I have read a statement from Sossinsky and Prasolov' s book "Knots, Lİnks, Braids and 3-Manifold", it says that two reduced word represent isotopic braids if and only if they have the same reduced form, page 54. My claim is that this statement is not true: take $b_2b_1b_2^{-1}b_3^{-1}b_3^{-1}$ and $b_3^{-1}b_2b_3b_2b_1b_2^{-1}b_2^{-1}b_3^{-1}b_2^{-1}$. They represent isotopic braids but they are not the same. What is the correct form of such a characterization?

  • $\begingroup$ By "reduced" they probably mean a word in the/a normal form (there are different normal forms for the Braid groups). Then the assertion is true, see the survey linked by Igor. $\endgroup$
    – Misha
    Mar 16, 2012 at 23:38
  • $\begingroup$ Not knowing how they define "reduced word" or "reduced form", I can only suggest looking closely at those definitions. Usually those terms mean that the words have no cancellation, which is equivalent to the words representing the same element in the free group, and that is not the same as equality in the braid group. $\endgroup$
    – Lee Mosher
    Mar 16, 2012 at 23:48
  • 2
    $\begingroup$ The title of this question is very vague. A better title would be something like 'The word problem in braid groups'. $\endgroup$
    – HJRW
    Mar 17, 2012 at 9:24

2 Answers 2


Here "reduced" refer to the so-called handle reduction algorithm introduced by Dehornoy.

So far I remember, this algorithm does not provide a normal form, therefore I agree that the statement is false: two reduced word may represent the same braid even if they are different. Indeed, it can be checked with the following Java applet: http://www.math.unicaen.fr/~tressapp/index.html (it applies handle reduction the other way as you does, i.e. it pushes handles to the left, so in that case the reduction will produce the same word. But if you swap $b_1$ and $b_3$ you still get isotopic braid, and now the applet will tell you that the words are already reduced).

The true statement is that a word represent the trivial braid if and only if its reduction is the empty word. Of course it still solve the word problem: checking if $w_1,w_2$ represent the same braid is the same as checking if $w_1w_2^{-1}$ represents the trivial braid.

Edit. For the sake of completeness let me recall why not leading to a normal form is not a negative point against this algorithm.

  1. When you use normal forms, testing equality is trivial, but multiplication and inversion are hard because you have to normalize the result each time. So it's somehow better to apply algebraic operation only at the word level, and to apply the algorithm only when you want to test equality.
  2. Of course you can still do it with a normal form algorithm. But since you ask less, you can hope to find a more efficient algorithm. Indeed, handle reduction is strongly believed to be of quadratic complexity, and is without any doubt the most efficient solution to the word problem in the braid group.
  3. Maybe the most important things is that handle reduction actually leads to a left (or right) ordering on the braid group, and to a bi-ordering on the pure braid group, which has quite a lot of group-theoretic consequences.
  • $\begingroup$ Because links tend to break after a few years, here is an explanation of the algorithm for direct computer implementation: matematita.science.unitn.it/applet/handlereduction/en/info.html. $\endgroup$
    – Anthony
    Aug 1, 2021 at 0:53
  • $\begingroup$ Regarding my previous comment: be careful that the webpage I linked doesn't talk about the (only) obstruction to this algorithm, which is mentioned in Dehornoy's paper (see page 6, paragraph below figure 1.4). Implementing it directly without this precaution makes the recursive pile loop endleslessly! $\endgroup$
    – Anthony
    Aug 1, 2021 at 13:35

For a very nice (if slightly typesetting-challenged) survey of this subject, see Patrick Dehornoy's survey.


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