Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 6 / 2

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.