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The SU(2) Ghost Sector Solves the Classical Master Equation

Example

Let $s u (2)$ have basis $T_{a}$ with $[T_{b}, T_{c}] = ϵ^{a}_{b c} T_{a}$ . Form the finite graded-commutative coordinate algebra $B = C [c_{1}^{*}, c_{2}^{*}, c_{3}^{*}] \otimes Λ (c^{1}, c^{2}, c^{3})$ , where each $c^{a}$ is odd of ghost number $1$ and each $c_{a}^{*}$ is even of ghost number $- 2$ . Use the canonical antibracket

(F, G) = \sum_{a = 1}^{3} (\frac{\partial_{r} F}{\partial c^{a}} \frac{\partial_{l} G}{\partial c_{a}^{*}} - \frac{\partial_{r} F}{\partial c_{a}^{*}} \frac{\partial_{l} G}{\partial c^{a}}) .

For $S_{C E} = \frac{1}{2} c_{a}^{*} ϵ^{a}_{b c} c^{b} c^{c}$ , the element $S_{C E}$ is even with ghost number zero and satisfies $(S_{C E}, S_{C E}) = 0$ . Hence the Hamiltonian differential $s F = (S_{C E}, F)$ is nilpotent. This is only the finite-dimensional non-Abelian ghost sector; it is not a Yang–Mills action or a quantum field theory.

Example references

Canonical BV bracket and degree convention

Master equation/nilpotence equivalence

Facts & Assumptions

Given: The displayed finite coordinate algebra, antibracket, and $S_{C E}$ .

[A1]

The structure constants $ϵ^{a}_{b c}$ of $s u (2)$ satisfy the Lie-algebra Jacobi identity.

[F1]

Classical BV Antibracket and Yang–Mills Master-Action Data: the canonical antibracket has ghost number $+ 1$ , shifted Jacobi, and the displayed left/right derivative convention.

[F2]

Classical Master Equation and Nilpotence of the BV Differential: on a canonical BV domain with no nonzero central element of ghost number one, the classical master equation is equivalent to nilpotence of $s F = (S, F)$ .

Verification

1.1A1F1algebra

The ghost number of $S_{C E}$ is $- 2 + 1 + 1 = 0$ , and its parity is even. The derivative convention gives $\partial_{r} S_{C E} / \partial c^{a} = - c_{e}^{*} ϵ^{e}_{a b} c^{b}$ , $\partial_{l} S_{C E} / \partial c^{a} = + c_{e}^{*} ϵ^{e}_{a b} c^{b}$ , and $\partial_{r} S_{C E} / \partial c_{a}^{*} = \partial_{l} S_{C E} / \partial c_{a}^{*} = \frac{1}{2} ϵ^{a}_{c d} c^{c} c^{d}$ . Since $c_{a}^{*}$ is even, substitution in the defining bracket gives $(S_{C E}, S_{C E}) = (- c_{e}^{*} ϵ^{e}_{a b} c^{b}) (\frac{1}{2} ϵ^{a}_{c d} c^{c} c^{d}) - (\frac{1}{2} ϵ^{a}_{c d} c^{c} c^{d}) (+ c_{e}^{*} ϵ^{e}_{a b} c^{b}) = - c_{e}^{*} ϵ^{e}_{a b} ϵ^{a}_{c d} c^{b} c^{c} c^{d}$ . The ghost product is totally antisymmetric, so the tensor contraction reduces to its antisymmetrization $ϵ^{e}_{a [b} ϵ^{a}_{c d]}$ , which is the Jacobi identity and vanishes by [A1].

2.1F2step 1.1∎

The coordinate algebra is finite and canonical, so its center consists of constants and has no nonzero ghost-number-one element. Step 1.1 proves the classical master equation, and [F2] therefore gives $s^{2} = 0$ on the entire algebra.

The SU(2) Ghost Sector Solves the Classical Master Equation

Example

Example references

Facts & Assumptions

Verification

Depends on

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