Proposition: A computed null defect

Let $d=2$, $X={\mathbb{R}}^2$ with coordinates $(x_1,x_2)$, $Z=\{0\}$, $m=1$, $r=0$. Let $T^{\circ }\in {\mathcal{D}}^{\prime }(X^{\circ })$ be the restriction to $X^{\circ }$ of $\delta (x_1)$, that is, $⟨T^{\circ },\psi ⟩=\int \psi (0,x_2)\mathrm{d}{x}_2$ for $\psi \in \mathcal{D}(X^{\circ })$. Consider the two extensions

and the presentation $a\in {{\mathcal{P}}_r}^{\mathrm{jet}}=C^{\infty }(U)$, $a(x,q)=x_1$. Then:

1. $H=\{a\in C^{\infty }(U):a(0,x_2,0)=0\text{for all }{x}_2\}$, and $a=x_1$ belongs to $H$;

2. $H_{T_1}=H$, so ${d_0}^{T_1}=0$;

3. $H_{T_2}=\{a\in H:({\partial }_{x_1}a)(0,0,0)=0\}$, the quotient $H/H_{T_2}$ is one-dimensional, and

   $${d_0}^{T_2}(x_1)={\mathsf{E}}_0\left(x_1{\partial }_{x_1}{\delta }_0\right)=-{\mathsf{E}}_0{\delta }_0\ne 0,\mathrm{supp}{d_0}^{T_2}(x_1)=\{(0,0,0)\}=Z\times \{0\}.$$

Proof. For $r=0$, ${\beta }_0(a)=a(\cdot ,0)$. (i): $a(\cdot ,0)T^{\circ }=0$ on $X^{\circ }$ means $\int a(0,x_2,0)\psi (0,x_2)\mathrm{d}{x}_2=0$ for all $\psi \in \mathcal{D}(X^{\circ })$, i.e. $a(0,x_2,0)=0$ for $x_2\ne 0$, hence for all $x_2$ by continuity. For $a=x_1$ this holds. (ii): the same computation on $X$ gives $H_{T_1}=H$. (iii): for $a\in H$,

so $a(\cdot ,0)T_2=-({\partial }_{x_1}a)(0,0,0){\delta }_0$. This vanishes if and only if $({\partial }_{x_1}a)(0,0,0)=0$, giving $H_{T_2}$; the linear functional $a\mapsto ({\partial }_{x_1}a)(0,0,0)$ is onto $\mathbb{C}$ on $H$ (take $a=x_1$), so $H/H_{T_2}$ is one-dimensional. For $a=x_1$ the coefficient is $-{\delta }_0$, and Lemma lem:extraction gives the nonvanishing and the support.