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Is it possible to construct a four-dimensional hypersphere where every point on the surface is equidistant from the center, yet with all three of the surface dimension being orthogonal to each other and the radius? Also, the radius is some finite value and -- according the data I'm trying to interpret -- the size of the circumference of the sphere appears to be a function of the radius. That is, is it possible to construct a four-dimensional hypersphere with a curvature of 0? If so, what would you call it?

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    $\begingroup$ Most likely the answer to your question is NO. But, depending on your definition of "curvature" and on the space in which your sphere sits, there might be room for exploration. $\endgroup$ – Ted Shifrin Dec 31 '17 at 16:47
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It would be just like one dimension down. A plane is in a sense the limit of a sphere as the radius runs off to infinity. The two axes in the plane are perpendicular to the radius. If you accept this as a construction, it works. In four dimensions you would have a three space that works the same way, being a limit of a hypersphere as the radius goes to infinity. That is useful for some purposes, but you need to be careful to justify the construction in your use.

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  • $\begingroup$ The data I have indicates that the radius would be large, but finite. Also, the three surface dimensions appear to be functions of the radius. $\endgroup$ – Quarkly Dec 31 '17 at 16:55
  • $\begingroup$ Note that a sphere in three space satisfies everything you are asking for except the curvature of zero. The curvature is the inverse of the radius. Nothing changes in 4D. What data are you speaking of? $\endgroup$ – Ross Millikan Dec 31 '17 at 17:02
  • $\begingroup$ The data is SNe Ia magnitude data. It fits better to a 4-d hypersphere than one defined by FLRW. My biggest issue is the curvature which we know to be very close to zero. $\endgroup$ – Quarkly Dec 31 '17 at 17:06
  • $\begingroup$ If you were to slice the Earth along the arctic circle, then measure the circle created by lopping off the top of the world, you'd find a circle that was smaller than $2\pi R$ (that is, $C = 2\pi R \sin\theta$) . It seems to me that $C=2\pi R$ is an attribute of Cartesian space. Is it possible to work in higher dimensions where the diameter of the 'sphere' is not the product of the diameter and $\pi$? $\endgroup$ – Quarkly Dec 31 '17 at 17:18
  • $\begingroup$ The circumference of a hypersphere is $\pi$ times the diameter in any Euclidean space, regardless of dimension. It is not in non-Euclidean spaces because the overall space has curvature. I think you are mixing up space curvature and the curvature of the hypersphere as embedded in Euclidean space. They are not the same thing. $\endgroup$ – Ross Millikan Dec 31 '17 at 17:28

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