# On Poincare group’s Casimir operators

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.

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

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.