Proposition: The remaining freedom in XV-D's example

In (R5), ${\mathcal{S}}_Z(T^{\circ })=\mathrm{span}\{{{\partial }_2}^j{\delta }_0:j\ge 0\}$, which is infinite-dimensional, and the descending extensions are exactly $\delta (x_1)+{\sum }_{j\le k}{c}_j{{\partial }_2}^j{\delta }_0$, $k\ge 0$, $c_j\in \mathbb{C}$.

Proof. An $s$ supported at $0$ has finite order $k$ on a compact neighborhood of $0$, so by Theorem thm:dimension(i) below $s={\sum }_{|\mu |\le k}{c}_{\mu }(-1{)}^{|\mu |}{\partial }^{\mu }{\delta }_0$ and $⟨s,h⟩={\sum }_{\mu }{c}_{\mu }{\partial }^{\mu }h(0)$. By (R5), $h\in \mathrm{Ann}(T^{\circ })\cap \mathcal{D}(X)$ iff $h$ is compactly supported with $h(0,x_2)=0$ for all $x_2$. For such $h$, ${\partial }^{\mu }h(0)=0$ whenever ${\mu }_1=0$. For ${\mu }_1\ge 1$, $h_{\mu }=x^{\mu }\chi$ ($\chi \in \mathcal{D}(X)$, $\chi =1$ near $0$) lies in $\mathrm{Ann}(T^{\circ })\cap \mathcal{D}(X)$ and ${\partial }^{\nu }{h}_{\mu }(0)=\mu !{\delta }_{\mu \nu }$ for all $\nu$. Testing on all $h_{\mu }$ with ${\mu }_1\ge 1$ gives $c_{\mu }=0$ for those $\mu$; conversely if these vanish, $⟨s,h⟩=0$ for all admissible $h$. The ${{\partial }_2}^j{\delta }_0$ are independent (test on ${x_2}^j\chi$). Since $T_1$ descends by (R5), Theorem thm:affine gives the last assertion.