If $L$ is a semisimple lie algebra then $L=[L,L]$. Is the opposite true?
1 Answer
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No. A Lie algebra satisfying that property is called perfect. For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $L$ and a nontrivial irreducible representation $V$ of $L$, and define a bracket on $L \times V$ by $$ [(X,v),(Y,u)] := ([X,Y],XuYv). $$ This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \times V) = V$.
(Note (YCor): this is a semidirect product, hence usually denoted $L\ltimes V$.)

3$\begingroup$ Maybe it's also helpful to mention the substantial though often isolated literature on perfect Lie algebras and related structure theory? There are some interesting connections with other questions, as in the paper by Benkart and Zelmanov in Invent. Math. 126 (1996), 145. $\endgroup$ Commented Apr 5, 2011 at 13:08

$\begingroup$ @JimHumphreys, you mention the substantial literature on this subject—where else would one look? $\endgroup$– LSpiceCommented Jun 11, 2015 at 18:49

1$\begingroup$ @LSpice: Besides the paper I mentioned, there are many others listed on MathSciNet under "perfect Lie algebra" though I'm not sure what would interest you. I guess my point was that the purely algebraic theory of Lie algebras (often in characteristic 0) has been studied by many people over the past century; thus, much is known. A random example is the paper by Baranov and Zalesskii: Plain representations of Lie algebras, J. London Math. Soc. (2) 63 (2001), no. 3, 571–591. You can also find some related posts here by searching MO for "perfect Lie algebra". $\endgroup$ Commented Jun 11, 2015 at 21:13

3$\begingroup$ Note that the onedimensional vector space $V$ for which $\rho : L \to \mathfrak{gl}(V)$ is zero is also a simple $L$module, and in this case $[L \times V, L \times V] = L \subsetneq L \times V$. You need to exclude that case, otherwise it works fine since $[L,V]$ contains a nonzero vector, hence is equal to $V$. $\endgroup$ Commented Feb 15, 2017 at 6:15

4$\begingroup$ A remark on this answer, is that every nonsemisimple perfect Lie algebra (fin. dim., in char 0) has a quotient of the form you're describing. So, in a sense, these are the minimal counterexamples (well, it's minimal if the representation is faithful, in the sense that in this case $L\ltimes V$ is not semisimple but all its proper quotients are semisimple). $\endgroup$– YCorCommented May 2, 2022 at 7:59