# Lagrangian (classical) BRST cohomology groups

I'm trying to understand BRST complex in its Lagrangian incarnation i.e. in the form mostly closed to original Faddeev-Popov formulation. It looks like the most important part of that construction (proof of vanishing of higher cohomology groups) is very hard to find in the literature, at least I was not able to do so. Let me formulate couple of questions on BRST, but in the form of exercises on Lie algebra cohomology.

Let $X$ be a smooth affine variety, and $g$ is a (reductive?) Lie algebra acting on $X$, I think we assume $g$ to be at least unimodular, otherwise BRST construction won't work, and also assume that map $g \to T_X$ is injective. In physics language this is closed and irreducible action of a Lie algebra of a gauge group of the space of fields $X$. Structure sheaf $\mathcal{O}_X$ is a module over $g$, and I could form Chevalley-Eilenberg complex with coefficients in this module $$C=\wedge g^* \otimes \mathcal{O}_X.$$

The ultimate goal if BRST construction is to provide "free model" of algebra of invarinats $\mathcal{O}_X^g$, it is nor clear what is "free model", but I think BRST construction is just Tate's procedure of killing cycles for Chevalley-Eilenberg complex above (Tate's construction works for any dg algebra, and $C$ is a dg algebra).

My first question is what exactly are cohomology of the complex $C$? In other words before killing cohomology I'd like to understand what exactly have to be killed. For me it looks like a classical question on Lie algebra cohomology and, perhaps, it was discussed in the literature 60 years ago.

It is not necessary to calculate these cohomology groups and then follow Tate's approach to construct complete BSRT complex (complete means I added anti-ghosts and lagrange multipliers to $C$ and modified the differential), but even if I start with BRST complex $$C_{BRST}=(\mathcal{O}_X \otimes \wedge (g \oplus g^*) \otimes S(g), d_{BRST}=d_{CE}+d_1),$$ where I could find a proof that all higher cohomology vanishes?

• What makes you think that higher BRST cohomology groups vanish? In general, they do not and some of them (in degree 1 and 2) may be obstructions to symmetry reduction after quantization (anomalies). Also, your description of the BRST construction seems a bit off. What is your main reference? Briefly, the CE differential is the same the BRST differential. A Koszul-Tate resolution is applied instead to the ideal generated by the Lagrangian equations of motion (which appear nowhere in your question) in the more general Batalin-Vilkoviski construction. – Igor Khavkine Nov 20 '13 at 12:57
• For example, here arxiv.org/pdf/1008.1411v1.pdf on p. 35, Reshetikhin claims that higher cohomology groups vanish. Perhaps, quantum BRST cohomology groups do not vanish. Reshetikhin's notes could be considered as my main reference if you want some. BRST differential is clearly different from CE differential. I know how to use Koszul-Tate for shell and I know BV formalism, but my question is about different things. Why my description is a bit off? – Sasha Pavlov Nov 20 '13 at 13:09
• By 'seems off', I only meant the remarks at the end of my first comment. Well, Reshetikhin makes that comment after assuming that the action of $G$ on $X$ is faithful. I know that, in general, the computation of higher BRST cohomology groups in field theory is rather technical and I don't know whether their vanishing follows from a simple hypothesis like faithfulness. You can find examples of such computations in the review by Barnich, Brandt and Henneaux (arxiv.org/abs/hep-th/0002245). – Igor Khavkine Nov 20 '13 at 14:55
• Incidentally, you can see in Reshetikhin's notes that $Q$ differs from the CE differential only by the term $\lambda_a \partial/\partial\bar{c}_a$. Since the variables $(\lambda,\bar{c})$ do not appear elsewhere in $Q$, they form a contractible pair. That is, if all dependence on them is eliminated and that last term is dropped from $Q$, the cohomology is unchanged. But the remaining part of $Q$ is just the CE differential. – Igor Khavkine Nov 20 '13 at 14:57
• I see what you mean, BRST complex is a tensor product of $C$ and a contractible complex, but higher cohomology of $C$ is almost never zero, even for one dimensional lie algebra fist cohomology are coinvariants. It means that BRST complex is far from being a free model in this case. What is the meaning of it? Just a fancy way to fix gauge via fermion, thats all? – Sasha Pavlov Nov 20 '13 at 15:45