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The concept of manifolds is freaking me out.

For me it seems like a manifold is just a subspace embedded in a higher dimension. In order to clear out my confuision I have created a list and I would be glad if someone could tell me if my guesses are right or not, and why they are wrong.

  1. Sphere in $\mathbb R^n$: is a smooth manifold
  2. Cube in $\mathbb R^n$: is not a smooth manifold but a topological manifold
  3. Line in $\mathbb R^n$: is not a manifold as there are no open neighborhoods for points on the line.
  4. Point in $\mathbb R^n$: not a manifold as there are no open neighborhoods.
Michael Hardy
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MrYouMath
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    A line in $\mathbb R^n $ is a manifold of dimension $1$. Manifolds are a generalization of curves and surfaces. – littleO May 23 '16 at 22:21
  • Why? But I cant take partial derivatives in all directions. Is it also a smooth manifold or just a topological manifold? – MrYouMath May 23 '16 at 22:23
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    A line is a smooth manifold because you can definitely map it to $\mathbb{R}$ in an infinitely differentiable way – M Turgeon May 23 '16 at 22:25
  • A line and point are both smooth manifolds of 1 and 0 dimensions respectively. –  May 23 '16 at 22:25
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    @MTurgeon: Ah. Then anything with a differentiable parametriziation is a manifold? So the Cube is not a smooth manifold (but a topological manifold) but the xy plane in $R^3$ is a smooth manifold. – MrYouMath May 23 '16 at 22:29
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    @MrYouMath: Having a differentiable parametrization is not enough; for example, the map $t \mapsto (\cos^{3} t, \sin^{3} t)$ is real-analytic but its image, the astroid curve, has cusps. Having a regular parametrization (differentiable, with differential of maximum rank at each point) suffices, however. – Andrew D. Hwang May 23 '16 at 23:00

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A sphere is a (2D) manifold that can admit a smooth structure.

A cube, I interpret to mean as $[0,1]^3$ is not even a topological manifold.

A line is a (1D) manifold that can admit a smooth structure.

A point is a (0D) manifold that can admit a smooth structure.

If you are looking for a manifold without a smooth structure, you will have a difficult time. Every manifold in 1,2,and 3 dimensions has a smooth structure.

Edit if by "cube" you mean "boundary of cube" then that is a 2d manifold with a smooth structure (as ALL 2d manifolds have smooth structures).

  • Why is a cube not even a topological manifold. Couldn't i define multiple paramteriziations and patch them together for the cube? – MrYouMath May 23 '16 at 22:45
  • @MrYouMath Can you find an open set around $(1,1,1)$ that is homeomorphic to a subset of $\Bbb{R}^3$? –  May 23 '16 at 22:46
  • @MrYouMath See https://en.wikipedia.org/wiki/Differentiable_manifold#Relationship_with_topological_manifolds btw –  May 23 '16 at 22:47
  • I'd have taken "cube" to mean the boundary of $[0,1]^3$ in $\mathbb R^3$. That is a $2$-manifold. $\qquad$ – Michael Hardy May 23 '16 at 23:16
  • How about $(0,1)^3$, to enrich the content we can say this is a topological manifold to a subset of $\mathbb R^3$ – Jale'de jaled Oct 27 '20 at 12:11
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To clear up some confusion, recall that a $k$-manifold $M$ is a second-countable, Hausdorff space which is locally homeomorphic to $\mathbb{R}^k$. The last requirement is the most important, at least it's the one that can be understood better since it says something about the intrinsic geometry. To make things a bit easier, lets look at all of your examples in the case where $k = 0,1,2,3$.

enter image description here

If you take any point of the sphere and draw an open set around it, it looks bent disk which can be unbent smoothly to get a regular disk that you can place in the plane.

enter image description here

The cube has some nice open sets which look like 3-disks, but if you look at any 3-disk about a vertex (i.e pointed boundary) then you can continuously transform it to a 3-disk.enter image description here

If you take any point on a line and make a neighborhood about it, it looks like an open interval in the plane.

enter image description here

The topology for a space with one point, it the trivial topology, i.e the point itself is an open set. The definition for manifold requires that for every point, there exists! and open set about it with a certain property. Thus taking the open set to be the point, that just looks like a number on the real line.

In these examples, whatever the neighborhoods on these objects are homeoomorphic to, that determines their dimension. So the sphere is a smooth 2-manifold, the cube is a smooth 3-manifold, the line is a smooth 1-manifold, and the point is a smooth 0-manifold.

Faraad Armwood
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  • "the cube is a 2-manifold" Do you mean the boundary of a cube? If so there is no 2d manifold without a smooth structure. https://en.wikipedia.org/wiki/Differentiable_manifold#Relationship_with_topological_manifolds –  May 23 '16 at 22:50
  • Yeah, I was only thinking boundary. – Faraad Armwood May 23 '16 at 22:52
  • Then it is indeed a smooth manifold. As there can be no non smooth 2d manifolds. –  May 23 '16 at 22:53
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    However, this differentiable structure will not be diffeomorphic to the sphere. It is homeomorphic to the sphere. – ncmathsadist May 23 '16 at 22:53
  • @ncmathsadist Good point. –  May 23 '16 at 22:55
  • Yeah, yeah I got a bit sloppy. Of course the restriction of $(\mathbb{R}^N, Id)$ to any of these sub-manifolds gives a smooth structure. – Faraad Armwood May 23 '16 at 22:57
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    @ZacharySelk I think that this could be phrased better. Every surface has a smooth structure, but when one is talking about subsets of $\Bbb R^n$, one usually takes "is a smooth manifold" to mean "is a smooth submanifold of $\Bbb R^n$". So poorly embedded submanifolds (like the cube) would not be called smooth manifolds in this context, even though they're smoothable. Another reason to not confuse this terminology is that topological manifolds may support more than one smooth structure up to diffeomorphism, in which case it would be strange to say that "$M$ is a smooth manifold" when... –  May 23 '16 at 22:58
  • ...the smooth structure is not well-defined. –  May 23 '16 at 22:58
  • @MikeMiller You are very correct. Thanks for pointing this out. –  May 23 '16 at 22:59
  • Very nice explanation. But are the surface of a sphere and the volume of the sphere both manifolds or does a manifold always have a lower dimension then the space in which it is embedded. – MrYouMath May 24 '16 at 07:59