In physics, it's common to use the relations \textbf{r}^\prime=\mathscr{R}\textbf{r};~~\text{and}~~\textbf{r}^\prime\cdot\boldsymbol{\sigma} =\mathscr{U}(\textbf{r}\cdot\boldsymbol{\sigma}) \mathscr{U}^{\dagger}\tag{1} to establish a two-to-one homomorphism between {\rm SU(2)} and {\rm SO(3)} where \textbf{r}\in \mathbb{R}^3, \mathscr{R}\in {\rm SO(3)}, \mathscr{U}\in {\rm SU(2)} and \boldsymbol{\sigma}=(\sigma_1,\sigma_2,\sigma_3) are three Pauli matrices. Both the relations of Eq.(1) represent rotation of coordinates in real three-dimensioanl space because both of them satisfy |\textbf{r}^\prime|^2=|\textbf{r}|^2. It's easy to see from (1) that corresponding to every 3\times 3 matrix \mathscr{R}\in {\rm SO(3)} there exist two 2\times 2 matrices \pm \mathscr{U}\in {\rm SU(2)} that represent the same rotation.
Question Note that the above proof of 2-to-1 homomorphism is based on fundamental representations of SO(3) and SU(2). But for any odd-dimensional representation of SU(2), if \mathscr{U} has determinant +1, -\mathscr{U} is not a representation of SU(2) since it has determinant -1. Hence, if \mathscr{U} is a member of an odd-dimensional representation of SU(2). \mathscr{U} is not. Does it mean that 2-to-1 homomorphism between SU(2) and SO(3) is not true in general?
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