Lemma: Grouped normal traces off the characteristic locus

On $S_{\mathrm{resp}}\cap (U\setminus N_U)$, every finite normal trace

is defined.

At a singular point of such a trace its tangential wavefront is bounded by a nonzero multiple of $i^{\ast }\mathrm{d}{q}_1$ or $i^{\ast }\mathrm{d}{q}_2$. No point in this domain has both $q_1=q_2=0$.

Proof. On $S_{\mathrm{resp}}$, $q_R=0$ and $a=z$. By the copied zero-locus statement (eq:zerolocus), simultaneous vanishing of $q_1,q_2$ would put the point in $N_U$. Thus away from $N_U$ at most one of the two cross-line factors is singular.

If neither is singular, the grouped distribution is smooth locally, and all normal traces exist.

If only $q_1=0$, the other line is smooth locally. The grouped distribution and its derivatives have wavefront bounded by nonzero multiples of $\mathrm{d}{q}_1$. The $y$ component of $\mathrm{d}{q}_1$ is nonzero because $y\ne a$. Every conormal of $S_{\mathrm{resp}}$ has zero $y$ component, since $q_R$ and $a-z$ do not depend on $y$. The restriction normal-set condition therefore holds.

The case with only $q_2=0$ is identical, using $y\ne b$. Differentiation does not enlarge the line-wavefront bound. The restriction theorem gives (eq:traces) and the tangential upper bound Hormander (Section 8.2).