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Quantum Groups at Roots of Unity and Tilting Quotients: Examples and Counterexamples

1 · Prerequisites

2 · Summary

This companion checks the A-page conventions in two rank-one cases. The companion A page proves the higher-rank Frobenius and Verlinde results.

Companion: quantum-groups-at-roots-of-unity-and-tilting-quotients.

Worked example

At a primitive fifth root $ζ$ , the rank-one root-power quotient has basis ${{\overline{F}}^{a} {\overline{K}}^{b} {\overline{E}}^{c} : 0 \leq a, b, c < 5}$ and therefore dimension $5^{3} = 125$ . The calculation is the specialization of the A-page basis theorem, not a general higher-rank dimension formula.

Counterexample

The identity-on-generators map from the rank-one De Concini–Kac specialization to the Lusztig divided-power specialization is not injective at order five: the source element $E^{5}$ is nonzero by its PBW basis and its image vanishes by $E^{5} = [5]_{ζ}! E^{(5)} = 0$ . This refutes preservation of injectivity under this specialization.

The source and calculation details are recorded in the batch-68 coverage ledger.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The rank-one small quantum group at a primitive fifth root has dimension 125

Example

Let $ζ$ be a primitive fifth root of unity and let $u_{ζ} ({s l}_{2})$ be the rank-one root-power quotient of De Concini–Kac, Lusztig, and small quantum-group forms at roots of unity. Its generators satisfy $E^{5} = F^{5} = 0$ and $K^{5} = 1$ .

Facts & Assumptions

Given: A primitive fifth root $ζ$ and the rank-one quotient $u_{ζ} ({s l}_{2})$ .

[F1]

For every odd $ℓ > 2$ , the quotient has basis ${{\overline{F}}^{a} {\overline{K}}^{b} {\overline{E}}^{c} : 0 \leq a, b, c < ℓ}$ and dimension $ℓ^{3}$ (The rank-one root-power quotient is a finite-dimensional Hopf algebra).

Verification

1.1givenF1algebra

Since $ζ$ has order five, $ζ^{5} = 1$ and $ζ^{2}$ also has order five. In particular $ζ - ζ^{- 1} \neq 0$ , so the rank-one root forms and the quotient in [F1] are defined at this parameter.

2.1givenF1step 1.1algebra∎

Substitute $ℓ = 5$ in the basis assertion [F1]. The exponents $a, b, c$ each range independently over ${0, 1, 2, 3, 4}$ , so this basis has $5 \cdot 5 \cdot 5 = 125$ elements. Hence $\dim_{C} u_{ζ} ({s l}_{2}) = 125$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

At order five, the rank-one form specialization map is not injective

Statement refuted

For every odd root order $ℓ > 2$ , the natural generator map $U_{ζ}^{D K} ({s l}_{2}) \to U_{ζ}^{L} ({s l}_{2})$ is injective.

Facts & Assumptions

Given: The asserted injectivity for all odd $ℓ > 2$ , and the rank-one root forms at a primitive fifth root.

[F1]

The identity-on-generators comparison map is well-defined and kills $E^{ℓ}$ although $E^{ℓ}$ is nonzero in its source, for each odd $ℓ > 2$ (The rank-one De Concini–Kac to Lusztig specialization map is not injective).

Counterexample

1.1givenF1algebra

The value $ℓ = 5$ is an odd integer greater than $2$ , and a primitive fifth root satisfies the root-order hypotheses in the refuted statement.

2.1givenF1step 1.1algebra∎

Apply [F1] at $ℓ = 5$ : the source element $E^{5}$ is nonzero and its image is zero. Therefore the map is not injective, contradicting the asserted universal injectivity.

Sources

Pavel Etingof and Mykola Semenyakin, A Brief Introduction to Quantum Groups