Timeline for Classification of smooth atlases
Current License: CC BY-SA 3.0
16 events
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| Mar 6, 2012 at 20:05 | answer | added | Ryan Budney | timeline score: 9 | |
| Mar 6, 2012 at 18:47 | history | edited | Cristi Stoica |
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| Feb 27, 2012 at 10:22 | history | bounty ended | Cristi Stoica | ||
| Feb 20, 2012 at 12:04 | history | edited | Cristi Stoica | CC BY-SA 3.0 |
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| Feb 20, 2012 at 9:40 | history | bounty started | Cristi Stoica | ||
| Feb 17, 2012 at 18:13 | comment | added | Cristi Stoica | @Tom Goodwillie: I added an example, to explain better what I mean, and to show that my questions are justified. | |
| Feb 17, 2012 at 18:07 | history | edited | Cristi Stoica | CC BY-SA 3.0 |
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| Feb 17, 2012 at 17:17 | history | edited | Cristi Stoica | CC BY-SA 3.0 |
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| Feb 17, 2012 at 17:14 | comment | added | Cristi Stoica | @Tom Goodwillie: You are right, any equivalence class of atlases contains a maximum atlas, which is the union of all other atlases in the equivalence class. And two maximal atlases are either identical, or not equivalent. So we can restate my questions in terms of maximal atlases instead of the equivalence classes of atlases. The questions remain. | |
| Feb 17, 2012 at 16:52 | comment | added | Tom Goodwillie | I still do not understand the question. What can you mean by equivalence between two atlases? It appears that you mean compatibility, in the sense that comparing the charts of one atlas and the charts of the other atlas leads to diffeomorphisms between open subsets of $\mathbb R^n$. But then if your atlases are maximal (meaning maximal w.r.t. this compatibility) then they are never equivalent; and if they are not required to be maximal then two atlases are equivalent if and only if they both determine the same maximal atlas. Am I missing the point? | |
| Feb 17, 2012 at 16:24 | history | edited | Cristi Stoica | CC BY-SA 3.0 |
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| Feb 17, 2012 at 16:19 | comment | added | Cristi Stoica | @Tom Goodwillie: Yes, $M$ is a topological manifold with more atlases, and I avoided saying that $\mathcal{A}'$ is a new atlas on the same differentiable manifold. I tried to avoid the term "smooth structures" because there is a confusion of terminology, as pointed here en.wikipedia.org/wiki/Smooth_structure (it is also pointed that the map f will provide a diffeomorphism betwen $M$ and $M'$). This is why I discussed in terms of smooth atlases. | |
| Feb 17, 2012 at 16:14 | comment | added | Tom Goodwillie | Of course, they are in your class of examples ($f$ gives a diffeomorphism between $M$ and $M'$). | |
| Feb 17, 2012 at 16:11 | comment | added | Tom Goodwillie | Your $\mathcal A'$ should not be called a new atlas on the same differentiable manifold, but rather a new atlas on the same topological manifold; or you can say that it makes a new differentiable manifold $M'$ with the same underlying topological manifold as $M$. Are you wondering when $M$ and $M'$ are diffeomorphic? | |
| Feb 17, 2012 at 16:08 | comment | added | Tom Goodwillie | The question is unclear. As I understand the terms, a differentiable manifold has a unique smooth maximal atlas. The same topological manifold can have more than one differentiable structure, that is, is can be more than differentiable manifold, that is, it can be given more than one maximal smooth atlas (where "smooth atlas" means that any two charts in the atlas are compatible with each other and "maximal" means that every chart compatible with the atlas is in the atlas). | |
| Feb 17, 2012 at 15:55 | history | asked | Cristi Stoica | CC BY-SA 3.0 |