# Adjoint representation of $\mathfrak{b}_2$ is undecomposable

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.
• Thank you! I can see that $\mathfrak{b}_2(\Bbb C)$ is nilpotent hence solvable. But is $\mathfrak{b}_2(\Bbb C)$ simple? – PerelMan Apr 15 at 20:40
• 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). – Dietrich Burde Apr 15 at 20:42
• 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) – PerelMan Apr 15 at 20:53