Theorem: The frozen cubic kernel leaves the polynomial microcausal class

Fix $m\ge 0$, the frozen $g$, and the frozen ${\sigma }_{\epsilon }$. Let $t_g:=\min \{t:(t,\mathbf{x})\in \mathrm{supp}g\}>0$. Choose $f\in {C_c}^{\infty }({M_T}^{\circ })$, $f\ge 0$, $f\not\equiv 0$, with $\mathrm{supp}f\subset (0,t_g)\times {\mathbb{R}}^3$ and $\cos (mt)>0$ on $\mathrm{supp}f$. Let ${\varphi }_m$ be XI-A's input, ${\varphi }_m=\sin (mt)/m$ for $m>0$ and ${\varphi }_m=t$ for $m=0$. Then the cubic kernel $u$ of ${\mathcal{C}}_3(f,g)$ in (eq:ukernel) satisfies

More precisely, with $\ell :=⟨u,\cdot \otimes v_2\otimes {\varphi }_m⟩$ for the explicit $v_2$ below, there are points $y$ in an arbitrarily thin collar around $\Sigma$ with $(y,\pm \mathrm{d}t)\in \mathrm{WF}(\ell )$. Consequently the ambient representative of ${\mathcal{C}}_3(f,g)$ is not in ${\mathcal{F}}_{\mathrm{mc},\mathrm{pol}}$, and XI-E's admission map $Q_H$ does not apply to it.

Proof. {Step 1: the kernel for this $f$.} Since ${\mathcal{D}}_f\subset \{t<t_g\}$ and $\mathrm{supp}g\subset \{t\ge t_g\}$, we have ${\mathcal{D}}_f\cap \mathrm{supp}g=\varnothing$, so the $g$-term of (eq:alphaexplicit) vanishes and $\alpha =-{\sigma }_{\epsilon }{C}^{\ast }b$ with $b={{\mathsf{E}}_0}^{\mathrm{A}}(f\mathrm{d}{\varphi }_m)$. Here ${\partial }_{\mu }{\varphi }_m=\cos (mt){{\delta }^0}_{\mu }$, so $b^0={{\mathsf{E}}_0}^{\mathrm{A}}(f\cos (mt))$, $b^i=0$, and

Hence $\mathrm{tr}\alpha =-{\sigma }_{\epsilon }{\partial }_t{b}^0$ and

Therefore $\beta ={{\mathsf{E}}_0}^{\mathrm{A}}\mathcal{P}\alpha$ has ${\beta }^{ij}=0$. Put

Because ${{\mathsf{E}}_0}^{\mathrm{A}}$ acts on sources of compact spatial support per slice and ${\int }_{{\mathbb{R}}^3}\Delta (\cdot ){\mathrm{d}}^{3}x=0$, spatial integration commutes with ${{\mathsf{E}}_0}^{\mathrm{A}}$ and reduces it to its zero-frequency form: $\int {{\mathsf{E}}_0}^{\mathrm{A}}F{\mathrm{d}}^{3}x={{\int }_s}^T(t^{\prime }-s)\bar{F}(t^{\prime })\mathrm{d}{t}^{\prime }$ with $\bar{F}=\int F{\mathrm{d}}^{3}x$. Thus

Then ${\partial }_t\bar{b}(t)=-{{\int }_t}^T\bar{F}\le 0$, strictly negative for $t<t_{-}:=\inf \{t:(t,\mathbf{x})\in \mathrm{supp}f\}$, while ${\sigma }_{\epsilon }(t^{\prime })>0$ for $t^{\prime }>0$. Hence

Step 2: homogeneous configurations and the reduced functional. The spatial projection of ${\mathcal{D}}_f$ lies in a ball $B_R$. Choose $\vartheta \in {C_c}^{\infty }({\mathbb{R}}^3)$ with $\vartheta =1$ on $B_{R+1}$, and set, for $\lambda \in \mathbb{R}$, $\chi \in {C_c}^{\infty }((-\delta ,\delta ))$ with $\chi (0)=1$ and $0<\delta <T$,

as purely spatial symmetric tensors (${\gamma }_{0\mu }=0$). Complex values are admissible by linearity. By finite propagation (X-A, continuity proposition), ${\mathcal{C}}_3$ depends on $\gamma$ only through $\gamma {|}_{{\mathcal{D}}_f}$, and on a neighbourhood of ${\mathcal{D}}_f$ the configuration $\gamma =v_2+\epsilon {v_1}^{\lambda }$ is spatially homogeneous with ${\gamma }_{0\mu }=0$. By Lemma lem:ADM, ${J_3}^{0i}[\gamma ]=0$ and ${J_3}^{00}[\gamma ]=q(\dot{\gamma },\dot{\gamma })$ there, so (eq:C3beta) reduces to

With $q(e,b_0)=\frac{1}{4}(1\cdot 0-1)=-\frac{1}{4}$,

using $\chi (T)=0$. By Riemann–Lebesgue the last integral tends to zero as $|\lambda |\to \infty$, so

Step 3: the covector. Since ${v_1}^{\lambda }$ has only the $(1,1)$ component, $⟨\ell ,{v_1}^{\lambda }⟩=⟨{\ell }^{11},\chi \vartheta e^{i\lambda y^0}⟩$, which is the Fourier transform of the compactly supported distribution $w:=\chi \vartheta {\ell }^{11}$ at the covector $-\lambda \mathrm{d}t$. By (eq:limit) it does not decay along either ray $\lambda \mathrm{d}t$, $\lambda \to \pm \infty$. By Hormander (Definition 8.1.2, Proposition 8.1.3), both $\pm \mathrm{d}t$ lie in the cone $\Sigma (w)$, which is the projection of $\mathrm{WF}(w)$ to the covector variable, and $\mathrm{WF}(w)\subset \mathrm{WF}({\ell }^{11})$. Hence there is $y\in \mathrm{supp}(\chi \vartheta )\subset \{|y^0|<\delta \}$ with $(y,\pm \mathrm{d}t)\in \mathrm{WF}({\ell }^{11})\subset \mathrm{WF}(\ell )$. These covectors are timelike, hence in ${\tilde{V}}_{+}$ and ${\tilde{V}}_{-}$ respectively. By Lemma lem:smear, $u$ violates (eq:mc). The last assertion is Definition def:domain.