Commutator of the Pauli-Lubanski vector operator and the generator of translations $P^\alpha$ I'm trying to obtain the commutation relation between the Pauli-Lubanski vector operator and the generators of the Lorentz Group:
$$[W^\mu,P_\sigma]=[\frac{1}{2}\epsilon^{\mu\nu\lambda\rho} P_\nu M_\lambda\rho,P_\sigma]\\ \hspace{2.3cm}= \frac{1}{2}\epsilon^{\mu\nu\lambda\rho}[ P_\nu M_{\lambda\rho},P_\sigma]\\ \hspace{3.9cm}=\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}P_\nu[ -\eta_{\lambda\sigma}P_\rho+\eta_{\rho\sigma}P_\lambda]\\ \hspace{5.cm}=\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\rho\sigma}P_\nu P_\lambda-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho $$
Now, I know a priori that this commutator is zero and I'm trying to change indices accordingly. For instance, in the first term I want to rename the dummy indices $\lambda$ to $\rho$ and vice-versa. This will permute the respective last indices in the Levi-Civita tensor. Since I'm only renaming indices, I assume that I don't have to put a minus sign when doing so:
$$=\frac{1}{2}\epsilon^{\mu\nu\rho\lambda}\eta_{\lambda\sigma}P_\nu P_\rho-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho$$
But now, to get the same form for the Levi-Civita tensor in both terms I permute the last two indices of that tensor in the first term, taking into account that it is an antisymmetric tensor:
 $$=-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho=-\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho$$ , which is not $0$.
Have I gone wrong somewhere?
If I do the same for the second term instead of the first I get the same result but positive. Since the results must be equal, the only possibility for something of the form $+\text{final result}=-\text{final result}$ is for $\text{final result}=0$. Does it make sense?
 A: Every term of the form 
$$\varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\sigma}$$ 
in your calculation  is zero, including the thing you said was not zero, because e.g. in $\varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\sigma}$ the $\varepsilon^{\mu \nu \rho \sigma}$ is anti-symmetric in $\rho$ and $\sigma$, $\varepsilon^{\mu \nu \rho \sigma} = - \varepsilon^{\mu \nu \sigma \rho}$, while $P_{\rho} P_{\sigma}$ is symmetric in ν and ρ, $P_{\rho} P_{\sigma} = P_{\sigma} P_{\rho}$.
The simplest way to show that this is zero is to prove that $A = - A$ so that $2A = 0$. For simplicity we consider the two-dimensional analogue
$$\varepsilon^{\mu \nu} P_{\mu} P_{\nu}$$
where $\mu, \nu = 0,1$. I want to show that $\varepsilon^{\mu \nu} P_{\mu} P_{\nu} = - \varepsilon^{\mu \nu} P_{\mu} P_{\nu}$. The calculation is as follows:
\begin{align}
\varepsilon^{\mu \nu} P_{\mu} P_{\nu} &= + \varepsilon^{\nu \mu} P_{\nu} P_{\mu} \ \ (1) \\
&= - \varepsilon^{\mu \nu} P_{\nu} P_{\mu} \ \ (2) \\
&= - \varepsilon^{\mu \nu} P_{\mu} P_{\nu}  \ \ (3) 
\end{align}
In line $(1)$ I used the fact that I can just re-label dummy indices whatever way I want, since they are dummy indices, and here I want to have them written in reverse order so I can later invoke anti-symmetry on $\varepsilon^{\mu \nu}$ and symmetry on $P_{\mu} P_{\nu}$. To see very explicitly why I can re-label dummy indices, just write it out:
\begin{align}
\varepsilon^{\mu \nu} P_{\mu} P_{\nu} &= \varepsilon^{0 \nu} P_{0} P_{\nu} + \varepsilon^{1 \nu} P_{1} P_{\nu} \\
&= (\varepsilon^{00} P_{0} P_{0} + \varepsilon^{0 1} P_{0} P_{1}) + (\varepsilon^{1 0} P_{1} P_{0} + \varepsilon^{1 1} P_{1} P_{1}) \\
&= (\varepsilon^{0\mu} P_{0} P_{\mu}) + (\varepsilon^{1 \mu} P_{1} P_{\mu}) \\
&= \varepsilon^{\nu \mu} P_{\nu} P_{\mu}.
\end{align}
Note I have done absolutely nothing but write it out so that there are no dummy indices, then collect the terms up with dummy indices again, but now using a different labelling.
In going from $(1)$ to $(2)$ I used the anti-symmetry of $\varepsilon^{\mu \nu}$ and in going from $(2)$ to $(3)$ I used the symmetry of $P_{\mu} P_{\nu}$. Now I have $A = - A$ so that $2A = 0$.
Another way to prove this result is to write it in the form $A = \frac{1}{2} A + \frac{1}{2} A = \frac{1}{2}A - \frac{1}{2} A = 0$ which is just a longer way of doing the above calculation, and clearly uses the above calculation in going from the $+$ to the $-$, but it is often used (e.g. to derive the angular momentum operators in special relativity/classical mechanics etc...) so it's good to be aware:
\begin{align}
\varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} &= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} +  \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} \ \ (1) \\
&=  \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} -  \frac{1}{2} \varepsilon^{\mu \rho \nu \sigma} P_{\nu} P_{\rho} \ \ (2) \\
&= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} -  \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\nu} \ \ (3) \\
&= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} -  \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho}  \ \ (4) \\
&= 0. 
\end{align}
In line $(1)$ I know that $\varepsilon^{\mu \nu \rho \sigma}$ is anti-symmetric while $P_{\nu} P_{\rho}$ is symmetric so that the whole thing is immediately zero, and I want to show this explicitly by turning it into something like $A = \frac{1}{2} A + \frac{1}{2} A = \frac{1}{2} A - \frac{1}{2} A = 0$, so I introduce the $1/2$ just to get two copies of it which I expect will cancel one another. In going $(1)$ to to $(2)$ I just used the anti-symmetry of $\varepsilon^{\mu \nu \rho \sigma}$ to write one of them with a $-$ sign. In going from $(2)$ to $(3)$ I then re-labelled the dummy indices so that I would have $\varepsilon^{\mu \nu \rho \sigma}$ in both terms, In going from $(3)$ to $(4)$ I then used commutativity of $P_{\mu}$ and $P_{\nu}$. Note $(4)$ is now in the form $A = \frac{1}{2}A - \frac{1}{2}A = 0$.
