CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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An infinite-dimensional dual algebra with a nonrational module

Statement refuted

For every infinite-dimensional coalgebra $C$ , every unital left $C^{*}$ -module is rational and is induced by a right $C$ -comodule under the evaluation action $f ▹ m = (id \otimes f) ρ (m)$ .

Example

Let $C = ⨁_{n \in N} k e_{n}$ with $Δ (e_{n}) = e_{n} \otimes e_{n}$ and $ε (e_{n}) = 1$ . Then $C$ is infinite-dimensional and its full dual $C^{*}$ is the space of all sequences $(a_{n})_{n \in N}$ with pointwise multiplication. Its regular left module $M = C^{*}$ is not rational and cannot be induced by a right $C$ -coaction.

Facts & Assumptions

Given: A field $k$ , the direct-sum vector space $C = ⨁_{n \in N} k e_{n}$ , its displayed structure maps, and the full algebraic dual.

[F1]

A coalgebra has coassociative $Δ$ and a two-sided counit $ε$ ; a coalgebra map preserves both structure maps. This item checks the axioms for its displayed example. (Coalgebra and coalgebra map over a field)

[F2]

$C^{*} = {Hom}_{k} (C, k)$ has candidate convolution $(f * g) (c) = (f \otimes g) Δ (c)$ , unit $ε$ , and rationality means each vector orbit has a finite tensor representative in $M \otimes C$ . (The dual convolution algebra and rational modules)

[F3]

A unital left module has an associative action and the unit acts as identity. (Unital left and right modules over a ring; unqualified module means left module)

[F4]

The full algebraic dual is the space of all linear functionals, with no support restriction on their values on the basis of a direct sum. (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ )

[F5]

In this example, “induced by a right coaction” means that some coaction $ρ : M \to M \otimes C$ gives the action by the evaluation formula $f ▹ m = ({id}_{M} \otimes f) ρ (m)$ , using the right-comodule and dual-module conventions in Right comodules, comodule maps, and subcomodules and The dual convolution algebra and rational modules.

[F6]

A vector space is infinite-dimensional exactly when it has no finite basis. If a space has a finite spanning set, it cannot contain a linearly independent subset equinumerous with $N$ . (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ )

Refutation

Given: The coalgebra, dual algebra, regular module, and evaluation action specified above.

1.1givenF1construct

Every vector of $C$ has finite support. On each basis vector, $(Δ \otimes id) Δ (e_{n}) = e_{n} \otimes e_{n} \otimes e_{n} = (id \otimes Δ) Δ (e_{n})$ and $(ε \otimes id) Δ (e_{n}) = e_{n} = (id \otimes ε) Δ (e_{n})$ . The structure maps are linear, so these equalities extend to $C$ and make the displayed structure a coalgebra.

2.1F2F4F6step 1.1algebra

A functional $f \in C^{*}$ is determined by the sequence $a_{n} = f (e_{n})$ , and every sequence occurs: for a finite-support vector $\sum_{n} λ_{n} e_{n}$ , define $f (\sum_{n} λ_{n} e_{n}) = \sum_{n} λ_{n} a_{n}$ . The convolution satisfies $(f * g) (e_{n}) = f (e_{n}) g (e_{n})$ , so under this identification $C^{*}$ is the pointwise algebra $k^{N}$ with unit $ε = (1, 1, \dots)$ . The basis vectors $e_{n}$ are linearly independent in $C$ , and the coordinate functionals $δ_{n}$ are linearly independent in $C^{*}$ because evaluating a finite linear relation at $e_{j}$ recovers its $j$ th coefficient. Both independent sets are equinumerous with $N$ . By [F6], neither $C$ nor $C^{*}$ has a finite spanning set, hence neither has a finite basis; both are infinite-dimensional by [F6].

3.1F2F3step 2.1construct

The regular action $f ▹ m : = f * m$ makes $M = C^{*}$ a unital left module because the pointwise product is associative and $ε$ is its unit. At its unit vector $m = ε$ , the orbit map is $C^{*} \to C^{*}$ , $f \mapsto f ▹ ε = f$ . Its image is all of the infinite-dimensional space $C^{*}$ . If $ε$ were rational, some finite tensor $t = \sum_{i = 1}^{r} m_{i} \otimes c_{i}$ would represent this orbit, so every $f ▹ ε = (id \otimes f) t$ would lie in the finite-dimensional span of $m_{1}, \dots, m_{r}$ , a contradiction. Thus $M$ is not rational.

4.1F2F3F4F5step 2.1step 3.1

If a right coaction $ρ : M \to M \otimes C$ induced the regular action, write its finite tensor $ρ (ε) = \sum_{i = 1}^{r} m_{i} \otimes c_{i}$ . The defining evaluation formula [F5] would give $f = f ▹ ε = \sum_{i = 1}^{r} f (c_{i}) m_{i}$ for every $f \in C^{*}$ , again forcing $C^{*}$ into the finite-dimensional span of the $m_{i}$ , contrary to step 2.1. Hence no such coaction induces this action.

5.1step 1.1step 2.1step 3.1step 4.1∎

Step 1.1 verifies the coalgebra, step 2.1 identifies its full dual algebra, step 3.1 disproves rationality, and step 4.1 rules out an inducing coaction.

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Sources

Daniel Kongsgaard and UCSD Math 207A, Hopf Algebras Lecture Notes, Spring 2020 (standard reference, not scraped)
Andrew Baker, Hopf algebras over fields: an introduction, SMSTC supplementary course notes 2026–7 (standard reference, not scraped)