 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 $A$:Contracting with $k^{[c}{v}^{d]}$ gives zero by $k^2=0$ and $k^a{A}_{ab}=0$, so $\mathcal{R}(k\wedge v)=0$. If ${\omega }_{cd}{k}^d=0$, every term in $R_{ab}^{(1)}{}^{cd}{\omega }_{cd}$ 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 spaceThe condition ${\omega }_{ab}{k}^b=0$ defines the three-dimensional spaceTheir 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

Solved by gpt-5.6-sol high.