Proposition: Recalled from XV-D

For every $T\in {\mathcal{D}}^{\prime }(X)$ and $a\in {{\mathcal{P}}_r}^{\mathrm{jet}}$, ${\mathsf{Rec}}_T(a)={\sum }_{|\gamma |\le r}{\mathsf{E}}_{\gamma }({\beta }_{\gamma }(a)T)$, and ${\sum }_{\gamma }{\mathsf{E}}_{\gamma }{b}_{\gamma }=0$ iff all $b_{\gamma }=0$. Consequently

$r{F}_0^T={\mathsf{Rec}}^{\circ }$, ${d_0}^T(a)={\sum }_{\gamma }{\mathsf{E}}_{\gamma }({\beta }_{\gamma }(a)T)$ with every coefficient supported in $Z$, $\ker {d_0}^T=H_T$, and ${d_0}^T=0$ iff $H_T=H$.

Proof. The normal form is the Leibniz rule applied to ${{\partial }_q}^{\alpha }(a_{\alpha }\varphi )(\cdot ,0)$ in (eq:insert). Extraction: with $\chi \in \mathcal{D}(V_q)$ equal to one near $0$ and ${\mathsf{J}}_{\gamma }\psi (x,q)=\frac{(-1{)}^{|\gamma |}}{\gamma !}{q}^{\gamma }\chi (q)\psi (x)$, one has $(-1{)}^{|\beta |}{{\partial }_q}^{\beta }{\mathsf{J}}_{\gamma }\psi (x,0)={\delta }_{\beta \gamma }\psi (x)$, so testing on ${\mathsf{J}}_{\gamma }\psi$ returns $⟨b_{\gamma },\psi ⟩$. The rest follows as in XVD: restriction of the normal form to $U^{\circ }$ gives $r{F}_0^T={\mathsf{Rec}}^{\circ }$ and $H_T\subseteq H$; for $a\in H$ each ${\beta }_{\gamma }(a)T$ restricts to zero on $X^{\circ }$.