= Solution
An independent physical solution is the <cross polarization>
$$
\boxed{\epsilon_\times^{\mu\nu}
=\begin{pmatrix}0&0&0&0\\0&0&h_\times&0\\0&h_\times&0&0\\0&0&0&0\end{pmatrix}.}
$$
It is symmetric, traceless and transverse to $k^\mu=(\omega,0,0,\omega)$. Unlike the pure coordinate mode, it has nonzero <linearized Riemann curvature operator>: for example, in the convention $R_{0i0j}^{(1)}=-\partial_t^2h_{ij}/2$,
$$
R_{0102}^{(1)}=\frac12\omega^2h_\times\cos[\omega(t-z)].
$$
Changing the curvature sign convention reverses this component but cannot make it vanish. <Gauge invariance of the linearized Riemann tensor> therefore proves that this nonzero mode cannot be removed by a <coordinate transformation>.
The principal stretch directions are rotated by $45$ degrees relative to the <plus polarization>. In the same proper-length convention as part (c), its first-order angular response is $\delta\ell/\Delta=-\tfrac12h_\times\sin(2\theta)\cos[\omega(t-z_0)]$. The two independent transverse patterns are the two propagating gravitational degrees of freedom.
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