Circular curvature zeros for a planar SU2 exponential ($r=2\pi k$)

For $g=\exp (-i(x{\sigma }_1+y{\sigma }_2)/2)$, write $r=\sqrt{x^2+y^2}$ and $n=(x/r,y/r,0)$. The Pauli matrix multiplication law gives $g=\cos (r/2)I-i\sin (r/2)n\cdot \sigma$. On the circles $r=2\pi k$, the angular derivative is zero, so the two Cartesian right logarithmic derivatives are proportional to the same radial generator. They commute, giving zero gauge curvature for every scaling constant. At the origin their commutator is $-i{\sigma }_3/2$, which need not vanish.