# Can you construct a hypersphere with a curvature of zero?

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?

• 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. – Ted Shifrin Dec 31 '17 at 16:47

• 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 $\pi$? – Quarkly Dec 31 '17 at 17:18
• 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. – Ross Millikan Dec 31 '17 at 17:28