Manuscript X-A@1 · X-A.tex · lemma · dc7f5c0a

Lemma: Operator expansion

Regulator preflight › The regulated coordinate problem · lines 876–901 · proof included or linked

X-A@1#lem:Dexp, X-A@1#eq:V, X-A@1#eq:W

$D_{ε} [q] = □ + κ V_{ε} [h] + κ^{2} W_{ε} [h, k] (mod κ^{3})$ , with

V_{ε} [h] ψ W_{ε} [h, k] ψ = \partial_{α} (σ_{ε} a_{1}^{α β} [h] \partial_{β} ψ) = σ_{ε} \partial_{α} (a_{1}^{α β} [h] \partial_{β} ψ) + σ_{ε}^{'} a_{1}^{0 β} [h] \partial_{β} ψ, = \partial_{α} (σ_{ε} (a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) = σ_{ε} \partial_{α} ((a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) + σ_{ε}^{'} (a_{1}^{0 β} [k] + a_{2}^{0 β} [h]) \partial_{β} ψ .

The regulator enters with at most one derivative; there are no $σ_{ε}^{''}$ terms and no $g$ terms.

Proof. Insert (eq:Aexp) into (eq:Aeps) and expand the divergence; $\partial_{α} σ_{ε} = δ^{0}_{α} σ_{ε}^{'}$ .

Displayed formulas in this passage

X-A@1#eq:V · line 879
$V_{ε} [h] ψ W_{ε} [h, k] ψ = \partial_{α} (σ_{ε} a_{1}^{α β} [h] \partial_{β} ψ) = σ_{ε} \partial_{α} (a_{1}^{α β} [h] \partial_{β} ψ) + σ_{ε}^{'} a_{1}^{0 β} [h] \partial_{β} ψ, = \partial_{α} (σ_{ε} (a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) = σ_{ε} \partial_{α} ((a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) + σ_{ε}^{'} (a_{1}^{0 β} [k] + a_{2}^{0 β} [h]) \partial_{β} ψ .$
X-A@1#eq:W · line 879
$V_{ε} [h] ψ W_{ε} [h, k] ψ = \partial_{α} (σ_{ε} a_{1}^{α β} [h] \partial_{β} ψ) = σ_{ε} \partial_{α} (a_{1}^{α β} [h] \partial_{β} ψ) + σ_{ε}^{'} a_{1}^{0 β} [h] \partial_{β} ψ, = \partial_{α} (σ_{ε} (a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) = σ_{ε} \partial_{α} ((a_{1}^{α β} [k] + a_{2}^{α β} [h]) \partial_{β} ψ) + σ_{ε}^{'} (a_{1}^{0 β} [k] + a_{2}^{0 β} [h]) \partial_{β} ψ .$