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We’ve defined Casimir operator for a group as an operator which commutes with all generators of that group. For the Poincare group we’ve found two Casimir operators: $p_\mu p^\mu$ and $W_\mu W^\mu$ where $W_\mu$ is the Pauli-Lubanski vector. In checking that they are indeed Casimir operators, can I say that, since $p_\mu p^\mu$ is a scalar, it automatically commutes with all the generators? And same for the second Casimir operator.

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    $\begingroup$ See what happens if you try $\frac12 M^{\mu\nu}M_{\mu\nu}$ (which is in fact a Casimir of the Lorentz subgroup) $\endgroup$ Dec 12, 2020 at 18:09
  • $\begingroup$ @Nihar Karve You should probably write your point, expanded, as an answer, a good one, IMO. $\endgroup$ Dec 12, 2020 at 21:10

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Unfortunately, Lorentz invariant operators are not automatically Casimir operators - you can see this since there are essentially infinite independent Lorentz scalars you can construct from $M_{\mu\nu}$ and $P_\mu$, whereas the dimension of the Cartan subalgebra of the Poincaré group can be shown to be finite. An example is $\frac12 M_{\mu\nu} M^{\mu\nu}$, which is actually a Casimir operator of the Lorentz subgroup - but in the full Poincaré group, this operator fails to commute with $P_\mu$, so it falls short of being a Casimir operator for the full group.

The essence of this lies in the fact that the commutator $[AB, C]$ equals $A[B, C] + [A, C]B$, which is not identically zero (perhaps you have gotten caught up in the terminology - it is identically zero for scalars as in numbers, not Lorentz scalars)

Thus the most straightforward method to prove their Casimir-ness is to simply crank through the commutation relations (a few tricks may be employed in the case of $W_\mu W^\mu$, but that is beyond the scope of this answer). The converse, proving that these are the only 2 Casimir operators for the Poincaré group, is much trickier - see this excellent answer by David Bar Moshe for an exposition.

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  • $\begingroup$ Thank you very much. Yes I got confused by terminology, I thought scalars numbers behave as Lorentz scalar. $\endgroup$
    – john
    Dec 13, 2020 at 8:08

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