Let’s learn about knots! I’m sure we’ve all experienced knots. For example, if you have a bunch of computer cables, chances are they have knots in them. Knots are actually quite tricky. Of course, first, we should say what a knot is, should we not?

A knot is a continuous map $S^1\to \R^3$ which is a homeomorphism onto its image. Sounds kind of complicated, but intuitively, just think of a knot as a circle embedded in three-dimensional space. Or you can forget everything about topology and start drawing them:

Everyone would agree that this is the simplest knot of all: it is a piece of string which is not knotted. It’s is called the “unknot”. You can think of it as an identity element.

Next up is a trefoil knot:

Knots are drawn like this so that if you lay a knot flat on a table, the breaks correspond to the part of the string that goes under another part of string. It seems intuitive that if you take a piece of string as in the trefoil and try to finesse it into the unknot, you can’t without breaking the string (breaking or cutting the string is not allowed).

Mathematically, we have to define when two knots are equivalent. Now, we have defined a knot as an embedding $K:S^1\to\R^3$. Therefore, we say that two knots $K_1$ and $K_2$ are equivalent if there exists a homeomorphism $f:\R^3\to\R^3$ such that $f\circ K_1 = K_2$. Basically, two knots are equivalent if we can move the string of one around to look like the string of another.

Back to the trefoil:

Intuitively we know this knot is not the unknot, but how can we prove this? For any knot $K$, we can calculate the fundamental group of $\R^3 – K$. If we do this with the unknot we get $\Z$ (interestingly, this is also the fundamental group of $K$ itself, but that is not very useful for knots because all knots are just circles and the fundamental group of the circle is just $\Z$).

We call the fundamental group $\pi_1(\R^3 – K)$ the knot group of $K$. There is a very easy way to write down a presentation of this group for any knot. Let’s do that together now. First, notice that when we draw a knot, we are actually drawing a finite number of line segments. Let’s label them:

We also have to choose an orientation on the knot (this is the direct you would travel if you were a really small ladybug on the knot):Now, you select all but one of the crossings draw little squares around them, and put orientations on those squares:

For each square, go around in the direction you chose. Every time you cross a piece of the knot, write down the letter of the segment you crossed. If the segment direction is coming in from the left (imagine traffic driving on the knot), put a $-1$ i the exponent of the letter. For example, for the square in the top right I drew, I would write $a^{-1}cbc^{-1}$. You can actually start anywhere in the square, and then you might get a cyclic shift of the word. It doesn’t matter. Now the fundamental group of the knot is the group generated by the letters denoting the segments together with the relations given by the words from traversing the squares. In our case, we get:

$$\langle~ a,b,c~|~ a^{-1}cbc^{-1}, aca^{-1}b^{-1}~\rangle.$$ You could have chosen the other crossing or even used all three: each crossing is a consequence of all the others. That’s why we omit it.

Now, we can simplify this presentation a little to make it easier to work with. I will leave it as an exercise to show that this presentation is equivalent to:

$$G = \langle~a,c~|~ cac=aca~\rangle.$$ If you get stuck on showing this, consult the post on Tietze transformations. Now, we already remarked that the knot group of the unknot is $\Z$. Therefore, we need to prove that $G$ as above is **not** the integers.

For a general knot, this may be quite difficult. In fact, I will probably talk more about knot invariants later. For this group, one strategy would be to find a nonabelian group $H$ such that there is a surjection $G\to H$. This would prove that $G$ is not abelian.

The smallest nonabelian group is $S_3$, the permutation group on three elements. If we can find two elements $x,y\in S_3$ such that $xyx = yxy$ and such that $x,y$ generate $S_3$, then that will demonstrate the existence of a surjection $G\to S_3$. For example, $x = (1 2)$ and $y=(1 3)$ are such elements. Moreover, the smallest subgroup containing these elements contains the identity and $(1 2)(1 3) = (1 3 2)$, so at least four elements. Therefore, the smallest subgroup containing $x$ and $y$ is actually $S_3$ by Lagrange’s theorem (the order of a subgroup divides the order of a group).

So there you have it: the trefoil is not the unknot.