 A Hermitian nilpotent BRST charge requires an indefinite auxiliary space

On a positive-definite Hilbert space, these two identities imply $\Vert Q\psi {\Vert }^2=⟨\psi ,Q^2\psi ⟩=0$, hence $Q=0$ on its invariant domain. A nontrivial covariant BRST charge therefore acts before reduction on an indefinite-inner-product gauge/ghost space. Exact vectors are null and orthogonal to closed vectors, and BRST cohomology quotients them out. Positivity of the reduced physical space requires the additional gauge-theory structure; it does not follow from abstract nilpotence alone.