Is the space on which the BRST $Q$ operator acts a Hilbert space?

Question

When I looked at the BRST symmetry in Yang-Mills-Theories I was puzzled by the statement:

Suppose we go back to canonical quantisation with a Hilbert space $\mathcal{H}$. The BRST symmetry leads to an associated operator $Q$ acting on $\mathcal{H}$.

As far as I understood it, the problem we have with Yang-Mills theories is that we cannot find a Hilbert space (since we cannot have a positive-definite norm) and that the Hilbert space actually is the cohomology space defined by $\mathcal{H_{phys}} = Ker(Q)/Im(Q)$

So the question, is the starting space really a Hilbert space as stated in the quote above?

Answer 1

Indeed, the space the BRST operator initially acts on is not a "Hilbert space" in the strict meaning of the word since there are negative- and null-norm states in that space. Something like "pseudo-inner product space" would be more accurate.