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The total winding homomorphism of the punctured disk

Definition

Let $n \geq 1$ , let $D^{2} = {z \in C : ∣ z ∣ \leq 1}$ , let $Q_{n} = (q_{1}, \dots, q_{n})$ be the base configuration of the boundary-fixed punctured disk (Boundary-fixed mapping class group of a punctured disk), put $X = D^{2} ∖ Q_{n}$ with basepoint $d$ , and identify $π_{1} (X, d) = F_{n} = ⟨ x_{1}, \dots, x_{n} ⟩$ with the free group (Free group on a set of generators) through the standard meridians $x_{i}$ of Standard meridians of a punctured disk (The punctured-disk fundamental group is free on the standard meridians). The total winding homomorphism is the unique group homomorphism (Monoid homomorphism and group homomorphism) $ω : F_{n} ⟶ Z, ω (x_{i}) = 1 (1 \leq i \leq n),$ whose existence and uniqueness come from the universal property of the free basis. On a word in the $x_{i}^{\pm 1}$ it is the sum of the exponents, and $ω (x_{1} \dots x_{n}) = n$ . It is surjective and its kernel is the subgroup of words of exponent sum $0$ .

Clauses. (1) $ω$ is well defined and independent of all choices, because the standard meridians form a free basis. (2) Invariance under the braid action: $ω \circ ρ (β) = ω$ for every $β \in B_{n}$ , where $ρ$ is the Artin representation of The Artin representation on a free group; and, under AC, $ω \circ h_{*} = ω$ for every homeomorphism representative $h$ of a braid mapping class.

Caveats. The functional $ω$ is the winding about the punctures in total, not about a single puncture: no winding functional about a single $q_{i}$ is used on this page.

Facts & Assumptions

Given: $n \geq 1$ , the punctured disk $X = D^{2} ∖ Q_{n}$ , the basepoint $d$ , the identification $F_{n} ≅ π_{1} (X, d)$ , $x_{i} \mapsto [x_{i}]$ , and a braid word $β \in B_{n}$ .

[F1]

The classes $[x_{1}], \dots, [x_{n}]$ form a free basis of $π_{1} (X, d)$ ; by the universal property of the free group, every function ${x_{1}, \dots, x_{n}} \to G$ into a group $G$ extends to a unique homomorphism $F_{n} \to G$ (Free group on a set of generators, The punctured-disk fundamental group is free on the standard meridians).

[F2]

The Artin representation $ρ : B_{n} \to Aut (F_{n})$ is the unique homomorphism with $ρ (σ_{i}) (x_{i}) = x_{i} x_{i + 1} x_{i}^{- 1}$ , $ρ (σ_{i}) (x_{i + 1}) = x_{i}$ and $ρ (σ_{i}) (x_{j}) = x_{j}$ for $j \notin {i, i + 1}$ ; the braid group $B_{n}$ of The braid group by Artin presentation is generated by $σ_{1}, \dots, σ_{n - 1}$ (The Artin representation on a free group, Artin automorphisms of the free group).

[F3]

Assuming AC, for every braid word $β$ the automorphism of $F_{n}$ induced by the mapping class of $β$ under the identification of [F1] equals $ρ (β)$ (The geometric action on meridians is the Artin representation, Boundary-fixed mapping class group of a punctured disk).

[F4]

AC is the assertion that every family of nonempty sets has a choice function (The Axiom of Choice).

Proof

technique · direct

1.1F1

Well-definedness and basic properties. By [F1] the assignment $x_{i} \mapsto 1$ extends to a unique homomorphism $ω : F_{n} \to Z$ , so $ω$ is well defined and independent of every choice made in the definition of the standard meridians. A word $x_{i_{1}}^{ε_{1}} \dots x_{i_{m}}^{ε_{m}}$ has image $\sum_{j = 1}^{m} ε_{j}$ by the homomorphism law, the kernel of $ω$ is therefore exactly the set of words of exponent sum $0$ , and $ω (x_{1}) = 1$ shows that $ω$ is surjective.

2.1F2step 1.1

Invariance under the Artin action. The set $H = {β \in B_{n} : ω \circ ρ (β) = ω}$ is a subgroup of $B_{n}$ : $ρ$ is a homomorphism, $ρ (1) = id$ gives $ω \circ ρ (1) = ω$ , and if $ω \circ ρ (β) = ω$ and $ω \circ ρ (β^{'}) = ω$ then $ω \circ ρ (β β^{'}) = (ω \circ ρ (β)) \circ ρ (β^{'}) = ω \circ ρ (β^{'}) = ω$ and $ω \circ ρ (β^{- 1}) = ω \circ ρ (β)^{- 1} = ω$ , because $ρ (β)$ is an automorphism. By [F2] it suffices to show $σ_{i} \in H$ for every $i$ . For $ρ (σ_{i})$ , the images of the basis elements $x_{i} \mapsto x_{i} x_{i + 1} x_{i}^{- 1}$ , $x_{i + 1} \mapsto x_{i}$ and $x_{j} \mapsto x_{j}$ all have exponent sum $1 = ω (x_{j})$ , so $ω \circ ρ (σ_{i})$ and $ω$ agree on the free basis and hence, by [F1], on all of $F_{n}$ . Therefore $H = B_{n}$ , which is the first assertion of clause (2).

3.1F3F4step 1.1step 2.1∎

Geometric representative. Assume AC and let $h$ be a homeomorphism representative of the braid mapping class of $β$ , i.e. a boundary-fixed homeomorphism whose mapping class is the image of $β$ (Boundary-fixed mapping class group of a punctured disk). By [F3] the automorphism $h_{*}$ induced on $π_{1} (X, d) = F_{n}$ equals $ρ (β)$ , so $ω \circ h_{*} = ω \circ ρ (β) = ω$ by step 2.1. AC is used only here, through [F3], and the statement of step 2.1 is choice free.

Depends on

Used by

The Burau infinite cyclic cover Definition
Braid mapping classes lift equivariantly to the Burau cover Lemma

Dependency tree · two levels

38 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

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), and section 4 (Theorem 4.1) (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)
Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6 (printed pp. 8-10) and section 4 (Garside structure, printed pp. 26-30) (standard reference, not scraped)