I have a problem with the follwing setup:
Let $\phi:SU(2) \to GL(3,\mathbb{C})$ be a representation and let $\psi:SU(2) \to P_2$. Where $P_2$ is spanned by the monomials $x^2,xy,y^2$ over $\mathbb{C}.$ The map acts by:
$$\psi(P)(f)(x,y) = f((x,y)\phi(P^{-1})^T)$$ I dont understand what $(x,y)\phi(P^{-1})^T$ means. Surely $\phi(P^{-1})^T$ is a $3 \times 3$ matrix so this is not defined? Thanks.
Recall that if $\phi:SU(2)\to GL(V)$ is a representation, then the dual representation $\phi^*:SU(2)\to GL(V^*)$ for $V^*=\hom(V,\mathbb{C})$ is defined by $$\phi^*(P) f=f\circ \phi(P)^{-1}$$ for $P\in SU(2)$ and $f\in V^*$. (This is so that $\langle f,v\rangle = \langle \phi^*(P)f,\phi(P)v\rangle$ for all $f\in V^*$, $v\in V$, and $P\in SU(2)$, where $\langle f,v\rangle$ is the canonical pairing.)
Let $V$ denote a $2$-dimensional representation of $SU(2)$. Let $C(V)$ denote the set of all continuous $\mathbb{C}$-valued functions defined on $V$, which includes, for example, polynomially defined functions, inside of which both $V^*$ and $P_2$ are subsets.
$C(V)$ is an $SU(2)$ representation in the same way $V^*$ is: $\phi^*(P) f=f\circ \phi(P)^{-1}$. Written with arguments, this looks more like $(\phi^*(P) f)(v)=f(\phi(P)^{-1}v)$ with $v\in V$. With $v$ as a column vector, this looks like $$(\phi^*(P) f)\begin{pmatrix}x\\y\end{pmatrix} = f\left(\phi(P)^{-1}\begin{pmatrix}x\\y\end{pmatrix}\right),$$ and by transposing the arguments, we can write $v$ as a row vector to match "$f(x,y)$" notation like so: $$(\phi^*(P) f)(x,y) = f((x,y)(\phi(P)^{-1})^T).$$
Hopefully this helps unravel the mystery.