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I am trying to prove that for a Lie algebra $\mathfrak{g}$:

$ad_{\mathfrak{g}}$ the adjoint representation of $\mathfrak{g}$ is irreducible iff $\mathfrak{g}$ is simple.

I tried to use the fact that stable ideals of $ad_{\mathfrak{g}}$ are ideals of $\mathfrak{g}$: Then if $\mathfrak{h}$ is a stable space under $\mathfrak{g}$ it is $\{0\}$ or the entire $\mathfrak{g}$. but I Couldn't go further.

Thank you for your help.

Conjecture
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1 Answers1

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Subrepresentations of the adjoint representation just correspond to ideals of $\mathfrak{g}$. Since $\mathfrak{g}$ is simple, they are trivial or the whole Lie algebra. Hence the adjoint representation is irreducible.

Dietrich Burde
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    Thank you! This is my attempt to prove the correspondence: $\mathfrak{h}$ ideal of $\mathfrak{g} \Leftrightarrow [\mathfrak{h},\mathfrak{g}] \subset \mathfrak{h} \Leftrightarrow \forall X\in \mathfrak{g}: ad_X(\mathfrak{h}) \subset \mathfrak{h} \Leftrightarrow \mathfrak{h}$ is subrepresentation of $ad_{\mathfrak{g}}$. is that correct? it's quite obvious but just to be sure I get it. – Conjecture Apr 15 '19 at 17:35
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    Yes, this is correct. – Dietrich Burde Apr 15 '19 at 18:03
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    Just a note here that is too obvious for some people to mention but I think is worth saying: it could also be the case that $ \mathfrak{g} $ is the one dimensional abelian Lie algebra which in some conventions is not considered simple. Then statement would be "adjoint rep is irreducible iff $ \mathfrak{g} $ is simple or $ \mathfrak{g} $ is the one dimenionsal abelian Lie algebra." – Ian Gershon Teixeira Mar 07 '22 at 20:26