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I'm a high school math teacher and I write math raps every year for my students. I'm working on my lyrics and I need help making sure something is mathematically accurate.

I'd like to make reference to a 3 dimensional space with zero curvature being Euclidean space. Is it correct to say that a space with zero curvature is Euclidean?

I've only really seen curvature applied to discussions of surfaces- and, for example, a cylinder has zero Gaussian curvature, but is clearly not "flat" (i.e. Euclidean)

I would appreciate any insights : )

PS: If you'd like to see any of my raps you can find them on my Youtube channel.

Mike Andrejkovics
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    It is partially correct. Such a surface is locally Euclidean. – Henricus V. Feb 11 '16 at 01:59
  • @HenryW - I know what you mean, but the term "locally Euclidean" is about topology only, and applies to all manifolds, regardless of their curvature. "Euclidean" by itself is a term that has often been used for manifolds without curvature. – Paul Sinclair Feb 11 '16 at 05:10
  • I don't know if this is appropriate for MathStackExchange, but since you all helped me fact check this thing, I thought you might enjoy seeing the final product. Here's the MathRap video: https://youtu.be/DzcO_RnejAA?list=PL59BEC8684FADD8D2 – Mike Andrejkovics Jun 05 '16 at 23:03

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It is acceptable to refer to any space with zero-curvature as "Euclidean". If you examine the behavior of geometry in the cylinder, but restricting your attention to a region that does not completely encircle it, you will discover that the behavior in it is completely indistinguishable from a similarly restricted region in the plane. (Note that distances are to be measured along the cylinder here, not by direct line through the middle.) This is obvious enough: split the cylinder along a lengthwise chord, and you can unroll it perfectly flat. Thus the cylinder actually is Euclidean.

It is a little harder to see that the torus is also Euclidean, as it doesn't have that same property of being able to be flattened completely when cut. But the thing is, this is a result of the "extrinsic geometry" of the surface - that is, how the surface has been made to sit in the surrounding space. But if we ignore the surrounding space and pretend that all we are aware of is the points of the surface itself, no such difference is spottable. Curvature and similar behavior is "intrinsic geometry". No matter how this surface is placed in space (even higher dimensional spaces), the intrinsic geometry does not change, and the cylinder and torus behave exactly like the plane, as long as you restrict your attention to a non-encircling region.

Paul Sinclair
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  • But the torus doesn't have zero curvature, right? – Gerry Myerson Feb 11 '16 at 06:08
  • @GerryMyerson - The torus has zero curvature. It's intrinsic behavior is exactly that of the plane, and since the plane has zero curvature, so does it. To see this, note that one construction of it is $\Bbb R^2 / \sim$, where $(x, y) \sim (w, z)$ if and only if $x -w$ and $y -z$ are integers. That is, the torus is just a square in the plane with its top glued to its bottom and its left glued to its right, without any flipping in either direction. – Paul Sinclair Feb 11 '16 at 15:55
  • So, we're not talking about Gaussian curvature here, right? http://www.win.tue.nl/~rvhassel/Onderwijs/Tensor-ConTeX-Bib/Examples-diff-geom/Torus-diff-geom/torus-together.pdf, http://math.stackexchange.com/questions/495232/are-there-any-surfaces-that-contain-both-positive-and-negative-gaussian-curvatur – Gerry Myerson Feb 11 '16 at 21:43
  • @GerryMyerson - The problem is that when I talk about the torus, I automatically assume it's "natural" metric, which is that of the "Clifford torus". This metric is flat. But this torus cannot be embedded isometrically in $\Bbb R^3$, so the 3D version of the torus inherits a different metric from the imbedding. That metric still has total curvature $0$ (integral of curvature over the manifold), but it varies from point to point. – Paul Sinclair Feb 12 '16 at 00:39