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 $Q_B$.

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    $\begingroup$ 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. $\endgroup$ Nov 13 '21 at 16:31
  • $\begingroup$ Are you asking how we know that the quotient $\ker Q_B/\mathrm{im}\ Q_B$ is a well-defined Hilbert space? $\endgroup$ Nov 14 '21 at 3:32
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    $\begingroup$ @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. $\endgroup$
    – aitfel
    Nov 14 '21 at 15:22

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