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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ Yes, this is the algebra I meant. This algebra is not nilpotent. $\endgroup$ – Dietrich Burde Apr 15 at 21:14
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    $\begingroup$ Then the action would be trivial. But the action of the adjoint representation is not trivial. $\endgroup$ – Dietrich Burde Apr 16 at 9:29

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