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

In the finite-dimensional setting, let $R(a,b)$ denote the irreducible ${\mathfrak{sl}}_3$ highest-weight representation of highest weight $a{L}_1-b{L}_3$, where $a,b$ are nonnegative integers. Thus $(a,b)$ are its Dynkin labels. The target $\lambda =L_1-2L_3$ has labels $(1,2)$. Write the highest-weight difference in simple roots asThe dominant weight multiplicity formula for sl3 gives zero unless $r,s$ are nonnegative integers, and otherwise givesFor completeness, this multiplicity is an actual count. In the sl3 interlacing character formula, use top row $(a+b,b,0)$ and bottom entry $s+3$; the target diagonal exponents are $(s+3,s+2,s)$. The middle row $(P,Q)$ obeys $P+Q=2s+5$ andEach allowable integral $P$ gives one Gelfand–Tsetlin basis vector. Subtracting the lower bound from the upper bound gives the minimum of $a,s,r,r+3,b,s+2$, which reduces to $\min (a,b,r,s)$. This proves the displayed multiplicity formula in the case at hand.

Since both summands have a nonzero weight space and their dimensions at $\lambda$ sum to three, their multiplicities must be one and two. For multiplicity one, the minimum is zero. The cases $a=0$ and $b=0$ give $(0,4+3j)$ and $(5+3j,0)$; the remaining cases $r=0$ or $s=0$ give $(1,2)$ and $(3,1)$. Hence the full family isFor multiplicity two, the minimum is one. The cases $a=1$ and $b=1$ give $(1,5+3j)$ and $(6+3j,1)$; the remaining cases $r=1$ or $s=1$ give $(2,3)$ and $(4,2)$. ThusThe minimum conditions exhaust all cases and each listed module has the stated multiplicity. All candidate unordered pairs are $\boxed{\{U,V\}\text{ with }U\in {\mathcal{C}}_1,V\in {\mathcal{C}}_2}$. There are infinitely many at this stage; the question imposes no upper bound on their highest weights.

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.