LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Conjugation by the half twist reverses Artin generators

Statement

Let $n \geq 2$ , let $Δ = Δ_{n} = T_{1} T_{2} \dots T_{n - 1}$ be the half twist of The Garside half twist and simple positive braids, where $T_{k} = σ_{k} σ_{k - 1} \dots σ_{1}$ and $U_{k} = σ_{1} \dots σ_{k}$ , and let $τ$ be the index-reversal automorphism of $B_{n}^{+}$ induced by $σ_{j} \mapsto σ_{n - j}$ (it is well defined because it permutes the defining relations of Positive braid monoid). Then:

(a) The conjugation identity. $σ_{i} Δ \equiv^{+} Δ σ_{n - i}$ for every $1 \leq i \leq n - 1$ .

(b) The mirror identity. $Δ σ_{i} \equiv^{+} σ_{n - i} Δ$ for every $i$ ; more generally $Δ w \equiv^{+} τ (w) Δ$ and $w Δ \equiv^{+} Δ τ (w)$ for every positive word $w$ .

(c) Index reversal fixes the half twist. $τ (Δ) = Δ$ , and the class of the reversed word $Δ^{r e v} = U_{n - 1} U_{n - 2} \dots U_{1}$ is $Δ$ as well; equivalently $ρ (Δ) = Δ$ for the reversal anti-automorphism $ρ$ of The positive braid monoid is left and right cancellative.

(d) The square of the half twist is central. $w Δ^{2} \equiv^{+} Δ^{2} w$ for every positive word $w$ ; in particular $σ_{i} Δ^{2} \equiv^{+} Δ^{2} σ_{i}$ for every $i$ .

(e) Sliding a generator through a triangular block. $σ_{j} T_{k} \equiv^{+} T_{k} σ_{j + 1}$ whenever $1 \leq j \leq k - 1$ . The restriction $j \leq k - 1$ is essential: when $k \leq n - 2$ , the analogous words $σ_{k} T_{k}$ and $T_{k} σ_{k + 1}$ have different supports, hence are distinct in $B_{n}^{+}$ .

For $n = 0, 1$ the alphabet is empty, $Δ = 1$ and all statements are trivial. No choice principle is used.

Facts & Assumptions

Given: A natural number $n \geq 2$ , the monoid $B_{n}^{+}$ with its generators $σ_{i}$ , the blocks $T_{k} = σ_{k} \dots σ_{1}$ and $U_{k} = σ_{1} \dots σ_{k}$ , the half twist $Δ = T_{1} \dots T_{n - 1}$ of length $N = n (n - 1) / 2$ , and the index-reversal map $τ$ .

[F1]

$B_{n}^{+} = Σ_{n}^{*} / ⁣ \equiv^{+}$ with $[u v] = [u] [v]$ , generated by the braid pairs $σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1}$ ( $1 \leq i \leq n - 2$ ) and the far-commutation pairs $σ_{i} σ_{j} = σ_{j} σ_{i}$ ( $∣ i - j ∣ \geq 2$ ); $ℓ ([w]) = ∣ w ∣$ with $ℓ (x) = 0$ only for $x = 1$ . Every defining pair preserves the set of generators occurring in a word (its support), so an equivalence derivation beginning in a sub-alphabet stays in that sub-alphabet (Positive braid monoid, Positive artin relations preserve homogeneous length).

[F2]

Cancellation (The positive braid monoid is left and right cancellative): $x a = x b \Rightarrow a = b$ and $a x = b x \Rightarrow a = b$ in $B_{n}^{+}$ ; reversal $ρ$ is an anti-automorphism with $ρ (σ_{j}) = σ_{j}$ , so $ρ [(T_{1} \dots T_{n - 1})] = [U_{n - 1} \dots U_{1}]$ .

[F3]

The blocks: $T_{1} = σ_{1}$ , $Δ = T_{1} \dots T_{n - 1}$ , and $Δ_{n} = Δ_{n - 1} T_{n - 1}$ for $n \geq 2$ , where $Δ_{n - 1} = T_{1} \dots T_{n - 2}$ is the half twist of the sub-alphabet ${σ_{1}, \dots, σ_{n - 2}}$ (The Garside half twist and simple positive braids).

Proof

technique · direct

1.1

The sliding lemma (e). Let $1 \leq j \leq k - 1$ and consider $σ_{j} T_{k} = σ_{j} σ_{k} σ_{k - 1} \dots σ_{1}$ . First move the leading $σ_{j}$ rightwards across $σ_{k}, σ_{k - 1}, \dots, σ_{j + 2}$ : each of these letters is at distance $\geq 2$ from $j$ , so far commutation [F1] applies, giving $σ_{k} \dots σ_{j + 2} σ_{j} σ_{j + 1} σ_{j} σ_{j - 1} \dots σ_{1}$ (the indices $k, \dots, j + 2$ are present by $j \leq k - 1$ , and the block $σ_{j} σ_{j + 1} σ_{j}$ is contiguous because the remaining letters of $T_{k}$ after $σ_{j + 2}$ are $σ_{j + 1}, σ_{j}$ ). Second, apply the three-term relation to that contiguous block: $σ_{j} σ_{j + 1} σ_{j} \equiv^{+} σ_{j + 1} σ_{j} σ_{j + 1}$ , giving $σ_{k} \dots σ_{j + 2} σ_{j + 1} σ_{j} σ_{j + 1} σ_{j - 1} \dots σ_{1}$ . Third, move the rightmost $σ_{j + 1}$ rightwards across $σ_{j - 1}, σ_{j - 2}, \dots, σ_{1}$ : each is at distance $\geq 2$ from $j + 1$ , so far commutation gives $σ_{k} \dots σ_{j + 2} σ_{j + 1} σ_{j} σ_{j - 1} \dots σ_{1} σ_{j + 1} = T_{k} σ_{j + 1}$ . Hence $σ_{j} T_{k} \equiv^{+} T_{k} σ_{j + 1}$ . For $j = k \leq n - 2$ , the two words $σ_{k} T_{k}$ and $T_{k} σ_{k + 1}$ have different supports: the first uses exactly $σ_{1}, \dots, σ_{k}$ , while the second also uses $σ_{k + 1}$ . By [F1] they are not equivalent; when $k = n - 1$ , $σ_{k + 1}$ is not a generator, so the comparison is not stated.

1.2

Word reversal fixes the half twist (c), first half. For $m \geq 2$ let $R_{m} : = U_{m - 1} U_{m - 2} \dots U_{1}$ , the reversal of the word $T_{1} T_{2} \dots T_{m - 1}$ that defines $Δ_{m}$ . We prove by induction on $m$ that $R_{m} = Δ_{m},$ together with the auxiliary identity $C (m) : U_{m - 1} Δ_{m - 1} = Δ_{m}$ . All these identities take place in the sub-monoid generated by $σ_{1}, \dots, σ_{m - 1}$ inside $B_{n}^{+}$ ; by [F1] every class containing a word over that sub-alphabet has a representative over it, so the computation may be performed in the sub-alphabet and read in $B_{n}^{+}$ . For $m = 2$ one has $R_{2} = U_{1} = σ_{1} = T_{1} = Δ_{2}$ , and $C (2)$ reads $U_{1} Δ_{1} = σ_{1} \cdot 1 = σ_{1} = Δ_{2}$ , both by [F3]. For the step $m \geq 3$ assume $R_{m - 1} = Δ_{m - 1}$ and $C (m - 1)$ . By the definitions of the blocks in the Given, $U_{m - 1} = U_{m - 2} σ_{m - 1}$ , $T_{m - 1} = σ_{m - 1} T_{m - 2}$ , while $Δ_{m - 1} = Δ_{m - 2} T_{m - 2}$ is the recursion of [F3], so $U_{m - 1} Δ_{m - 1} = U_{m - 2} σ_{m - 1} Δ_{m - 2} T_{m - 2} = U_{m - 2} Δ_{m - 2} σ_{m - 1} T_{m - 2} = Δ_{m - 1} σ_{m - 1} T_{m - 2} = Δ_{m - 1} T_{m - 1} = Δ_{m} .$ The middle equality holds because every letter of the word $Δ_{m - 2}$ lies in ${σ_{1}, \dots, σ_{m - 3}}$ and hence is at distance $\geq 2$ from $σ_{m - 1}$ , so finitely many far-commutation relations of [F1] interchange $σ_{m - 1}$ with $Δ_{m - 2}$ ; the fourth equality is $C (m - 1)$ ; the last is the recursion of [F3]. This proves $C (m)$ , and then $R_{m} = U_{m - 1} R_{m - 1} = U_{m - 1} Δ_{m - 1} = Δ_{m}$ by the induction hypothesis and $C (m)$ . Since $R_{m}$ is the reversed word of $Δ_{m}$ and $ρ$ is the reversal anti-automorphism of [F2], this is the assertion $ρ (Δ_{m}) = Δ_{m}$ of (c).

F1F2F3

2.1

Interior case of a simultaneous induction for (a). The interior and boundary cases below together prove the full assertion (a) by induction on $m$ : the induction hypothesis at level $m - 1$ includes its boundary case, the interior argument proves every lower index at level $m$ , and the boundary argument then proves the final index. For every $m \geq 2$ and every $1 \leq i \leq m - 2$ , we prove $σ_{i} Δ_{m} \equiv^{+} Δ_{m} σ_{m - i}$ . For $m = 2$ the range is empty, so the assertion is vacuous. Assume the full assertion (a), including its boundary case, has been proved at level $m - 1$ by the preceding induction stage; by [F3] write $Δ_{m} = Δ_{m - 1} T_{m - 1}$ . The induction hypothesis applied to the sub-alphabet ${σ_{1}, \dots, σ_{m - 2}}$ states $σ_{i} Δ_{m - 1} \equiv^{+} Δ_{m - 1} σ_{m - 1 - i}$ for every $1 \leq i \leq m - 2$ , and every word in that equivalence is a word over ${σ_{1}, \dots, σ_{m - 2}}$ ; since the displayed derivation is valid in that sub-alphabet (support is preserved by [F1]), the same equivalence holds in $B_{n}^{+}$ . Multiplying by $T_{m - 1}$ on the right and using multiplicativity gives $σ_{i} Δ_{m - 1} T_{m - 1} \equiv^{+} Δ_{m - 1} σ_{m - 1 - i} T_{m - 1}$ . Now apply the sliding lemma 1.1 with $j = m - 1 - i$ and $k = m - 1$ : the constraint $1 \leq j \leq k - 1$ is exactly $1 \leq m - 1 - i \leq m - 2$ , that is $1 \leq i \leq m - 2$ , which holds; we obtain $σ_{m - 1 - i} T_{m - 1} \equiv^{+} T_{m - 1} σ_{m - i}$ . Hence $σ_{i} Δ_{m} \equiv^{+} Δ_{m - 1} T_{m - 1} σ_{m - i} = Δ_{m} σ_{m - i}$ , as required.

F1F3step 1.1

3.1

The remaining case $i = m - 1$ of (a). For $m = 2$ the unique index is $i = 1 = m - 1$ , and $σ_{1} Δ_{2} \equiv^{+} σ_{1} σ_{1} \equiv^{+} Δ_{2} σ_{1}$ holds because both sides are the very same word. For $m \geq 3$ , apply the anti-automorphism $ρ$ of [F2] to the case $i = 1$ of step 2.1, which is in range because $1 \leq m - 2$ ; recall $ρ (σ_{j}) = σ_{j}$ and $ρ (Δ_{m}) = Δ_{m}$ by step 1.2, and $ρ (x y) = ρ (y) ρ (x)$ . The equivalence $σ_{1} Δ_{m} \equiv^{+} Δ_{m} σ_{m - 1}$ gives $Δ_{m} σ_{1} = ρ (Δ_{m}) ρ (σ_{1}) = ρ (σ_{1} Δ_{m}) \equiv^{+} ρ (Δ_{m} σ_{m - 1}) = ρ (σ_{m - 1}) ρ (Δ_{m}) = σ_{m - 1} Δ_{m}$ , and since $m - (m - 1) = 1$ this is case $i = m - 1$ of (a). Together with step 2.1 this proves (a) for every $i \in {1, \dots, m - 1}$ .

F1F2step 1.2step 2.1

4.1

The mirror identity (b). Applying (a) with the index $m - i$ (which lies in ${1, \dots, m - 1}$ ) gives $σ_{m - i} Δ_{m} \equiv^{+} Δ_{m} σ_{m - (m - i)} = Δ_{m} σ_{i}$ , which is the first assertion. For the second, argue by induction on the length of a positive word $w$ (The principle of mathematical induction): $ε$ gives $Δ ε = ε Δ$ ; if $w = w^{'} σ_{i}$ and $Δ w^{'} \equiv^{+} τ (w^{'}) Δ$ , then $Δ w = Δ w^{'} σ_{i} \equiv^{+} τ (w^{'}) Δ σ_{i} \equiv^{+} τ (w^{'}) σ_{m - i} Δ = τ (w) Δ$ , using the first assertion. Applying the same argument to $τ (w)$ and using $τ \circ τ = i d$ gives $Δ τ (w) \equiv^{+} w Δ$ , that is, $w Δ \equiv^{+} Δ τ (w)$ .

F1step 3.1

5.1

$τ$ fixes the half twist (c), second half. Put $w : = Δ_{m}$ in the identity $Δ_{m} w \equiv^{+} τ (w) Δ_{m}$ of step 4.1: $Δ_{m} Δ_{m} \equiv^{+} τ (Δ_{m}) Δ_{m}$ . Right cancellation [F2] gives $τ (Δ_{m}) = Δ_{m}$ . Together with the equality $ρ (Δ_{m}) = Δ_{m}$ of step 1.2 this establishes (c).

F2step 1.2step 4.1

5.2

Centrality of $Δ^{2}$ (d). For a positive word $w$ , by step 4.1 applied to $w$ and to $τ (w)$ , and using $τ \circ τ = i d$ , we get $Δ^{2} w = Δ (Δ w) \equiv^{+} Δ (τ (w) Δ) = (Δ τ (w)) Δ \equiv^{+} (w Δ) Δ = w Δ^{2}$ , which is (d). For $w = σ_{i}$ this is the stated generator case.

F1step 4.1

6.1

Assembly. Part (a) is steps 2.1 and 3.1, part (b) is step 4.1, part (c) is steps 1.2 and 5.1, part (d) is step 5.2 and part (e) is step 1.1. Every induction above is over $N$ (length of a word, or the level $m$ ), all computations are finite, and no choice principle is used; the case $n \leq 1$ of the statement is the trivial empty-alphabet case. ∎

step 1.1step 1.2step 2.1step 3.1step 4.1step 5.1step 5.2

Remarks

No source fact is assumed. GM Section 4, printed p. 27, reports all of (a) and the reversed-word equality as Garside's "elementary arguments" without reproducing the slides; here every move is reconstructed. The reversed-word equality $ρ (Δ) = Δ$ is proved in step 1.2 by the auxiliary recursion $U_{m - 1} Δ_{m - 1} = Δ_{m}$ , the case $i \leq m - 2$ of (a) is derived in step 2.1 from the sliding lemma 1.1, the case $i = m - 1$ is obtained in step 3.1 by transporting the case $i = 1$ through the reversal anti-automorphism $ρ$ , and the mirror identity, $τ (Δ) = Δ$ and the centrality of $Δ^{2}$ are proved in steps 4.1--5.2 from (a) and cancellation.
Note how the two triangular blocks are used: the recursion $Δ_{n} = Δ_{n - 1} T_{n - 1}$ couples a letter $σ_{i}$ with the index $n - i$ , and the sliding lemma 1.1 realises exactly that shift inside a block. Reversal $ρ$ is not the same operation as $τ$ : $ρ$ fixes each letter but reverses products, $τ$ permutes letters without reversing products.
Conventions: all identities are in the monoid $B_{n}^{+}$ , so no inverse of $Δ$ and no conjugation in a group are used; the phrase "conjugation by $Δ$ " in the title refers to the two-sided sliding $σ_{i} Δ = Δ σ_{n - i}$ , which is a conjugation identity only after the Ore embedding of The group of fractions of the positive braid monoid is the Artin braid group. No choice principle is used.

Depends on

Used by

Dependency tree · two levels

14 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

J. Gonzalez-Meneses, Basic results on braid groups, Section 4, printed pp. 27-28 (sigma_i Delta = Delta sigma_{n-i} and its consequences) (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter I, Reference Structure 2 and formula (1.6), printed pp. 5-7 (standard reference, not scraped)