1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.