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Canonical Antibracket Generates the Declared BRST Field Variations

Statement

Let $B$ be the finite-dimensional algebra that is polynomial in even canonical coordinates and exterior in odd canonical coordinates $Φ^{A}, Φ_{A}^{*}$ . Give each homogeneous minimal coordinate $Φ^{A}$ parity $ϵ_{A} \in {0, 1}$ and ghost number $g_{A}$ , and give its antifield parity $ϵ_{A} + 1 (m o d 2)$ and ghost number $- 1 - g_{A}$ . Use the canonical antibracket

$(F, G) = \sum_{A} (\frac{\partial_{r} F}{\partial Φ^{A}} \frac{\partial_{l} G}{\partial Φ_{A}^{*}} - \frac{\partial_{r} F}{\partial Φ_{A}^{*}} \frac{\partial_{l} G}{\partial Φ^{A}}) .$

Suppose each $s_{0} Φ^{A}$ is independent of all antifields, has parity $ϵ_{A} + 1 (m o d 2)$ and ghost number $g_{A} + 1$ , and let $S_{0}$ be even of ghost number zero and independent of antifields. Define $S : = S_{0} + \sum_{A} (- 1)^{ϵ_{A}} Φ_{A}^{*} s_{0} Φ^{A}$ , with the displayed product order. Then $(S, Φ^{A}) = s_{0} Φ^{A}$ for each minimal coordinate. For a local-field version on a nonempty open $U \subseteq R^{n}$ , use fields $Φ^{A} (x)$ and antifields $Φ_{A}^{*} (x)$ with the same parity and ghost-number assignments, replace coordinate derivatives in the bracket by variational derivatives and the sum by $\int_{U} d^{n} x$ , and set $S_{U} : = S_{0, U} + \int_{U} d^{n} x \sum_{A} (- 1)^{ϵ_{A}} Φ_{A}^{*} (x) s_{0} Φ^{A} (x)$ . If this domain makes the variational derivatives and integrations by parts well-defined and its boundary term vanishes, then $(S_{U}, Φ^{A} (x)) = s_{0} Φ^{A} (x)$ pointwise.

Statement references

Gomis, París, and Samuel, §§4.2 and 4.6

Qiu and Zabzine, §§3.1–3.2 and 4.2

Facts & Assumptions

Given: The canonical BV coordinates, derivative convention, and antifield-linear action term stated above.

[A1]

$S_{0}$ and every $s_{0} Φ^{A}$ are independent of the antifield coordinates; $∣ s_{0} Φ^{A} ∣ = ϵ_{A} + 1 (m o d 2)$ .

[F1]

Classical BV Antibracket and Yang–Mills Master-Action Data: the canonical bracket is $(F, G) = \sum_{A} (\partial_{r} F / \partial Φ^{A} \partial_{l} G / \partial Φ_{A}^{*} - \partial_{r} F / \partial Φ_{A}^{*} \partial_{l} G / \partial Φ^{A})$ .

[F2]

Classical BV Antibracket and Yang–Mills Master-Action Data: the candidate action contains $(- 1)^{ϵ_{A}} Φ_{A}^{*} s_{0} Φ^{A}$ in that order, and $S_{0}$ is independent of antifields.

Proof

1.1F1algebra

For $G = Φ^{A}$ , $\partial_{l} G / \partial Φ_{B}^{*} = 0$ and $\partial_{l} G / \partial Φ^{B} = δ_{B}^{A}$ , so the first bracket term vanishes and the second gives $(S, Φ^{A}) = - \partial_{r} S / \partial Φ_{A}^{*}$ .

2.1A1F1F2step 1.1algebra∎

The right-derivative Leibniz rule is $\partial_{r} (X Y) / \partial z = X (\partial_{r} Y / \partial z) + (- 1)^{∣ z ∣ ∣ Y ∣} (\partial_{r} X / \partial z) Y$ . Apply it with $X = Φ_{A}^{*}$ , $Y = s_{0} Φ^{A}$ , and $z = Φ_{A}^{*}$ ; by [A1] the derivative of $Y$ is zero, while $∣ z ∣ = ∣ Y ∣ = ϵ_{A} + 1$ . Thus $\partial_{r} (Φ_{A}^{*} s_{0} Φ^{A}) / \partial Φ_{A}^{*} = (- 1)^{(ϵ_{A} + 1) (ϵ_{A} + 1)} s_{0} Φ^{A} = (- 1)^{ϵ_{A} + 1} s_{0} Φ^{A}$ . Multiplying by $(- 1)^{ϵ_{A}}$ and using [F2], we obtain $\partial_{r} S / \partial Φ_{A}^{*} = - s_{0} Φ^{A}$ , so step 1.1 yields $(S, Φ^{A}) = s_{0} Φ^{A}$ . In the local version $δ_{r} S_{U} / δ Φ_{A}^{*} (x)$ has the same value $- s_{0} Φ^{A} (x)$ because the antifield occurs without derivatives, and the functional bracket with $Φ^{A} (x)$ is its negative.

Depends on

Classical BV Antibracket and Yang–Mills Master-Action Data