Posted by: Dave Richeson | May 4, 2009

The maypole braid group

174913998_9ccb5aa8d8_b Over the weekend I attended a May Day party thrown by one of my colleagues.

During the party they had a traditional maypole dance. An example of a maypole dance is shown at left. A maypole is a tall pole with colorful ribbons attached to the top that are fanned out in a cone shape. The dancers hold the ribbons and dance according to the rules called by the party’s host (like a square dance). Some walk clockwise and others walk counterclockwise. When one dancer encounters another she goes under the other’s ribbon or lets the other dancer go under hers (according to the rules of the dance). As this is happening the ribbons wind around the pole, making a decorative pattern, such as those shown below.

151673381_879f609868_o
330029108_35e63eef59_o

Having just taught the last day of my knot theory class the day before the party I couldn’t help but look at the maypole with a mathematical eye.

One of the topics in my knot theory course was the (Artin) braid group. A braid is a collection of n strings hanging from a rod, wound around each other, and attached to another rod below so that each string is always going downward (as in the picture below). Note that there can be no knotting in the strands because that would require the strings to move upward temporarily.

braid

This is a bona fide group. The group operation is concatenation—i.e, to mutiply two braids, attach the second braid to the bottom of the first. The identity element is the braid in which all string run straight downward without any twisting. It is not difficult to see that each braid has an inverse. Moreover, the braid group is generated by elementary braids of the form shown below (called \sigma_i if the ith strand crosses over the (i+1)st and \sigma_i^{-1} if the (i+1)st crosses over the ith.

generators

For example, the braid notation for the braid given at the top is \sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}\sigma_2^{-1}\sigma_1.

The braid group B_n is nonabelian (for n>2), but generators do commute if they are far enough apart (as illustrated below). That is, \sigma_i\sigma_j=\sigma_j\sigma_i if |i-j|\ge 2 (this would also hold if either of the generators was replace by its inverse). This is called far commutativity.

commute

There are three other relations among the generators; all three corresponds to the third Reidemeister move and are commonly called the braid relations. One of them is shown below; we can write it symbolically as \sigma_{i+1}\sigma_i\sigma_{i+1}=\sigma_i\sigma_{i+1}\sigma_i.

r3

It turns out that these are the only relations, and thus it is possible to study the braid group purely symbolically.

Artin’s Theorem. The braid group B_n is the group generated by \sigma_1,\ldots,\sigma_{n-1} subject to the following relations:

  1. Far commutativity: \sigma_i\sigma_j=\sigma_j\sigma_i for |i-j|\ge 2
  2. Braid relations
    • \sigma_{i+1}\sigma_i\sigma_{i+1}=\sigma_i\sigma_{i+1}\sigma_i
    • \sigma_{i+1}^{-1}\sigma_i\sigma_{i+1}=\sigma_i\sigma_{i+1}\sigma_i^{-1}
    • \sigma_{i+1}\sigma_i\sigma_{i+1}^{-1}=\sigma_i^{-1}\sigma_{i+1}\sigma_i

The braiding of the maypole ribbons is an interesting twist (har har) on this classic idea. Instead of the string being attached to two line segments, they are attached to two circles and we do not allow the strings to pass through the axis running through the centers of the two circles (the maypole itself!). In particular, we now draw the projection of the braid on the surface of a cylinder. Below is an example of such a projection.

cylbraid

The braiding pattern that the party host coordinated went as follows. There were an even number of ribbons, so he had the dancers count off by twos. The 1’s walked counterclockwise and the 2’s walked clockwise. The 1’s went under the 2’s ribbons initially, then they went over them the next time, and they continued alternating from then onward. The resulting pattern is shown below. I’ve drawn the pattern on an annulus, which is topologically the same as a cylinder. (We can also think of this as the view of the maypole dancers from above.)

circlebraid1

Clearly this maypole dancing generates a group, but it is not the usual braid group B_n. What is it?

Let’s call the maypole braid group C_n. Can we define this group using generators and relations? At first glance it looks as if we can simply generalize the ordinary braid group with the added modification that the nth string is adjacent to the first string. In particular, there is a new generator \sigma_n in which the nth string crosses over the first string. Then the same relations hold as in Artin’s theorem, but the arithmetic on the indices is performed modulo n.

Using this notation the group element representing the circular braid above is (\sigma_1^{-1}\sigma_3^{-1}\sigma_5^{-1}\sigma_7^{-1}\sigma_2\sigma_4\sigma_6\sigma_8)^2. (I’m taking the inside circle to be the top of the maypole and the eastern ribbon to be ribbon #1.)

However, this does not describe all possible maypole braids. In particular, the three braids below cannot be obtained as products of the \sigma_i^{\pm 1}. There’s a really easy way to prove this—can you see it? I’ll post the solution tomorrow. [Now posted.]

circlebraid2 circlebraid3circlebraid4

It seems to me that we need one more generator in order to obtain the group C_n. Let’s call the new generator \tau; \tau corresponds to the motion in which for all i, the ith strand moves to the (i+1)st place (mod n). In other words, all the dancers rotate one space counterclockwise. There is a corresponding clockwise movement \tau^{-1}. All three of the circular braids shown above can be generated by the \sigma_i and \tau. The first is \tau^2 and the second is \sigma_2^{-1}\sigma_4^{-1}\sigma_6^{-1}\sigma_8^{-1}\tau. I leave the third one as an exercise. I’ll post the solution tomorrow. [Now posted.]

This new generator does come with new relations. In particular \sigma_i\tau=\tau\sigma_{i+1} (where the indices are modulo n).

Thus it seems to me that the group C_n is generated by \sigma_1,\ldots,\sigma_n,\tau and the following relations. (Again, we assume that the arithmetic for the indices are all computed mod n.)

  1. Rotation relation: \sigma_i\tau=\tau\sigma_{i+1}.
  2. Far commutativity: \sigma_i\sigma_j=\sigma_j\sigma_i for |i-j|\ge 2
  3. Braid relations
    • \sigma_{i+1}\sigma_i\sigma_{i+1}=\sigma_i\sigma_{i+1}\sigma_i
    • \sigma_{i+1}^{-1}\sigma_i\sigma_{i+1}=\sigma_i\sigma_{i+1}\sigma_i^{-1}
    • \sigma_{i+1}\sigma_i\sigma_{i+1}^{-1}=\sigma_i^{-1}\sigma_{i+1}\sigma_i
Happy May Day!

[Photo credits: Pete Ashton, yksin, f_heaney]

About these ads

Responses

  1. […] braid group (solutions) Yesterday I wrote about the maypole braid group and left two questions for homework. Here are the […]

  2. […] Richeson celebrates May Day by devising and investigating the maypole braid group. (Nice pictures on this one […]

  3. As a high school math teacher AND a homeschool parent, I am constantly reminding children that “Math is everywhere, man.” (ht: Johnny Cash). I chuckled when I read “I couldn’t help but look at the maypole with a mathematical eye.”

  4. hello! im a musician and i was very intrigued and inspired to find your explanation of maypole using mathematics. eversince ive been introduced to hopfstadter, its been a real pleasure to delve into the life phenomenon (as a simple musician) and though im kinda rubbish in math, i thought you may enjoy a very small thoughts of mine that sprang from your post. best, c


Categories

Follow

Get every new post delivered to your Inbox.

Join 222 other followers

%d bloggers like this: