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The classical Lie-action BRST differential squares to zero

Statement

For every classical Lie-action BRST input of Classical Lie-action BRST cochain complex and every n≥0, dBRSTn+1dBRSTn=0. Consequently, the quotient HBRSTn(g,A):=ker⁡(dBRSTn)/im⁡(dBRSTn−1) is a well-defined vector space, with CBRST−1=0.

Facts & Assumptions

Given: A characteristic-zero field k, a Lie algebra g, a unital commutative k-algebra A, and a Lie algebra representation of g on A by derivations.

[F1]

Classical Lie-action BRST cochain complex: the BRST cochains are Hom⁡k(Λng,A) and the operator is the zero-based two-sum Chevalley–Eilenberg formula for the representation ρ.

[F2]

The Chevalley–Eilenberg differential squares to zero: for a Lie algebra over a field and a representation on a module, the displayed Chevalley–Eilenberg differential satisfies dn+1dn=0 for every n.

Proof

1.1givenF1F2

The assumptions in [F1] make A a g-module because ρ is a Lie algebra homomorphism, so [ρ(x),ρ(y)]=ρ([x,y]). The cochain spaces and both signed sums in dBRST are term-by-term those in [F2].

2.1givenF1F2step 1.1

Therefore [F2] applies to these exact inputs and gives dBRSTn+1dBRSTn=0 for every n≥0.

3.1givenstep 2.1algebra∎

If z∈im⁡(dBRSTn−1), write z=dBRSTn−1w. Step 2.1 gives dBRSTnz=dBRSTndBRSTn−1w=0, so the image is contained in the kernel and the displayed quotient is well-defined.

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