LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

The reduced Burau module is free of rank n minus one

Statement

Let $M_{r e d} = H_{1} (\tilde{X}; Z)$ be the reduced Burau module over $Λ_{1} = Z [t^{\pm 1}]$ of The reduced Burau homology module, with the $Λ_{1}$ -action induced by the deck generator $t$ . Then $M_{r e d}$ is a free $Λ_{1}$ -module of rank $n - 1$ . The freeness is realised on the lifted spine $Σ$ of The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model: transporting the isomorphism $H_{1} (\tilde{X}) ≅ H_{1} (Σ)$ along the cellular computation there, $M_{r e d}$ has the $Λ_{1}$ -basis given by the absolute cycle classes $ϵ_{i} - ϵ_{n}$ , $1 \leq i \leq n - 1$ , where $ϵ_{i} = e_{i}^{(0)}$ is the level- $0$ lifted spine edge in the notation of that lemma (the generators of the cellular chain module $C_{1} (Σ)$ declared at level $0$ ). In particular the free rank equals $n - 1$ , and the deck generator acts by $t \cdot (ϵ_{i} - ϵ_{n}) = ϵ_{i}^{(1)} - ϵ_{n}^{(1)}$ , the level- $1$ classes. No choice principle is used.

Facts & Assumptions

Given: $n \geq 1$ , the Burau cover $p : \tilde{X} \to X$ with deck generator $t$ , the lifted spine $Σ$ of The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model with its level- $k$ edge classes $e_{i}^{(k)}$ , and the reduced Burau module $M_{r e d} = H_{1} (\tilde{X}; Z)$ with its $Λ_{1}$ -action $t \mapsto (T_{t})_{*}$ .

[F1]

The lifted spine has the cellular chain complex $C_{1} (Σ) = ⨁_{i = 1}^{n} Λ_{1} e_{i}$ , $C_{0} (Σ) = Λ_{1} v$ , $\partial_{1} e_{i} = (t - 1) v$ , where $e_{i} = e_{i}^{(0)}$ and $v = v_{0}$ ; consequently $H_{1} (Σ) = \ker \partial_{1} = ⨁_{i = 1}^{n - 1} Λ_{1} (e_{i} - e_{n})$ is a free $Λ_{1}$ -module of rank $n - 1$ , and the deck action on the cellular chains satisfies $t \cdot e_{i}^{(k)} = e_{i}^{(k + 1)}$ (The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model).

[F2]

There is an isomorphism of $Λ_{1}$ -modules $Φ : H_{1} (Σ) \to H_{1} (\tilde{X}) = M_{r e d}$ , induced by the deck-equivariant homotopy equivalence $(\tilde{X}, p^{- 1} d) \to (Σ, Σ^{0})$ , where the module structures are those induced by the deck actions (The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model, The reduced Burau homology module).

[F3]

$Λ_{1}$ is an integral domain and $t - 1 \neq 0$ (Units, powers and the domain property of the Laurent polynomial ring).

Proof

technique · direct

1.1F1F3

The homology of the spine. By [F1] the cellular chain module is free on $e_{1}, \dots, e_{n}$ over $Λ_{1}$ with $\partial_{1} e_{i} = (t - 1) v$ , and $H_{1} (Σ)$ is the kernel of $\partial_{1}$ ; as computed in [F1] this kernel is exactly the direct sum of the rank-one free submodules $Λ_{1} (e_{i} - e_{n})$ , $1 \leq i \leq n - 1$ , so $H_{1} (Σ)$ is free of rank $n - 1$ with basis the classes of $e_{i} - e_{n}$ .

2.1F2step 1.1

Transport to $M_{r e d}$ . The $Λ_{1}$ -module isomorphism $Φ$ of [F2] carries the basis classes of $e_{i} - e_{n}$ in $H_{1} (Σ)$ to linearly independent $Λ_{1}$ -generators of $M_{r e d}$ : the inverse image of any $Λ_{1}$ -linear relation among the images would be a relation among the $e_{i} - e_{n}$ in the free module $H_{1} (Σ)$ . Hence $M_{r e d}$ is free of rank $n - 1$ with the transported basis, as asserted.

3.1F1F2step 2.1

The deck action on the basis. Since $t \cdot e_{i} = e_{i}^{(1)}$ in the cellular chain module by [F1], the cycle $t \cdot (e_{i} - e_{n})$ is $e_{i}^{(1)} - e_{n}^{(1)}$ ; as these are the level- $1$ classes, the deck generator acts on the spine basis by $t \cdot (ϵ_{i} - ϵ_{n}) = ϵ_{i}^{(1)} - ϵ_{n}^{(1)}$ , and by $Λ_{1}$ -linearity of $Φ$ the same formula holds for the transported basis of $M_{r e d}$ .

4.1step 1.1step 2.1step 3.1∎

Conclusion. The module $M_{r e d}$ is free of rank $n - 1$ with basis the classes $ϵ_{i} - ϵ_{n}$ ( $1 \leq i \leq n - 1$ ), and the deck generator acts by the level-one classes; no choice principle was used, the whole argument being the transport of the cellular computation of [F1] along the deck-equivariant isomorphism of [F2].

Depends on

Used by

The reduced Burau representation Definition
Unreduced and reduced Burau matrices for three strands Example
The Burau determinant recovers the Alexander polynomial of a closed braid Proposition
The reduced and unreduced Burau representations have the same kernel Proposition
The topological and matrix Burau representations agree Theorem

Dependency tree · two levels

36 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

Stephen J. Bigelow, The Burau representation is not faithful for n = 5, Geometry & Topology 3 (1999) 397-404, Definition 1.1, Theorems 1.2 and 1.4 (printed pp. 397-399) (standard reference, not scraped)
Vasudha Bharathram, Joan S. Birman and Tara E. Brendle, The Burau representation is faithful for n = 4, arXiv:2607.05283v1 (6 July 2026), section 2 (printed pp. 1-5) (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey (background on Burau matrices, the cyclic cover and absolute homology) (standard reference, not scraped)