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Substance
Organization not yet worked out, refer to menu > blog > subsections for more
The closure of a braid connects the ends of the braid as shown to create a knot or knotted links. Mathematically, for a braid \(b\) the resulting closure is denoted by \(\beta(b)\)
We can determine if the closure of a braid is a knot rather than links by considering the "canonical epimorphism" - which to my understanding essentially means a mapping of sorts - of the braid group \(B_{n}\) onto the permutation group \(S_{n}\). Consider the braid
The mapping \(B_{n} \mapsto S_{n}\) is thus a mapping of \(b \mapsto \sigma(b)\) where \(b_{i} \mapsto s_{i}\) and \(s_{i}\) denotes the switching of the \(i\)th and \((i+1)\)th numbers (or elements at said indices) in the sequence. Thus, for the braid b above, we have \(\sigma(b) = s_{1}s_{2}s_{1}s_{2}\) or we can reduce \(b\) to \(b = b_{2}b_{1}\) and get \(\sigma(b) = s_{2}s_{1}\). In either case, \(\sigma(b) = (3, 1, 2)\). Since this permutation can be obtained from the original point by moving each digit to the right once (w/ the last digit cycling to the front), this is a cyclic permutation. This means that the braid's closure yields a knot because the cyclic nature of the permutation ensures that we will pass through all points with a single strand when drawing the closure (this is much easier to make sense of if one draws the closure and follows the flow of the lines). More precisely, "the closure of \(\beta(b)\) of the braid \(b\) is a knot if and only if the permutation \(\sigma(b)\) associated to the braid generates the cyclic subgroup of order \(n\), \(\mathbb{Z}/n\mathbb{Z}\), in the permutation group \(S_{n}\)" (pg. 55).
Detour into permutations...
The permutation of \((1, 2, 3)\) to \((3, 1, 2)\) is denoted by \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} The identity permutation is essentially the non-permutation of \((1, 2, 3)\) and the order of a permutation \(\sigma\) is the number \(n\) such that \(\sigma^{n} = 1\) or the identity. The multiplication of permutations is performed like so: \begin{equation} \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \end{equation} \begin{equation} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \end{equation} Thus, for our example the permutation has an order \(n\). The cyclic nature comes from the fact that we pass through all points in the process: \begin{align} 1 \mapsto 3 \mapsto 2 \mapsto 1 \\ 2 \mapsto 1 \mapsto 3 \mapsto 2 \\ 3 \mapsto 2 \mapsto 1 \mapsto 3 \end{align}
...end detourâ€‹
The closure of braids is not injective, meaning that there isn't always a 1-to-1 map between braids and their closures. However, Alexander's Braiding Theorem does state that the closure operation is surjective meaning that all knots and links constitute the closure of some braid.
One way to prove this is via the Vogel algorithm which is as follows:
which is illustrated in the examples below.
The algorithm always yields a set of nested Seifert circles in a finite number of steps thus proving that any knot is the closure of some braid (also note that the final change infinity move appears to take whatever single un-nested Seifert circle there is and braid it around so as to create a final outside Seifert circle around the already nested ones hence the lack of a loop back to the while loop). The act of drawing the steps as well as various braid closures should give one an intuitive understanding of this fact although I do hope to update this post with a proof eventually (time permitting or should it become necessary for further work).
The Markov Theorem states that the closures of two braids can be proven to be isotopic if and only if one braid can be transformed to the other by a finite sequence of the Markov moves which are defined as: 1. \(b \leftrightarrow aba^{-1}\) with \(a, b \in B_{n}\) 2. \(b \leftrightarrow bb_{n}^{\pm 1}\) with \(b \in B_{n}\) and \(b_{n} \in B_{n + 1}\)
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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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