I've seen some visualizations of manifolds. It seems that they are all "curved" shapes. Is there a "non-curved" manifold?
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7Why not plain (plane?) old $\mathbb{R}^2$? – Brian Tung Jun 08 '20 at 04:45
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1There are no compact manifolds that can be represented in Euclidean space as non-curved. But remove the “compact” condition, any connected open subset of $\mathbb R^n$ is a “flat” manifold. – Thomas Andrews Jun 08 '20 at 04:48
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2(All of the pictured manifolds are compact.) – Thomas Andrews Jun 08 '20 at 04:52
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@ThomasAndrews Can you share an intuitive explanation of "compact" regarding to manifolds? – niebayes Jun 08 '20 at 05:02
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1Indeed there are two versions of "curved" manifolds. One intrinsic and one extrinsic. The "torus" $\mathbb S^1 \times \mathbb S^1$ in $\mathbb R^4$ is not curved intrinsically. – Arctic Char Jun 08 '20 at 05:04
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Yes, you can get a flat, compact manifold embedded in $\Bbb R^3$. This is a consequence of the Nash embedding theorem. In particular, a couple of years ago, this was done in practice with a torus, and the results are quite visually interesting. Here are the first three steps in the construction:
It may look curved, but it does actually give a $C^1$ isometric embedding of the flat torus $S^1\times S^1$ in three-dimensional Euclidean space. ($C^1$, i.e. continuously differentiable, is important because otherwise distances along the surface are not necessarily well-defined.)
Arthur
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1I think the main point is that there does not exists $C^3$ embedded flat surface in $\mathbb R^3$. Your answer is so technical for someone who do not even know what compact means. – Arctic Char Jun 08 '20 at 22:16