Timeline for Number of real forms of a (not semisimple, solvable) Lie algebra
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 1 at 17:28 | comment | added | JRojo | Thank you very much again, this self-contained argument saves the day for me (I know very little of the general theory of Lie algebras). It was very helpful your answer. | |
| Jun 28 at 15:04 | comment | added | YCor | @JRojo yes, as I said, it's a theorem in Bourbaki. However, in the present case, one can argue more directly. In the complexification $\mathfrak{g}\subset\mathfrak{g}_\mathbf{C}$, if one takes an element outside the union of two hyperplanes, its centralizer is a Cartan subalgebra (here, 2-dimensional abelian complement to the derived subgroup). These two hyperplanes intersect $\mathfrak{g}$ at most in hyperplanes. So there exists some element in $\mathfrak{g}$ not in these hyperplanes, and hence its centralizer in $\mathfrak{g}$ is an abelian complement to the derived subalgebra. | |
| Jun 28 at 14:44 | vote | accept | JRojo | ||
| Jun 28 at 14:43 | comment | added | JRojo | Everything makes sense, the only I would ask is a little explanation as to the fact that $\mathfrak{a}+\mathfrak{h}=\mathfrak{g}$, is it always true that a cartan subalgebra plus the derived subalgebra fill the total algebra? | |
| Jun 28 at 13:39 | comment | added | JRojo | okey that was silly, any 2-dimensional nilpotent algebra is abelian | |
| Jun 28 at 11:04 | comment | added | JRojo | Thank you very much for the answer. A couple of questions: the fact that $\mathfrak{h}$ is 2-dimensional follows since in $\mathfrak{g}$ there are not nilpotent subalgebras of dimension $\ge 3$ (at the complex level), right? Another: why do you get an abelian subalgebra of $\mathfrak{gl}_2(K)$? The algebra $\mathfrak{h}$ is nilpotent, in principle it is not abelian, right? | |
| Jun 26 at 13:17 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
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| Jun 26 at 11:55 | history | answered | YCor | CC BY-SA 4.0 |