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The reduced Burau representation detects every power of the fourth power of the half twist in B3

Statement

Assume AC (inherited through the definition of the reduced representation and the agreement theorem with the topological representation). Let $Δ = σ_{1} σ_{2} σ_{1}$ be the half twist of $B_{3}$ and let ${\overset{ˉ}{ρ}}_{3} : B_{3} \to {GL}_{2} (Λ_{1})$ be the reduced Burau representation in the basis $(g_{1}, g_{2})$ of The topological and matrix Burau representations agree, over $Λ_{1} = Z [t^{\pm 1}]$ . Then ${\overset{ˉ}{ρ}}_{3} (Δ^{2}) = t^{3} I_{2}, hence {\overset{ˉ}{ρ}}_{3} (Δ^{4 k}) = t^{6 k} I_{2}$ for every $k \in Z$ , and $t^{6 k} I_{2} = I_{2}$ if and only if $k = 0$ . Consequently no nonzero power $Δ^{4 k}$ , $k \neq 0$ , lies in the kernel of ${\overset{ˉ}{ρ}}_{3}$ , and the cyclic subgroup $⟨ Δ^{4} ⟩ = {Δ^{4 k} : k \in Z}$ maps isomorphically onto its image under ${\overset{ˉ}{ρ}}_{3}$ .

Facts & Assumptions

Given: AC; the half twist $Δ = σ_{1} σ_{2} σ_{1}$ of $B_{3}$ ; the reduced representation ${\overset{ˉ}{ρ}}_{3}$ in the basis $(g_{1}, g_{2})$ ; the ring $Λ_{1} = Z [t^{\pm 1}]$ with its element $t$ .

[F1]

In the basis $(g_{1}, g_{2})$ , ${\overset{ˉ}{ρ}}_{3} (σ_{1}) = (\begin{array}{cc} - t & t \\ 0 & 1 \end{array})$ , ${\overset{ˉ}{ρ}}_{3} (σ_{2}) = (\begin{array}{cc} 1 & 0 \\ 1 & - t \end{array})$ , and ${\overset{ˉ}{ρ}}_{3} (Δ^{2}) = t^{3} I_{2}$ for the full twist $Δ^{2} = (σ_{1} σ_{2} σ_{1})^{2}$ (The topological and matrix Burau representations agree).

[F2]

${\overset{ˉ}{ρ}}_{3} : B_{3} \to {GL}_{2} (Λ_{1})$ is a group homomorphism; hence ${\overset{ˉ}{ρ}}_{3} (β^{m}) = {\overset{ˉ}{ρ}}_{3} (β)^{m}$ for every $β \in B_{3}$ and every $m \in Z$ , where negative powers are taken in the group ${GL}_{2} (Λ_{1})$ (The reduced Burau representation).

[F3]

$t^{m} \neq 1$ in $Λ_{1}$ for every integer $m \neq 0$ (Units, powers and the domain property of the Laurent polynomial ring, clause (b)).

[F4]

$Δ^{2} = (σ_{1} σ_{2} σ_{1})^{2}$ is the full twist of $B_{3}$ and $Δ^{4 k} = (Δ^{2})^{2 k}$ for every $k \in Z$ (The Garside half twist and simple positive braids).

Proof

technique · direct

1.1F1F2F4algebra

The full twist in the reduced representation. Write $M_{1} = {\overset{ˉ}{ρ}}_{3} (σ_{1}) = (\begin{array}{cc} - t & t \\ 0 & 1 \end{array})$ and $M_{2} = {\overset{ˉ}{ρ}}_{3} (σ_{2}) = (\begin{array}{cc} 1 & 0 \\ 1 & - t \end{array})$ as in [F1]. Since ${\overset{ˉ}{ρ}}_{3}$ is a homomorphism and $Δ^{2} = (σ_{1} σ_{2} σ_{1})^{2}$ by [F4], ${\overset{ˉ}{ρ}}_{3} (Δ^{2}) = (M_{1} M_{2} M_{1})^{2}$ in the composition order of the conventions. Direct computation gives $M_{1} M_{2} = (\begin{array}{cc} 0 & - t^{2} \\ 1 & - t \end{array})$ and $M_{1} M_{2} M_{1} = (\begin{array}{cc} 0 & - t^{2} \\ - t & 0 \end{array})$ , so $(M_{1} M_{2} M_{1})^{2} = (\begin{array}{cc} (- t^{2}) (- t) & 0 \\ 0 & (- t) (- t^{2}) \end{array}) = t^{3} I_{2}$ , confirming the displayed value ${\overset{ˉ}{ρ}}_{3} (Δ^{2}) = t^{3} I_{2}$ .

2.1F2F4step 1.1algebra

All powers of the fourth power. For $k \geq 0$ , ${\overset{ˉ}{ρ}}_{3} (Δ^{4 k}) = {\overset{ˉ}{ρ}}_{3} ((Δ^{2})^{2 k}) = ({\overset{ˉ}{ρ}}_{3} (Δ^{2}))^{2 k} = (t^{3} I_{2})^{2 k} = t^{6 k} I_{2}$ by [F2], [F4] and step 1.1. For $k < 0$ set $k^{'} = - k > 0$ ; the matrix $t^{3} I_{2}$ is invertible with inverse $t^{- 3} I_{2}$ , and $Δ^{4 k} = (Δ^{2})^{- 2 k^{'}}$ , so ${\overset{ˉ}{ρ}}_{3} (Δ^{4 k}) = ({\overset{ˉ}{ρ}}_{3} (Δ^{2}))^{- 2 k^{'}} = (t^{- 3} I_{2})^{2 k^{'}} = t^{6 k} I_{2}$ , the same formula. Hence ${\overset{ˉ}{ρ}}_{3} (Δ^{4 k}) = t^{6 k} I_{2}$ for every $k \in Z$ .

3.1F3step 2.1∎

Detection. The scalar matrix $t^{6 k} I_{2}$ equals $I_{2}$ if and only if $t^{6 k} = 1$ in $Λ_{1}$ , and by [F3] applied to $m = 6 k$ this holds if and only if $6 k = 0$ , that is, if and only if $k = 0$ . Therefore $Δ^{4 k} \in \ker {\overset{ˉ}{ρ}}_{3}$ is possible only for $k = 0$ : no nonzero power of $Δ^{4}$ lies in the kernel. Consequently the restriction of the homomorphism ${\overset{ˉ}{ρ}}_{3}$ to the cyclic subgroup $⟨ Δ^{4} ⟩ = {Δ^{4 k} : k \in Z}$ has trivial kernel, so it is injective and maps that subgroup isomorphically onto its image.

Depends on

Used by

The reduced Burau representation is faithful for at most three strands 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)
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)