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The reduced and unreduced Burau representations have the same kernel

Statement

Assume AC (inherited through the topological definition of the reduced representation and the agreement theorem). Let $n \geq 2$ , let $ρ_{n}^{m a t}$ and ${\overset{ˉ}{ρ}}_{n}$ be the unreduced and reduced Burau representations over $Λ_{1} = Z [t^{\pm 1}]$ , and let $K = Frac (Λ_{1}) ≅ Q (t)$ be the fraction field of the integral domain $Λ_{1}$ , so that $Λ_{1} \subseteq K$ canonically. Then $\ker ρ_{n}^{m a t} = \ker {\overset{ˉ}{ρ}}_{n},$ that is, a braid acts trivially in the unreduced representation if and only if it acts trivially in the reduced one. The proof uses the field splitting: over $K$ , $K^{n} = \ker σ \oplus K v$ with $σ = (1, t, \dots, t^{n - 1})$ and $v = (1, \dots, 1)^{T}$ , both summands $B_{n}$ -invariant, the line $K v$ trivial, and $\ker σ$ the reduced module after extension of scalars; hence trivial action is equivalent on the two summands. Caveat: this splitting is a field statement, obtained by inverting the nonzero elements of $Λ_{1}$ . It is not asserted over $Λ_{1}$ , and no integral complement or integral splitting of the exact sequence of The unreduced module fits an exact sequence with the reduced module is claimed.

Facts & Assumptions

Given: AC; $n \geq 2$ ; the ring $Λ_{1} = Z [t^{\pm 1}]$ and its fraction field $K$ ; the free module $W = Λ_{1}^{n}$ with its standard basis $e_{1}, \dots, e_{n}$ ; the unreduced representation $ρ_{n}^{m a t}$ on $W$ and the reduced representation ${\overset{ˉ}{ρ}}_{n}$ on $M_{r e d}$ ; the column vector $v = (1, \dots, 1)^{T}$ and the row vector $σ = (1, t, \dots, t^{n - 1})$ .

[F1]

$ρ_{n}^{m a t} : B_{n} \to {GL}_{n} (Λ_{1})$ and ${\overset{ˉ}{ρ}}_{n} : B_{n} \to {GL}_{n - 1} (Λ_{1})$ are group homomorphisms: the first acts on $W$ by the matrices $B_{i}$ of The unreduced Burau matrices extended multiplicatively, and the second is the matrix of the action on $M_{r e d}$ in its fixed basis (The unreduced Burau matrices satisfy the Artin relations, The reduced Burau representation); the braid group is generated by $σ_{1}, \dots, σ_{n - 1}$ (The braid group by Artin presentation).

[F2]

$U = H_{1} (\tilde{X}, p^{- 1} d; Z)$ is free with basis $e_{1}, \dots, e_{n}$ ; the action of $B_{n}$ on $U$ has matrix $ρ_{n}^{m a t} (β)$ in this basis; the inclusion $M_{r e d} \to U$ is $B_{n}$ -equivariant with image $\ker \partial_{*}$ , and under the basis identification $U ≅ W$ the submodule $\ker \partial_{*}$ is carried onto $\ker σ = {x \in W : σ (x) = 0}$ , the connecting map being $\partial_{*} (e_{i}) = t^{i - 1} (t - 1)$ . Hence the action on $M_{r e d}$ corresponds to the restriction of $ρ_{n}^{m a t}$ to $\ker σ$ (The topological and matrix Burau representations agree, clause (1); The unreduced module fits an exact sequence with the reduced module; The reduced Burau representation).

[F3]

$B_{i} v = v$ and $σ B_{i} = σ$ for every $i$ , hence $ρ_{n}^{m a t} (β) v = v$ and $σ ρ_{n}^{m a t} (β) = σ$ for every $β \in B_{n}$ by [F1]; and $σ (v) = 1 + t + \dots + t^{n - 1}$ is a nonzero element of $Λ_{1}$ (The invariant vector and the invariant covectors of the unreduced Burau (a), (b), (c)).

[F4]

$Λ_{1}$ is an integral domain, its fraction field $K = Frac (Λ_{1})$ is a field containing $Λ_{1}$ , and the structure map $Λ_{1} \to K$ is injective: a nonzero element of a domain is not equivalent to $0$ in its localisation at the nonzero elements (Units, powers and the domain property of the Laurent polynomial ring (a), The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain, Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[F5]

$M_{r e d}$ is free of rank $n - 1$ and, under the identification of [F2], $\ker σ$ is a direct summand of $W$ : since $σ (e_{1}) = 1$ , every $x \in W$ is $x = (x - σ (x) e_{1}) + σ (x) e_{1}$ with $x - σ (x) e_{1} \in \ker σ$ (The reduced Burau module is free of rank n minus one, The unreduced module fits an exact sequence with the reduced module).

Proof

technique · direct

1.1F1F4F5algebra

Extension of scalars is injective on endomorphism rings. Let $V$ be a free $Λ_{1}$ -module of finite rank with basis $x_{1}, \dots, x_{m}$ and let $T : V \to V$ be $Λ_{1}$ -linear. The coordinate isomorphism $K \otimes_{Λ_{1}} V \to K^{m}$ sends $a \otimes \sum_{j} λ_{j} x_{j}$ to $(a λ_{j})_{j}$ , with inverse $(c_{j})_{j} \mapsto \sum_{j} c_{j} \otimes x_{j}$ . Then $T = {i d}_{V}$ if and only if $T_{K} : = {i d}_{K} \otimes_{Λ_{1}} T$ is the identity of $K \otimes_{Λ_{1}} V$ : the forward direction is immediate, and if $T_{K} = i d$ then $1 \otimes (T - i d) (x_{j}) = 0$ in $K \otimes_{Λ_{1}} V ≅ K^{m}$ for each $j$ , and reading this in the coordinates of $V$ and using the injectivity of $Λ_{1} \to K$ of [F4] gives $(T - i d) (x_{j}) = 0$ for all $j$ , so $T = {i d}_{V}$ . Applying this to $W$ (free of rank $n$ by [F1]) and to $M_{r e d}$ (free of rank $n - 1$ by [F5]), for every $β \in B_{n}$ we have $ρ_{n}^{m a t} (β) = I_{n}$ if and only if its extension acts as the identity on $K^{n} = K \otimes_{Λ_{1}} W$ , and ${\overset{ˉ}{ρ}}_{n} (β) = I_{n - 1}$ if and only if $β$ acts as the identity on $K \otimes_{Λ_{1}} M_{r e d}$ .

1.2F1F3F4F5algebra

The field splitting and its invariance. Extend $σ$ to the $K$ -linear functional $σ_{K}$ on $K^{n}$ with the same coordinates. By [F3], $σ (v) = 1 + t + \dots + t^{n - 1}$ is a nonzero element of $Λ_{1}$ , hence a nonzero element of $K$ , so $σ_{K} (v) \neq 0$ and $\ker σ_{K} \cap K v = 0$ ; every $x \in K^{n}$ has the decomposition $x = (x - σ_{K} (x) σ_{K} (v)^{- 1} v) + σ_{K} (x) σ_{K} (v)^{- 1} v$ , whose first term lies in $\ker σ_{K}$ . Thus $K^{n} = \ker σ_{K} \oplus K v$ . Both summands are $B_{n}$ -invariant: $K v$ because $ρ_{n}^{m a t} (β) v = v$ for every $β$ by [F3], and $\ker σ_{K}$ because the row-vector identity $σ ρ_{n}^{m a t} (β) = σ$ extends to $K$ and gives $σ_{K} (ρ_{n}^{m a t} (β) x) = σ_{K} (x)$ for every $x \in K^{n}$ ; on $K v$ the action is the identity. Moreover $K \otimes_{Λ_{1}} \ker σ = \ker σ_{K}$ inside $K^{n}$ : the split decomposition $W = \ker σ \oplus Λ_{1} e_{1}$ of [F5] extends to $K^{n} = (K \otimes \ker σ) \oplus K e_{1}$ , with the extended inclusion injective because it still has a left inverse. The functional $σ_{K}$ vanishes on the first summand and sends $a e_{1}$ to $a$ , so its kernel is exactly that first summand.

2.1F1F2F5step 1.1step 1.2

Transport to the reduced action. The identification $M_{r e d} ≅ \ker σ$ of [F2] is $Λ_{1}$ -linear and $B_{n}$ -equivariant, so after extension of scalars it gives $K \otimes_{Λ_{1}} M_{r e d} ≅ K \otimes_{Λ_{1}} \ker σ = \ker σ_{K}$ , and the action of every $β \in B_{n}$ on $K \otimes_{Λ_{1}} M_{r e d}$ corresponds to the restriction of $ρ_{n}^{m a t} (β)_{K}$ to $\ker σ_{K}$ . Combining with step 1.1 applied to the free module $M_{r e d}$ , for every $β \in B_{n}$ : ${\overset{ˉ}{ρ}}_{n} (β) = I_{n - 1}$ if and only if the action of $β$ on $K \otimes_{Λ_{1}} M_{r e d}$ is the identity, if and only if $ρ_{n}^{m a t} (β)_{K}$ is the identity on $\ker σ_{K}$ .

3.1step 1.1step 1.2step 2.1∎

Equality of kernels. Let $β \in B_{n}$ . Since $K^{n} = \ker σ_{K} \oplus K v$ by step 1.2 and $ρ_{n}^{m a t} (β)_{K}$ is the identity on $K v$ , the map $ρ_{n}^{m a t} (β)_{K}$ is the identity on $K^{n}$ if and only if it is the identity on $\ker σ_{K}$ . Therefore step 1.1 for $W$ and step 2.1 give $β \in \ker ρ_{n}^{m a t}$ if and only if $ρ_{n}^{m a t} (β) = I_{n}$ if and only if $ρ_{n}^{m a t} (β)_{K} = {i d}_{K^{n}}$ if and only if $ρ_{n}^{m a t} (β)_{K}$ is the identity on $\ker σ_{K}$ if and only if ${\overset{ˉ}{ρ}}_{n} (β) = I_{n - 1}$ if and only if $β \in \ker {\overset{ˉ}{ρ}}_{n}$ . Hence $\ker ρ_{n}^{m a t} = \ker {\overset{ˉ}{ρ}}_{n}$ as subgroups of $B_{n}$ . The argument used the field splitting of step 1.2; no integral complement or splitting over $Λ_{1}$ is asserted.

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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), Introduction and section 2 (printed pp. 1-5) (standard reference, not scraped)