Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 2 / 2 / b / Solution

Use the comultiplication $\Delta E=E\otimes K+1\otimes E$, $\Delta F=F\otimes 1+K^{-1}\otimes F$, $\Delta K=K\otimes K$ of the quantum enveloping algebra of sl2. Choose a basis $u_0,u_1$ for $V_{1,1}$ with $F{u}_0=u_1$, $E{u}_1=u_0$, and $K{u}_i=q^{1-2i}{u}_i$. Choose a basis $w_0,w_1,w_2$ for $V_{1,2}$ withand all terminal raising or lowering actions zero. This is one of the corrected bases from part (a).

In the tensor product of modules, putThe vector $z_0$ is killed by $E$ and has $K$ eigenvalue $q^3$. Direct application of the comultiplication givesand the string formula gives $E{z}_1=[3]z_0$, $E{z}_2=[2]z_1$, $E{z}_3=z_2$. Also $K{z}_i=q^{3-2i}{z}_i$. Thus $W_3=⟨z_0,z_1,z_2,z_3⟩$ is a submodule isomorphic to $V_{1,3}$.

The other highest vector and its lowered partner areIndeed $E{h}_0=([2]q^2-q^2[2])u_0\otimes w_0=0$, $K{h}_0=q{h}_0$, $K{h}_1=q^{-1}{h}_1$, $E{h}_1=h_0$ and $F{h}_1=0$. Therefore $W_1=⟨h_0,h_1⟩$ is a submodule isomorphic to $V_{1,1}$.

To verify that this is a direct sum, in the $q$ weight space the determinant of the coefficients of $z_1,h_0$ is $-q^2-q^{-1}[2]=-[3]\ne 0$. In the $q^{-1}$ weight space, the coefficient determinant of $z_2,h_1$ is $-q[2]-q^{-2}=-[3]\ne 0$. The extreme weight spaces are spanned by $z_0,z_3$. Hence these six vectors form a basis of the entire tensor product of modules. Each summand is a simple module: projecting a nonzero vector onto one of its distinct $K$ weight spaces and applying the nonzero raising and lowering coefficients recovers its entire basis. Consequently