Proof that the Reidemeister Moves do not Destroy a Colouring

It must be shown that if a colouring of a knot is given and a Reidemeister move is applied to the knot, then the knot still has a colouring. This will be done by examining th ethree reidemeister moves separately.

For moves of type I things are quite simple. Suppose that the following part of a coloured knot is given and that the shown arcs are coloured a and b. If a move of type I is applied to it then thetwo arcs become a single arc and it is not clear whether the new single arc should be assigned the colour a or the colour b. Fortunately, no choice has to be made because, by using the defintion of a colouring, it is possible to show that a= b. To see that this is so, botice that at the crossing shown the equation a + b = 2? a mod m must be satisfied. Regardless of the value of m, by subtracting a from both sides of the equation it follows that b = a mod m. A Reidemeister move of type I which adds an arc by flipping the string around; namely which transforms into is even easier to handle. In this case, simply colour both new arcs the same colour as the original arc from which they both originated.

Moves of type II are a bit more complicated. Suppose that part of a knot has been coloured as follows  It suffices to show that the new knot after the Reidemeister move of type II has been applied it can be coloured by giving the new arc colour a and keeping the colour of the old arc the same. One difficulty which might arise is that thepart which was oroiginally coloured c is now coloured a and this migth cause problems in the rest of the knot. However, it will be shown that a = c , so that this problem does not occuur. Notice that from the definition of a colouring the equations a + b =2? d mod m and c + b =2? d mod m. As in the case of type I moves, the value m is not important because, from the two equations it follows that c + b = a + b mod m and, hence, c = a mod m . The other problem which might arice is that, by eliminating the colour b the number of colours has been reduced to 1 (recall that a colouring musthave at least two distinct colours). However, if this is the case then d = a mod m and, recalling that a + b =2? d mod m, it follows that b = d mod m. But now it follows that all all fours colours a , b , c and d are the same modulo m . Hence the total numer of colours used is can not be reduced by using only a and d . The Reidemeisetr moves of type II in the other direction are more easily handled, Simply assign the new arc the colour it had before.

Finally, Reidemeister moves of type III must be considered. Consider part of a knot, shown on the left, with the colouring given by a , b , c , d , e and f . A Reidemeister move of type II will transform this part of the knot as follows and, if the arcs shown are assigned the indicated colouring, the question becomes: "Is it possible to find a value x so that the resulting assignment is a valid colouring?". By considering the crossing at which the colours a , c and x meet, it is clear that x must be equal to 2 ? a - c . It must be check that this value of x does not violate the colouring equation at the crossing where the colours e , b and x meet. This equation is:

(*) 2 ? a - c + e = 2 ? b

From the colouring equation at the crossing where the colours e , d and a meet, it follows that 2 ? a = d + e and by considering the colouring equation at the crossing where the colours c , d and f meet, it follows that 2 ? f = d + c . By subtracting the second equation from the first, one gets that -c + e = 2 ? a - 2 ? f . By substituting this into (*) one gets that 2 ? a + ( 2 ? a - 2 ? f ) = 2 ? b and this reduces to 4 ? a = 2 ? f + 2 ? b . This equation holds by looking at the crossing where the colours a , f and b meet. This shows that this assignment of x results in a good colouring. The other Reidemeister move of type III is handled similarly.


Mike Zabrocki
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Department of Mathematics and Statistics
York University
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