 Solution

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