential equations on manifolds. Moreover, without Hausdorff you do not have smooth hat functions that are useful in extending local smooth tensor fields to global smooth ones. The definition of tangent vectors as derivatives over the ring of globally defined smooth functions turns out to be a bit cumbersome (even if is possible and that way is effectively followed in complex algebraic geometry, for instance, or also for real analytic manifolds).

Finally, in absence of paracompactness you cannot construct smooth partitions of the unity and defining a notion of integral is difficult.

(modified after Ricky Demer's comment)

It is important to know that a Hausdorff, second countable, locally homeomorphic to $\mathbb R^n$ space is paracompact. Conversely, a Hausdorff, locally homeomorphic to $\mathbb R^n$, paracompact space is secondly countable iff its connected components are countable (so, in particular when the space is connected).

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I think disconnected unions of uncountably many copies of $\mathbb{R}^n$ are counterexamples to your last paragraph. $\;$ –  Ricky Demer Apr 24 at 20:30
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@Ricky Demer It is proved in Kobayashi-Nomizu's book on differential geometry (I think the first volume) and it is quoted at the end of Appendix A of Wald's textbook on GR. I cannot check now, tomorrow I will do. Maybe it is necessary to include the hypotheses of connectedness... –  Valter Moretti Apr 24 at 21:34
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Indeed, it is necessary that the connected components are countable to get second countability from paracompactness (for a topological manifold). Thanks for your remark. –  Valter Moretti Apr 24 at 22:25
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@RickyDemer Not really relevant, but cute and funny: Henrik Schlicht of U Copenhagen classifies Lie groups into three classes, the "good", the "bad" and the "ugly", with Sergio Leone's last epithet going to example like yours: see math.ku.dk/~schlicht/Liegroups/gbu.pdf : I certainly giggled heartily at that one. –  WetSavannaAnimal aka Rod Vance Apr 25 at 3:46

To add to Jerry Schirmer's answer: Roger Penrose's "Road To Reality" has a really grand discussion, with sketches in section 12.2 of what non-Hausdorff would mean. I'd really recommend you should read and ponder it. Unless the Hausdorff axiom is fulfilled, "branching" would arise at the edge of a transition region where "co-ordinate patches" (i.e. charts) overlap. Distinct points "welded together" by indivsible open sets could make jumps between distinct points "continuous" and the notion of "topology" could not be consistent with that of the individual overlapping patches. That is, the charts are homoeomorphisms (diffeomorphisms in relativity) between open neighbourhoods of the origin in $\mathbb{R}^N$ and subsets of the manifold and thus, when we bestow the local topology of $\mathbb{R}^N$ on each patch individually through the co-ordinate function, the topology of each patch separately makes it indivdually Hausdorff, which would lead to an "inconsistency" if the patches were glued together in a way making them non-Hausdorff. If you tried to make calculus definitions on such a beast, I can't see how you would do it without introducing some kind of an equivalence relationship to deem all distinct points in indivisible open sets to be the "same": in other words, you would end up forcing the Hausdorff axiom before you began your serious work. Now this could simply be lack of mathematical imagination on my part, but I don't believe I've ever seen manifolds seriously worked with without the Hausdorff axiom, even though I've seen some authors say things like (pretentiously IMO) "it is customary to assume the Hausdorff axiom" (like: OK buster, stop trying to sound superior by sounding like we make the Hausdorff assumption to make things easy for us "trivial beings" and that you know all about non-Hausdorff manifolds and give me a concrete example of how they might work!). See the inimitble Sir Penrose's account below.

Penrose Hausdorff Discussion

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