Timeline for Can one show the equivalence of the abstract and classical Jordan decompositions for simple Lie algebras without complete reducibility?
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7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 21, 2018 at 19:40 | answer | added | Alexander Premet | timeline score: 7 | |
| Sep 15, 2013 at 22:27 | answer | added | Michaël Le Barbier | timeline score: 6 | |
| Sep 15, 2013 at 14:20 | answer | added | Ben Webster♦ | timeline score: 5 | |
| Sep 15, 2013 at 14:05 | answer | added | Jim Humphreys | timeline score: 9 | |
| Sep 15, 2013 at 4:22 | comment | added | Marguax | One can give a proof using just basic notions for linear algebraic groups over any field $k$ of char. 0 (no unitarian trick or root systems or serious structure theory lurking in the shadows). It isn't suitable for your purposes, but doesn't use anything about the theory of simple Lie algebras. Since char($k$)=0 and $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ by simplicity, Corollary 7.9 in Ch. II of Borel's book on algebraic groups implies $\mathfrak{g}={\rm{Lie}}(G)$ for a (smooth) Zariski-closed $k$-subgroup $G \subset {\rm{GL}}_n$. Use 4.4 in Ch. I (valid in any characteristic) to conclude. | |
| Sep 15, 2013 at 1:31 | comment | added | darij grinberg | Do you consider Cartan's criterion elementary enough? I always found its proof an unmotivated (if short and very readable, by now) tour-de-force. | |
| Sep 15, 2013 at 0:33 | history | asked | Terry Tao | CC BY-SA 3.0 |