 Solution

An alpha-plane is a totally null complex two-plane with tangent vectors $u^A{\pi }^{A^{\prime }}$ for a fixed primed spinor. A beta plane has tangent vectors ${\lambda }^A{u}^{A^{\prime }}$ for a fixed unprimed spinor. The metric tensor pairs two vectors of either family to zero, since the repeated fixed spinor contracts with an symplectic form. These are the two chiral families of totally null planes; labels for their self-duality depend on the orientation convention.

Globally in the twistor correspondence, writewhere $\ell$ is a line and $H$ a three-dimensional hyperplane in twistor space. Both compact families are Complex projective spaces ${\mathbb{CP}}^2$.

For distinct ${\ell }_1,{\ell }_2$, any $W\in A_{{\ell }_1}\cap A_{{\ell }_2}$ must contain their two-dimensional span. Since $W$ itself has dimension two, it must equal that span. Conversely the span belongs to both planes. Thus two distinct compact alpha-planes meet in exactly one point:In affine coordinates, independent ${\pi }_1,{\pi }_2$ determine a unique matrix $x$ from the two incidence equations. If the primed spinors are proportional while the projective twistors are distinct, no finite affine solution exists. The compact intersection still exists, at conformal infinity; saying the planes are disjoint without this affine qualification would be wrong for the printed space $M$.