Theorem: Exact null defect of an extension lift

Let $T$ extend $T^{\circ }$ and let $a\in H$. Then:

1. ${d_0}^T(a)={\sum }_{|\gamma |\le r}{\mathsf{E}}_{\gamma }({\beta }_{\gamma }(a)T)$, and every coefficient ${\beta }_{\gamma }(a)T$ is supported in $Z$;

2. $\mathrm{supp}{d_0}^T(a)\subseteq Z\times \{0\}$; in particular ${d_0}^T(a)\in Y$ and ${d_0}^T(a){|}_{U^{\circ }}=0$;

3. $\ker {d_0}^T=H_T$; hence ${d_0}^T$ induces an injective linear map $H/H_T\to Y$, and ${d_0}^T=0$ if and only if $H_T=H$;

4. ${d_0}^T(a)=0$ on all of $U$ if and only if ${\beta }_{\gamma }(a)T=0$ in ${\mathcal{D}}^{\prime }(X)$ for all $|\gamma |\le r$; otherwise the defect is a nonzero distribution supported in $Z\times \{0\}$, and there is no open $G\subset U\setminus (Z\times \{0\})$ with ${d_0}^T(a){|}_G\ne 0$.

Proof. (i) is (eq:normalform). For $a\in H$, $({\beta }_{\gamma }(a)T){|}_{X^{\circ }}={\beta }_{\gamma }(a){|}_{X^{\circ }}{T}^{\circ }=0$ by (eq:Hpunct), so $\mathrm{supp}({\beta }_{\gamma }(a)T)\subseteq Z$. (ii) follows from Lemma lem:extraction(ii) and the definition of $Y$. (iii): $\ker {d_0}^T=H\cap \ker {\mathsf{Rec}}_T=H\cap H_T=H_T$ by Theorem thm:kernels. (iv) is Lemma lem:extraction(i)–(ii).