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Incidence alpha planes detect positive-orientation anti-self-duality
Statement
Let be a smooth finite-rank complex vector bundle and let be a smooth connection on , with curvature . For real two-forms , let be the Hodge star characterized by , and extend it complex-linearly to the form factor of . At a finite point of the positively oriented stereographic chart in the incidence geometry, evaluate complex-bilinearly on and . Then for every if and only if the self-dual part vanishes at that point. Equivalently, restricts to zero on every alpha-plane in the full projective family , including as represented in the second incidence chart. Vanishing on the single plane at alone is insufficient.
Statement references
Facts & Assumptions
Given: The positive orientation, metric, and horizontal vectors of the incidence definition; the complexification of a real two-form is evaluated complex bilinearly.
The chart uses the positive orientation , metric , and horizontal vectors , . Positive incidence geometry for the coupled Penrose transform
Proof
Write and use antisymmetry. Bilinearity gives . If this polynomial vanishes for every complex , each coefficient vanishes; conversely, vanishing of the three coefficients makes the polynomial zero.
In the oriented Euclidean orthonormal coframe , the complex two-forms , , and span the complexified self-dual two-forms: applying the Euclidean star to each gives itself. The metric is a positive conformal rescaling of the Euclidean metric, and on two-forms in dimension four the star is unchanged by such a rescaling. The three coefficients in step 1.1 are the components of paired with this basis, up to nonzero constants and exchange of the first two basis vectors, so their vanishing is equivalent to , entrywise in . In the second chart, put and use , . On the overlap these are , and their curvature evaluation extends as to . Thus the finite-parameter condition implies vanishing also at infinity, proving the full-family equivalence. At alone it tests only ; for example the nonzero self-dual form vanishes on that plane.