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as well as their inverse \(b_{i}^{-1}\) which has the opposite crossing. For example, the braid group \(B_{4}\) has generators
and inverses
The multiplication of 2 braids of \(n\) strands is then given by putting both braids end to end. For example, the identity \(b_{1}b_{1}^{-1}\) in \(B_{3}\) is identified as the braid
Note to self: The next 3 images in this post are incorrect due to file name issues. Fix eventually.
It is easy to see that this is equivalent to the identity which is 3 unbraided strands. Another way to determine the equivalence is to consider the permutation of the point (1, 2, 3).
It is pretty clear that braid multiplication is associative, ie. \(a(bc) = (ab)c\). The braid group is characterized by the Artin relations:
1. \(b_{i}b_{i}^{-1} = 1\) 2. \(b_{i}b_{i+1}b_{i} = b_{i+1}b_{i}b_{i+1}\) \(b_{i}b_{i+1}b_{i}^{-1} = b_{i+1}^{-1}b_{i}b_{i+1}\) \(b_{i}^{-1}b_{i+1}b_{i} = b_{i+1}b_{i}b_{i+1}^{-1}\) 3. \(b_{i}b_{j} = b_{j}b_{i}\) when \(|i - j| \ge 2\) relation 2 is called the braid relation and can be visualized as
for the \(B_{3}\) group. The third relation is called far commutativity visualized as
for the \(B_{4}\) group.
Word Problem and Duhornoy Algorithm
A word \(w \in B_{n}\) is any sequence of multiplications of the generators of \(B_{n}\). A reduced word is described as \(w\) in the form where "any occurrence of the letter \(b_{i}\) is separated from any occurrence of the letter \(b_{i}^{-1}\) by at least one occurrence of a letter \(b_{j}^{\pm 1}\) with \(j < i\)" (pg. 53). Thus, the RHS of the last 2 braid relations are reduced while the LHS of the same are not reduced. Based on the property that any element of \(B_{n}\) (as per the Artin relations) may be represented by a reduced word, we have the Duhornoy algorithm which takes a non-reduced word to its reduced form.
An example of an application of the algorithm is given on page 54 (where the part of the word to be reduced is given in parenthesis in each step). Intermediate steps have been given in the places where I felt the book was not clear enough on what's actually happening. Each step constitutes an application of the appropriate Artin relation. \begin{split} w & = (b_{1}^{-1}b_{2}b_{1})b_{3}^{-1}b_{1}^{-1}b_{3}^{-1}b_{1} \\ R(w) & = b_{2}(b_{1}b_{2}^{-1}b_{3}^{-1}b_{1}^{-1})b_{3}^{-1}b_{1} \\ & = b_{2}b_{2}^{-1}b_{1}^{-1}b_{3}^{-1}b_{2}b_{3}^{-1}b_{1} \\ R^{2}(w) & = (b_{2}b_{2}^{-1})b_{3}^{-1}b_{1}^{-1}b_{2}^{-1}b_{3}b_{2}b_{3}^{-1}b_{1} \\ R^{3}(w) & = b_{3}^{-1}b_{1}^{-1}(b_{2}^{-1}b_{3}b_{2})b_{3}^{-1}b_{1} \\ R^{4}(w) & = b_{3}^{-1}(b_{1}^{-1}b_{3}b_{2}b_{3}^{-1}b_{3}^{-1}b_{1}) \\ & = b_{3}^{-1}(b_{2}b_{3}b_{1}b_{3}^{-1}b_{3}^{-1}b_{2}^{-1}) \\ R^{5}(w) & = b_{3}^{-1}b_{2}b_{3}b_{2}b_{1}b_{2}^{-1}b_{2}^{-1}b_{3}^{-1}b_{2}^{-1} \end{split} The Duhornoy algorithm functions a bit like an invariant but is stronger in the sense that we can say definitively that if two braids have the same reduced word then they constitute the same element of the braid group \(B_{n}\). In fact, this is an "if and only if" situation. It is with great regret that I report the absence of such 3D modelling software (see fig. 2 above) or an alternative head. Furthermore, I believe Mattel's marketing department to be in need of a new demographic. Unrelated but fine example of inflation *gasp, surprised emoji*.
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