= 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 $e,f,h$ with
$$
[h,e]=2e,\qquad[h,f]=-2f,\qquad[e,f]=h.
$$
For every $n\geq0$, its $(n+1)$-dimensional irreducible module $V(n)$ has basis $v_0,\ldots,v_n$ and action
$$
hv_i=(n-2i)v_i,\qquad
fv_i=v_{i+1},\qquad
ev_i=i(n-i+1)v_{i-1},
$$
with 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 $e_\alpha,f_\alpha$ with
$$
h_\alpha=[e_\alpha,f_\alpha],\qquad
\alpha(h_\alpha)=2.
$$
After 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 $L_{-\alpha}$, $\mathbb Ch_\alpha$, 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
$$
\boxed{\dim L_\alpha=1.}
$$
For $\beta\notin\mathbb Z\alpha$, the space
$$
W=\bigoplus_{k=-q}^pL_{\beta+k\alpha}
$$
is 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 give
$$
\beta(h_\alpha)-2q=-(p+q),\qquad
\beta(h_\alpha)+2p=p+q,
$$
and therefore $\beta(h_\alpha)=q-p$. Since $[L_\alpha,L_{-\alpha}]=\mathbb Ch_\alpha$, every $x$ in this bracket line satisfies
$$
\boxed{\beta(x)=\frac{q-p}{2}\alpha(x).}
$$
Back to article page
