How can we systematically derive Casimir operators for graded Lie algebras?

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SUMMARY

The discussion focuses on systematically deriving Casimir operators for graded Lie algebras, specifically through the use of the SUSY Casimir operator, which is modified from the Pauli-Lubanski vector. The method involves starting with the commutation relations [T_a, T_b] = f_ab^c T_c, forming the adjoint representation ad_T_a = J_a, and using the Killing form g_ab = Tr(J_aJ_b) to find the dual basis elements. The Casimir invariant is then expressed as C = J_a J^a = g^ab J_a J_b. The discussion also notes that modifications are necessary for superalgebras, such as replacing the trace with the supertrace.

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The discussion is beneficial for theoretical physicists, mathematicians specializing in algebraic structures, and researchers working on quantum field theories involving graded Lie algebras and superalgebras.

petergreat
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I followed a book on SUSY where the SUSY Casimir operator, modified from the Pauli-Lubanski vector, is stated and then proven to commute with all SUSY generators. However, I'm wondering whether there's a systematic approach of deriving the Casimir operator, other than trial-and-error? Is there a general method of deriving Casimir operators for an arbitrary graded Lie algebra?
 
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Generally, the idea is to start with [T_a, T_b] = f_ab^c T_c, form adjoint rep ad_T_a = J_a, then form the Killing form g_ab = Tr(J_aJ_b), find its inverse g^ab such that g^ab g_bc = del^a_c, form the dual basis elements J^a = g^ab J_b, and then finally, the Casimir invariant is given by C = J_a J^a = g^ab J_a J_b. Looking at the definition of g_ab, we see that this is a symmetric metric, as one would hope for. Other representations of C can be formed as well; it'd simply be C = g^ab T_a T_b.

More details here: http://en.wikipedia.org/wiki/Casimir_invariant

Edit: What I said applies to algebras. I assume you would need to modify this some for superalgebras; Tr -> Str and so on. I don't think anything out of the ordinary happens beyond that, but I'm not positive. I'm less familiar with superalgebras.
 

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