Proof of the commutator ## [P^2,P_\mu]=0 ##

[RicardoMP]

Homework Statement

I want to prove that the Casimir operator of the Poincaré algebra $P^2$ satisfies $[P^2,P_\mu]=0$.

Relevant Equations

The most relevant equation is $[P_\mu,P_\nu]=0$.

I want to make certain that my proof is correct:
Since $P^2 = P_\nu P^\nu=P^\nu P_\nu$, then $[P^2,P_\mu]=[P^\nu P_\nu,P_\mu]=P^\nu[P_\nu,P_\mu]+[P^\nu,P_\mu]P_\nu=[P^\nu,P_\mu]P_\nu=g^{\nu\alpha}[P_\alpha,P_\mu]P_\nu=0$, since $g^{\nu\alpha}$ is just a number, I can bring it out of the commutator, thus giving me the desired result. Is this last step correct?

 

[PeroK]

It looks good to me. You could also simply expand $P^{\nu} = g^{\nu \alpha}P_{\alpha}$ to see that $P^{\nu}$ and $P_{\mu}$ commute. And, in general, if $A$ commutes with $B$ and $C$ then $A$ commutes with $BC$.