Timeline for Can every manifold be given an analytic structure?
Current License: CC BY-SA 2.5
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| Sep 24, 2021 at 7:37 | comment | added | YCor | @TimCampion This might be worth a separate question. In the statement "every $C^k$-structure has a unique $C^\ell$-smoothing ($k<\ell\le\omega$)" I would understand that given two $C^\ell$-smoothings there is a $C^k$-diffeomorphism carrying one $C^\ell$-structure to the other one. But is this true? and if so, does one expect more uniqueness (e.g., that the diffeomorphism be isotopic — in which sense?— to the identity? that the diffeomorphism can be taken arbitrarily — in which sense— close to identity?) | |
| Apr 27, 2021 at 18:38 | answer | added | Moishe Kohan | timeline score: 8 | |
| Jun 13, 2019 at 20:54 | review | Suggested edits | |||
| Jun 13, 2019 at 21:47 | |||||
| Apr 28, 2013 at 21:16 | comment | added | Tim Campion | Oh-- and nobody has said anything about uniqueness of smoothings from $C^k$ to $C^{k+1}$ (although existence follows from the existence of a $C^\infty$ smoothing). Does it hold, in whatever relevant sense? I'd have to guess from the rest of the discussion that it does, but nobody has actually said it... | |
| Apr 28, 2013 at 21:09 | comment | added | Tim Campion | At a minimum, uniqueness of smoothings ought to mean that the "smooth" diffeo type determines the "rough" diffeo type. That is, if $Y,Y'$ are smoothings of $X$, then there exists SOME "smooth" diffeo $Y \to Y'$. You'd have to think that this "smooth" diffeo is somehow close/homotopic to the canonical "rough" diffeo $Y \to Y'$ coming from the smoothing maps. Does the regularity of the homotopy just fall out of the more general theory of smoothing of maps? In any case, the embedding of the "smooth" diffeo group into the "rough" one should be dense, and a (weak?) homotopy equivalence ... right? | |
| Apr 28, 2013 at 20:23 | comment | added | Tim Campion | E.g. When "rough" = $C^\infty$ and "smooth" = $C^\omega$: Let $\phi$ be a $C^\infty$, nondecreasing "step" function on $\mathbb{R}$ (constant outside an interval, but not globally constant). Then $f(x) = x + \phi(x)$ is a $C^\infty$ diffeomorphism of $\mathbb{R}$ which is not $C^\omega$. Similarly, when "rough" = $C^1$ and "smooth" = $C^2$ or better: $f(x) = x + \int_0^x |t|dt$ is a $C^1$-diffeo of $\mathbb{R}$ which is not $C^2$. | |
| Apr 28, 2013 at 20:13 | comment | added | Tim Campion | Caveat: the notion of uniqueness has to be kind of subtle. Obviously a "smoothing" of a "rough" manifold $X$ is a "rough" diffeomorphism $X \to Y$, where $Y$ is a "smooth" manifold. What does it mean for the smoothing to be unique? Naively one might ask that for any other smoothing map $X \to Y'$, the induced map $Y \to Y'$ is "smooth". This would imply that any "rough" diffeo between "smooth" manifolds is automatically "smooth". This is false for the notions of "rough" and "smooth" that we're considering. | |
| Apr 28, 2013 at 17:22 | comment | added | Tim Campion | In perusing this question, I was amused to notice that a complete answer is spread out across the three answers below: $C^0$ manifolds cannot in general be smoothed to $C^1$, and if they can, then the smoothing may not be unique in dimensions 4 and up. In contrast, for $k \geq 1$, a $C^k$ manifold admits a unique smoothing to a $C^\infty$ structure, and a $C^\infty$ manifold admits a unique smoothing to a $C^\omega$ structure. | |
| Dec 13, 2009 at 22:17 | history | edited | Kim Morrison | CC BY-SA 2.5 |
added 7 characters in body
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| Dec 13, 2009 at 20:45 | answer | added | Greg Kuperberg | timeline score: 47 | |
| Dec 13, 2009 at 20:04 | vote | accept | Theo Johnson-Freyd | ||
| Dec 13, 2009 at 20:03 | answer | added | Ryan Budney | timeline score: 89 | |
| Dec 13, 2009 at 19:53 | answer | added | Mariano Suárez-Álvarez | timeline score: 18 | |
| Dec 13, 2009 at 19:44 | history | asked | Theo Johnson-Freyd | CC BY-SA 2.5 |