Proposition: Finite normal order can retain infinite tangential freedom

Let $S=\{r=0\}$ in $Y\times {\mathbb{R}}^c$, where $c\ge 1$ and $Y$ contains a positive-dimensional open coordinate set. The order-zero space

is infinite-dimensional. A finite normal derivative-order bound therefore does not imply finite-dimensional ambiguity.

Proof. Choose arbitrarily many independent compact smooth functions $a_j$ on $Y$. If ${\sum }_j{b}_j{a}_j(y)\delta (r)=0$, test with $\psi (y)\chi (r)$ where $\chi (0)=1$. This gives ${\sum }_j{b}_j{a}_j=0$, forcing every $b_j=0$.