Manuscript XV-B@2 · XV-B.tex · lemma · 4d3f12e8

Lemma: The elementary degree-mixing relation

Finite normal jets and complete reconstruction-null tests › Why highest presentation degree is not intrinsic · lines 952–983 · proof included or linked

XV-B@2#lem:null, XV-B@2#eq:null

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In one normal variable,

q δ^{'} (q) + δ (q) = 0.

Suppose an adapted reconstruction convention satisfies

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

and multiplication by $q$ is reconstructed as multiplication by the same normal coordinate. Then

Rec ((q i θ + 1) M_{R}) = 0.

If $R \neq = 0$ , the degree-zero and degree-one summands reconstruct separately to nonzero opposite distributions.

Proof. For $ψ \in D (R)$ ,

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

Tensoring $q δ^{'} = - δ$ with $R$ gives the identity. The tensor $δ \otimes R$ is nonzero when $R \neq = 0$ , as is seen by choosing separated tests with nonzero pairings.

Displayed formulas in this passage

XV-B@2#eq:null · line 965
$Rec ((q i θ + 1) M_{R}) = 0.$
unlabeled · line 955
$q δ^{'} (q) + δ (q) = 0.$
unlabeled · line 959
$Rec (M_{R}) = δ (q) \otimes R, Rec (i θ M_{R}) = δ^{'} (q) \otimes R,$
unlabeled · line 975
$⟨ q δ^{'}, ψ ⟩ = - \frac{d}{d q} (q ψ (q))_{q = 0} = - ψ (0) .$

Lemma: The elementary degree-mixing relation

Displayed formulas in this passage

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