Manuscript XI-A@1 · XI-A.tex · theorem · b8211ecc

Theorem: Fixed-background Noether identity

Noether identity and the inherited Route II › The correct adjoint gauge operator · lines 818–860 · proof included or linked

XI-A@1#thm:noether, XI-A@1#eq:noether, XI-A@1#eq:noether-gf

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Off shell, the invariant action satisfies

N_{ν} (e) = κ (E_{ϕ} \partial_{ν} ϕ - B_{ν}) .

The gauge-fixed action instead satisfies

N_{ν} (e + e_{gf}) = κ (E_{ϕ} \partial_{ν} ϕ - B_{ν}) + N_{ν} (e_{gf}) .

On $U$ , $B_{ν} = 0$ . This removes the switching breaking, not the gauge-fixing contribution.

Proof. By symmetry of $e$ and integration by parts,

δ_{ξ} S_{inv} = \int ξ^{ν} [- N_{ν} (e) + κ E_{ϕ} \partial_{ν} ϕ] d^{4} x .

Take the infinitesimal coefficient of the imported (eq:RouteII). The flat scalar variation is

δ_{ξ} L_{η} = κ (T_{0}^{μ}_{ν} \partial_{μ} ξ^{ν} + \partial_{μ} (ξ^{μ} L_{η})) .

After integrating these two terms, the coefficient of $κ ξ^{ν}$ is exactly (eq:B). The fundamental lemma for arbitrary compact interior $ξ$ proves (eq:noether). Adding the separate gauge-fixing variation proves (eq:noether-gf). On $U$ , both $1 - g$ and $\partial g$ vanish.

Displayed formulas in this passage

XI-A@1#eq:noether · line 821
$N_{ν} (e) = κ (E_{ϕ} \partial_{ν} ϕ - B_{ν}) .$
XI-A@1#eq:noether-gf · line 829
$N_{ν} (e + e_{gf}) = κ (E_{ϕ} \partial_{ν} ϕ - B_{ν}) + N_{ν} (e_{gf}) .$
unlabeled · line 841
$δ_{ξ} S_{inv} = \int ξ^{ν} [- N_{ν} (e) + κ E_{ϕ} \partial_{ν} ϕ] d^{4} x .$
unlabeled · line 849
$δ_{ξ} L_{η} = κ (T_{0}^{μ}_{ν} \partial_{μ} ξ^{ν} + \partial_{μ} (ξ^{μ} L_{η})) .$

Theorem: Fixed-background Noether identity

Displayed formulas in this passage

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