Manuscript IX-A@1 · IX-A.tex · lemma · 55aa6f6d

Lemma: Well-defined boundary trace modulo bulk linear kernels

The sharp adjoint and an intrinsic boundary trace › The linear boundary quotient · lines 721–752 · proof included or linked

IX-A@1#lem:charge, IX-A@1#eq:charge, IX-A@1#eq:chargedef

Back to manuscript

There is a well-defined linear map

b : S_{1} ⟶ C_{c}^{\infty} (Σ; T^{*} M ∣_{Σ})

given by

b (m + q_{a}) = (G_{A}^{0} a) ∣_{Σ}, m \in M_{1} .

It commutes with complex conjugation and vanishes on $M_{1}$ .

Proof. Suppose $m + q_{a} = m^{'} + q_{a^{'}}$ . Then

q_{a - a^{'}} = m^{'} - m \in M_{1} .

Proposition prop:adjoint shows that the possible wavefront covectors of $q_{a - a^{'}}$ are normal to the spacelike surface $Σ$ , and hence timelike. Membership in $M_{1}$ excludes them. Lemma lem:boundary therefore gives

(G_{A}^{0} (a - a^{'})) ∣_{Σ} = 0.

Thus (eq:chargedef) does not depend on the decomposition. Compactness of the trace support follows from finite propagation from the compact source. Reality of the operators proves the conjugation statement.

Displayed formulas in this passage

IX-A@1#eq:charge · line 724
$b : S_{1} ⟶ C_{c}^{\infty} (Σ; T^{*} M ∣_{Σ})$
IX-A@1#eq:chargedef · line 729
$b (m + q_{a}) = (G_{A}^{0} a) ∣_{Σ}, m \in M_{1} .$
unlabeled · line 738
$q_{a - a^{'}} = m^{'} - m \in M_{1} .$
unlabeled · line 745
$(G_{A}^{0} (a - a^{'})) ∣_{Σ} = 0.$

Lemma: Well-defined boundary trace modulo bulk linear kernels

Displayed formulas in this passage

Nearby in this section