Manuscript XVI-I@1 · XVI-I.tex · proposition · a7046a8d

Proposition: Identification and inverse identity

An ambient propagator in the selected normalization › The tensor time-ordered kernel · lines 673–715 · proof included or linked

XVI-I@1#prop:DF, XVI-I@1#eq:timeordered

The scalar distribution $d$ is

d (r) = {W_{0} (r), W_{0} (- r), r^{0} > 0, r^{0} < 0,

with its globally defined distributional meaning fixed by (eq:DF). Consequently

D_{F}^{T} = D_{F}, (- □ P) D_{F} = i id

on ambient compact tensor tests. In components,

D_{F}^{I K} (y, w) = P^{I K} d (y - w) .

Proof. For $t < 0$ , the spatial Fourier expression is

\frac{e ^{- i ∣ p ∣ t}}{2∣ p ∣} + i \frac{sin ( t ∣ p ∣ )}{∣ p ∣} = \frac{e ^{i ∣ p ∣ t}}{2∣ p ∣} .

For $t > 0$ , the advanced summand vanishes. This proves (eq:timeordered) and symmetry under $r \mapsto - r$ .

The temporal derivative of the two Fourier branches has jump $- i$ at $t = 0$ . Hence $□ d = - i δ$ . Using $P^{2} = id$ gives

(- □ P) (P d) = i δ id .

Displayed formulas in this passage

XVI-I@1#eq:timeordered · line 676
$d (r) = {W_{0} (r), W_{0} (- r), r^{0} > 0, r^{0} < 0,$
unlabeled · line 686
$D_{F}^{T} = D_{F}, (- □ P) D_{F} = i id$
unlabeled · line 693
$D_{F}^{I K} (y, w) = P^{I K} d (y - w) .$
unlabeled · line 700
$\frac{e ^{- i ∣ p ∣ t}}{2∣ p ∣} + i \frac{sin ( t ∣ p ∣ )}{∣ p ∣} = \frac{e ^{i ∣ p ∣ t}}{2∣ p ∣} .$
unlabeled · line 712
$(- □ P) (P d) = i δ id .$