 Solution

Let $v_k=x^{n-k}{y}^k$. In the summand $V_{2n-2j}$ from part (d), a highest-weight vector isThe raising operator $x{\partial }_y$ annihilates this sum because consecutive terms cancel. Interchanging the tensor factors sends $w_j$ to $(-1{)}^j{w}_j$. Since each summand in part (d) has multiplicity one, the flip acts by that scalar on the whole summand. The alternating square therefore consists exactly of the odd-$j$ summands. This flip parity in the SU2 tensor square gives