Cartan Killing metric and Casimir operators

I'm a little confused about Casimir operators and Cartan-Killing metric.

The Lorentz group is a semi-simple group and its Cartan-Killing metric is non-degenerate, say $g_{ab}$; it is invertible and therefore is a "good" metric, which I can raise and lower indexes with and I can define Casimir operator through as $C=g_{ab}A^aA^b$, with A belonging to the algebra.

Now, if i turn to Poincaré group, it si non-semi-simple and so it has a degenerate metric, due to the translations Abelian invariant subgroup.

Now, it is not so clear to me how, if I don't have an invertible metric, I can define Casimir operators also for the Poincaré group (using Pauli-Lubanski operator $W^{\mu}$ and translation operator). Maybe there is something I'm not considering about the structure of $W^{\mu}$?

• related: Heuristic derivation of $W^\mu=\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}P_\nu J_{\sigma\rho}$ using combination of physical and mathematical arguments. In shot, the vector $W^\mu$ only exists for $d=4$, so its kind of "an accident". The quartic Casimir exists for all $d$ though. – AccidentalFourierTransform May 21 '17 at 16:51
• Ok thank you :) I still don't get how you can define the Casimir operator without a metric. Casimir operator would be $W^{\mu}W_{\mu}$ so I indeed need a metric tensor to evaluate this. – user129511 May 21 '17 at 17:24
• A Casimir operator is just an operator that commutes with all the other generators, i.e. an element in the center of the universal envelopping algebra (most "general" associative algebra that contains the Lie algebra. Center= Ideal of elements that commute with everybody). In the special case of a Lie algebra with a non-degenerate Killing form, there is an expression for the Casimir. In general there are several Casimirs and I am not sure there is a formula for them and one does not need a Killing form to define one. I'm actually also interested in that. – Noix07 May 25 '17 at 13:54