Timeline for Looking for a simple proof that R^2 has only one smooth structure
Current License: CC BY-SA 2.5
16 events
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| Apr 12 at 2:04 | comment | added | The Amplitwist | Reposting a link mentioned in a previous comment so that it appears in the "Linked" questions list: Does every smooth manifold of infinite topological type admit a complete Riemannian metric? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 23, 2015 at 17:58 | comment | added | user78462 | I am a bit confused by the proof, as what it gives is that U is diffeomorphic to a subset of itself which in turn is homeomorphic to a ball...how then you conclude that the thus produced homeomorphism is a diffeomorphism? I do not think that the disk you end up with by the simple fact that it is "small" as you want must be diffeomorphic to the standard unit ball. Please, if you have time, would you clarify your argument? As it is pointed out, this would also show that $R^4$ has only one differential structure: given something homeomorphic to a 4-ball you have a Morse function with just one cri | |
| Mar 15, 2011 at 16:52 | vote | accept | Hugo Chapdelaine | ||
| Mar 15, 2011 at 11:34 | history | edited | Hugo Chapdelaine | CC BY-SA 2.5 |
added 1 characters in body
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| Mar 15, 2011 at 9:42 | answer | added | Johannes Ebert | timeline score: 14 | |
| Mar 15, 2011 at 2:25 | comment | added | Mariano Suárez-Álvarez | @Ken, if $\gamma:\mathbb R\to M$ is a geodesic in a $1$-dimensional manifold $M$, then $\gamma$ is a local diffeo by the inverse function theorem: it's differential is not zero, so it is an isomorphism by the fact that the dimension of the tangent vector spaces is just $1$. | |
| Mar 15, 2011 at 1:02 | comment | added | Maxime Bourrigan | It seems to me that some of the modern proofs of the Uniformisation theorem have few to no topological prerequisites. I wouldn't be surprised if some of them did prove that any smooth structure on R2 can be refined to a complex structure equivalent to C or the unit disk, and is therefore diffeomorphic to R2. | |
| Mar 14, 2011 at 23:50 | comment | added | Ken Knox | Dan- thanks for the clarification. The definition of smooth structure in use here is what I would call a 'diffeomorphism type'. | |
| Mar 14, 2011 at 23:35 | comment | added | Hugo Chapdelaine | Hi Ken, the example you gave, say $X$, is equivalent to the usual smooth structure on $\mathbf{R}$. Indeed the map $\sqrt[3]:X\rightarrow\mathbf{R}$ is a diffeomorphism! | |
| Mar 14, 2011 at 23:22 | comment | added | Dan Ramras | Ken, I think there's a confusion regarding the meaning of "smooth structure". One meaning is a maximal atlas of smooth charts. Another is a diffeomorphism class of such maximal atlases. Your example gives two distinct maximal atlases of smooth charts on $\mathbb{R}$. But these two maximal atlases give rise to diffeomorphic manifolds (note, though, that the diffeomorphism is not given by the identity map). I think this is explained in the first chapter of Spivak's book. | |
| Mar 14, 2011 at 22:29 | comment | added | Ken Knox | @Mariano I don't see how one gets a local diffeo here. For instance if I choose $(\mathbb{R}$, $f(x) = x^3$, d$x^2)$ as my manifold, smooth structure, and metric, then the unit speed geodesic emanating from $0$ is just the identity map. This does not give a local diffeo at $0$ from $(\mathbb{R}, \textrm{Id})$ to $(\mathbb{R}, f)$ since the transition map is not differentiable at the origin. | |
| Mar 14, 2011 at 22:06 | comment | added | Mariano Suárez-Álvarez | A different way to phrase that argument is: we can pick a complete Riemannian metric on our $1$-dimensional smooth manifold $M$, see e.g. mathoverflow.net/questions/18844; the geodesic through one of its points is a surjective map $\mathbb R\to M$, according to Hopf-Rinow. The map is locally a diffeo by the inverse function theorem, and if it were not injective it would be periodic, by uniqueness of geodesics---but in that case $M$ would be compact. | |
| Mar 14, 2011 at 22:02 | comment | added | Ken Knox | Perhaps it is worth clarifying your definition of smooth structure? For instance the smooth structure determined by $(\mathbb{R}, f)$ where $f(x) = x^3$ is usually given as an example showing that $\mathbb{R}$ does not have only one smooth structure. | |
| Mar 14, 2011 at 21:55 | comment | added | Mariano Suárez-Álvarez | It would be strange (or interesting...) that a similar argument worked for $n=2$ and $n=3$, because nothing similar can work for $n=4$ :) | |
| Mar 14, 2011 at 21:45 | history | asked | Hugo Chapdelaine | CC BY-SA 2.5 |