The Geometry of $SU(2)$

The unitary group in two complex dimensions is defined by \begin{equation} \SU(2) = \{ M\in\CC^{2\times 2} : M^\dagger M = I, \det M = 1 \}          \end{equation} A $2\times2$ matrix has 4 complex components, the constraint $M^\dagger M=I$ imposes 4 real conditions, and the determinant restriction adds just one more (since the other conditions already imply that $|\det M|=1$). Thus, we expect there to be three independent “rotations” in $\SU(2)$. A set of generators is given by \begin{align} R_x &= R_x(\alpha) = \begin{pmatrix} \cos\left(\frac\alpha2\right) & -i\sin\left(\frac\alpha2\right) \\ -i\sin\left(\frac\alpha2\right) & \cos\left(\frac\alpha2\right) \end{pmatrix}\\ R_y &= R_y(\alpha) = \begin{pmatrix} \cos\left(\frac\alpha2\right) & -\sin\left(\frac\alpha2\right) \\ \sin\left(\frac\alpha2\right) & \cos\left(\frac\alpha2\right) \end{pmatrix}\\ R_z &= R_z(\alpha) = \begin{pmatrix} e^{-i\alpha/2} & 0 \\ 0 & e^{i\alpha/2} \end{pmatrix} \label{su2gen} \end{align} where the factor of $2$ is conventional. These matrices are closely related to the Pauli matrices \begin{equation} \sigma_m = 2i\frac{\partial R_m}{\partial\alpha} \Bigg|_{\alpha=0} \end{equation} which are explicitly given by \begin{align} \sigma_x &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\\ \sigma_y &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\\ \sigma_z &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{align} The Pauli matrices have some important properties: They are Hermitian ($\sigma_m^\dagger=\sigma$), tracefree ($\tr\,\sigma_m=0$), they each square to the identity matrix ($\sigma_m^2=I$), and they each have determinant $-1$ ($\det\sigma_m=-1$). We will have more to say about the Pauli matrices later.

How are we to interpret such complex transformations?

A complex vector $v\in\CC^2$ can also be viewed as an element of $\RR^4$, by treating the real and imaginary parts of the components of $v$ as independent. So consider \begin{equation} v = \begin{pmatrix} a\\ b \end{pmatrix} = \begin{pmatrix} a_1+a_2i\\ b_1+b_2i \end{pmatrix} \end{equation} with $a_1,a_2,b_1,b_2\in\RR$. The matrix \begin{align} \XX &= vv^\dagger = \begin{pmatrix} |a|^2 & a\bar{b} \\ b\bar{a} & |b|^2 \end{pmatrix} \\ \end{align} has vanishing determinant, since \begin{equation} \det\XX = |a|^2|b|^2 - |a\bar{b}|^2 = 0 \end{equation} If we further assume that $v$ is normalized, that is, that \begin{equation} v^\dagger v = |a|^2 + |b|^2 = 1 \end{equation} then we can write \begin{equation} \XX = \begin{pmatrix} \frac12+z & x-iy \\ x+iy & \frac12-z \end{pmatrix} \\ \end{equation} where \begin{align} x+iy &= (a_1b_1+a_2b_2) + (a_1b_2-a_2b_1)i \\ z &= \frac12 (|a|^2 - |b|^2) \end{align}

Since $\det M=1$ for $M\in\SU(2)$, the transformation \begin{equation} \XX \longmapsto M\XX M^\dagger \label{MXM} \end{equation} preserves the determinant of $\XX$, which is now \begin{equation} \det\XX = 0 = \frac14 - (x^2+y^2+z^2) \end{equation} But the transformation (\ref{MXM}) also preserves the trace of $\XX$, since \begin{equation} \tr\XX = \tr(vv^\dagger) = v^\dagger v = |v|^2 \end{equation} and $M\in\SU(2)$. Such transformations must therefore preserve $x^2+y^2+z^2$. Thus, elements of $\SU(2)$, acting on $\RR^3$ via (\ref{MXM}), induce transformations in $\SO(3)$. Since $-M$ induces the same $\SO(3)$ transformation as $+M$, we cannot quite conclude that $\SU(2)\cong\SO(3)$. We say instead that $SU(2)$ is the 2-to-1 cover of $\SO(3)$. Since the double-cover of $\SO(3)$ is called $\textrm{Spin}(3)$, we can write this relationship as \begin{equation} \SU(2) \cong \textrm{Spin}(3) \end{equation} Under this correspondence, $R_x\in\SU(2)$ does indeed correspond to a counterclockwise rotation by $\alpha$ in the $xy$ plane, and similarly for $R_y$ and $R_z$, which is why we used the same names as we did for $SO(3)$, and why we introduced the factor of $2$ in (\ref{su2gen}).

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