= Solution
Write the <flat connection> as $d+A_x\,dx+A_y\,dy$ in a global unitary frame. Solve the matrix ordinary differential equation
$$
\partial_x u(x,y)=-A_x(x,y)u(x,y),\qquad u(0,y)=I.
$$
Smooth dependence on parameters gives a smooth solution on the entire open square. Since $A_x$ is skew-Hermitian and trace free, differentiating $u^*u$ and $\det u$ shows that $u(x,y)\in SU(2)$. Under the <bundle gauge transformation> $u$, the transformed connection has
$$
A'_x=u^{-1}A_xu+u^{-1}\partial_xu=0.
$$
Its <vector-bundle curvature> is still zero. The remaining <vector-bundle curvature> equation is $\partial_xA'_y=0$, so $A'_y=B(y)$ is independent of $x$. Now solve $v'(y)=-B(y)v(y)$ with $v(0)=I$. Again $v\in SU(2)$; this second <bundle gauge transformation>, independent of $x$, leaves $A'_x=0$ and makes $A''_y=0$. Thus \b[the original connection is gauge equivalent to the trivial connection]. Both transformations exist on the whole square, not just in a small coordinate neighborhood.
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