Timeline for Is every finite-dimensional Lie algebra the Lie algebra of an algebraic group?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2017 at 4:57 | answer | added | YCor | timeline score: 3 | |
| May 5, 2010 at 18:43 | answer | added | Jim Humphreys | timeline score: 15 | |
| May 5, 2010 at 16:47 | comment | added | BCnrd | A very nice fact over fields $k$ of char. 0: for any linear algebraic $k$-group G and Lie $k$-subalgebra h in g = Lie(G), [h,h] = Lie(G') for a (unique) connected closed $k$-subgroup $G'$ in $G$. In particular, if h is a semisimple Lie $k$-subalgebra of g (so h = [h,h]) then it is the Lie algebra of a connected closed $k$-subgroup of $G$. See 7.9 in Borel's book on linear algebraic groups (and 7.7 for a nec/sufficient condition in general, in char. 0). So as always, it's the commutative/solvable stuff that creates all the headaches. | |
| May 5, 2010 at 14:51 | answer | added | Allen Knutson | timeline score: 8 | |
| May 5, 2010 at 6:17 | answer | added | Nicolás | timeline score: 0 | |
| Oct 21, 2009 at 13:30 | answer | added | JS Milne | timeline score: 42 | |
| Oct 15, 2009 at 19:25 | vote | accept | Anton Geraschenko | ||
| Oct 15, 2009 at 14:38 | answer | added | Enno | timeline score: 3 | |
| Oct 15, 2009 at 6:28 | answer | added | William Slofstra | timeline score: 28 | |
| Oct 7, 2009 at 21:17 | answer | added | Theo Johnson-Freyd | timeline score: 0 | |
| Oct 5, 2009 at 20:02 | answer | added | Ben Webster♦ | timeline score: 1 | |
| Oct 5, 2009 at 19:48 | history | asked | Anton Geraschenko | CC BY-SA 2.5 |