2
$\begingroup$

Hello,

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 :)

$\endgroup$
2
6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.