# Low Dimensional Topology

## January 3, 2011

### Coloured knots: Part I

Filed under: 3-manifolds,Knot theory — dmoskovich @ 2:30 pm

Last Thursday, I uploaded a preprint to arXiv titled “Surgery presentations for knots coloured by metabelian groups”. The eventual goal of this series of blog post is to advertise its main results, and to try to convince you to read it.
Fix a group $G$. A $G$-coloured knot is pair $(K,\rho)$ consisting of a knot $K$ in $S^3$, together with a representation $\rho$ of its knot group $\pi\stackrel{\textup{\tiny def}}{=}\pi_1(\overline{S^3\setminus N(K)})$ onto $G$. There’s a natural equivalence relation on $G$-coloured knots called $\rho$-equivalence, and I classify $G$-coloured knots up to $\rho$-equivalence for certain families of metabelian groups $G$.

So now I have to tell you why one might care about $G$-coloured knots, why one might want to classify them up to $\rho$-equivalence, what my result is, and what I would like to do with it. This post addresses the first of these questions.

### Why coloured knots?

If I want you to read my preprint, I first have to interest you in coloured knots. Let’s face it- classical knots are rather special topological objects. They’re smooth (or PL) embeddings of a specific 1-manifold (the circle) in a specific 3-manifold (the 3-sphere). A $G$-coloured knot is even more special, because it’s a pair $(K,\rho)$ where the knot group of $K$ has to surject onto this specific group $G$. Now I don’t know about you, but I’d be bored stiff reading most papers about something even more specific than knots, such as a paper about torus knots or a paper about twist knots. So why should coloured knots be any different? Well…

• Coloured knots are classical:
Let’s face it- only a very small part of knot theory is actually about knots. Most of knot theory is really about knot groups. It’s been this way since 1908, when Tietze used Poincaré’s concept of the fundamental group to prove the existence of non-trivial knots. The basic premise of algebraic topology, I suppose, is that you’re not losing much information by studying homotopy groups instead of topological spaces- and this certainly holds true for knots, because a knot is determined by its complement (plus peripheral data, which we’ll ignore here), which is in turn determined by its fundamental group, at least for prime knots, by Waldhausen.
The key problem in knot theory is to classify knots, and what I’ve just told you, roughly, is that knots are classified by their knots groups; and calculating knot groups by way of the Wirtinger presentation is a piece of cake.

One’s first reaction upon discovering the Wirtinger presentation might be elation. It looks as though knot theory is “solved” because knot groups are the perfect invariant; and we can all go home. That elation quickly turns to despair, however, when we realize that all we’ve done is to replace one hard problem with another. If anything, groups are more complicated objects than knots, all the more so for group presentations.
So what can you do? On the desk in front of you lies a table of knot diagrams of up to 10 crossings, painstakingly compiled by Tait, Little, and Kirkman over a period of 25 years during the 19th century. They compiled this table by resolving double-points in all possible ways for all possible 4-valent graphs with up to 10 vertices, and then eliminating doubles by literally building physical models of the knots and playing around with bits of string to get them to look alike. But maybe they weren’t persistent enough, and maybe there are still doubles in the knot table (in fact, indeed there were). How are you going to eliminate those doubles, or conversely to prove that a pair of knots in the knot table are indeed distinct?
With the reader’s indulgence, I will henceforth simplify history for dramatic effect. The scene cuts to 1927, where we see 29 year-old James Waddell Alexander II sitting hunched over his desk. “Fine,” he sighs, “I can’t do much with the knot group itself- it’s too complicated. But maybe it contains simpler subgroups which I can say something about.”
For added dramatical effect, the scene suddenly cuts to 1972, Osaka City University, where Akio Kawauchi (later to become the father of Japanese knot theory) is reading Algebraic Topology by Edwin Spanier. Homological algebra was new at the time, certainly to a young graduate student, and young Kawauchi was planning to stun the world by using homological algebra to solve all of knot theory’s problems. He was in an optimistic mood that day, feeling he had enough knowledge to realize his dream, and it was time to begin reaping the harvest. He calculated the homology of a knot complement (not its fundamental group) using a Mayer-Veitoris sequence (there are other ways to do it of course). And then, in an instant, his excitement was transformed to utter despair… it’s $\mathbb{Z}$. It’s always $\mathbb{Z}$, for any knot.
Homology is useless!! Kawauchi threw his textbook at the wall in disgust. He would have to stun the knot theory world in another way.
Of course this was all known to Alexander in 1927, and to Schreier (and others) before him. The first stage $\pi/\pi^{(1)}$ in the derived series of $\pi$ gave no distinguishing information at all.
$\cdots\pi^{(2)}\triangleleft \pi^{(1)}\triangleleft \pi^{(0)}=\pi \quad\text{with\ } \pi^{(n+1)}\stackrel{\textup{\tiny def}}{=} [\pi^{(n)},\pi^{(n)}].$
But what about $\pi^{\text{mb}}\stackrel{\textup{\tiny def}}{=}\pi/\pi^{(2)}$? Again, we run into trouble. The group $\pi/\pi^{(2)}$ is metabelian, not abelian. It’s embarassing but true that pretty much all mathematics that we understand only works in a commutative setting. In particular, linear algebra doesn’t work over non-commutative rings; so homology becomes useless. We can’t expect to sensibly distinguish between two presentations of metabelian groups. It looks as though we’re no better off then where we started!
But, then again, perhaps things are not quite so bad as they seem. You see, $\pi^{\text{mb}}$ isn’t just any metabelian group. As Kawauchi rediscovered the hard way in his youth, its abelianization is $\mathbb{Z}$, and that’s true for the fundamental group too. So there’s a map $\mathrm{Link}\colon\thinspace \pi\twoheadrightarrow \mathbb{Z}$, which corresponds topologically to the linking pairing, sending any path in the knot complement to its linking number with the knot $K$. So now, from any knot, we obtain a pair $(K,\mathrm{Link})$ which is the degenerate example of a coloured knot. It’s degenerate because the colouring $\mathrm{Link}$ is canonically given, so we add no information by specifying it. If $A$ is an abelian group, an $A$-coloured knot is just a knot.
Given the $\mathbb{Z}$-colouring $\mathrm{Link}$, we look at $H_1(C_\infty)\stackrel{\textup{\tiny def}}{=}[\pi^{\text{mb}},\pi^{\text{mb}}]$, which is an abelian group! Equivalently, $H_1(C_\infty)=\textrm{Ab}(\pi/\mathbb{Z})$. Topologically, this quotient corresponds to the first homology group of the infinite cyclic cover of the knot complement. The group $\mathbb{Z}$, for which we choose a generator $t$, acts on $H_1(C_\infty)$ by covering transformations. This makes $H_1(C_\infty)$ into a $\mathbb{Z}[t,t^{-1}]$-module called the Alexander module. Using the Alexander module, Alexander was able to distinguish most of the knots in the 19th century knot table.
Most, but not all. He could not, for example, distinguish the $9_2$-knot from the $7_4$-knot.
The Alexander module is the biggest thing that ever happened in knot theory. That’s both a good thing and a bad thing. Good because it’s a splendid idea, but bad because it means that we haven’t had a comparable breakthrough since 1928. That’s pretty embarassing if you think about it. For all of our grand technologies and great minds and prestigious universities, just about everything we can calculate is contained in the Alexander module. That’s knot theory’s dirty secret. Stick to the Alexander module, and you can say pretty much everything. Step outside its comfortable bounds, and you can say just about nothing.
Reidemeister quickly realized what the problem was, and with typical energy, he rushed to find a solution. To calculate invariants, you need a commutative setting, right? So you’re looking to calculate the homology of some space. Very well then. Choose a non-abelian group $G$. It doesn’t have to be too complicated- let’s take it to be $D_6$, the symmetries of a triangle, AKA the dihedral group of order 6, AKA the symmetric group on 3 letters. The group $D_6$ is a metabelian group itself, having as its presentation:
$D_{6}\stackrel{\textup{\tiny def}}{=}\ \left\langle t,s\left|\rule{0pt}{9.5pt}\ t^{2}=s^{3}=1,\ tst=s^{-1}\right.\right\rangle.$
We write our metabelian groups as $\mathcal{C}_m\ltimes_\phi A$ where $A$ is an abelian group, whose operation is written as addition, and where $\mathcal{C}_m$ acts on $A$ by conjugation $\phi(a)=t^{-1}at$. In this language, $D_6=\mathcal{C}_2\ltimes_{[-1]}\mathbb{Z}/3\mathbb{Z}$.
Consider $D_6$–coloured knots, which are pairs $(K,\rho)$ where $\rho$ is a map from $\pi$ onto $D_6$. The key fact about $D_6$ is indeed that it is itself metabelian, because that guarantees that $\rho$ respects the derived series of $\pi$ in the sense that it factors through
$\beta\colon\thinspace \pi\to \mathbb{Z}\ltimes_\phi H_1(C_\infty),$
mapping the meridian of $K$ onto $t$ via $\mathrm{Link}$, and mapping $\pi^{(1)}$ onto $H_1(C_\infty)$.

You have not one, but two covering spaces associated to $\rho$.

1. The regular cover, associated to $\ker\rho$. This is the cover you get by pretending that $D_6$ were $\mathcal{C}_2\times \mathbb{Z}/3\mathbb{Z}$, forgetting the action. It is formed by first taking the 2-fold covering of $S^3$ branched over $K$, and then taking the 3-fold cover of that. To remind you how that goes:
Take a Seifert surface of $K$, cut the knot complement open along it, and compactify. So now you have a region $R$ whose boundary consists of a part which comes from $\overline{N(K)}$, and another two parts $F^\pm$ which come from a bicollar of the Seifert surface. Take a $\mathcal{C}_2$-worth of copies of $R$, indexed $R_0,R_1$, and glue $S_+$ in $R_n$ to $S_-$ in $R_{n+1}$ for all $n\in\mathcal{C}_2$. This gives the 2-fold covering space. Form the branched covering space $\tilde{M}$ by gluing the tubular neighbourhood of the knot back in. So there’s a knot in $\tilde{M}$ which projects down onto $K$. Take the 3-fold cover $\hat{M}$ of $\tilde{M}$ associated to the map from $H_1(C_\infty)$ to $\mathbb{Z}/3\mathbb{Z}$, and you get a 6-fold cover of $S^3$ branched over $K$, which contains a 3-component link $\hat{K}$ which projects down onto $K$.
2. The irregular cover, associated to the $\rho$-preimage of $\{1,t\}$. Recall that $\mathcal{C}_2$ acts on $\mathbb{Z}/3\mathbb{Z}$ by conjugation, taking $s$ to $s^{-1}$ and fixing $0\in \mathbb{Z}/3\mathbb{Z}$. This is an involution on $\mathbb{Z}/3\mathbb{Z}$, and we quotient $\hat{M}$ by it to obtain a 3-fold covering space $M$ of $S^3$, branched over $K$. This quotient action exchanges two components of $\hat{K}$ and fixes the third, so the link $\tilde{K}\subset M$ which projects down onto $K$ has two components.

Despite its discouraging name, it’s the irregular cover which is the natural and useful one. Its first homology group is strictly further down the derived series of $\pi$ than the Alexander module. It looks tempting to try exactly what Alexander did in the abelian case, and to try to extract information from $H_1(M)$ as a $\mathbb{Z}[D_6]$-module. Nobody has done that yet, for reasons I will not go into here (but which are basically L-theoretic reasons, I think- and these are issues which can be overcome in my opinion, although not fully “head on”). But you don’t even need all of the homology. You see, just about any non-trivial information about $(M,\tilde{K})$ sees beyond the Alexander module. What’s the simplest information you can think of? Well, if $M$ is simple enough (for instance if it’s $S^3$, which happens uncannily often as we’ll discuss below), the simplest information which I can think of is the linking number between the two components of $\tilde{K}$.
So that’s exactly what Reidemeister did immediately, in 1929. It took him no time at all to see all of this stuff (which I think is pretty amazing), and he calculated these linking numbers in irregular dihedral covering spaces, and distinguished knots with the same Alexander polynomial. Thus, Reidemeister was the first to consider invariants of coloured knots. Later on, others, most notably amateur mathematician and New York lawyer Kenneth Perko, refined these methods to complete a table of knots up to 11 crossings.
And so I’m claiming that invariants of coloured knots are the next natural thing after Alexander modules. If you want to step out of your comfort zone and to try something radical and new (in the mathematical sense), then coloured knots might be for you!

• Coloured knots are pretty:
“What’s all this about colours?” you ask me. “All I see is algebra sprinkled with a bit of geometric topology. When I think colours, I think red, white, and blue!”
“Very well,” replies Ralph Fox, speaking from back in 1954. “If it’s colours that you want, then colours you shall have! Although for purely artistic reasons, I would rather the colours be red, green, and blue.”
Let us look once again at our $D_6$-coloured knot. Its knot group is generated by its Wirtinger generators, which are all conjugate to one another. Therefore, $D_6$ is generated by images of Wirtinger generators, which must all be conjugate to one another. This forces each Wirtinger generator to map to $t$, $ts$, or $ts^2$ (these are the reflections of the triangle, and are of order 2). Call $t$ “red”, call $ts$ “green”, and call $ts^2$ “blue”. Then a knot is “3-colourable” if and only if you can colour the arcs of its knot diagram red, green, and blue so that at least two coloured are used, and at each crossing, either all three colours meet, or only a single colour meets itself. In fact, I could do the same thing for any dihedral group. Consider
$D_{2n}\stackrel{\textup{\tiny def}}{=}\ \left\langle t,s\left|\rule{0pt}{9.5pt}\ t^{2}=s^{n}=1,\ tst=s^{-1}\right.\right\rangle,$
for any odd number $n$. Wirtinger generators would all map to elements of the form $ts^i\in D_{2n}$. I could replace $ts^i$ by $i$ for $i=1,2,\ldots,n$, or I could colour code them…
Here’s an example of a knot coloured by $D_{10}$:
• Coloured knots are useful:
I’m going to prove to you that you should be interested in coloured knots… by intimidation!
First, a paper tiger. The biggest problem in 3-dimensional topology, at least until recently, has been the Poincaré Conjecture. And in the early 1970’s, people were out to disprove it. A brave young Mexican mathematician by the name of José Maria Montesinos led the assault. His first result was that it’s not unusual for a 3-manifold to be a 3-fold irregular covering space of $S^3$ branched over a knot. In fact, any 3-manifold is a 3-fold irregular covering space of $S^3$ branched over some knot, if you take the knot to be complicated enough. Thus, a $D_6$-coloured knot represents a 3-manifold (and a covering link inside it- but let’s forget the covering link); and any 3-manifold can be represented in this way.
This in itself is compelling enough. Coloured knots represent 3-manifolds. But it was Montesinos’s second result which people found most exciting. Montesinos proved that there is a rather simple local move on knots and links, which preserves the covering space (although it plays havoc with the covering link).

And so, a bold strategy emerged to find a counterexample to the Poincaré Conjecture. Take some strange $D_6$-coloured knot, obtain a representation of its 3-fold irregular cover, and simplify that presentation. You can choose your knot so that the fundamental group of this 3-fold irregular cover vanishes, and you can choose a really whacky one where this happens. Next, simplify your knot with Montesinos’s moves. If you can’t end up with a trivial link with 2 components (Montesinos’s move changes the number of components), then you are done, because $S^3$ is the 3-fold irregular cover over a $D_6$-coloured 2-component unlink.
Hopes were high, and coloured knots were on the top of the world, until Joan Birman came and spoiled the party. Joan Birman is a New York mathematician, most of whose papers were written after her children were grown up and she had time to devote to Mathematics. A heroine of mathematics for older folks.
“I’m afraid that there’s an inherent difficulty in this approach,” Birman told Montesinos and the Low Dimensional Topology world in 1974, in the typical way in which Joan Birman talks. “It is unlikely that you will find a counterexample with your method.”
It was a 3-page note in JAMS. And in an instant, coloured knots fell from the top of the stack, plunk in the mud.
Joan Birman had been working with Hugh Hilden on the Heegaard genus of covering spaces (this work has been useful for me more than once particularly when writing reviews), and she made what was by that point the simple observation, that irregular dihedral covers of 2-bridge knots were always $S^3$, and that there was no hope of a counterexample to the Poincare conjecture for coloured knots of bridge index less than 5. Knots of high bridge index are perfectly respectable mathematical objects… except that their fundamental groups are far beyond what anyone can calculate, and they’re off all the knot tables. You would need a super dooper computer to stand a chance, much more super than anything likely to be available within the next few decades. Thus, this attempt to refute the Poincaré Conjecture fails rather badly.
After this embarrasing debacle, coloured knots fell from favour for a while. Knot theory itself has a history of dashed hopes and disappointment. Once upon a time, more than a century ago, it was to be the theory of atoms. And then along came Neils Bohr.
People didn’t want to talk much about coloured knots for about 15 years. And then, along came Xiao-Song Lin from UC Riverside, in sunny California. Xiao-Song had an idea, which came again from homological algebra. If you can twist coefficients in homological algebra (a rather natural thing to do), why shouldn’t you it to the homology of the infinite cyclic cover? And then what happens to the Alexander polynomial? It becomes a twisted Alexander polynomial. And twisted Alexander polynomials have been popular and useful ever since. Why? Because they see beyond the Alexander module. The typical property of a coloured knot invariant.
But what is a twist of coefficients? Why, it corresponds exactly to a colouring of a knot! A twisted Alexander polynomial isn’t a knot invariant at all. It is indeed an invariant of a coloured knot! Amazingly, this is swept under the rug in a lot of publications- people make believe that the twisted Alexander polynomial is an invariant of knots. But it isn’t, and that’s nothing to be ashamed of. It’s an invariant of coloured knots. Say that out loud. Say it and be proud.
The twisted Alexander polynomial is an invariant of coloured knots.
This is my proof by intimidation. Coloured knot invariants are worthwhile, as exemplified by twisted Alexander polynomials. There’s a lot more going on beneath the Alexander module, and it’s for coloured knots to dredge that information out, and to lay it out in front of us. To find gold, don’t dig where everybody else has dug already. Dig a new mine with coloured knots!.

1. Of course I like it. What’s your mailing address? I have a short write-up of the simpler examples that you might enjoy.

Comment by Ken Perko — October 11, 2013 @ 9:00 pm

• Wow- thanks!! I’d love to see these. My current postal address is:
Daniel Moskovich,
Division of Mathematical Sciences
Nanyang Technological University
Singapore 637371

Comment by dmoskovich — October 13, 2013 @ 8:38 am

• Did you get my letter? Ken Perko

Comment by Ken Perko — November 5, 2013 @ 11:44 am

• Not yet, but sometimes international mail takes time. Thanks for sending it! I’ll let you know when it arrives.

Comment by dmoskovich — November 6, 2013 @ 6:46 pm

• Your letter has just arrived!
I’ve only just opened the envelope, but it has already really made my day- it’s vastly entertaining and tremendously interesting, and I look forward to reading through all the various parts of it closely!

Comment by dmoskovich — November 7, 2013 @ 2:02 am

2. Enjoy. I rather thought it fell within the range of your interests, and you could probably teach it at the junior high school level in Singapore. Also, don’t miss my recent comments on “The Weisstein Pair” on Elwes’ “Revenge” blog and on Mathoverflow, that fountain of misinformation. And who is the mysterious “Zarrax”?

Comment by Ken Perko — November 7, 2013 @ 2:57 pm

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