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Given a braid group $$ B_n \simeq \left\langle x_1,\ldots,x_{n-1} \middle| \begin{array}{l} x_ix_j = x_jx_i, \;\text{for } |i-j|>1 \\ x_ix_{i+1}x_i = x_{i+1}x_ix_{i+1} \end{array} \right\rangle $$ for $n \geq 5$

  1. What is the centralizer of $x_1$?

or, less general question,

  1. Given a word $w = w(x_2,\dots,x_{n-1}) \in B_n$ such that $[x_1, w]=1$, does it imply that $w = w(x_3,\dots,x_{n-1})$?
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    $\begingroup$ The following may be helpful: departamento.us.es/da/prepubli/nsprep20.pdf, there is a computation for your first question in section 6. $\endgroup$
    – Jon Bannon
    Jan 25, 2018 at 18:33
  • $\begingroup$ Thank you. Indeed, the minimal summit graph for $x_1$ is small and can be used to describe the centralizer of $x_1$. Can we use to show (2)? $\endgroup$ Jan 25, 2018 at 19:10
  • $\begingroup$ Why $n\ge 5$? The answer for $n\le 4$ is "contained" in the answer for $n\ge 5$. If you know the answer for $n=4$ (notably a positive answer to (2), it would be interesting information. $\endgroup$
    – YCor
    Jan 25, 2018 at 21:09

1 Answer 1

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The answer to question 2 is "yes". It follows from the explicit description of the centralizers for elements $A_{i,j}=a_i\ldots a_j$, which was published in the paper below.

G.Gurzo, "Systems of generators for the normalizers of certain elements of the braid group". Mathematics of the USSR-Izvestiya(1985),24(3):439.

If reading in Russian is easy for you, the original version is freely accessible here: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=im&paperid=1454&option_lang=rus (see Theorem 1 at page 487).

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