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PB_3 as F_2 by Z, with its section action

Example

Assume the Axiom of Choice and let, with the standard pure braid generators of Standard geometric pure braid generators A_ij for $n = 3$ ,

$a : = A_{12}, b : = A_{13}, c : = A_{23} \in P B_{3} .$

Then $⟨ b, c ⟩$ is the kernel of the forgetting homomorphism $φ : P B_{3} \to P B_{2}$ , it is free on $b, c$ , and $P B_{2} = ⟨ a ⟩$ is infinite cyclic. With the far-right section of A choice-free continuous section of planar coordinate forgetting — adjusted at the basepoint along a path in the fibre, so that on a small representative of $a$ it fixes the third point — the extension splits and

$P B_{3} ≅ ⟨ b, c ⟩ ⋊ ⟨ a ⟩ ≅ F_{2} ⋊ Z .$

Writing $w : = b c$ , the action of the positive generator $a$ on the free kernel is

$x ⟼ w^{- 1} x w (x \in F_{2}),$

for this section and the first-under-second convention. This does not assert a direct-product decomposition.

Facts & Assumptions

Given: the Axiom of Choice, the canonical base configuration $Q = (q_{1}, q_{2}, q_{3})$ of The elementary geometric half twist, its support disc, and its opposite, so that $h = \frac{1}{16}$ , $q_{1} = - \frac{1}{8}$ , $q_{2} = 0$ and $q_{3} = \frac{1}{8}$ ; the half twists $σ_{1}, σ_{2}$ and the geometric braid group $G_{3}$ of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism; the open and closed ordered configuration spaces $F_{3} (int D^{2}) \subseteq F_{3} (D^{2})$ and $F_{2} (int D^{2}) \subseteq F_{2} (D^{2})$ at the base configurations $Q$ and $Q^{'} : = (q_{1}, q_{2})$ ; the pure braid groups $P B_{3}$ , $P B_{2}$ and the forgetting homomorphism $φ : P B_{3} \to P B_{2}$ of The Fadell-Neuwirth short exact sequence for pure braids.

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

In ZF, AC implies DC (AC implies DC implies countable choice).

[F2]

The standard generators are $A_{i j} = [W_{i j}]$ with $W_{i j} = σ_{j - 1} \dots σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} \dots σ_{j - 1}^{- 1}$ under first-under-second stacking, and the same letters denote their images under the isomorphism $Ψ$ in $P B_{n}$ ; for $n = 3$ this gives $A_{12} = [σ_{1}^{2}]$ , $A_{13} = [σ_{2} σ_{1}^{2} σ_{2}^{- 1}]$ and $A_{23} = [σ_{2}^{2}]$ (Standard geometric pure braid generators A_ij).

[F3]

The geometric three strand relation holds: $[σ_{1}] [σ_{2}] [σ_{1}] = [σ_{2}] [σ_{1}] [σ_{2}]$ in $G_{3}$ (The geometric three strand braid relation).

[F4]

$G_{3}$ is a group with product induced by stacking, $[γ] [β] = [γ ⋆ β]$ , and the classes $[σ_{1}], [σ_{2}]$ generate $G_{3}$ ; the endpoint permutation $π_{g e o} : G_{3} \to S_{3}$ is a homomorphism with $π_{g e o} ([σ_{i}])$ the transposition of $i$ and $i + 1$ , and $G_{3}^{p u r e} = \ker π_{g e o}$ (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, The Artin presentation surjects onto the geometric braid group, The elementary geometric half twist, its support disc, and its opposite).

[F5]

The map $Ψ : G_{n}^{p u r e} \to P B_{n}$ , $Ψ ([β]) = (ι_{*}^{F} [z_{β}])^{- 1}$ , is a group isomorphism, where $z_{β}$ is the coordinate path of $β$ and $ι_{*}^{F}$ is the open-to-closed isomorphism on fundamental groups of configuration spaces (Pure geometric braids and ordered configuration loops, The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[F6]

Under AC the sequence $1 \to F_{2} \overset{κ}{\to} P B_{3} \overset{φ}{\to} P B_{2} \to 1$ is short exact with $im κ = \ker φ$ , and under the identification of $κ$ the elements $A_{13}, A_{23}$ are a free basis of $\ker φ$ , each represented by the clockwise meridian of the corresponding puncture (The Fadell-Neuwirth short exact sequence for pure braids, The $A_{i n}$ are meridian generators of the forgetful free kernel).

[F7]

At the canonical two-point configuration $Q^{(2)} = (- 1 / 12, 1 / 12)$ , $P B_{2}$ is infinite cyclic generated by $A_{12}^{(2)}$ (The two-strand pure braid group is infinite cyclic). Here the quotient is instead $P B_{2} = π_{1} (F_{2} (D^{2}), Q^{'})$ , with $Q^{'} = (- 1 / 8, 0)$ . The path $η (r) = ((1 + r / 3) q_{1} + r / 12, (1 + r / 3) q_{2} + r / 12)$ runs from $Q^{'}$ to $Q^{(2)}$ . Basepoint transport gives an isomorphism $τ : π_{1} (F_{2} (D^{2}), Q^{(2)}) \to π_{1} (F_{2} (D^{2}), Q^{'})$ , $τ ([ℓ]) = [η * ℓ * \overset{ˉ}{η}]$ (Conjugating loop classes by a path is an isomorphism of fundamental groups, Higher homotopy basepoint transport and moving homotopies). Write $a^{'} : = τ (A_{12}^{(2)})$ for the quotient generator; occurrences of $a$ in the cyclic quotient factor of the Example mean $a^{'}$ . Step 1.3 checks that forgetting sends $a \in P B_{3}$ to $a^{'}$ , using the actual coordinate-forgetting map of The Fadell-Neuwirth short exact sequence for pure braids.

[F8]

The extension of [F6] splits: there is a homomorphic section $s : P B_{2} \to P B_{3}$ of $φ$ , and then $P B_{3} ≅ F_{2} ⋊ P B_{2}$ compatibly with $κ$ and $φ$ , the action of $h \in P B_{2}$ on the free kernel being $h \cdot x = s (h) x s (h)^{- 1}$ ; the section is the based version of the far-right explicit section, and for a section the action depends on that section (The pure braid extension splits as a semidirect product, Splitting lemma for groups: a section, a complement, and a semidirect-product decomposition are equivalent).

[F9]

The far-right section of the planar forgetful map is $s (z_{1}, \dots, z_{n - 1}) = (z_{1}, \dots, z_{n - 1}, 1 + ∣ z_{1} ∣ + \dots + ∣ z_{n - 1} ∣)$ for $F_{n} (C) \to F_{n - 1} (C)$ , transported to the disc by the radial homeomorphism $h (w) = w / (1 + ∣ w ∣)$ with inverse $h^{- 1} (z) = z / (1 - ∣ z ∣)$ ; fixing a base configuration $q$ and a path $α$ in the fibre from $q$ to $s^{'} (q^{'})$ , the formula $σ ([β]) = [(α * (s^{'} \circ β)) * \overset{ˉ}{α}]$ defines a homomorphism $σ : π_{1} (F_{n - 1} (int D^{2}), q^{'}) \to π_{1} (F_{n} (int D^{2}), q)$ with ${\tilde{p}}_{*} \circ σ = id$ , and the section $s$ of [F8] is $ι_{*}^{F} \circ σ \circ (ι_{*}^{F})^{- 1}$ (A choice-free continuous section of planar coordinate forgetting).

[F10]

For $i = 1, 2$ the support disc of $σ_{i}$ is $U_{i} = B (m_{i}, \frac{3}{2} h)$ with $m_{1} = - \frac{1}{16}$ and $m_{2} = \frac{1}{16}$ ; the two strands of the braid $σ_{1}^{2}$ stay in $U_{1}$ at every time, and $U_{1} \subseteq B (0, \frac{5}{2} h) = B (0, \frac{5}{32})$ , while $q_{3} = \frac{1}{8} \notin U_{1} \cup {q_{1}, q_{2}}$ and the real interval $[\frac{1}{8}, \frac{37}{64}]$ is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ (The elementary geometric half twist, its support disc, and its opposite).

Verification

technique · direct

1.1A1F1F2F6F8

Choice bookkeeping and the standing identifications. By [F1] the Axiom of Choice [A1] yields DC, so the short exact sequence and the splitting of [F6] and [F8] are available. Throughout, $a, b, c$ denote the elements of $P B_{3}$ named in the statement, that is $a = Ψ ([σ_{1}^{2}])$ , $b = Ψ ([σ_{2} σ_{1}^{2} σ_{2}^{- 1}])$ and $c = Ψ ([σ_{2}^{2}])$ , The quotient generator is $a^{'} = τ (A_{12}^{(2)})$ of [F7]; only the letter $a$ is reused for that cyclic factor, and $b, c$ remain kernel elements in $P B_{3}$ .

1.2F2F3F4F5

The product $a b c$ is central. Put $s : = [σ_{1}]$ and $t : = [σ_{2}]$ in $G_{3}$ . By [F2] the geometric classes of the three words are $s^{2}$ , $t s^{2} t^{- 1}$ and $t^{2}$ , so in the group $G_{3}$ of [F4] the bracketing is immaterial and $s^{2} \cdot t s^{2} t^{- 1} \cdot t^{2} = s^{2} t s^{2} t = s (s t s) s t = s (t s t) s t = (s t)^{3},$ where the middle step uses the relation $t s t = s t s$ of [F3]; put $Δ : = s t s$ . The same relation gives $Δ = t s t$ , hence $Δ s = t Δ$ and $Δ t = s Δ$ , and therefore $Δ^{2} s = Δ (t Δ) = (Δ t) Δ = (s Δ) Δ = s Δ^{2}, Δ^{2} t = Δ (s Δ) = (Δ s) Δ = (t Δ) Δ = t Δ^{2} .$ So $Δ^{2}$ commutes with $s$ and $t$ ; since $s$ and $t$ generate $G_{3}$ by [F4], $Δ^{2}$ is central in $G_{3}$ . Also $π_{g e o} (Δ^{2}) = π_{g e o} (s t)^{3} = i d$ because $π_{g e o} (s t)$ is a product of the two distinct transpositions $(1 2)$ and $(2 3)$ , a three-cycle whose cube is the identity; hence $Δ^{2} \in G_{3}^{p u r e}$ and $z : = Ψ ([Δ^{2}])$ is defined. By [F5] the map $Ψ$ is a homomorphism, so with the identification of [F2] $a b c = Ψ ([σ_{1}^{2}]) Ψ ([σ_{2} σ_{1}^{2} σ_{2}^{- 1}]) Ψ ([σ_{2}^{2}]) = Ψ ([σ_{1}^{2} σ_{2} σ_{1}^{2} σ_{2}^{- 1} \cdot σ_{2}^{2}]) = Ψ ([Δ^{2}]) = z .$ Since $Ψ$ is an isomorphism onto $P B_{3}$ and $Δ^{2}$ is central in $G_{3}$ , the element $z = a b c$ is central in $P B_{3}$ .

1.3F5F6F7F10

Forgetting and the transported quotient generator. Let $z = (z_{1}, z_{2})$ be the two moving coordinates of the rank-three word $σ_{1}^{2}$ , a loop at $Q^{'}$ . Then $φ (a) = (ι_{*}^{F} [z])^{- 1}$ by [F5] and the naturality of coordinate forgetting in [F6]. Define $g_{r} (v) = (1 + r / 3) v + r / 12$ and $H (t, r) = (g_{r} (z_{1} (t)), g_{r} (z_{2} (t)))$ . The map $g_{r}$ is injective, so these coordinates stay distinct. By [F10], $∣ z_{i} (t) ∣ \leq 5 / 32$ , whence $∣ g_{r} (z_{i} (t)) ∣ \leq (4 / 3) (5 / 32) + 1 / 12 = 7 / 24 < 1$ . Thus $H$ is a homotopy through ordered configurations with basepoint track $η$ from [F7]. At $r = 1$ the midpoint $- 1 / 16$ is sent to $0$ , while the relative diamond displacement $ρ$ is multiplied by $4 / 3$ , changing its scale from $1 / 16$ to $1 / 12$ . Therefore $H (-, 1)$ is exactly the raw coordinate loop of the canonical rank-two full twist. The moving-basepoint identity of [F7], and its compatibility with open-to-closed inclusions, give $ι_{*}^{F} [z] = τ (ι_{*}^{F} [z_{σ_{1}^{2}}^{(2)}])$ . Since $τ$ preserves inverses, $φ (a) = τ (A_{12}^{(2)}) = a^{'}$ . By [F7], $a^{'}$ generates this quotient $P B_{2} ≅ Z$ .

2.1step 1.2

The conjugation by $a$ is conjugation by $w^{- 1}$ . Put $w : = b c \in \ker φ$ . By step 1.2, $a b c = z$ with $z$ central, so $a = z (b c)^{- 1} = z w^{- 1}$ , and for every $x \in P B_{3}$ , in particular for every $x \in \ker φ$ , $a x a^{- 1} = z w^{- 1} x w z^{- 1} = w^{- 1} x w .$

2.2F5F8F9F10step 1.3

The far-right section sends $a^{'}$ to $a$ . Write $z_{2} (t) = (z_{1} (t), z_{2} (t))$ for the coordinate path of $σ_{1}^{2}$ in $F_{2} (int D^{2})$ , a loop at $Q^{'}$ whose two entries lie in $U_{1}$ for every $t$ by [F10], and put $u_{i} (t) : = h^{- 1} (z_{i} (t))$ , so that the far-right lift of [F9] is $ς (t) : = (z_{1} (t), z_{2} (t), h (1 + ∣ u_{1} (t) ∣ + ∣ u_{2} (t) ∣)),$ a loop at $s^{'} (Q^{'})$ in $F_{3} (int D^{2})$ . Since $∣ z_{i} (t) ∣ \leq \frac{5}{32}$ by [F10], we have $∣ u_{i} (t) ∣ \leq \frac{5}{27}$ and, because $r \mapsto \frac{r}{1 + r}$ is increasing on $[0, \infty)$ , $h (1 + ∣ u_{1} (t) ∣ + ∣ u_{2} (t) ∣) \in [\frac{1}{2}, \frac{37}{64}],$ an interval in the positive real axis. The value of the far-right section at $Q^{'}$ is $s^{'} (Q^{'}) = (q_{1}, q_{2}, \frac{8}{15})$ : here $h^{- 1} (q_{1}) = (- 1 / 8) / (7 / 8) = - \frac{1}{7}$ and $h^{- 1} (q_{2}) = 0$ , so the third coordinate is $h (1 + \frac{1}{7}) = h (\frac{8}{7}) = \frac{8}{15}$ . Let $α (t) : = (q_{1}, q_{2}, \frac{1}{8} + t (\frac{8}{15} - \frac{1}{8}))$ be the path in the fibre from $Q$ to $s^{'} (Q^{'})$ , whose third coordinate runs along the real interval $[\frac{1}{8}, \frac{8}{15}]$ ; this interval, and likewise $[\frac{1}{2}, \frac{37}{64}]$ , is contained in $[\frac{1}{8}, \frac{37}{64}]$ , which is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ by [F10]. By [F9] the section of [F8] is built from $σ = φ_{\overset{ˉ}{α}} \circ s_{*}^{'}$ with $σ ([z_{2}]) = [(α * ς) * \overset{ˉ}{α}]$ , and we claim this class is the class of the third-strand-fixed loop $ω_{0} (t) : = (z_{1} (t), z_{2} (t), q_{3})$ in $π_{1} (F_{3} (int D^{2}), Q)$ . To see this, let $v : I \to R$ denote the third coordinate path of $γ : = (α * ς) * \overset{ˉ}{α}$ , so that $v (0) = v (1) = q_{3}$ , $v$ takes values in $[\frac{1}{8}, \frac{37}{64}]$ , and $v$ coincides with the third coordinate of $α$ on the first quarter, with $h (1 + ∣ u_{1} ∣ + ∣ u_{2} ∣)$ on the second quarter and with the reversed third coordinate of $α$ on the last half. Let $ψ_{0}$ be the map that is $0$ on $[0, \frac{1}{4}]$ , $4 t - 1$ on $[\frac{1}{4}, \frac{1}{2}]$ and $1$ on $[\frac{1}{2}, 1]$ , and put $ψ_{r} (t) : = (1 - r) ψ_{0} (t) + r t$ for $(r, t) \in I \times I$ . Then $G (r, t) : = (z_{1} (ψ_{r} (t)), z_{2} (ψ_{r} (t)), (1 - r) v (t) + r q_{3})$ is continuous, lies in $F_{3} (int D^{2})$ pointwise: each $ψ_{r} (t)$ lies in $[0, 1]$ , so both $z_{i} (ψ_{r} (t))$ lie in $U_{1}$ and are distinct, while the third coordinate is a convex combination of two points of $[\frac{1}{8}, \frac{37}{64}]$ and therefore also lies in that interval, which is disjoint from $U_{1} \cup {q_{1}, q_{2}}$ by [F10]; moreover $ψ_{r} (0) = 0$ , $ψ_{r} (1) = 1$ and $v (0) = v (1) = q_{3}$ , so $G (r, 0) = G (r, 1) = Q$ for every $r$ . Finally $G (0, t) = γ (t)$ and $G (1, t) = ω_{0} (t)$ . Hence $G$ is a path homotopy relative to ${0, 1}$ from $γ$ to $ω_{0}$ and $σ ([z_{2}]) = [ω_{0}]$ . Now $(ι_{*}^{F})^{- 1} (a^{'}) = [z_{2}]^{- 1}$ in $π_{1} (F_{2} (int D^{2}), Q^{'})$ : by [F5] the element $a^{'} = φ (a) \in P B_{2}$ is $(ι_{*}^{F} [z_{2}])^{- 1}$ and $ι_{*}^{F}$ is an isomorphism. Since $σ$ is a homomorphism, $σ ([z_{2}]^{- 1}) = [ω_{0}]^{- 1}$ , and since the element $a = A_{12}$ of $P B_{3}$ is by [F5] the class $(ι_{*}^{F} [ω_{0}])^{- 1}$ (the coordinate path of the braid $σ_{1}^{2}$ in $G_{3}$ is $ω_{0}$ ), the section $s$ of [F9] satisfies $s (a^{'}) = ι_{*}^{F} (σ ((ι_{*}^{F})^{- 1} (a^{'}))) = ι_{*}^{F} ([ω_{0}]^{- 1}) = a .$

3.1F6F8step 1.3step 2.1step 2.2∎

The action and the semidirect product. By step 2.2 the section of [F8] satisfies $s (a^{'}) = a$ , and by step 2.1 $a x a^{- 1} = w^{- 1} x w$ for every $x$ in the free kernel $\ker φ = ⟨ b, c ⟩$ of [F6]. Since the action of the section is $h \cdot x = s (h) x s (h)^{- 1}$ by [F8], the positive generator $a^{'}$ of $P B_{2} = ⟨ a^{'} ⟩ ≅ Z$ of step 1.3 acts by $x \mapsto w^{- 1} x w$ . The splitting of [F8] therefore exhibits $P B_{3} ≅ ⟨ b, c ⟩ ⋊ ⟨ a ⟩ ≅ F_{2} ⋊ Z$ with that action. The action is conjugation by the element $w^{- 1}$ of the free kernel, so it depends on the normalised section, no triviality of the action is claimed, and no direct-product decomposition is asserted.

Remarks

The computation of step 1.2 is a direct rank-three calculation inside $G_{3}$ : the full twist $Δ^{2} = (s t)^{3}$ is written as the product of the three standard generators $a, b, c$ , and centrality of that product is read off the single braid relation. The later centre theorem for $P B_{n}$ is not used, and neither is any Artin-presentation injectivity: only the surjectivity of The Artin presentation surjects onto the geometric braid group enters, through the generation of $G_{3}$ by $s$ and $t$ .
The section used above is the normalised far-right section: the fibre path is the straight segment from $q_{3}$ to $s^{'} (Q^{'})_{3}$ on the positive real axis, and on the small representative $σ_{1}^{2}$ of $a$ the far-right lift is homotopic to the loop with constant third coordinate, which is why $s (a^{'}) = a$ . Any other section $s^{'}$ of $φ$ has $s^{'} (a^{'}) = k a$ for some $k \in \ker φ$ , so its action is $x \mapsto k (w^{- 1} x w) k^{- 1}$ , an inner automorphism of the free kernel; the displayed formula is the one for this section, and clearing it of the normalisation would require a separate conjugation bookkeeping.
The action is by an inner automorphism of the free kernel, because $w \in ⟨ b, c ⟩$ itself. This example nevertheless asserts only the semidirect-product decomposition with the action of the chosen section; the classical direct-product decomposition $P B_{3} ≅ F_{2} \times Z$ is not derived here.

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, section 2.1, printed pp. 11-14 (the split pure braid tower and the explicit cross-section) (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 1.3, author manuscript pp. 5-7 (the free kernel and the splitting of the pure braid sequence) (standard reference, not scraped)