# Question on BRST closed vectors which are also co-closed

I'm studying the BRST quantization formalism from this reference.

I have a question though, about page 44.

The author introduces a co-BRST operator on the extended Hilbert space (which also include ghosts), denoted by $^*\Omega$.

At the bottom of page 44, he says that any BRST closed vector $\Omega \psi = 0$ is determined uniquely by requiring it to be co-closed as well. I have trouble understanding this remark.

He proves it in the following way. Let $^* \Omega \psi = 0$; then $$^* \Omega (\psi + \Omega \chi) = 0 \quad \Leftrightarrow \ ^*\Omega \Omega \chi = 0 \qquad \Leftrightarrow \Omega \chi = 0.$$

I'm not sure how it follows from this that a BRST-closed vector $\Omega \psi = 0$ is determined uniquely by requiring it to be co-closed. Suppose I write an element of the cohomology as $\psi^{'} = \psi + \Omega \chi$. If this vector is BRST closed, then $\psi^{'} \in \ker \Omega$. If I require it to be co-closed as well, does this mean that I can just drop the $\Omega \chi$ part? The author also says:

Thus, if we regard the BRST transformations as gauge transformations on states in the extended Hilbert space generated by $\Omega$, then $^*\Omega$ represents a gauge-fixing operator determining a single particular state out of the complete BRST orbit. States which are both closed and co-closed are called (BRST) harmonic.

How can I see precisely that the co-BRST operator represents a gauge-fixing operator?

• Look up Hodge theory! – Ryan Thorngren Apr 22 '18 at 12:08

This assertion stems from the assumption that on the extended Hilbert space $\mathcal{H}_{ext}$ there exists a non-degenerate inner product, please see the last paragraph on page 43.

Due to this assumption:

$$^*\Omega\Omega \chi = 0 \implies (^*\Omega\Omega \chi, \chi) = (\Omega \chi, \Omega\chi) = 0 \implies \Omega\chi = 0$$

Thus if you choose a vector which is both closed and co-closed, you cannot add to it an exact vector and at the same preserve the closeness + co-closeness property, thus it is a unique choice among all the closed vectors describing the same physical state (i.e., differing by an exact vector).

The gauge fixing here is a gauge fixing of the BRST symmetry, i.e., the hypersurface in the extended Hilbert space defined by:

$$\mathcal{S} = \{\psi \in \mathcal{H}_{ext} \quad \mid \quad ^*\Omega \psi = 0 \}$$

meets every BRST orbit (i.e., $\psi + \Omega \chi$ ) of the same physical state ($\psi$) exactly once.