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In the two-component spinor convention with primed symmetric gauge curvature self-dual, an Abelian Maxwell field is anti-self-dual when its primed symmetric gauge curvature two-component spinor vanishes. For a locally exact field $F=dA$, the Bianchi identity gives $dF=0$; the duality relation then also gives $d\ast F=0$. Thus it obeys the source-free Maxwell equations. In complexified Lorentzian signature the Hodge star eigenvalues are $\pm i$; the two-component spinor condition specifies which component is meant without imposing a Euclidean eigenvalue on that signature.

// Destination: general-relativity.bigb

The Abelian Ward splitting by Laurent series gives this gauge potential through the Cauchy integral formula, with $f_{,B}=\partial f/\partial {\omega }^B$ evaluated on the twistor line. The integrand has total projective degree zero. Holding the contour fixed locally and differentiating gives $F_{A{A}^{\prime }B{B}^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}(2\pi i{)}^{-1}{\oint }_{\Gamma }{f}_{,AB}{\pi }_{D^{\prime }}d{\pi }^{D^{\prime }}$, using ${\pi }_{A^{\prime }}{\iota }_{B^{\prime }}-{\pi }_{B^{\prime }}{\iota }_{A^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}({\iota }_{C^{\prime }}{\pi }^{C^{\prime }})$. Its unprimed two-component spinor is symmetric, so the primed gauge curvature vanishes and the field obeys the anti-self-dual Maxwell equations. These signs use ${ϵ}_{0^{\prime }{1}^{\prime }}=1$, ${ϵ}^{0^{\prime }{1}^{\prime }}=-1$ and raising by left multiplication; other consistent raising conventions can reverse both displayed overall signs.

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.

On complexified Minkowski spacetime, use the twistor incidence relationThe harmless factor $i$ is a conventional choice; omitting it corresponds to rescaling the complexified coordinates. For fixed $x$, projectivizing the nonzero two-component spinors $\pi$ gives a twistor line $L_x\simeq {\mathbb{CP}}^1$. The two stated patches cover all these incidence lines; they describe the affine-space-time twistor domain with the line $\pi =0$ removed from full projective twistor space.

Since $E{|}_{L_x}$ is holomorphically trivial, choose a global bundle frame on that line. Comparing its columns with the bundle frames in the two patch trivializations gives invertible matrices $H(x,\pi )$ and $\tilde{H}(x,\pi )$, holomorphic on their respective patches, withThis is the holomorphic splitting of a twistor patching matrix. It follows directly from existence of the global bundle frame, rather than from an arbitrary factorization of an arbitrary matrix: nontrivial line bundles would obstruct such a nonsingular splitting.

The factors can be chosen holomorphically in $x$ near any given $x_0$. An explicit local argument is useful. Normalize the known splitting at $x_0$ and set $Q(x,\lambda )=\tilde{H}(x_0,\lambda {)}^{-1}F(x,\lambda )H(x_0,\lambda )$ on a fixed circular contour in the overlap. It is close to the identity when $x$ is close to $x_0$. Let $P_{+}$ select strictly positive Laurent powers, and seek $q=1+$ positive powers satisfyingIn a holomorphic contour-function norm the operator has norm below one for sufficiently small variation, so its Neumann series converges and is holomorphic in $x$. Then $Qq=\tilde{q}$ has only nonpositive powers and is holomorphic on the exterior patch including infinity. Both factors are close to the identity and invertible. They extend over their full patches through $Q$ and its inverse. Taking $H=H(x_0)q$ and $\tilde{H}=\tilde{H}(x_0)\tilde{q}$ gives the required local family. This is also the elementary Laurent series reason that $H^1({\mathbb{CP}}^1,\mathcal{O})=0$ leaves no infinitesimal obstruction.

Raise primed two-component spinor indices with the antisymmetric two-component spinor form and define $V_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$, holding $\pi$ fixed. Incidence givesThus $V_{A}F=0$ for a pulled-back function depending only on twistor coordinates. Differentiating the splitting and multiplying by the inverse factors givesThe left expression is holomorphic on one two-component spinor patch, the right on the other, and each is homogeneous of degree one in $\pi$. They therefore glue to a global matrix-valued section of $\mathcal{O}(1)$ on the line.

Such a section is linear in its homogeneous coordinates. To see this explicitly, set $\lambda ={\pi }_{0^{\prime }}/{\pi }_{1^{\prime }}$ on $U$. The normalized expression is an entire matrix function of $\lambda$; on the other patch its quotient by $\lambda$ is holomorphic at infinity. Hence it has at most linear growth. Cauchy estimates force all its derivatives of order at least two to vanish. Entry by entry it is $a_A(x)+\lambda b_A(x)$, or invariantlyThe coefficients are holomorphic in $x$. This proves existence of the desired space-time one-form, with no remaining dependence on $\pi$ in its coefficients.

A second splitting must differ by right multiplication $H\mapsto Hg(x)$ and $\tilde{H}\mapsto \tilde{H}g(x)$. Indeed the comparison matrix is holomorphic on both patches of a compact complex projective line and therefore constant along the line. The resulting transformation isthe gauge connection transformation law. Local coefficient matrices consequently glue to a holomorphic gauge connection on the bundle whose fiber at $x$ is the space of holomorphic sections of $E{|}_{L_x}$. If a single global matrix-valued one-form is wanted on all $M_{\mathbb{C}}\simeq {\mathbb{C}}^4$, choose a basis at the origin and parallel-transport it along the straight radial paths to each $x$. Linear transport exists on every finite path and varies holomorphically with $x$; this gives a global holomorphic bundle frame. The assertion does not require flat connection in all four directions, only a fixed choice of paths for a trivialization. On more general space-time domains the intrinsic object is a gauge connection with these gauge-related local one-forms.

Finally put $D_{A{A}^{\prime }}={\partial }_{A{A}^{\prime }}+{\Phi }_{A{A}^{\prime }}$ and ${\mathcal{D}}_A(\pi )={\pi }^{A^{\prime }}{D}_{A{A}^{\prime }}$. From the defining identity,Thus the columns of $H^{-1}$ are an invertible parallel bundle frame in the two alpha-plane directions. Their matrix commutator must annihilate that bundle frame, soUse the spinor decomposition of gauge curvaturewhere both gauge curvature two-component spinors are symmetric. The second term vanishes under contraction with ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}$. The first is a quadratic polynomial in the two two-component spinor components; vanishing for every two-component spinor forces each of its coefficients to vanish. Thereforewhich is precisely the ASDYM equation in the convention where the primed symmetric two-component spinor is the self-dual part. Equivalently, the gauge connection is flat when restricted to every alpha-plane. This completes the forward Penrose-Ward correspondence, including both the existence of the gauge connection and its anti-self-duality of gauge curvature.

A holomorphic vector bundle on an appropriate twistor domain, trivial on its space-time incidence lines, determines a complex anti-self-dual gauge gauge connection. Split its patching matrix on each line and apply the tangent vector fields ${\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$. They annihilate the unsplit twistor patching matrix, so the two logarithmic derivatives of the factors agree and define a global degree-one function of $\pi$. It is consequently ${\pi }^{A^{\prime }}{A}_{A{A}^{\prime }}$. The inverse splitting bundle frame is parallel along each alpha-plane; commuting these parallel equations proves that the primed gauge curvature two-component spinor vanishes. Reality conditions recover unitary Euclidean gauge fields; Euclidean instanton bundles need triviality on the real twistor lines, not on every possible complex line.

A global holomorphic section of $\mathcal{O}(1)$ on ${\mathbb{CP}}^1$ is a linear homogeneous polynomial in the two projective coordinates. On one patch its coefficient is an entire function $p(\lambda )$; regularity on the other patch says $p(\lambda )/\lambda$ is regular at infinity. Hence $p$ has at most linear growth. Cauchy estimates make its second and higher derivatives zero, so $p=a+b\lambda$. The same argument entrywise applies to matrix-valued sections and makes the logarithmic derivative in the Penrose-Ward correspondence linear in the primed two-component spinor.

In complexified four-dimensional space-time, antisymmetry giveswhere both gauge curvature two-component spinors are symmetric. The two summands are the two duality components. In the convention where the primed symmetric part is self-dual, the Anti-self-dual Yang-Mills equations are ${\phi }_{A^{\prime }{B}^{\prime }}=0$. Equivalently ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}{F}_{A{A}^{\prime }B{B}^{\prime }}=0$ for every primed two-component spinor $\pi$, since a quadratic polynomial vanishing for every two-component spinor has all symmetric coefficients zero.