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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