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Use the signature-$(2,2)$ twistor Hermitian formA real Minkowski point $x$ is a Hermitian matrix $x^{A{A}^{\prime }}$. Every vector of its incidence subspace has the form $(ix\pi ,\pi )$. Substitution givesIf the two null geodesics meet at $x$, their null twistors $X,Y$ both belong to this totally isotropic subspace. In addition to their individual nullness, they must obeyFor independent $X,Y$, this says that every point of the projective line $\mathbb{P}\mathrm{span}(X,Y)$ is null: it lies entirely in projective null twistor space. It is the twistor line of their meeting point. In the finite affine setting, distinct null directions at the meeting point make their primed spinors independent; proportional primed spinors would give proportional incident twistors and the same null geodesic.

The meaningful twistor-geometric interpretation of “colinear in $PN$” is that the entire projective line through two distinct projective null twistors lies in projective null twistor space. Mere membership of two points in a projective line would impose no condition, because every pair of distinct points in ${\mathbb{CP}}^3$ spans one.

Indeed, individual nullness givesThis vanishes for every $a,b\in \mathbb{C}$ precisely when $h(X,Y)=0$. Thus $W=\mathrm{span}(X,Y)$ is a maximal totally isotropic two-plane, a real point of Minkowski conformal compactification under the Lorentzian reality structure on twistor space. The two corresponding alpha-planes meet at this unique $W$. Hence their compact intersection is a real spacetime point, and their real null geodesics meet there.

There are two qualifications. The point can be at conformal infinity, so a finite affine meeting is not automatic. If $X,Y$ represent the same projective point instead, their alpha-planes coincide rather than having a unique intersection. If “colinear” is read only as spanning a line in ${\mathbb{CP}}^3$, individual nullness alone does not force a real intersection; the additional orthogonality condition above is indispensable.

For a concrete counterexample to the weaker reading, take $X=(0,0,1,0)$ and $Y=(1,0,0,1)$ in $({\omega }^0,{\omega }^1,{\pi }_{0^{\prime }},{\pi }_{1^{\prime }})$ coordinates. Each is a null twistor, but $h(X,Y)=1$. Their two incidence equations give $x=-i\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, which is not Hermitian. Thus their alpha-planes meet at a complex point that is not a real Minkowski point.

A real spacetime point $p$ is represented by its twistor line $L_p\cong {\mathbb{CP}}^1$, lying entirely in projective null twistor space. Each point of $L_p$ is a null twistor specifying one real null geodesic through $p$. Their union is the null cone with vertex $p$. Thus the twistor description of the null cone is the family of null geodesics parametrized by the single line $L_p\subset PN$.

Equivalently, the spacetime points on that cone are precisely those whose twistor lines meet $L_p$:For distinct finite points this follows immediately from the incidence equations: a shared nonzero $\pi$ obeys $(q-p{)}^{A{A}^{\prime }}{\pi }_{A^{\prime }}=0$, which is equivalent to $\det (q-p)=0$, the null-separation condition. The flag description extends it to the compact spacetime. Projective null twistors encode whole unoriented null geodesics, so this includes both future and past generators; a time orientation and an affine patch distinguish the two halves. The line itself is the sphere of null directions at the vertex, not a three-dimensional image of the cone embedded pointwise into twistor space.

The five-real-dimensional space of projective null twistors. A projective line lies entirely in it precisely when its corresponding two-plane is totally isotropic for the twistor Hermitian form. Such lines represent real points of Minkowski conformal compactification. Two independent null twistors lie on one such line exactly when $h(X,Y)=0$.