PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Quartic graph superficial degree from exact counting

Statement

Let $D \in Z_{\geq 1}$ and let $Γ$ be a finite quartic graph in the sense of Finite quartic graph and superficial scaling conventions. With $V$ vertices, $I$ internal edges, $E$ external half-edges, and loop count $L$ , the incidence and loop-count identities are $4 V = 2 I + E$ and $L = I - V + 1$ . Its superficial degree under the stated Euclidean scalar power-counting convention is $ω_{D} (Γ) = D L - 2 I = D + (D - 4) V - \frac{D - 2}{2} E$ . In $D = 4$ this reduces to $ω_{4} (Γ) = 4 - E$ . A zero value is a logarithmic superficial degree by the convention in the definition; these identities do not assert convergence or divergence of an actual graph integral.

Facts & Assumptions

Given: $D \in Z_{\geq 1}$ and a finite connected multigraph $Γ$ with $V \geq 1$ four-slot vertices, $I$ internal edges, and $E$ external half-edges, using degree $D$ per loop integration and degree $- 2$ per internal scalar propagator.

[F1]

Finite quartic graph and superficial scaling conventions: the internal-edge ends and the attached external half-edges are in bijection with the four incidence slots at each vertex; $L : = I - V + 1$ is the cycle rank; one formal loop integration contributes degree $D$ , each internal scalar propagator contributes degree $- 2$ , and $ω_{D} (Γ) : = D L - 2 I$ . The value $ω_{D} = 0$ is called logarithmic superficial degree and is only formal graph bookkeeping.

Proof

Given: The graph and power-counting data in [F1].

1.1givenF1algebra

Counting the four slots at every vertex gives $4 V$ slots. Each of the $I$ internal edges occupies two slots, including a self-loop, and each of the $E$ external half-edges occupies one. Hence $4 V = 2 I + E$ .

1.2givenF1construct

Start with one vertex and repeatedly add an internal edge from the reached vertices to a vertex not yet reached. Connectedness provides such an edge until all $V$ vertices are reached. Each added edge introduces one new vertex, so the selected edges form a spanning tree with $V - 1$ edges; for $V = 1$ it is empty. Exactly $I - (V - 1) = I - V + 1 = L$ internal edges lie outside this tree, including any unselected loops or parallel edges, so the cycle rank in [F1] is nonnegative and counts the non-tree internal edges.

2.1F1step 1.1step 1.2algebra

By [F1], $ω_{D} = D L - 2 I$ . Substitute $L = I - V + 1$ from step 1.2 to obtain $ω_{D} = (D - 2) I - D V + D$ . From step 1.1, $I = 2 V - E / 2$ , so $ω_{D} = (D - 2) (2 V - E / 2) - D V + D = D + (D - 4) V - \frac{D - 2}{2} E$ .

3.1F1step 2.1algebra∎

Setting $D = 4$ in step 2.1 gives $ω_{4} = 4 - E$ . The defining convention in [F1] calls its zero value logarithmic superficial degree; it makes no claim about an actual integral's convergence or divergence.

Depends on

Finite quartic graph and superficial scaling conventions

Quartic graph superficial degree from exact counting

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources