Jae Choon Cha
POSTECH
Knots and diagrams
Recall that a map $f\colon X\to Y$ is called an embedding if $f\colon X\to f(X)$ is a homeomorphism. When it is the case, we can identify the subspace $f(X)$ with the space $X$.
Definition. A knot in $\R^3$ is the image of an embedding $S^1\to \R^3$. That is, a knot is a simple closed curve in $\R^3$.
The following diagrams illustrate knots:
The first one could also be described as follows:
The knot in the second picture above is called a trefoil knot.
Knot diagrams
Often, it is convenient to describe a knot as a picture as shown above, from which it is easy to visualize a simple closed curve in $\R^3$. A more precise description is as follows. A vertex in a graph has valence $k$ if there are $k$ edges attached to the vertex. (If an edge starts and finishes at the same vertex, count it twice.) Consider a finite graph embedded in the $xy$plane in which every vertex has valence 4. The idea is to view this as the shadow, or projection, of a knot to the $xy$plane. Note that, near each vertex of the graph, it looks like the following picture:
This may be viewed as the intersection of the shadow, or projection, of two arcs. For each vertex, as additional data, we choose one of these arcs and call it "over." Call the other "under." In a diagram, we represent our choice by drawing a vertex as follows:
In other words, for each vertex of the graph, we assign one of the above two. The above illustration of the trefoil knot is an example.
When a vertex is shown as one of the above choices, we call it a crossing, instead of the above.
Definition. A knot diagram is a graph with 4valence vertices which is embedded in the $xy$plane, together with a choice of over and under for each vertex.
After replacing vertices with crossings as above, a knot diagram consists of connected arcs. Each connected arc in a knot diagram is called a strand. For instance, in the above diagram of the trefoil knot, there are three strands. Since there are two ends of strands involved at each crossing (i.e., those representing "under" arcs), and since each strand has two ends, it is easily seen that the number of strands in a knot diagram is equal to the number of crossings.
Now the knot given by a knot diagram is defined as follows. Regard the graph as a subset in the $xy$plane, and fix $\epsilon>0$. For each crossing, choose a neighborhood, say $U$, of the vertex in the $xy$plane in which the graph looks like the left hand side of the picture below. Inside $U\times[\epsilon,\epsilon]$, Leave the "over" strand as it is, and alter the "under" strand as shown in the right hand side of the picture below. Perform this for all crossings to obtain a knot in $\R^3$.
Equivalence of knots
Any two knots are homeomorphic as spaces, since every knot is homeomorphic to $S^1$ by definition. But, there are many distinct embeddings of $S^1$ into the same space $\R^3$. That is, as a subspace of $\R^3$, there may be many different knots. For example, the trivial knot and trefoil knot look very different.
We formulate the notion of the "same" knots as follows.
Definition. Two knots $K$ and $J$ in $\R^3$ are equivalent if there is a homeomorphism $h\colon \R^3 \to \R^3$ such that $h(K)=J$.
A knot is unknotted, or trivial, if it is equivalent to the trivial knot defined $U$ above.
This is obviously an equivalence relation. Often, it is easy to see that two knots are equivalent by drawing pictures. See the following example.
Examples.

Let $J$ be the knot given by the first (leftmost) knot diagram below. It is obvious that the second knot diagram gives a knot which is equivalent to $J$, since the first knot diagram is taken to the second by a homeomorphism of the $xy$plane. The second and third diagrams also give equivalent knots. To see this, just grab the top strand in the second diagram, pull it to your side (i.e., out of the paper/blackboard/screen, toward the positive $z$direction), and then push it down, to get the third diagram. The third and fourth diagrams give equivalent knots in the same way as we did for the first and second. Finally, the fourth one, namely the rightmost diagram, gives a knot equivalent to the trefoil knot shown above. So, it follows that the knot $J$ we started with is equivalent to the trefoil knot. In other words, $J$ is the same knot as the trefoil knot.
▶︎allrunning… 
While the following diagram might look pretty complicated, it turns out that it describes a trivial knot. Can you find a sequence of pictures converting this diagram to the above diagram of the unknot $U$?
▶︎allrunning…
We will address the problem of showing two given knots are equivalent or not. To show that two knots are equivalent, it is sufficient to provide pictures of continuous deformation from one diagram of another. On the other hand, proving that two knots are not equivalent is more difficult. Because there are too many ways to alter a given diagram, it is just impossible to check all the possibilities.
So, we will develop some more systematic methods, using fundamental groups. The following statement is very useful.
Theorem. If two knots $K$ and $J$ are equivalent, then $\pi_1(\R^3\sm K)$ and $\pi_1(\R^3\sm J)$ are isomorphic.
Proof. Suppose $h\colon \R^3\to \R^3$ is a homeomorphism such that $h(K)=J$. Then the restriction
is a homeomorphism, so the induced homomorphism
is an isomorphism. $\quad\square$
Thus, if one could prove that the groups $\pi_1(\R^3\sm K)$ and $\pi_1(\R^3\sm J)$ are not isomorphic, then it would follow that $K$ and $J$ are not equivalent. This leads us to the study of the fundamental group $\pi_1(\R^3\sm K)$ of the knot complement.
Fundamental group of the unknot complement
Theorem. For the trivial knot $U$, $\pi_1(\R^3\sm U) \cong \Z$.
There are several ways to prove this. One of the most well known methods is to consider $\pi_1(S^3\sm U)$. Indeed, it is an easy exercise to show that if we regard $\R^3$ as a subspace of $S^3=\R^3\cup\{\infty\}$ which is the onepoint compactification of $\R^3$, then the inclusion induces an isomorphism $\pi_1(\R^3\sm K) \cong \pi_1(S^3\sm K)$ for any knot $K$ in $\R^3$. For the trivial knot case, it is known that $S^3\sm U$ is homotopy equivalent to $S^1$. To see this, one may use the following:
Under the above identification, there is an embedding $S^1=S^1\times\{0\} \hookrightarrow S^1\times D^2 \subset S^3$. One can verify that its image is equivalent to the trivial knot $U$. From this, it follows that
This proves the claim that $S^3\sm U$ is homotopy equivalent to $S^1$, so $\pi_1(S^3\sm U) \cong \Z$.
In what follows, we will present an alternative proof.
Proof. Let $Z=\{(0,0,z)\mid z\in \R\}\subset \R^3$ be the $z$axis, and let $P=\{(x,0,z)\mid x>0\}$ be the positive half of the $xz$plain. Note that $P\cap U$ is the single point $(1,0,0)$. Let $A=\R^3\sm (U\cup Z)$. Observe that $A$ is obtained by rotating the punctured plane $P\sm U$ around the $z$axis. It follows that $A\cong (P\sm U) \times S^1$. Taking $\pi_1$, we obtain
Since $P\sm U\cong \R^2\sm \{0\} \simeq S^1$, $\pi_1(P\sm U)$ is isomorphic to an infinite cyclic group $\langle \mu \mid \cdot\rangle$, where the generator $\mu$ is represented by a loop in $P$ that runs around the point $P\cap U$ once. Also, $\pi_1(S^1)$ is an infinite cyclic group too. It follows that
where $\lambda$ is represented by a loop in the $xy$plane which runs round the $z$axis once. (This corresponds to a generator of the $\pi_1(S^1)$ factor!)
Now, let $B=\{(x,y,z) \mid x^2+y^2 < 1/4 \}$, a $\frac12$neighborhood of our $Z$. Since $B \cong \R\times \operatorname{int} D^2$, $B$ is contractible. Also,
It follows that $\pi_1(A\cap B)$ is isomorphic to the infinite cyclic group $\langle \lambda \mid\cdot \rangle$. Now, apply the Seifertvan Kampen theorem:
From the above computation, it follows that the pushout $\pi_1(\R^3\sm U)$ is given by
This completes the proof that $\pi_1(\R^3\sm U)\cong \Z$. Moreover, it is generated by $\mu$ defined above. $\quad\square$
The trefoil knot is nontrivial
Let $K$ be the trefoil knot, given by the following knot diagram:
Our goal is to compute $\pi_1(\R^3\sm K)$. For this purpose, first observe that the trefoil knot can be drawn on a torus $T$ in $\R^3$, as illustrated below. Here, thick dashed arcs represent parts of our knot $K$ which lies on the "back" side of the torus $T$, while the nondashed thick arcs represent the parts of $K$ on the "front" side.
Observe that $\R^3\sm T$ is the disjoint union of two regions, one bounded and the other unbounded. Let $A$ be the bounded region, and let $B$ be the unbounded region. Fix suffiently small $\epsilon>0$, and identify an $\epsilon$neighborhood of $T\subset \R^3$ with $T\times(\epsilon,\epsilon)$ in such a way that $T$ is identified with $T\times \{0\}$. That is, we identify a "thicking" of $T$ with the product $T\times(\epsilon,\epsilon)$. Now, let
Then $U$ and $V$ are open subsets of $\R^3\sm K$ such that $U\cup V=\R^3\sm K$ and
Moreover, $U\simeq \overline A$ and $V\simeq \overline B$. (Here, $\overline A$, $\overline B$ designates the closure.) So, by applying the Seifertvan Kampen theorem, we obtain the following pushout diagram which consists of inclusioninduced homomorphisms:
Since $\overline A\cong S^1\times D^2$ (it is just a solid torus bounded by $T$!), we have $\pi_1(\overline A)=\langle x \mid\cdot \rangle$, an infinite cyclic group, where the generator $x$ is represented by the core circle of the solid torus $A$. See the picture below. Also, $\overline B =\R^3\sm A \simeq \R^3\sm U$, the unknot complement. So, $\pi_1(\overline B)=\langle y \mid\cdot \rangle$, an infinite cyclic group, where the generator $y$ is represented by the curve shown in the picture below.
Now, we need to compute $\pi_1(T\sm K)$. By cutting $T$ along $K$, one obtains leftmost picture, which shows the space $T\sm K$. The second picture shows the same (homeomorphic) space, with wider cut. Making the cut wider, or equivalently making the space narrower, one gets the third picture, which clearly shows that $T\sm K$ is homeomorphic to an annulus $S^1\times(1,1)$, which is homotopy equivalent to $S^1$.
So, $\pi_1(T\sm K)$ is isomorphic to an infinite cyclic group $\langle z \mid \cdot \rangle$, where the generator $z$ is represented by the loop on $T\sm K$ shown below:
Observe, from the above pictures, that the $z$ is homotopic, rel $\{0,1\}$, to the loop $x^2$ in the space $\overline A$. On the other hand, in $\overline B$, $z$ is homotopic to $y^{3}$ rel $\{0,1\}$. This tells us that the image of $z$ under $\pi_1(T\sm K) \to \pi_1(\overline A)$ is equal to $x^2$, and the image of $z$ under $\pi_1(T\sm K) \to \pi_1(\overline B)$ is equal to $y^{3}$. So, by the above Seifertvan Kampen pushout diagram, we obtain the following result, which we state as a theorem:
Theorem. For the trefoil knot $K$, $\pi_1(\R^3\sm K) = \langle x, y \mid x^2 y^3 \rangle$.
Using this, we want to compare the trefoil $K$ with the trivial knot $U$. We already have $\pi_1(U)\cong \Z$. So, we are naturally led to the following question? Is $\pi_1(\R^3\sm K)$ an infinite cyclic group?
Recall that the abelianization was very useful for the fundamental groups of orientable surfaces. So, perhaps we should try to apply the same idea. Namely, to show that two groups $G$ and $\Gamma$ are not isomorphic, it suffices to show that the abelianizations $G_{ab}=G/G'$ and $\Gamma_{ab}=\Gamma/\Gamma'$ are not isomorphic. The advantage here is that abelian groups are often easier to compare. For the case of the trivial knots, $\pi_1(\R^3 \sm U)_ {ab} = \Z_ {ab} = \Z$ since $\Z$ is already abelian. For the trefoil $K$, unfortunately, it can be shown that $\pi_1(\R^3 \sm K)_ {ab}$ is an infinite cyclic group too! We leave it as an exercise. So, what this tells us is that the abelinization is not powerful enough to distinguish the groups $\pi_1(\R^3 \sm U)$ and $\pi_1(\R^3 \sm K)$.
But it is too early to be disappointed. We can still proceed, as follows:
Lemma. For the trefoil knot $K$, $\pi_1(\R^3 \sm K)$ is not abelian.
Proof. Define
by $\phi(x)=x$, $\phi(y)=y$. Since the relator $x^2y^3$ is killed by this, $\phi$ is a welldefined group homomorphism. Since the generators of $D_6$ are contained in the image of $\phi$, $\phi$ is surjective. So, if $\pi_1(\R^3\sm K)$ was abelian, then $D_6$ would be abelian. Since $D_6$ is not abelian, it follows that $\pi_1(\R^3\sm K)$ is not abelian. $\quad\square$
Now we are ready to prove the following result:
Theorem. The trefoil knot $K$ is not equivalent to the trivial knot $U$.
Proof. By the above lemma, $\pi_1(\R^3 \sm K)$ is not abelian. It follows that $\pi_1(\R^3\sm K)$ is not isomorphic to $\pi_1(\R^3 \sm U)=\Z$, which is abelian. Therefore $K$ is not equivalent to $U$. $\quad\square$
So, now we know that not all knots are equivalent. Our arguments and results obtained above illustrate that the information from the fundamental group of the knot complement is very useful. But, in general, for knots other than the trivial knot and the trefoil knot, how could one compute the fundamental group of the complement? Our next subject will address this problem.