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Braid groups have an infinite cyclic group center, generated by the square of the fundamental braid.

Geometrically, the fundamental braid has the property that any two strands cross positively exactly once.

For a braid group of about four strands it is easy to show that the square of the fundamental braid commutes with the generators, using the braid relations (I have used the Artin presentation). How can this be generalized? Is there an intuitive reasoning of why this property holds for all braid groups? Maybe an inductive proof or appeal to the geometric nature of every brand crossing twice in the square of the fundamental braid?

Also, is there a way of showing that there are not any other word that commutes with every generator?

Single Malt
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2 Answers2

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A little late, but I happened to see this, and hope the following is still of use to someone. Let's think of the strands as lying in a ribbon. The fundamental braid is just a twist of this ribbon over $180^\circ$ around its long axis. Its square is a twist of the ribbon over $360^\circ$, so that all strands are back where they started. This operation clearly commutes with any braid group element; except for a full twist in the ambient space nothing really happened. It generates a copy of $\mathbb{Z}$ in the center of the braid group. While I don't have a simple proof that it's the whole center, I find it intuitively quite clear from this picture that that should be the case.

Jules Lamers
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Better late than never! This answer is a compromise between Jules Lamer's intuitive explanation and the full algebraic solution (with the option to go fully algebraic if you want; see 'further comments' below).

Let $n\geq2$ and denote the generators of $B_{n}$ by $\{\sigma_{i}\}_{i=1}^{n-1}$. To recap, the fundamental braid is given by $\beta_{n}=b_{n-1}b_{n-2}\cdots b_{1}$ where $b_{k}=\sigma_{1}\cdots\sigma_{k}$.

Proposition $\beta^{2}_{n}\in Z(B_{n})$.

The lemma below is useful for showing the desired proposition. To grasp the intuition, draw the fundamental braid for some small $n$ and observe how when pre- or post-composing with a generator, the generator can 'slide along' the fundamental braid.

Lemma $\beta_{n}\sigma_{i}^{\pm1}=\sigma_{n-i}^{\pm1}\beta_{n}$.

Given any nontrivial braidword $w$, let $w'$ denote the word $w$ but where each letter $\sigma_{i}$ is replaced with $\sigma_{n-i}$. Note that $w''=w$. Applying the lemma, $w\beta^{2}_{n}=\beta_{n}w'\beta_{n}=\beta_{n}^{2}w''$. This shows the proposition.

Further comments

  • The above lemma is Lemma 5.3 in this paper, where I've proved it in Appendix B.

It turns out I was unwittingly playing with the $n$-strand "fundamental braid", which I've instead called the "superselection braid". Further identities pertaining to $\beta_{n}$ can be found in this appendix. For instance, it's a palindrome as a braidword, and can also be recursively parsed into products of fundamental sub-braids. (No plug intended, honest!).

Aside (physics application): the action of $\beta_{n}^{2}$ on some 2d open disc $D$ equates to Dehn twisting $\partial D$; the distinct eigenvalues of this action on the state space of some enclosed, localised identical particles index the superselecton sectors (i.e. possible total anyon charges) of the enclosed system. The properties of the fundamental braid can be used to show that these sectors define a commutative fusion ring.

  • Can't think of an obvious proof for uniqueness right now. The standard reference for $Z(B_{n})=\langle\beta_{n}^{2}\rangle$ seems to be Garside's exposition (1967) (see Section 3.4) of Chow's earlier result (1948).
Sachin Valera
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