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?

  • 5
    $\begingroup$ 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. $\endgroup$ – Peter Kravchuk Apr 13 '18 at 6:26
  • $\begingroup$ 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).... ??? $\endgroup$ – Frobenius Apr 13 '18 at 11:00
  • $\begingroup$ @Frobenius Consider the spin-1 representation. It's 3-dimensional. $\endgroup$ – SRS Apr 13 '18 at 13:18
  • $\begingroup$ 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. $\endgroup$ – Frobenius Apr 13 '18 at 14:28
  • 2
    $\begingroup$ @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. $\endgroup$ – 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.