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INTUITIVE DESCRIPTION For this introduction, we will take ''n'' = 4; the generalization to other values of ''n'' will be straightforward. Consider two sets of four items lying on a table, with the items in each set being arranged in a vertical line, and such that one set sits next to the other. (In the illustrations below, these are the black dots.) You are given four strands, and you are to connect each item of the first set with an item of the second set so that a one-to-one correspondence results. Such a connection we call a ''braid''. Often some strands will have to pass over or under others, and this is crucial: the following two connections are ''different'' braids: : On the other hand, two such connections which can be made to look the same by "pulling the strands" are considered ''the same'' braid: : We require that all strands move from left to right; knots like the following are ''not'' considered braids: :
Now given two braids, we can ''compose'' them by drawing the first next to the second, identifying the four items in the middle, and connecting corresponding strands: : Another example: : The composition of the braids σ and τ is written as στ. The set of all braids on four strands is denoted by ''B''4. The above composition of braids is indeed a Group operation. The Neutral Element is the braid consisting of four parallel horizontal strands, and the Inverse of a braid consists of that braid which "undoes" whatever the first braid did. (Our first two example braids above are inverses of each other.) GENERATORS AND RELATIONS Consider the following three braids:
It turns out that every braid can be written as a composition of a number of these braids and their inverses. In other words, these three braids Generate the group ''B''4. To see this, take an arbitrary braid and scan it from left to right; whenever you encounter a crossing of strands ''i'' and ''i'' + 1 (counting from the top at the point of the crossing), write down σ''i'' or σ''i''−1, depending on whether strand ''i'' moves under or over strand ''i'' + 1. When you reach the right end, you have written your braid as a product of the σ's and their inverses. It is clear that :σ1σ3 = σ3σ1, while the following two relations are not quite as obvious: :σ1σ2σ1 = σ2σ1σ2 :σ2σ3σ2 = σ3σ2σ3 (one can appreciate these best by drawing the braid on a piece of paper). It can be shown that all other relations among the braids σ1, σ2 and σ3 already follow from these relations and the group axioms. So one can abstractly define the group ''B''''n'' via the following Presentation :
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