Timeline for Looking for a simple proof that R^2 has only one smooth structure
Current License: CC BY-SA 2.5
12 events
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| Mar 15, 2011 at 19:11 | comment | added | Johannes Ebert | If I remember correctly, the first step in the most standard proof of the uniformzation theorem is to show that a simply connected surface can be exhausted by relatively compact simply connected subsets (maybe with nice boundary). For that, a proper function and some algebraic topology is needed, in other words: quite a bit of what I sketched above. Only after this is done, the geometric analysis part of the proof begins. | |
| Mar 15, 2011 at 18:18 | comment | added | Johannes Ebert | ...once it is second countable, which we have to assume in the first place. Otherwise, it won't be diffeomorphic to $R^2$ anyway. | |
| Mar 15, 2011 at 16:52 | vote | accept | Hugo Chapdelaine | ||
| Mar 15, 2011 at 15:21 | comment | added | Igor Belegradek | Maxime, any smooth manifold admits a proper smooth function constructable via a partition of unity. | |
| Mar 15, 2011 at 14:23 | comment | added | Maxime Bourrigan | OK, my confusion stemmed from the fact that I don't usually suppose Morse functions to be proper. To be complete, how do you prove that a surface homeomorphic to the plane admits a proper function? | |
| Mar 15, 2011 at 14:14 | comment | added | Johannes Ebert | When I say ''Morse function'', I mean: 1) $f$ has the singularities as asserted, 2) f is bounded from below; 3) $f$ is proper. The distance function on a small exotic $R^4$ (i.e. embedded into the standard $R^4$) is not a Morse function in this sense, because it is not proper. | |
| Mar 15, 2011 at 14:12 | comment | added | Johannes Ebert | Assume that the minimum of $f$ is at $p$ and $f(p)=0$. For any $b>a >0$, the manifold $f^{-1}[a,b]$ is diffeomorphic to $[a,b] \times f^{-1}(a)$, see Milnor, Morse theory, Thm 3.1. Moreover, $f$ corresponds to the projection onto $R$ under this diffeomorphism (the phrase "use the flow lines" corresponds to this). Do this for many intervals; you get a diffeo $f^{-1}[a,\infty) \cong f^{-1}(a) \times [a,\infty)$. By the Morse lemma, $f^{-1}[0,a]$ is a disc when $a$ is small enough. These things can be glued together to give the asserted diffeo. | |
| Mar 15, 2011 at 13:06 | comment | added | Maxime Bourrigan | BTW, your proof confuses me, wouldn't the square of the norm be a Morse function with exactly one minimum on a small exotic R4 ? How do you "cook up" your diffeomorphism? | |
| Mar 15, 2011 at 12:49 | comment | added | Maxime Bourrigan | It is definitely far from simple and indeed was the source of great development in the XIXth and XXth century (if you allow me some advertisement pro domo, this book explains a part of this story: amazon.fr/…) but some modern proofs of the whole package aren't that long... | |
| Mar 15, 2011 at 11:40 | comment | added | Hugo Chapdelaine | Thanks Johannes, I think that your argument fulfils my requirement! And yes indeed, the proof of the uniformization theorem of simply connected Riemann surfaces is far from trivial and extremely deep! | |
| Mar 15, 2011 at 9:55 | history | edited | Johannes Ebert | CC BY-SA 2.5 |
added 95 characters in body; added 27 characters in body
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| Mar 15, 2011 at 9:42 | history | answered | Johannes Ebert | CC BY-SA 2.5 |