How to show that the polarization tensor of the gravitational plane wave in vacuum is transverse 
This is Ta-Pei Cheng's book p.254. I am trying to understand (13.28). The Lorentz gauge condition (13.18) is like below. How can I use the (13.18) condition to show (13.28)? I am trying to calculate but the trace term $h$ makes everything twisted...

 A: It seems to me that Ta-Pei Cheng is simply wrong. The Lorenz gauge does not automatically imply transverseness of the wave. As a simple counter example take equation 13.25 with
$$ k_\mu = (-1,1,0,0) $$
and
$$ \epsilon_{\mu\nu} =\begin{pmatrix} 1 & 0 & 0 &0 \\0&-1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$.
One easily confirms that this satisfies both the wave equation 13.23 and the Lorenz gauge condition 13.18. However, it is manifestly not transverse since $k^\mu\epsilon_{\mu\nu} = (1,-1,0,0)$.
It is however true, that one can use the residual gauge freedom in the Lorenz gauge to transform plane wave to the traceless transverse (TT) gauge. See e.g. Poisson&Will Sec.11.1.5.
A: $$
\partial^\mu h_{\mu\nu}=g^{\mu\sigma}\partial_\sigma\epsilon_{\mu\nu}e^{ik_\lambda x^\lambda}
\\=g^{\mu\sigma}\epsilon_{\mu\nu}ik_\alpha\delta^\alpha_\sigma e^{ik_\lambda x^\lambda}
\\=g^{\mu\sigma}\epsilon_{\mu\nu}ik_\sigma e^{ik_\lambda x^\lambda}
\\=\epsilon_{\mu\nu}ik^\mu e^{ik_\lambda x^\lambda}
\\\Rightarrow k^\mu\epsilon_{\mu\nu}=0\qquad\text{(transverse)}
$$
If the divergence of the perturbation is zero, that is, if it is transverse (Lorenz condition), then the polarization tensor is also transverse.
