Subtlety in the proof of 2-to-1 homomorphism between $SU(2)$ and $SO(3)$

Question

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-dimensional 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?

Answer 1

TL;DR: The status of the group isomorphism$^1$ $SO(3)\cong SU(2)/\mathbb{Z}_2$ and OP's eq. (1) are not jeopardize by the existence of non-faithful $SU(2)$ representations, cf. above comments by Peter Kravchuk and DanielC.

In more details:

1. Let $\rho$ denote the $n$-dimensional irreducible Lie group representation $\rho: SU(2)\to GL(n,\mathbb{C})$, and (with a slight misuse of notation) let $\rho$ also denote the corresponding $n$-dimensional irreducible Lie algebra representation $\rho: su(2)\to gl(n,\mathbb{C})$.

2. Then $$\rho(\pm {\bf 1}_{2\times 2})~=~(\pm 1)^{n+1}{\bf 1}_{n\times n},$$ and $$ {\rm ker}(\rho)~:=~ \rho^{-1}(\{{\bf 1}_{n\times n}\})~=~\left\{\begin{array}{ll} \{{\bf 1}_{2\times 2}\} & \text{for } n \geq 2\text{ even}, \cr \{\pm{\bf 1}_{2\times 2}\} & \text{for } n \geq 3\text{ odd}, \cr SU(2) & \text{for } n=1 \end{array} \right. $$ i.e. odd-dimensional representations are not faithful.

3. It is possible to apply $\rho$ to both sides of OP's eq. (1) without contradictions. Eq. (1) is also discussed in my Phys.SE answer here.

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$^1$ Define the Lie group $$SU(2)~:=~\{ g \in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid g^{\dagger}\eta g=\eta \wedge \det (g) = 1 \}$$ with a $2\times 2$ positive definite, real metric
$$\eta_{ab}~=~{\rm diag}(+1,+1).$$ The corresponding Lie algebra is the set of traceless anti-hermitian $2\times 2$ matrices $$su(2)~:=~\{ x \in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid x^{\dagger}=-\eta x\eta^{-1} \wedge {\rm tr} (x) = 0 \}.$$

The group isomorphism $SU(2)/\mathbb{Z}_2\cong SO(3)$ can be explicitly constructed by

• using a bijective isometry $$(\mathbb{R}^3, ||\cdot||^2)~\cong~ (su(2),\det(\cdot))$$ between the 3-dimensional Euclidean space and the Lie algebra $su(2)$ (a traceless anti-hermitian $2\times 2$ matrix has non-negative determinant);

• using the adjoint representation ${\rm Ad}: SU(2)\to GL(su(2))\cong GL(3,\mathbb{R})$ given by $${\rm Ad} (g)~:=~ gxg^{-1}, \qquad g\in SU(2), \qquad x\in su(2); $$

• identifying $$ O(3)~\cong~ O(su(2)).$$

One may show that $${\rm Im}({\rm Ad})~\cong~ SO(3) \qquad\text{and}\qquad {\rm ker}({\rm Ad})~=~\{\pm{\bf 1}_{2\times 2}\} .$$ An equivalent proof uses quarternions, cf. this Phys.SE post.