3
$\begingroup$

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 $SO(3)$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$\pmatrix{A+Bi&C+Di\cr -C+Di&A-Bi\cr}\mapsto \pmatrix{2(CD-BA)&A^2+C^2-B^2-D^2&2(AD+BC)\cr B^2+C^2-A^2-D^2&2(AB+CD)&2(AC-BD)\cr 2(AC+BD)&2(AD-BC)&A^2+B^2-C^2-D^2}$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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:

$$e^H\,X\,e^{-H}=X;\;\forall\,X\in\mathfrak{su}(2)\tag{2}$$

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.

$\endgroup$

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.