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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-08
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A connected graded bialgebra has a unique antipode, given by the reduced-coproduct recursion

Statement

Let k be a commutative ring and let H=⨁n≥0Hn be a connected graded bialgebra over k (Graded coalgebras, bialgebras and Hopf algebras over a commutative ring), with multiplication m, unit u, comultiplication Δ and counit ε, so that u:k→∼H0. Then there is a unique k-linear map S:H→H satisfying

m(S⊗id)Δ=uε=m(id⊗S)Δ.

Thus H is a graded Hopf algebra with antipode S. The map preserves the grading, S(Hn)⊆Hn. For every homogeneous x∈Hn with n≥1, the reduced coproduct

Δ~(x):=Δ(x)−x⊗1H−1H⊗x

lies in ⨁i=1n−1Hi⊗kHn−i, and the recursion is

S(x)=−x−m(S⊗id)Δ~(x)=−x−m(id⊗S)Δ~(x).

Equivalently, if Δ~(x)=∑ixi′⊗xi′′, then S(x)=−x−∑iS(xi′)xi′′=−x−∑ixi′S(xi′′); this value is independent of the finite tensor expression used. No choice principle is used.

Facts & Assumptions

Given: A commutative ring k and a connected graded bialgebra H over k with the structure maps in the statement.

[F1]

The multiplication is associative and unital; Δ and ε are unital algebra maps; Δ is degree-zero and coassociative; ε vanishes in positive degrees and satisfies both counit identities; and connectedness means u:k≅H0 (Graded coalgebras, bialgebras and Hopf algebras over a commutative ring).

[F2]

Every tensor is a finite sum of elementary tensors, with the defining additivity and balance relations (The tensor product M⊗RN from the additive group underlying the free Z-module on M×N, elementary tensors, and finite tensor sums).

[F3]

A balanced bilinear map induces a unique homomorphism from the module tensor product, so maps on tensor factors defined on elementary tensors are well defined (Universal property of the tensor product for balanced maps into abelian groups).

[F4]

The canonical tensor associator rebrackets (M⊗RN)⊗SP as M⊗R(N⊗SP) (Associativity of tensor products for compatible bimodules).

[F5]

The convolution product f∗g=m(f⊗g)Δ on Hom⁡k(H,H) is associative with unit uε (Graded coalgebras, bialgebras and Hopf algebras over a commutative ring).

Proof

technique · direct
1.1F1givenalgebra

If x∈Hn with n>0, gradedness places Δ(x) in ⨁i+j=nHi⊗kHj. Since ε vanishes in positive degree and H0=k1H, the two counit identities force the bidegree (0,n) and (n,0) components to be 1H⊗x and x⊗1H. Hence Δ~(x)∈⨁i=1n−1Hi⊗kHn−i, with Δ~(x)=0 when n=1; for c∈k, Δ(u(c))=u(c)⊗1H and ε(u(c))=c.

1.2F1F3F4F5algebra

For f,g,h∈Hom⁡k(H,H), the convolution bracketings are (f∗g)∗h=m(m⊗id)(f⊗g⊗h)(Δ⊗id)Δ and f∗(g∗h)=m(id⊗m)(f⊗g⊗h)(id⊗Δ)Δ. Under the canonical rebracketing [F4], coassociativity of Δ and associativity of m identify these maps. Hence convolution is associative, and its unit is e:=uε by [F5].

2.1F1F2F3step 1.1construct

Set Sℓ∣H0=Sr∣H0=idH0. Inductively, once both maps are defined on ⨁j<nHj, define for x∈Hn by Sℓ(x):=−x−m(Sℓ,<n⊗id)Δ~(x) and Sr(x):=−x−m(id⊗Sr,<n)Δ~(x), where Sℓ,<n and Sr,<n are their restrictions to ⨁j<nHj. By step 1.1 both tensor factors of Δ~(x) have degree below n, so [F3] makes the displayed maps well defined on the tensor element itself, independently of any chosen finite expression as elementary tensors; they are k-linear in x and extend to Hn. Induction over n therefore defines k-linear maps Sℓ,Sr:H→H.

3.1F1step 1.1step 2.1algebra

For homogeneous x∈Hn with n>0, step 2.1 gives m(Sℓ⊗id)Δ(x)=Sℓ(x)+x+m(Sℓ,<n⊗id)Δ~(x)=0=uε(x). For x=u(c)∈H0, Sℓ is the identity and Δ(u(c))=u(c)⊗1H, so the same convolution identity equals uε(x). By linearity, Sℓ∗id=uε on H.

3.2F1step 1.1step 2.1algebra

For homogeneous x∈Hn with n>0, the right recursion in step 2.1 gives m(id⊗Sr)Δ(x)=x+Sr(x)+m(id⊗Sr,<n)Δ~(x)=0=uε(x). The identity holds on H0 because Sr∣H0=idH0 and Δ(u(c))=u(c)⊗1H. Thus id∗Sr=uε on all of H.

4.1F5step 3.1step 3.2step 1.2algebra

By steps 3.1 and 3.2, Sℓ∗id=e=id∗Sr. Associativity and the unit from step 1.2 yield Sℓ=Sℓ∗e=Sℓ∗(id∗Sr)=(Sℓ∗id)∗Sr=e∗Sr=Sr. Their common value S is therefore a two-sided convolution inverse of id, so it satisfies both displayed antipode identities.

5.1F5step 1.2step 4.1algebra

If T is any other antipode, then T∗id=e=id∗S. Associativity and the unit imply T=T∗e=T∗(id∗S)=(T∗id)∗S=e∗S=S, so the antipode is unique.

6.1F1step 1.1step 2.1step 4.1algebra∎

The recursion preserves degree: if x∈Hn, step 1.1 places every reduced-coproduct term in Hi⊗Hn−i with 1≤i<n; induction gives S(Hi)⊆Hi, and the graded multiplication then puts every recursive product in Hn. The base case is S∣H0=id. The construction uses induction on n and canonical maps on Δ~(x), never selected tensor representatives; thus no form of the axiom of choice is used. This proves the graded antipode claim.

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