With $U^2=N^2=0$ and $U\mathbin\cdot N=-1$, define the <screen-space projector>
$$
\boxed{P^a{}_b=\delta^a_b+U^aN_b+N^aU_b}.
$$
Direct contraction gives $P^a{}_bU^b=P^a{}_bN^b=0$, while a vector orthogonal to both $U$ and $N$ is unchanged. Moreover $P^a{}_cP^c{}_b=P^a{}_b$ and $P^a{}_a=2$, so its image is precisely the two-dimensional transverse space.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 1 / a / ii / Solution

{title2=$P^a{}_b$}

Given null vectors $k$ and $n$ with $k\mathbin\cdot n=-1$, the screen-space projector is
$$
P^a{}_b=\delta^a_b+k^an_b+n^ak_b.
$$
It annihilates $k$ and $n$ and projects onto the two-dimensional spacelike space transverse to them.


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