 Solution

An independent physical solution is the cross polarizationIt 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 $R_{0i0j}^{(1)}=-{\partial }_t^2{h}_{ij}/2$,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 $\delta \ell /\Delta =-\frac{1}{2}{h}_{\times }\sin (2\theta )\cos [\omega (t-z_0)]$. The two independent transverse patterns are the two propagating gravitational degrees of freedom.