# Alexander invariant of torus knot

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:

$$(t^q+t^{2q}+...+t^{pq})\alpha=(t^p+t^{2p}+...+t^{qp})\beta$$

$$(t^q+t^{2q}+...+t^{(-s)q})\alpha=(t^p+t^{2p}+...+t^{rp})\beta$$

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

$$\Delta(t)=\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$$

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.

• Convert the matrix to its Smith normal form. In this particular case, this is doable over the integers. May 10, 2014 at 5:56
• 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. May 10, 2014 at 17:36
• May 10, 2014 at 18:19
• I want to get the relation $\Delta(t)\alpha=0$ by eliminate the gennerator $\beta$.So we get the $\Lambda-$module $H_1(\tilde{X})\cong(\alpha\mid \Delta(t)\alpha=0)\cong \Lambda/(\Delta(t))$ May 11, 2014 at 1:19
• yeah. We can get the result by other methods ,like free calculus . May 11, 2014 at 1:26

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 give 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.

• 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. Jul 14 at 19:07
• 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. Jul 14 at 19:13
• Thank you for your comment @ShiyuLiang, I edited my answer. Now it should be more precise/correct Jul 29 at 12:58