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The invariant vector and the invariant covectors of the unreduced Burau

Statement

Let $v = (1, 1, \dots, 1)^{T} \in Λ_{1}^{n}$ and let $σ = (1, t, t^{2}, \dots, t^{n - 1})$ be the row vector, so that $σ (x) = \sum_{i = 1}^{n} t^{i - 1} x_{i}$ on column vectors $x = (x_{1}, \dots, x_{n})^{T}$ . For the unreduced matrices $B_{i}$ of The unreduced Burau matrices:

(a) $B_{i} v = v$ for every $i$ , hence $ρ_{n}^{m a t} (β) v = v$ for every $β \in B_{n}$ and the line $Λ_{1} v$ is a $B_{n}$ -invariant submodule of $Λ_{1}^{n}$ ;

(b) a row vector $τ$ satisfies $τ B_{i} = τ$ for every $i$ if and only if $τ = c σ$ for some $c \in Λ_{1}$ , so the invariant covectors form the free rank-one $Λ_{1}$ -module $Λ_{1} σ$ ;

(c) $σ (v) = 1 + t + \dots + t^{n - 1}$ , which is nonzero in the integral domain $Λ_{1}$ (Units, powers and the domain property of the Laurent polynomial ring). No choice principle is used.

Facts & Assumptions

Given: $n \geq 1$ , the ring $Λ_{1} = Z [t^{\pm 1}]$ , the column vector $v = (1, \dots, 1)^{T}$ , the row vector $σ = (1, t, \dots, t^{n - 1})$ , and the matrices $B_{1}, \dots, B_{n - 1}$ with $ρ_{n}^{m a t}$ as in The unreduced Burau matrices satisfy the Artin relations.

[F1]

$B_{i}$ has the block $(\begin{array}{cc} 1 - t & t \\ 1 & 0 \end{array})$ in rows and columns $i, i + 1$ and is the identity elsewhere; its action on the basis vectors is $e_{i} \mapsto (1 - t) e_{i} + e_{i + 1}$ , $e_{i + 1} \mapsto t e_{i}$ , and $e_{j} \mapsto e_{j}$ otherwise (The unreduced Burau matrices).

[F2]

$ρ_{n}^{m a t} : B_{n} \to {GL}_{n} (Λ_{1})$ is the homomorphism extending $σ_{i} \mapsto B_{i}$ (The unreduced Burau matrices satisfy the Artin relations), and $B_{n}$ is generated by $σ_{1}, \dots, σ_{n - 1}$ (The braid group by Artin presentation).

[F3]

$Λ_{1}$ is an integral domain; the polynomial $p = 1 + t + \dots + t^{n - 1}$ is nonzero in $Z [t]$ for $n \geq 1$ (its leading coefficient is $1$ in degree $n - 1$ ), and the localisation map $Z [t] \to Λ_{1}$ is injective because no power $t^{k} \neq 0$ annihilates a nonzero polynomial in the domain $Z [t]$ (Units, powers and the domain property of the Laurent polynomial ring, Equality, vanishing, and the kernel of the localisation map, Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

Proof

technique · direct

1.1F1F2

Clause (a). For a fixed $i$ compute the two affected coordinates of $B_{i} v$ using the column convention: the $i$ -th entry is $(1 - t) \cdot 1 + t \cdot 1 = 1$ , and the $(i + 1)$ -st entry is $1 \cdot 1 + 0 \cdot 1 = 1$ ; all other entries coincide with those of $v$ , so $B_{i} v = v$ . Since the $σ_{i}$ generate $B_{n}$ by [F2] and $ρ_{n}^{m a t}$ is a homomorphism, $ρ_{n}^{m a t} (β) v = v$ for every braid word $β$ ; hence the line $Λ_{1} v$ is mapped into itself by every $ρ_{n}^{m a t} (β)$ .

1.2F1algebra

Clause (b). For a row vector $τ = (τ_{1}, \dots, τ_{n})$ compute the affected coordinates of $τ B_{i}$ : the $i$ -th entry is $τ_{i} (1 - t) + τ_{i + 1} \cdot 1$ and the $(i + 1)$ -st entry is $τ_{i} t + τ_{i + 1} \cdot 0$ ; all other entries are unchanged. Hence $τ B_{i} = τ$ holds if and only if $τ_{i} (1 - t) + τ_{i + 1} = τ_{i}$ and $τ_{i} t = τ_{i + 1}$ , both of which are equivalent to $τ_{i + 1} = t τ_{i}$ . If $τ B_{i} = τ$ for every $i$ , the recurrence gives $τ_{i} = t^{i - 1} τ_{1}$ for $i = 1, \dots, n$ by induction, so $τ = τ_{1} σ$ ; conversely $σ B_{i} = σ$ by the same formulas, and then $τ B_{i} = τ$ for $τ = c σ$ . This proves clause (b).

1.3F3

Clause (c). Evaluating the row vector $σ$ on $v$ gives $σ (v) = \sum_{i = 1}^{n} t^{i - 1} = 1 + t + \dots + t^{n - 1}$ , the image of the nonzero polynomial $p$ under the localisation map, which is injective by [F3]; hence $σ (v)$ is a nonzero element of the domain $Λ_{1}$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion. Clause (a) is step 1.1, clause (b) is step 1.2 and clause (c) is step 1.3; no choice principle was used.

Depends on

Used by

An invariant line need not have an invariant complement over a Laurent ring Counterexample
Specializing Burau at t = 1 recovers permutation data Example
Unreduced and reduced Burau matrices for three strands Example
The reduced and unreduced Burau representations have the same kernel Proposition
The unreduced module fits an exact sequence with the reduced module Proposition
The topological and matrix Burau representations agree Theorem

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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)
Stephen J. Bigelow, The Burau representation is not faithful for n = 5, Geometry & Topology 3 (1999) 397-404, Definition 1.1 (printed pp. 397-398) (standard reference, not scraped)