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Let $R$ be the orthogonal matrix corresponding to an operation in $O(3)$. If R is a proper rotation, then both vectors $\vec{V}$ and axial vectors $\vec{A}$ are transformed in the same way

$$ \vec{V} \rightarrow \vec{V}' = R\vec{V}$$ $$ \vec{A} \rightarrow \vec{A}' = R\vec{A}$$

if $R$ is an improper rotations, then $$ \vec{V} \rightarrow \vec{V}' = R\vec{V}$$ $$ \vec{A} \rightarrow \vec{A}' = -R\vec{A}$$

And let $\mathcal{V}$ be the vector representation of $O(3)$ and $\mathcal{P}$ be the axial-vector representation of $O(3)$. Thus we can write, $$\vec{V} \rightarrow \vec{V}' = \mathcal{V}(R)\vec{V}$$ $$\vec{A} \rightarrow \vec{A}' = \mathcal{P}(R)\vec{A}$$

My question is, considering $\mathcal{V}$ and $\mathcal{P}$ as representations of $C_\mathrm{4v}$ group, how can I find their composition in terms of irreps ?

$$\begin{array}{c|ccccc|cc} \hline C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline \mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 \\ \mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\ \mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\ \mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy \\ \mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline \end{array}$$

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