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A nonrepresentation does not give a BRST differential

Statement refuted

The two-sum Chevalley–Eilenberg formula squares to zero for every linear map ρ:g→End⁡k(A), even when ρ is not a Lie algebra representation.

Facts & Assumptions

Given: The two-sum formula from Classical Lie-action BRST cochain complex, evaluated algebraically even though the proposed action below is not a representation.

[F1]

Classical Lie-action BRST cochain complex: on a 0-cochain f∈A, (df)(z)=ρ(z)f; on a 1-cochain g, (dg)(z0,z1)=ρ(z0)g(z1)−ρ(z1)g(z0)−g([z0,z1]).

Counterexample

Take k=C, the two-dimensional abelian Lie algebra g=Cu⊕Cv, and the commutative algebra A=C[x,y]. Let ρ(u)=∂x,ρ(v)=x∂y. Both maps are derivations, but their commutator is not the zero map.

1.1givenF1algebra

The maps ∂x and x∂y are derivations by the polynomial product rule, and [∂x,x∂y]f=∂x(x∂yf)−x∂y(∂xf)=∂yf. Since [u,v]=0 but ∂y≠0, ρ is not a Lie algebra homomorphism and is not an admissible input in Classical Lie-action BRST cochain complex.

2.1givenF1step 1.1algebra∎

For the 0-cochain f=y, [F1] gives (df)(u)=0 and (df)(v)=x. Applying [F1] again gives d2f(u,v)=ρ(u)(x)−ρ(v)(0)−0=∂xx=1. Thus the formula fails to square to zero when the representation hypothesis is dropped, refuting the statement.

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