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 $c$.