Manuscript X-B-ST@1 · X-B-ST.tex · section · d03c150c

Bare insertion representatives

The frozen classical handoff › Bare insertion representatives · lines 483–518

X-B-ST@1#eq:barecoeff, X-B-ST@1#eq:tests

Fix $f \in C_{c}^{\infty} (O; R)$ , let $K_{f} = supp f$ , and put

B^{μ} [h, k] = w^{ν} [h] \partial_{ν} w^{μ} [h] - v^{μ} [h, k] .

The bare coefficients, on a smooth scalar configuration $ψ$ , are

O^{(0)} (ψ) O^{(1)} [h] (ψ) O^{(2)} [h, k] (ψ) = \int f ψ d^{4} x, = - \int f w^{μ} [h] \partial_{μ} ψ d^{4} x, = \int f (B^{μ} [h, k] \partial_{μ} ψ + \frac{1}{2} w^{μ} [h] w^{ν} [h] \partial_{μ} \partial_{ν} ψ) d^{4} x .

Integration by parts fixes the representatives

c_{0} c_{2} [h, k] = f, c_{1} [h] = \partial_{μ} (f w^{μ} [h]), = - \partial_{μ} (f B^{μ} [h, k]) + \frac{1}{2} \partial_{μ} \partial_{ν} (f w^{μ} [h] w^{ν} [h]) .

Write

F_{c} (ψ) = \int c ψ d^{4} x, O_{κ} = F_{c_{0}} + κ F_{c_{1}} + κ^{2} F_{c_{2}} .

The test $f$ is one fixed label-space density. It is not transformed during the background comparison.

Displayed formulas in this passage

X-B-ST@1#eq:barecoeff · line 492
$O^{(0)} (ψ) O^{(1)} [h] (ψ) O^{(2)} [h, k] (ψ) = \int f ψ d^{4} x, = - \int f w^{μ} [h] \partial_{μ} ψ d^{4} x, = \int f (B^{μ} [h, k] \partial_{μ} ψ + \frac{1}{2} w^{μ} [h] w^{ν} [h] \partial_{μ} \partial_{ν} ψ) d^{4} x .$
X-B-ST@1#eq:tests · line 504
$c_{0} c_{2} [h, k] = f, c_{1} [h] = \partial_{μ} (f w^{μ} [h]), = - \partial_{μ} (f B^{μ} [h, k]) + \frac{1}{2} \partial_{μ} \partial_{ν} (f w^{μ} [h] w^{ν} [h]) .$
unlabeled · line 487
$B^{μ} [h, k] = w^{ν} [h] \partial_{ν} w^{μ} [h] - v^{μ} [h, k] .$
unlabeled · line 512
$F_{c} (ψ) = \int c ψ d^{4} x, O_{κ} = F_{c_{0}} + κ F_{c_{1}} + κ^{2} F_{c_{2}} .$