and arbitrary displacement four-vector $\xi^\mu$, $x^\mu\rightarrow x^\mu+\xi^\mu$. Because any scalar, can be decomposed directly into a sum of spherical harmonics and any vector can be decomposed into a divergence part and a divergence-free part. Therefore we have $$\begin{aligned}
\xi_t &=\sum_{l,m}T_{l,m}(t,r)Y_l^m(\theta,\phi)\,,\\
\xi_r &=\sum_{l,m}R_{l,m}(t,r)Y_l^m(\theta,\phi)\,,\\
\xi_a &=\sum_{l,m}\Bigl[\Theta_{l,m}(t,r)\partial_a Y_l^m(\theta,\phi)+\Xi_{l,m}(t,r)E_a^{~b}\partial_b Y_l^m(\theta,\phi)\Bigr]\,.\end{aligned}$$ Notice that $\Theta_{l,m}(t,r)\partial_a Y_l^m(\theta,\phi)$ is the divergence part of the 2-vector $\xi_a$ and $\Xi_{l,m}(t,r)E_a^{~b}\partial_b Y_l^m(\theta,\phi)$, the divergence-free part. Only the latter contributes to the odd-parity perturbations. Therefore we consider an infinitesimal displacement of the following form $$\begin{aligned}
\xi_t &=0\,,\\
\xi_r &=0\,,\\
\xi_a &=\sum_{l,m}\Xi_{l,m}(t,r)E_a^{~b}\partial_b Y_l^m(\theta,\phi)\,.\end{aligned}$$ Under this transformation, the metric perturbations transform as $$\begin{aligned}
h_{0,(l,m)} &\rightarrow h_{0,(l,m)} +\dot\Xi_{l,m}(t,r)\,,\\
h_{1,(l,m)} &\rightarrow h_{1,(l,m)} +r^2\Bigl(\frac{\Xi_{l,m}(t,r)}{r^2}\Bigr)'\,,\\
h_{2,(l,m)} &\rightarrow h_{2,(l,m)} +2 \Xi_{l,m}(t,r)\,.\end{aligned}$$ We see therefore that the only way to fix completely the gauge is to consider $h_2=0$, any other choice, will fix partially the gauge which is usually refereed to as the “Regge-Wheeler” gauge [@Regge:1957td]. The additional advantage of using this gauge is that the equations of motion are independent of $m$, which can therefore be set to zero. Therefore the spherical harmonics can be replaced by Legendre polynomials when fixing $m=0$ $$\begin{aligned}
Y_l^m=\sqrt{\frac{(2l+1)(l-m)!}{4\pi(l+m)!}}P_l^me^{im\phi} \xrightarrow[m=0]{} \sqrt{\frac{(2l+1)}{4\pi}}P_l(\cos \theta)\,,\end{aligned}$$ where $P_l^m$ are the associated Legendre polynomials and $P_l$, the Legendre polynomials. 