Theorem: Non-extension by the present product tools

1. The square ${\Delta }_{\mathrm{F}}\cdot {\Delta }_{\mathrm{F}}$ fails the ordinary wavefront product criterion at every point $(x,x)$ of the diagonal.

2. The corresponding Wick square ${\Delta }_{+}\cdot {\Delta }_{+}$, and its finite differentiated products, pass that criterion.

3. A general quadratic–quadratic time-ordered contraction requires an identified-endpoint product of two Feynman lines, with the vertex differential operators applied. The undifferentiated case is exactly ${\Delta }_{\mathrm{F}}\cdot {\Delta }_{\mathrm{F}}$. The present file selects neither a diagonal extension nor its normalization.

4. Consequently no value of $T_2(V,V)$ for the required general quadratic sector, no $R(V;V)$, and no general time-ordered family is released by the one-contraction check.

This is not a theorem that renormalized extensions or time-ordered families do not exist.

Proof. At $z=(x,x)$, Lemma lem:DF gives both

in $\mathrm{WF}({\Delta }_{\mathrm{F}})$, for every $\xi \ne 0$. They sum to zero. This is precisely the failure of the sufficient product criterion Hormander (Theorem 8.2.10).

For ${\Delta }_{+}$, opposite wavefront covectors would require a nonzero first-slot covector and its negative both in the same fixed half null cone. This is impossible. Differentiation cannot enlarge the wavefront set, so the same argument handles its finite differentiated products.

For the general quadratic pair the formal two-contraction term is

The kernels of the local differential Hessians have the form $-L_x\delta (x-x^{\prime })$. Contracting them identifies the two first endpoints and the two second endpoints of the tensor product of Feynman lines. Before differentiation, that operation is the pullback of

to $x_1=x_2,y_1=y_2$.

For an order-zero compact coefficient $L=\lambda (x)$, which is a test case for a proposed general quadratic rule, the expression is literally the weighted square $\lambda (x)\lambda (y){\Delta }_{\mathrm{F}}(x,y{)}^2$. Thus a general prescription cannot be obtained from the present unextended product criterion. Differential vertices require separately justified differentiated versions of the same endpoint operation; their formal notation supplies no such theorem.

Away from the diagonal the Hadamard Feynman wavefront orientation permits the usual square: for positive time separation the kernel agrees locally with ${\Delta }_{+}$, for negative separation with ${\Delta }_{+}^{\mathrm{T}}$, and at equal-time separated points the relevant distributions are smooth. What is missing is the diagonal extension and a normalization for it. Those are additional renormalization data Radzikowski,BF. None is chosen here.

The original quadratic–linear pair avoids the two-line operation because ${F_c}^{(2)}=0$. Its successful check therefore supplies no value for the missing quadratic sector and no family.