All regular formal temporal causal-factorization coefficients

Taking vacuum expectations in the commutator of [F3] and using the covariance in [F2] gives $W_{\mu }(f,g)-W_{\mu }(g,f)=-i{\mathrm{\Delta }}_{\mu }(f,g)$ for real compact tests. Complex bilinearity extends this equality to complex product tests, and density [F6] extends it to the lifted bidistributions: ${\tilde{W}}_{\mu }-{\tilde{W}}_{\mu }^{\mathsf{T}}=-i{\tilde{\mathrm{\Delta }}}_{\mu }$. If $\mathrm{\Psi }$ is supported where $x-y$ is spacelike, the relative test $z\mapsto \int \mathrm{\Psi }(y+z,y)dy$ is compactly supported in the spacelike open set. By [F3] its pairing with ${\mathrm{\Delta }}_{\mu }$ is zero. Thus ${\tilde{W}}_{\mu }-{\tilde{W}}_{\mu }^{\mathsf{T}}$ vanishes there. The conclusion is conditional on the current QFT-MF02 statements.

Put $C=A-B$. Every $z=x-y\in C$ either has $z^0>0$ or is spacelike: if $z^0\le 0$ and it is not spacelike, then $z\in C_{-}$, excluded by $A{\succcurlyeq }_{\mathrm{o}\mathrm{u}\mathrm{t}}B$; this also excludes the zero difference. Thus $C$ lies in the open set $\mathrm{\Omega }=U_{+}\cup U_{\mathrm{s}\mathrm{p}}$, where $U_{+}=\{z:z^0>0\}$ and $U_{\mathrm{s}\mathrm{p}}=\{z:(z^0{)}^2-\mathrm{\mid }\mathbf{z}{\mathrm{\mid }}^2<0\}$. By [F7], the cover $\{U_{+},U_{\mathrm{s}\mathrm{p}}\}$ of $\mathrm{\Omega }$ has a locally finite smooth partition of unity subordinate to these sets. Any compactly supported relative test in $\mathrm{\Omega }$ meets only finitely many partition functions. On each $U_{+}$ piece [F1] gives $H_{\mu }=W_{\mu }$; on each $U_{\mathrm{s}\mathrm{p}}$ piece [F1] gives agreement with both Wightman orderings, which coincide there by step 1.1. Therefore ${\tilde{H}}_{\mu }-{\tilde{W}}_{\mu }$ vanishes on $\mathrm{\Omega }$. The open set $O=\{(x,y):x-y\in \mathrm{\Omega }\}$ contains $A\times B$, and the lift [F4] sends every test supported in $O$ to a relative test supported in $\mathrm{\Omega }$. Hence ${\tilde{H}}_{\mu }-{\tilde{W}}_{\mu }$ vanishes on an open neighborhood of $A\times B$.

For any finite number $r$ of cross pairs, expand ${\tilde{H}}_{\mu }^{\otimes r}-{\tilde{W}}_{\mu }^{\otimes r}$ by telescoping; each summand contains one factor ${\tilde{H}}_{\mu }-{\tilde{W}}_{\mu }$. Its support is disjoint from the corresponding open neighborhood of $A\times B$ by [F8] and step 2.1, so it pairs to zero with derivative-kernel tests supported in $(A\times B{)}^r$. Therefore every finite cross contraction is unchanged when the time-ordered kernel is replaced by $W_{\mu }$.

In the finite contraction formula for $T_{p+q}^H(\underset{p\text{ inputs}}{\underbrace{V,\dots ,V}},\underset{q\text{ inputs}}{\underbrace{U,\dots ,U}})$, split ${\sum }_{i<j}{D}_{\mathsf{H}}^{ij}$ into contractions within the first $p$ slots, within the last $q$ slots, and between the two blocks. Two such operators may share a field block, but they contract distinct variable copies. If they share block $i$, symmetry of its coefficient kernel lets one exchange the labels of the two contracted variables; the tensor-product pairings then agree in either order by [F8]. Thus the within-first, within-last, and cross-block sums commute. Their first two exponentials give $T_p^H(V,\dots ,V)$ and $T_q^H(U,\dots ,U)$. By the finite product rule on the block coefficients, the $r$-th term of the cross-block exponential is ${\eta }^r\mathrm{/}r!$ times the independent $r$-pair contraction of the $r$-th derivative kernels of these two block outputs. After replacing the cross kernels by $W_{\mu }$ using step 3.1, this is exactly the finite star-product formula [F5]. Every remaining field variable comes from an original coefficient kernel, so within-block contractions retain support in $A$ or $B$. Hence $T_{p+q}^H(\underset{p\text{ inputs}}{\underbrace{V,\dots ,V}},\underset{q\text{ inputs}}{\underbrace{U,\dots ,U}})=T_p^H(V,\dots ,V){\star }_{W_{\mu }}{T}_q^H(U,\dots ,U)\mathrm{.}$ For $p=0$ or $q=0$, the empty block is $T_0^H=1$ and the same identity follows directly.

In the degree-$n$ coefficient of $S_z^H(V+U)$, symmetry of $T_n^H$ groups terms with $p$ copies of $V$ and $q$ copies of $U$, $p+q=n$, with factor $\left(\genfrac{}{}{0px}{}{n}{p}\right)$. Dividing by $n!$ gives $1\mathrm{/}(p!q!)$. Substitution of step 4.1 therefore gives exactly the degree-$n$ coefficient of $S_z^H(V){\star }_{W_{\mu }}{S}_z^H(U)$. Since $n$ was arbitrary, the causal-factorization identity holds coefficientwise, with no convergence in $z$.

If $A,B$ are pairwise spacelike separated, step 1.1 gives ${\tilde{W}}_{\mu }={\tilde{W}}_{\mu }^{\mathsf{T}}$ on $A\times B$. Swapping the two inputs in the finite ${\star }_{W_{\mu }}$ formula reverses each cross pairing and changes no within-input term, so $F{\star }_{W_{\mu }}G=G{\star }_{W_{\mu }}F$. Applying this coefficientwise to the formal series proves their commutation.