= Solution
After the trace gauge, $h_{ab}=\bar h_{ab}=A_{ab}e^{ik\cdot x}$. The <linearized Riemann curvature operator> is built from terms containing two factors of $k$ and one factor of $A$:
$$
R^{(1)}_{abcd}
=-\frac12\left(
k_c k_bA_{ad}+k_d k_aA_{bc}
-k_d k_bA_{ac}-k_c k_aA_{bd}
\right)e^{ik\cdot x}.
$$
Contracting with $k^{[c}v^{d]}$ gives zero by $k^2=0$ and $k^aA_{ab}=0$, so $\mathcal R(k\wedge v)=0$. If $\omega_{cd}k^d=0$, every term in $R^{(1)}_{ab}{}^{cd}\omega_{cd}$ also contains such a contraction and vanishes.
To count the kernel, use the null basis $k,\ell,e_1,e_2$. The forms $k\wedge v$ span the three-dimensional space
$$
\langle k\wedge\ell,\ k\wedge e_1,\ k\wedge e_2\rangle.
$$
The condition $\omega_{ab}k^b=0$ defines the three-dimensional space
$$
\langle k\wedge e_1,\ k\wedge e_2,\ e_1\wedge e_2\rangle.
$$
Their intersection has dimension two, so their sum is a four-dimensional subspace of $\ker\mathcal R$. Since $\dim\Lambda^2=6$, the rank-nullity theorem gives
$$
\operatorname{rank}\mathcal R\leq2.
$$
Solved by gpt-5.6-sol high.
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