Commutator of the Pauli-Lubanski with a vector $P^{\mu}$ The Pauli-Lubanski vector is defined as $W_{\mu} = -\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}J^{\nu\rho}P^{\sigma}$, where $\varepsilon_{\mu\nu\rho\sigma}$ is the 4-dimensional Levi-Civita symbol.
The commutation reltaions are $$\left[P^{\alpha}, J^{\nu \rho}\right]=i\left(g^{\nu \alpha} P^{\rho}-g^{\rho \alpha} P^{\nu}\right)$$ $$\left[P^{\alpha}, P^{\beta}\right]=0.$$
Calculating explicitly (applying the first commutator to $P^{\alpha}J^{\nu\rho}$) $$[W_{\mu},P^{\alpha}] = \frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}(P^{\alpha}J^{\nu\rho}P^{\sigma} - J^{\nu\rho}P^{\sigma}P^{\alpha}) $$
$$= \frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}[(J^{\nu\rho}P^{\alpha} + i\left(g^{\nu \alpha} P^{\rho}-g^{\rho \alpha} P^{\nu}\right))P^{\sigma} - J^{\nu\rho}P^{\sigma}P^{\alpha}]$$
Because $[P^{\alpha}, P^{\beta}]=0$, we can switch the order of the $P$'s. We then end up with $$[W_{\mu},P^{\alpha}] = \frac{i}{2}\varepsilon_{\mu\nu\rho\sigma}\left(g^{\nu \alpha} P^{\rho}-g^{\rho \alpha} P^{\nu}\right)P^{\sigma}.$$ Is this correct? If it is, we can then argue that, because we're using Minkowski metric, that would require $\alpha=\nu=\rho$, but, because the Levi-Civita symbol requires all its indices to be different, the commutator has to be zero. Is this correct?
 A: There are two terms in your last expression:
$$\frac{i}{2}\varepsilon_{\mu\nu\rho\sigma} g^{\nu \alpha} P^{\rho} P^{\sigma}$$
and
$$-\frac{i}{2}\varepsilon_{\mu\nu\rho\sigma} g^{\rho \alpha} P^{\nu} P^{\sigma}$$
The indices need to be treated separately in each; this means that in the first, $\nu = \alpha$ and in the second $\rho = \alpha$ for the reason you gave. So this means that there may still be non-zero terms remaining which have $\alpha \ne\rho$ from the first term, for example $\frac{i}{2}\varepsilon_{1234}g^{22}P^3P^4$ when $\mu=1$ and $\alpha=2$.
However, we are left with a term involving contracting $\varepsilon_{\mu \nu \rho \sigma}$ with $P^\rho P^\sigma$; since the former is totally anti-symmetric and the latter is clearly symmetric under swapping $\rho$ and $\sigma$, this vanishes (to see this, swap $\rho$ with $\sigma$, causing it to pick up a minus sign from the anti-symmetry of $\varepsilon$, then re-label $\rho\leftrightarrow\sigma$, returning you to the same expression but with a minus sign, meaning that it must equal minus itself). The second term is zero for the same reason, so we arrive at the same answer but for a different reason.
