Using the Colouring Invariant to Determine Whether Two Knots are Equivalent

To begin, the family of equations arising from the crossing of the first knot will be defined in Maple. See the attached diagram for the definitions of the variable.

> 2*a = b + c, 2*b = d + a, 2*c = b + e, 2*d = a + e, 2*e = d + c;

The nest step is to try to solve this system of equation modulo some integer. Since it is not yet clear modulo which integer the equations should be solved, let us begin by simply try to find a non-trivial solution.

> solve(",a,b,c,d,e);

This solution is not very useful since it only yields the trivial solution when all the variables are equal and a colouring must have at least two distinct colour. Perhaps trying to solve a subset of the equations will provide more information.

> solve(2*a = b + c, 2*b = d + a, 2*c = b + e, 2*d = a + e,a,b,c,d,e);

> solve(2*a = b + c, 2*b = d + a, 2*c = b + e,a,b,c,d,e);

So, by solving for only three of the equations a non-trivial solution has been found in trems of the variable a and c. Notice that the equations for which a solution has been found will be satisfied modulo any integer. However, this is not necessarily true for the equations which have been ommited. To analyse these, values will be assigned to .b, d and e and then the two remaining equations will be reevaluated.

> assign(");

> 2*d = a + e;


> 2*e = d + c;

Since both of these equations are easily seen to reduce to 7(c-a) it follows that the two equations left will be satisfied mod 7 by any assignment of values to a and c. (For example a = 1 and c= 2 will work). Hence the first knot is colourable modulo 7. A similar analysis will now be performed on the second knot. In order to clear the assignment of variable, Maple will be restartaed.

> restart;
The next step is to define the equations for the second knot using the notation in the attached diagram .

> 2*a = b + c, 2*c = d + a, 2*d = a + e, 2*e = d + f, 2*f = e + g, 2*g = f + h, 2*h = g + b;

> solve(",a,b,c,d,e,f,g,h);

Once again the solution is not very useful so the same approach as in the previous case is used.

> solve(2*a = b + c, 2*c = d + a, 2*d = a + e, 2*e = d + f, 2*f = e + g, 2*g = f + h,a,b,c,d,e,f,g,h);

Here many nontrivial solutions are found in terms of f and g. However, these f and g must both be even in order to get integer solutions. It is important to note that the solutions found for these equations are solutions modulo any number. Hence it suffcies to find a number m such that the equation left out is also satsified modulo m . Proceed as before, assigning values and then evaluating this last equation.

> assign(");
> 2*h = g + b;


By inspection, one sees that multiplying by 2 and moving the lefthand side to the right will yield an expression which has 13 as factor. Hence the second knot is 13 colourable. To check that the two knots are not equivalent it therefore suffices to show that the first knot is not 13 colourable. This can be done by a simple minded search of all possible colourings modulo 13 of the 5 arcs of the knot using do loops. In this case, there are only 371293 possibilities so this can be done in a reasonable time on the machines available to this course. (However, in more complicated cases a better approach is to solve for some of the equations in terms of some small set of variables, as was done for the first knot. Thgen, a do loop routine can assign values to only the variables from this small set and generate the values of the other variables using the solutions found for them.) The routine will check whether the 5 equations are satisfied for each possibility. Before proceeding, Maple will have to be restarted.

> restart;

> CheckColour := proc()
> for a from 0 to 12 do;
> for b from 0 to 12 do;
> for c from 0 to 12 do;
> for d from 0 to 12 do;
> for e from 0 to 12 do;
> if 2*a = b + c mod 13 and 2*b = d + a mod 13 and 2*c = b + e mod 13 and 2*d = a + e mod 13 and 2*e = d + c mod 13 then 
> print([a,b,c,d,e]); fi; od; od; od; od; od; end;
Running this procedure shows that the only colouring modulo 13 is the trivial colouring. Hence the first knot is not 13 colourable and, consequently, the two knots are not equivalent.


Mike Zabrocki
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Department of Mathematics and Statistics
York University
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