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I know that if two topological spaces $X$ and $Y$ are homeomorphic then so are their one point compactifications $X^*$ and $Y^*$. If $X$ and $Y$ (say both are smooth manifolds) are diffeomorphic what do we know about $X^*$ and $Y^*$? Are they diffeomorphic too or only homeomorphic? If the latter is true, are there further condition under which $X^*$ and $Y^*$ will be diffeomorphic?

  • Generally exotic $\Bbb R^4$'s won't have a smooth $S^3$ hyper surface near infinity (an example of one of these would disprove SPC4), so there really isn't any reasonable way to define a canonical smooth structure on the one point compactification. Unless you have some sort of canonical smooth structure in mind this question probably has zero content. – PVAL-inactive May 01 '17 at 20:43
  • Since the answers address the mistaken part of the question, it is not good to remove it. But the mistaken part doesn't make the question nonsensical. It's actually a pretty interesting question. – Daniel Fischer May 02 '17 at 20:07

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One point compactification of a manifold does not always have to be a manifold. For more details read this:

One-point compactification of manifold

But even when it is, I don't think it has to be unique. Please correct me if I'm wrong (manifolds is not my field of study).

For example take $M$ to be any exotic sphere of dimension $n\neq 4$. Then when you remove a point you obtain $\mathbb{R}^n$ as a topological space but since $n\neq 4$ then it is $\mathbb{R}^n$ as a manifold (the only exotic Euclidean space is $\mathbb{R}^4$).

Reversing, you have standard $\mathbb{R}^n$ which has at least two, non-diffeomorphic compactifications: the standard and exotic sphere.

freakish
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  • Exotic spheres were actually one of the motivations for my question (isn't the first dimension where exotic spheres appear dimenion 7? So you don't have to exclude dimension 4?). I was reading this post https://math.stackexchange.com/questions/1270099/proof-of-reebs-theorem-without-using-morse-lemma and I was wondering where one loses the differentiable structure. It seems like this happens in the end when passing to the one point compactification (because right before that $M-{pt}$ is diffeomorphic to $\mathbb{R}^n$ and then in the next step $M$ is only homeomorphic to $S^n$). –  May 01 '17 at 13:54
  • @MM95 I believe that it is still an open question whether there are exotic 4-spheres or not. Anyway, yeah, you can't deduce much about one-point compactification. – freakish May 01 '17 at 13:58
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You can't put a differentiable structure on the one-point compatification, so your question is not well defined. For example, consider the union of several open segments $(0,1)$, the one-point compactification will be a wedge of circle which is not a smooth manifold. I also think that one point compactification of $\mathbb R^2 \backslash D$, where $D$ is a union of disks, can't be a smooth manifold.

  • You can under a certain condition: https://math.stackexchange.com/questions/240339/one-point-compactification-of-manifold The condition seems to be satisfied by OP's example. – freakish May 01 '17 at 13:35
  • Is there an intuitive explanation why it's not possible to put a smooth structure on a one point compactification? –  May 01 '17 at 13:56
  • @MM95 : do you see why the wedge of circles is not a manifold ? –  May 01 '17 at 14:09
  • There is no chart around the wedge point. But this is just one particular example. And I was hoping to get an intuition why it's not possible at all to put a smooth structure on a one-pt compactification (because you wrote "You can't put a differentiable structure on the one-point compatification, so your question is not well defined") –  May 01 '17 at 14:14
  • So a visual picture of a one point compactification is "adding the same point at every "end" of a space. A end of a space can be defined the following way : assume you have a family of increasing connected compact $K_i$ with $ \cup_i K_i = M$. Then an end is a connected component of the complement of $K_i$. If you have more that two end, you will have exactly the same problem that the wedge. –  May 01 '17 at 14:17
  • Thank you for the answer. I learned that the one point compactification of the complex plane is a Riemann surface, so in particular a smooth surface of real dimension 2. I'm a bit confused. Isn't this an example of a one-point compactification with a differentiable structure? –  May 01 '17 at 18:18
  • I didn't say that all one-point compactification can't be a differentiable manifolds (yourself you know that for $\mathbb R^n$ the one-point compactification is $S^n$ which has a smooth structure). But if a manifold has more that two ends, then its one-point compactification will not be a manifold. –  May 01 '17 at 18:20