Given a knot $K$, which is an embedding $S^1\to \R^3$, we have see how to compute the fundamental group of $K$, defined as $\pi_1(\R^3 – K)$. For example, we have computed the fundamental group of the trefoil knot and the fundamental group of the cinquefoil knot. The fundamental group of the trefoil can be given by

$$G_1 = \langle~a,c~|~ cac=aca~\rangle.$$ Of course, that was not the Wirtinger presentation we obtained in the trefoil post, but in that post we showed how to simplify it to this one. Similarly, we computed the fundamental group of the cinquefoil knot as the presentation

$$G_2 = \langle~d,x~|~dx^2d = x^3~\rangle.$$ We note that there is a surjective homomorphism $G_1\to S_3$ where $S_n$ denotes the symmetric group on $n$ letters. However, there is no such homomorphism $G_2\to S_3$. In fact, $n=5$ is the smallest integer such that there exists a surjective homomorphism of $G_2$ onto a nonabelian subgroup of $S_n$. Therefore, this tells us that the trefoil and cinquefoil are not the trivial knot, and more importantly, they are not equivalent as knots.

Now, actually finding a surjective homomorphism of a knot group onto a nonabelian group is not always easy. And even if one can be found in an algorithmic way sometimes, actually carrying out that algorithm is actually a lot of work. And even if you *can* get a computer to do it, brute-forcing the result with a computer seems somewhat inelegant to my mind.

Therefore, one might like a simpler way to distinguish a large class of knots. The *Alexander polynomial* of a knot is one such way. The Alexander polynomial is a generator of a principal ideal in the ring $\Z[t,t^{-1}]$ of Laurent polynomials, which is also the integral group ring of the integers. In this little tutorial, I will show you how to compute the Alexander polynomial of a knot. The theory of why this works will only be hinted upon, and the interested reader can consult Crowell and Fox’s Introduction to Knot Theory for an outstanding exposition of the details.

## Derivations

Now, we need to introduce the concept of a derivation. It is quite simple: if $G$ is a group and $A$ a $\Z G$-module, then a derivation $D:\Z G\to A$ is a abelian group homomorphism that satisfies $D(gh) = D(g) + gD(h)$ (not a typo!) for all $g,h\in G$. Therefore, if we know $D$ on a set of generators for $G$, we can compute $D$ on all of $\Z G$ via this product rule and extending linearly.

If $F$ is a free group on generators $\{ x_i \}_{i\in I}$ where $I$ is any index set, then there is a unique derivation $\partial d/\partial x_i:\Z F\to\Z F$ satisfying $\partial x_j/\partial x_i = \delta_{ij}$ where $\delta_{ij}$ is the indicator function. That is, $\delta_{ij} = 1$ if and only if $i=j$ and it is equal to zero otherwise.

## The Alexander Polynomial

We will assume that you have a Wirtinger presentation of your knot group. My post on the trefoil knot goes into great detail on how to compute it, so refer to that post if you forgot how. For example, let us use the presentation we derived for the trefoil:

$$G = \langle~ a,b,c~|~ a^{-1}cbc^{-1}, aca^{-1}b^{-1}~\rangle.$$ In general you will have a presentation with $n$ generators and $n-1$ relators:

$$\langle~ x_1,\dots, x_n~|~ r_1,\dots, r_{n-1}~\rangle.$$ Once you have this presentation for the knot group, we do the following:

- Compute the matrix $(n-1)\times n$ matrix $A$ where the $i,j$-entry of $A$ is
- Substitute a symbol $t$ for each generator $x_i$ in the matrix $A$ to give a matrix with entries in $\Z[t,t^{-1}]$.
- Choose one of the two $(n-1)\times (n-1)$ submatrices and compute its determinant
- Normalize the determinant by multiplying by $\pm t^k$ for some $k$ and choice of $\pm$ so that resulting polynomial has only positive powers and positive constant term. This is the Alexander polynomial of the knot.

$$A_{ij} = \frac{\partial}{\partial x_j}(r_i).$$

N.B. Step 3 gives a generator of a *principal ideal* in $\Z[t,t^{-1}]$ that is an invariant of the knot. Because the only units of $\Z[t,t^{-1}]$ are $\pm t^k$, the final step gives a standard representative such that two knot ideals will be equal if and only if their normalized representatives are. *Hence, if two knots have different Alexander polynomials, they will be inequivalent knots. On the other hand, the Alexander polynomial is not a complete invariant of knots in the sense that there exists pairs of inequivalent knots with the same Alexander polynomial.*

Let’s do an example. Recall we have

$$G = \langle~ a,b,c~|~ a^{-1}cbc^{-1}, aca^{-1}b^{-1}~\rangle,$$ which is the Wirtinger presentation of the trefoil knot. We first need to compute the derivations for Step 1:

$$\begin{align*}

\frac{\partial}{\partial a}(a^{-1}cbc^{-1}) &= -a^{-1}\\

\frac{\partial}{\partial b}(a^{-1}cbc^{-1}) &= a^{-1}c\\

\frac{\partial}{\partial c}(a^{-1}cbc^{-1}) &= a^{-1}(1 – cbc^{-1})\\

\frac{\partial}{\partial a}(aca^{-1}b^{-1}) &= 1 – aca^{-1}\\

\frac{\partial}{\partial b}(aca^{-1}b^{-1}) &= -aca^{-1}b^{-1}\\

\frac{\partial}{\partial c}(aca^{-1}b^{-1}) &= a.\\

\end{align*}$$ The matrix we obtain by substituting $t$ for each generator is the following matrix:

$$\begin{pmatrix}

-t^{-1} & 1 & t^{-1} – 1\\

1-t & -1 & t

\end{pmatrix}$$ The determinant of the leftmost $2\times 2$ block is

$$t^{-1} – t + t.$$ Normalizing by multiplying by the unit $t$ gives the Alexander polynomial of the trefoil:

$$f(t) = 1 – t + t^2.$$ N.B: We could have chosen $t^{-1}$ instead of $t$ for the generator in the group ring $\Z[t,t^{-1}]$. However, due to a special dual property of the knot group, that would give the same polynomial, so the knot polynomial is truly well-defined!