## Ghostly Lie algebra cohomology

Let $\mathfrak{g}$ be our Lie algebra and $V_\rho$ a representation space with representation map $\rho : \mathfrak{g} \to \mathrm{End}(V_\rho)$. $V_\rho$ is, by the action through the representation, naturally a $\mathfrak{g}$-module (people missing the ring structure in $\mathfrak{g}$ - just embed it into the universal enveloping algebra). We define the associated Chevally-Eilenberg complex as the complex of $V_\rho$-valued differential forms on $\mathfrak{g}$:

$$ \dots \overset{\mathrm{d}}{\to} \Lambda^{p-1}\mathfrak{g}^* \otimes V_\rho \overset{\mathrm{d}}{\to} \Lambda^{p}\mathfrak{g}^* \otimes V_\rho\overset{\mathrm{d}}{\to} \Lambda^{p+1}\mathfrak{g}^* \otimes V_\rho\overset{\mathrm{d}}{\to}$$

whose cohomology we call the *Lie algebra cohomology* of $\mathfrak{g}$ with coefficients in $V_\rho$. Now, the algebraist is disturbed: There's an ugly differential in our complex, spoiling the fun! Let's build an operator expression for it:

Recall that, on $\Lambda^p \mathfrak{g}^*$, we have two natural operations:

*Contraction*, which is

$$\iota : \Lambda^p \mathfrak{g}^* \times \mathfrak{g} \to \Lambda^{p-1} \mathfrak{g}^*, (\omega,G) \mapsto \omega(G)$$

and the *wedge product*, which is

$$\wedge : \Lambda^p \mathfrak{g}^* \times \mathfrak{g}^* \to \Lambda^{p+1} \mathfrak{g}^*, (\omega,k)\mapsto \omega \wedge k$$

and these define two operators $\iota_G$ and $\wedge_k$ acting upon $p$-forms. Now, observe that (or check that, if you're bored) their concatenation

$$ \iota_G \wedge_k \omega = (\omega \wedge k)(G) + (-1)^{\mathrm{deg}(\omega)}\omega \wedge \beta(G)$$

is an odd derivation on $\Lambda^p \mathfrak{g}^*$.

Now, choose any canonically dual basis of $\mathfrak{g}$ resp. $\mathfrak{g}^*$, let's call them $T_a$ resp. $S^a$, and write

$$ \mathrm{d} = \wedge_{S^a}\rho(T_a) - \frac{1}{2}\wedge_{S^a}\wedge_{S^b}\iota_{[T_a,T_b]}$$

Using it on the basis elements of $\Lambda^{p}\mathfrak{g}^* \otimes V_\rho$, we can show by direct computation that this is indeed the *differential* from the Chevalley-Eilenberg complex, and thus an operator expression for the differential. Defining $c^a := \wedge_{S^a}$ as the *ghost* and $b_a := \iota_{T_a}$ as the *anti-ghost* yields that the Chevalley-Eilenberg differential is indeed the BRST operator

$$ Q = \mathrm{d} = c^a\rho(T_a) - \frac{1}{2}f^c_{ab} c^a c^b b_c $$

## What does $Q$ compute in physics?

Classically, we apply this approach to symplectic manifolds/phase spaces $\mathcal{M}$ that possess a (symplectomorphic) group action by a Lie group $G$, and we construct the equivariant moment map

$$ \mu : \mathcal{M} \to \mathfrak{g}^*$$

defined by being equivariant under the coadjoint action of $G$ on $\mathfrak{g}^*$ and fulfilling $\mathrm{d}(\mu(\dot{})(g)) = \omega(\rho(g),\dot{})$ with $\omega$ as the symplectic form. If the action of $G$ represents a gauge symmetry, we would like to obtain the *coisotropic reduction* $\tilde{\mathcal{M}} := \mathcal{M}/ G$ containing no redundancies. Define the submanifold $M_0 := \mu^{-1}(0)$ and observe that the poisson algebra of functions on $\tilde{\mathcal{M}}$ fulfills

$$ C^\infty(\tilde{\mathcal{M}}) = H^0(\mathfrak{g};C^\infty(M_0))$$

since the zeroth cohomology of a Lie algebra with coefficients in a module consists of precisely the elements of the module that are invariant under the group action *and* because the natural projection $\pi : \mathcal{M}_0 \to \tilde{\mathcal{M}}$ provides a pullback from functions on the reduction to $\mathcal{M}_0$. We don't want to rehash the derivation of the Koszul complex here, suffice it to say that $H^0(\mathfrak{g};C^\infty(M_0))$ can be computed by looking at the complex

$$ \dots \to \Lambda^2 \mathfrak{g} \otimes C^\infty(\mathcal{M}) \to \Lambda \mathfrak{g} \otimes C^\infty(\mathcal{M}) \to C^\infty(\mathcal{M}) \to 0$$

and computing $H^0 = C^\infty(\mathcal{M}_0)$ and $H^p = 0$ otherwise, leading to the projective resolution

$$ \dots \to \Lambda^2 \mathfrak{g} \otimes C^\infty(\mathcal{M}) \to \Lambda \mathfrak{g} \otimes C^\infty(\mathcal{M}) \to C^\infty(\mathcal{M}) \to C^\infty(\mathcal{M}_0) \to 0 $$

yielding, since the tensor product is left exact, a projective resolution for $\Lambda^p \mathfrak{g}^* \otimes C^\infty(\mathcal{M}_0)$

This yields a bicomplex $C^{p,q} := \Lambda^p \mathfrak{g}^* \otimes \Lambda^q \mathfrak{g} \otimes C^\infty(\mathcal{M})$, from which a usual graded complex $\mathcal{C}^p$ may be constructed by $\mathcal{C}^p := \bigoplus_{r + s = p}C^{r,s}$, which is the infamous BRST complex, and which can be written as $\mathcal{C}^p = \Lambda^p(\mathfrak{g} \oplus \mathfrak{g}^*) \otimes C^\infty(\mathcal{M})$

With some algebraic magic involving the *Poisson superalgebra* structure of this complex, one can retrace the steps for deriving an explicit from for the differential from the ghostly cohomology for Lie algebras, and obtain that, here,

$$ \mathrm{d} = \{Q,\dot{}\}$$

with $Q \in \mathcal{C}^1$ being the classical BRST operator, and this time the ghosts and antighosts are the images of the generators of $\mathfrak{g}$ and $\mathfrak{g}^*$ under the natural embedding of these into $\Lambda(\mathfrak{g} \oplus \mathfrak{g}^*$).

A lengthier, but still quick and very readable discussion of this can be found in Josê Figueroa-O'Farrill's lecture notes on *"BRST Cohomology"*.

This post imported from StackExchange Physics at 2014-09-24 20:50 (UTC), posted by SE-user ACuriousMind