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I understand the notation [tex][\mathcal{L},\mathcal{L}]=0[/tex] to be "the whole Lie algebra commutes with itself", but the precise meaning of [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex] I'm unsure of.

Does it mean that every element of [tex]\mathcal{L}[/tex] can be generated by taking the commutator of two elements within [tex]\mathcal{L}[/tex]?

I've been asked to show that if [tex]\mathcal{L}[/tex] is simple, and [tex][\mathcal{L},\mathcal{L}] \not= 0[/tex], then [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

Am I right in thinking if [tex]I[/tex] is an ideal, then by definition [tex][I,\mathcal{L}] = I[/tex]. Since [tex]\mathcal{L}[/tex] is simple, then [tex]I = 0[/tex] or [tex]I =\mathcal{L}[/tex]. The only possible combinations of ideals are therefore [tex][0,\mathcal{L}]=0[/tex], [tex][\mathcal{L},\mathcal{L}]=0[/tex] and [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

ThoughI don't feel it constitutes much of a proof. I'd be very grateful for any help given.

Does it mean that every element of [tex]\mathcal{L}[/tex] can be generated by taking the commutator of two elements within [tex]\mathcal{L}[/tex]?

I've been asked to show that if [tex]\mathcal{L}[/tex] is simple, and [tex][\mathcal{L},\mathcal{L}] \not= 0[/tex], then [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

Am I right in thinking if [tex]I[/tex] is an ideal, then by definition [tex][I,\mathcal{L}] = I[/tex]. Since [tex]\mathcal{L}[/tex] is simple, then [tex]I = 0[/tex] or [tex]I =\mathcal{L}[/tex]. The only possible combinations of ideals are therefore [tex][0,\mathcal{L}]=0[/tex], [tex][\mathcal{L},\mathcal{L}]=0[/tex] and [tex][\mathcal{L},\mathcal{L}]=\mathcal{L}[/tex].

ThoughI don't feel it constitutes much of a proof. I'd be very grateful for any help given.

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