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Spinor helicity, rational factorization and tree recursion — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A constant contact amplitude needs the boundary term

Statement refuted

The boundary-free finite-pole formula $A (0) = - \sum_{α} {Res}_{z = z_{α}} (A (z) / z)$ applies to every rational shifted function with only simple finite poles, even when $A (z)$ does not vanish at infinity.

Statement references

Tree recursion under an explicit vanishing-boundary hypothesis

Elvang and Huang, Scattering Amplitudes, §3.1

Marzolla, The four-dimensional on-shell three-point amplitude in spinor-helicity formalism and BCFW recursion relations, §5

Britto, Cachazo, Feng, and Witten, Direct Proof of Tree-Level Recursion Relation in Yang-Mills Theory, §§2–3.1

Facts & Assumptions

Given: The proposed rule omits the boundary term and a nonzero constant $c \in C^{\times}$ supplies a formal scalar four-point contact coefficient $A (z) = c$ . This is a rational coefficient example, not a claim about a complete interacting theory.

[F1]

Tree recursion under an explicit vanishing-boundary hypothesis: the finite-pole identity is proved only under the additional hypothesis $A (z) \to 0$ as $z \to \infty$ ; the pole sum is empty when there are no finite poles.

[A1]

Fix $c \in C^{\times}$ and set $A (z) = c$ for all $z \in C$ . Here a formal scalar four-point contact coefficient means only this shift-independent constant contribution; no full field-theory model is assumed.

[A2]

For a rational function $f$ , define ${Res}_{z = \infty} f$ as minus the coefficient of $z^{- 1}$ in its Laurent expansion at infinity; the boundary correction in the residue identity is $- {Res}_{z = \infty} (A (z) / z)$ .

Counterexample

1.1givenF1A1algebra

The constant $A (z) = c$ is rational, regular at $0$ , and has no finite poles. Hence every finite-pole residue vanishes and the proposed boundary-free sum equals $0$ .

2.1givenA1A2step 1.1algebra∎

But $A (0) = c \neq 0$ . At infinity $A (z) / z = c / z$ , so by [A2] the residue at infinity is $- c$ and the boundary correction is $c$ . Omitting it gives the contradiction.

Remarks

The cited §2.3 example is the derivative-free $λ ∣ ϕ ∣^{4}$ interaction, whose four-scalar contact amplitude is constant. This counterexample assumes only the explicitly stipulated shift-independent coefficient $A (z) = c$ ; it makes no claim that contact terms with derivatives have constant shifted coefficients.