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Why positive curvature implies finite universe?

by Dmitry67
Tags: curvature, finite, implies, positive, universe
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Dmitry67
#1
Nov13-12, 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 half-sphere (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?
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TrickyDicky
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Nov13-12, 04:57 AM
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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, self-connectedness, compactness... and others.
bcrowell
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Nov13-12, 10:32 AM
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Quote Quote by Dmitry67 View Post
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?
Embedding is irrelevant. Differential geometry has a broad class of theorems that relate curvature to topology. One of them is (IIRC, could be getting the details wrong) that if you have a simply connected three-dimensional Riemannian space with constant, positive curvature, it has the topology of a three-sphere. In a cosmology with positive spatial curvature, this applies to the surfaces of constant cosmological time (i.e., the preferred time coordinate defined by the symmetry of the spacetime).

Bill_K
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Nov14-12, 10:58 AM
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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 3-space (the Ricci tensor) has 6 components.
PAllen
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Nov14-12, 11:37 AM
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I wonder if some form of Myer's theorem is known for psuedo-riemannian 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).
PAllen
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Nov14-12, 12:10 PM
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I wonder if some form of Myer's theorem is known for psuedo-riemannian 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).
And this appears to be an appropriate generalization:

http://intlpress.com/JDG/archive/1979/14-1-105.pdf
bcrowell
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Nov14-12, 06:16 PM
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Quote Quote by Bill_K View Post
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 3-space (the Ricci tensor) has 6 components.
I think one way of stating it is that for any two orthogonal directions a and b, [itex]R^a_{bab}[/itex] (no implied sum) has a certain sign. But I'm pretty sure that a positive curvature can't be sufficient to prove anything about finite volume. E.g., a hyperboloid of one sheet would be a counterexample in two dimensions.
Dmitry67
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Nov16-12, 05:19 AM
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Whats about the following toy model.

I have sphere in 3-dimensional 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 self-consistent to me, even it can't be correctly embedded. The question is, is it consistent with GR?
PAllen
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Nov16-12, 11:01 AM
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Quote Quote by Dmitry67 View Post
Whats about the following toy model.

I have sphere in 3-dimensional 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 self-consistent to me, even it can't be correctly embedded. The question is, is it consistent with GR?
Have you looked at the references I provided? The one from post #6 seems to be about the best possible answer: if the spacetime meets certain cosmologically plausible conditions (much more general than FRW models), and spatial curvature is positive (in the sense I described in post #5) and bounded below by some number, then not only can you say it is spatially closed, you can put an upper bound on its diameter.
Dmitry67
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Nov16-12, 12:08 PM
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But look my previous post: my toy Universe has finite diameter, but it is not spacially finite!!!
PAllen
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Nov16-12, 01:20 PM
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Quote Quote by Dmitry67 View Post
But look my previous post: my toy Universe has finite diameter, but it is not spacially finite!!!
Sorry, but the statement that it is a sphere and yet geodesics don't close is a contradiction. A sphere is defined by topology and geometry, not coordinate conventions. A geometric sphere has, as a property, that all geodesics are closed curves.

OK, so you propose that there is a 'funny 2-surface' 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.
Bill_K
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Nov16-12, 02:11 PM
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I have sphere in 3-dimensional 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 idea fails at the North and South Poles, where the local geometry topology will not be Euclidean. You cannot "unwrap" a sphere.


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