Suppose is a solution of the linearized SUGRA and the corresponding ghost number 2 vertex. We have:
Now let us deform the worldsheet action with the integrated vertex corresponding to another linearized solution . This will deform also the BRST operator:
With the new BRST operator:
However it is true that . The question is whether is BRST exact or not?
This was a bit sloppy, the actual question is, if we deform the worldsheet action
with nilpotent parameters and , if exists the operator such that:
is a good sigma-model. The obstacle is some ghost number three cohomology class linear in and symmetric in .
For more details see my paper on beta-deformations
If the ghost number three cohomology were zero, we would be sure that there is no obstruction. But it is actually nonzero. How do we know that there is no obstruction?