Lorentz covariance of Pauli-Lubanski pseudo-vector The Pauli-Lubanski pseudo-vector is defined as:
$$W_{\mu}=\frac{1}{2}\epsilon_{\mu \nu \lambda \rho}J^{\nu \lambda}P^{\rho}$$
Where the rotation and translation operators transform as:
\begin{align}
U(\Lambda,a)P^{\mu}U^{-1}(\Lambda,a)&=\Lambda^{\mu}_{\nu}P^{\nu}\\
U(\Lambda,a)J^{\mu \nu}U^{-1}(\Lambda,a)&=(\Lambda^{-1})^{\mu}_{\lambda}(\Lambda^{-1})^{\nu}_{\rho}(J^{\lambda \rho}-a^{\lambda}P^{\rho}+a^{\rho}P^{\lambda})
\end{align}
I'm working on how the Pauli-Lubanski transforms under Lorentz, to show that it's in fact Lorentz covariant. I saw a solution in one of the answers of this post Calculating the commutator of Pauli-Lubanski operator and generators of Lorentz group the following:
\begin{align}
U(\Lambda,a)W_{\mu}U^{-1}(\Lambda,a)&=\frac{1}{2}\epsilon_{\mu \nu \lambda \rho}(\Lambda^{\nu}_{\alpha}\Lambda^{\rho}_{\beta}J^{\alpha \beta}+a^{\nu}\Lambda^{\rho}_{\alpha}P^{\alpha}-a^{\rho}\Lambda^{\nu}_{\rho}P^{\rho})\Lambda^{\sigma}_{\delta}P^{\delta}\\
&=\frac{1}{2}\Lambda^{\alpha}_{\mu}\epsilon_{\alpha \nu \rho \sigma}J^{\nu \rho}P^{\sigma}\end{align}
However, I'm having trouble handling the $\Lambda's$ in order to get from step one to step two. Could you offer me some guidance?
 A: Since you're interested in the transformation law of $W^{\mu}$ under the Lorentz group, let us set $a^{\mu}=0$ in your equations. Also, notice that you've got the first one wrong; it should read
$$
U(\Lambda)P^{\mu}U(\Lambda)^{-1}=(\Lambda^{-1})^{\mu}_{\nu}P^{\nu}
$$
Using the latter together with the transformation law for $J^{\mu\nu}$, we write
$$
U(\Lambda)W_{\mu}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}U(\Lambda)J^{\nu\lambda}P^{\rho}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}U(\Lambda)J^{\nu\lambda}U(\Lambda)^{-1}U(\Lambda)P^{\rho}U(\Lambda)^{-1}=\frac{1}{2}\epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}\ J^{\tau\sigma}P^{\alpha}
$$
Now, observe that from the definition of the determinant
$$
\epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\mu}_{\beta}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}=\det(\Lambda^{-1})\ \epsilon_{\beta\tau\sigma\alpha}
$$
follows the equality
$$\epsilon_{\mu\nu\lambda\rho}(\Lambda^{-1})^{\nu}_{\tau}(\Lambda^{-1})^{\lambda}_{\sigma}(\Lambda^{-1})^{\rho}_{\alpha}=\det(\Lambda^{-1})\ \Lambda_{\mu}^{\beta}\ \epsilon_{\beta\tau\sigma\alpha}$$
Therefore
$$
U(\Lambda)W_{\mu}U(\Lambda)^{-1}=\det(\Lambda^{-1})\Lambda_{\mu}^{\beta}\ \frac{1}{2}\ \epsilon_{\beta\tau\sigma\alpha}\ J^{\tau\sigma}P^{\alpha}=\det(\Lambda^{-1})\Lambda^{\beta}_{\mu}W_{\beta}
$$
Rearranging the indices and using the defining property of the $\Lambda$'s, $\Lambda^{T}\eta\Lambda=\eta$, we find
$$
U(\Lambda)W^{\mu}U(\Lambda)^{-1}=\det(\Lambda^{-1})(\Lambda^{-1})_{\beta}^{\mu}W^{\beta}
$$
Since $\det(\Lambda^{-1})=\pm 1$ depending on the Lorentz transformation, we find
$$
U(\Lambda)W^{\mu}U(\Lambda)^{-1}=\pm(\Lambda^{-1})_{\beta}^{\mu}W^{\beta}
$$
which is the transformation law for a pseudo-vector.
