Proposition: Two non-tadpole graph types

Under Assumption ass:graphs, the connected ${\kappa }^2\hslash$ coefficient of (eq:fullcoefficient) has exactly two contraction types:

1. ${\mathsf{L}}_{\alpha }$: two lines between the $h\varphi$ part of $O_1$ and one $h\varphi \varphi$ vertex;

2. ${\mathsf{L}}_{\beta }$: the $F_f$ scalar line attaches to one $h\varphi \varphi$ vertex, and two lines, one metric and one scalar, join that vertex to a second $h\varphi \varphi$ vertex.

Each type has one external scalar leg. Its unextended flat contraction coefficient contains no $h_{\mathrm{cl}},k_{\mathrm{cl}}$, ghost field, gravitational cubic vertex, or background gauge-fixing coefficient.

There are no connected contraction terms at ${\kappa }^2{\hslash }^r$ with $r\ge 2$ under these assumptions.

Proof. For $n$ interaction vertices and $E$ contraction lines, the quantum power is $E-n$. Connectedness requires $E\ge n$.

For $n=0$, the insertion is normal ordered, so no quantum contraction occurs.

For $n=1$, the order-one insertion has either one scalar leg or one scalar and one metric leg. Quantum order one requires $E=2$. Both lines must end on the two-leg insertion. The only order-one vertex with the required species is $h\varphi \varphi$. It leaves one scalar leg. A single $V_2$ attached to the one-leg $F_f$ has at most one line and cannot contribute at quantum order one.

For $n=2$, the insertion is $F_f$ and has degree one. Quantum order one requires $E=3$. One line attaches $F_f$ to a scalar-bearing vertex; the other two lines must run between the two interaction vertices. The vertex attached to $F_f$ must therefore have two remaining legs. The background $\varphi \varphi$ vertex does not. The $h\varphi \varphi$ vertex does, leaving $h\varphi$; the only second vertex matching both species is another $h\varphi \varphi$ vertex. Again one scalar leg remains.

The same degree bounds give $E\le 2$ for $n=1$ and $E\le 3$ for $n=2$. Thus $E-n\le 1$. Orders $n\ge 3$ require at least ${\kappa }^3$.