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 \(c\).

BRST

Finally, I want to discuss gauge-invariant of the gauge-fixed theory. (!?) We saw in the previous posts that if we have a gauge theory with connection \(A\) and matter fields \(\psi\), in order to derive sensible Feynman rules we have to introduce a gauge-fixing function \(G\) as well as Fermionic fields \(c, \bar{c}\), the ghosts. (Note: last time I used \(\eta, \bar{\eta}\) for the ghosts but I want to match the more standard notation, so I've switched to \(c, \bar{c}\)).

Usually it is convenient to use the gauge-fixing function \(G(A) = \partial^\mu A_\mu\). Under an infinitesimal gauge-transformation \(\lambda\), \(A\) transforms as
\[ A \mapsto -\nabla \lambda, \]
so \(G(A)\) transforms as
\[ G(A) \mapsto G(A) - \partial^\mu \nabla_\mu \lambda. \]
Hence the term in the Lagrangian involving the ghosts is
\[ -\bar{c}^a \partial^\mu \nabla_\mu^{ab} c^b, \]
and our gauge-fixed Lagrangian is
\[ \mathcal{L} = -\frac{1}{4} |F|^2 + \bar{\psi}(iD\!\!\!/-m)\psi + -\frac{|\partial^\mu A_\mu|^2}{2\xi}
 - \bar{c}^a \nabla_\mu^{ab} c^b \]
Introducing an auxiliary filed \(B^a\), this is of course equivalent to
\[ \mathcal{L} = -\frac{1}{4} |F|^2 + \bar{\psi}(iD\!\!\!/-m)\psi + \frac{\xi}{2} B^a B_a
 + B^a \partial^\mu A_{\mu a} - \bar{c}^a \nabla_\mu^{ab} c^b. \]
Now, there are two questions one might ask: (1) how can we tell that this is a gauge-theory? i.e., what remains of the original gauge symmetry? and (2) does the resulting theory depend in any way on the choice of gauge-fixing function?

The answer to both of these questions is BRST symmetry. The field \(c\) is Lie-algebra valued, so we could think of it as being an infinitesimal gauge transformation. Rather, for \(\epsilon\) a constant odd variable, \(\epsilon c\) is even and an honest infinitesimal gauge transformation. Under this transformation, we have
\[ \delta_\epsilon A = -\nabla (\epsilon c) = -\epsilon \nabla c. \]
Then we define a graded derivation \(\delta\) by
\[ \delta A = - \nabla c. \]
We have a grading by ghost number, where \(\mathrm{gh}(A) = 0, \mathrm{gh}(\psi) = 0, \mathrm{gh}(c) = 1, \mathrm{gh}(\bar{c}) = -1\). We would like to extend \(\delta\) to a derivation of degree \(+1\) that squares to 0. First, we should figure out what \(\delta c\) is. We compute:
\begin{align}
0 &= \delta^2 A \\
&= \delta(-\nabla c) \\
&= -\partial \delta c - (\delta A) c - A (\delta c) + (\delta c) A - c (\delta A) \\
&= -\partial \delta c + (\nabla c) c + c (\nabla c) - [A, \delta c] \\
&= -\nabla(\delta c) + \nabla(c^2).
\end{align}
From this, we see that \(\nabla(\delta c) = \nabla(c^2)\), so we can set
\[ \delta c = c^2 = \frac{1}{2}[c, c]. \]
Then \(\delta^2 c = 0\) is just the Jacobi identity for the group's Lie algebra! Finally, we would like to extend \(\delta\) to act on \(\psi\), \(B\), and \(\bar{c}\) so that \(\delta \mathcal{L} = 0\), and \(\delta^2 = 0\). Since the action on \(A\) is by infinitesimal gauge transformation, this leaves the curvature term of \(\mathcal{L}\) invariant. Similarly, the \(\psi\) term is invariant if we simply take
\[ \delta \psi = c \cdot \psi \]
where dot denotes the infinitesimal gauge transformation. Using the known rules for \(\delta\), we find that
\[ \delta \mathcal{L} = \frac{\xi}{2} \left(\delta B B + B \delta B \right) + \delta B \cdot \partial^\mu A_\mu
 - B \cdot \partial^\mu \nabla_\mu c - \delta\bar{c} \cdot \partial^\mu \nabla_\mu c \]
By comparing coefficients, we find (together with what we've already computed)
\begin{align}
\delta A &= -\nabla c \\
\delta \psi &= c \cdot \psi \\
\delta c &= \frac{1}{2}[c,c] \\
\delta \bar{c} &= B \\
\delta B &= 0.
\end{align}
This is the BRST differential. Now, suppose that \(\mathcal{O}(A, \psi)\) is a local operator involving the physical fields \(A\) and \(psi\). Then by construction,\(delta O\) is the change of \(O\) under an infinitesimal gauge transformation. Hence, we find

An operator \(\mathcal{O}\) is gauge invariant \(\iff \delta\mathcal{O} = 0\).

Now, suppose the functional measure \(\mathcal{D}A \mathcal{D}\psi \mathcal{D}B \mathcal{D}c \mathcal{D}\bar{c}\) is gauge-invariant, i.e. is BRST closed. (This assumption is equivalent to the absence of anomalies, but we'll completely ignore this in today's post.) Then we have
\[ \langle \delta \mathcal{O} \rangle = 0 \]
for any local observable \(\mathcal{O}\). This just follows from integration by parts (this is where we have to assume the measure is \(\delta\)-closed). Now, why is this significant? First, this tells us that the space of physical observables is
\[ H^0(C^\ast_{\mathrm{BRST}}, \delta) \]
where \(C^\ast_{\mathrm{BRST}}\) is the cochain complex of local observables, graded by ghost number.

Now, the real power of the BRST formalism is the following. We find that the gauge-fixed Lagrangian can be written as
\[ \mathcal{L}_{gf} = \mathcal{L}_0 +\delta \left(\bar{c} \frac{B}{2} + \bar{c}\Lambda\right) \]
where \( \Lambda = \partial^\mu \nabla_\mu A \) is our gauge-fixing function, and \(\mathcal{L}_0\) is the original Lagrangian without gauge-fixing. Now the point is, any two choices of gauge fixing differ by terms which are BRST exact, and hence give the same expectation values on the physical observables \(H^0\). So we have restored gauge invariance, while obtaining a gauge-fixed perturbation theory!

Fadeev-Popov Ghosts, continued

Last time I sketched how we can represent an integral over a submanifold \(M \subset \mathbb{R}^n\) by an integral of the form
\[ \int_{\mathbb{R}^n} f(x) \delta(G(x)) \exp\left(\bar{\eta}G(x+\eta) \right) d\eta d\bar{\eta} dx. \]
Here, \(\eta, \bar{\eta}\) are Fermionic variables called Fadeev-Popov ghosts, which are introduced to cancel an unwanted determinant factor. The function \(G(x)\) singles out the submanifold \(M\) as \(M = G^{-1}(0)\).

Now suppose that we start with a vector (or affine) space \(V\), which is acted on by a group \(H\). We would like to undstand integrals over the quotient \(V / H\) in terms of integrals over \(V\). Suppose there is some function \(G(x)\) on \(V\) satisfying the following property:

For each level \(w\) of \(G\), the subspace \(M_w := G^{-1}(w)\) intersects the orbits transversely, and furthermore every \(H\)-orbit intersects \(M_w\) exactly once. (*)

We call such a function a gauge-fixing function, and a level \(w\) a gauge-fixing. By assumption, we have \(V/H \cong M_w\) for any \(w\). Hence using the integral we derived last time, we can integrate over \(M_w\) for any particular choice of \(w\), and this ought to be the same as integrating over \(V/H\). The problem, however, is that in the QFT setting it's not clear what the Feynman rules should be for such a path integral. The final trick is that since the answer should be independent of \(w\), and by integrating over all possible \(w\) we obtain a Lagrangian from which we can derive sensible Feynman rules.

We have some integral

\[Z = \int \delta(G(x) - w) \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta\right\} dx d\eta d\bar{\eta} \]
which is independent of \(w\). So we add a Gaussian weight an integrate over \(w\):
\begin{align}
Z' &= \int \delta(G(x) -w) \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta - \frac{1}{2\xi} |w|^2\right\}
 dx dw d\eta d\bar{\eta}\\
&= \int \exp\left\{\frac{i}{\hbar}S(x) + \bar{\eta}dG \eta - \frac{1}{2\xi} |G(x)|^2 \right\} dx d\eta d\bar{\eta}.
\end{align}
Here, \(\xi\) is an arbitrary real positive constant, and we denote the new integral by \(Z'\) to indicate that it differs from the old path integral \(Z\) by (at most) an overall constant. Now the important thing is that the new action appear in the integrand of \(Z'\) is gauge-fixed and hence there is no problem whatsoever in deriving sensible, meaningful Feynman rules. The gauge-fixing term \(|G(x)|^2\) serves to make the action non-degenerate, so that propagators are well-defined, while the term involving the Fermions \(\eta, \bar{\eta}\) generates new Feynman rules that "cancel" the superfluous degrees of freedom due to gauge redundancy.

The question remains, what if we choose some other gauge-fixing function? i.e., what happens if we perturb \(G(x)\) to some new function satisfying property (*)? We'll answer this using the BRST formalism.

Fadeev-Popov Ghosts

Today I want to review the Fadeev-Popov procedure, with a view toward BRST and eventually BV.

Gauge-Invariance and Gauge-Fixing

First we'll review the Fedeev-Popov method, to motivate the introduction of ghosts. Suppose we have a gauge theory involving a \(G\)-connection \(A\) and some field \(\phi\) charged under \(G\). Under a gauge transformation \(g(x)\), \(\phi\) transforms as
\[ \phi \mapsto g \cdot \phi. \]
We would like that the covariant derivative transforms in the same way, i.e.
\[ \nabla \phi \mapsto g \nabla \phi. \]
In terms of the connection 1-form \(A\), the covariant derivative is
\[ \nabla = d + A. \]
Let \(\nabla'\) denote the gauge-transformed covariant derivative, and \(\phi'\) the gauge-transformed field. Then we want
\[ \nabla' \phi' = g \nabla \phi. \]
We compute
\begin{align}
\nabla' \phi' &= (d + A')(g \phi) \\
&= dg \phi + g d\phi + A'(g\phi) \\
&= gg^{-1}dg \phi + gd\phi + g g^{-1} A' g\phi \\
&= g(d\phi + g^{-1} A' g \phi + g^{-1} dg \phi \\
&= g(d\phi + A\phi)
\end{align}
Comparing terms, we see that
\[ A = g^{-1} A' g + g^{-1} dg, \]
so upon re-arranging we have
\[ A' = g A g^{-1} - dg g^{-1} \]

This causes a problem: at critical points of the action, the Hessian of the action is degenerate in directions tangent to the gauge orbits. This means that the propagator is not well-defined, and there is no obvious way to derive the Feynman rules for perturbation theory. The solution is to take the quotient by gauge-transformations. To do this, we pick some gauge-fixing function \(G(A)\) which ought to be transverse to the orbits. Then we can restrict to the space \(G(A) = 0\), on which the Hessian of the action is non-degenerate, leading to a well-defined propagator. Formally, the path integral is
\[ Z = \int_{\{G(A) = 0\}} \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi  \]
Formally, this suggests that the path integral should be something like
\[ Z = \int \delta(G(A)) \exp{\frac{i}{\hbar} S[A, \phi]} \mathcal{D}A \mathcal{D}\phi, \]
but this is not quite right! To understand the source of the problem, we'll first study the finite-dimensional case and then use this to solve the problem in infinite-dimensions.

The Fadeev-Popov Determinant

Suppose we are on \(\mathbb{R}^n\), and we would like to integrate a function \(f(x)\) over a submanifold \(M\) defined by \(M = G^{-1}(0)\) for some smooth function \(G: \mathbb{R}^n \to \mathbb{R}^k\). Naively, we might expect that the answer is
\[ \int_M f(x) \stackrel{?}{=} \int f(x) \delta(G(x)) dx. \]
To see why this is not correct, write the delta function as
\[ \delta(G(x)) = \frac{1}{(2\pi)^k}\int e^{ip\cdot G(x)} d^k p. \]
We can regularize this by taking the limit as \(\epsilon \to 0\) of
\[ \frac{1}{(2\pi)^k}\int \exp\left\{ip\cdot G(x) -\frac{\epsilon}{2} |p|^2\right\} d^k p \]
This integral is Gaussian, so we obtain explicitly
\[ \left(\frac{2\pi}{\epsilon} \right)^{\frac{k}{2}} \exp\left\{-\frac{1}{2\epsilon} |G(x)|^2 \right\}. \]
So our original guess becomes
\[ \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int f(x) \exp\left\{ -\frac{1}{2\epsilon} |G(x)|^2 \right\} dx .\]
As \(\epsilon \to 0\), this integral localizes on the locus \(\{G(x) = 0\}\), as desired, but does not give the right answer! To see this, let \(u\) be a coordinate on \(M = G^{-1}(0)\) and \(v\) coordinates normal to \(M\). Then we have
\[ G(x) = G(u,v) = v^T H(u) v + o(|v|^3) \]
where \(H(x)\) is the Hessian of \(|G|^2\) at the point \(x = (u, 0)\). So the integral becomes (as \(\epsilon \to 0\))
\begin{align}
I_\epsilon &= \left(\frac{1}{2\pi \epsilon}\right)^\frac{k}{2} \int\int
 f(u, v) \exp\left\{ -\frac{1}{2\epsilon} v^T H(u) v \right\} du dv \\
&= \int_M \frac{f(u)}{\sqrt{\det H(u)}} du.
\end{align}
This is not correct. We have to account for the determinant of the Hessian. Now, the Hessian is given by
\begin{align} H_{ij} &= \frac{1}{2} \frac{\partial^2 |G|^2}{\partial v^i \partial v^j} \\
&= \frac{\partial}{\partial v^i} \left( G^a \partial_j G^a \right) \\
&=  \left(\partial_i G^a \partial_j G^a + G^a \partial_{ij} G^a \right) \\
&=  \partial_i G^a \partial_j G^a
\end{align}
where we have used the fact that \(G = 0\) on \(x = (u, 0)\). Hence we see that
\[ \det H = (\det A)^2 \]
where \(A\) is the \(k \times k\) matrix with entries \(\partial_i G^a\). Hence
\[ \sqrt{\det H} = \det A. \]
Now there is a straightforward way to eliminate the determinant. We introduce Fermionic coordinates \(\eta^i, \theta^i\), \(i = 1, \ldots, k\). Then by Berezin integration, we have
\[ \int e^{\eta^i G^i(0, \theta^j)} d\theta d\eta = \det A. \]
So in the end, we find
\[ \int_M f(x) d\mu = \int_{\mathbb{R}^n}
f(x) \delta(G(x)) \exp\left(\eta \cdot G(x+ \theta) \right)
 dx d\theta d\eta. \]

The Weyl and Wigner Transforms

Today I'd like to try to understand better how deformation quantization is related to the usual canonical quantization, and especially how the latter might be used to deduce the former, i.e., given an honest quantization (in the sense of operators), how might be reproduce the formula for the Moyal star product?

We'll fix our symplectic manifold once and for all to be \(\mathbb{R}^2\) with its standard symplectic structure, with Darboux coordinates \(x\) and \(p\). Let \(\mathcal{A}\) be the algebra of observables on \(\mathbb{R}^2\). For technical reasons, we'll restrict to those smooth functions that are polynomially bounded in the momentum coordinate (but of course the star product makes sense in general). Let \(\mathcal{D}\) be the algebra of pseudodifferential operators on \(\mathbb{R}\). We want to define a quantization map

\[ \Psi: \mathcal{A} \to \mathcal{D} \]

such that

\[ \Psi(x) = x \in \mathcal{D} \]

\[ \Psi(p) = -i\hbar \partial \]
Out of thin air, let us define
\[ \langle q| \Psi(f) |q' \rangle = \int e^{ik(q-q')} f(\frac{q+q'}{2}, k) dk \]
This is the Weyl transform. Its inverse is the Wigner transform, given by
\[ \Phi(A, q, k) = \int e^{-ikq'} \left\langle q+\frac{q'}{2} \right| A \left| q - \frac{q'}{2} \right\rangle dq' \]
Note: I am (intentionally) ignoring all factors of \(2\pi\) involved. It's not hard to work out what they are, but annoying to keep track of them in calculations, so I won't.

Theorem For suitably well-behaved \(f\), we have \( \Phi(\Psi(f)) = f\).

Proof Using the "ignore \(2\pi\)" conventions, we have the formal identities
\[ \int e^{ikx} dx = \delta(k), \ \ \int e^{ikx} dk = \delta(x). \]
The theorem is a formal result of these:
\begin{align} \Phi(\Psi(f))(q, k) &= \int e^{-ikq'} \left\langle q + \frac{q'}{2} \right| \Psi(f) \left| q - \frac{q'}{2} \right\rangle \\\
&= \int e^{-ikq'} e^{ik'q'} f(q, k) dk' dq' \\\
&= f(q,k).
\end{align}

One may easily check that \(\Psi(x) = x\) and \(Psi(k) = -i\partial\), so this certainly gives a quantization. But why is it particularly natural? To see this, let \(Q\) be the operator of multiplication by \(x\), and let \(P\) be the operator \(-i\partial\). We'd like to take \(f(q,p)\) and replace it by \(f(Q, P)\), but we can't literally substitute like this due to order ambiguity. However, we could work formally as follows:
\begin{align}
f(Q, P) &= \int \delta(Q-q) \delta(P - p) f(q,p) dq dp \\\
&= \int e^{ik(Q-q) + iq'(P-p)} f(q,p) dq dq' dp dk.
\end{align}
In this last expression, there is no order ambiguity in the argument of the exponential (since it is a sum and not a product), and furthermore the expression itself make sense since it is the exponential of a skew-adjoint operator. So let's check that this agrees with the Weyl transform. Using a special case of the Baker-Campbell-Hausdorff formula for the Heisenberg algebra, we have
\[ e^{ik(Q-q) + iq'(P-p)} = e^{ik(Q-q)} e^{iq'(P-p)} e^{-ikq'/2} \]
Let us compute the matrix element:
\begin{align}
\langle q_1 | P | q_2 \rangle &= \int \langle q_1 | p_1 \rangle
\langle p_1 | P | p_2 \rangle \langle p_2 | q_2 \rangle dp_1 dp_2 \\\
&= \int e^{iq_1p_1 - iq_2 p_2} p_2 \delta(p_2 - p_1) dp_1 dp_2 \\\
&= \int e^{i p(q_1-q_2)} p dp.
\end{align}
Hence we find that the matrix element for the exponential is
\begin{align} \langle q_1 |e^{ik(Q-q) + iq'(P-p)} | q_2 \rangle
&= e^{-ikq'/2 + ik(q_1-q)} \langle q_1 | e^{iq'(P-p)} | q_2 \rangle \\\
&=  \int e^{-ikq'/2 + ik(q_1-q) -iq'p} e^{iq'p'' + ip''(q_1-q_2)} dp'' \\\
&= \delta(q' + q_1 - q_2)  e^{-ikq'/2 + ik(q_1-q) -iq'p}
\end{align}
Plugging this back into the expression for \(f(Q, P)\) we find
\begin{align}
 \langle q_1 | f(Q. P) | q_2 \rangle &= \int \delta(q' + q_1 - q_2)  e^{-ikq'/2 + ik(q_1-q) -iq'p}
f(q,p) dq dq' dp dk \\\
&= \int  e^{ ik(q_1/2 +q_2/2-q) -ip(q_1-q_2)} f(q,p) dq dp dk \\\
&= \int e^{ip(q_1-q_2)} f(\frac{q_1+q_2}{2}, p) dp,
\end{align}
which is the original expression we gave for the Weyl transform.