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Interpret triviality as holomorphic triviality. Choose a global holomorphic bundle frame $s_1,\dots ,s_k$ for $E$. Let $H_{\alpha }$ have as its columns the coordinates of these sections in the prescribed local frame over $U_{\alpha }$, with the convention that a fiber coordinate vector transforms as $s_{\beta }=F_{\alpha \beta }{s}_{\alpha }$. ThenEach $H_{\alpha }$ is holomorphic and invertible because its columns form a basis in every fiber. This proves the requested holomorphic splitting of transition functions on a trivial bundle. Smooth or topological triviality alone would not justify holomorphic splitting.

For the line bundle on projective twistor space, pull its patching function back to the twistor line $L_x$ using the incidence convention of this problem, ${\omega }^A=x^{A{A}^{\prime }}{\pi }_{A^{\prime }}$, without an extra factor of $i$. Triviality on each line gives nonvanishing linewise holomorphic functions $H_0,H_1$ with $F_{01}=H_1{H}_0^{-1}$. The factors can be chosen as holomorphic functions of $x$ locally: a degree-zero scalar transition function has zero winding on the overlap annulus, hence admits a logarithm there; splitting its positive and negative Laurent series parts gives factors whose coefficients are Cauchy integrals that are holomorphic functions of $x$. The explicit splitting is worked out below. Their only common linewise ambiguity is multiplication by a nonzero function $g(x)$: the ratio is a global holomorphic function on the compact complex projective line, hence constant on that line.

Take ${ϵ}_{0^{\prime }{1}^{\prime }}=1$, ${ϵ}^{0^{\prime }{1}^{\prime }}=-1$ and raise a two-component spinor by left multiplication with ${ϵ}^{A^{\prime }{B}^{\prime }}$. The derivative along an alpha-plane is ${\ell }_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$. It annihilates the unsplit twistor function, becauseThusis holomorphic on the whole line and has projective weight one. A section of $\mathcal{O}(1)$ is linear in the homogeneous coordinates: in an affine coordinate, holomorphic regularity at infinity bounds its growth to at most first order, so the Cauchy integral formula excludes all higher powers. ThereforeThe inverse splitting functions obey $({\ell }_A+G_A)H_{\alpha }^{-1}=0$. Commuting these two equations provesThe spinor decomposition of gauge curvature iswhere both $\phi$ two-component spinors are symmetric. Vanishing of the contraction for every $\pi$ forces ${\phi }_{A^{\prime }{B}^{\prime }}=0$. This is exactly the anti-self-dual Maxwell equations. Since the Abelian gauge curvature is $dA$, it is closed; its duality condition also makes it co-closed, giving the source-free Maxwell equations. Changing both splitting factors by $g(x)$ changes $A$ by $d\log g$, the expected Abelian gauge transformation. This derives the rank-one Penrose-Ward correspondence in the needed local form.

For the exponential patching function, choose a representative $f$ of the Čech cohomology class, of homogeneous degree zero. On the overlap of the two line patches write $f_x(\zeta )=f(x\pi ,\pi )={\sum }_{n\in \mathbb{Z}}{f}_n(x){\zeta }^n$. To compute the gauge potential explicitly, use a constant two-component spinor basis in which ${\iota }_{A^{\prime }}=(1,0)$ and the finite patch ${\pi }_{0^{\prime }}=\zeta ,{\pi }_{1^{\prime }}=1$. Then ${\pi }^{0^{\prime }}=-1$, ${\pi }^{1^{\prime }}=\zeta$ andFor a positively oriented contour in the overlap annulus, split its Laurent series asThe first is holomorphic inside the contour, the second outside including infinity; $H_{\alpha }=e^{h_{\alpha }}$ split $e^f$. Since $(-{\partial }_{B{0}^{\prime }}+\zeta {\partial }_{B{1}^{\prime }})f_x=0$, coefficient comparison gives ${\partial }_{B{0}^{\prime }}{f}_n={\partial }_{B{1}^{\prime }}{f}_{n-1}$. Applying that operator separately to $h_0,h_1$ cancels all nonconstant terms and leavesThus $-A_{B{0}^{\prime }}+\zeta A_{B{1}^{\prime }}=G_B$ yields $A_{B{1}^{\prime }}=0$ and $A_{B{0}^{\prime }}=-{\partial }_{B{1}^{\prime }}{f}_{-1}$. MoreoverSubstitution gives $A_{B{0}^{\prime }}=-(2\pi i{)}^{-1}\oint f_{,B}d\zeta$, which in homogeneous form is preciselyThis is the Abelian Ward splitting by Laurent series. The factors have weights $-1,-1,2$, respectively, so the integrand is invariant under a change of homogeneous representative. The contour avoids ${\iota }_{C^{\prime }}{\pi }^{C^{\prime }}=0$; the basis chosen above places that point at infinity. One can keep the contour fixed for $x$ in a sufficiently small domain, allowing differentiation under the integral.

Finally differentiate the contour potential for an anti-self-dual Maxwell field. By the twistor incidence relation, ${\partial }_{A{A}^{\prime }}{f}_{,B}={\pi }_{A^{\prime }}{f}_{,AB}$. ConsequentlyThe second equality uses the elementary two-spinor identity ${\pi }_{A^{\prime }}{\iota }_{B^{\prime }}-{\pi }_{B^{\prime }}{\iota }_{A^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}({\iota }_{C^{\prime }}{\pi }^{C^{\prime }})$ in the declared convention. The remaining two-component spinor is symmetric in $A,B$, and no primed symmetric gauge curvature part survives. This directly verifies $dA$ is anti-self-dual. In complexified Lorentzian signature duality has Hodge star eigenvalues $\pm i$; here the vanishing primed two-component spinor fixes the stated ASD convention rather than incorrectly imposing the Euclidean equation $\ast F=-F$ on that signature.