e-MATH

Knots and their polynomials

The value of the Jones polynomial does not depend on the orientation chosen.

Let us focus on one crossing in an oriented knot diagram:

J(undercrossing).

If we reverse the orientation and rotate the diagram 180 degrees, that crossing appears exactly as it did before.

So reversing the orientation in a diagram and then applying the skein relation to a crossing is the same as rotating the original diagram 180 degrees, applying the skein relation, rotating the products of the relation another 180 degrees, and then reversing the orientation.

Symbolically, letting ``-'' represent the orientation change, Rot a 180-degree rotation and Sk the effect of applying the skein relation to the knot diagram K, we have

Sk (- K) = - (Rot Sk Rot K)

where we apply operations, as usual, from right to left. Suppose our analysis involves 2 applications of the skein relation.

Sk2 Sk1 (- K)

= Sk2 (- Rot Sk1 Rot K)

= - (Rot Sk2 Rot Rot Sk1 Rot K)

= - (Rot Sk2 Sk1 Rot K)


since two consecutive 180-degree rotations of the entire figure have the same effect as no rotation at all, and similarly for more than two applications.

Suppose that it takes n applications of the skein relation to reduce all the diagrams to unknots, for which the Jones polynomial (symbolically J{ }) does not depend on orientation. Then

J{(- K)} = J{Skn ... Sk1 (- K)}

= J{ - (Rot Skn ... Sk1 Rot K)}

= J{Rot Skn ... Sk1 Rot K}

(since the Jones polynomial of an unknot-diagram does not depend on the orientation)

= J{Skn ... Sk1 K} = J{K}
(since a rotated knot is topologically the same).

So the Jones polynomial of the reversed-orientation diagram is the same as the Jones polynomial of the original diagram.

So why is an orientation required? Because to apply the skein relation correctly at a crossing, we need the two strands to be coherently oriented. Otherwise the calculation derails, as a little experimentation will show.

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