Artin Presentation Completeness and Braid Combing — Examples

The two halves of $W$ are trivial. In $B_4$ the relation ${\sigma }_3{\sigma }_2{\sigma }_3={\sigma }_2{\sigma }_3{\sigma }_2$ replaces the first three letters of $W$, and then two free deletions give ${\sigma }_2{\sigma }_3{\sigma }_2{\sigma }_2^{-1}{\sigma }_3^{-1}{\sigma }_2^{-1}\equiv {\sigma }_2{\sigma }_3{\sigma }_3^{-1}{\sigma }_2^{-1}\equiv {\sigma }_2{\sigma }_2^{-1}\equiv 1$; the same relation with index $1$ gives ${\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}\equiv {\sigma }_2{\sigma }_1{\sigma }_2{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}\equiv {\sigma }_2{\sigma }_1{\sigma }_1^{-1}{\sigma }_2^{-1}\equiv 1$ for the last six letters. Hence $W$ is equivalent to the empty word, and ${\phi }_4(W)=1$ because ${\phi }_4$ is a homomorphism: $W$ traces the trivial geometric braid.

The position sequence. Since the geometric image of $W$ is trivial, the tracked point returns to its initial position, $j_{12}=j_0=4$. The letter ${\sigma }_k^{\pm 1}$ interchanges positions $k$ and $k+1$ and fixes the others, so the tracked point passes from position $4$ to $3$ at the first letter ${\sigma }_3$, from $3$ to $2$ at ${\sigma }_2$, stays at $2$ under the next ${\sigma }_3$, passes to $3$ at ${\sigma }_2^{-1}$ and back to $4$ at ${\sigma }_3^{-1}$; the remaining letters ${\sigma }_2^{-1}$ and the six letters with index at most $2$ act only on the first three positions, so the tracked point stays at $4$. The sequence is therefore $4,3,2,2,3,4,4,4,4,4,4,4,4$.

The twelve combing factors. With the positions of step 2.1 and ${\alpha }_4=1$, ${\alpha }_3={\sigma }_3$, ${\alpha }_2={\sigma }_2{\sigma }_3$ from [F2], [F3] writes $W$ as the product of the twelve factors $F_k={\alpha }_{j_{k-1}}^{-1}{\sigma }_{i_k}^{{\epsilon }_k}{\alpha }_{j_k}$: $F_1={\sigma }_3^2,F_2={\sigma }_3^{-1}{\sigma }_2^2{\sigma }_3,F_3={\alpha }_2^{-1}{\sigma }_3{\alpha }_2,F_4={\sigma }_3^{-1}{\sigma }_2^{-2}{\sigma }_3,F_5={\sigma }_3^{-1}{\sigma }_3^{-1},$ and $F_6={\sigma }_2^{-1}$, $F_7={\sigma }_1$, $F_8={\sigma }_2$, $F_9={\sigma }_1$, $F_{10}={\sigma }_2^{-1}$, $F_{11}={\sigma }_1^{-1}$, $F_{12}={\sigma }_2^{-1}$.

Six-case reduction. By [F4], applied with the pair $(j,k,\epsilon )$ of each factor: $F_1={\alpha }_4^{-1}{\sigma }_3{\alpha }_3=x_3$ and $F_2={\alpha }_3^{-1}{\sigma }_2{\alpha }_2=x_2$ (case $j=k+1$, $\epsilon =1$); $F_3={\alpha }_2^{-1}{\sigma }_3{\alpha }_2$ has $k=3>j=2$ and reduces to ${\sigma }_{k-1}={\sigma }_2$; $F_4={\alpha }_2^{-1}{\sigma }_2^{-1}{\alpha }_3=x_2^{-1}$ and $F_5={\alpha }_3^{-1}{\sigma }_3^{-1}{\alpha }_4=x_3^{-1}$ (case $j=k$, $\epsilon =-1$); and $F_6,\dots ,F_{12}$ all have $k\le 2<j-1=3$ (with $j=4$, so that ${\alpha }_j=1$ on both sides of the letter) and reduce to the letters ${\sigma }_2^{-1},{\sigma }_1,{\sigma }_2,{\sigma }_1,{\sigma }_2^{-1},{\sigma }_1^{-1},{\sigma }_2^{-1}$ themselves. Hence $W\equiv x_3{x}_2{\sigma }_2{x}_2^{-1}{x}_3^{-1}{\sigma }_2^{-1}{\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}\mathrm{.}$

Collecting the $x$-letters. By [F5] with $i=2$: ${\sigma }_2{x}_2{\sigma }_2^{-1}=x_2^{-1}{x}_3{x}_2$, hence ${\sigma }_2{x}_2^{-1}=(x_2^{-1}{x}_3{x}_2{)}^{-1}{\sigma }_2=x_2^{-1}{x}_3^{-1}{x}_2{\sigma }_2$; and ${\sigma }_2{x}_3{\sigma }_2^{-1}=x_2$, hence ${\sigma }_2{x}_3^{-1}=x_2^{-1}{\sigma }_2$. Substituting these two identities into the word of step 4.1, $x_3{x}_2({\sigma }_2{x}_2^{-1})x_3^{-1}{\sigma }_2^{-1}(\text{rest})\equiv x_3{x}_2{x}_2^{-1}{x}_3^{-1}{x}_2({\sigma }_2{x}_3^{-1}){\sigma }_2^{-1}(\text{rest})\equiv x_3{x}_2{x}_2^{-1}{x}_3^{-1}{x}_2{x}_2^{-1}{\sigma }_2{\sigma }_2^{-1}(\text{rest}),$ where $(\text{rest})={\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}$; deleting the adjacent pair ${\sigma }_2{\sigma }_2^{-1}$ gives $W\equiv W_1{W}_2$ with $W_1=x_3{x}_2{x}_2^{-1}{x}_3^{-1}{x}_2{x}_2^{-1}$ and $W_2={\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}$.

Both factors are trivial. The word $W_1$ reduces to the empty word by the free cancellations $x_2{x}_2^{-1}$ and $x_3{x}_3^{-1}$: $x_3{x}_2{x}_2^{-1}{x}_3^{-1}{x}_2{x}_2^{-1}\equiv x_3{x}_3^{-1}{x}_2{x}_2^{-1}\equiv 1$. The word $W_2$ is trivial in the rank-3 subgroup: the braid relation ${\sigma }_1{\sigma }_2{\sigma }_1={\sigma }_2{\sigma }_1{\sigma }_2$ gives $W_2\equiv {\sigma }_2{\sigma }_1{\sigma }_2{\sigma }_2^{-1}{\sigma }_1^{-1}{\sigma }_2^{-1}\equiv {\sigma }_2{\sigma }_1{\sigma }_1^{-1}{\sigma }_2^{-1}\equiv 1$. Thus the combing algorithm decomposes $W$ into a factor that is freely trivial in the $x$-letters and a factor that is trivial on the lower rank, which is exactly the mechanism of Every trivial braid word combs as W_1W_2 and of the completeness theorem; the example illustrates that a word can fail to be freely trivial after combing while both of its combed factors are accounted for. ∎

The combing words. By [F1], $x_2={\sigma }_2^2$ and $x_1={\alpha }_2^{-1}{\sigma }_1^2{\alpha }_2={\sigma }_2^{-1}{\sigma }_1^2{\sigma }_2$; both displays are literal word computations from the definition, with ${\alpha }_3=1$ the empty word.

The standard generators. By [F2], $A_{23}=[W_{23}]$ with $W_{23}={\sigma }_2^2$ (both outer blocks of the display empty) and $A_{13}=[W_{13}]$ with $W_{13}={\sigma }_2{\sigma }_1^2{\sigma }_2^{-1}$.

The two identities, by free cancellation. Substituting the words of steps 1.1 and 1.2, $A_{23}^{-1}{A}_{13}{A}_{23}={\sigma }_2^{-2}\left({\sigma }_2{\sigma }_1^2{\sigma }_2^{-1}\right){\sigma }_2^2\equiv {\sigma }_2^{-1}{\sigma }_1^2{\sigma }_2=x_1$ and $A_{23}=x_2$; the only moves are deletions of the adjacent inverse pairs ${\sigma }_2^{-1}{\sigma }_2$ and ${\sigma }_2^{-1}{\sigma }_2$, in the middle of the first display. This is the identity $x_1=P_1^{-1}{A}_{13}{P}_1$ with $P_1=A_{23}$ of the combing lemma at $n=3$.

The kernel is freely generated by the two combing words. Assume AC, so that the kernel $K$ of $P{B}_3\to P{B}_2$ is free with basis $A_{13},A_{23}$ by [F3]. Identify $K$ with the abstract free group $F(a_1,a_2)$ through $a_1\mapsto A_{13}$, $a_2\mapsto A_{23}$, and define the endomorphism $\theta \text{\hspace{0.17em}⁣}:F\to F$ on that basis by $\theta (a_1):=a_2{a}_1{a}_2^{-1}$, $\theta (a_2):=a_2$. Let $\rho \text{\hspace{0.17em}⁣}:F\to F$ be the homomorphism with $\rho (a_1):=a_2^{-1}{a}_1{a}_2$ and $\rho (a_2):=a_2$. Then $\theta (\rho (a_1))=\theta (a_2{)}^{-1}\theta (a_1)\theta (a_2)=a_2^{-1}(a_2{a}_1{a}_2^{-1})a_2=a_1$ and $\theta (\rho (a_2))=a_2$, so $\theta \circ \rho$ is the identity on the free basis and hence on $F$; therefore $\rho$ is injective and the elements $\rho (a_1)=A_{23}^{-1}{A}_{13}{A}_{23}$, $\rho (a_2)=A_{23}$ are a free basis of the subgroup they generate. By step 2.1 these are ${\mathrm{\Psi }}_3({\phi }_3(x_1))$ and ${\mathrm{\Psi }}_3({\phi }_3(x_2))$ under the identification of [F4], and they generate $K$: each of $A_{13}=A_{23}\rho (a_1)A_{23}^{-1}$ and $A_{23}=\rho (a_2)$ lies in $⟨\rho (a_1),\rho (a_2)⟩$, while both $\rho (a_i)$ lie in $K$. Hence ${\mathrm{\Psi }}_3({\phi }_3(x_1)),{\mathrm{\Psi }}_3({\phi }_3(x_2))$ are a free basis of $K$; the argument is a free-group computation and uses no choice principle beyond the freeness of $K$ supplied by [F3].

Conclusion. Combining steps 2.1 and 3.1: the two combing words satisfy $x_2=A_{23}$ and $x_1=A_{23}^{-1}{A}_{13}{A}_{23}$ by free cancellation, and the free kernel of $P{B}_3\to P{B}_2$, freely generated by $A_{13},A_{23}$, is equally freely generated by $x_1,x_2$. Under the fibre-inclusion identification, write ${\mu }_i$ for the clockwise meridian corresponding to $A_{i3}$ by [F3]. Step 2.1 gives the fibre classes ${\mu }_2^{-1}{\mu }_1{\mu }_2$ and ${\mu }_2$ for $x_1$ and $x_2$ respectively. In the free group these first-meridian classes differ: the word ${\mu }_2^{-1}{\mu }_1{\mu }_2$ is reduced and is not ${\mu }_1$; conjugation changes the based stem class. ∎

The assignment $x\mapsto g$ induces a surjection of $P$ onto $G$. By [F2] one has $g^2=1_G$, hence $g^4=(g^2{)}^2=1_G$; the evaluation of the single defining relator $x^4$ of $P$ under the function $u(x):=g$ is therefore $g^4=1_G$, so [F1] provides a unique homomorphism $\phi \text{\hspace{0.17em}⁣}:P\to G$ with $\phi (x)=g$. Its image contains $g$, and $g$ generates $G$ by [F2], so $\phi$ is surjective by the surjectivity criterion of [F1].

$x^2$ is nontrivial in $P$. The evaluation of the relator $x^4$ under the function $u^{\mathrm{\prime }}(x):=[1{]}_4$ is $[1{]}_4^4=[4{]}_4=[0{]}_4$ by [F4], so [F1] provides a homomorphism $\psi \text{\hspace{0.17em}⁣}:P\to \mathbb{Z}\mathrm{/}4$ with $\psi (x)=[1{]}_4$. Then $\psi (x^2)=[1{]}_4^2=[2{]}_4\mathrm{\ne }[0{]}_4=\psi (1_P)$ by [F4], because a homomorphism carries the identity to the identity; hence $x^2\mathrm{\ne }{1}_P$ in $P$.

The surjection is not injective and the groups are not isomorphic. By step 1.1, $\phi (x^2)=\phi (x{)}^2=g^2=1_G$ using [F2], while $x^2\mathrm{\ne }{1}_P$ by step 1.2, so the nontrivial element $x^2$ lies in $\ker \phi$ and $\phi$ is not injective. In fact the two groups have different sizes: by [F3] every element of $P$ is one of $x^0,x^1,x^2,x^3$, and $\psi$ from step 1.2 is surjective onto the four-element group $\mathbb{Z}\mathrm{/}4$ by [F4] and the criterion of [F1], so $P$ has exactly four elements; by [F2] every element of $G$ is $1_G$ or $g$, and the function $v(g):=[1{]}_2$ evaluates the relator $g^2$ to $[2{]}_2=[0{]}_2$, so [F1] yields a homomorphism $G\to \mathbb{Z}\mathrm{/}2$ with $g\mapsto [1{]}_2\mathrm{\ne }[0{]}_2$, whence $g\mathrm{\ne }{1}_G$ and $G$ has exactly two elements. Since $4\mathrm{\ne }2$, no bijection and hence no group isomorphism $P\to G$ exists: the presentation $⟨x\mid x^4=1⟩$ does not present $G$, even though its defining relator holds in $G$ and the image of its generator generates $G$. This refutes the inference under examination.

The braid application. For the Artin presentation, [F5] verifies exactly the two hypotheses of the refuted inference — the Artin relators evaluate to the identity among the geometric half-twist classes, and these classes generate $G_n$ — and von Dyck yields precisely the surjection ${\phi }_n$, with no injectivity. The counterexample of step 2.1 shows that this pattern of hypotheses does not in general force the von Dyck map to be an isomorphism, so the published surjectivity argument alone cannot establish that the Artin presentation presents the geometric braid group. What it leaves open is the triviality of $\ker {\phi }_n$; the combing kernel argument of [F6] supplies exactly that missing obligation, under AC.

Conclusion. The counterexample $P\to G$ refutes the general inference, and the parallel von Dyck pattern of the Artin surjection shows that the geometric completeness statement needs the separate kernel computation of [F6]; the counterexample's own arithmetic is choice-free, while AC is carried only because the completeness supplier it contrasts with is AC-dependent. ∎