 Affine transverse trace in a diagonal linearized plane wave is pure gauge

For a perturbation $h_{xx}=2a(t-z)$, $h_{yy}=2b(t-z)$ with all other components zero, the linearized Vacuum Einstein equations give $(a+b{)}^{\prime \prime }=0$. Put $s=a+b$ and choose ${\xi }_x=sx/2$, ${\xi }_y=sy/2$, ${\xi }_t=-s^{\prime }(x^2+y^2)/4$, ${\xi }_z=s^{\prime }(x^2+y^2)/4$. Under the linearized coordinate gauge transformation, the mixed time/transverse and longitudinal/transverse components cancel, while the other time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The remaining components are $h_{xx}=a-b$, $h_{yy}=-(a-b)$, $h_{xy}=0$, a transverse-traceless gauge perturbation of plus polarization. This explicitly removes a nonzero affine trace without assuming it vanishes by boundary conditions. If $(a-b{)}^{\prime \prime }=0$ too, all linearized curvature vanishes.