Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 19F (Representation Theory)

For $z=(x,y{)}^T$, define $(\rho (g)f)(z)=f(g^{-1}z)$. This preserves the space of degree-$n$ homogeneous polynomials and satisfies $\rho (g_1)\rho (g_2)=\rho (g_1{g}_2)$. The diagonal torus of SU(2) acts on the monomials $x^{n-j}{y}^j$ with distinct weights $2j-n$.

If a nonzero subspace $W$ is invariant, averaging against each torus character projects $W$ onto the individual weight spaces. Therefore $W$ contains at least one monomial. Invariance also implies invariance under differentiated Lie-algebra operators and their complex linear combinations. The complexified algebra contains the operators $x{\partial }_y$ and $y{\partial }_x$, up to harmless signs. They carry each monomial to its adjacent monomial with a nonzero coefficient unless already at the appropriate endpoint. Repeated application produces every monomial, so $W=V_n$. The representation is irreducible.

Use the Clebsch-Gordan decomposition for SU2, which states that $V_m\otimes V_n\cong {⨁}_{j=0}^{\min (m,n)}{V}_{m+n-2j}$. Here this givesTo determine symmetry, realize the tensor product as polynomials in two pairs of variables. A highest-weight vector in the $j$th summand can be chosen proportional to $(x_1{y}_2-y_1{x}_2{)}^j{x}_1^{3-j}{x}_2^{3-j}$. Exchanging the two factors multiplies it by $(-1{)}^j$, and hence multiplies the entire irreducible summand by that sign. Even $j$ belong to the symmetric square, odd $j$ to the exterior square. ThusTheir dimensions are respectively $7+3=10$ and $5+1=6$, as required for a four-dimensional space.

The dual representation on $V^{\ast }$ is $(g\ell )(v)=\ell (g^{-1}v)$. The bilinear form $\epsilon (z,w)=\det (z,w)$ on ${\mathbb{C}}^2$ is nondegenerate and invariant under SU(2), because every group matrix has determinant one. Thus it gives an equivariant isomorphism ${\mathbb{C}}^2\to ({\mathbb{C}}^2{)}^{\ast }$. Taking the $n$th symmetric power gives an equivariant isomorphism between its $n$th symmetric power and its dual. Since homogeneous polynomials realize $V_n={\mathrm{Sym}}^n(({\mathbb{C}}^2{)}^{\ast })$, this proves