Timeline for Differential Topology over $\mathbb{Q}$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| May 24, 2015 at 16:51 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
Minor phrasing improved
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| May 24, 2015 at 16:50 | vote | accept | Asaf Shachar | ||
| May 22, 2015 at 7:40 | comment | added | Simon Henry | Sorry, I got confused you're right. I was thinking about $\sigma$-compacity and all the other equivalent properties and I forgot that paracompactness is a weaker assumption. | |
| May 22, 2015 at 0:54 | comment | added | David Roberts♦ | @SimonHenry - really? The disjoint union of uncountably-many copies of $\mathbb{R}$ is paracompact is it not? Or are you assuming second countability in your definition of 'ordinary manifold'? | |
| May 21, 2015 at 15:10 | comment | added | Simon Henry | Countablity is the natural analogue of paracompactness: an ordinary manifold is paracompact if and only if it has a countable atlas. | |
| May 21, 2015 at 15:08 | comment | added | Gerald Edgar | For example, an uncountable disjoint union of copies of $\mathbb Q^n$. So the requirement may as well be that the whole space is countable (as Simon assumes). | |
| May 21, 2015 at 15:04 | comment | added | Asaf Shachar | You are probably right about paracompactness. I thought I should add a requirement that the manifold will be second-countable, but was unsure it was necessary in this case. | |
| May 21, 2015 at 15:01 | answer | added | Simon Henry | timeline score: 11 | |
| May 21, 2015 at 15:00 | comment | added | Asaf Shachar | Yes. I meant only pointwise differentiable. This is the weakest notion of differentiability I could think of. As I said, the chain rule implies that even w.r.t this weak notion of equivalence, manifolds of different dimensions are inequivealent. Of course I see no apparent reason for not choosing to work in a smooth (infinitely differentiable) category or any other degree of differentiability between them. (That is, to use stronger equivalence relations). I will consider any ability to distinguish between two manifolds in every smooth category whatsoever. | |
| May 21, 2015 at 14:53 | comment | added | Gerald Edgar | Also, we need a definition of "rational manifold". Without paracompactness you could have some sort of "long line" not of the same cardinal as the "short line" $\mathbb Q$. | |
| May 21, 2015 at 14:39 | comment | added | Gerald Edgar | So "differentiable" means differentiable at each point of the domain? So the differential may vary from point to point? And need not depend continuously on the point? | |
| May 21, 2015 at 13:58 | history | edited | Simon Rose | CC BY-SA 3.0 |
Very minor grammatical change.
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| May 21, 2015 at 13:25 | history | asked | Asaf Shachar | CC BY-SA 3.0 |