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.

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March 14, 2007 - Posted by | Algebra, Group Examples, Group theory, Knot theory

8 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 | Reply

  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 | Reply

  3. I have been looking through some knot theory books, and I have started leafing through some abstract algebra books. I am curious about chirality tests for knots is 3 space, and, in particular, for what integer values n will there be achiral knots. I found out there is a proof that there are some whenever n is even, and none when n is prime. For composite odd numbers n, there are some that have achiral knots. In the tables I have found to date, there are no 9 crossing knots listed as being achiral. One achiral knot with 15 crossings was found in 1998 by M. Thistlewaite and friends. Has one been found for 21,33,35, or 39 yet? I suspect there will be at least one or more for each of those cases. However, I do not think there will be any for 25, 27, 49, 81, 121, 125, …. Has any progress been made on this front? I’m rather curious. I have an idea I am investigating, but I am just starting and working my way into group theory.

    Jim aka Maddy

    Comment by Jim Balliette | February 13, 2009 | Reply

  4. I have to say, Jim, I really don’t know much about chirality results like that, but I can ask around.

    Comment by John Armstrong | February 13, 2009 | Reply

  5. Thanks John,

    I think it is a rather tough question. It has been long since I looked at group theory, so I am having to start from scratch. I have some background, but it is mostly dormant. I did some analysis on Hoste’s 15 crossing knot, but I am not sure I found anything that will help me. Perhaps I should look up Professor Thistlewaite and see if he has anything. I thought I would run a test on my computer, but the size of the sets I need to manipulate are too big. The symmetric groups get rather large rather quickly! I am hitting a website here and one there, but I am not finding what I am looking for as of yet.

    Thanks for maintaining this site!

    Jim

    Comment by Jim Balliette | February 14, 2009 | Reply

  6. Ok John,

    there is a 9 crossing achiral knot. I was bothered by the fact I didn’t find one in the tables I was using. I’m going to compare the number of chiral knots with n crossings to the number of Sylow Subgroups of a cyclic group of order n. I believe there is relationship, a strong one, but I could be way off. If I’m right, then there WILL be chiral knots for prime powers, though I’m not sure about nonalternating chiral knots when an odd prime power is involved. I am going to try to draw a 21 crossing non-alternating chiral knot this weekend. This should be a challenge!

    Hope you are having a great time this weekend!

    Jim

    Comment by Jim Balliette | February 14, 2009 | Reply

  7. I seem to be using the word chiral when I mean ACHIRAL or AMPHICHIRAL. Dang. Those other ones are rather easy to draw.

    Jim

    Comment by Jim Balliette | February 14, 2009 | Reply

  8. Ooops…I misread the book. There is NOT a 9 crossing achiral knot. Perhaps powers of odd primes will not have a non-alternating achiral knot, or maybe no achiral knots either. If so, what is the connection between this, if any, and cyclic groups of prime power order having unbranched normal decomposition series? ( I hope I am using the right terminology!! )

    My attempt to draw the knot was a failure, which I expected. I ended up guessing quite a bit and created a 3 link consisting of a free link and an alternating link of two components. I don’t have enough data to see what is happening yet. Still, trying to find a pattern is challenging and a wee bit of fun.

    So far I have been able to distinguish all of the knots I have tried from the unknot using a very simple test. Every attempt to make a nasty unknot that I cannot tell is the unknot has failed, but I haven’t made unknots with too many crossings yet. I am convinced the method is invariant for Reidemeister moves Type I and Type II, but type III is a bit harder to examine. This is interesting, but it doesn’t mean the method will always work. However, if it is a test that will work, all of the computer code needed to do it is already written. I do my tests in GAP, but I suppose the test could be done using any computer algebra system. A very competent typist can type in the data using a single line. Unfortunately, I am NOT a good typist, so typing in the data in my case takes anywhere from 5 to 50 times!

    Ok, off to start over with more care to be VERY systematic. I sure hope I don’t find out I made a typing error that destroys all of the fun stuff I think I am seeing.

    Jim

    Comment by Jim Balliette | February 15, 2009 | Reply


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