Smooth Hadamard changes intertwine the polynomial microcausal products

Define ${\mathcal{C}}_g^{-}:={\tau }^{\ast }{\mathcal{C}}_g^{+}$, where $\tau (x,y)=(y,x)$; its first covectors are past-directed. Put $K=W^{\mathrm{\prime }}-W$. The common antisymmetric part gives $K-K^{\mathsf{T}}=0$, so $K$ is symmetric. By [F1]–[F2], $\mathrm{WF}(K)\subseteq \mathrm{WF}(W)\cup \mathrm{WF}(W^{\mathrm{\prime }})={\mathcal{C}}_g^{+}$. Symmetry also gives $\mathrm{WF}(K)={\tau }^{\ast }\mathrm{WF}(K)\subseteq {\mathcal{C}}_g^{-}$. A nonzero null covector cannot be both future and past, so ${\mathcal{C}}_g^{+}\cap {\mathcal{C}}_g^{-}=\mathrm{\varnothing }$ and $\mathrm{WF}(K)=\mathrm{\varnothing }$. By [F2], $K$ is smooth.

By [F3], ${\mathrm{\Gamma }}_K$ is defined on the class and preserves microcausality. On a polynomial of degree $d$, ${\mathrm{\Gamma }}_K^{j}F=0$ for $2j>d$, so ${\alpha }_{K}F$ and ${\alpha }_{-K}F$ are finite sums. The coefficient of ${\mathrm{\Gamma }}_K^q$ in either composition is multiplied by ${\sum }_{j=0}^q\left(\genfrac{}{}{0px}{}{q}{j}\right)(-1{)}^{q-j}=(1-1{)}^q$, which is zero for $q>0$ and one for $q=0$. Hence ${\alpha }_K{\alpha }_{-K}={\alpha }_{-K}{\alpha }_K=I$.

On the regular polynomial subspace of a Euclidean chart, [F4]–[F5] fix the contraction ordering and the factor $\mathrm{\hslash }\mathrm{/}2$. For the general class, label the two functional arguments by $1,2$ and define $D_L(F\otimes G)$ by multiplying their first derivative distributions by $L$ and integrating the contracted pair with ${\mu }_g^{\otimes 2}$. Let ${\mathrm{\Gamma }}_K^{(1)}$ and ${\mathrm{\Gamma }}_K^{(2)}$ contract two derivative slots within the first and second arguments using the same density. The finite Leibniz expansion of the second derivative gives ${\mathrm{\Gamma }}_K\circ m=m\circ ({\mathrm{\Gamma }}_K^{(1)}+{\mathrm{\Gamma }}_K^{(2)}+2D_K),$ where $m$ is pointwise multiplication. The factor $2$ consists of the two cross terms, equal because $K$ is symmetric. Every term is defined by [F3] and is the canonical labelled matching of [F6]. Its two smooth cross terms are equal by the proved slot-permutation rule and symmetry of $K$, without extending the regular coefficient-kernel topology to singular inputs.

Expand each coefficient into its original $F$ and $G$ kernel slots. Every $D_W$ edge joins the first original block to the second; every additional edge from ${\mathrm{\Gamma }}_K^{(1)}$, ${\mathrm{\Gamma }}_K^{(2)}$ or $D_K$ is smooth. Each derivative slot is removed at most once. Hence [F6] proves the full matching and every intermediate product/pushforward are admissible and independent of edge order. Finite Leibniz and permutation identities count exactly the same labelled matchings in either order. Thus these operators commute on the coefficient expressions used here, and the binomial exponential law is a finite polynomial identity in commuting degree-lowering operations. It makes no assertion about a singular ${\mathrm{\Gamma }}_W$ on one microcausal functional. Finite polynomial degree makes every coefficient sum finite. Thus ${\alpha }_K(F{\star }_{W}G)=m{e}^{\frac{\mathrm{\hslash }}{2}({\mathrm{\Gamma }}_K^{(1)}+{\mathrm{\Gamma }}_K^{(2)}+2D_K)}{e}^{\mathrm{\hslash }{D}_W}(F\otimes G)=m{e}^{\frac{\mathrm{\hslash }}{2}({\mathrm{\Gamma }}_K^{(1)}+{\mathrm{\Gamma }}_K^{(2)})}{e}^{\mathrm{\hslash }{D}_{W+K}}(F\otimes G)={\alpha }_{K}F{\star }_{W+K}{\alpha }_{K}G$. Here $D_{W+K}=D_W+D_K$ by bilinearity and $W+K=W^{\mathrm{\prime }}$. These are coefficientwise identities of finite sums, not analytic convergence claims. Together with step 1.2 this proves the stated intertwiner and bijectivity.