There are two Casimirs of the Poincare group:

$$ C_1 = P^\mu P_\mu, \quad C_2 = W^\mu W_\mu $$

with the Pauli-Lubanski vector $W_\mu$. This implies the Poincare group has rank 2.

Is there a way to show that there really are no other Casimir operators other than trying to build all possible combinations of generators and seeing them fail?

To state it differently: Can one determine the rank of the Poincare group without explicit construction of the Casimir operators?


For semisimple groups (and the Poincaré group is not such), the number of Casimirs (i.e., the number basic generators center of the universal enveloping algebra) is equal to the dimension its Cartan subalgebra (maximal commuting subalgebra) which is the rank of the algebra. This is called Chevalley's theorem.

The Poincaré group however is not semisimple (it is a reductive Lie group given by a semidirect product of a semisimple (Lorentz) and an Abelian group (translations)), thus this theorem is not valid in this case (even if both ranks are equal to 2 in the case of the Poincaré group, there is no general theorem for that).

For such groups the center of the universal enveloping algebra can be characterized by the Harish-Chandra isomorphism , which is less constructive than the Chevalley's theorem.

There is a way to "understand" why the number of Casimirs of the Poincaré group is 2. The Poincaré group is a Wigner-İnönü contraction of the de-Sitter group $SO(4,1)$ which is semisimple and of rank 2.

The Casimirs of the Poincaré group can be obtained from the Casimirs of $SO(4,1)$ explicitly in the contraction process. This is not a full proof because group contractions are singular limits, but at least it is a way to understand the case of the Poincaré group.

  • $\begingroup$ Does the contraction not contain any more Casimirs than the original group? How does one see this? $\endgroup$
    – label
    May 29 '20 at 18:23

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