= A Hermitian nilpotent BRST charge requires an indefinite auxiliary space
{title2=$Q^\dagger=Q,\quad Q^2=0$}
On a positive-definite Hilbert space, these two identities imply $\|Q\psi\|^2=\langle\psi,Q^2\psi\rangle=0$, 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.
Back to article page
