## MATH 422 Lecture Note #11 (2018 Spring)Wirtinger presentation

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

Our goal is to present a systematic method to compute a presentation of the fundamental group of the knot complement, from a knot diagram.

Start with a given knot diagram, and let $n$ be the number of crossings. In what follows, when we provide an illustration, the following knot diagram will be used as an example. Note that $n=4$ for this example.

running…

The knot given above is called the figure eight knot.

Recall that a knot diagram defines a knot, which looks like the following picture in a neighborhood of each crossing.

running…

We may assume that the knot lies in the product $\R^2\times[1,-1]$, identifying the $xy$-plane with $\R^2$. The picture above is understood as part of the knot in $U\times [-1,1]$, where $U$ is a neighborhood of a crossing in the plane. We assume that the lower arc, which is parallel to the $xy$-plane, lies in $U\times \{-1\}$. Outside these neighborhoods of crossings, the knot lies in $\R^2\times \{0\}$. A global picture is shown below.

running…

Consider $K\cap \R^2\times[-1/2,\infty)$. This consists of arcs embedded in $\R^2\times[-1/2,\infty)$, which correspond to the strands of the original diagram. See the picture below. Note that the number of strands is equal to $n$, the number of crossings in the diagram.

running...

It is obvious that the above is homeomorphic to the following:

running...

More precisely, there is a homeomorphism of $\R^2\times[-1/2,\infty)$ onto itself which takes the $n$ arcs in the second last picture to the $n$ arcs in the last picture. It is easily seen that, in the last picture, the complement $\R^2\times[-1/2,\infty) \sm (n\text{ arcs})$ is homotopy equivalent to the graph with one vertex and $n$ edges. See the following picture:

running...

So, the fundamental group $\pi_1(\R^2\times[-1/2,\infty) \sm K)$ is the free group on $n$ generators $x_1,x_2,\ldots,x_n$, where the generators are represented by the loops based at the basepoint $*$, as shown in the following picture. Note that the basepoint is in the upper half space, with large $z$ coordinate.

running...

By the above, we have

$\pi_1(\R^2\times[-1/2,\infty) \sm K) = \langle x_1,\ldots,x_n \mid \cdot \rangle.$

To specify the loops $x_i$ precisely, we use the following convention. Orient the knot diagram, by starting from a point on the knot and traveling on the knot along a fixed direction. Add an arrow on each strand along the direction. Then, our $x_i$ is defined to be the loop which links the $i$th strand along the right hand direction.

Now, for each crossing, take a copy of $D^2\times (0,1)$. Attach this to $\R^2\times[-1/2,\infty) \sm K$, by identifying $S^1\times(0,1)$ with the annulus in $\R^2\times\{-1/2\} \sm K$ which is shown in grey in the picture below. The space obtained by attaching $n$ copies of $D^2\times(0,1)$, one for each crossing, is homeomorphic to $\R^2\times[-2,\infty)\sm K$, which is homotopy equivalent to $\R^3\sm K$.

running...

We will investigate how the fundamental group changes when we attach a copy of $D^2\times(0,1)$ around a crossing. Let $A=\R^2\times[-1/2,\infty) \sm K$, and $B$ be the copy of $D^2\times (0,1)$ which is attached around a crossing. We will compute $\pi_1(A\cup B)$ using the Seifert-van Kampen theorem. Note that $A\cap B$ is the grey annulus shown in the above picture along which $B$ is attached. A technical issue is that $B$ does not contain the basepoint we used above, which is far above the $xy$-plane. So, we proceed as follows. Choose a point on the grey annulus $A\cap B$, and take a vertical arc connecting the point to the basepoint in the space $A$. See the above picture. Let $B'$ be the union of $B$ and this arc. We will use $B'$ instead of $B$. Note that $B'$ is contractible, and thus $\pi_1(B')$ is trivial. Also, $A\cap B'$ is the union of the arc and the attaching annulus, which is homotopy equivalent to $S^1$. So $\pi_1(A\cap B') = \langle z \mid \cdot \rangle$ is an infinite cyclic group, where its generator $z$ is represented by the core circle of the attaching annulus. Enlarge $A$ and $B'$ slightly to make them open subsets in $R^3$, and apply the Seifert van Kampen theorem, to obtain the following pushout diagram:

running...

From this it follows that $\pi_1(A\cup B')$ is equal to the quotient of $\pi_1(A)$ modulo the normal subgroup generated by the image of $z\in \pi_1(A\cap B')$ in $\pi_1(A)$. That is, the image of $z$ in $\pi_1(A)$ gives a new relator.

To understand the image of $z$ under $\pi_1(A\cap B') \to \pi_1(A)$, consider the following picture, which is the top view of a crossing.

running...

There are two cases, depending on the orientation of the strands (the thick arcs in the picture) near the crossing. The thin short arrows represent our generators $x_i$ of $\pi_1(A)$, which correspond to the strands of the knot diagram. Since there are three strands involved in a crossing, we have three generators denoted by $x_a$, $x_b$, $x_c$ in the picture. Recall that the orientations of the loops $x_i$ are determined from the orientation we chose for the knot, by the right hand rule. The arrow direction of $x_i$ in the picture shows this. The loop $z$ in the picture represents our generator $z$ of $\pi_1(A\cap B')=\langle z \mid\cdot \rangle$. The dot on $z$ represents the endpoint of the arc chosen to define $B'$. So, precisely speaking, the loop $z$ starts from the basepoint, goes down along the arc from the basepoint to the dot in the picture below, runs along the circle denoted by $z$, and then comes back to the basepoint along the arc again. Here, we remark that the orientation of the circle $z$ and the location of the dot can be chosen arbitrarily. (Different choices may give us different relators in the presentation we will obtain below, but such different presentations will give the same (isomorphic) group, of course.)

From the above picture, it is seen that the image of $z$ in $\pi_1(A)$ is equal to $x_{\vphantom{1}a}^{\vphantom{1}} x_{\vphantom{1}c}^{\vphantom{1}} x_b^{-1} x_{c^{\vphantom{1}}}^{-1}$, if we choose an appropriate starting point and orientation for $z$. So, the attachment of $D^2\times (0,1)$ gives us the word $x_{\vphantom{1}a}^{\vphantom{1}} x_{\vphantom{1}c}^{\vphantom{1}} x_b^{-1} x_{c^{\vphantom{1}}}^{-1}$ as a relator. Performing this for each crossing, we prove the following result.

Theorem. Suppose that $K$ is represented by a knot diagram with $n$ crossings. Note that each crossing in the diagram is of either one of the two forms shown in the above picture. Let $x_{a_j}$, $x_{b_j}$ and $x_{c_j}$ be the three generators involved in the $j$th crossing ($j=1,\ldots,n$) as shown in the above picture. Then

$\pi_1(\R^3\sm K) = \langle x_1,\ldots,x_n \mid x_{\vphantom{1}a_1}^{\vphantom{1}} x_{\vphantom{1}c_1}^{\vphantom{1}} x_{b_1}^{-1} x_{c_1\vphantom{1}}^{-1}, \ldots, x_{\vphantom{1}a_n}^{\vphantom{1}} x_{\vphantom{1}c_n}^{\vphantom{1}} x_{b_n}^{-1} x_{c_n\vphantom{1}}^{-1} \rangle.$

This presentation of the fundamental group of the knot complement is called a Wirtinger presentation.

We remark once again that the relators in the Wirtinger presentation may be altered by changing the choice of the orientation of the loop $z$ and its starting point indicated by the dot in the above picture. But, for any choice, the result is a presentation for $\pi_1(\R^3\sm K)$. Indeed, an alternative choice changes the relator by conjugation and inversion.

Example. We will compute the fundamental group of the figure eight knot complement.

running…

Since there are four strands in the diagram, we have four generators, which are denoted by $x$, $y$, $z$, $w$ in the diagram. By reading words around crossings above, we obtain the following relators, which correspond to the top-left, middle, bottom, and rightmost crossings:

$r_1=yxz^{-1}x^{-1},\quad r_2=wzx^{-1}z^{-1},\quad r_3=yw x^{-1}w^{-1},\quad r_4=wyz^{-1}y^{-1}$

So, by writing the four generators and the four relators above, you will get a presentation of $\pi_1(\R^3\sm K)$.

Before writing down the presentation, we will make a very useful observation. First, we have a product of certain conjugates of the four relators (or the inverses) which becomes trivial, as shown below:

$r_1^{-1} \cdot (yz^{-1}\cdot r_2^{-1} \cdot zy^{-1}) \cdot (yz^{-1}y^{-1} \cdot r_3 \cdot yzy^{-1}) \cdot (yz^{-1}y^{-1} \cdot r_4 \cdot yzy^{-1}) = 1 \tag{P}$

The identity (P) can be verified by a straightforward cancellation computation. From this, it follows that any one of the four relators is expressed as a product of conjugates of the other three relators or their inverses. In other words, the normal subgroup generated by $r_1,r_2,r_3,r_4$ is equal to the normal subgroup generated by any three of them. This says that we can eliminate any one of the four relators from the presentation. So, in particular, by removing $r_4$, we have the following Wirtinger presentation for the figure eight knot:

$\pi_1(\R^3\sm K) = \langle x,y,z,w \mid yxz^{-1}x^{-1},\, wzx^{-1}z^{-1},\, yw x^{-1}w^{-1} \rangle$

It is an interesting question how one could find the identity (P). Was it just lucky, or was it due to a judicious guess? Also, is this peculiar to the case of the figure eight knot?

#### One of the Wirtinger relators can always be eliminated

Indeed, the identity (P) has a topological interpretation. To understand this, consider the following diagram:

running…

Note that there are four loops labeled by $\gamma_i$ around crossings, and each $\gamma_i$ is connected to the point $p$ along an arc $\beta_i$. Recall that our basepoint $*$ of $\R^2\times [-1/2,\infty)$ is in the upper half space above the $xy$-plane, with large positive $z$-coordinate. Fix a vertical path $\alpha$ from $*$ to $p$. All the paths and loops $\alpha$, $\beta_i$, $\gamma_i$ we consider here are regarded as ones in the space $\R^2\times [-1/2,\infty)$.

From the picture, it is seen that the product

$(\beta_1^{\vphantom{1}} \gamma_1^{\vphantom{1}} \beta_1^{-1}) \cdot (\beta_2^{\vphantom{1}} \gamma_2^{\vphantom{1}} \beta_2^{-1}) \cdot (\beta_3^{\vphantom{1}} \gamma_3^{\vphantom{1}} \beta_3^{-1}) \cdot (\beta_4^{\vphantom{1}} \gamma_4^{\vphantom{1}} \beta_4^{-1})$

is homotopic, to rel $\{0,1\}$, to the constant path $c_p$ in $\R^2\times [-1/2,\infty)$. Indeed, the product $\prod \beta_i^{\vphantom{1}}\gamma_i^{\vphantom{1}}\beta_i^{-1}$ is homotopic, rel $\{0,1\}$ in $\R^2\times [-1/2,\infty)$, to the outmost loop $\gamma$ shown in the above picture. Why? To see this, take the disk bounded by the loop $\gamma$, remove the smaller inner disks bounded by the $\gamma_i$, and then cut it along the arcs $\beta_i$. You get a disk embedded in $\R^2\times [-1/2,\infty)$, which is a desired homotopy $\prod \beta_i^{\vphantom{1}}\gamma_i^{\vphantom{1}}\beta_i^{-1} \simeq \gamma$ rel $\{0,1\}$. Since $\gamma$ is null-homotopic rel $\{0,1\}$ in $\R^2\times [-1/2,\infty)$, the product $\prod \beta_i^{\vphantom{1}}\gamma_i^{\vphantom{1}}\beta_i^{-1}$ is null-homotopic too!

From this, it follows that the product of conjugates

$(\alpha\beta_1^{\vphantom{1}} \gamma_1^{\vphantom{1}} \beta_1^{-1}\alpha^{-1}) \cdot (\alpha\beta_2^{\vphantom{1}} \gamma_2^{\vphantom{1}} \beta_2^{-1}\alpha^{-1}) \cdot (\alpha\beta_3^{\vphantom{1}} \gamma_3^{\vphantom{1}} \beta_3^{-1}\alpha^{-1}) \cdot (\alpha\beta_4^{\vphantom{1}} \gamma_4^{\vphantom{1}} \beta_4^{-1}\alpha^{-1})$

is trivial in $\pi_1(\R^2\times [-1/2,\infty))$, which is the free group on the generators $x_i$. This is, indeed, the identity (P). One can easily verify that the loop $\alpha\beta_i^{\vphantom{1}} \gamma_i^{\vphantom{1}} \beta_i^{-1}\alpha^{-1}$ is equal to the $i$th factor of the product in the left hand side of the identity (P).

Our topological interpretation of the identity (P) does not depend on that our diagram is that of the figure eight knot. That is, the same argument applies to any knot diagram. Therefore, we showed the following:

Addendum to the theorem. Suppose $K$ is given by a diagram with $n$ crossings. Then, in the Wirtinger presentation, any one of the $n$ relators can be eliminated.