BRST and Lie Algebra Cohomology

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 \(c\).