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Why positive curvature implies finite universe? 
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#1
Nov1312, 04:20 AM

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This post in influenced by 3 new threads in our cosmology forum. Recent observational data favors positive curvature of our Universe.
The question I have, however, is why positive curvature implies spatially finite Universe? Yes, it might look quite obvious if we embed curved space into higher dimensional flat space. But we can do it, but not must do it  we can work with GR without embedding, am I correct? Take Klein bottle as example. You can't correctly embed it into 3D space without having intersections with itself. Still it is a valid mathematical object, when you forget about intersections. I tend to believe that 'intersections' require additional axiom, saying that "when 2 different points of space can be mapped to the same point in the higher dimensional space, then it is the same point in lower dimensional space too". Without this axiom, space with positive curvature can be infinite  compare a circle and an infinite spring  they both have the same positive curvature... Also without embedding there is yet another option. For the space with constant positive curvature at least 2 finite configurations exist: sphere and halfsphere (sphere cut in half, with diametrically opposite points interconnected on a 'cut' side). So without embedding, full information about curvature doesn't give us the volume! So how do we know that our favorite 'balloon' is not cut in half? 


#2
Nov1312, 04:57 AM

P: 3,009

Well, one has to assume some topological conditions other than positive curvature to infer spatial finitenes, like (I'm not sure about all of them): orientability, selfconnectedness, compactness... and others.



#3
Nov1312, 10:32 AM

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PF Gold
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#4
Nov1412, 10:58 AM

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Thanks
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Why positive curvature implies finite universe?
Seems like you should still be able to conclude there's a finite volume under certain assumptions, even if the curvature is not constant. But what exactly does positive or negative curvature mean in that case? Since the curvature of a 3space (the Ricci tensor) has 6 components.



#5
Nov1412, 11:37 AM

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PF Gold
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I wonder if some form of Myer's theorem is known for psuedoriemannian geometry in 4 dimensions:
http://en.wikipedia.org/wiki/Myers_theorem (Positive Ricci curvature for above states that the Ricci tensor contracted with any unit tangent vector (twice) is positive). 


#6
Nov1412, 12:10 PM

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PF Gold
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http://intlpress.com/JDG/archive/1979/141105.pdf 


#7
Nov1412, 06:16 PM

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#8
Nov1612, 05:19 AM

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Whats about the following toy model.
I have sphere in 3dimensional space. Every point on it is defined by 2 polar coordinates P and Q. Let's map P and Q into a plane. We assume that point with P=2*pi is the same as point with P=0, and Q=2*pi the same as Q=0. But this is an extra assumption. Let's say that area covered by P and Q is infinite and is not looped to itself. So, after travelling all around the globe, traveler finds himself in some different place, not at the same point... This model looks selfconsistent to me, even it can't be correctly embedded. The question is, is it consistent with GR? 


#9
Nov1612, 11:01 AM

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#10
Nov1612, 12:08 PM

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But look my previous post: my toy Universe has finite diameter, but it is not spacially finite!!!



#11
Nov1612, 01:20 PM

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PF Gold
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OK, so you propose that there is a 'funny 2surface' with constant positive curvature everywhere, that is not closed. Well, that contradicts Meyer's theorem. Forget embedding, this means that if you go through the steps to actually set it up as manifold with the proposed properties, you will fail. I suspect where you must fail is that in defining the open sets for your magical surface, you get a contradiction  some point must be near and not near some other point, at the same time. 


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