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.


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.

  • 2
    $\begingroup$ 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. $\endgroup$ – PerelMan Apr 15 at 17:35
  • 1
    $\begingroup$ Yes, this is correct. $\endgroup$ – Dietrich Burde Apr 15 at 18:03

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