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According to (3.6) a Dirac state $\Psi(q)$ is related to a BRST invariant state $\Upsilon$ by $$ \Psi(q)=\int d\mu ~d\Pi~ d\bar\Pi~\Upsilon(q,-i\mu,\Pi,\bar\Pi). … \tag{*}$$ But how does it follow that $\Upsilon$ is actually BRST invariant? …

Consider the BRST operator $\Omega$ (which has ghost number $+1$) and the gauge fermion operator $\rho$ which has ghost number $-1$. Given an exact state $|\Phi\rangle$ (i.e. …

The BRST operator $\Omega$ has ghost number $+1$ and has an expansion in powers of $c$: \begin{equation} \Omega=\sum_{n\geq 0}\left(c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\right)_W~M^{ … W.r.t. the positive-definite inner product, however, (supposing it exists), the adjoint of $\Omega$ is known as the co-BRST operator $^*\Omega$ (which I suppose has ghost number $-1$). …

The BRST operator $\Omega$ can be expanded in powers of the ghost fields $c^{\alpha}$ and their conjugates $b_{\alpha}$ (which satisfy $\{c^\alpha,b_\beta\}=\delta^{\alpha}_{\beta}$): $$ \Omega=c^{\alpha …

In the classical Hamiltonian BRST formalism, there arise structure functions $\Omega^{\beta_1...\beta_n}_{\alpha_1... … These are such that the BRST charge $$ \Omega=\sum_{n\geq0}c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\Omega^{\beta_1...\beta_n}_{\alpha_1... …