Does an N-Cube have Surface Area?


by dimensionless
Tags: ncube, surface
dimensionless
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#1
Dec16-09, 04:45 PM
P: 464
Let's say I have a four dimensional cube. Would it have a true surface area? I'm wondering if maybe it would have a surface volume rather than a surface area.
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wofsy
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#2
Dec16-09, 05:07 PM
P: 707
its boundary is not a surface but does have a 3d volume
Rasalhague
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#3
Dec16-09, 05:59 PM
P: 1,400
Would this n-1 dimensional boundary be a hypersurface?

wofsy
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#4
Dec16-09, 06:13 PM
P: 707

Does an N-Cube have Surface Area?


Quote Quote by Rasalhague View Post
Would this n-1 dimensional boundary be a hypersurface?
Depend what you mean by hypersurface. Explain.
HallsofIvy
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#5
Dec17-09, 06:22 AM
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In n dimensional geometry, a "hypersurface" is the n-1 dimensional boundary of a bounded n-dimensional region.

As for dimensionless's original question, its really a matter of convention whether you call the 3 measure of the boundary of a 4 dimensional region "area" or "volume". That's why most people just talk about n or n-1 dimensional "measure".
dimensionless
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#6
Dec17-09, 07:13 AM
P: 464
Quote Quote by wofsy View Post
its boundary is not a surface but does have a 3d volume
Does that mean that a light wave in 4D would have a flux through a volume rather than a surface area?
wofsy
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#7
Dec17-09, 07:30 AM
P: 707
Quote Quote by dimensionless View Post
Does that mean that a light wave in 4D would have a flux through a volume rather than a surface area?
In general there would be an exact analogue of flux but with light there is a Lorentz metric and I am not sure how that would work.
g_edgar
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#8
Dec17-09, 08:34 AM
P: 608
Solution of the wave equation is quite different in even dimensions vs. odd dimensions.
dimensionless
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#9
Dec17-09, 10:46 AM
P: 464
Quote Quote by g_edgar View Post
Solution of the wave equation is quite different in even dimensions vs. odd dimensions.
Why would that be?
Rasalhague
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#10
Dec17-09, 11:09 AM
P: 1,400
Quote Quote by wofsy View Post
Depend what you mean by hypersurface. Explain.
I had in mind an (n - 1)-dimensional "bit" of the given n-dimensional space. HallsofIvy's "the n-1 dimensional boundary of a bounded n-dimensional region" sounds like what I was thinking but more precisely worded that I'd have managed. Wikipedia calls a surface a "two dimensional topological manifold". Would a hypersurface then be an (n - 1)-dimensional topological manifold (and is every manifold at least a topological manifold)?
wofsy
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#11
Dec17-09, 03:28 PM
P: 707
Quote Quote by Rasalhague View Post
I had in mind an (n - 1)-dimensional "bit" of the given n-dimensional space. HallsofIvy's "the n-1 dimensional boundary of a bounded n-dimensional region" sounds like what I was thinking but more precisely worded that I'd have managed. Wikipedia calls a surface a "two dimensional topological manifold". Would a hypersurface then be an (n - 1)-dimensional topological manifold (and is every manifold at least a topological manifold)?
Every manifold is at least topological but may have additional structure such as a differentiable structure.

A submanifold of dimension n-1 is a called a hypersurface. You may be aware that you can have submanifolds of lower dimension as well. For instance in 4 space the Klein bottle can be embedded as 2 dimensional surface.


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