Proposition: Two-direction equivalence of relations

Use Theorem thm:wpA-Sbv and fix $N$. For constants $r_{\lambda }$:

1. if ${\sum }_{\lambda }{r}_{\lambda }{n}_{\lambda }=0$ as a polynomial, then ${\sum }_{\lambda }{r}_{\lambda }{q^{\circ }}_{\lambda }=0$ in ${\mathcal{D}}^{\prime }(\Omega )$;

2. if ${\sum }_{\lambda }{r}_{\lambda }{q^{\circ }}_{\lambda }=0$ in ${\mathcal{D}}^{\prime }(\Omega )$, then ${\sum }_{\lambda }{r}_{\lambda }{n}_{\lambda }=0$ as a polynomial.

Relations mixing different $N$ split into homogeneous components, because ${q^{\circ }}_{\lambda }$ is homogeneous of degree $-4-N$. Hence the nullspace of the numerator map is exactly the relation space of $\pi$.

Proof. (1) By Proposition prop:num, $\sum r_{\lambda }{q^{\circ }}_{\lambda }=c^2(\sum r_{\lambda }{n}_{\lambda })T_{2+N}=0$. (2) Restrict to the nonempty open set $\{Q<0\}\subset \Omega$ (it contains $(0,1,0,0)$). There $T_{2+N}=Q^{-2-N}$ is smooth and nowhere zero, so $c^2(\sum r_{\lambda }{n}_{\lambda })Q^{-2-N}=0$ pointwise. Hence the polynomial $\sum r_{\lambda }{n}_{\lambda }$ vanishes on a nonempty open set and is therefore zero. For mixed $N$, dilations $z\mapsto tz$ act on $\Omega$, and distinct homogeneity degrees separate.