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A central positive braid is a power of delta squared for n greater than two

Statement

Let $n > 2$ , let $B_{n}$ be the braid group of The braid group by Artin presentation with its positive braid monoid $B_{n}^{+}$ , its atoms $σ_{1}, \dots, σ_{n - 1}$ , its half twist $Δ$ and its divisibility order $≼_{L}$ (Left and right divisibility for positive braids). If $z \in B_{n}^{+}$ is central in $B_{n}$ , i.e. $z x = x z$ for every $x \in B_{n}$ , then $z = Δ^{2 k} for some k \geq 0.$

In particular the only central powers of $Δ$ that are positive are the even ones. Nothing here uses a choice principle. The hypothesis $n > 2$ is essential: for $n = 2$ every power $Δ^{p}$ with $p \in Z$ is central, and this is the subject of The center of b two is all of b two.

Facts & Assumptions

Given: A natural number $n > 2$ , the positive braid monoid $B_{n}^{+} \subseteq B_{n}$ with atoms $σ_{1}, \dots, σ_{n - 1}$ and half twist $Δ = Δ_{n}$ , and a central element $z \in B_{n}^{+}$ .

[F1]

Index reversal and centrality of $Δ^{2}$ . $σ_{i} Δ = Δ σ_{n - i}$ for every $i \in {1, \dots, n - 1}$ ; consequently $σ_{k} Δ^{m} = Δ^{m} σ_{n - k}$ for odd $m$ and $σ_{k} Δ^{m} = Δ^{m} σ_{k}$ for even $m$ , for every integer $m$ : induction gives the formulas for $m \geq 0$ , and the inverse of $σ_{i} Δ = Δ σ_{n - i}$ gives $σ_{i} Δ^{- 1} = Δ^{- 1} σ_{n - i}$ , from which induction gives the negative powers. Also $Δ^{2}$ commutes with every positive word, and $Δ$ has length $N = n (n - 1) / 2$ (Conjugation by the half twist reverses Artin generators, The Garside half twist and simple positive braids).

[F2]

Left normal form. Every $x \in B_{n}$ has a unique expression $x = Δ^{p} A$ with $p \in Z$ , $A \in B_{n}^{+}$ and $Δ ̸ ≼_{L} A$ ; moreover $p$ is the largest integer with $Δ^{p} ≼_{L} x$ (Left garside normal form is unique).

[F3]

Atom lcms. For adjacent indices $∣ i - j ∣ = 1$ the atoms have left-lcm $σ_{i} \lor_{L} σ_{j} = σ_{i} σ_{j} σ_{i}$ , and this element left-divides every common left multiple of $σ_{i}$ and $σ_{j}$ ; distinct atoms are incomparable in $≼_{L}$ (Artin atoms have explicit left and right lcms and complements).

[F4]

Atom criterion for $Δ$ . If $m \in B_{n}^{+}$ satisfies $σ_{i} ≼_{L} m$ for every $i$ , then $Δ ≼_{L} m$ . Also $Δ$ is a left multiple of each atom (Delta is the lcm of the artin atoms and has the same left and right divisors).

[F5]

Distinct atoms and cancellation. $π (σ_{1}) = s_{1}$ and $π (σ_{n - 1}) = s_{n - 1}$ are distinct permutations of $S_{n}$ when $n > 2$ (Reduced adjacent-transposition words have well-defined positive lifts); $B_{n}^{+}$ is left and right cancellative, and $ℓ$ is additive with $ℓ (w) = 0$ only for $w = 1$ (The positive braid monoid is left and right cancellative, Positive artin relations preserve homogeneous length).

Proof

technique · direct

1.1

The normal form of a central positive braid. By [F2] write $z = Δ^{p} A$ with $p \in Z$ , $A \in B_{n}^{+}$ and $Δ ̸ ≼_{L} A$ ; we show $A = 1$ and $p$ even. Since $z$ is positive and $A$ is positive, additivity of $ℓ$ gives $ℓ (z) = ℓ (Δ^{p}) + ℓ (A)$ when $p \geq 0$ , so the case $A = 1$ is the case $z = Δ^{p}$ ; in general we first prove $A = 1$ .

F2F5

1.2

A is central when $p$ is even, and satisfies a twisted identity when $p$ is odd. If $p$ is even then $Δ^{p} = (Δ^{2})^{p / 2}$ is central by [F1], so $A = Δ^{- p} z$ is central as well: $A σ_{j} σ_{i} = σ_{j} σ_{i} A$ for all $i, j$ . If $p$ is odd, centrality of $z$ gives $z σ_{n - j} σ_{n - i} = σ_{n - j} σ_{n - i} z$ for all $i, j$ ; inserting $z = Δ^{p} A$ , using [F1] to move $Δ$ past the two atoms, namely $σ_{n - j} σ_{n - i} Δ^{p} = Δ^{p} σ_{j} σ_{i}$ for odd $p$ , and cancelling the factor $Δ^{p}$ on the left (in the group) yields the twisted identity $A σ_{n - j} σ_{n - i} = σ_{j} σ_{i} A$ for all $i, j$ .

F1F2F5

2.1

Propagation from one atom prefix. Suppose $A \neq 1$ and choose $i$ with $σ_{i} ≼_{L} A$ (possible because a positive word for $A$ of length $ℓ (A) \geq 1$ has an atom as its first letter). Let $j$ satisfy $∣ i - j ∣ = 1$ . In the even case of step 1.2, the element $E : = σ_{j} σ_{i} A$ satisfies $E = A σ_{j} σ_{i}$ , so $σ_{i} ≼_{L} E$ (since $A = σ_{i} c$ gives $E = σ_{i} c σ_{j} σ_{i}$ ) and $σ_{j} ≼_{L} E$ (trivially); by [F3] the lcm $σ_{j} σ_{i} σ_{j}$ left-divides the common multiple $E$ , and cancelling the prefix $σ_{j} σ_{i}$ with [F5] gives $σ_{j} ≼_{L} A$ . In the odd case of step 1.2 the same argument applies with $E : = σ_{j} σ_{i} A = A σ_{n - j} σ_{n - i}$ : here $σ_{i} ≼_{L} A σ_{n - j} σ_{n - i} = E$ (because $A = σ_{i} c$ gives $E = σ_{i} (c σ_{n - j} σ_{n - i})$ ) and $σ_{j} ≼_{L} E$ trivially, so $σ_{j} σ_{i} σ_{j} ≼_{L} E$ and cancellation gives $σ_{j} ≼_{L} A$ . Hence in both cases every $j$ adjacent to a member of $S : = {k : σ_{k} ≼_{L} A}$ also lies in $S$ .

F3F5step 1.2

3.1

Every atom divides $A$ . The graph on ${1, \dots, n - 1}$ joining consecutive integers is connected for $n \geq 2$ ; by step 2.1 the nonempty set $S$ has no boundary, so $S = {1, \dots, n - 1}$ : every atom left-divides $A$ . By [F4] this forces $Δ ≼_{L} A$ , contradicting the normal form choice $Δ ̸ ≼_{L} A$ of step 1.1. Therefore $A = 1$ and $z = Δ^{p}$ .

F4step 1.1step 2.1

4.1

The exponent is even. With $z = Δ^{p}$ central and $p$ odd, [F1] gives $σ_{k} Δ^{p} = Δ^{p} σ_{n - k}$ while centrality of $z$ gives $σ_{k} Δ^{p} = Δ^{p} σ_{k}$ ; cancelling $Δ^{p}$ in the group, $σ_{n - k} = σ_{k}$ for every $k$ . For $k = 1$ this says $σ_{n - 1} = σ_{1}$ , contradicting the distinctness of the images $s_{n - 1} \neq s_{1}$ in $S_{n}$ when $n > 2$ by [F5]. Hence $p$ is even, $p = 2 k$ .

F1F5step 3.1

5.1

The exponent is nonnegative. Since $z = Δ^{2 k} \in B_{n}^{+}$ and $2 k = p$ : if $k < 0$ , then $Δ^{- 2 k} \in B_{n}^{+}$ has $ℓ (Δ^{- 2 k}) = (- 2 k) N > 0$ and $Δ^{- 2 k} z = 1$ would give $0 = ℓ (1) = ℓ (Δ^{- 2 k}) + ℓ (z) > 0$ by additivity and $ℓ (1) = 0$ , a contradiction. Hence $k \geq 0$ and $z = Δ^{2 k}$ .

F5step 4.1

6.1

Assembly. Step 1.2 separates the even and the odd exponent of the normal form of $z$ , step 3.1 forces the positive tail $A$ to be trivial, step 4.1 rules out odd exponents using the distinct atoms $σ_{1} \neq σ_{n - 1}$ , and step 5.1 gives the sign of the exponent. The two hypotheses used beyond the normal form and the atom calculus are the $Δ$ -sliding identity and the locality of the atom lcms; no geometric input and no choice principle is used. ∎

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

Remarks

Why the cases $p$ even and $p$ odd differ. For even $p$ the factor $Δ^{p}$ is central and $A$ inherits centrality; for odd $p$ the best available identity is the twisted one $A σ_{n - j} σ_{n - i} = σ_{j} σ_{i} A$ , obtained from $σ_{k} Δ^{p} = Δ^{p} σ_{n - k}$ . Both identities suffice to propagate an atom prefix to adjacent atoms, which is all the argument needs. This is the case distinction in Garside's proof of Theorem 4.2 as reproduced in J. González-Meneses, Basic results on braid groups, printed pp. 30--31.
Where positivity is used. Positivity of $z$ enters only to write the maximal-power decomposition with a positive tail and to conclude $k \geq 0$ ; the propagation argument itself needs only the left normal form of $z$ and the atom calculus.
Nothing here uses the Axiom of Choice or any weaker choice principle.

Depends on

Used by

The center of b n is generated by the full twist for n greater than two Theorem

Dependency tree · two levels

28 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, Theorem 4.2, printed pp. 30-31 (standard reference, not scraped)