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I am reading the Zee book on group and at pg.442 when explaining the Poincare' algebra an the infinity dimensional representation, it states that since $P_{\mu}P^{\mu}$ is a Lorentz invariant then it commutes with all the generators of the Lorentz algebra $so(1,3)$.

Since $P_{\mu}$ are differential operator, is there any simple proof of that statement?

In the Wu-Ki Tung book pg.182 (I am using it only occasionally as a reference) an operator transform as follow under a Lorentz transformation:

$T \to \Lambda T \Lambda^{-1}$

Where that comes from?

I know that using element of $SL(2,\Bbb C)$ a Lorentz transformation can be defined as $T \to X T X^\dagger$ where $X \in SL(2,\Bbb C)$, so may be that's the way to prove that statements

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    $\begingroup$ I'm not sure what you think there is to prove - the $P_\mu$ as generators of translation are part of the Poincaré algebra, and $P_\mu P^\mu$ being Lorentz invariant holds already on the level of the Poincaré algebra (it's a Casimir operator) before choosing any specific representation as differential operators, matrices or whatever else. $\endgroup$
    – ACuriousMind
    Nov 16 '21 at 20:57
  • $\begingroup$ I am fine with $P_{\mu} P^{\mu}$ beeing invariant but I don't understand how easily follow that it commutes with the generators $J^{\mu\nu}$. I have got it by doing explicit calculation using the commutators. But I have the same problem with the Pauli-Lubanski operator $W^{\mu}$ and with that doing explicitly calculation using commutators is much more challenging $\endgroup$
    – Andrea
    Nov 16 '21 at 21:03
  • $\begingroup$ The $J^{\mu\nu}$ generate the Lorentz transformation - how could something that does not commute with them be invariant under them? (try writing down the infinitesimal version of $P \mapsto \Lambda P \Lambda^{-1}$) $\endgroup$
    – ACuriousMind
    Nov 16 '21 at 21:05
  • $\begingroup$ How do I check if an operator is Lorentz invariant? Sorry may be a silly question but I'm not sure how to do that with operators $\endgroup$
    – Andrea
    Nov 16 '21 at 21:09
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    $\begingroup$ Zee's being silly. An operator being Lorentz invariant means, by definition, that it commutes with $J^{\mu\nu}$. So his claim is rather vacuous. What you have to do is to compute the commutator explicitly; just use the Lorentz algebra to compute $[J^{\mu\nu},P_\rho P^\rho]$ and check that, indeed, it vanishes. $\endgroup$ Nov 16 '21 at 21:51

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