The first challenge is about the uniqueness of the notation. The answer is almost identical to the case of knots. Given some regular link presentation, there is a multitude of possible notations that correspond to that presentation. In the case of knots, one could obtain (at most) 2N such notations, where N is the crossing number. To obtain a different notation, one would have to select a different starting point. In the case of links, however, the number of possible notations is usually higher than twice the crossing number (unless the link consists of exactly one component). One may vary the starting point, but one may also select a different combination of segments that are going to "touch" each other and generate the artificial crossings. The number of possible notations cannot exceed the value of (2N)**k (2N to the power of k), where N is the number of crossings (including the artificial ones) and k is the number of link components. Almost always it is much smaller than this value.
In the case of knots, if one notation consisted of pairs (x,y) and the other of pairs (x',y'), then (x'-x)mod(2N) = (y'-y)mod(2N), and this difference was the same for all pairs of the notations. In the case of links there is no such similar formula, which creates an additional difficulty when we create the appropriate algorithm to tabulate the links. We shall explain this issue in more detail in this page.
The second challenge is about the existence of a link presentation, once we are given a particular notation. The answer comes out to be similar to the case for the knots: the notation must contain all numbers between 1 and 2N exactly once. In addition, it must be "drawable". Odd numbers must be paired to even, but even this condition is not sufficient. If for example, N = 5 and the numbers 1,3,5,7,9 are paired to 4,6,8,10,2 respectively, the notation is not "drawable".
When discussing the problem of knots, we resolved the question of "drawability" by considering all possible loops, and requiring that any two loops should either have one or more segments in common, or intersect at an even number of crossings. This condition, while accurate and "finite", would occupy most of the running time of the program. We were thus forced to consider "equal drawability" moves: two notations connected through such moves, are either both "drawable", or none. By applying this logic, we ended up having very few notations to check explicitly.
We are now going to adopt an even better method to determine the "drawability" of a notation. It is going to be presented in this page. It can be summarized as follows. A notation is "drawable" if and only if it yields consistent values for the signs of the crossings, and the "genus" of the resulting presentation is zero (satisfies Euler's theorem).
With respect to the third challenge: Reidemeister moves are still valid, and they are the only moves needed to connect ambient isotopic presentations. For links, however, there is an additional complication. The reader may recall that there were three types of Reidemeister of moves, who would affect the notation as follows.
A first type move would add or remove a pair (i,i'). A second type move would add or remove the pairs (i,j) and (i',j'). A third type move would replace the three pairs (i,j), (i',k), (j,k) with the pairs (i',j'), (i,k'), (j',k'). In the first two types of moves, some numbers corresponding to other pairs would have to be increased or reduced by 2 or 4. In all three types of moves, the absolute values |i'-i|, |j'-j|, |k'-k| were either equal to 1, or to 2N-1 (in the latter case one of the two numbers was equal to 1 and one equal to 2N).
In the case of links, things are not exactly the same. Dus to the addition of artificial crossings, which alter the numbering but do not affect the Reidemeister moves, we would have to alter the condition that |i'-i|, |j'-j|, |k'-k| = 1 or 2N-1, with the condition that i,j,k have to be "neighbors" of i,j,k respectively. While "visually" it is simple to find out which number is "neighbor" to which (just redraw the original link presentation without the artificial crossings), there is a specific method to determine the "neighbors" through the notation, without having to draw the link. This method is going to be presented here.
Finally, as far as the invariants are concerned. "Colorization" invariants may be calculated in a way similar to the one used for knots. As far as the HOMFLYPT polynomial is concerned, a new complication arises. One may remember that each "chiral" knot had effectively two such polynomials, related by P_1(x,y) = P_2(y,x), where each polynomial corresponded to each of the chiral versions of the knot. If one inverted the orientation of the entire knot, the polynomial would not change.
In the case of links, there is an additional complication. One might alter the orientation of some (but not all) of the link components. Then the value of the polynomial would (almost certainly) change, but not in a simple way. For a link consisting of k components, the number of such combinations would be 2**(k-1) (two to the power of k-1). One would need to calculate explicitly each of these polynomials. If one fails to do so, one might end up comparing two links with entirely unrelated polynomials, but would actually consist of the same points (even though some of their components would have opposite orientations).
As noted above, particular parts of this section are going to be discussed in more detail in the following pages:
Or, you may prefer to click here for the rest of the discussion.
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