 Solution

A finite-dimensional Lie algebra representation is a homomorphism $\rho :L\to \mathfrak{gl}(V)$. It is irreducible when $V$ has no invariant subspaces other than $0$ and $V$.

The algebra ${\mathfrak{sl}}_2$ has basis $e,f,h$ withFor every $n\ge 0$, its $(n+1)$-dimensional irreducible module $V(n)$ has basis $v_0,\dots ,v_n$ and actionwith out-of-range vectors zero. Any nonzero invariant subspace contains a weight vector; repeated application of $e$ reaches $v_0$, and repeated application of $f$ then generates the whole module, proving irreducibility. The adjoint module of ${\mathfrak{sl}}_2$ is $V(2)$, so every ideal is an invariant subspace and ${\mathfrak{sl}}_2$ is simple.

For a root $\alpha$, nondegeneracy of the Killing pairing between $L_{\alpha }$ and $L_{-\alpha }$ allows choices $e_{\alpha },f_{\alpha }$ withAfter rescaling, $(e_{\alpha },f_{\alpha },h_{\alpha })$ obey the ${\mathfrak{sl}}_2$ relations, giving a copy of ${\mathfrak{sl}}_2$ in $U_{\alpha }$.

Restrict the adjoint representation of $L$ to this copy. Finite-dimensional ${\mathfrak{sl}}_2$ theory shows that the $\alpha$-string through zero has one nontrivial summand $V(2)$, with weight spaces $L_{-\alpha }$, $\mathbb{C}{h}_{\alpha }$, and $L_{\alpha }$. Any additional vector in $L_{\alpha }$ would generate another weight-two summand and another independent zero-weight coroot, contradicting nondegeneracy of the root--coroot pairing on $H$. Hence

For $\beta \notin \mathbb{Z}\alpha$, the spaceis stable under $e_{\alpha },f_{\alpha },h_{\alpha }$. Adjacent raising and lowering maps are nonzero until the endpoints, and every root space is one-dimensional, so $W$ is a simple ${\mathfrak{sl}}_2$-module of highest weight $p+q$. Its lowest and highest $h_{\alpha }$-weights giveand therefore $\beta (h_{\alpha })=q-p$. Since $[L_{\alpha },L_{-\alpha }]=\mathbb{C}{h}_{\alpha }$, every $x$ in this bracket line satisfies