Proposition: Basis independence and graph factor

Equation (eq:Kdef) is independent of the basis of $V$. It is the contraction of the source polarization and vertex Hessian by the two fiber edge tensors (eq:edgefiber).

For the ordinary, non-divided second derivatives used here, its numerical graph factor is $2/2!=1$.

Proof. Applying the source bilinear map to the first vectors of two copies of $\mathsf{E}$, and the vertex bilinear form to their second vectors, gives exactly (eq:Kdef).

For an arbitrary invertible basis change ${\tilde{B}}_i=B_j{A}_{ji}$,

The two inverse matrices from the dual vectors cancel the two matrices from the Hessian. The double sum is unchanged. Equivalently, the fiber tensor $\mathsf{E}$ is intrinsically defined.

The second tensor derivative of the quadratic insertion is $2\mathfrak{s}$, whereas the vertex derivative is the ordinary Hessian $L_h$. The second term of the cross-contraction exponential has coefficient $1/2!$. Thus the product of these two numerical factors is one. No additional division for symmetric spacetime indices is made: the Gram inverse already accounts for the independent components. The identical-edge convention is the one displayed in FR (Eq. (82)).