LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Regular tensor-algebra multiplication is detected at the unit tensor

Statement

Let $A$ be a unital algebra over a field $k$ . For $n = 2, 3$ , give $A^{\otimes n}$ its iterated componentwise tensor-product algebra structure and unit $1_{A}^{\otimes n}$ . If $X, Y \in A^{\otimes n}$ and left multiplication by $X$ and $Y$ agree as maps $A^{\otimes n} \to A^{\otimes n}$ , then $X = Y$ . In particular, evaluation of a left-multiplication operator at $1_{A}^{\otimes n}$ recovers the tensor that defines it.

Facts & Assumptions

Given: A field $k$ , a unital $k$ -algebra $A$ , and $n \in {2, 3}$ .

[F1]

The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ : the tensor product of unital algebras has componentwise multiplication and unit $1 \otimes 1$ .

[F2]

Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors: the canonical associator identifies the two parenthesizations of three tensor factors on elementary tensors.

Proof

technique · direct

1.1givenF1F2construct

By [F1], $A \otimes A$ has componentwise multiplication and unit $1_{A} \otimes 1_{A}$ . Applying [F1] once more to $(A \otimes A) \otimes A$ gives the triple componentwise product and unit $(1_{A} \otimes 1_{A}) \otimes 1_{A}$ ; [F2] identifies this with the other parenthesization $1_{A} \otimes (1_{A} \otimes 1_{A})$ . Thus for either allowed $n$ , $A^{\otimes n}$ is a unital algebra with the stated unit.

2.1givenstep 1.1algebra∎

If $L_{X} = L_{Y}$ as left-multiplication maps on $A^{\otimes n}$ , evaluate both maps at the unit $1_{A}^{\otimes n}$ . The unit laws give $L_{X} (1_{A}^{\otimes n}) = X$ and $L_{Y} (1_{A}^{\otimes n}) = Y$ , so equality of the maps implies $X = Y$ . Conversely, if $X = Y$ , the two left-multiplication maps are equal by substitution.

Depends on

Used by

A quasitriangular structure gives and is recovered from a module braiding Theorem

Dependency tree · two levels

11 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

Kalle Kytölä, Introduction to Hopf Algebras and Representations (Helsinki lecture notes, Spring 2011) (standard reference, not scraped)