= Trace of a raising-lowering product in an irreducible sl2 module
{title2=$\operatorname{tr}(XY)=k(k+1)(k+2)/6$}
For a highest-weight vector $v$ of weight $k$ in an irreducible finite-dimensional <sl2 Lie algebra> module, the <basis> $v,Yv,\ldots,Y^kv$ diagonalizes $XY$, with eigenvalue $(j+1)(k-j)$ on $Y^jv$. Summing those eigenvalues gives $\operatorname{tr}(XY)=k(k+1)(k+2)/6$. The formula $XY^nv=n(k-n+1)Y^{n-1}v$ requires $v$ to be a <highest-weight vector>, not an arbitrary vector of the module.
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