Does an N-Cube have Surface Area?

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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