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The finite dual is a right regular Hopf module

Statement

Let H be a finite-dimensional Hopf algebra over a field k. For f∈H∗ and h,x∈H, define

(f↼h)(x)=f(xS(h)).

For f∈H∗, define ρ(f)=∑f0⊗f1∈H∗⊗H by the requirement

∑p(f1)f0(x)=∑p(x1)f(x2)(p∈H∗, x∈H).

This requirement determines a unique tensor. With this right action and right coaction, H∗ is a right-right Hopf module. Its coinvariants are exactly the left integral functionals: (H∗)co⁡H=∫H∗ℓ.

Facts & Assumptions

Given: A finite-dimensional Hopf algebra H over k; its dual Hopf algebra H∗; f∈H∗; and h,x∈H, p,q∈H∗.

[F1]

Left and right integrals and integral functionals: left integral functionals satisfy p∗f=p(1)f for every p∈H∗, with convolution dual to ΔH.

[F2]

Right-right Hopf modules and coinvariants: a right-right Hopf module has a right H-action and right H-coaction satisfying ρ(mh)=∑m0h1⊗m1h2; its coinvariants satisfy ρ(m)=m⊗1.

[F3]

Finite tensor duality and basis-independent coevaluation: for finite-dimensional spaces, tensor duality identifies bilinear pairings with unique tensors, and the finite dual basis separates tensors.

[F4]

Uniqueness and anti-multiplicativity and anti-comultiplicativity of the antipode: the antipode satisfies S(1)=1, S(hg)=S(g)S(h), and Δ(S(h))=∑S(h2)⊗S(h1). The antipode equations in Hopf algebras and antipodes give ∑S(h1)h2=∑h1S(h2)=ε(h)1.

[F5]

The dual of a finite-dimensional Hopf algebra is a Hopf algebra: since H is finite-dimensional, its full linear dual has the transposed Hopf structure, with convolution product and counit f↦f(1).

Proof

1.1givenF3algebra

Fix a basis (ei) of H with dual basis (ei∗). The tensor ρ(f)=∑iei∗⊗(∑(ei)1f((ei)2)) has the required pairing, since x=∑iei∗(x)ei. Uniqueness follows from finite tensor separation [F3]. The assignment f↦ρ(f) is linear because its defining pairing is linear in f and the representing tensor is unique.

1.2givenF2F3F4algebra

The formula defines a right action: ((f↼h)↼g)(x)=f(xS(g)S(h))=f(xS(hg))=(f↼(hg))(x) by anti-multiplicativity [F4], and f↼1=f. Pair each side of ρ(f↼h)=∑(f0↼h1)⊗f1h2 with x,p. The left side gives ∑p(x1)f(x2S(h)). The right side gives ∑p1(x1S(h2))p2(h3)f(x2S(h1))=∑p(x1S(h2)h3)f(x2S(h1))=∑p(x1)f(x2S(h)), because Δp=∑p1⊗p2 is dual to multiplication, Δ(xS(h1))=∑x1S(h2)⊗x2S(h1) by anti-comultiplicativity [F4], and ∑S(h2)h3=ε(h2)1. The pairings agree; finite tensor separation [F3] proves compatibility.

2.1step 1.1F3F5algebra

The counit identity follows by evaluating ρ(f) against p=ε: ∑ε(f1)f0(x)=∑ε(x1)f(x2)=f(x). For coassociativity, pair both (ρ⊗id)ρ(f) and (id⊗Δ)ρ(f) with x,p,q. The first pairing is (p∗(q∗f))(x) by the defining equation twice; the second is ((p∗q)∗f)(x). Convolution is associative because Δ is coassociative, so the pairings agree for all x,p,q, and finite tensor separation [F3] gives equality of the tensors.

3.1step 1.1F1F5algebra∎

The coaction coinvariant condition is equivalent, after pairing with every p,x, to (p∗f)(x)=p(1)f(x). Since εH∗(p)=p(1), this is exactly the definition of a left integral functional in [F1]; hence (H∗)co⁡H=∫H∗ℓ.

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