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The unreduced Burau matrices

Definition

Let $U = H_{1} (\tilde{X}, p^{- 1} d; Z)$ be the unreduced Burau module over $Λ_{1} = Z [t^{\pm 1}]$ (The unreduced Burau relative homology module) and use the relative lifted-edge basis $e_{1}, \dots, e_{n}$ of the lifted spine $Σ$ , where for the design's conventions $e_{i}$ is the relative class of the $i$ -th lifted spine edge taken at deck level $i - 1$ : $e_{i} = t^{i - 1} ϵ_{i}$ with $ϵ_{i}$ the level- $0$ class of The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model. The unreduced Burau matrices are the matrices $B_{1}, \dots, B_{n - 1} \in {GL}_{n} (Λ_{1})$ that are the identity outside rows and columns $i, i + 1$ and whose $2 \times 2$ block in rows and columns $i, i + 1$ is $B_{i} = (\begin{array}{cc} 1 - t & t \\ 1 & 0 \end{array}),$ acting on column vectors: the first column is the image of $e_{i}$ and the second the image of $e_{i + 1}$ , so $e_{i} \mapsto (1 - t) e_{i} + e_{i + 1}$ and $e_{i + 1} \mapsto t e_{i}$ . Each $B_{i}$ is invertible with $B_{i}^{- 1} = t^{- 1} (\begin{array}{cc} 0 & t \\ 1 & t - 1 \end{array}),$ since $\det B_{i} = - t$ is a unit of $Λ_{1}$ . The assignment $σ_{i} \mapsto B_{i}$ defines a representation of the presented braid group only through The unreduced Burau matrices satisfy the Artin relations ↗; its topological meaning on $U$ is The topological and matrix Burau representations agree ↗. Convention: matrix multiplication follows the library composition order, so a braid word $σ_{i_{1}} \dots σ_{i_{k}}$ acts by $B_{i_{1}} \dots B_{i_{k}}$ on column vectors, the rightmost letter acting first; this displayed block and that order control every later matrix.

Facts & Assumptions

Given: $n \geq 1$ , the unreduced Burau module $U$ with its $Λ_{1}$ -module structure, the relative lifted-edge basis $ϵ_{1}, \dots, ϵ_{n}$ of the lifted spine $Σ$ , and the ring $Λ_{1} = Z [t^{\pm 1}]$ .

[F1]

The deck-equivariant homotopy equivalence of pairs identifies $U = H_{1} (\tilde{X}, p^{- 1} d; Z)$ with $H_{1} (Σ, Σ^{0})$ , which is the free $Λ_{1}$ -module $⨁_{i = 1}^{n} Λ_{1} ϵ_{i}$ on the relative classes of the level- $0$ edges; the deck generator acts by $t \cdot ϵ_{i} = ϵ_{i}^{(1)}$ , the level- $1$ class (The cyclic cover retracts onto the lifted flower and has a deck-equivariant spine model, The unreduced Burau relative homology module).

[F2]

$t$ is a unit of $Λ_{1}$ , so every power $t^{i - 1}$ is a unit and $- t$ is a unit; multiplication by a unit carries a basis to a basis (The Laurent polynomial ring as the principal localisation of Z[t] at t, Units, powers and the domain property of the Laurent polynomial ring).

[F3]

${GL}_{n} (Λ_{1})$ is the group of invertible $n \times n$ matrices over $Λ_{1}$ , the matrix of a $Λ_{1}$ -linear map in a fixed basis has as its columns the images of the basis vectors, and the determinant of a square matrix is computed from its entries; a $2 \times 2$ matrix $(\begin{array}{cc} a & b \\ c & d \end{array})$ has determinant $a d - b c$ (Invertible square matrices and similarity over a commutative ring, The Leibniz formula gives $\det (\begin{array}{cc} a & b \\ c & d \end{array}) = a d - b c$ , A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

Proof

technique · direct

1.1F1F2

The relative lifted-edge basis. By [F1] the classes $ϵ_{1}, \dots, ϵ_{n}$ form a $Λ_{1}$ -basis of $U$ ; since $e_{i} = t^{i - 1} ϵ_{i}$ differs from $ϵ_{i}$ by the unit $t^{i - 1}$ by [F2], the family $e_{1}, \dots, e_{n}$ is again a $Λ_{1}$ -basis of $U$ . In particular every element of $U$ has a unique expression $\sum_{i = 1}^{n} c_{i} e_{i}$ with $c_{i} \in Λ_{1}$ .

1.2F2F3algebra

Invertibility. Put $C_{i} : = t^{- 1} (\begin{array}{cc} 0 & t \\ 1 & t - 1 \end{array}) = (\begin{array}{cc} 0 & 1 \\ t^{- 1} & (t - 1) t^{- 1} \end{array})$ , the identity outside the same block. Multiplying the two $2 \times 2$ blocks in the order $B_{i} C_{i}$ gives $(\begin{array}{cc} 1 - t & t \\ 1 & 0 \end{array}) (\begin{array}{cc} 0 & 1 \\ t^{- 1} & (t - 1) t^{- 1} \end{array}) = (\begin{array}{cc} 1 & (1 - t) + (t - 1) \\ 0 & 1 \end{array}) = I_{2}$ , and the reverse order gives $I_{2}$ by the same computation, so $B_{i} C_{i} = I_{n} = C_{i} B_{i}$ : $B_{i}$ is invertible with inverse $C_{i}$ , and $\det B_{i} = (1 - t) \cdot 0 - t \cdot 1 = - t$ by the $2 \times 2$ formula, a unit of $Λ_{1}$ by [F2].

2.1F3step 1.1

The block action. For $i \in {1, \dots, n - 1}$ let $B_{i}$ be the identity outside rows and columns $i, i + 1$ with the displayed $2 \times 2$ block. By the column convention of [F3] the images of the basis vectors are the columns of $B_{i}$ , namely $e_{i} \mapsto (1 - t) e_{i} + e_{i + 1}$ , $e_{i + 1} \mapsto t e_{i}$ and $e_{j} \mapsto e_{j}$ for $j \notin {i, i + 1}$ ; these three formulas determine $B_{i}$ uniquely as a $Λ_{1}$ -linear map on $U$ followed by the chosen basis.

3.1F1F3step 1.1step 2.1∎

Conventions and what is deferred. The assignment $σ_{i} \mapsto B_{i}$ is defined for $1 \leq i \leq n - 1$ ; that it extends to a homomorphism on the presented braid group of The braid group by Artin presentation requires the Artin relations verified in The unreduced Burau matrices satisfy the Artin relations ↗, and that its action on $U$ is realised by the geometric half twists is The topological and matrix Burau representations agree ↗; both are proved later on this page and are not used here. Matrices act on column vectors, and a word $σ_{i_{1}} \dots σ_{i_{k}}$ acts by $B_{i_{1}} \dots B_{i_{k}}$ with the rightmost letter first, matching the library's leftmost-outermost composition convention of [F3].

Depends on

Used by

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Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey (background on Burau matrices, the cyclic cover and absolute homology) (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)