```

<----c------
|	       |
|	       c
|	       |
c	  --a-->--------a---
|	  |    |C	   |
|	  |    b	   |
|----->-c--|----d----  a
A |    |D	|  |
d    <----b---|--|
|		|B
-----<---d-----

This is a 4-crossing knot. The 4 crossings are located at the
points A, B, C, D. The number of strands is equal to the number
of crossings. Strand a starts at the undercrossing A, passes
through the overcrossing C and ends at the undercrossing B;
strand b starts at undercrossing B, passes through the over-
crossing D and ends at undercrossing C; strand c starts at
undercrossing C, passes through overcrossing A and ends at
undercrossing D; strand d starts at undercrossing D, passes
through overcrossing B and ends at undercrossing A. On the
figure one may see points of each strand marked by the number
of the strand.

We shall now apply some color tests on this knot.

First, the linear color test defined by n=3, t=2. It is defined
by the matrix
1   2   3
-------------
1| 1   3   2
2| 3   2   1
3| 2   1   3

With respect to strands a and b, there are two possibilities:
either they are colored the same, or not. If they are colored
the same, then one may see that strands c and d must also have
the same color in order to satisfy the matrix relations at
crossings C and D respectively. If this occurs, then the matrix
relations are satisfied at all points. One thus obtains 3
successful mappings.

Let now a and b have different colors. If a is colored 1 and b
is colored 2, then the condition at crossing C requires c to be
colored 3. The condition at D will required d to be colored 1,
but in such case the condition at B is not satisfied. Therefore
it is impossible to have any additional successful mappings.

Since the trivial knot admits 3 successful mappings while the
trefoil 9, we have shown up to this point that the knot drawn
above (the "figure 8" knot) is distinct to the trefoil; it may
or may not be trivial.

Let us now apply the 4-color test defined by the matrix

1   2   3   4
---------------
1| 1   4   2   3
2| 3   2   4   1
3| 4   1   3   2
4| 2   3   1   4

In additional to the four trivial mappings, where all strands
are mapped to the same color, it is possible now to have 12
additional mappings, such as for instance a -> 1, b -> 2,
c -> 4, d -> 3 . Therefore the "figure 8" knot through the
4-color test has been proved non-trivial.

We shall now calculate the Alexander polynomials of this knot.

The equations one obtains are the following.

Crossing A: ta + (1-t)c - d = 0
Crossing B: ta - b + (1-t)d = 0
Crossing C: (1-t)a - b + tc = 0
Crossing D: (1-t)b + tc - d = 0

If one removes for example the last equation and sets d=0, one
obtains the system

ta+(1-t)c = 0, ta-b = 0, (1-t)a-b+tc = 0.

The determinant of this system is equal to t^2 - 3t + 1, which
is exactly the Alexander-Conway polynomial. This is different
from 1 (the trivial knot's polynomial) and t^2-t+1 (the trefoil's
polynomial), and thus the "figure 8" knot is inequivalent to
both these knots.

Since P(t=2)=-1, which is not 0 mod 3, the figure 8 knot cannot
have non-trivial mappings for the n=3, t=2 linear test. For t=4
however, P(t=4)=5=0 mod 5 and thus this knot admits non-trivial
mappings for the n=5, t=4 test.

If one attempts to calculate the secondary Alexander polynomial,
one easily notices that the GCD of all the relevant determinants
is 1. (Take for instance the first two rows and columns, the
determinant is t, and GCD(t,t^2-3t+1)=1 ). Therefore no Alexander
polynomial other than the Alexander-Conway is non-trivial, and
thus for any n-color linear test there are at most n^2 successful
mappings.
```
Charilaos Aneziris, aneziris@hades.ifh.de Copyright 1995 COPYRIGHT STATEMENT Educational institutions are encouraged to reproduce and distribute these materials for educational use free of charge as long as credit and notification are provided. For any other purpose except educational, such as commercial etc, use of these materials is prohibited without prior written permission.