Manuscript X-B@3 · X-B-v3.tex · section · 2d415d58

The sign ledger

B2: one retarded map, its sign, support, and quantum order › The sign ledger · lines 902–927

X-B@3#eq:Hessian

Compactly supported integration by parts gives

V_{1, g}^{'} [φ] = - L_{1, g} [h] φ, V_{1, g}^{''} = - L_{1, g} [h] .

The free action has $S_{0}^{''} = - P_{m}$ , hence

Δ_{S}^{R} = - E_{m}^{R}, Δ_{S}^{A} = - E_{m}^{A} .

The positive Euler coefficient is not the variational Hessian. Using $+ E_{m}^{R}$ as the inverse of $S_{0}^{''}$ would reverse the response and violate the normalization sheet.

For a quadratic insertion as in (A1), the classical retarded response of $F_{c}$ is

- ⟨ c, Δ_{S}^{R} V^{'} ⟩ = - ⟨ c, E_{m}^{R} L φ ⟩ .

This also follows by differentiating the retarded solution of $(S_{0} + λV)^{'} = 0$ .

Displayed formulas in this passage

X-B@3#eq:Hessian · line 905
$V_{1, g}^{'} [φ] = - L_{1, g} [h] φ, V_{1, g}^{''} = - L_{1, g} [h] .$
unlabeled · line 911
$Δ_{S}^{R} = - E_{m}^{R}, Δ_{S}^{A} = - E_{m}^{A} .$
unlabeled · line 921
$- ⟨ c, Δ_{S}^{R} V^{'} ⟩ = - ⟨ c, E_{m}^{R} L φ ⟩ .$

The sign ledger

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