Manuscript XV-A@1 · XV-A.tex · lemma · f53f3515

Lemma: Null presentation

Reconstruction descent and presentation degree · lines 1153–1185 · proof included or linked

XV-A@1#lem:null, XV-A@1#eq:null

In one normal variable $q$ ,

q δ^{'} (q) + δ (q) = 0.

Accordingly, in an adapted reconstruction convention satisfying

Rec (M_{R}) = δ (q) R (r), Rec (i θ M_{R}) = δ^{'} (q) R (r),

one has

Rec ((q i θ + 1) M_{R}) = 0.

If $R \neq = 0$ , the two summands reconstruct separately to nonzero distributions with opposite signs.

Proof. For $ψ \in D (R)$ ,

⟨ q δ^{'}, ψ ⟩ = ⟨ δ^{'}, q ψ ⟩ = - \frac{d}{d q} (q ψ (q))_{q = 0} = - ψ (0) .

Hence $q δ^{'} = - δ$ . Tensoring this identity with $R (r)$ gives (eq:null). The tensor $δ (q) \otimes R (r)$ is nonzero when $R$ is nonzero: choose separated tests whose two pairings are nonzero.

Displayed formulas in this passage

XV-A@1#eq:null · line 1165
$Rec ((q i θ + 1) M_{R}) = 0.$
unlabeled · line 1156
$q δ^{'} (q) + δ (q) = 0.$
unlabeled · line 1160
$Rec (M_{R}) = δ (q) R (r), Rec (i θ M_{R}) = δ^{'} (q) R (r),$
unlabeled · line 1175
$⟨ q δ^{'}, ψ ⟩ = ⟨ δ^{'}, q ψ ⟩ = - \frac{d}{d q} (q ψ (q))_{q = 0} = - ψ (0) .$

Lemma: Null presentation

Displayed formulas in this passage

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