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The geometric action on meridians is the Artin representation

Statement

Assume AC. Let $G_{n}$ be the geometric braid group, $φ_{n} : B_{n} \to G_{n}$ the published surjection of The Artin presentation surjects onto the geometric braid group, and $Ψ : G_{n} ⟶ Mod (D^{2}, Q_{n}; \partial D^{2})$ the isomorphism of Braid group as boundary-fixed punctured-disk mapping classes. Identifying $π_{1} (D^{2} ∖ Q_{n}, d)$ with $F_{n}$ by the standard meridians of Standard meridians of a punctured disk, the automorphism of $F_{n}$ induced by the mapping class $Ψ (φ_{n} (β))$ equals $ρ (β)$ for every braid word $β$ . In particular the geometric half twist $σ_{i}$ acts as the Nielsen automorphism of Artin automorphisms of the free group.

Facts & Assumptions

Given: AC, the number $n$ , the punctured disk $X = D^{2} ∖ Q_{n}$ with basepoint $d$ , the geometric braid group $G_{n}$ with its surjection $φ_{n}$ and the isomorphism $Ψ$ , and the standard meridian loops $x_{1}, \dots, x_{n}$ with the identification $F_{n} ≅ π_{1} (X, d)$ , $x_{j} \mapsto [x_{j}]$ .

[F1]

The published identification. $Ψ$ is a group isomorphism, and for $1 \leq i \leq n - 1$ the image of the standard positive geometric half twist $σ_{i}$ under $Ψ \circ φ_{n}$ is the mapping class of the explicit boundary-fixed homeomorphism $H_{i}$ constructed in the published proof, which is supported in the support disc $U_{i}$ of the adjacent pair and exchanges $q_{i}$ and $q_{i + 1}$ ; the construction rotates the support disc about the midpoint $m_{i}$ through the half turn whose total angle is $π$ . (Braid group as boundary-fixed punctured-disk mapping classes, The Artin presentation surjects onto the geometric braid group.)

[F2]

Positivity and the size of the support disc. The support disc $U_{i} = {w : ∥ w - m_{i} ∥_{2} < 3 h / 2}$ contains exactly the two base points $q_{i}, q_{i + 1}$ , each at distance $h$ from $m_{i}$ ; the standard positive half twist turns the moving pair anticlockwise about $m_{i}$ , the label $i$ passing below its midpoint $m_{i}$ and the label $i + 1$ above. Hence the homeomorphism $H_{i}$ of [F1] acts on $U_{i}$ as the half rotation of the pair about $m_{i}$ that carries $q_{i}$ through the lower half-plane to $q_{i + 1}$ and $q_{i + 1}$ through the upper half-plane to $q_{i}$ . (The elementary geometric half twist, its support disc, and its opposite.)

[F3]

Standard meridians and their freedom of radius. The stems $s_{j}$ are the straight segments from $d = (0, 1)$ to the points $q_{j}$ of the standard-meridian definition; the loops $x_{j} = s_{j} c_{j} s_{j}^{- 1}$ are based at $d$ , and the class $[x_{j}]$ is independent of the admissible radius of the circle $C_{j}$ ; the assignment $x_{j} \mapsto [x_{j}]$ is an isomorphism $F_{n} \to π_{1} (X, d)$ (Standard meridians of a punctured disk, The punctured-disk fundamental group is free on the standard meridians).

[F4]

Functoriality of the induced map. A pointed continuous map induces a group homomorphism on $π_{1}$ , homotopic pointed maps induce the same homomorphism, ${id}_{*} = id$ , and $(g \circ f)_{*} = g_{*} \circ f_{*}$ ; hence the operation $[f] \mapsto f_{*}$ is a well-defined group homomorphism from the mapping class group of boundary-fixed homeomorphisms to $Aut (π_{1} (X, d))$ (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[F5]

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

[F6]

A convenient slit system for reading loops. Use the vertical downward segments $ℓ_{j}$ from $q_{j}$ to the lower outer boundary, instead of the standard upper tethers. Their interiors are pairwise disjoint, and every standard truncated tether avoids them. Remove small puncture disks and open along the remaining parts of these downward segments. The thin-strip disk construction of The standard stem system cuts the punctured disk open to a disk applies to these disjoint straight cuts as well: it gives a compact disk, with each removed circle opened into a boundary arc. A transverse based loop is therefore read by its signed slit crossings. To justify the rule, split it at the crossings and contract each intervening path in that disk; gluing back one paired side gives its standard lasso, reached from $d$ above the cuts. A crossing from left to right over the downward slit is positive, since a positive small meridian crosses it in that direction. Thus it contributes $x_{j}$ , and the opposite crossing contributes $x_{j}^{- 1}$ . All loop segments remain in the holed disk; no puncture tip is traversed.

[F7]

The boundary class equals $x_{1} \dots x_{n}$ (The oriented boundary loop represents the ordered product of the standard meridians). A boundary-fixed homeomorphism fixes that class. In the published half-rotation formula of [F1], $H_{i}$ is $(r, θ) \mapsto (r, θ + π χ (r))$ about $m_{i}$ , with $0 \leq χ \leq 1$ , $χ = 1$ for $r \leq 5 h / 4$ , and $χ = 0$ for $r \geq 11 h / 8$ .

Proof

1.1F1F3F4

The induced action is a homomorphism on braids. By [F1] the composite $Ψ \circ φ_{n} : B_{n} \to Mod (D^{2}, Q_{n}; \partial D^{2})$ is a group homomorphism, and by [F4] the assignment $[f] \mapsto f_{*}$ is a group homomorphism to $Aut (π_{1} (X, d))$ , which under the identification of [F3] is $Aut (F_{n})$ . Hence $Θ : β ⟼ (the automorphism induced by Ψ (φ_{n} (β)))$ is a group homomorphism $B_{n} \to Aut (F_{n})$ .

1.2F1F5

Reduction to the half twists. By [F5] $B_{n}$ is generated by $σ_{1}, \dots, σ_{n - 1}$ and $ρ$ is the unique homomorphism carrying $σ_{i}$ to the displayed substitution; two homomorphisms from the presented group $B_{n}$ that agree on all generators agree on $B_{n}$ . It therefore suffices to prove $(H_{i})_{*} = ρ (σ_{i})$ for every $i$ , where $(H_{i})_{*}$ is the automorphism of $F_{n}$ induced by the homeomorphism $H_{i}$ of [F1].

1.3F1F2F6F7construct

The transported right tether avoids every other downward slit. Write the right stem as $z (t) = (1 - t) d + t q_{i + 1}$ and put $y = 1 - t$ . Whenever it meets $U_{i}$ , $0 < y < 3 h / 2$ . Its horizontal coordinate relative to $m_{i}$ is $h - y q_{i + 1} > h - (3 h / 2) (1 / 4) = 5 h / 8 > 0$ , since $∣ q_{i + 1} ∣ < 1 / 4$ . Thus its polar angle about $m_{i}$ satisfies $0 < θ < π / 2$ . Under $H_{i}$ , its angle becomes $Θ = θ + π χ (r)$ , so $0 < Θ < 3 π / 2$ . The downward slit from $q_{i + 1} = m_{i} + (h, 0)$ could be crossed only at a point with horizontal coordinate $h$ relative to $m_{i}$ and negative vertical coordinate; that would require an angle in $(3 π / 2, 2 π)$ , impossible in this range. Every other slit except $ℓ_{i}$ has horizontal distance at least $3 h$ from $m_{i}$ and avoids $U_{i}$ . Outside $U_{i}$ the stem stays above the real axis and is unchanged. Hence the transported right tether can cross only $ℓ_{i}$ . This argument uses actual truncated tethers ending at a small circle, so no concatenation passes through a deleted puncture.

2.1F1F3step 1.2

Every meridian with $j \notin {i, i + 1}$ is fixed. The $x$ -coordinates of $m_{i}$ and $q_{j}$ differ by $∣ 2 (i - j) + 1 ∣ h \geq 3 h$ , and $∣ x (q_{j}) ∣ < 1 / 4$ ; hence the distance from $m_{i}$ to the line through $d$ and $q_{j}$ , namely $∣ x (m_{i}) - x (q_{j}) ∣ / \sqrt{1 + x (q_{j})^{2}}$ , exceeds $3 h \cdot 4 / \sqrt{17} > 3 h / 2$ , so the stem $s_{j}$ avoids the open support disc $U_{i}$ and $dist (q_{j}, U_{i}) \geq 3 h - 3 h / 2 = 3 h / 2 > 0$ . Choose, by the radius independence of [F3], a lasso $x_{j}^{'}$ homotopic rel $d$ to $x_{j}$ whose circle has radius less than $3 h / 2$ around $q_{j}$ ; then $x_{j}^{'}$ avoids $U_{i}$ , and since $H_{i}$ is the identity outside $U_{i}$ by [F1], $H_{i} \circ x_{j}^{'} = x_{j}^{'}$ pointwise. Hence $(H_{i})_{*} [x_{j}] = [x_{j}] = ρ (σ_{i}) (x_{j})$ .

3.1F3F5F6F7step 2.1step 1.3algebra

The right meridian and then the left meridian. Choose the circle for $x_{i + 1}$ sufficiently small to lie wholly in the core $r < 5 h / 4$ . It maps under $H_{i}$ to a positive round circle about $q_{i}$ , and crosses $ℓ_{i}$ once positively and no other downward slit. By step 1.3 the preceding tether has a crossing word $x_{i}^{m}$ for some integer $m$ , after a small general-position perturbation supported away from the other slits. Its returning tether gives $x_{i}^{- m}$ . [F6] therefore reads the actual based loop $H_{i} \circ x_{i + 1}$ as $x_{i}^{m} x_{i} x_{i}^{- m} = x_{i}$ . The other meridians are fixed by step 2.1, and [F7] says the whole ordered product is fixed. Cancelling its unchanged prefix and suffix gives $(H_{i})_{*} (x_{i}) (H_{i})_{*} (x_{i + 1}) = x_{i} x_{i + 1}$ . Substitution of $(H_{i})_{*} (x_{i + 1}) = x_{i}$ yields $(H_{i})_{*} (x_{i}) = x_{i} x_{i + 1} x_{i}^{- 1}$ . These are exactly the formulas of [F5].

4.1F1F5step 1.1step 1.2step 2.1step 1.3step 3.1∎

Comparison and conclusion. Steps 2.1 and 3.1 show $(H_{i})_{*} (x_{j}) = ρ (σ_{i}) (x_{j})$ for every $j$ and every $i$ , and two endomorphisms agreeing on the free basis $x_{1}, \dots, x_{n}$ agree as automorphisms; hence $(H_{i})_{*} = ρ (σ_{i})$ . By step 1.2 the homomorphism $Θ$ of step 1.1 and the Artin representation $ρ$ agree on the generators of $B_{n}$ , so $Θ (β) = ρ (β)$ for every braid word $β$ , which is the first assertion; the case $β = σ_{i}$ is the second. AC is used through the published isomorphism and half-twist identification of [F1] and the slit-disk construction of [F6].

Remarks

The proposition is the point where the algebraic convention of Artin automorphisms of the free group is matched to the frozen geometric conventions of this library: the positive (anticlockwise) half twist induces the substitution $x_{i} \mapsto x_{i} x_{i + 1} x_{i}^{- 1}$ , $x_{i + 1} \mapsto x_{i}$ , which is Artin's substitution with the letter and its inverse interchanged. The mirror (clockwise) half rotation realizes Artin's original formulas verbatim.
The read-off in steps 2.1, 3.1 and 4.1 uses only the support disc and the stems; it shows at the same time that the induced action fixes $x_{1} \dots x_{n}$ , as it must, since $H_{i}$ fixes $\partial D^{2}$ pointwise.

Depends on

Used by

The Artin representation is faithful Theorem

Dependency tree · two levels

54 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

Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, printed pp. 8-10 (the action of sigma_i on the generators x_i, x_{i+1}, Figure 4) (standard reference, not scraped)
Emil Artin, Theory of Braids, Annals of Mathematics 48 (1947), pp. 101-126, equations (14)-(15), printed pp. 113-114 (standard reference, not scraped)
The published identification of the positive half twist with the supported half rotation, thm-braid-group-is-the-boundary-fixed-mapping-class-group-of-the-punctured-disk, statement parts 1 and 2 (standard reference, not scraped)