I know this has been done on this site in a different manner but I'm wondering if it's possible to show the 2:1 Lie group homomorphism between $SU(2)$ and $SO(3)$ using exponentials of the generators of $SU(2)$ and $SO(3)$.

  • $\begingroup$ The algebra or product relationships of these two groups are similar, but globally they are different... $\endgroup$ – Xiaodong Qi Oct 7 '15 at 3:30
  • $\begingroup$ If you like this question you may also enjoy reading this answer. $\endgroup$ – Qmechanic Oct 7 '15 at 6:35
  • $\begingroup$ Related: physics.stackexchange.com/q/106102/2451 $\endgroup$ – Qmechanic Oct 7 '15 at 9:53

Given an element $\phi$ of $SU(2)$, let the first row of $\phi$ be $(P,Q)$ where $P$ and $Q$ are complex numbers.

Let $q(\phi)$ be the quaternion $P+Qj$.

Now for any $(x,y,z)\in {\mathbb R}^3$, consider the quaternion $q(\phi)(xi+yj+zk)q(\phi)^{-1} = (ai+bj+ck)$. The map $(x,y,z)\mapsto (a,b,c)$ is a rotation of ${\mathbb R}^3$ and hence an element of $SO(3)$. This element is the image of $\phi$.

Alternatively, you can write $q(\phi)=\cos(\theta)+v\sin(\theta)$ where $v$ is a quaternion in the span of $i,j,k$ (and hence identified with an element of ${\mathbb R}^3$). Then the image of $\phi$ is a rotation by $2\theta$ around $v$.

If you grind through what that means --- and if I've done this right (you really should check it) then the map is given explicitly by $$\pmatrix{A+Bi&C+Di\cr -C+Di&A-Bi\cr}\mapsto \pmatrix{2(CD-BA)&A^2+C^2-B^2-D^2&2(AD+BC)\cr B^2+C^2-A^2-D^2&2(AB+CD)&2(AC-BD)\cr 2(AC+BD)&2(AD-BC)&A^2+B^2-C^2-D^2}$$


A more physical construction:

Let $R_3(\theta, \bf{n})$ be the matrix of a rotation of angle $\theta$ around axis $\bf{n}$ in $\mathbb{R}^3$. Then if $\hat{J}_i$, $i=1,2,3$ are corresponding SO(3) generators, $$ \left[\hat{J}_i, \hat{J}_j \right] = i \epsilon_{ijk} \hat{J}_k $$ we have $$ R_3(\theta, {\bf n} ) = exp\left(-i\;\theta \;n^i \hat{J}_i\right) $$

Now, if $\hat{\sigma}_i$, $i=1,2,3$ are the Pauli matrices, define the SU(2) element $$ R_2(\theta, {\bf n} ) = exp\left(-i\;\frac{\theta}{2} \;n^i \hat{\sigma}_i\right) $$ Furthermore, for any ${\bf x} \in \mathbb{R}^3$ define $$ X = x^i \hat{\sigma}_i $$

Then if ${\bf x}$ transforms into $\overline{{\bf x}}$ under rotation $R_3(\theta, {\bf n} )$, $$ \overline{{\bf x}} = R_3(\theta, {\bf n} )\;{\bf x} $$ one also has that under $R_2(\theta, {\bf n} )$ the matrix $X$ transforms into $$ \overline{X} = R_2^\dagger(\theta, {\bf n} ) \;X \;R_2(\theta, {\bf n} ) = \overline{x}^i \hat{\sigma}_i $$ The homomorphism follows thereof.


I'll sketch how you do it.

In a general Lie group $\mathfrak{G}$ setting, the mapping $\mathrm{Ad}$ is clearly a homomorphism, since the action of the image of $\gamma_1\,\gamma_2$ under $\mathrm{Ad}$ is

$$X\mapsto \gamma_1\,\gamma_2\,X\,(\gamma_1\,\gamma_2)^{-1} = \gamma_1\,\gamma_2\,X\,\gamma_2^{-1}\,\gamma_1^{-1} = \mathrm{Ad}(\gamma_1)\circ\mathrm{Ad}(\gamma_2)\,X$$

so your question boils down to finding out what the kernel of the homomorphism is. So we seek to find what the assertion $\mathrm{Ad}(\gamma_1) = \mathrm{Ad}(\gamma_2)$; equivalently, what the assertion $\mathrm{Ad}(\gamma_1\,\gamma_2^{-1}) = \mathrm{id}$ implies about the relationship between $\gamma_1$ and $\gamma_2$. Thus we have:

$$\gamma_1\,\gamma_2^{-1}\,X\,\gamma_2\,\gamma_1^{-1} = X;\;\forall\,X\in\mathfrak{su}(2)\tag{1}$$

and, since $\exp:\mathfrak{su}(2)\to SU(2)$ is surjective (as $\exp$ is for all compact groups), we know then that $\gamma_1\,\gamma_2^{-1} = e^H$ where $H\in\mathfrak{su}(2)$ is the generator of a transformation fulfilling:


For $SU(2)$ we can use the Rodrigues formula:

$$\exp(H) = \cos(\|H\|)\,\mathrm{id} + \frac{\sin(\|H\|)}{\|H\|}\,H;\,\forall\,H\in\mathfrak{su}(2)\tag{3}$$

where $\|H\| = \frac{1}{2}\sqrt{\mathrm{tr}(H^\dagger\,H)}$. Now write $H$ and $X$ as general superpositions of the Pauli matrices $H=i\,(h_x\,\sigma_x+h_y\,\sigma_y+h_z\,\sigma_z)$ and $X=i\,(x_x\,\sigma_x+x_y\,\sigma_y+x_z\,\sigma_z)$ where $h_j$ and $x_j$ are real. So now, substitute (3) into (2) and the substitute the Pauli matrix substitutions, and you will ultimately conclude that $\| H\|\in\{0,\,\pi\}$, whence from (3), $\gamma_1 = \pm\,\gamma_2$. So the cosets of the homomorphism are of the form $\{\gamma,\,-\gamma\}$, for any $\gamma\in SU(2)$.

Afterword: With not too much trouble, you can understand, with techniques like the above, that the kernel of the homomorphism $\mathrm{Ad}$ for any Lie group is in fact the center of the group. So you always, in a problem like this, work out a way to find the group's center.


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