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I've read several definitions of $LCS$ and derived series of Lie algebra, but I'm not sure if i get it right: In case of $LCS$, the relationship is given as $g_{k+1}=[g,g_k]$, does $``g_k"$ stand for $k$-th generator? And could someone kindly provide an example of such relationship for an $su(N)$ in matrix representation?

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  • $\begingroup$ This is a recursive definition missing the definition of the first value $g_0$ which I believe should be the Lie algebra $g$ itself. So $g_0 = g$, $g_1 = [g,g]$ and $g_2 = [g_1,g] = [[g,g],g]$, a decreasing sequence of subspaces of $g$. Note that you cannot really ask for an 'example' in the case $g = su(N)$ because for fixed $g$ there is only one lower central series. $\endgroup$ Jul 15, 2015 at 6:45
  • $\begingroup$ $g=su(N)$ is a quite boring example: since $su(N)=[su(N),su(N)]$ this series is constantly equal to $su(N)$ itself. If you compute it in the case of strictly upper triangular matrices (try the $3\times 3$ and $4\times 4$case) you''ll understand things better $\endgroup$
    – N. Ciccoli
    Jul 15, 2015 at 6:52
  • $\begingroup$ Does su(n)=[su(n),su(n)] mean it is neither solvable nor nilpotent? (Sorry if my questions are primitive, i just started studying lie algebras) $\endgroup$
    – Kosm
    Jul 15, 2015 at 8:10
  • $\begingroup$ Yes, this is correct. $\mathfrak{su}(n)$ is not solvable, hence not nilpotent, because it is perfect, i.e., equals its own commutator ideal. $\endgroup$ Jul 15, 2015 at 14:34
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    $\begingroup$ Ok, thanks all. I'll keep studying. $\endgroup$
    – Kosm
    Jul 16, 2015 at 6:53

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