Compact non-Abelian Lie-group structure of SU(2)

If $U,V\in SU(2)$, then $(UV{)}^{\ast }(UV)=V^{\ast }{U}^{\ast }UV=I_2$ and $\det (UV)=\det U\det V=1$. The identity belongs to $SU(2)$, and $U^{-1}=U^{\ast }$ is unitary with determinant $\overline{\det U}=1$. Thus the set is a group. Its standard smooth Lie-group structure and smooth matrix inclusion into $U(2)$ are supplied by [F2] under the definition [F3]. The map ${\rho }_2$ is a group homomorphism because it is the matrix inclusion, and it is unitary because $U^{\ast }U=I_2$.

In real coordinates the equations $U^{\ast }U=I_2$ and $\det U=1$ are continuous equations, so $SU(2)$ is closed. Unitarity makes each column a unit vector, hence every real and imaginary entry lies in $[-1,1]$; the set is bounded in ${\mathbb{R}}^8$. Apply [F1] to conclude compactness. The inherited Euclidean metric makes this subspace Hausdorff: for distinct $U,V$, the open balls of radius $\mathrm{\parallel }U-V\mathrm{\parallel }\mathrm{/}3$ are disjoint, since a point in both would give $\mathrm{\parallel }U-V\mathrm{\parallel }<2\mathrm{\parallel }U-V\mathrm{\parallel }\mathrm{/}3$ by the triangle inequality.

The matrices $A=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$ and $B=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ are unitary with determinant one, while $AB=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$ and $BA=\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)$ differ. Therefore $SU(2)$ is non-Abelian.

For $V\in SU(2)$, write its first column as $(\alpha ,\gamma )$. Its second unit column orthogonal to the first is $\lambda (-\bar{\gamma },\bar{\alpha })$ for $\mathrm{\mid }\lambda \mathrm{\mid }=1$; determinant one gives $\lambda (\mathrm{\mid }\alpha {\mathrm{\mid }}^2+\mathrm{\mid }\gamma {\mathrm{\mid }}^2)=\lambda =1$. Hence $V=\left(\begin{array}{cc}\alpha & \beta \\ -\bar{\beta }& \bar{\alpha }\end{array}\right)$ with $\mathrm{\mid }\alpha {\mathrm{\mid }}^2+\mathrm{\mid }\beta {\mathrm{\mid }}^2=1$, and $\frac{1}{2}{\mathrm{ReTr}}_{2}V=\mathrm{Re}\alpha \in [-1,1]$. Also ${\mathrm{Tr}}_2(V^{-1})={\mathrm{Tr}}_2(V^{\ast })=\overline{{\mathrm{Tr}}_{2}V}={\mathrm{Tr}}_{2}V$, while cyclicity gives ${\mathrm{Tr}}_2(gV{g}^{-1})={\mathrm{Tr}}_{2}V$. Thus the normalized trace is real, bounded, conjugation invariant, and inverse invariant.