# The Unapologetic Mathematician

## The Category of Braids

Before I jet off for a week of knot homology — where I’m looking forward to seeing Scott Morrison, Jacob Rasmussen, and Ben Webster of the Secret Blogging Seminar — I thought I’d finally start tying in (sorry) this category theory stuff with knot theory.

As I said yesterday, the analogue of commutativity for a monoidal category is called a braiding. The name comes from a deep connection between braided monoidal categories and braid groups. Specifically, there is a (strict) braided monoidal category of braids: $\mathcal{B}raid$. The objects of this category are the natural numbers, and $\hom_{\mathcal{B}raid}(m,n)$ is the group $B_n$ if $m=n$ and empty otherwise. We think of a braid with $n$ strands as taking a collection of $n$ points and shuffling them around, and we draw a diagram like this:

As a side note, I always intend diagrams like these to be read from bottom to top, though other authors go the other way.

First, the monoidal structure. We don’t need any of the structural isomorphisms because we’re making a strict monoidal structure, but we do need a bifunctor $\underline{\hphantom{X}}\otimes\underline{\hphantom{X}}$. On the objects it’s just addition. For morphisms we need a way of taking a braid with $m$ strands and one with $n$ strands and making a braid with $m+n$ strands. We’ll just stand both of our braids next to each other like this:

Now we need a braiding. That is, for each pair of objects $(m,n)$ we need a morphism (a braid) from $m+n$ to $n+m$ that “commutes” with all the other morphisms. We’ll just pass the $m$ strands from the left over the $n$ strands from the right. Here’s what I mean for $m=3$ and $n=2$:

This is natural because if we stick on a braid below the leftmost three strands on the bottom, the Reidemeister moves will let us pull it up and over the other two strands until it’s sitting on top of the three strands at the right on the top. Similarly, a braid can be pulled along the undercrossing strands. Since our monoidal structure is strict, the hexagon identities degenerate into triangles, and the proof is just the following diagram:

So we indeed have a strict braided monoidal category. It turns out to be extremely important because of the following theorem of Joyal and Street:

If $\mathcal{M}$ is a braided monoidal category with underlying category $\mathcal{M}_0$ (forget the monoidal structure and the braiding to just have a regular old category), then the category of braided monoidal functors from $\mathcal{B}raid$ to $\mathcal{M}$ is equivalent to $\mathcal{M}_0$.

This is sort of like a coherence theorem for braided monoidal categories: for each natural isomorphism built from $\alpha$, $\lambda$, $\rho$, and $\beta$ there is an “underlying braid”, and two such isomorphisms are equal if and only if they have the same underlying braid. I’ll defer the proof of this for now, but you should think about it a bit. The details aren’t that bad, but the basic idea just leaps out at you after a bit.

July 3, 2007

## The Fundamental Involutory Quandle

As I discussed last time, coloring a knot with any abelian group is secretly using the dihedral quandle associated to that group. This is an involutory quandle with action $a\triangleright b=2a-b$. The reason knot coloring works out so nicely is that the axioms of (involutory) quandles line up with the Reidemeister moves.

But for the moment we’re stuck with picking this or that involutory quandle and counting how many colorings it gives for a given knot. Different quandles give different coloring numbers, and we’d like to find a better way of thinking of them all at once. We’re going to construct a new involutory quandle from a knot that captures all of them.

Take any diagram of the knot we’re interested in. Remember the knot table if you want to pick one out. Now each arc in the diagram has to get some color, no matter what quandle we’re using to color it. Instead of picking a color from a specific quandle, let’s just slap a label like $x$, $y$, or $z$ on each arc. Be sure to use a different label for each different arc.

Now those labels will generate an involutory quandle. We can throw them together with the two quandle compositions to get “words” like $x\triangleright((y\triangleright z)\triangleright z)$. These words, of course, are subject to the normal quandle equivalences, but we need more relations for our purposes. At each crossing the values in a coloring have to satisfy a certain relation, so we’re going to build that right into our quandle. If the arcs labeled $x$ and $z$ meet under the crossing arc labeled $y$, then we must have $z=y\triangleright x$.

This seems to depend on the choice of a diagram, though. Well, it sort of does, but any Reidemeister move gives an isomorphism of quandles relating the two sides. For example, performing the first one splits an arc into two pieces. Say label $x$ becomes $x_1$ and $x_2$. Then the relations we introduce say that $x_1\triangleright x_1=x_2$. But the axioms of quandles say that $x_1\triangleright x_1=x_1$, so $x_1=x_2$ and we can just drop one of these generators and the relation we’ve now “used up”. Try to find the isomorphisms for the other two moves. This justifies calling the quandle we’ve constructed (up to isomorphism) “the” fundamental involutory quandle $Q(K)$ of the knot $K$.

So what’s a coloring? A coloring assigns an element of some quandle to each arc of the knot diagram. But arcs in the diagram are just generators of the fundamental quandle. That is, a coloring is a function that takes generators of the fundamental quandle to a selected target quandle. If it plays nicely with the relations between the generators, it will be a quandle homomorphism. In fact it does, precisely because we picked the relations between the generators to be exactly those required by colorings. A given relation comes from a crossing, and every coloring of a knot obeys the same restrictions at crosings.

In the end we’ve found that the set of all colorings of $K$ by an involutory quandle $Q$ is the set of quandle homomorphisms $\hom_{\mathbf{Quan}}(Q(K),Q)$, so the number of $Q$-colorings is the cardinality of this set. If we have a good understanding of quandles and their homomorphisms, we can read off coloring numbers by involutory quandles from the fundamental involutory quandle.

May 16, 2007 Posted by | Knot theory, Quandles | 1 Comment

## More knot coloring

Two weeks ago I went over how to color knots with three colors. Of course, being mathematicians, we want to generalize the hell out of this.

Really, as is often the case with this sort of thing, “color” is just a metaphor. We labeled each arc of a diagram with one element of a set with three elements and put a condition on the labels attached to the arcs at each crossing. So, what sets do we know of with three elements. I’ll give you some time to think.

Back? Did you come up with “the abelian group $\mathbb{Z}_3$“? If so, great. If not, this is one of the first examples of groups I mentioned waaaaaay back in January. Basically, it’s made of the numbers in the set $\{0,1,2\}$. We add and subtract as usual, but we loop around from $2$ to ${}0$. For example, $2+2=1$ and $0-1=2$.

So let’s imagine we’re “coloring” a knot with $\mathbb{Z}_3$. What’s the condition at a crossing? Imagine we’ve got an overcrossing arc colored $a$, and we approach it along an undercrossing arc colored $b$. If $a=b$ then we have two arcs of the same color, and we have to leave along an arc with the same color. On the other hand, if $a\neq b$ we have to use the third color. So how do we recognize the third color? It turns out that there’s an easy way to do this: the third color is $2a-b$. In fact, this also works if $a=b$. Try writing down all nine combinations of $a$ and $b$ and seeing that $2a-b$ is always a valid coloring for the third arc.

Now there’s nothing in our notion of coloring that relies on the number $3$. All we’re really using in this new condition is the fact that we have an abelian group. So take your favorite abelian group $G$ and try coloring knots with it, requiring that if an arc colored $b$ undercrosses an arc colored $a$, it comes out the other side colored $2a-b$.

Does this new form of $G$-coloring depend on the knot diagram we use, or only on the knot. Well, it turns out just to depend on the knot. To show this, we again use Reidemeister moves.

In each of these, I’ve colored some of the end and worked across the diagram seeing what I’m required to color the other edges. In the first move, for example, if I color the top of the left $a$, the bottom will also get $a$. If I color the top of the right $a$, the arc crosses under itself so the bottom must be colored $2a-a$. But we see that $2a-a=a$, so this is the same end coloring. The same sort of argument works for the second and third moves.

So for every abelian group $G$, we have an invariant: the number of ways of $G$-coloring the arcs of any diagram of the knot. The colorings we did the last time were $\mathbb{Z}_3$-colorings.

But why does this rule work so well? It turns out that we’re not interested in $G$ as an abelian group. We take the underlying set of $G$ and equip it with the operation $a\triangleright b=2a-b$. This makes $G$ into an involutory quandle, called the “dihedral quandle” of $G$. The name comes from the fact that $b\mapsto2a-b$ is like “reflection through $a$“, and such reflections generate all symmetries of regular polygons in the plane. A polygon in the plane, of course, is a sort of degenerate polyhedron in space with only two sides: “dihedron”.

Anyhow, recall the axioms for an involutory quandle:

• $a\triangleright a=a$
• $a\triangleright(a\triangleright b)=b$
• $a\triangleright(b\triangleright c)=(a\triangleright b)\triangleright(a\triangleright c)$

First check that these axioms really do hold for the operation we defined on $G$. Then imagine using any other involutory quandle $Q$ to color knots. Go back to the Reidemeister diagrams and do just what we did for $G$-coloring, but use the quandle operation instead: if an arc colored $b$ undercrosses one colored $a$, it leaves colored $a\triangleright b$. Show that the number of $Q$-colorings is an invariant of the knot, not just of its diagrams.

May 2, 2007 Posted by | Knot theory | 10 Comments

## Some of my own stuff

I’m talking tomorrow in the Geometry, Symmetry, and Physics seminar here at Yale about the work that spun out of my realization of March 16. This ties into knot colorings, but goes far beyond that starting point. I won’t be able to say this with just the tools I’ve developed in the main line of my writings, so like the Atlas stuff it may not be comprehensible (yet!) to anyone beyond professionals.

So here’s the most general statement. Let $\mathcal{C}$ be an algebraic category and $X$ a co-$\mathcal{C}$ object in the category of pointed topological pairs up to homotopy. Write $P_n$ for the plane with $n$ marked points, and $C$ for the cube. Then every tangle $T:m\rightarrow n$ gives rise to a cospan in $\mathcal{C}$:

$\hom(X,P_m)\rightarrow\hom(X,(C,T))\leftarrow\hom(X,P_n)$

where the $\hom$-objects are taken in the category of pointed pairs up to homotopy. This then gives rise to an anafunctor from the comma category $(\hom(X,P_m),\mathcal{C})$ to the comma category $(\hom(X,P_n),\mathcal{C})$. This assignment is a monoidal functor from the category of tangles to the category of (categories, anafunctors). When $\mathcal{C}$ is the category of quandles, this functor categorifies the extension to tangles of the coloring number invariant of knots and links.

## Coloring knots (again)

A few weeks ago I mentioned the knot coloring problem, and left you to play with it. Now I’m going to say what’s going on.

First let’s remember what it means to color a knot. We take a knot and draw a knot diagram to represent it. Then we color each arc of the diagram — from the undercrossing at one end to the undercrossing at the other — either red, green, or blue. At every crossing three arcs come together, and we require that either all three get the same color or all three get different colors. We can always just give every arc the same color, but we’re interested in when we can use all three colors.

Of course you’re now screaming (or you should be) that we had to choose a diagram for the knot, so how do we know that the answer doesn’t depend on which choice we made? Luckily we have a way to tell if two knot diagrams represent the same knot: Reidemeister moves! So how do colorings behave when we do a Reidemeister move?

Let’s go through them one at a time. The first move looks like this:

Of course, it doesn’t have to be red. There’s a similar diagram for green and blue. So, if we have a strand colored red we can twist it, coloring both arcs red. On the other hand, if we have a twisted strand, both arcs have to be the same color. Otherwise the crossing wouldn’t be colored right. We can then untwist the strand.

Here’s the second move:

On the left we have two strands of different colors. After performing the move we can give the new arc the third color. If the strands were both the same color we could give the new arc that same color again. On the other side, any coloring of the right side of the move will give the same color to the top and bottom ends of each strand. We can then undo the move and still have a valid coloring.

Finally the third move:

Any coloring of the ends of the three strands that can be extended to a valid coloring of the middle on the left side can be extended to a valid coloring on the right side, and vice versa. For example, both sides require that the strand running through the middle get the same color at both ends.

Now the first and third moves don’t change the number of colors that appear. The second one seems like it might, though. If we have a coloring on the right using all three colors, maybe the left only has two? If we’re dealing with a knot (rather than a link with more than one loop) then eventually the red and green strands will have to meet up. When they do, it’s at a crossing, and then we’ll need to use blue. So if any diagram of a knot has a coloring with all three colors then all of them do. “Three-colorability” is a property of the knot itself, not just of a diagram.

Actually, we can do even better. Pick a diagram on one side of a Reidemeister move and color the ends of each strand. Either this coloring can be extended to a valid coloring of the interior of the diagram or it can’t, and if it can there’s only one way to do it. The extendible colorings of the ends of one side of a move are exactly the same as the extendible colorings of the other side.

The upshot is that we can ask about how many colorings a knot has, or even a multi-loop link. Some of them may not use all three colors, but every diagram of the same link has the same number of valid colorings. Every knot has at least three (monochromatic) colorings, so a knot is three-colorable (using all three colors) if and only if the number of valid colorings of any diagram is bigger than 3.

April 18, 2007 Posted by | Knot theory | 1 Comment

## Reidemeister moves

I’ve finally beaten my copy of Maple into submission and sketched a few pictures, so finally I can press ahead with knot theory.

The early days of knot theory were heavily topological, and there’s still a large part that works primarily with the tools of algebraic topology. However there’s a lot we can do combinatorially at a very low level. Mostly this works because of the Reidemeister theorem relating the study of knots to the study of knot diagrams.

Remember that a knot is actually a closed loop floating around in three-dimensional space, and a link is just the same but with more than one loop. If we imagine the loops are made of string and the space is above a table we can imagine dropping a knot down so it lies mostly flat on the table, bumping up in order for one string to cross over another. If we draw this curve in the plane, noting which strand crosses over another at a crossing, we get a knot diagram.

Now we’re considering two knots to be “the same” if we can move the one curve in space to the other without cutting the string. We need a similar notion of equivalence for knot diagrams so that two diagrams represent the same knot if and only if they are equivalent. It turns out that we can extract the right notion of equivalence from the method we use to produce diagrams!

What we’re going to do is actually somewhat different from the original method of Reidemeister, but I think it makes a lot more intuitive sense. I’m also going to play a little fast and loose with the fine analytic details, but if you’re expert enough to see what might go wrong you should also expert enough to tidy up the arguments.

We want to make our diagrams as simple as possible, in a sense that should become clear. We can’t eliminate all “double points” where two strands have to cross over each other in the plane. If we could we wouldn’t really have knots. However we can ask that there be no “triple points” where three strands cross each other. If we have a triple point we can push the top strand off the crossing of the other two strands to get three double points. But we have a choice of how to do this, and both ways are equally valid simplifications of the diagram. To handle this, we introduce the “third Reidemeister move”:

What does this diagram mean? If we have a knot diagram with three strands forming a triangle like the one on the left — and no other strands in the area, so the diagram really does look like the left-hand side — we can replace the “local” diagram on the left with the one on the right. We can also go the other way. If we imagine this as pulling the top strand back and forth over the crossing, we see that these are the two different ways of simplifying a triple point.

Okay, now we want to ask that at any double point the strands actually cross each other and don’t just brush against each other. If we have such a bad point we can tweak it a bit, and again there are two choices. One way pulls the strands off of each other entirely, and the other creates two honest crossings. Corresponding to this choice we have the “second Reidemeister move”:

Again, we interpret the diagram as we did above.

Finally, we want to avoid “cusps” where the string stops, moves down a bit, then doubles back on itself. This would create annoying corners in our diagrams. We handle it by twisting the string a bit, either smoothing out the cusp or turning it into a loop with an actual crossing. Corresponding to this choice we have the “first Reidemeister move”:

The Reidemeister theorem says that this is all we need to do. Any two diagrams represent “the same” knot if and only if they are related by a finite sequence of these three Reidemeister moves, along with pushing around diagrams in the plane. This makes it easy (well.. easier) to create knot invariants: define a function on knot diagrams so that we get the same answer before and after applying any Reidemeister move. Then if two diagrams represent the same knot they’ll get the same value of the function.

I want to step a bit aside from the main stream at this point to push some terminology that I like. If we define some sort of function of knot diagrams so that applying a Reidemeister move just gives you an equivalent value of the function rather than the same value, I like to call that a “knot covariant” (as opposed to “invariant”). Those of you who also read The n-Category Café will surely see the motivation.

April 3, 2007 Posted by | Knot theory | 7 Comments

## Coloring knots

Today I’m going to be talking to the graduate students about various topics relating to coloring knots. I think I’ll leave you with a little project to play with.

First, go to Bar-Natan’s table of knots. Notice how all the diagrams seem to be made up of arcs meeting up where one strand of the knot crosses under another. Pick a knot diagram and try to color each arc either red, green, or blue, subject to the following rule: at any crossing, the three arcs that meet (two for the undercrossing strand and one for the overcrossing) must either be all the same color or all different colors.

Which knots can you color using all three colors at least once? If that’s too easy for you, how many ways can you color a given knot? If that’s too easy for you, you’ve almost surely seen this before.

To get you started, I’ve tricolored the trefoil knot using all three colors.

March 30, 2007 Posted by | Knot theory | 3 Comments

## Slides for Bracket Extension talk

I’ve posted the slides for my talk. There are a few typos that I noticed as I was speaking, but nothing that makes it incomprehensible. I actually started the lecture on page 10 to save time, and because the run-up is pretty standard material.

March 17, 2007 Posted by | Knot theory | 1 Comment

## New result

Never believe anyone who says that drinking never helps anything, for tonight over many a pint o’ Guinness I hit upon the solution to a problem that’s been nagging at me for some time. The answer is so incredibly simple that I’m feeling stupid for not thinking of it before. So here it is:

Cospans in the comma category of quandles over a given quandle Q from the free quandle on $m$ letters to the free quandle on $n$ letters categorify the extension of link colorings by Q to tangles.

This gives me a new marker to aim at. I can explain this, but it will require more preliminaries before it’s really accessible to the GILA (Generally Interested Lay Audience) as yet. Those who know quandles — a topic I’ll be covering early next week — and category theory and some knot theory should be able to piece together the meaning now. For the rest of you.. stay tuned.

March 17, 2007 Posted by | Knot theory | 1 Comment

## Braid groups

Okay, time for a group I really like.

Imagine you’re playing the shell game. You’re mixing up some shells on the surface of a table, and you can’t lift them up. How can you rearrange them? At first, you might think this is just a permutation group all over again, but not quite. Let’s move two shells around each other, taking a picture of the table of the table each moment, and then stack those pictures like a flip-book movie. I’ve drawn just such a picture.

We read this movie from the bottom to the top. The shell on the right moves behind the shell on the left as they switch places. It could also have moved in front of the left shell, though, and that picture would show the paths crossing the other way. We don’t want to consider those two pictures the same.

So why don’t we want to? Because the paths those shells trace out look a lot like the strands of knots! The two dimensions of the table and one of time make a three-dimensional space we can use to embed knots. Since these pictures just have a bunch of strands running up and down, crossing over and under each other as they go we call them “braids”. In fact, these movies form a group. We compose movies by running them one after another. We always bring the $n$ shells back where they started so we can always start the next movie with no jumps. We get the identity by just leaving the shells alone. Finally, we can run a movie backwards to invert it.

There’s one such braid group $B_n$ for each number $n$ of shells. The first one, $B_1$ is trivial since there’s nothing to do — there’s no “braiding” going on with one strand. The second one, $B_2$ is just a copy of the integers again, counting how many twists like the one pictured above we’ve done. Count the opposite twist as $-1$. Notice that this is already different from the symmetric groups, where $S_2$ just has the two moves, “swap the letters” or “leave them alone”.

Beyond here the groups $B_n$ and $S_n$ get more and more different, but they’re also pretty tightly related. If we perform a braiding and then forget which direction we made each crossing we’re just left with a permutation. Clearly every permutation can arise from some braiding, so we have an epimorphism from $B_n$ onto $S_n$. In fact, this shows up when we try to give a presentation of the braid group.

Recall that the symmetric group has presentation:

$
$s_is_{i+1}s_is_{i+1}s_is_{i+1} (1\leq i\leq n-2)>$

The generator $s_i$ swaps the contents of places $i$ and $i+1$. The relations mean that swapping twice undoes a swap, widely spaced swaps can be done in either order, and another seemingly more confusing relation that’s at least easily verified. The braid group looks just like this, except now a twist is not its own inverse. So get rid of that first relation:

$$

The fact that we get from the braid group to the symmetric group by adding relations reflects the fact that $S_n$ is a quotient of $B_n$. It’s interesting to play with this projection and compute its kernel.

March 14, 2007