The Unapologetic Mathematician

Mathematics for the interested outsider

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.
Braid Crossing

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_1,...,s_{n-1}|s_i^2 (1\leq i\leq n-1),s_is_js_i^{-1}s_j^{-1} (|i-j|\geq 2),
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:

<s_1,...,s_{n-1}|s_is_js_i^{-1}s_j^{-1} (|i-j|\geq 2),s_is_{i+1}s_is_{i+1}s_is_{i+1} (1\leq i\leq n-2)>

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 - Posted by John Armstrong | Algebra, Group Examples, Group theory, Knot theory | | 2 Comments

2 Comments »

  1. One question occurs to me: when can we find a neat free resolution of the braid group? Do we know which braid groups have finite resolutions? Do we know the complexities of braid group resolutions?

    Comment by Mikael Johansson | March 15, 2007

  2. [...] a “braided monoidal category”, for a very good reason I’ll talk about tomorrow (hint). Now if by chance the braiding is its own inverse, we call it a “symmetry”, and call [...]

    Pingback by Braidings and Symmetries « The Unapologetic Mathematician | July 2, 2007

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