Use the standard basis
$$
e=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad
f=\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad
h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}
$$
of $\mathfrak{sl}_2$. For the invariant <Trace form of a Lie algebra representation>[trace form] $B(x,y)=\operatorname{tr}(xy)$, the dual basis is $(f,e,h/2)$, so
$$
\Omega=ef+fe+\frac12h^2.
$$
On a highest-weight vector $v$ in the $(n+1)$-dimensional irreducible module, $ev=0$ and $hv=nv$. Since $efv=[e,f]v=n v$ and $fev=0$,
$$
\Omega v=\left(n+\frac{n^2}{2}\right)v
=\frac{n(n+2)}2v.
$$
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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-102/1/v/solution