Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2015 at 11:33 | comment | added | Allen Knutson | The version I'd heard (possibly due to Mazur?) of Mariano's argument was about $K^\times/\mathbb R^\times$, the sphere (rather than projective space). If $K > \mathbb R$, then this sphere is a compact connected abelian group, hence (by their classification) a torus. But for $\dim K > 2$ spheres are simply connected hence not tori. Unfortunately this doesn't show the uniqueness of $\mathbb C$. | |
| Aug 20, 2015 at 23:51 | comment | added | ParaH2 | @MarianoSuárez-Alvarez I'm still laughing !! :) | |
| Jan 30, 2013 at 22:54 | comment | added | Johannes Ebert | @Ryan: yes, in a sense it is a nice proof. Lefschetz fixed point theorem is a hard result, which depends either on Poincare duality or on simplicial approximation. Most topological proofs I know are considerably more elementary (and use the topology of the complex plane, which is more obviously related to the problem than self-maps of $CP^n$). | |
| Dec 30, 2012 at 5:35 | comment | added | Ryan Reich | This is a perfectly good proof, and is it so much more sophisticated than any of the others? Many of the favorite proofs of this theorem are similarly topological. | |
| Dec 21, 2012 at 20:04 | comment | added | Mariano Suárez-Álvarez | A slightly different approach is to suppose there is a finite extension $K/\mathbb C$. Then $P(K)$, the projectivisation of $K$ as a complex vector space is a compact abelian Lie group (with multiplication induced by that of the group $K^\times$). Such a thing must be a torus, so its $H^1$ is not zero. Yet it is a complex projective space, and one easily sees that its $H^1$ is zero. | |
| Dec 21, 2012 at 19:13 | history | answered | Johannes Ebert | CC BY-SA 3.0 |