Constant noncommuting potentials with nonzero curvature

Regard $M_2(\mathbb{C})$ as ${\mathbb{R}}^8$. The conditions $g^{\ast }g=I$ and $\det g=1$ defining $SU(2)$ are closed equations in the matrix entries, so $SU(2)$ is closed. Each column of a unitary matrix has norm one, hence all entries have modulus at most one, so $SU(2)$ is bounded. By [F2], it is compact. The matrices $D=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)$ and $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ are unitary with determinant one, but $DJ=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\mathrm{\ne }\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)=JD$. Thus $SU(2)$ is noncommutative, and [F1] supplies its Lie-group structure. This verifies the group conditions in [A1]. The declared axiom base is ${\mathrm{A}\mathrm{C}}_{\omega }$ by [A2], while [F2] uses no choice. Each Pauli matrix is Hermitian and traceless, so each $T_a$ is anti-Hermitian and traceless; [F6] puts them in $\mathfrak{s}\mathfrak{u}(2)$. Direct multiplication gives ${\sigma }_1{\sigma }_2=i{\sigma }_3$ and ${\sigma }_2{\sigma }_1=-i{\sigma }_3$, hence $[T_1,T_2]=(-i\mathrm{/}2{)}^2[{\sigma }_1,{\sigma }_2]=-(1\mathrm{/}4)(2i{\sigma }_3)=T_3$. Since $T_3\mathrm{\ne }0$, this also verifies that $T_1$ and $T_2$ do not commute.

The product $P=M\times SU(2)$ with right multiplication on its second coordinate is smooth, free, and transitive on each fibre, and the product chart is equivariant; [F10] therefore makes it a smooth right principal bundle. By [F8], the displayed $\omega$ is smooth. A fundamental vertical vector generated by $Z\in \mathfrak{s}\mathfrak{u}(2)$ at $(z,g)$ is $(0,gZ)$, and $({\theta }_L{)}_g(gZ)=Z$, so $\omega (Z_P)=Z$. For a fixed $h\in SU(2)$, the right-translated tangent is $(X,\xi h)$; [F8] gives $({\theta }_L{)}_{gh}(\xi h)=h^{-1}({\theta }_L{)}_g(\xi )h={\mathrm{Ad}}_{h^{-1}}(({\theta }_L{)}_g(\xi ))$, and [F9] gives ${\mathrm{Ad}}_{(gh{)}^{-1}}={\mathrm{Ad}}_{h^{-1}}{\mathrm{Ad}}_{g^{-1}}$. Therefore ${\omega }_{(z,gh)}(X,\xi h)={\mathrm{Ad}}_{h^{-1}}{\omega }_{(z,g)}(X,\xi )$. Together with reproduction, [F3] shows it is a principal connection form. Its pullback along $s(z)=(z,e)$ is $A$.

The coefficients of $A$ are constant, so $dA=0$. Evaluating the bracket on $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$, [F4] gives $\frac{1}{2}[A\wedge A]({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)=[T_1,T_2]=T_3$ by step 1.1. Since a 2-form on ${\mathbb{R}}^2$ is determined by this value on the coordinate basis, $F=T_{3}dx\wedge dy$.

For the defining representation, the differential ${\rho }_{\ast }$ sends each tangent matrix $T\in \mathfrak{s}\mathfrak{u}(2)$ to the same endomorphism $T$ of ${\mathbb{C}}^2$, since $\rho$ is the matrix inclusion. Applying [F7] to a smooth $\psi :M\to {\mathbb{C}}^2$ gives the two displayed covariant derivatives. The constant matrices commute with the coordinate derivatives, and ${\mathrm{\partial }}_x{\mathrm{\partial }}_y\psi ={\mathrm{\partial }}_y{\mathrm{\partial }}_x\psi$; expansion gives $[{\mathrm{\nabla }}_{{\mathrm{\partial }}_x},{\mathrm{\nabla }}_{{\mathrm{\partial }}_y}]\psi =(T_1{T}_2-T_2{T}_1)\psi =T_3\psi$. This verifies the stated operator commutator for every smooth section in the global trivialization.