Two knots are considered to be the same if they can be transformed one into the other in 3 dimensional space without resorting to scissors. For example, most people would agree that the following pairs of knots are equivalent.
is transformed to this
and then this
one gets this
and then this
One method for distinguishing between different knots is to assign what are known as invariants to knots. As their name implies, invariants in mathematics are invariant under certain transformations. In the case of knots, invariants should be invariant under the Reidemeister moves; in other words, an invariant of a knot should not change its value when a Reidemeister move is applied to a knot. If this is the case, the Reidemeister's Theorem immediately implies that if two knots are equivalent then they have the same invariant. Notice however, that there is no reaon to believe the converse, knots with the same invariant need not be equivalent. An easy way to see this is to assign to all knots the number 1. This is clearly an invariant, but it is not very useful since it does not help use distinguish any pair of knots. The question of finding useful invariants though, is of considerable interest.
One such invariant is known as the the colouring invariant of a knot. To explain how this invariant is defined suppose that a projection of a knot is given. It has already been observed that if there are k crossings in this projection, then these also decompose the string into k disjoint arcs. The integer m is said to be a colouring invariant of this knot if it is possible to colour the arcs using at least two different integers (in other words, assign to each arc an integer and so that at least two arcs are given different integers) in such a way that at any crossing, if a and b are the integers assigned to the two arcs which end (or begin, depending on your point of view) at the crossing and c is the colour of the arc which passes on top at the crossing then a + b = 2c mod m. The mod m is very important here because it allow knots to be classified into several distinct equivalence classes. Without this, there would be only two possibilities: either the knot can be coloured or not. This would mean that knots are divided into only two classes. Using colouring mod m allows a much finer analysis. Finally, notice that the requirement that at least two colours are used is also crucial since any knot can be coloured trivially using only one colour so that all the equations are satisfied.
Of course simply calling something an invariant does not make it an
invariant. It is necessary to supply a proof
that the colouring invariant is an invariant before anything
further can be done with it. However, before looking at this proof, it
is probably instructive to look at an example of a knot and a
colouring of it.
In the projection shown, the knot at the left has 4 arcs and, of courses, 4 crossings as
well. The crossings have been labeled a ,
b , c and d while the
arcs have been coloured with integers from 0 to 2.
Each of the crossings corresponds to an equation which
must be satisfied by any colouring of the arcs. For example, at
crossing
a the arc passing below from the bottom is coloured
0, the arc passing below from the top is coloured 2 and the arc
passing on top of the crossing is coloured 1. At this crossing, the
equation 0 + 2 = 1×2 mod m must be satisfied.
But what is the value of m? It turns out that in this case
m = 3 will work. Since 2 = 1×2 mod m for any value
of m the equation
at crossing a is certainly satisfied. However, the equation
at crossing b uses arithmetic modulo 3 in an
essential way. the equation at this crossing is
0 + 1 = 2×2 mod m and it is satisfied mod m
only if m=3. The equation at crossing c is also
0 + 1 = 2×2 mod 3
while the equation at crossing d is
0 + 2 = 2×1 mod 3.
Use the colouring invariant to determine whether or not the following
two knots are equivalent.
When looking over the
provided solution pay attention to the fact that the solution is a
not perfect but this fcat is not hidden and is mentioned
in the summary.