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Popular articles on mathematics often explain the difference between homeomorphism and diffeomorphism with statements like - "A rectangle is homeomorphic to the circle but not diffeomorphic to it". However, this is not really true, since all compact one dimensional smooth manifolds are diffeomorphic to $S^1$ - the corners of a rectangle are just artifacts of a particular embedding into the plane.

So, my question is - What is the best way to "visualize" manifolds which are homeomorphic but not diffeomorphic, like the usual sphere versus exotic spheres, say ?

It should be possible to embed exotic spheres in some $\mathbb{R}^n$. Now if this is a smooth embedding, then the image should not have any edges or corners - unlike a square or tetrahedron.

So then, how can one visualize the "failure of diffeomorphism" ? I ask this because typically, examples of "continuous but not differentiable" functions involve the image having a corner or edge.

Anindya
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  • Related question: https://mathoverflow.net/questions/97073 – Sam Hopkins Oct 16 '23 at 22:30
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    You can embed exotic 7-spheres in $\Bbb R^{13}$. (Probably lower, but irrelevant to the point I want to make.) How does that help you intuit them, though? The first example is in 4-manifolds, which I can barely pretend to be able to visualize. Instead, we develop tools to understand how to construct and distinguish these, and our intuition comes from our tools --- the constructions we know how to do, and the obstructions we know how to make. – mme Oct 16 '23 at 22:31
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    For instance, to understand exotic (n+1)-spheres, these can be obtained by gluing two standard n-balls by a funny diffeomorphism of the boundary. So exotic (n+1)-spheres shed light on exotic diffeomorphisms of the n-sphere. So we can intuit these as "funny regluings of the hemispheres". But this is pretty weak intuition: why does such an exotic diffeomorphism exist, how do you understand it? (Actually, I think you're more likely to understand exotic diffeomorphism of S^6 using what you know about exotic 7-spheres...) – mme Oct 16 '23 at 22:35
  • Might the cube and the $2$-sphere serve? – Michael Hardy Oct 17 '23 at 00:11
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    @MichaelHardy I think this would not satisfy OP as they have already (rightly) discounted the example of a circle and a square. – Connor Malin Oct 17 '23 at 00:59
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    The exotic 7-spheres are all Brieskorn spheres, links of singularities, so embed in $S^9$. https://en.wikipedia.org/wiki/Exotic_sphere#Brieskorn_spheres This gives them implicitly. – Ian Agol Oct 17 '23 at 04:10
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    Technically, your assertion about the square is wrong. The square does have a differentiable structure, but not as a smooth manifold. The differentiable structure on the square demands that a map out of the square is smooth whenever it admits local extensions (to open subsets of the plane) that are smooth. With this definition, you can prove the square is not diffeomorphic to a smooth 1-manifold. i.e. the point here is that subsets of smooth manifolds have smooth structures, but they aren't always smooth submanifolds. – Ryan Budney Oct 17 '23 at 05:17
  • @ConnorMalin : But there are complications at the vertices of a cube unlike the ones at the corners of a square. Whether that means there is no diffeomorphism is another question. – Michael Hardy Oct 17 '23 at 16:41
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    What does "Is it really the pres" mean? – LSpice Oct 17 '23 at 23:58
  • @LSpice - sorry, it looks like my question didn't upload in full. – Anindya Oct 18 '23 at 04:44
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    The question was - "Is it really the presence of higher dimensional equivalents of edges and corners in exotic spheres that prevents a diffeomorphism from existing ?" – Anindya Oct 18 '23 at 04:45
  • In retrospect, I realize this can't be the case. An smooth embedding of a manifold into R^n doesn't have edges or corners – Anindya Oct 18 '23 at 04:46
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    For the clutching map interpretation of exotic spheres, we appear to be getting close to having explicit descriptions of the diffeomorphisms involved. It might take another year or two, but there are quite a few interesting candidates, starting with the recent work of Watanabe. – Ryan Budney Oct 18 '23 at 07:19
  • @RyanBudney: Could you please elaborate a bit more about the "explicit descriptions of diffeomorphisms" (or did you mean homeomorphism) ? That would help a lot - an explicit homeomorphism between a regular and exotic sphere plus a demonstration of why it isn't a diffeomorphism – Anindya Oct 18 '23 at 17:34
  • Exotic sphere in dimensions $n \geq 6$ are obtained by gluing two discs together along their boundary, via a diffeomorphism. At this point there have not been concrete descriptions (at the point-set level) of those maps. I am talking about recent progress in describing maps such as this. – Ryan Budney Oct 18 '23 at 18:41

1 Answers1

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Milnor’s original construction of an exotic $\mathbb{S}^7$ is fairly explicit and it’s reasonable to think that the resulting space is too weirdly twisted to actually be a sphere in the usual sense. Showing that it is not a standard sphere is very non-trivial and one of Milnor's key insights. However, it's not to hard to cook up a Morse function on this space with only two critical points, and from that you can see that the space has to be a sphere topologically. Intuitively, the space is topologically a sphere but because of the twisting there is no consistent way to identify ``directions" compared to a standard sphere if one travels from the North to South Pole. Here's the basic gist of how this goes.

For a normal sphere, we can consider $\mathbb{B}^4$ as the the unit ball in $\mathbb{R}^4$, and let $\mathbb{S}^3$ be its boundary. The boundary of $\mathbb{B}^4 \times \mathbb{S}^3$ would then be $\mathbb{S}^3 \times \mathbb{S}^3$. We can then take two copies of $\mathbb{B}^4 \times \mathbb{S}^3$ and glue them together along the boundary. Putting a standard flip into the $\mathbb{S}^3$ part and doing the gluing yields a sphere in the usual sense.*

However, if we consider $\mathbb{R}^4$ as the space of quaternions $\mathbb{H}$ and $\mathbb{S}^3$ as the space of unit quaternions. Take two copies of $\mathbb{R}^4 \times \mathbb{S}^3$ and identify the subsets $\left(\mathbb{R}^4-\{0\}\right) \times \mathbb{S}^3$ under the diffeomorphism $$ (u, v) \rightarrow\left(u^{\prime}, v^{\prime}\right)=\left(\frac{u} {\|u\|^2}, \frac{u^2 v u^{-1}}{\|u\|}\right) $$ (using quaternion multiplication). Here, we have "twisted" the identification using quaternionic multiplication (and also extended the gluing to not just occur on the boundary). So it is pretty reasonable to think that this space is not a sphere in the usual sense.

But then you might stumble on the function $$f=\frac{\Re(v)}{\left(1+\|u\|^2\right)^{1 / 2}}=\frac{\Re\left(u^{\prime \prime}\right)}{\left(1+\left\|u^{\prime \prime}\right\|^2\right)^{1 / 2}}$$ where $u^{\prime \prime}=v^{\prime} u^{\prime}$ and $\Re$ indicates the real part of the quaternion. A bit of effort will show that this is a Morse function with only two critical points, and thus the space must be a sphere in the sense of topology, which doesn't see the smooth structure.

*If you glue the boundaries without the flip, the resulting space is $\mathbb{S}^4 \times \mathbb{S}^3$. Thanks to @HenrikRüping for pointing this out.

Gabe K
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    Probably dumb question: what if we try to do something similar with $\mathbb{C}$ in place of $\mathbb{H}$? – Sam Hopkins Oct 17 '23 at 02:25
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    Not a dumb question at all. But since multiplication in $\mathbb{C}$ is commutative, we just get the usual three-dimensional sphere. – Gabe K Oct 17 '23 at 02:31
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    "Doing so in the most natural way yields a sphere in the usual sense."

    I would say the most natural way gives $S^4\times S^3$, we still need a flip.

     – HenrikRüping Oct 17 '23 at 06:57
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    What if we take octonions instead of quaternions? – Vít Tuček Oct 17 '23 at 10:16
  • @HenrikRüping Thanks for pointing that out, I'll make that edit. – Gabe K Oct 17 '23 at 10:44
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    @VítTuček In that case, you will get exotic 15-dimensional spheres. For a reference, see Shimada, 1957 "Differentiable structures on the 15-sphere and Pontrjagin classes of certain manifolds." – Gabe K Oct 17 '23 at 10:54
  • Is this the Gromoll-Meyer sphere? – Gro-Tsen Oct 17 '23 at 22:28
  • Nice answer ! Could you please elaborate a bit more on what you mean by "because of the twisting there is no consistent way to identify directions" ? – Anindya Oct 18 '23 at 05:11
  • @Gro-Tsen I’m not entirely sure. One can create other exotic spheres by conjugating by other powers of $u$, and I think one of those is the Gromoll-Meyer sphere since I believe those are the exotic spheres which admit positive curvature. But I’m not sure which one it is. – Gabe K Oct 18 '23 at 11:27
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    @Anindya So typically when we think of a Morse function with only two critical points, we can essentially interpret it as the latitude of our manifold and each level set will be a $n-1$ dimensional sphere. Gluing it all together the space will be a sphere. But here Milnor has twisted up the geometry so that the typical polar angles parametrizing the level sets do not extend smoothly from pole to pole. – Gabe K Oct 18 '23 at 11:53