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It was some time since I've been doing some basic differential geometry, and when I was recently revisiting it I was confused about the following: a unit cube surface $C$ in the ambient space $\Bbb R^3$ is not smooth in a sense that you can't put a tangen plane at its vertices, contrary to the unit sphere $S$ where each point allows for such a surface.

Now, both $C$ and $S$ inherit topology as subspaces of the ambient space, and this topology is the same, they are homeomorphic. I am not sure however whether we can push on them any smooth structure from the ambient space. My point is: is there a way to say that the smooth structure imposed on $C$ truly makes it $C$, and not just a smooth structure from $S$ push onto it?

Maybe it is only possible when there's some extra structure imposed on these spaces, like the Riemann metric. Overall it kinda boils down to the question: is there an intrinsic characterization of the $C$ as a smooth manifold which distinguishes it from $S$?

I have now added clarification that I wonder about intrinsic differences between $S$ and $C$ as smooth manifolds. As topological manifolds they are obviously the same (as stated in the OP), and as Riemann manifolds $C$ and $S$ will be different, but so will be $S$ and an ellipsoid, so nothing that will be caused by the "corners".

SBF
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    This makes little sense: In the first paragraph you say that $C$ as a submanifold of $\mathbb{R}^3$ is not smooth, and then you go ahead and talk about "the smooth structure imposed on $C$" in the second paragraph anyways. Also you should tell us what a manifold is to you. – Ben Steffan May 09 '25 at 12:51
  • @BenSteffan actually I did not say that $C$ is not smooth as a submanifold, I have only stated that it has corners. And that's not the only point where I find your comment not very constructive – SBF May 09 '25 at 13:13
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    You say "a unit cube surface $C$ in the ambient space $\mathbb{R}^3$ is not smooth in a sense..." in the first paragraph and then "I am not sure however we can push on them any smooth structure from the ambient space." Well, what smooth structure "imposed on $C$" are you considering, then? Also what does "as topological manifolds the are the same" mean? That they're homeomorphic? In fact, $S$ and $C$ admit unique smooth structures up to diffeomorphism (an upshot of the classification of surfaces), so in what sense do you want to consider smooth structures as "the same" or "different"? – Ben Steffan May 09 '25 at 13:25
  • @BenSteffan do you mean that I can define a smooth structure on $C$ as a subset of $\Bbb R^3$? How exactly can I do this – SBF May 09 '25 at 13:32
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    There are ways to express a question like yours with sufficient mathematical precision, but they would require deeper tools from differential topology. For instance, perhaps one could express this question using the language of a smooth atlas on a manifold. Without something like that, though, I cannot see how to make much sense out of the question. – Lee Mosher May 09 '25 at 13:32
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    "As Riemannian manifolds they will be different"- one of them has a natural Riemannian structure, while the other (the cube) does not. But $C$ does have a natural "intrinsic metric." (Distance is computed by minimizing lengths of paths connecting points.) One can ask if this intrinsic metric is isometric to some Riemannian metric. This question has negative answer. Maybe this is what you were trying to ask. See here for further discussion of the cube surface as an example. – Moishe Kohan May 09 '25 at 14:09
  • @MoisheKohan You're right, a Riemann manifold will require a smooth structure to build upon. So would that be correct to say that $C$ as a metric space can't be modeled by a Riemann manifold, and hence it is intrinsically not smooth? – SBF May 09 '25 at 14:29
  • Exactly. I suggest you read my answer in the link (on curvature concentrated at 8 points which explains why this metric cannot come from a Riemannian metric). – Moishe Kohan May 09 '25 at 14:34
  • See also my answer https://math.stackexchange.com/questions/106508/existence-of-a-riemannian-metric-inducing-a-given-distance/673742#673742 – Moishe Kohan May 09 '25 at 14:38
  • @MoisheKohan yeah I did and it's an elegant idea to replace the curvature with a measure allowing atoms. I however can see that on the three levels (topology, smoothness, metric) we have $C = S$ on the first level, $C \neq S$ on the third level, but I specifically wonder about the second level. I guess the point is that we can bring the topology/metric from the ambient space to $C$ and $S$, but I am not sure whether there's a "subspace atlas" that we can bring from the ambient space to them. – SBF May 09 '25 at 14:41
  • @MoisheKohan and thanks for your previous comments, I hope you maybe know the answer to my last question – SBF May 09 '25 at 14:45
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    There is no notion of a subspace atlas unless you have a smooth submanifold. There is a notion of a Lipschitz atlas that can be used here. Then one can ask if this subspace atlas is Lipschitz-isomorphic to a smooth one. This question has positive answer. One has to go to dimension 4 to get exotic examples of Lipschitz structures. – Moishe Kohan May 09 '25 at 14:59

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I won't even point at proofs in this answer, but just take a stab at straightening out the different concepts at play in your question.

You can use a homeomorphism between the two spaces to transport the differentiable structure from the sphere to the cube. But where did this diff-structure on the sphere come from? It's the one it inherited as a sub-diff'ble manifold of $\mathbb R^3$ (at least that's what we default to assuming if no other diff structure is mentioned).

So what about the cube? Doesn't it inherit a sub-diff manifold structure from $\mathbb R^3$ too? And the answer is "no, because corners". (That claim is the one I'm most skipping the proof of.)

With those terms in place, I'd answer your question with "the cube's non-smoothness isn't intrinsic, but is due to the way it sits in $\mathbb R ^3$", but note that I've had to provide definitions for your terms, which I think are what most people would come up with, but may not be what you had been thinking of.

JonathanZ
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  • thanks, I'd be glad if you point me to the roadmap of how to show that a cube does not inherit a sub-diff manifold structure from $\Bbb R^3$ – SBF May 11 '25 at 08:27
  • I've never followed up the details myself, so I can't point authoritatively to a proof-path, but to me it looks like MoisheKohan's "curvature concentrated at 8 points which explains why this metric cannot come from a Riemannian metric" is the way to go. I also think I've seen an answer here on math.se that does the simpler square vs. circle example using more elementary methods, but I suck at searching. – JonathanZ May 11 '25 at 14:03