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

Take a basis $x,y$ of the defining representation of the sl2 Lie algebra, with $Hx=x$, $Hy=-y$, $Ey=x$, $Ex=0$, $Fx=y$ and $Fy=0$. A monomial basis of the symmetric power ${\mathrm{Sym}}^{4}V$ isFor the exterior power $U={\bigwedge }^2{\mathrm{Sym}}^{4}V$, put $w_{ij}=v_i\wedge v_j$ for $i<j$. These ten elements form a basis, and their weights of a representation add:Thus the complete weight-space decomposition, with explicit bases, isThe weight diagram therefore has multiplicities $1,1,2,2,2,1,1$ at weights $6,4,2,0,-2,-4,-6$, respectively.

The induced raising operator and lowering operator act bySince $E{w}_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible sl2 Lie algebra module ${\mathrm{Sym}}^{6}V$. By complete reducibility of semisimple Lie algebra representations, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the weight diagram leaves weights $2,0,-2$ once each. HenceThese are the only two irreducible isomorphism types occurring as submodules.

For the three-dimensional summand, calculate $E{w}_{03}=3w_{02}$ and $E{w}_{12}=w_{02}$. Thereforeis a nonzero highest-weight vector of weight $2$. The classification of finite-dimensional sl2 representations shows that the submodule $W$ it generates is the irreducible module ${\mathrm{Sym}}^{2}V$, of dimension three.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.

Let $x_1,x_2,x_3$ be the coordinate basis of the defining representation $W$ of ${\mathfrak{sl}}_3$. Write $s_{ij}=x_i{x}_j$ for $1\le i\le j\le 3$. The six $s_{ij}$ form a basis of the symmetric power ${\mathrm{Sym}}^{2}W$, with weight $L_i+L_j$. A basis for the tensor product of Lie algebra representations $T$ consists of all 36 ordered tensorsThe order of the two factors matters; there is no quotient identifying their interchange.

Each weight of a representation has the form $k_1{L}_1+k_2{L}_2+k_3{L}_3$, where the nonnegative integers satisfy $k_1+k_2+k_3=4$. Its weight multiplicity is the number of ways to split this triple into two nonnegative triples, each summing to two. The possible patterns, including every permutation, areFor example, $(2,2,0)$ splits as $(2,0,0)+(0,2,0)$, the reverse split, or $(1,1,0)+(1,1,0)$, giving multiplicity three. For $(2,1,1)$, the four splits pair $(2,0,0)$ with $(0,1,1)$ and $(1,1,0)$ with $(1,0,1)$, each in both orders. The multiplicities sum to $3+12+9+12=36$, verifying the dimension.

The following weight diagram uses Dynkin labels $(p,q)=(k_1-k_2,k_2-k_3)$; the circle numbers give multiplicities and orange points mark the Closed dominant Weyl chamber.

Take the simple roots $L_1-L_2$ and $L_2-L_3$. A weight represented by $(k_1,k_2,k_3)$ is in the Closed dominant Weyl chamber exactly when $k_1\ge k_2\ge k_3$. Among nonnegative triples of total four, the possibilities are $(4,0,0),(3,1,0),(2,2,0),(2,1,1)$. Using $L_1+L_2+L_3=0$, their explicit weight space bases areEach listed tensor has the required summed weight, and the monomial basis from part (i) shows linear independence and exhausts all possible tensors of that weight. Thus these are all the requested bases.

Use the positive simple-root raising operators $E_{12}=x_1{\partial }_{x_2}$ and $E_{23}=x_2{\partial }_{x_3}$. On the weight space of $2L_1-L_3$, the first maps both $s_{11}\otimes s_{12}$ and $s_{12}\otimes s_{11}$ to $s_{11}\otimes s_{11}$, while the second annihilates both. Consequentlyis a nonzero highest-weight vector; the remaining positive-root operator $E_{13}=[E_{12},E_{23}]$ also annihilates it. Its Dynkin labels are $(2,1)$, and the generated submodule $Z$ is the irreducible highest-weight representation with that highest weight.

Put $F_1=E_{21}=x_2{\partial }_{x_1}$ and $F_2=E_{32}=x_3{\partial }_{x_2}$, acting on both tensor factors. Direct calculation givesDefine the linearly independent tensorsThey both have weight $L_1$, andTherefore $A,B\in Z_{L_1}$. To prove that they exhaust this weight space, its weight differs from the highest weight by $(L_1-L_2)+(L_2-L_3)$. The Poincaré-Birkhoff-Witt theorem expresses all descendants at that difference using $F_2{F}_{1}v$ and $F_1{F}_{2}v$: the third possible operator $E_{31}=[F_2,F_1]$ is their difference. Hence the space has dimension at most two, and the two independent vectors already found give

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.