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A conjugate-permuting automorphism that does not fix the boundary word is not in the braid image

Statement refuted

For $n \geq 2$ let $ε \in Aut (F_{n})$ be $ε (x_{1}) = x_{2}, ε (x_{2}) = x_{1}, ε (x_{j}) = x_{j} (j \geq 3) .$ Then $ε (x_{i})$ is conjugate (indeed equal) to a generator for every $i$ , but $ε (x_{1} x_{2} x_{3} \dots x_{n}) = x_{2} x_{1} x_{3} \dots x_{n} \neq x_{1} x_{2} \dots x_{n};$ hence $ε$ is not in the image of the Artin representation. No choice principle is used.

Facts & Assumptions

Given: the free group $F_{n} = ⟨ x_{1}, \dots, x_{n} ⟩$ with $n \geq 2$ , the basis permutation $ε$ of the statement, and the Artin representation $ρ : B_{n} \to Aut (F_{n})$ of The Artin representation on a free group.

[F1]

The permutation is an automorphism. A map of the free basis $x_{1}, \dots, x_{n}$ extends uniquely to a group homomorphism $F_{n} \to F_{n}$ , and the same is true of its inverse permutation, so $ε$ is an automorphism with $ε^{- 1} = ε$ . (Free group on a set of generators, Reduced words form the free group on an alphabet.)

[F2]

Necessary condition. For every braid word $β$ and every $i$ , the element $ρ (β) (x_{i})$ is conjugate in $F_{n}$ to one of the generators and $ρ (β) (x_{1} \dots x_{n}) = x_{1} \dots x_{n}$ ; this necessary direction uses no choice principle. (Artin automorphisms permute meridian conjugacy classes and fix the boundary word.)

[F3]

The two conditions. An automorphism satisfying the two properties of [F2] is called peripheral-boundary-preserving (Peripheral-boundary-preserving automorphisms of F_n); the counterexample shows that the first condition alone does not suffice.

Counterexample

Take the basis permutation $ε$ of the statement and compare it with the necessary condition of [F2].

1.1F1F3

$ε$ permutes peripheral conjugacy classes. By [F1], $ε$ is an automorphism and $ε (x_{i})$ is the generator $x_{2}$ for $i = 1$ , the generator $x_{1}$ for $i = 2$ , and the generator $x_{i}$ for $i \geq 3$ ; in each case $ε (x_{i})$ is a generator, hence conjugate to a generator (via the empty word).

1.2F1algebra

$ε$ changes the boundary word. By [F1], $ε (x_{1} x_{2} x_{3} \dots x_{n}) = ε (x_{1}) ε (x_{2}) ε (x_{3}) \dots ε (x_{n}) = x_{2} x_{1} x_{3} \dots x_{n} .$ The words $x_{2} x_{1} x_{3} \dots x_{n}$ and $x_{1} x_{2} x_{3} \dots x_{n}$ are both reduced; for $n \geq 2$ they differ in their first two letters, so by reduced-word uniqueness they represent different elements of $F_{n}$ : $ε (x_{1} \dots x_{n}) \neq x_{1} \dots x_{n}$ .

1.3F1F3constructalgebra

The reverse nonimplication. For the complementary witness described in the Remarks of Peripheral-boundary-preserving automorphisms of F_n, use [F1]'s free-group universal property to define $A (x_{1}) = x_{1}^{- 1}$ , $A (x_{2}) = x_{1}^{2} x_{2}$ , and $A (x_{j}) = x_{j}$ for $j \geq 3$ . Applying $A$ twice gives $A^{2} (x_{1}) = x_{1}$ and $A^{2} (x_{2}) = x_{1}^{- 2} x_{1}^{2} x_{2} = x_{2}$ , with all other generators fixed; hence $A^{2} = id$ and $A$ is an automorphism. Moreover, $A (x_{1} \dots x_{n}) = x_{1}^{- 1} x_{1}^{2} x_{2} \dots x_{n} = x_{1} \dots x_{n}$ . The same universal property gives a homomorphism $h : F_{n} \to (Z, +)$ with $h (x_{1}) = 1$ and $h (x_{j}) = 0$ for $j \neq 1$ . Conjugation preserves $h$ , but $h (A (x_{1})) = - 1$ , whereas every positive basis generator has $h$ -value $0$ or $1$ . Thus $A (x_{1})$ is not conjugate to any positive basis generator: $A$ satisfies the boundary condition and fails the peripheral condition.

2.1F2step 1.2

$ε$ is outside the braid image. By [F2] every automorphism in the image of $ρ$ fixes the ordered product $x_{1} \dots x_{n}$ . Step 1.2 shows that $ε$ does not, so $ε \neq ρ (β)$ for every braid word $β$ : the automorphism $ε$ permutes the meridian conjugacy classes but is not induced by a braid.

3.1F2step 1.1step 1.2step 2.1step 1.3∎

Conclusion. Steps 1.1 and 1.2 show that the peripheral condition does not imply boundary preservation; step 1.3 proves the reverse nonimplication. Thus the two conditions are independent for $n \geq 2$ , and step 2.1 establishes the stated exclusion of $ε$ from the braid image. Only the choice-free necessary direction [F2], explicit free-group homomorphisms, and finite word computations were used.

Remarks

The example also shows that the condition on $δ$ cannot be checked in the abelianisation: $ε (δ) = x_{2} x_{1} x_{3} \dots x_{n}$ has the same abelianised class as $δ$ , so condition (2) is genuinely stronger than the abelianised equality. For $n = 2$ , $ε (δ) = x_{2} x_{1}$ is even conjugate to $δ = x_{1} x_{2}$ , since $x_{2} x_{1} = x_{2} (x_{1} x_{2}) x_{2}^{- 1}$ , while for $n \geq 3$ the cyclic reduced words of $x_{2} x_{1} x_{3} \dots x_{n}$ and $x_{1} x_{2} x_{3} \dots x_{n}$ differ; in every case $ε$ changes $δ$ itself.
The complement of this example is the sufficiency theorem thm-every-peripheral-boundary-preserving-free-group-automorphism-is-an-artin-automorphism: once both conditions hold, the automorphism is induced by a braid word.

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Sources

Emil Artin, Theory of Braids, Annals of Mathematics 48 (1947), pp. 101-126, condition (16) and Theorem 16, printed p. 114 (standard reference, not scraped)
Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, Theorem 1.3 conditions (1) and (2), printed p. 9 (standard reference, not scraped)