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I found a paper by A. Belopolsky : "Picture changing operators in supergeometry and superstring theory" where he says there exists a possible physical state in one picture that doesn't exist in another picture. What A.B. does is essentially defining the picture formalism on the BRST complex and the de-Rham complex saying that the main idea would be that there is an even and an odd ghost number, picture changing involving the change in the odd ghost number. What strikes me here is the splitting of the complex on the odd number. I am not sure I get the full idea about what this means? Is it a sort of split-exact complex? And then why are the isomorphisms on the cohomology such a big deal? Shouldn't that be natural?

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    $\begingroup$ I added a hyperlink to the paper. $\endgroup$ Jan 20, 2014 at 17:31
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    $\begingroup$ Ok, more detailed: a picture in string theory is an independent set of (super)ghosts that are used to obtain a specific ground state. A state can be represented differently according to the way in which the ground state was constructed. This means that each physical state can be regarded from various different points of view, each forming a "picture". A specific number of superghost operators will annihilate a ground state but the "picture number" gives precisely the number of applications of ghosts needed until one annihilates the ground state. $\endgroup$
    – user33923
    Jan 20, 2014 at 19:48
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    $\begingroup$ Now, there is a algebra forming here and the operators are related in a chain complex via external differential operators (to be assimilated to the BRST operators or the de-Rham differentials, depending on the case). An exact split sequence is (for the abelian case) defined like this: en.wikipedia.org/wiki/Splitting_lemma Is the trick in this paper just the fact that A.B. generalizes this for supergeometries? Why is that such a big deal? How does the abelian condition restrict the validity of this idea? $\endgroup$
    – user33923
    Jan 20, 2014 at 19:49
  • $\begingroup$ wow, that many answers! $\endgroup$
    – user33923
    Jan 29, 2014 at 13:04

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