Let $\mathfrak{g}=\mathfrak{b}_2(\Bbb C)$.

Prove that adjoint representation $ad_\mathfrak{g}$ of $\mathfrak{g}$ is undecomposable into a direct sum of irreducible representations.

My attempt:

I guess undecomposable means irreducible? In that case:

I proved that commutator ideal $D(\mathfrak{g}) \neq \mathfrak{g}$ which means that $\mathfrak{g}$ is not simple.

On the other hand:

If we prove the aim of my original question, that would mean that $ad_\mathfrak{g}$ has subrepresentations and by the correspondance that $\mathfrak{g}$ is not simple.

I can't find my mistake, thanks for your help.


No, "undecomposable" means that it cannot be written as a direct sum of irreducible representations. By Lie's Theorem, every irreducible representation of a complex solvable Lie algebra is $1$-dimensional. Hence the adjoint representation is reducible, but obviously not a direct sum of trivial representations.

  • $\begingroup$ Thank you! I can see that $\mathfrak{b}_2(\Bbb C)$ is nilpotent hence solvable. But is $\mathfrak{b}_2(\Bbb C)$ simple? $\endgroup$ – PerelMan Apr 15 at 20:40
  • $\begingroup$ What is $\mathfrak b_2(\Bbb C)$ for you? I thought that you mean the $2$-dimensional non-abelian Lie algebra, which is solvable but not nilpotent (and also not simple). $\endgroup$ – Dietrich Burde Apr 15 at 20:42
  • $\begingroup$ Oh sorry for confusion, for me $\mathfrak{b}_2(\Bbb C)$ if the subalgebra of $\mathfrak{gl}(\Bbb C)$ of upper triangular matrices of size $2\times2$. I thought it was standard notation (like a Borel algebra) $\endgroup$ – PerelMan Apr 15 at 20:53
  • 1
    $\begingroup$ Yes, this is the algebra I meant. This algebra is not nilpotent. $\endgroup$ – Dietrich Burde Apr 15 at 21:14
  • 1
    $\begingroup$ Then the action would be trivial. But the action of the adjoint representation is not trivial. $\endgroup$ – Dietrich Burde Apr 16 at 9:29

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.