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A module and comodule need not satisfy Hopf compatibility

Statement refuted

Every right $H$ -module and right $H$ -comodule on the same vector space automatically form a right-right Hopf module.

Facts & Assumptions

Given: A field $k$ , the group algebra $H = k C_{2}$ with basis $1, g$ and $g^{2} = 1$ , and the one-dimensional vector space $M = k m$ . Set $Δ (g) = g \otimes g$ , $ε (g) = 1$ , and $S (g) = g$ . Give $M$ the right action $m \cdot h = ε (h) m$ and right coaction $ρ (m) = m \otimes g$ .

[F1]

Hopf-module compatibility would require $ρ (m \cdot h) = \sum m_{0} \cdot h_{1} \otimes m_{1} h_{2}$ (Right-right Hopf modules and coinvariants).

[F2]

In $k C_{2}$ , $1$ and $g$ are distinct basis elements, $g^{2} = 1$ , and $Δ (g) = g \otimes g$ with $ε (g) = 1$ ; these maps satisfy the Hopf axioms as verified below.

Proof

1.1F2algebra

The multiplication on $H$ is the group multiplication on $C_{2}$ extended bilinearly, hence associative and unital. The maps $Δ, ε$ are unital algebra maps because $Δ (g^{2}) = 1 \otimes 1 = (g \otimes g)^{2}$ and $ε (g^{2}) = 1 = ε (g)^{2}$ . Coassociativity and counitality hold on the basis $1, g$ since both are group-like, and $S (1) = 1, S (g) = g$ satisfies $g S (g) = 1 = S (g) g$ and the same identity on $1$ . Thus $H$ is a Hopf algebra over $k$ , in every characteristic.

1.2F2algebra

The rule $m \cdot h = ε (h) m$ is a unital right module action because $ε$ is a unital algebra map: $(m \cdot a) \cdot b = ε (a) ε (b) m = ε (a b) m = m \cdot (a b)$ . The map $ρ (m) = m \otimes g$ is a right coaction because $Δ (g) = g \otimes g$ and $ε (g) = 1$ . Therefore each structure separately satisfies its axioms.

2.1F1F2step 1.2algebra∎

For $h = g$ , the left side of [F1] is $ρ (m \cdot g) = ρ (m) = m \otimes g$ , while the right side is $(m \cdot g) \otimes g^{2} = m \otimes 1$ . Since $m \neq 0$ and $1, g$ are linearly independent, these tensors differ. Thus the two structures do not satisfy Hopf-module compatibility, refuting the assertion.

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