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Rooted forests, Hopf antipodes and subtraction recursion — Examples

1 · Prerequisites

Rooted forests, Hopf antipodes and subtraction recursion

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Nested subdivergence has a nontrivial convolution counterterm

Example

Let $K = Q$ and let $ϵ$ be formal. Let $τ$ be the one-vertex rooted tree and $t$ the two-vertex ladder with one admissible edge cut, so $P_{C} (t) = R_{C} (t) = τ$ . Define a character (a unital algebra homomorphism) $ϕ$ by $ϕ (τ) = a ϵ^{- 1} + b$ and $ϕ (t) = c ϵ^{- 2} + d ϵ^{- 1} + e$ , for $a, b, c, d, e \in Q$ ; assign zero to other tree generators and extend multiplicatively to forests. With $R$ the strictly negative-power projection, the subtraction recursion gives $ϕ_{-} (t) = - R (ϕ (t) + ϕ_{-} (τ) ϕ (τ)) = \frac{a^{2} - c}{ϵ^{2}} + \frac{a b - d}{ϵ} .$ The preparation term includes $- a^{2} ϵ^{- 2} - a b ϵ^{- 1}$ ; subtracting only $R ϕ (t)$ therefore generally gives a different counterterm.

The diagram places the root above its child, with the arrow pointing away from the root. The dashed edge is the chosen admissible cut; the root-side trunk is $R_{C} (t) = τ$ and the child-side pruned forest is $P_{C} (t) = τ$ .

Example references

Admissible-cut Hopf algebra for subdivergence forests

Laurent-series target and pole projection for subtraction

Laurent subtraction factors a character into counterterm and finite characters

Facts & Assumptions

Given: The rooted-tree cut convention, Laurent projection, and character factorization cited below.

[F1]

Admissible-cut Hopf algebra for subdivergence forests: $Δ (t) = t \otimes 1 + \sum_{C admissible} P_{C} (t) \otimes R_{C} (t)$ , where the sum includes the empty edge cut and each $P_{C} (t)$ is the child-side forest while $R_{C} (t)$ is the root-side tree.

[F2]

Laurent-series target and pole projection for subtraction: $R$ retains exactly the strictly negative powers of $ϵ$ and kills every nonnegative power.

[F3]

Laurent subtraction factors a character into counterterm and finite characters: for positive-degree $h$ , $ϕ_{-} (h) = - R (ϕ (h) + \sum_{\tilde{Δ} (h)} ϕ_{-} (h_{(1)}) ϕ (h_{(2)}))$ , and $ϕ_{-} (1) = 1$ .

Verification

Given: The character values in the Example, the one-edge ladder $t$ , and [F1]–[F3].

1.1F1F2F3construct

The only nonempty admissible cut of the two-vertex ladder contributes $τ \otimes τ$ to the reduced coproduct, while $τ$ itself has no edge cut. Therefore [F3] gives $ϕ_{-} (τ) = - R ϕ (τ) = - a ϵ^{- 1}$ .

2.1F2F3step 1.1algebra

Substituting this value gives $ϕ_{-} (τ) ϕ (τ) = (- a ϵ^{- 1}) (a ϵ^{- 1} + b) = - a^{2} ϵ^{- 2} - a b ϵ^{- 1}$ . Hence the preparation is $(c - a^{2}) ϵ^{- 2} + (d - a b) ϵ^{- 1} + e$ , and applying $- R$ yields $ϕ_{-} (t) = (a^{2} - c) ϵ^{- 2} + (a b - d) ϵ^{- 1}$ .

3.1F2step 2.1algebra∎

The primitive-only counterterm is $- R ϕ (t) = - c ϵ^{- 2} - d ϵ^{- 1}$ , so the recursive counterterm differs by $a^{2} ϵ^{- 2} + a b ϵ^{- 1}$ . For $a = 1$ and $b = 0$ this difference is $ϵ^{- 2} \neq 0$ in $Q ((ϵ))$ , proving that the primitive-only subtraction generally fails in this nested example.