Ordering Knot Notations

Let A and B two sets of pairs denoting regular knot projections. Set A pairs number 1 to the even number 2a_1, while set B pairs 1 to 2b_1. If a_1 < b_1 , we say that A is ordered ahead of B. If S is the set of notations, then A and B belong to S and (A,B) belongs to O. If of course a_1 > b_1 , it is the other way round.

It is possible of course that a_1 = b_1 . In such a case one compares a_2 to b_2, where 2a_2 is paired to 3 in A, while 2b_2 is paired to 3 in B. Unless a_2 = b_2 , one establishes the relation between A and B in a way similar to the one of the previous paragraph, by checking whether a_2 < b_2 or a_2 > b_2 . If a_2 = b_2 , then one proceeds similarly by obtaining and comparing a_3 to b_3 and so on.

It is possible however that a_i = b_i for all i in {1,2,...,N}. In such a case one proceeds as follows.

If (1,2a_1) belongs to A while (2a_1,1) belongs to B, then A is ordered ahead of B. If (1,2a_1) or (2a_1,1) belongs to both A and B, one checks whether (3,2a_2) or (2a_2,3) belongs to A or B. If (3,2a_2) belongs to A and (2a_2,3) to B, A is ordered ahead of B. If necessary, one proceeds in a similar manner until the end.

If even at this point the issue has not been settled, this means that A and B consist of identical elements and thus A and B.

From now on each regular knot projection will be denoted by the notation that is ahead of all others. In other words, if P is denoted by A_1, A_2, ..., A_f, it will be denoted by A_k where (A_k,A_j) belongs to O for all j and thus A_k is ordered ahead of all other notations.

It may seem at this point that all knot projections have been classified. If one however uses a notation to draw a projection, at some points one will be faced with a "chirality" ambiguity: should a crossing look -|->, or should it look like <-|- ? The number of such ambiguities is equal to the number of "prime components" of the knot shadow. A reader familiar with the notions of connected sums and prime manifolds should be able to understand what is meant and what exactly is happening and why. For the benefit of the general public however, there is a page explaining this problem, and you may read it by clicking here. In this page we explain how to obtain the number of prime components from the notation, and how the ambiguity is overcome.

Charilaos Aneziris, charilaos_aneziris@standardandpoors.com

Copyright 1995


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