Does an N-Cube have Surface Area?

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The discussion centers on the concept of surface area in relation to a four-dimensional cube, questioning whether it should be termed "surface area" or "surface volume." It explores the definition of a hypersurface as the n-1 dimensional boundary of an n-dimensional space, emphasizing that terminology can vary based on convention. The conversation also touches on the implications for light waves in four dimensions, suggesting that flux may be conceptualized through volume rather than surface area. Additionally, it clarifies that every manifold is at least a topological manifold, with submanifolds of lower dimensions existing within higher-dimensional spaces. Overall, the thread delves into the complexities of dimensional geometry and the appropriate language to describe these concepts.
dimensionless
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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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its boundary is not a surface but does have a 3d volume
 
Would this n-1 dimensional boundary be a hypersurface?
 
Rasalhague said:
Would this n-1 dimensional boundary be a hypersurface?

Depend what you mean by hypersurface. Explain.
 
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".
 
wofsy said:
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?
 
dimensionless said:
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.
 
Solution of the wave equation is quite different in even dimensions vs. odd dimensions.
 
g_edgar said:
Solution of the wave equation is quite different in even dimensions vs. odd dimensions.

Why would that be?
 
  • #10
wofsy said:
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)?
 
  • #11
Rasalhague said:
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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