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i am very interested in knot theory, especially in knot groups and knot polynomials. Therefore i am reading the book of Crowell and Fox (Introduction to knot theory). I want to compute some Alexanderpolynomials (with the technique they use in this book, thus Fox calculu, Abelization, free groups etc.) I can do this for special knots for example trefoil and cinquefoil (this are also exercises in this book). But now i want to compute the Alexanderpolynomial of the torus knot $T_{p,q}$ for $p$ and $q$ coprime. Therefore they want to prove that the following formula holds: $$\Delta(T_{p,q})=\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$$ if the knot group is $G(T_{p,q})= \langle x,y:x^p=y^q\rangle$ (this is not so difficult to prove). But here i can give the solution (or the way to solve it -.-). I have also make computations but they they are not good. Can someone help me with this? Thank you for help :)

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Have you looked at Rolfsen's book "Knots and Links"? – Lee Mosher May 5 '13 at 14:11
Crossposted to math.SE:… – Arturo Magidin May 5 '13 at 22:52

This is an exercise in many topology books. Here is a reference with a complete proof: Look up Example 9.15 in the book "Knots" by G. Burde and H. Zieschang. The Jacobian of the presentation $G(T_{p,q})=\langle x,y \mid x^py^{-q}\rangle$ is computed. It is $$ \left( \frac{t^{pq}-1}{t^q-1}, -\frac{t^{pq}-1}{t^p-1}\right). $$ The greatest common divisor of it is the Alexander polynomial.

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I think that the computation of the Alexander polynomial of torus knots and more general algebraic knots goes back to Bureau. There is a general trick called the Seifert-Torres formula that allows you to compute the desired Alexander polynomial of the $(p,q)$-torus knot and much more. For a particularly an elegant proof based on the concept of Reidemeister torsion I refer to Turaev's most excellent survey Reidemeister torsion in knot theory, Russian Math. Surveys, vol. 41 (1986), 119-182. The concept of Reidemeister torsion is what hides behind the Alexander polynomial so it's worth having a look at this concept.

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Thank you for this link. I found another link for it, where you do not need a subscription: – Dietrich Burde May 5 '13 at 14:53
Thanks for the free access link. – Liviu Nicolaescu May 5 '13 at 15:38

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