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

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?

• The homomorphism is defined for groups irrespective of the representations, and $SU(2)$ by definition is a group of 2x2 matrices, so your question is irrelevant for the proof. – Peter Kravchuk Apr 13 '18 at 6:26
• A Special Unitary operator $\:U \in SU(2)\:$ is represented by a $\:2 \times 2\:$ complex matrix having the general form$$U = \begin{bmatrix} \alpha & \beta \\ -\beta^* & \alpha^* \end{bmatrix} \qquad \alpha,\beta \in \mathbb{C} \qquad \det(U)=\alpha \alpha^* +\beta\beta^* =\left\|\alpha\right\|^2 + \left\|\beta\right\|^2=1 \tag{01}$$ ....3-dimensional representation of SU(2).... ??? – Frobenius Apr 13 '18 at 11:00
• @Frobenius Consider the spin-1 representation. It's 3-dimensional. – SRS Apr 13 '18 at 13:18
• Yes, of course. But the spin-1 representation has nothing to do with the (matrix) representation of the special unitary operators $\:U \in SU(2)$. I don't think you are confused. Something else is in your mind but may be you don't ask the right question. – Frobenius Apr 13 '18 at 14:28
• @Peter Kravchuk gave you the answer. Representations of groups are one thing, groups themselves (and group elements in particular) are another. You are mixing the two. – DanielC Apr 14 '18 at 9:12

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.

--

$$^1$$ The group isomorphism $$SU(2)/\mathbb{Z}_2\cong SO(3)$$ can be explicitly constructed by considering the 3-dimensional Euclidean space $$(\mathbb{R}^3, ||\cdot||^2)~\cong~ (su(2),\det(\cdot))$$ and the adjoint representation $${\rm ad}: SU(2)\to GL(su(2))\cong GL(3,\mathbb{R})$$ given by $${\rm ad} (g)~:=~ gxg^{-1}, \qquad g\in SU(2), \qquad x\in su(2).$$ 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 on quarternions, cf. this Phys.SE post.