Derived series of Lie algebra in reverse

There exists Lie algebras which are not a derived Lie algebra at all. More precisely, a filiform nilpotent Lie algebra is characteristically nilpotent (i.e., all derivations are nilpotent) if and only if it is not a derived algebra - see here.

Here is such an example, of a characteristically nilpotent Lie algebra $L$ of dimension $13$. The Lie brackets, with respect to a basis $(e_1,\ldots ,e_{13})$ are given as follows:

\begin{align*} [e_1,e_i] & = e_{i+1},\quad 2\le i\le 12 \\[0.2cm] [e_2,e_3] & = e_5 \\ [e_2,e_4] & = e_6 \\ [e_2,e_5] & = \frac{9}{10}e_7-e_9 \\ [e_2,e_6] & = \frac{4}{5}e_8-2e_{10} \\ [e_2,e_7] & = \frac{5}{7}e_9-\frac{335}{126}e_{11}+ \frac{22105}{15246}e_{13}\\ [e_2,e_8] & = \frac{9}{14}e_{10}-\frac{125}{42}e_{12}\\ [e_2,e_9] & = \frac{7}{12}e_{11}-\frac{4421}{1452}e_{13}\\ [e_2,e_{10}] & = \frac{8}{15}e_{12}\\ [e_2,e_{11}] & = \frac{27}{55}e_{13}\\[0.5cm] [e_3,e_4] & = \frac{1}{10}e_{7}+e_{9}\\ [e_3,e_5] & = \frac{1}{10}e_{8}+e_{10}\\ [e_3,e_6] & = \frac{3}{35}e_9+\frac{83}{126}e_{11}-\frac{22105}{15246}e_{13}\\ [e_3,e_7] & = \frac{1}{14}e_{10}+\frac{20}{63}e_{12}\\ [e_3,e_8] & = \frac{5}{84}e_{11}+\frac{697}{10164}e_{13}\\ [e_3,e_{9}] & = \frac{1}{20}e_{12}\\ [e_3,e_{10}] & = \frac{7}{165}e_{13}\\ [e_4,e_5] & = \frac{1}{70}e_9+\frac{43}{126}e_{11}+\frac{22105}{15246}e_{13}\\ [e_4,e_6] & = \frac{1}{70}e_{10}+\frac{43}{126}e_{12}\\ [e_4,e_7] & = \frac{1}{84}e_{11}+\frac{7589}{30492}e_{13}\\ [e_4,e_8] & = \frac{1}{105}e_{12}\\ [e_4,e_9] & = \frac{1}{132}e_{13}\\[0.5cm] [e_5,e_6] & = \frac{1}{420}e_{11}+\frac{313}{3388}e_{13}\\ [e_5,e_7] & = \frac{1}{420}e_{12}\\ [e_5,e_8] & = \frac{3}{1540}e_{13}\\[0.5cm] [e_6,e_7] & = \frac{1}{2310}e_{13} \end{align*}

Tags:

Lie Algebras

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