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. $\endgroup$ May 10, 2014 at 5:56
  • 2
    $\begingroup$ 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, 2014 at 17:36
  • 1
    $\begingroup$ See mathoverflow.net/questions/129717/…. $\endgroup$ May 10, 2014 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 $H_1(\tilde{X})\cong(\alpha\mid \Delta(t)\alpha=0)\cong \Lambda/(\Delta(t))$ $\endgroup$ May 11, 2014 at 1:19
  • $\begingroup$ yeah. We can get the result by other methods ,like free calculus . $\endgroup$ May 11, 2014 at 1:26

1 Answer 1


Putting everything together, and with the big help of "Knots" by Burde, I found the following solution. Let $f_n(t)\colon= 1+t+\cdots+t^{n-1}$, we note that $\frac{t^m-1}{t-1}=f_m(t)$. @Jacob.Z.Lee's third point gives the following presentation matrix for $H_1(\tilde{X})$: $$ \begin{bmatrix} \frac{t^{pq}-1}{t^q-1} & \frac{t^{pq}-1}{t^p-1} \\ \frac{t^{-qs}-1}{t^q-1} & \frac{t^{pr}-1}{t^p-1}. \end{bmatrix} $$ The polynomials $f_p(t)$ and $f_q(t)$ are coprime (see the complex roots that they have and consider that $p,q$ are coprime) and so, applying the Euclidian algorithm to them, we find $\alpha(t),\beta(t) \in \mathbb{Z}[t] \subset \mathbb{Z}[t,t^{-1}]$ such that

$$\alpha(t) f_p(t) + \beta(t)f_q(t)= 1.$$

By explicit calculations, we see that

$$ \alpha(t) \frac{t^{pq}-1}{t^q-1} + \beta(t) \frac{t^{pq}-1}{t^p-1} = \Delta_{p,q}(t). $$

Now, we see that

$$\begin{bmatrix} \frac{t^{pq}-1}{t^q-1} & \frac{t^{pq}-1}{t^p-1} \\ \frac{t^{-qs}-1}{t^q-1} & \frac{t^{pr}-1}{t^p-1} \end{bmatrix} \cdot \begin{bmatrix} \alpha(t) & f_q(t) \\ \beta(t) & -f_p(t)\end{bmatrix} \cdot \begin{bmatrix} 1 &0 \\ 0 & t^{-qs}\end{bmatrix} \cdot \begin{bmatrix} 0 &1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} 0 & \Delta_{p,q}(t) \\ -1 & q(t) \end{bmatrix}.$$

Note that every matrix in the previous equation is inventible in $\mathbb{Z}[t,t^{-1}]$. Let $a,b$ be the two generators of $\Lambda^2$ as a $\Lambda$-module. The last matrix in the previous equations gives, as a presentation matrix for $H_1(\tilde{X})$ the relations $$\left\{ \begin{matrix} \Delta_{p,q}(t) \cdot b =0 \\ -a + q(t) \cdot b = 0, \end{matrix} \right.$$ and this is the conclusion.

  • $\begingroup$ Some typos: 1. The variable in the definition of $f_n(t)$ should be $t$ instead of $x$. 2. In the the exponent of $t^{qs}$ of the entry at row 2 column 1 of coefficient matrix should have a negative sign. $\endgroup$ Jul 14, 2022 at 19:07
  • $\begingroup$ 3. It seems to me that in the final expression there should be a negative sign for $a$. None of them affects the result though. $\endgroup$ Jul 14, 2022 at 19:13
  • $\begingroup$ Thank you for your comment @ShiyuLiang, I edited my answer. Now it should be more precise/correct $\endgroup$ Jul 29, 2022 at 12:58

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