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

Question

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.

Answer 1

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.