I am very interested in knot theory, especially in knot groups and knot polynomials. As is well known, it is easy to calculate the Alexander polynomial from the fundamental group $\pi_{1}(K)$ of a knot $K$ by free calculus. But I now am reading the book of Rolfsen (Knots and Links) which gets the polynomial by calculating the Alexander invariant from $\pi_{1}(K)$. I want to compute the Alexander polynomial of the torus knot $T_{p,q}$ for $p$ and $q$ coprime by the method in Rolfsen's book. There is a hint in his book as following:

  • The knot group has presentation $G(T_{p,q})=( u,v\mid u^p=v^{q})$ where $u\mapsto q,v\mapsto p$ under abelianization.

  • Choose integer $r,s$ satisfying $pr+qs=1,r>0,s<0$.Let $x=u^{s}v^{r},a=ux^{-q},b=vx^{-p}$ to obstain the presentation with $x\mapsto 1,a\mapsto 0,b\mapsto 0$:

$$G(T_{p,q})=(x,a,b\mid (ax^{q})^p=(bx^p)^{-q},x=(ax^q)^s(bx^p)^r)$$

  • Let $ C=[G,G]$ then $C/[C,C]$ has a $\Lambda-$module presentation with generators $\alpha,\beta$ and relations:



  • $H_1(\tilde{X})\cong \Lambda/(\Delta(t))$ where


I know the (1)-(3),but I do not know how to get the (4) from (1)-(3). I need to know $\beta=(?)\alpha$ by eliminating the generator $\beta$ from the two relations.Can someone help me with this? Thanks a lot.

  • 1
    $\begingroup$ Convert the matrix to its Smith normal form. In this particular case, this is doable over the integers. $\endgroup$ – Alex Degtyarev May 10 '14 at 5:56
  • $\begingroup$ Just a remark: torus knot complements are mapping tori of finite-order automorphisms, from which one may deduce the Alexander polynomial by taking the characteristic polynomial. $\endgroup$ – Ian Agol May 10 '14 at 17:36
  • $\begingroup$ See mathoverflow.net/questions/129717/…. $\endgroup$ – Dietrich Burde May 10 '14 at 18:19
  • $\begingroup$ I want to get the relation $\Delta(t)\alpha=0$ by eliminate the gennerator $\beta$.So we get the $\Lambda-$module $H_1(\tilde{X})\cong(\alpha\mid \Delta(t)\alpha=0)\cong \Lambda/(\Delta(t))$ $\endgroup$ – Jacob.Z.Lee May 11 '14 at 1:19
  • $\begingroup$ yeah. We can get the result by other methods ,like free calculus . $\endgroup$ – Jacob.Z.Lee May 11 '14 at 1:26

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