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Doing some studies on Group theory, I worked Frederic Schuller's lectures on youtube where he classifies all semisimple Lie algebras by the Dynkin's diagrams; I should say it was interesting.

  • Trying to do the same work on my own on the Poincaré algebra, I stumbled on the fact that Poincaré's algebra is non-semisimple (that's caused by the abelian subalgebra of translations), this realisation made me search for a classification that would treat this kind of algebra nonetheless I didn't find anything (If you have an idea please share with me).

  • The fact that the Poincaré algebra was non-semisimple made me look for the killing form $K(a,b)= tr(ad(a)\circ ad(b))$ of the algebra and then calculate the determinant to find it equal to zero, but I, personally, found it hard to do because we have two types of bases $P_i$ and $M_{ab}$ and I found myself doing a case by case not really being sure of what am doing. My question is how to calculate the Killing form of the Poincaré algebra?

  • My last remark is about the Cartan subalgebra (CSA) of Poincaré algebra. The CSA has a hard definition (a nilpotent subalgebra of a Lie algebra that is self-normalising) trying to project this definition on the Poincaré algebra was difficult for me and I confused myself with too much information about centralizers and maximal abelian subalgebra, can you guide me on that? that would be great.

I will edit my post if I find myself misunderstood.

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  1. The infinite-dimensional, unitary reps are classified and constructed using the method of induced representations (basically the famous Wigner paper). The finite-dimensional, non-unitary or indecomposable, reps are not classified. To get a taste of what you’re getting into, check out this nice paper on the simpler E(2) case. I can’t find the reference but I remember reading the E(2) is “wild” so quite likely Poincaré is also wild.

  2. The Killing form is always computed using the trace of the adjoint. It’s hard to do here because you have lots of generators, but there’s no trick to save time here.

  3. The Cartan subalgebra kinda makes sense only for semi-simple cases. In principle you can start with Cartan of the semi-simple case (that would be the same as the Cartan of $\mathfrak{so}(4)$). Moreover, the Cartan subalgebra is really defined for the complexification of the algebra, so doesn’t do much for you if you have to go back to the real form.

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  • $\begingroup$ thank you for your time. $\endgroup$
    – intelligible no
    Commented Jun 9, 2021 at 5:26
  • $\begingroup$ correct if am wrong but, all what am going to say is equivalent : irreps of the Poincaré algebra comes from the CSA in other words from the set of casimirs, from the set of matrix representation of some elements of the algebra that can be diagonalizable, from the stabilizer subgroups $\endgroup$
    – intelligible no
    Commented Jun 9, 2021 at 5:42
  • $\begingroup$ can i say that the root set of the Poincaré algebra is all the eigenvalues of the casimirs? $\endgroup$
    – intelligible no
    Commented Jun 9, 2021 at 5:48
  • $\begingroup$ The root vectors are not eigenvalues of any operators: the roots are linear combinations of the generators. Moreover, Casimirs are not in the algebra itself but in the enveloping algebra (polynomials in the generators). Thus, the Cartan subalgebra is not spanned by Casimirs. $\endgroup$
    – ZeroTheHero
    Commented Jun 9, 2021 at 11:39
  • $\begingroup$ As a simple example, in $\mathfrak{su}(2)$ the Cartan element is usually taken to be $\hat L_z$, with positive and negative eigenvalues (the weights) ranging from $-\ell$ to $\ell$ for irrep of dimension $2\ell+1$. The Casimir is $L^2$, with positive only eigenvalues, and in fact a single eigenvalue for all states in the irrep, irrespective of their weights and whether they are eigenstates of $\hat L_z$ or not. $\endgroup$
    – ZeroTheHero
    Commented Jun 9, 2021 at 11:42

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