# What does cohomology of $Q_B$ mean in BRST quantization in Polchinski?

While proving no-ghost theorem ($$4.4$$ Polchinski) the term cohomology of $$Q_B$$ is used quite a lot of time. From what I understand this has to be a set since "cohomology of $$Q_B$$" is isomorphic to Hilbert space of OCQ or light cone quantization. Above $$(4.2.18)$$ following statement is made about the meaning of cohomology:

There is a natural construction for a nilpotent operator, and is known as cohomology of $$Q_B$$

By above statement isomorphism to Hilbert space loses its meaning. After reading Wikipedia I understand that cohomology is a collection of quotient sets and to define them you need chains (sequence of maps of sets satisfying certain conditions). I can't fit these ideas together to give meaning to cohomology of $$Q_B$$.

• Are you familiar with De Rham cohomology? If so, the cohomology of $Q$ is formally the same as the cohomology of $\mathrm d$ -- things that are killed by, modulo things that are spit out by. Nov 13 '21 at 16:31
• Are you asking how we know that the quotient $\ker Q_B/\mathrm{im}\ Q_B$ is a well-defined Hilbert space? Nov 14 '21 at 3:32
• @NiharKarve no! not that. That's the part of no ghost theorem. I think user accidentalFourierTransform has answered my problem, I need to study a bit of deRham cohomology. Nov 14 '21 at 15:22