Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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The Grothendieck ring of a tensor category is always commutative

Statement

False claim. The Grothendieck ring of a tensor category is always commutative.

Facts & Assumptions

Given: A tensor category.

[F1]

Duality gives an anti-isomorphism, reversing product order (Duality induces an anti-isomorphism on the Grothendieck ring).

Refutation

technique · direct
1.1givenconstruct

Let G=S3 and let VecG be the category of finite-dimensional G-graded vector spaces. Its simple objects δg are indexed by g∈G, and δg⊗δh≅δgh; it is a tensor category.

2.1step 1.1F1∎

Thus K0(VecG)≅Z[G]. Since S3 is nonabelian, for example (12)(23)≠(23)(12), this ring is not commutative. The order reversal in [F1] is consistent with, but does not remove, this counterexample.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources