Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there written anywhere a full definition of BRST cohomology? All I have found so far is BRST cohomology in _______. As far as I can see, BRST cohomology is the cohomology of a complex in which the differential has at least a piece that "looks like" the Chevalley Eilenberg differential.

That seems to include all known examples, but is hardly a precise definition.

share|improve this question
    
Hi Jim. Alas, there was a time when in the Physics literature every differential complex was called a BRST complex :( Which BRST cohomology are you interested in? The original of Becchi, Rouet, Stora and Tyutin? –  José Figueroa-O'Farrill May 7 '13 at 21:23
3  
What exactly is _______? :-) –  Mariano Suárez-Alvarez Jul 22 '13 at 1:41
add comment

1 Answer

It is difficult to give a precise definition, because there are many cohomology theories which go by the name of BRST. It might be helpful to give a couple of examples, not necessarily in chronological order.

In classical (i.e., non-quantum) physics, there is a notion of BRST cohomology which encodes symplectic reduction à la Marsden-Weinstein. This was the subject of my answer to an earlier question. It is the cohomology of a double complex and one of the differentials is the Chevalley-Eilenberg differential. However, it does admit generalisations (e.g., to coisotropic reduction) where there is no Chevalley-Eilenberg differential, while still being referred to as BRST cohomology.

The original BRST cohomology arose in quantum field theory, where it arose as an "invariance" of the gauge-fixed Fadde'ev-Popov action for a gauge theory and plays an important role in proving the renormalisability of four-dimensional gauge theories. The BRST differential again has a part which is the Chevalley-Eilenberg differential of the Lie algebra of the gauge group. Again, there are generalisations (to theories with "open algebras") where there is still a BRST cohomology, but now there is no longer any Chevalley-Eilenberg differential.

In the context of two-dimensional conformal field theories (e.g., string theory), the BRST cohomology can be identified with a certain (relative) semi-infinite cohomology in the sense of Feigin.

In all cases precise definitions can be given, but they are different.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.