= Solution
Keep $\ell=e^0-e^3$ and set $P=p'+p^2=A''/A$, $Q=q'+q^2=B''/B$. Differentiation gives $d(p e^1)=P\ell\wedge e^1$ and $d(q e^2)=Q\ell\wedge e^2$. In <Cartan's second structure equation>, the quadratic terms for these components vanish. The apparently possible transverse component cancels:
$$
\omega^1{}_0\wedge\omega^0{}_2+\omega^1{}_3\wedge\omega^3{}_2=pq\,e^1\wedge e^2-pq\,e^1\wedge e^2=0.
$$
Thus the potentially nonzero mixed-index <curvature 2-forms> are
$$
\boxed{\begin{aligned}
\Theta^0{}_1=\Theta^1{}_0=\Theta^3{}_1&=P\ell\wedge e^1,&\Theta^1{}_3&=-P\ell\wedge e^1,\\
\Theta^0{}_2=\Theta^2{}_0=\Theta^3{}_2&=Q\ell\wedge e^2,&\Theta^2{}_3&=-Q\ell\wedge e^2.
\end{aligned}}
$$
All other components vanish. Equivalently, the independent lower-index forms are $\Theta_{01}=\Theta_{13}=-P\ell\wedge e^1$ and $\Theta_{02}=\Theta_{23}=-Q\ell\wedge e^2$, with the rest supplied by lower-index antisymmetry. These signs use $\Theta^a{}_b=\tfrac12R^a{}_{bcd}e^c\wedge e^d$ and the curvature convention in question 2(c). This is the <curvature of a diagonal plane wave in Rosen coordinates>.
Back to article page
