Question about a torodial universe


by Vorde
Tags: torodial, universe
Vorde
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Jan23-12, 03:42 PM
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Does the fact that on 2D torus's, one dimension on the torus always has to be 'smaller' than the other dimension hold on the higher dimension versions?
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Drakkith
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Jan23-12, 09:10 PM
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You can have a 2d torus?
Vorde
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Jan23-12, 09:48 PM
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I mean 2D in the way you could call the surface of a sphere a 2-sphere, I guess I was trying to sound 'mathy'. I just mean a doughnut surface by 2D torus.

yenchin
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Jan24-12, 12:14 AM
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Question about a torodial universe


I am not sure what you mean...even in 2D, you can identify the opposite sides of a *square* to have torus topology. So what do you mean by "smaller"?
marcus
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Jan24-12, 01:02 AM
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Quote Quote by yenchin View Post
I am not sure what you mean...even in 2D, you can identify the opposite sides of a *square* to have torus topology. So what do you mean by "smaller"?
Good point, thanks. 2D torus so constructed is often given as an example of a flat differential manifold. Zero intrinsic curvature. No boundary. Not embedded in any surrounding space.
Sometimes people talk about the "Pac Man" game screen as a square with left edge identified with right edge, and top with bottom.

It is not true that one circumference or girth has to be bigger or smaller than the other. The construction works just as well with a square as with a rectangle.

The original question was about higher dimensional analogs and the same is true, one can for instance start with a cube and make the same "Pac Man" identifications. Going out thru the right side is the same as coming in thru the left. Ditto front and back, ditto top and bottom.

that's a 3D torus and it does not have to "live" in any higher dimensional space, and it is boundaryless. Standard differential geometry.

Thanks for making the same point in the other thread about the 1D torus analog---the 1D "ring" made by taking a line segment and identifying the endpoints. Does not require a 2D surround. Need not be immersed in any higher dim'l space.
Chalnoth
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Jan24-12, 02:12 AM
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Quote Quote by marcus View Post
It is not true that one circumference or girth has to be bigger or smaller than the other. The construction works just as well with a square as with a rectangle.
Right. This limitation only comes into play when you embed the torus in three dimensions. I believe you can get around this limitation by embedding it in four dimensions instead. Or just by not embedding it at all and only dealing with the two dimensions inherent to the torus.


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