Manuscript X-J-ST@1 · X-J-ST.tex · proposition · 8f0d1a36

Proposition: Retained third-order master identity

The retained classical interface and its recovery › The finite action being retained · lines 561–596 · proof included or linked

X-J-ST@1#prop:retainedCME

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Equation (eq:SI) satisfies

\frac{1}{2} (S_{λ}^{[3]}, S_{λ}^{[3]}) = 0 (mod κ^{4}), Q_{λ}^{[3]} F = (S_{λ}^{[3]}, F), Q_{λ}^{[3]}^{2} = 0 (mod κ^{4}) .

No term in this paper is added to that action.

Proof. The signed lift obeys $(H_{s}, Φ^{A}) = s Φ^{A}$ on every generator. Its self-bracket is the cotangent lift of $s^{2}$ , so it vanishes by Lemma lem:snil.

Naturality gives

s E [q] = κ \partial_{μ} (c^{μ} E [q]), s L_{q} = κ \partial_{μ} (c^{μ} L_{q}), s^{2} Ψ_{q ˉ} = 0.

Thus the full formal reference Hamiltonian has zero master square modulo divergences. Its nonnegative truncation has zero coefficients through order three.

The only new cross term from $V_{a}$ starts at order four because $s_{0} ϕ = 0$ , and $(V_{a}, V_{a}) = 0$ . The odd Jacobi identity proves the differential claim.

Displayed formulas in this passage

unlabeled · line 564
$\frac{1}{2} (S_{λ}^{[3]}, S_{λ}^{[3]}) = 0 (mod κ^{4}), Q_{λ}^{[3]} F = (S_{λ}^{[3]}, F), Q_{λ}^{[3]}^{2} = 0 (mod κ^{4}) .$
unlabeled · line 580
$s E [q] = κ \partial_{μ} (c^{μ} E [q]), s L_{q} = κ \partial_{μ} (c^{μ} L_{q}), s^{2} Ψ_{q ˉ} = 0.$

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section: The finite action being retained