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Use anti-Hermitian gauge potentials and the convention $D_i={\partial }_i+A_i$ for the gauge covariant derivative. Commuting these operators defines the gauge field strength:It is antisymmetric in its two spacetime indices, so $F_{xx}=F_{yy}=0$ and $F_{yx}=-F_{xy}$. In two dimensions there is therefore only one independent component, although that component is itself Lie-algebra valued.

Put $B_x=({\partial }_{x}g)g^{-1}$ and $B_y=({\partial }_{y}g)g^{-1}$. Differentiating $g^{-1}$ gives ${\partial }_i{g}^{-1}=-g^{-1}({\partial }_{i}g)g^{-1}$. Equality of mixed derivatives then gives the right Maurer-Cartan equationConsequently the scaled right Maurer-Cartan gauge potential has curvatureThus $\alpha =0$ and $\alpha =-1$ give zero curvature for every smooth $g$. At zero the potential is zero. At minus one it is a pure gauge potential: transforming the zero connection by $g$ gives $A=-dg{g}^{-1}$. If the gauge Lie algebra is abelian, the commutator vanishes for every $\alpha$. If it contains $X,Y$ with $[X,Y]\ne 0$, choose $g(x,y)=e^{xX}{e}^{yY}$; at the origin $B_x=X$ and $B_y=Y$. Thus in that case the two displayed values are the only choices flat for every $g$. If the potential is instead defined using $D=\partial -A$ and field components ${\partial }_x{A}_y-{\partial }_y{A}_x-[A_x,A_y]$, the same calculation reads $\alpha (1-\alpha )[B_x,B_y]$ and the nonzero pure-gauge value is $+1$. The sign convention must be specified.

For the specified SU(2) exponential, let $r=\sqrt{x^2+y^2}$ and, away from the origin, $n=(x/r,y/r,0)$. The Pauli matrix multiplication law gives $(n\cdot \sigma {)}^2=I$, so summing the exponential series yieldsThe continuous extension at the origin is $I$. For $r>0$, the second term is zero precisely when $\sin (r/2)=0$, and henceThese are infinitely many distinct circles.

At the origin, differentiating the exponential at zero gives $B_x=t_1$ and $B_y=t_2$. Since $[t_1,t_2]=t_3=-i{\sigma }_3/2$, the gauge field strength there isTo evaluate it on the circles, use polar coordinates. On $r=2\pi k$ the angular derivative of $g$ vanishes, while $B_r=-i(n\cdot \sigma )/2$. Therefore $B_x=(x/r)B_r$ and $B_y=(y/r)B_r$ commute, and $F_{xy}=0$ everywhere on every such circle, for every $\alpha$.

One can also see these circular curvature zeros for a planar SU2 exponential from a formula valid away from the origin. Write $n=(\cos \vartheta ,\sin \vartheta ,0)$, $m=(-\sin \vartheta ,\cos \vartheta ,0)$, and $t_n=-i(n\cdot \sigma )/2$, $t_m=-i(m\cdot \sigma )/2$. Direct differentiation givesand henceBoth coefficients vanish at the positive circle radii, and the expression tends to $t_3$ at the origin, agreeing with the direct calculation.

Write $B_i=({\partial }_{i}g)g^{-1}$ and use $D=d+A$. The right Maurer-Cartan equation gives ${\partial }_x{B}_y-{\partial }_y{B}_x=[B_x,B_y]$. Thus $A_i=\alpha B_i$ has the displayed gauge curvature. Both $\alpha =0$ and $\alpha =-1$ are universally flat; the latter is a pure gauge potential. Commuting $B_x,B_y$ give additional flat cases. For the opposite convention $D=d-A$, the corresponding component formula instead has $\alpha (1-\alpha )$.