The Artin Action on a Free Group — Examples

The table. Substituting the frozen formulas for $n=3$ gives the table displayed above: $\rho ({\sigma }_1)$ moves only $x_1,x_2$, and $\rho ({\sigma }_2)$ moves only $x_2,x_3$.

The composite $\rho ({\sigma }_1)\rho ({\sigma }_2)\rho ({\sigma }_1)$. Composing the table (rightmost letter first) gives $x_1\mapsto x_1{x}_2{x}_3{x}_2^{-1}{x}_1^{-1},x_2\mapsto x_1{x}_2{x}_1^{-1},x_3\mapsto x_1\mathrm{.}$ Indeed: applying $\rho ({\sigma }_1)$ first gives $(x_1,x_2,x_3)\mapsto (x_1{x}_2{x}_1^{-1},x_1,x_3)$; applying $\rho ({\sigma }_2)$ gives $(x_1{x}_2{x}_3{x}_2^{-1}{x}_1^{-1},x_1,x_2)$; and applying $\rho ({\sigma }_1)$ again gives $(x_1{x}_2{x}_3{x}_2^{-1}{x}_1^{-1},x_1{x}_2{x}_1^{-1},x_1)$.

The composite $\rho ({\sigma }_2)\rho ({\sigma }_1)\rho ({\sigma }_2)$. Composing in the opposite order gives $x_1\mapsto x_1{x}_2{x}_3{x}_2^{-1}{x}_1^{-1},x_2\mapsto x_1{x}_2{x}_1^{-1},x_3\mapsto x_1\mathrm{.}$ Indeed: applying $\rho ({\sigma }_2)$ first gives $(x_1,x_2{x}_3{x}_2^{-1},x_2)$; applying $\rho ({\sigma }_1)$ gives $(x_1{x}_2{x}_1^{-1},x_1{x}_3{x}_1^{-1},x_1)$; and applying $\rho ({\sigma }_2)$ again gives $(x_1{x}_2{x}_3{x}_2^{-1}{x}_1^{-1},x_1{x}_2{x}_1^{-1},x_1)$.

Comparison. The two composites of steps 2.1 and 2.2 agree on each of $x_1,x_2,x_3$, hence on the whole free basis; by [F1] they are equal as automorphisms, which verifies the braid relation in $B_3$.

Base case $m=0$. For $m=0$ the formula reads $x_k={\delta }_0{x}_{k\oplus 0}{\delta }_0^{-1}=x_k$, which holds since $U^0=\mathrm{id}$ and ${\delta }_0=1$.

Induction hypothesis. Assume that for some $m$ with $0\le m<n$ the formula $U^m(x_k)={\delta }_m{x}_{k\oplus m}{\delta }_m^{-1}$ holds for every $k$.

The action of $U$ on the generators and on ${\delta }_m$. By [F1], applying the factors of $U$ from the right (that is, $\rho ({\sigma }_{n-1})$ first) to a basis letter gives $U(x_k)=x_1{x}_{k\oplus 1}{x}_1^{-1}(1\le k\le n),$ with the wrap convention $x_{n\oplus 1}=x_1$: for $k<n$ the factors $\rho ({\sigma }_{n-1}),\dots ,\rho ({\sigma }_{k+1})$ fix $x_k$, the factor $\rho ({\sigma }_k)$ sends $x_k\mapsto x_k{x}_{k+1}{x}_k^{-1}$, and the factors $\rho ({\sigma }_{k-1}),\dots ,\rho ({\sigma }_1)$ successively replace the left and right occurrences of $x_k$ by $x_{k-1},\dots ,x_1$, leaving $x_1{x}_{k+1}{x}_1^{-1}$; for $k=n$ the factors send $x_n\mapsto x_{n-1}\mapsto \cdots \mapsto x_1$. Hence, multiplying the $m$ images and telescoping the inner conjugations, $U({\delta }_m)=U(x_1)\cdots U(x_m)=(x_1{x}_2{x}_1^{-1})(x_1{x}_3{x}_1^{-1})\cdots (x_1{x}_{m+1}{x}_1^{-1})={\delta }_{m+1}{x}_1^{-1}\mathrm{.}$

The induction step. By the induction hypothesis of step 1.2 and the fact that $U$ is an automorphism, $U^{m+1}(x_k)=U\left({\delta }_m{x}_{k\oplus m}{\delta }_m^{-1}\right)=U({\delta }_m)U(x_{k\oplus m})U({\delta }_m{)}^{-1}\mathrm{.}$ Substituting step 1.3 and using $(k\oplus m)\oplus 1=k\oplus (m+1)$ gives $U^{m+1}(x_k)=\left({\delta }_{m+1}{x}_1^{-1}\right)\left(x_1{x}_{k\oplus (m+1)}{x}_1^{-1}\right)\left(x_1{\delta }_{m+1}^{-1}\right)={\delta }_{m+1}{x}_{k\oplus (m+1)}{\delta }_{m+1}^{-1}\mathrm{.}$

Discharge and conclusion. Steps 1.1 and 2.1 establish the displayed formula for every $0\le m\le n$ by induction; at $m=n$ it reads $U^n(x_k)=\delta x_{k\oplus n}{\delta }^{-1}=\delta x_k{\delta }^{-1}$. By [F2] $\rho ({\mathrm{\Delta }}^2)=U^n$, so $\rho ({\mathrm{\Delta }}^2)(x_k)=\delta x_k{\delta }^{-1}$ for every $k$; by [F3] the element $\delta$ is the boundary word, so the full twist acts by conjugation by it. For $n=1$ there is no generator, ${\mathrm{\Delta }}^2$ is the empty product, $U=\mathrm{id}$ and $\delta =x_1$, and the identity $\rho (1)(x_1)=x_1=\delta x_1{\delta }^{-1}$ holds; for $n=0$ the assertion is vacuous. All computations are finite substitutions in the free basis, and no choice principle is used.

$\epsilon$ permutes peripheral conjugacy classes. By [F1], $\epsilon$ is an automorphism and $\epsilon (x_i)$ is the generator $x_2$ for $i=1$, the generator $x_1$ for $i=2$, and the generator $x_i$ for $i\ge 3$; in each case $\epsilon (x_i)$ is a generator, hence conjugate to a generator (via the empty word).

$\epsilon$ changes the boundary word. By [F1], $\epsilon (x_1{x}_2{x}_3\cdots x_n)=\epsilon (x_1)\epsilon (x_2)\epsilon (x_3)\cdots \epsilon (x_n)=x_2{x}_1{x}_3\cdots x_n\mathrm{.}$ The words $x_2{x}_1{x}_3\cdots x_n$ and $x_1{x}_2{x}_3\cdots x_n$ are both reduced; for $n\ge 2$ they differ in their first two letters, so by reduced-word uniqueness they represent different elements of $F_n$: $\epsilon (x_1\cdots x_n)\mathrm{\ne }{x}_1\cdots x_n$.

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 $j\ge 3$. 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 $A^2=\mathrm{id}$ and $A$ is an automorphism. Moreover, $A(x_1\cdots x_n)=x_1^{-1}{x}_1^2{x}_2\cdots x_n=x_1\cdots x_n$. The same universal property gives a homomorphism $h:F_n\to (\mathbb{Z},+)$ with $h(x_1)=1$ and $h(x_j)=0$ for $j\mathrm{\ne }1$. 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.

$\epsilon$ is outside the braid image. By [F2] every automorphism in the image of $\rho$ fixes the ordered product $x_1\cdots x_n$. Step 1.2 shows that $\epsilon$ does not, so $\epsilon \mathrm{\ne }\rho (\beta )$ for every braid word $\beta$: the automorphism $\epsilon$ permutes the meridian conjugacy classes but is not induced by a braid.

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 $n\ge 2$, and step 2.1 establishes the stated exclusion of $\epsilon$ from the braid image. Only the choice-free necessary direction [F2], explicit free-group homomorphisms, and finite word computations were used.

The value of $\rho ({\sigma }_1^2)$ on $x_1$. By [F1], $\rho ({\sigma }_1)(x_1)=x_1{x}_2{x}_1^{-1}$ and $\rho ({\sigma }_1)(x_2)=x_1$, so, using that $\rho ({\sigma }_1)$ is an automorphism, $\rho ({\sigma }_1^2)(x_1)=\rho ({\sigma }_1)\left(x_1{x}_2{x}_1^{-1}\right)=\rho ({\sigma }_1)(x_1)\rho ({\sigma }_1)(x_2)\rho ({\sigma }_1)(x_1{)}^{-1}=x_1{x}_2{x}_1{x}_2^{-1}{x}_1^{-1}\mathrm{.}$

The two automorphisms differ. The words $x_1{x}_2{x}_1{x}_2^{-1}{x}_1^{-1}$ and $x_1$ are both reduced; they differ, so by [F3] they represent different elements of $F_3$. Hence $\rho ({\sigma }_1^2)(x_1)\mathrm{\ne }{x}_1=\rho (1)(x_1)$, so $\rho ({\sigma }_1^2)\mathrm{\ne }\rho (1)$. Since $\rho$ is a well-defined function on $B_3$ with $\rho (1)=\mathrm{id}$, the word ${\sigma }_1^2$ does not represent the trivial braid: ${\sigma }_1^2\mathrm{\ne }1$ in $B_3$.

The two endpoint permutations agree. By [F2], ${\pi }_3({\sigma }_1)=(12)$, so ${\pi }_3({\sigma }_1^2)=(12{)}^2=1={\pi }_3(1)$: the braid ${\sigma }_1^2$ and the identity braid induce the same trivial permutation of the three strands.

Conclusion. Steps 1.2 and 1.3 exhibit the distinct braids $1$ and ${\sigma }_1^2$ in $B_3$ with equal endpoint permutation; the word ${\sigma }_1^2$ is moreover pure. Therefore the endpoint permutation of a braid does not determine the braid. The computation used finitely many substitutions and the choice-free suppliers [F1]-[F3], so no choice principle is used.