Conway's classification of rational tangles

This is a guest post by Glen Lim largely based on this paper by Goldman and Kauffman.

A rather big branch of topology is knot theory, which, as the name suggests, is the study of knots. Intuitively, a knot is a closed loop sitting in 3D space, and is generally allowed to move around in the way that (elastic) physical strings can. Mathematically, a knot is an embedding $S^1 \rightarrow \mathbb{R}^3$ (or rather, $S^3$ for technical reasons), and two knots $k,k': S^1\rightarrow \mathbb{R}^3$ are equivalent if they are ambiently isotopic: there's a continuous map $\mathbb{R}^3 \times I \rightarrow \mathbb{R}^3$ such that $F(\bullet,t)$ is a homeomorphism for each $t$, $F(\bullet,0)$ is the identity, and $k' = F(\bullet,1)\circ k$. There's a slight warning here: another natural notion of equivalence is isotopy: i.e. there's a homotopy $H: S^1 \times I \rightarrow \mathbb{R}^3$ from $k$ to $k'$ such that each $H(\bullet,t)$ is an embedding. This isn't what we want, because, in the words of a renowned knot theorist, this definition allows for knots to shrink and go *poof*, becoming unknotted:

The stronger notion of ambient isotopy allows us to move knots around while maintaining the intuitive physical restraints, so strands can't pass through each other and we can't make a knot vanish by pulling tight.

By the way, the pictures above are examples of knot diagrams: projections of a knot into $\mathbb{R}^2$ such that the preimage of each point has size at most two. Unfortunately, knot diagrams are very much unique; for example, the following is a diagram of the unknot.

In general, the question of whether two knot diagrams represent the same knot is hard. As an attempt to tackle this problem, Conway looked at a related mathematical object: rational tangles.

A tangle is the same thing as a knot, except that it can have multiple strands, and some of the strands can have free ends. We think of these has being embeddings of a disjoint union of closed intervals and $S^1$s embedded in $D^3$, with points being sent to $\partial D^3 = S^2$ iff they are endpoints of the intervals. We want the endpoints to be fixed, so we consider tangles to be equivalent if they differ by an ambient isotopy relative boundary, i.e. we add the extra condition that $F(\cdot,t)$ restricts to $\mathrm{id}_{S^2}$ on the boundary. In this post, all tangles will have endpoints on the equator, so we can draw them with pictures like this:

Moreover, all tangles we care about will have exactly four endpoints, which we draw in the northwest, northeast, southwest and southeast positions.

A rational tangle is a specific type of tangle which can be built out of the following basic building blocks:

We can combine tangles via horizontal and vertical composition:

A rational tangle is a tangle that can be built by starting from a $t_{\pm 1}$ and repeatedly applying $t_{\pm 1} + \bullet$, $\bullet + t_{\pm 1}$, $t_{\pm 1} + \bullet$ or $t_{\pm 1} +' \bullet$. In other words, we start with one of the $t_{\pm 1}$ and then repeatedly pick two adjacent endpoints and twist them.

Firstly, we should note that we may WLOG only twist the two right strands or the two bottom strands. This follows from

Theorem (flip theorem): For a rational tangle $A$, $A$ rotated by $180^\circ$ along the horizontal or vertical axis is equivalent to $A$.

Proof: This is true for $t_{\pm 1}$. We proceed inductively; WLOG consider $A + t_1$ (the other cases are analogous). The rotation about the horizontal axis is clear from the induction hypothesis:

For the other: 

Flip the green rectangle so as to move the twist to the left, then apply the induction hypothesis.

This directly implies

Corollary: For a rational tangle $A$, $t_1 + A \sim A + t_1$ (and likewise for $+'$)

Proof: This follows from the last diagram.

Hence, we may restrict ourselves to only the two operations $\bullet + t_{\pm 1}$ and $\bullet +' t_{\pm 1}$. Letting $t_{\pm n}$ be the $n$-fold horizontal composition of  $t_{\pm 1}$ and $t_{\pm n}'$ be the $n$-fold vertical composition of $t_{\pm 1}$, each rational tangle is equivalent to a basic tangle: a tangle of form $(\cdots (((t_{n_1} +' t_{m_1}') + t_{n_2}) +' t_{m_2}') + \cdots)$ (a basic horizontal tangle) or $(\cdots (((t_{n_1}' + t_{m_1}) +' t_{n_2}') + t_{m_2}) + \cdots)$ (a basic vertical tangle). Here's an example of a basic horizontal tangle:

The miraculous observation is that basic tangles of this form look like continuous fractions, which represent rational numbers. The idea is that the $t_n$ correspond to $n\in\mathbb{Q}$ and the $t_n'$ correspond to $\frac1n\in\mathbb{Q}$.

For this to work, we need some notion of additive and multiplicative inverses. Given a rational tangle $A$, let $-A$ be its mirror image in the plane, and let $\frac1A$ be $-A$ rotated in the plane by $90^\circ$ (in either direction; rotation by $180^\circ$ is the composition of the flips in the horizontal and vertical axes, so gives the same tangle by flip theorem). For example, we then have $-t_n = t_{-n}$ and $\frac1{t_n} = t_n'$. Moreover, by definition, $-\bullet$ and $\frac1\bullet$ are involutions.

Moreover, these satisfy the expected arithmetic properties:

• $A +' B \sim \frac1{\frac1A + \frac1B}$
• $-(A+B) \sim (-A) + (-B)$
• $-\frac1A \sim \frac1{-A}$

Then, any basic tangle can be written as a continuous fraction of $t_n$: for example, the one drawn above is $$((t_{-3} +' t_{-2}') +  t_3) +' t_1' \sim \frac1{t_1 + \frac1{t_3 + \frac1{t_{-2} + \frac1{t_{-3}}}}}.$$ Naturally, we may assign this to a rational number by replacing each $t_n$ by $n$. (More accurately, we should assign it to an element of $\mathbb{Q} \cup \{\infty\}$: $t_1 +' t_{-1}$, i.e. two vertical lines, should be assigned to $\frac1{1 -1} = \infty$.)

We can thus define the fraction $F(A)$ of a rational tangle $A$ as follows: write $A$ as a basic tangle, then apply the assignment above. Certainly, this assignment is surjective, as each rational number (and $\infty$) can be written as a continuous fraction. The other direction requires some basic properties of $F$: for a basic tangle $A$, $-A$ and $\frac1A$ are basic tangles, and we have

• $F(\frac1A) = \frac1{F(A)}$
• $F(-A) = -F(A)$
• $F(A + t_n) = F(A) + n$ for all $n$

These are easy to check. The middle one needs the properties of $-$ written above.

Using these, we may prove

Theorem (Conway): if two basic tangles $A,B$ have $F(A)=F(B)$, then they are ambiently isotopic.

Proof: Given a rational number $q$, there's a unique way to write it as a continued fraction such that all the terms (except possibly for the first) are positive. This corresponds to a basic tangle $T_q$. Given any other basic tangle $A$, we'll show that $A$ is ambiently isotopic to $T_{F(A)}$. We'll do this inductively, using the following deep fact in algebra: $a - \frac1b = (a-1) + \frac1{1 + \frac1{b-1}}$ for every integer $a$ and rational $b$. In the world of tangles, this corresponds to $t_n - \frac1A \sim t_{n-1} + \frac1{t_1 + \frac1{A-t_1}}$. Let's check this: WLOG $n=1$. Then, we want to show $t_1 -\frac1A \sim t_1 +' (A-t_1)$, i.e.

(Recall that $-\frac1A$ is just $A$ rotated anticlockwise by $90^\circ$.) Now, given an arbitrary continued fraction expansion for a rational number, we can for all its signs to be $+$ by inductively applying our identity from the inside to the outside. Since we can do the exact same moves to our tangle, the result follows.

Hence, we have proven a correspondence between rational tangles and $\mathbb{Q} \cup \{\infty\}$. Except...we have not. What remains is to show that $F$ is well-defined, i.e. we can't have two basic tangles $A,B$ which are ambiently isotopic but $F(A) \ne F(B)$. 

This may be done using a famous knot invariant known as the Kauffman bracket. Fix a formal variable $t$. Given a knot/tangle diagram, we impose the following relations:

• The skein relation: given a crossing, we may resolve it into a linear combination of the two different ways to uncross it (the purple circle represents a tiny neighbourhood around the crossing):

  

• Normalisation: we can delete an unknot at the expense of multiplying by $(-t^2-t^{-2})$.

Orient each component arbitrarily, so each crossing is either positive or negative:

Define the writhe $w(A)$ to be the number of positive crossings minus the number of negative crossings. Then, $(-t^3)^{-w(A)} <A>$ is invariant under ambient isotopy. The proof of this relies on Reidemeister's theorem, which says that two ambiently isotopic knots/tangles can be related by a series of Reidemeister moves, drawn below.

It then suffices to check that this is invariant on applying any of the three moves. It turns out that R2 and R3 don't change $<A>$, while $R1$ introduces a factor of $-t^3$, which is cancelled out by the writhe term.

Now, given a tangle $A$, applying the relations above repeatedly lets us write $<A>$ in the form $\alpha<t_\infty> + \beta<t_0>$ (where $t_\infty$ is two vertical lines and $t_0$ is two horizontal lines). Here, $\alpha,\beta$ are Laurent polynomials in $t$. Now, note that:

Theorem: $R(A) = \frac{\alpha}{\beta}$ is Reidemeister invariant.

Proof: This is clear for R2 and R3. For R1, the $-t^3$ term cancels between the numerator and the denominator, instead of with the writhe.

Now, set $t$ to be a square root of $i$ (the choice of root doesn't matter, since $\alpha,\beta$ are either both even or both odd). It remains to show that $F(A) = -iR(A)$ for all basic tangles $A$. Note that $-iR(\bullet)$ obeys the following properties:

• $-iR(\frac1A) = \frac1{\overline{-iR(A)}}$ (here $\overline{\bullet}$ means complex conjugate).
• $-iR(A+B) = -iR(A) - iR(B)$.

The first bullet point holds because we set $t^2=i$. For the second, note that our choice of $t$ means that removing an unknot corresponds to multiplication by $-t^2-t^{-2} = 0$. Then, use the fact that $t_0+t_0 = t_0, t_0+t_\infty = t_\infty+t_0=t_0$, and $t_\infty+t_\infty$ has an unknot.

Now, $-iR(t_{\pm 1}) = \pm 1$ by direct computation, so we get $-iR(A) = F(A)$ for every basic tangle $A$ by inducting on the continued fraction, and we are actually done this time.

This concludes Conway's classification of rational tangles: they are exactly in bijection with $\mathbb{Q} \cup \{\infty\}$.