Proposition: Collapse rank

1. In the formal reading, $\mathrm{Collapse}$ is injective on ${\mathcal{C}}_{\mathrm{coh}}$.

2. In the $\chi$-plateau reading, $\mathrm{Collapse}$ is injective (reported computation).

3. In the double-plateau reading, $\dim \ker \mathrm{Collapse}=22$ (with $4$ at $D=3$ and $18$ at $D=4$), $\dim \mathrm{im}\mathrm{Collapse}=3299$, and $\ker \mathrm{Collapse}\cap {\mathcal{C}}_{\mathrm{reg}}=0$ (reported computation).

Injectivity of $\mathcal{G}$ does not by itself imply injectivity after collapse. In the plateau reading, $\ker \mathcal{G}\subseteq \ker \mathrm{Collapse}$ grows from $21$ to $22$.

Proof. (1) Factor ${\bar{a}}_b={\chi }_{\mathrm{I}}{\hat{\bar{a}}}_b$ and expand the Leibniz rule for ${{\partial }_w}^{\gamma }{\bar{a}}_b$: write $E_C={\sum }_{\rho }{\chi }_{\rho }{F}_{\rho }$ with ${\chi }_{\rho }={\partial }^{\rho }{\chi }_{\mathrm{I}}$ and $F_{\rho }={\sum }_{\gamma \ge \rho }\left(\genfrac{}{}{0ex}{}{\gamma }{\rho }\right){\sum }_{N,b}{C}_{N,b,\gamma }{{\partial }_h}^{\gamma -\rho }{\hat{\bar{a}}}_b$, where ${\partial }_h$ acts on the $h$-jet only (the displayed $F_{\rho }$ use the switch-free ${\hat{\bar{a}}}_b$). In the formal reading the ${\chi }_{\rho }$ are independent, so $E_C=0$ forces $F_{\rho }=0$ for all $\rho$. Proceed by descending induction on $k$. If $C_{\cdot ,\cdot ,\gamma }=0$ for $|\gamma |>k$, then for $|\rho |=k$ one has $F_{\rho }={\sum }_{N,b}{C}_{N,b,\rho }{\hat{\bar{a}}}_b=0$. Since $U_N={\mathcal{H}}_{4+N}$ (Theorem thm:M3) and monomials of different $N$ are disjoint, $C_{\cdot ,\cdot ,\rho }=0$. (2), (3): Lemma lem:M4comp.