Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 102 / 1 / v / Solution

Use the standard basisof ${\mathfrak{sl}}_2$. For the invariant trace form $B(x,y)=\mathrm{tr}(xy)$, the dual basis is $(f,e,h/2)$, soOn a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=nv$ and $fev=0$,Centrality and Schur lemma make this the eigenvalue on the whole module. Thus $\lambda =n(n+2)/2$, the Casimir eigenvalue for sl2. If the form is instead the Killing form, which is four times the trace form on ${\mathfrak{sl}}_2$, the corresponding Casimir and eigenvalue are divided by four.