Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 3 / 1 / c / ii / Solution

Let $\nu =-L_3$, whose Dynkin labels are $(0,1)$. Since $\lambda -\nu =L_1-L_3$ is the sum of the two simple roots, its boundary distances in the dominant weight multiplicity formula for sl3 are $r+1,s+1$. ConsequentlyFor a module in ${\mathcal{C}}_1$, this multiplicity is one in the two families having $a=0$ or $b=0$, and two for $R(1,2)$ and $R(3,1)$. For a module in ${\mathcal{C}}_2$, it is two in the families having $a=1$ or $b=1$, and three for $R(2,3)$ and $R(4,2)$. In each case, increasing $r,s$ by one increases the minimum only if neither $a$ nor $b$ already limits it.

To obtain total multiplicity five at $\nu$, the two summands must therefore contribute two and three. The only surviving pairs, up to interchange, areTheir multiplicities at $\lambda$ remain one and two, so all four pairs satisfy both conditions.