MATH 422 Lecture Note #12 (2018 Spring)
Alexander polynomial

Jae Choon Cha
POSTECH $\def\id{\operatorname{id}} \def\rel{\text{ rel }} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\R{\mathbb{R}} \def\C{\mathbb{C}} \def\sm{\smallsetminus} \def\vp{{\vphantom{1}}}$

We have learned how to compute the fundamental group of knot complement, using the Wirtinger presentation. The next step, which is the goal of this lecture, is to extract useful information on knots from the fundamental group $G=\pi_1(\R^3\sm K)$ of the complement.

The abelianization is always infinite cyclic

Recall that the abelianization of the fundamental group was very useful in studying surfaces. We showed that the surfaces $\Sigma_g$ and $\Sigma_{g'}$ are not homeomorphic if $g\ne g'$, using the abelianization of the fundamental group. For knots, unfortunately, it is not the case, as seen in the following result.

Theorem. Suppose $K\subset \R^3$ is a knot. Then $\pi_1(\R^3\sm K)_{ab}$ is an infinite cyclic group, and any generator in the Wirtinger presentation becomes a generator of $\pi_1(\R^3\sm K)_{ab}$.

Proof. Recall that a relator in the Wirtinger presentation is of the form $x_{a\vp}^\vp x_{c\vp}^\vp x_b^{-1}x_{c\vp}^{-1}$, where $b\equiv a\pm1 \bmod n$ and $n$ is the number of strands. In the abelianization, this relator gives us $x_a=x_b$. That is, $x_i$ and $x_{i+1 \bmod n}$ are equal in the abelianization. The assertion follows from this. $\quad\square$

So, we need to extract deeper information than the abelianization. For this purpose, we will investigate the commutator subgroup $G'$, which is eliminated in the abelianization. In our approach, we will use the fact that $G_{ab}$ is infinite cyclic for $G=\pi_1(\R^3\sm K)$ as a crucial advantage. The above theorem is indeed useful.

Definition of the Alexander polynomial

Suppose $G$ is a group such that $G/G'$ is an infinite cyclic group, where $G'=[G,G]$ is the commutator subgroup. Let $G''=(G')'$, the commutator subgroup of the commutator subgroup. It is easily seen that $G''$ is a normal subgroup of not only $G'$ but also $G$. We will study the abelian group $G'/G''$, namely, the abelianization of $G'$, and the vector space of fractions $G'/G'' \otimes \Q$.

Using that $G/G'$ is an infinite cyclic group, we can give an additional structure on $G'/G''\otimes \Q$. Choose a generator $t$ of the infinite cyclic group $G/G'$. Note that there are exactly two choices, since $t$ and $t^{-1}$ are the only generators. In addition, choose a representative $\mu\in G$ of $t\in G/G'$, namely, $\mu G' = t$. Consider the conjugation homomorphism $L_\mu\colon G'/G'' \to G'/G''$ defined by $L_\mu(gG'')=\mu g \mu^{-1} G''$ for $g\in G'$. This is a well-defined group homomorphism. Indeed, if $h=gk$ with $k\in G''$, then

\[\mu h \mu^{-1} = \mu gk \mu^{-1} = \mu g \mu^{-1} \mu k \mu^{-1} \in \mu g \mu G''\]

since $G''$ is a normal subgroup. Also, $L_\mu$ is an isomorphism, since $(L_\mu)^{-1} = L_{\mu^{-1}}$. Moreover, if $\nu$ is another representative of the coset $t=\mu G'$, then $L_\nu=L_\mu$. To see this, note that $\nu=\mu k$ for some $k\in G'$. Then for $g\in G'$, we have $kgk^{-1}g^{-1} =[k,g] \in G''$, and thus $kgk^{-1}$ and $g$ are equal in $G'/G''$. It follows that $\nu g \nu^{-1} = \mu k g k^{-1} \mu ^{-1}$ and $\mu g \mu ^{-1}$ are equal in $G'/G''$.

This tells us that the isomorphism $L_\mu \colon G'/G'' \to G'/G''$ is determined by the generator $t\in G/G'$, independent of its representative $\mu\in G$. So, we denote $L_\mu$ by $t\colon G'/G'' \to G'/G''$. It induces an isomorphism on the vector space of fractions $G'/G''\otimes \Q$, which we also denote by

\[t\colon G'/G''\otimes \Q \to G'/G''\otimes \Q.\]

Definition. Suppose $K$ is a knot in $\R^3$. Let $G=\pi_1(\R^3\sm K)$. Choose a generator $t$ of the infinite cyclic group $G_{ab}=G/G'$.
Suppose the vector space $G'/G''\otimes\Q$ is finite dimensional over $\Q$. Then, the Alexander polynomial $\Delta_K(t)$ of $K$ is defined to be the characteristic polynomial of the linear isomorphism $t\colon G'/G''\otimes\Q \to G'/G''\otimes\Q$.

Recall that the characteristic polynomial of a linear transformation $\phi\colon V\to V$ is defined to be $\det(tI-A)$, where $A$ is a square matrix representing the transformation $\phi$ with respect to a choice of basis for $V$. Here we use $t$ as the variable of the polynomial. In general, the characteristic polynomial is well-defined up to nonzero scalar multiplication. Also, recall that if $\phi$ is invertible and has characteristic polynomial $p(t)$, then the characteristic polynomial of $\phi^{-1}$ is equal to $t^d p(t^{-1})$, where $d$ is the degree of $p(t)$.

Applying this to the knot case, namely when $G=\pi_1(\R^3\sm K)$, we obtain the following facts:

  1. When a generator $t$ of $G/G'$ is fixed, $\Delta_K(t)$ is well defined up to multiplication by nonzero rational numbers.

  2. If the generator $t$ of the infinite cyclic group $G/G'=\pi_1(\R^3\sm K)_{ab}$ is replaced by $t^{-1}$, then $\Delta_K(t)$ is replaced by $t^d \Delta_K(t^{-1})$, where $d=\deg \Delta_K(t)$.

For brevity, we write $p(t) \doteq q(t)$ if $p(t)=\lambda t^k q(t)$ for some nonzero $\lambda\in \Q$ and some integer $k$.

The following lemma says, briefly, that the pair $(G'/G'', t\colon G'/G'' \to G'/G'')$ is an invariant under isomorphisms of $G$.

Lemma. Suppose $G$ is a group such that $G/G'$ is an infinite cyclic group generated by $t\in G/G'$. If $f\colon G\to \Gamma$ is a group isomorphism, then the following hold.

  1. $f$ induces an isomorphism $f_{ab}\colon G/G' \to \Gamma/\Gamma'$. In particular, $s=f_{ab}(t)$ is a generator of the infinite cyclic group $\Gamma/\Gamma'$.

  2. $f$ induces an isomorphism $f_*\colon G'/G'' \to \Gamma'/\Gamma''$. Furthermore, the isomorphisms $t\colon G'/G'' \to G'/G''$ and $s\colon \Gamma'/\Gamma'' \to \Gamma'/\Gamma''$ satisfy $f_* \circ t = s\circ f_*$.

▶︎
all
running…

Proof. The proof consists of straightforward verifications. From the definition of $G'$, it follows that $f(G')=\Gamma'$. Applying this to $f|_{G'}$, it also follows that $f(G'')=\Gamma''$. By applying the first isomorphism theorem, it follows that $f$ induces isomorphisms $f_{ab}\colon G/G'\to \Gamma/\Gamma'$ given by $f_{ab}(gG')=f(g)\Gamma'$ for $g\in G$ and $f_*\colon G'/G'' \to \Gamma'/\Gamma''$ given by $f_*(gG'') = f(g)\Gamma''$ for $g\in G'$. This proves (1) and the first statement of (2). If $\mu G'=t$, then for $\nu= f(\mu)$, $s=f_{ab}(t) = f(\mu)\Gamma' = \nu\Gamma'$ by the definitions. So, $s\colon \Gamma'/\Gamma'' \to \Gamma'/\Gamma''$ is given by $s(h\Gamma') = \nu h\nu^{-1} \Gamma'$. Now it is routine to verify $f_* \circ t = s\circ f_*$: for $g\in G'$,

\[f_*(t(gG'')) = f_*(\mu g \mu^{-1}G'') = f(\mu) f(g) f(\mu)^{-1} \Gamma'' = \nu f(g) \nu^{-1} \Gamma'' = s(f(g)\Gamma'') = s(f_*(gG'')).\]

This completes the proof. $\quad\square$

Applying the lemma to the knot case with $G=\pi_1(\R^3\sm K)$, we immediately obtain the following:

Theorem. Suppose $K$ and $J$ are knots in $\R^3$. Choose generators of $\pi_1(\R^3\sm K)_{ab}$ and $\pi_1(\R^3\sm J)_{ab}$ to define the Alexander polynomials $\Delta_K(t)$ and $\Delta_J(t)$. If $K$ and $J$ are equivalent, then either $\Delta_K(t)\doteq \Delta_J(t)$ or $\Delta_K(t)\doteq \Delta_J(t^{-1})$.

Remark. It turns out that $\Delta_K(t)\doteq \Delta_K(t^{-1})$ holds for every knot $K$, although we do not provide a proof here. So, in fact, $\Delta_K(t)\doteq \Delta_J(t)$ holds whenever $K$ and $J$ are equivalent.

Computation from Wirtinger presentation

Suppose $G=\pi_1(\R^3\sm K)$ is given by a Wirtinger presentation:

\[G = \langle x_1,\ldots,x_n \mid r_1,\ldots,r_{n-1}\rangle\]

where each relator $r_j$ is of the form

\[r_j = x_{a_j\vp}^\vp x_{c_j\vp}^\vp x_{b_j}^{-1} x_{c_j\vp}^{-1}.\]

Here $a_j$, $b_j$, $c_j$ are the indices of the strands involved in the $j$th crossing. See the previous lecture note.

To define the Alexander polynomial, we need to fix a generator $t$ of $G_{ab}=G/G'$. Recall that any one of the $x_j$ represents a generator of $G/G'$. We use this generator as our $t\in G/G'$. In addition, we use $\mu=x_n\in G$ as a representative of $t=\mu G'$.

Our strategy is to construct a presentation of $G'$, using covering spaces. Recall that if $X$ is a space with $\pi_1(X)=G$, then there is a regular covering $p\colon E\to X$ such that $G'=\pi_1(E)$, since $G'$ is a normal subgroup. Moreover, the covering transformation group $G(E|X)$ is isomorphic to $G_{ab} = G/G'$, which is an infinite cyclic group generated by $t$. We will construct such an infinite cyclic cover, and compute $\pi_1(E)$ to obtain $G'$.

Of course, we already have a space $X$ whose fundamental group is equal to $G$. Namely, the knot complement $\R^3\sm K$ itself. But, to make the fundamental group computation for the covering space easier, we will use a different space $X$ whose $\pi_1$ is equal to our $G$. (But, I would like to remark that understanding the infinite cyclic cover of $\R^3 \sm K$ is a very important and enjoyable subject, which you may study later! It is a pity that this subject is out of the scope of this semester.)

To construct our $X$, start with a graph $X^1$ which has one vertex $*$ and $n$ edges attached to the vertex. For later purpose, we will use the following description. For a positive integer $n$, let $[n] = \{1,\ldots,n\}$, the space of $n$ points with discrete topology. Then $X^1$ can be understood as the quotient space

\[X^1 = S^1\times[n] / (1,i) \sim (1,i') \quad\text{for }i,i'\in [n].\]

Here $1\in S^1$ is the usual basepoint. Our basepoint $*$ of $X^1$ is (the equivalence class of) the point $(1,i)$.

We already know that $\pi_1(X^1)$ is the free group generated by the edges. Call the edges $x_1,\ldots,x_n$, so that

\[\pi_1(X^1) = \langle x_1,\ldots,x_n \mid \cdot \rangle.\]

To realize the Wirtinger relations, we will attach $n-1$ disks to $X^1$, along the words $r_1,\ldots,r_{n-1}$. A precise desciption is as follows. Recall that a word in $x_1,\ldots x_n$ represents a loop in the graph $X^1$. Let $\alpha_j\colon I\to X^1$ be the loop in $X^1$ represented by $r_j$. Define

\[X = \big(X^1 \cup (D^2\times[n-1])\big) / (e^{2\pi s\sqrt{-1}},j) \sim \alpha_j(s) \quad\text{for }s\in [0,1],\, j\in[n-1].\]

Here, $(e^{2\pi s\sqrt{-1}},j)$ is a point in $S^1\times\{j\}\subset D^2\times[n-1]$, and $\alpha_j(s)$ is a point in $X^1$.

By a standard application of the Seifert-van Kampen theorem, we have

\[\pi_1(X) = \langle x_1,\ldots,x_n \mid r_1,\ldots, r_{n-1} \rangle = G.\]

Now, we will construct a covering $p\colon E\to X$, which satisfies $\pi_1(E)\cong p_*\pi_1(E) = G' \subset G=\pi_1(X)$. Recall, once again, that such a cover must have infinite cyclic covering transformation group, since $G(E|X) = G/G'\cong \Z$. This provides a motivation for the following construction. Let

\[E^1 = \R\times[n] / (k,i) \sim (k,i') \quad\text{for }k\in \Z,\, i,i\in [n],\]

and take $e_0=(0,i)\in E^1$ as the basepoint.

▶︎
all
running…

Obviously $E^1\to X^1$ defined by $(s,i) \mapsto (e^{2\pi s\sqrt{-1}},i)$ is a covering map. (This may be regarded as a pinched version of the $n$ copies of the standard infinite cyclic cover $\R\to S^1$.) For $k\in \Z$ and $(s,i)\in E^1$, define $(s,i)+k = (s+k,i)$. This addition map $(s,i) \mapsto (s,i)+k$ is a covering transformation on $E^1$.

To extend this to a covering of $X$, attach infinitely many disks indexed by the set $[n-1]\times \Z$ as follows:

\[E = \big( E^1 \cup (D^2\times[n-1]\times\Z)\big) / \tilde\alpha_j(s)+k \sim (e^{2\pi s\sqrt{-1}},j,k) \quad\text{for }s\in [0,1],\, j\in[n-1],\, k\in \Z\]

where $\tilde\alpha_j\colon I \to E^1$ is the lift of $\alpha_j\colon I \to X^1$ such that $\tilde\alpha_j(0)=e_0$. Extend the covering $p\colon E^1\to X^1$ by defining $p(x,j,k) = (x,j)$ for $(x,j,k)\in D^2\times[n-1]\times \Z$. This gives a covering map $p\colon E\to X$.

We will show that $p_*\pi_1(E)\cong \pi_1(E) = G' \subset G=\pi_1(X)$ for our covering $p\colon E\to X$. The following lemma is useful for this purpose.

Lemma. The subgroup $G'$ is generated by $\{y_{i,k} \mid i\in [n], k\in \Z\}$, where

\[y_{i,k} = x_n^k (x_i x_n^{-1}) x_n^{-k}.\]

Proof. Recall that $G/G'$ is equal to $\langle x_1,\ldots,x_n \mid x_1 x_n^{-1},\ldots,x_{n-1}x_n^{-1} \rangle$. It follows that $G'$ is the normal subgroup generated by the elements $x_i x_n^{-1}$, and thus $G'$ is the subgroup generated by the conjugates $gx_ix_n^{-1}g^{-1}$, where $g\in G$.

Let, temporarily, $H$ be the subgroup generated by the elements $y_{i,k}$. For later use, note that $x_i^{-1}x_n = x_n^{-1}(x_ix_n^{-1})^{-1}x_n = y_{i,-1}^{-1} \in H$. Also, $x_n^\ell H x_n^{-\ell} = H$ for any $\ell\in \Z$. We have $H\subset G'$ by the previous paragraph. For the converse, we will show that $gx_ix_n^{-1}g^{-1} \in H$ for all $g$ and $i$, by induction on the length of the word $g$. When $g$ is trivial, $gx_ix_n^{-1}g^{-1} = x_i x_n^{-1} \in H$. Suppose $g=x_j^{\pm1} h$ and $hx_ix_n^{-1}h^{-1}\in H$. Then, since $x_n^{\pm1} (hx_ix_n^{-1}h^{-1}) x_n^{\mp1}\in H$ by the above, we have

\[x_j^{\pm1}h \cdot (x_ix_n^{-1}) \cdot h^{-1} x_j^{\mp1}= (x_j^{\pm1} x_n^{\mp1}) \cdot x_n^{\pm1} (hx_ix_n^{-1}h^{-1}) x_n^{\mp1}\cdot (x_j^{\pm1} x_n^{\mp1})^{-1} \in H.\]

This shows that $G'\subset H$. $\quad\square$

Now, consider $\pi_1(E)$, which we identify with the subgroup $p_*\pi_1(E)\subset G=\pi_1(X)$. Regarding $E^1$ as a graph with pinched points $(k,i)$ as vertices ($k\in\Z$), the line $\R\times\{n\}$ is a maximal tree, and every lift of $x_i$ gives a generator for $i\in [n-1]$. It follows that $\pi_1(E^1)\subset G$ is the free group generated by $\{y_{i,k} \mid i\in[n-1],\,k\in \Z\}$. The attached disks $D^2\times[n-1]\times \Z$ give relators, due to the Seifert-van Kampen theorem again. According to the above construction of $E$, the disk attachments give relators $x_n^k r_j x_n^{-k}$, for $j\in[n-1]$, $k\in \Z$. They can be rewritten in terms of the generators $y_{i,k}$, as shown below:

\[\begin{aligned} x_n^k r_j x_n^{-k} & = x_{n\vp}^k \cdot x_{a_j\vp}^\vp x_{c_j\vp}^\vp x_{b_j}^{-1} x_{c_j\vp}^{-1} \cdot x_{n\vp}^{-k} \\ & = x_{n\vp}^k (x_{a_j\vp}^\vp x_{n\vp}^{-1}) x_{n\vp}^{-k} \cdot x_{n\vp}^{k+1} (x_{c_j\vp}^\vp x_{n\vp}^{-1})x_{n\vp}^{-k-1} \cdot x_{n\vp}^{k+1} (x_{n\vp}^\vp x_{b_j}^{-1})x_{n\vp}^{-k-1} \cdot x_{n\vp}^k (x_{n\vp}^\vp x_{c_j\vp}^{-1}) \cdot x_{n\vp}^{-k} \\ &= y_{a_j,k} \cdot y_{c_j,k+1} \cdot y_{b_j,k+1}^{-1} \cdot y_{c_j,k}^{-1} \end{aligned}\]

So, we have shown the following:

\[\pi_1(E) = p_*\pi_1(E) = \big\langle y_{i,k} \,\big|\, y_{a_j,k} \cdot y_{c_j,k+1} \cdot y_{b_j,k+1}^{-1} \cdot y_{c_j,k}^{-1} \big\rangle\]

where $i\in[n-1]$, $j\in[n-1]$, $k\in \Z$.

In addition, since $G'$ is the subgroup in $G$ generated by $y_{i,k}$ by the above lemma, the subgroup $\pi_1(E) = p_*\pi_1(E)$ is exactly the same as $G'$. So, we have proven the following:

Theorem.

\[G' = \big\langle y_{i,k}^\vp \,\big|\, y_{a_j,k}^\vp \cdot y_{c_j,k+1}^\vp \cdot y_{b_j,k+1}^{-1} \cdot y_{c_j,k}^{-1} \big\rangle\]

where $i\in[n-1]$, $j\in[n-1]$, $k\in \Z$.

Now, it is straightforward to compute the vector space $G'/G''\otimes \Q$.

Theorem.

  1. The vector space of fractions for $G'/G''$ is given by $G'/G''\otimes \Q = V/W$, where $V$ is the vector space with basis $\{y_{i,k}\mid i\in [n-1],\, k\in \Z\}$, and $W$ is the subspace of $V$ spanned by

    \[\{y_{a_j,k} + y_{c_j,k+1} - y_{b_j,k+1} - y_{c_j,k} \mid j\in[n-1],\, k\in \Z\}.\]

  2. The linear isomorphism $t\colon G'/G''\otimes \Q \to G'/G''\otimes \Q$ is given by

    \[t(y_{i,k}) = y_{i,k+1}.\]

Proof. By applying our previous result which computes the vector space of fractions from a group presentation (see Lecture Note #9), (1) is immediately obtained. To show (2), recall that $t$ on $G'/G''\otimes \Q$ is the conjugation by $\mu=x_n$, that is, $t(g) = x_n g x_n^{-1}$. Applying this to $y_{i,k}$, we have $t(y_{i,k}) = x_n x_n^k (x_i x_n^{-1}) x_n^{-k} x_n^{-1} = y_{i,k+1}$. $\quad\square$

Recipe for the computation of the Alexander polynomial

Using the above theorem, we can compute the Alexander polynomial. It is convenient to use the following notation. Let $y_i = y_{i,0}$. Due to the above theorem, we have $y_{i,k} = t^k y_i$. Also, the following equation holds in $G'/G''\otimes \Q$.

\[y_{a_j,k} - y_{c_j,k} = y_{b_j,k+1} - y_{c_j,k+1}\]

It can be rewritten as follows:

\[t^k(y_{a_j} - y_{c_j}) = t^{k+1}(y_{b_j} - y_{c_j}) \tag{R}\]

To compute the Alexander polynomial, carry out the following steps:

  1. Find a basis for $G'/G''\otimes \Q$. Recall that in the above description of the vector space $G'/G''\otimes \Q$, we have an infinite set of generators $\{t^k y_i\}$. Using the above equations, we can always reduce the number of generators to find a finite basis for $G'/G''\otimes \Q$, while we do not present a proof. We remark that $G'/G''\otimes \Q$ is often spanned by the vectors $y_i$. (But they are not necessarily linearly independent in general.)

  2. Find a matrix for $t\colon G'/G''\otimes \Q \to G'/G''\otimes \Q$, with respect to the basis found in Step 1. The equation (R) is often useful in this step as well. Compute the characteristic polynomial for the matrix.

Note that $y_n= x_nx_n^{-1}$ is trivial. This is often very useful.

Perhaps this method is best understood by an example.

Example: trefoil knot. We will compute the Alexander polynomial of the trefoil knot. Here is a diagram of the trefoil knot, which we denote by $K$:

▶︎
all
running…

From the diagram, it is straightforward to compute a Wirtinger presentation.

\[G = \pi_1(\R^3\sm K) = \langle x,y,z \mid xyz^{-1}y^{-1}, zxy^{-1}x^{-1}\rangle\]

Here our $x$, $y$, $z$ play the role of $x_1$, $x_2$, $x_3$ above. By the above theorem, $G'/G''$ is generated by the vectors $y_{i,k}=t^ky_i$, $i\in\{1,2\}$, $k\in \Z$. In our case, the relations (R) are as follows:

\[\begin{aligned} t^k(y_1-y_2) &= t^{k+1}(y_3-y_2) \\ t^k(y_3-y_1) &= t^{k+1}(y_2-y_1) \end{aligned}\]

Since $y_3=0$ in $G'/G''\otimes\Q$, these relations are indeed as follows:

\[\begin{aligned} t^k(y_2-y_1) &= t^{k+1}y_2 \\ t^k y_1 &= t^{k+1}(y_1-y_2) \end{aligned}\]

We claim that $G'/G''\otimes \Q$ is a vector space of dimension 2, and $\{y_1,y_2\}$ is a basis. The idea of the proof is simple: we will eliminate generators using the above relations. When $k=0$, we have $ty_2 = -y_1+y_2$ and $ty_1-ty_2 = y_1$. It follows that $t y_1 = y_2$. So, $t y_1, t y_2$ are in the span of $y_1,y_2$. Therefore we can eliminate $ty_1,ty_2$ together with the two relations we used. In general, for arbitrary $k$, $t^{k+1}y_1, t^{k+1}y_2$ are in the span of $t^{k}y_1, t^{k}y_2$, so by induction, $t^k y_1, t^k y_2$ are in the span of $y_1,y_2$ for all $k> 0$. A similar argument shows that $t^k y_1, t^k y_2$ are in the span of $y_1,y_2$ for all $k<0$. Moreover, in this elimination process, all the relations are eliminated together with the generators $\{t^ky_i\mid i\in\{1,2\},\, k\ne 0\}$. This shows that $y_1,y_2$ form a basis for $G'/G''\otimes \Q$.

Now it is easy to compute the characteristic polynomial. Let $T$ be the matrix representing $t\colon G'/G''\otimes \Q \to G'/G''\otimes \Q$ with respect to the basis $\{y_1,y_2\}$. The above relations, for $k=0$, says that $T$ satisfies the following matrix equation:

\[\begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} = T\cdot \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}\]

From this it follows that

\[\Delta_K(t) \doteq \det\left( t\cdot \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} \right) = -t^2+t-1 \doteq t^2-t+1.\]

We remark that the Alexander polynomial $\Delta_K(t)=t^2-t+1$ of the trefoil knot has the property $\Delta_K(t) \doteq \Delta_K(t^{-1})$, as aforementioned.

Recall that $\pi_1(\R^3\sm U)=\Z$ for the trivial knot. It follows that the commputator subgroup $\pi_1(\R^3\sm U)'$ is trivial, so $\Delta_U(t)\doteq 1$. Since this is different from the above Alexander polynomial of the trefoil knot, we have obtained an alternative proof of the following fact: the trefoil knot is not trivial.

Example: figure eight knot. Recall that the following is called the figure eight knot.

▶︎
all
running…

We already computed a Wirtinger presentation. (See Lecture Note #11.)

\[G = \pi_1(\R^3\sm K) = \langle x_1,x_2,x_3,x_4 \mid yxz^{-1}x^{-1},\, wzx^{-1}z^{-1},\, yw x^{-1}w^{-1} \rangle\]

Once again, we will use the above theorem, to compute $G'/G''\otimes \Q$. In this case, we have generators $y_{i,k}=t^k y_i$ for $i\in\{1,2,3\}$, $k\in \Z$, and the following relations.

\[\begin{aligned} t^k(y_2 - y_1) &= t^{k+1}(y_3-y_1) \\ t^k(y_4 - y_3) &= t^{k+1}(y_1-y_3) \\ t^k(y_2 - y_4) &= t^{k+1}(y_1-y_4) \end{aligned}\]

Here, note that $y_4=0$. So, the last relation becomes $t^{k} y_2 = t^{k+1} y_1$. Moreover, we can eliminate $y_2$ from the first two relations, to get the following relations which involve $y_1$ and $y_3$ only:

\[\begin{aligned} t^k y_1 &= t^{k+1}(2y_1-y_3) \\ t^k y_3 &= t^{k+1}(y_3-y_1) \end{aligned}\]

From the above two relations, it follows that $t^k y_1$, $t^k y_3$ are contained in the span of $y_1, y_3$ for all $k$, as we did in the trefoil knot case. Moreover, the relation $t^{k} y_2 = t^{k+1} y_1$ tells us that $t^k y_2$ is also contained in the span of $y_1, y_3$ for all $k$. So, $y_1$ and $y_3$ form the basis for the vector space $G'/G''\otimes \Q$. Interestingly, $y_2$ is not a part of the basis!

Now, let $T$ be the matrix representing $t\colon G'/G''\otimes \Q \to G'/G''\otimes \Q$ with respect to the basis $\{y_1,y_3\}$. The two relations shown above, which involve $y_1$ and $y_3$ only, tells us that $T$ satisfies the following:

\[\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = T\cdot \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}\]

So, the Alexander polynomial of the figure eight knot $K$ is given by

\[\Delta_K(t) \doteq \det \left( t\cdot \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = t^2-3t+1.\]

Once again, observe that $\Delta_K(t) \doteq \Delta_K(t^{-1})$ is satisfied.

Note that this Alexander polynomial is different from the Alexander polynomials of the trivial knot and the trefoil knot. So, we have proven the following theorem:

Theorem.

  1. The figure eight knot is nontrivial.

  2. The figure eight knot is not equivalent to the trefoil knot.