Timeline for Zero-divisors in a graded Lie algebra
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jun 5, 2013 at 11:33 | vote | accept | Mark Grant | ||
Jun 4, 2013 at 12:32 | comment | added | Mark Grant | @Dietrich: yes, thanks. I should have mentioned that we assume that $\mathfrak{g}$ is infinite-dimensional. | |
Jun 4, 2013 at 12:31 | answer | added | Simon Wadsley | timeline score: 4 | |
Jun 4, 2013 at 11:53 | comment | added | Dietrich Burde | @Mark, I see. If $\mathfrak{g}$ is finite-dimensional, then it is nilpotent by Jacobson's theorem, and we may take a nonzero element $a\in Z(\mathfrak{g})$ and any $b$ with $a$ and $b$ linearly independent to obtain $[a,b]=0$. | |
Jun 4, 2013 at 11:43 | comment | added | Mark Grant | @Dietrich: Perhaps. I changed one word to make the question clearer. | |
Jun 4, 2013 at 11:42 | history | edited | Mark Grant | CC BY-SA 3.0 |
changed a word
|
Jun 4, 2013 at 11:38 | comment | added | Dietrich Burde | The Yang-Mills Lie algebra $\mathfrak{y}\mathfrak{m}(2)$ over $\mathbb{Q}$ has a grading concentrated in even degrees. It is isomorphic to the rational Heisenberg Lie algebra $h_1$ with basis $(e_1,e_2,e_3)$ and $[e_1,e_3]=[e_2,e_3]=0, [e_1,e_2]=e_3$. Did I misunderstand something ? | |
Jun 4, 2013 at 10:30 | history | asked | Mark Grant | CC BY-SA 3.0 |