Knowlpedia

Theorem

Vanishing products from separated auxiliary supports

Closed support separation makes all differentiated cross products vanish, including after evaluation along a smooth auxiliary map.

Statement

Let Fγ(x,Y)F_\gamma(x,Y) be globally with

supp⁡Fγ⊂Kγ×Rγ.\operatorname{supp}F_\gamma\subset K_\gamma\times R_\gamma.

Assume that, for distinct labels, Kγ∩Kγ′≠∅K_\gamma\cap K_{\gamma'}\ne\varnothing implies Rγ∩Rγ′=∅R_\gamma\cap R_{\gamma'}=\varnothing. Then every pair of mixed derivatives satisfies

(∂αFγ)(∂βFγ′)=0(γ≠γ′).(\partial^\alpha F_\gamma)(\partial^\beta F_{\gamma'})=0 \qquad(\gamma\ne\gamma').
Why derivatives and evaluation preserve the conclusion

A derivative of a smooth function has contained in the support of the function. The assumed product supports are disjoint, proving the identity. After at Y=Ψ(x)Y=\Psi(x), the chain rule expresses each derivative using these same mixed derivatives, so cross products still vanish.

For a finite or locally finite sum this removes all cross-label terms from a quadratic differential expression. It says nothing about different harmonics with the same label. Smooth extension across the support boundaries is essential: differentiating a discontinuous cutoff can create boundary distributions, outside the hypotheses here.