Timeline for Is there an analog of Clifford Theorem in the setting of Lie algebras?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 24, 2011 at 20:35 | comment | added | Salvatore Siciliano | Dear Torsten, thank you for your answer which confirm what I suspected. Really, I was mainly interested to the modular case, so your counterexample is just what I needed. | |
| Jun 24, 2011 at 20:22 | vote | accept | Salvatore Siciliano | ||
| Jun 24, 2011 at 18:35 | comment | added | Torsten Ekedahl | Yes, you're right, from my point of view the representation extends to the algebraic closure (the group generated by the exponentials). | |
| Jun 24, 2011 at 18:25 | comment | added | Victor Protsak | Yes, your fourth paragraph spells out some details omitted from my proof sketch. The Lie correspondence works in characteristic 0 without extra assumptions (and by Ado's theorem, the Lie algebra is linear, hence one can exponentiate), thus I don't think that algebraicity of the Lie algebra is an issue. The correspondence does, however, break down for infinite-dimensional representations, as in your first example. | |
| Jun 24, 2011 at 17:40 | history | answered | Torsten Ekedahl | CC BY-SA 3.0 |