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Consider a semisimple Lie algebra S and a vector space V - considered as Abelian Lie algebra - with a non-zero irreducible representation $$\rho: S \rightarrow End(V).$$ $L$ and $V$ are finite-dimensional over the base field $\mathbb{R}$ or $\mathbb{C}$.

Then the semidirect product $$L:= V \rtimes_\rho S$$ is a perfect Lie algebra, i.e. $[L,L]=L$.

  • How to prove this result?
Jo Wehler
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  • Hints: What is $[S,S]$? What is $[S,V]$? conclusion? – YCor Dec 17 '16 at 23:55
  • @YCor I know $[S,S]=S$ and $[S,V]=V$. But given $(v,s)\in L$ and knowing $s=[s_1,s_2]$ how to find $v_1,v_2\in V$ with $\rho(s_1)(v_2)-\rho(s_2)(v_1)=v$? – Jo Wehler Dec 18 '16 at 06:52
  • check out the definition of $[L,L]$: it is by definition the set of sums of elements in the image of the bracket map, not only the image itself. So the problem is easier than you thought. – YCor Dec 18 '16 at 07:11
  • @YCor I apologize, but even with your hint I do not recognize the solution: My problem is the representation of a given element $v\in V$ by using fixed(!) elements $s_i$ from L. – Jo Wehler Dec 18 '16 at 08:02
  • No, your problem is "how to prove that $L$ is a perfect Lie algebra" (which means that every element is a sum of finitely many commutators). If your problem is whether every element of $L$ is a single commutator, or anything else, this is more difficult and you should edit your question. – YCor Dec 18 '16 at 08:11
  • @YCor I agree: My problem is perfectness, which means to represent an element $(v,s)$ as a sum of commutators. But to start, I assume $s=[s_1,s_2]$. How to represent an element $(v,s)$ as a sum of commutators, using $s_1$ and $s_2$? – Jo Wehler Dec 18 '16 at 08:20
  • You have $[S,S]=S$ and $[S,V]=V$, so $[S+V,S+V]=S+V$, that's it. What you're trying to prove is more complicated. – YCor Dec 18 '16 at 08:39
  • @YCor Thanks, I was lacking the equivalence of $\rho(S)(V)=V$ and $[S,V]_L=V$. – Jo Wehler Dec 18 '16 at 09:12

1 Answers1

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Solution thanks to the comments of @YCor:

Assume $L=V \rtimes_{\rho}S$. Then $$L=V+S$$ with $S\subset L$ subalgebra and $V\subset L$ ideal. By definition of the commutator in $L$ holds

$$[S,S]_S=[S,S]_L.$$ Semisimpleness of $S$ implies $$S=[S,S]_S.$$ Moreover $$\rho(S)(V)=[S,V]_L.$$ Irreducucibility of $\rho \neq 0$ implies $$V=\rho(S)(V)=[S,V]_L.$$ Then $$L=S+V=[S,S]_L+[S,V]_L\subset [L,L]_L$$ The other inclusion is obvious.

Jo Wehler
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