Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 4 / d

Table of contents
- Solution d

Solution

4/3

pattern has winding number

4

, so the Satellite formula for the Alexander polynomial gives

Δ_{T_{3, 2}^{(4, 3)}} (t) ≐ Δ_{T_{3, 2}} (t^{4}) Δ_{T_{4, 3}} (t) .

Using the stated torus knot formula in each factor,

Δ_{T_{3, 2}^{(4, 3)}} (t) = \frac{( 1 - t ^{4} ) ( 1 - t ^{24} )}{( 1 - t ^{12} ) ( 1 - t ^{8} )} \frac{( 1 - t ) ( 1 - t ^{12} )}{( 1 - t ^{4} ) ( 1 - t ^{3} )} .

Up to a Laurent unit this may be simplified to

(1 - t) (1 - t^{24}) / ((1 - t^{8}) (1 - t^{3}))

.

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