Consider the first element in the derived series of a Lie algebra $L$, defined by $L^{(1)}:=[L,L]$. For a given Lie algebra $\tilde{L}$, is there always a Lie algebra $L$ such that $L^{(1)}=\tilde{L}$? If yes, how do I construct $L$?
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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*}
Dietrich Burde
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Thanks for your answer. Is it possible that there is a typo in the example? Using $[e_2,e_9] = \frac{7}{12}e_{11}-\frac{4421}{1452}e_{13}$ the Jacobi identity is fulfilled. – user71769 Dec 29 '18 at 20:40
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(a) there are documented filiform char nilp Lie algebras of dimension 7 (3 isomorphism classes over complex numbers), why do you give such a complicated one? One is $12=3$, $13=4$, $14=5$, $15=6$, $16=7$, $23=6$, $24=25=43=7$. (b) I don't have access to your link, it would be better include reference info too. – YCor Dec 29 '18 at 20:54
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@user71769 Yes, thank you, it was a typo. – Dietrich Burde Dec 29 '18 at 22:23
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@YCor I had this example just at hand (because it also is an example for other statements). Of course you are right, there are easier ones. – Dietrich Burde Dec 29 '18 at 22:24
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Thanks. The full reference info is Castro-Jiménez, F. J.; Núñez-Valdés, J. Gröbner basis in the classification of characteristically nilpotent filiform Lie algebras of dimension 10. Algorithms in algebraic geometry and applications (Santander, 1994), 115–133, Progr. Math., 143, Birkhäuser, Basel, 1996. (Link: p121) – YCor Dec 29 '18 at 23:07
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@YCor Do you know whether those 7d algebras are minimal examples (of say, complex) Lie algebras not realizable as a derived algebras? – Travis Willse Dec 30 '18 at 00:51
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3@Travis yes: over an arbitrary field of char not 2,3, every Lie algebra of dimension $\le 6$ has a grading in positive integers, and hence is a derived Lie algebra. – YCor Dec 30 '18 at 05:20
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Suppose that $\mathbb{k}$ is a field of characteristic zero. It follows from Lie’s theorem and Engel’s theorem that a finite-dimensional $\mathbb{k}$-Lie algebra $\mathfrak{g}$ is solvable if and only if its derived Lie algebra $[\mathfrak{g}, \mathfrak{g}]$ is nilpotent.
Every finite-dimensional, solvable, non-nilpotent $\mathbb{k}$-Lie algebra is therefore not the derived Lie algebra of another $\mathbb{k}$-Lie algebra. Examples for this are $\mathfrak{t}(n, \mathbb{k})$ with $n \geq 2$.
Jendrik Stelzner
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