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The topology of spacetimes 
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#55
Jan313, 08:11 AM

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IMHO, offtopic or not, this is by far the best thread in the Relativity section for quite some time. :)



#56
Jan313, 08:15 AM

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I have moved the offtopic posts to a new thread so we can keep discussing this.



#57
Jan313, 10:00 AM

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This leads to a slightly subtle definition of distance in a Riemannian manifold. The distance between points p and q is the greatest bound on the lengths of all "nice" paths between p and q. In my example, 2 is greatest lower bound of the lengths of paths between, even though there is no path of length 2, and 2 is the distance between (1 , 0) and (1 , 0). 


#58
Jan313, 10:23 AM

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#59
Jan313, 10:32 AM

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#61
Jan313, 10:51 AM

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But this is not an iffcondition. For example, (0,1) also has lengthminimizing geodesics but is not complete. 


#62
Jan313, 11:01 AM

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"Riemannian Manifolds: An Introduction to Curvature" by Lee has interesting stuff (again!) about this on pages 108111. For example: 


#63
Jan313, 11:16 AM

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http://www.math.washington.edu/~lee/...ian/errata.pdf 


#64
Jan313, 11:33 AM

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#65
Jan313, 12:13 PM

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A topological manifold is it metrizable, i.e. can its topology be described by a distance (may we use an atlas and [itex]\mathbb{R}^n[/itex] euclidean distance)? If yes, which distance? If not, what is then the topology of space time? Secondly, how can we use the fact that spacetime is not only a topological manifold, but a (pseudo)Riemannian one, to help us on this task? 


#66
Jan313, 01:44 PM

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But the important thing here is to separate the distance function from the topology, it is true that a metric distance function can induce a topology on a metrizable space, but this is not the case with manifolds, wich carry their own topology. 


#67
Jan313, 02:01 PM

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This is confusing. I thought we agreed the distance function on the manifold doesn't distinguish Riemannian metric tensor from semiriemannian metric tensor since they act locally on the tangent space rather than on the global manifold, and differentiable manifolds topological requirements only allow them to be metric spaces (can't be semimetric nor pseudometric spaces by definition, first of all because they are required to be Haussdorf). So I think Dalespam's question are relevant here. This is related to what I commented in posts #41, #44 and #52. So far have been ignored, care to give it a try and address them? Thanks George. 


#68
Jan313, 07:09 PM

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Is a Hausdorff space necessarily a metric space? Wikipedia just says thart pseudometric spaces are typically not Hausdorff, but that seems to allow that Hausdorff spaces can be neither metric nor pseudometric. If that is possible, then wouldn't it be possible that Hausdorff manifolds with pseudoRiemannian metric tensors need not be metric spaces?



#69
Jan313, 07:23 PM

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#70
Jan313, 07:32 PM

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#71
Jan313, 07:39 PM

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#72
Jan313, 09:37 PM

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I mean, we do have an invariant [itex]ds^2[/itex], but this may be positive, negative or null. Then, what is done in physics is that we define our distances simply by [tex] \int_a^bds=\int_a^b\sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\nu}}\geq 0 [/tex] and this only makes sense if b is inside the lightcone defined by a, so that (by using the proper convention sign [itex]\pm[/itex]) we always get a distance properly defined. This may be a simple trick by using the lightcone, but it can be supported by causality arguments that you want to include in our mode, I think. Does this seems good to you? I have some problems, as I'm thinking of the lightcone as being on the manifold, and not in the tangent space as has been argued. 


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