Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 1 / a

Use $h=\mathrm{diag}(1,-1)$, $e=E_{12}$ and $f=E_{21}$ in the sl2 Lie algebra, so $hx=x$, $hy=-y$, $ey=x$ and $fx=y$. On the symmetric power, a monomial $x^{3-k}{y}^k$ has weight $3-2k$. The tensor product of Lie algebra representations adds weights, because $h$ acts as $h\otimes I+I\otimes h$.

Put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$ for $0\le k\le 3$. These eight vectors are a basis of $U$; their weights are $4-2k$ and $2-2k$. Thus the weight-space decomposition and a weight basis for each summand areIn particular $\boxed{U=U_4\oplus U_2\oplus U_0\oplus U_{-2}\oplus U_{-4}}$, with dimensions summing to eight. Here the subscripts are the eigenvalues of the chosen Cartan element $h$.

Write $L(n)$ for the irreducible representation of ${\mathfrak{sl}}_2$ with highest weight $n$, dimension $n+1$ and weights $n,n-2,\dots ,-n$. The weight diagram below records multiplicities $1,2,2,2,1$ at weights $4,2,0,-2,-4$.

The unique top weight four forces one copy of $L(4)$, by complete reducibility of semisimple Lie algebra representations. Removing its five weights, each with multiplicity one, leaves weights $2,0,-2$ with multiplicity one, exactly the character of $L(2)$. HenceThe two lower rows draw their separate weight diagrams. The explicit invariant copies constructed next establish the Clebsch-Gordan splitting of a defining module and its cubic symmetric power within $U$ itself.

In the notation of part (i), the tensor raising operator and lowering operator act bywith terms beyond the stated index range omitted. Both $v_4=A_0$ and $v_2=B_0-A_1$ are killed by $e$, and have respective weights four and two. They are the desired highest-weight vectors.

For $v_4$, successive lowering givesThus a basis of the five-dimensional summand, in descending weight order, isFor $v_2$, the successive vectors are $f{v}_2=2(B_1-A_2)$, $f^2{v}_2=2(B_2-A_3)$ and $f^3{v}_2=0$. HenceThese spans are invariant under $h,e,f$. The sl2 highest-weight lowering formula gives $e{f}^r{v}_n=r(n-r+1)f^{r-1}{v}_n$, so all interior raising coefficients in either chain are nonzero. Any nonzero submodule therefore contains the highest vector and then the whole chain; each span is irreducible.

At the shared weights $2,0,-2$, the two displayed summand vectors are independent: their coefficient determinants in the original weight bases are $4,2,4$. At weights $4,-4$ only the five-dimensional summand occurs. Consequently the eight displayed vectors form a basis of $U$, proving that these are complementary submodules, not only modules with the right dimensions.