Manuscript XIX-D@1 · XIX-D.tex · lemma · 1f47e33b

Lemma: A normal null identity mixes presentation degrees

The inherited characteristic map remains a separate gate › Reconstruction descent is necessary but not admission · lines 1540–1559 · proof included or linked

XIX-D@1#lem:null

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For any nonzero distribution $R$ ,

q δ^{'} (q) \otimes R + δ (q) \otimes R = 0.

The two summands are separately nonzero.

Proof. For a test $ψ$ ,

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

Hence $q δ^{'} = - δ$ . Tensor with $R$ ; nonvanishing of $δ \otimes R$ follows by choosing separated tests with nonzero pairings.

Displayed formulas in this passage

unlabeled · line 1543
$q δ^{'} (q) \otimes R + δ (q) \otimes R = 0.$
unlabeled · line 1551
$⟨ q δ^{'}, ψ ⟩ = - \frac{d}{d q} (q ψ (q))_{q = 0} = - ψ (0) .$

Nearby in this section

proposition: Proposition: Descent through a reconstruction map
section: Reconstruction descent is necessary but not admission
section: Reconstruction descent is necessary but not admission