Manuscript XIV-J@1 · XIV-J.tex · lemma · 1c0e72ca

Lemma: A normal identity invisible to polynomial-degree bookkeeping

J2a: reconstruction descent is a necessary interface › An exact null presentation · lines 935–965 · proof included or linked

XIV-J@1#lem:null, XIV-J@1#eq:null

In an adapted $K_{5}$ chart,

Rec ((q i θ + 1) M_{R} (κ)) = 0.

Nevertheless the two reconstructed summands are separately nonzero:

Rec (q i θ M_{R}) = - δ (q) R (r), Rec (M_{R}) = δ (q) R (r) .

Consequently the polynomial-degree splitting of a presentation does not by itself define an intrinsic joint-symbol splitting on $K_{int}$ .

Proof. For a one-dimensional test $ψ$ ,

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

Hence $q δ^{'} = - δ$ . Tensoring with the existing nonzero distribution $R (r)$ and using (eq:Rec) gives all displayed identities.

Displayed formulas in this passage

XIV-J@1#eq:null · line 938
$Rec ((q i θ + 1) M_{R} (κ)) = 0.$
unlabeled · line 944
$Rec (q i θ M_{R}) = - δ (q) R (r), Rec (M_{R}) = δ (q) R (r) .$
unlabeled · line 957
$⟨ q δ^{'} (q), ψ ⟩ = - \partial_{q} (q ψ (q))_{q = 0} = - ψ (0) .$