LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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The bar boundary squares to zero and is augmented

Statement

For the bar maps of The augmented two-sided bar complex, $d_{n - 1} d_{n} = 0$ for $n \geq 2$ and $ε d_{1} = 0$ . Thus the augmented bar sequence is a chain complex of both left and right $A^{e}$ -modules.

Facts & Assumptions

Given: A unital associative algebra $A$ over a field $k$ , with the bar terms, faces, augmentation and outer $A^{e}$ -actions defined in the cited item.

[F1]

The maps are the alternating sum of adjacent-slot multiplication faces (The augmented two-sided bar complex).

[F2]

Every adjacent-multiplication face is linear on both $A^{e}$ sides (The augmented two-sided bar complex).

[F3]

The augmentation is multiplication and is linear on both $A^{e}$ sides (The augmented two-sided bar complex).

Proof

technique · direct

For $n \geq 1$ , write $\partial_{i}^{(n)}$ for the face that multiplies slots $i$ and $i + 1$ , so $d_{n} = \sum_{i = 0}^{n} (- 1)^{i} \partial_{i}^{(n)}$ .

1.1F1givenalgebra

If $0 \leq i < j - 1$ , the two faces multiply disjoint pairs of slots; doing the later one first and reindexing it by one gives $\partial_{i}^{(n - 1)} \partial_{j}^{(n)} = \partial_{j - 1}^{(n - 1)} \partial_{i}^{(n)}$ . The products are independent and retain their order, so the identity holds on the tensor terms.

1.2F1givenalgebra

If $j = i + 1$ , both composites multiply the consecutive triple $a_{i}, a_{i + 1}, a_{i + 2}$ into one slot, giving $a_{i} (a_{i + 1} a_{i + 2}) = (a_{i} a_{i + 1}) a_{i + 2}$ by associativity; thus the same face identity holds for every $i < j$ .

1.3F3givenalgebra

In degree one, $ε d_{1} (a_{0} \otimes a_{1} \otimes a_{2}) = (a_{0} a_{1}) a_{2} - a_{0} (a_{1} a_{2}) = 0$ by associativity, so $ε d_{1} = 0$ .

2.1step 1.1step 1.2F1algebra

In the double sum for $d_{n - 1} d_{n}$ , terms indexed by $i < j$ pair with $(j - 1, i)$ , exactly the terms with first index at least the second. Steps 1.1 and 1.2 identify the composites, and $(- 1)^{i + j} = - (- 1)^{(j - 1) + i}$ , so every term cancels and $d_{n - 1} d_{n} = 0$ for every $n \geq 2$ .

3.1step 2.1step 1.3F2F3algebra∎

By [F2] and [F3], all these maps are linear for both outer $A^{e}$ actions; steps 2.1 and 1.3 therefore give the augmented chain-complex identities in both module categories.

Remark

For $0 \leq i < j - 1$ , the bar faces multiply disjoint adjacent pairs; their composites agree after the later face is reindexed by one. For $j = i + 1$ , associativity on the overlapping triple gives the same face identity. These are the internal adjacent-multiplication cases used in the proof above.

Depends on

The augmented two-sided bar complex

Used by

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

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 9: Hochschild and Cyclic Homology, §9.1.3 (standard reference, not scraped)
Mikhail Khovanov, Triply-graded link homology and Hochschild homology of Soergel bimodules, Hochschild homology section (standard reference, not scraped)