BRST and Lie algebra cohomology

This post has been migrated from my old blog, the math-physics learning seminar.

We saw in previous posts that gauge-fixing is intimately related to BRST cohomology. Today I want to explain the underlying mathematical formalism, as it is actually something very well-known: Lie algebra cohomology. Let (\mathfrak{g}) be a Lie algebra and (M) a (\mathfrak{g})-module. We will construct a cochain complex that computes the Lie algebra cohomology with values in (M), (H^i(\mathfrak{g}, M)). Out of thin air, we define

[ C^\ast(\mathfrak{g}, M) = M \otimes \wedge^\ast \mathfrak{g}^\ast. ]

The grading is just the grading induced by the grading on (\wedge^\ast \mathfrak{g}^\ast), which we identify with the BRST ghost number. Let (e_i) be a basis for (M) and (T_a) be a basis for (\mathfrak{g}), with canonical dual basis (S^a). The differential is defined on generators to be

[ d e_i = \rho(T_a) e_i \otimes S^a ]

[ d S^a = \frac{1}{2} f^a_{bc} S^b \wedge S^c ]

where (\rho: \mathfrak{g} \to \mathrm{End}(M)) is the representation and (f^a_{bc}) are the structure constants of the group. This differential is then extended to satisfy the graded Leibniz rule, and is easily verified to satisfy (d^2 = 0) (this is just the Jacobi identity). The Lie algebra cohomology is just the cohomology of this cochain complex. Essentially by definition, we see that

[ H^0(\mathfrak{g}, M) = {m \in M \ | \ \xi \cdot m = 0 \ \forall \ \xi \in \mathfrak{g} }, ]

i.e. (H^0(\cdot) = (\cdot)^\mathfrak{g}) is the invariants functor. In fact, this can be taken to be the defining property of Lie algebra cohomology:

Theorem (H^k(\mathfrak{g}, M) = R^k (M)^\mathfrak{g}).

Returning to field theory, we see (modulo some hard technicalities!) that, roughly, (\mathfrak{g}) is the Lie algebra of infinitesimal gauge transformations, and (M) is the algebra of functions on the space of all connections. The ghost and anti-ghost fields can then be seen to be the multiplication and contraction operators. To wit, we can take (c^a) to be the operator

[ c^a: f \mapsto S^a \wedge f ]

and take (\bar{c}^a) to be the operator

[ \bar{c}^a: f \mapsto \frac{\partial}{\partial S^a} f = T_a \lrcorner f.]

Then we have

[ [c^a, \bar{c}^b] = \delta^{ab} ]

so that (\bar{c}) is indeed the antifield of (c).