Some beginner facts on representaions of $\mathfrak{sl}_3(\mathbb{C})$

Question

Beginning to learn about representations of $\mathfrak{sl}_3(\mathbb{C})$. One starts with a subspace $$\mathfrak{h}=\{\begin{pmatrix} a_1 & 0 & 0\\ 0 &a_2& 0\\ 0 & 0 & a_3\\ \end{pmatrix}:\space a_1+a_2+a_3=0\}\subset \mathfrak{sl}_3(\mathbb{C})$$ Let $V$ be an arbitrary representation of a Lie algebra $\mathfrak{sl}_3$. One takes $\alpha\in \mathfrak{h}^*$ and introduces the space $$V_{\alpha}=\{v\in V:\space \forall H\in \mathfrak{h}\space Hv=\alpha(H)v\}$$
The question is why $V=\bigoplus V_{\alpha}$? Could you please explain this phenomenon?

Answer 1

Talking about the case $\dim V<\infty$ only.

Basically this is because the matrices in the spaces $\mathfrak{h}$ commute with each other. Consequently so do the matrices $\alpha(H), H\in\mathfrak{h}$. With a bit more work you can show that those matrices are also semisimple, i.e. diagonalizable. Whenever you have a commuting set of diagonalizable matrices over an algebraically closed field, they are simultaneously diagonalizable. In other words there is a basis $\mathcal{B}$ of $V$ such that all the matrices $\alpha(H)$ are diagonal w.r.t. $\mathcal{B}$.

Obviously the diagonal elements of $\alpha(H)$ are linear combinations of the entries of $H$ (the representation $\alpha$ is a linear mapping). The subspaces $V_\alpha$ are then spanned by subsets of $\mathcal{B}$. The subset includes those basis elements that correspond to the same linear combination, i.e. an element of $\mathfrak{h}^*$.