The group $SU(2)$ acts on the space of degree $d$ homogeneous polynomials on $\mathbb{C}^2$ as
$$(g \cdot f)(z) = f(g^{-1} z)$$
where $g \in SU(2)$ and $z = (z_1, z_2) \in \mathbb{C}^2$. To compute the action on the polynomial $z_1$, we can consider it as the map $\pi_1 : \mathbb{C}^2 \to \mathbb{C}$ defined by $\pi_1(z_1, z_2) = z_1$. Then, for
$$g = \begin{bmatrix}\alpha & -\overline{\beta}\\ \beta & \overline{\alpha}\end{bmatrix}$$
we have
$$(g \cdot \pi_1)(z) = \pi_1(g^{-1} z) = \pi_1 \left( \begin{bmatrix}\overline{\alpha} & \overline{\beta}\\ -\beta & \alpha\end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \right) = \overline{\alpha}z_1 + \overline{\beta}z_2$$
This is the same result obtained in Singer's Linearity, Symmetry and Prediction in the Hydrogen Atom, p. 138. So, to me, it seems like when computing the action of $g$ on some polynomial we replace $z_1 \mapsto \overline{\alpha}z_1 + \overline{\beta}z_2$ in the polynomial's expression.
However, in Woit's Quantum Theory, Groups and Representations, p. 113, it is said that one must replace $z_1 \mapsto \alpha z_1 + \beta z_2$ (here I adapted his notation to mine). So my question is: which one is right, and why?
Best Answer
The first action you describe $(g\cdot f)(z):=f(g^{=1}z)$ is a group action.
The second formula seems to describe $(g\cdot f)(z):=f(g^Tz)$, also a group action.
(One could also have used $(f\cdot g)(z):=f(gz)$ as a right group action.)
Indeed, the first two actions are equivalent, because they are connected by an inner automorphism of $SU(2)$, namely $A\mapsto (A^T)^{-1}=JAJ^{-1}$ where $J=[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]$. Note that $(A^T)^{-1}$ is not an inner automorphism for $n\times n$ matrices for $n>2$.