Monday, July 06, 2015
The Fascination of Braids
by Carl Pierer
Braids are fairly simple to picture. A few interleaved strands of string, say, gives a complex and mesmerising object. They are aesthetically appealing, as their widespread use as ornament testifies. While most will be familiar with the standard braid used for braiding hair, there is basically no limit to complexity and beauty. Yet, braids are more than merely nice, artistic adornments for clothes and jewellery. The more and deeper you delve into braids and their complex interconnections, the more fascinating they become. Trying to look at them with a mathematical eye opens up pathways and connections to many deep and beautiful fields of pure mathematics.
Studied as mathematical objects, braids need to be defined rigorously. However, for present purposes it is enough to specify what we intuitively have in mind when thinking about braids. (Fig. 1) may well serve as an example. A braid consists of a certain number of strands n, say, together with a specification of how and where these strands cross each other. Furthermore, these strands (if they are not crossing) run parallel and we may adopt the convention that they are running from top to bottom. To avoid ambiguity, we require further that there are no two crossings at the same horizontal level. It is clear that for the braid to have any crossings at all, it must have at least two strands. If a braid does not have any crossings, it is called the trivial braid.
Braids also have a close connection to (mathematical) knots. Mathematical knots are simply everyday knots where the loose ends have been joined together. If you take an overhand (or trefoil) knot (Fig. 2) and join the loose ends, you have a mathematical knot. Imagine an extension cords where one end has been plugged into the other. One important feature is that there is no way of undoing this mathematical knot by pulling or stretching or any other deformation that does not break the connection. If you reconsider our
initial braid (Fig. 1), imagine that the ends have been joined up as in (Fig. 3). As all loose ends are now joined up, we can consider (Fig. 3) as a mathematical knot. Indeed, every mathematical knot can be expressed as a braid. Unfortunately, this is not a one to one correspondence. So, two different braids may end up as the same
knot. Or one and the same braid may give you two different knots. One problem is that there is no particular reason to join up ends of the braid as we did in (Fig. 3). If we have an even number of strands, we can equally well join up the endings that are next to each other (as in Fig. 4), which would give a very different beast altogether. Indeed, this is not a knot anymore but a so-called link, because it consists of 4 different components. A knot is a one component link. However, this connection between braids and knots is not the main concern here.
It may seem a bit strange that these entities should be studied in mathematics. After all, this does not have an awful lot to do with numbers. However, a question that arises naturally, perhaps more so in connection with knots than with braids, is whether a given braid (or knot) is trivial. That is, can we deform this braid into the trivial braid without "breaking" it? Again, the permissible moves are pulling and stretching of the strands but no cutting or twisting. It's easiest if you imagine that the ends of the strands have been fixed on a rigid frame. The justification for these rules are to be found in the more complicated mathematics behind it. Yet, intuitively, whether a braid is pulled very tight or stretched loose should not matter – it is still the "same" braid.
This is exactly where mathematics ties in. Often mathematics is concerned with establishing that two things are the "same" (given a certain understanding of what it means to be the "same") and it provides very powerful tools for doing so. For the even trickier question of how to establish that two things are different, mathematics alone probably possesses the methods to do so. If you want to show that two things are the "same", then you need to demonstrate how to get from the first to the second. If you want to say that the two things are different, it is not enough to say that you couldn't find a way to get from the first to the second. You need to prove that there is no such way. This, essentially, is mathematical territory.
In order to study the mathematical properties of braids, it is useful to encode braids using symbols.
Recall the few requirements imposed on braids: a certain number of strands plus a prescription of how and where they cross. The symbol of choice then is the Greek letter sigma: σ. A subscript identifies which strand this sigma is referring to, so in general, σi means the ith strand is crossing over the i+1th strand. A negative power, σi-1 (read: sigma i inverse), means the ith strand crosses under the i+1th strand. With this notation, we can associate every braid with a so called braid word, a sequence of σI's. The braid in (Fig. 5) for example, can be written as σ1 σ2 σ1.The problem of deciding whether two braids are the same then comes to deciding whether one braid word can be change into the other, using only permitted operations. Therefore, this problem is also known as the word problem.
Looking at (Fig. 5), we can see that it is just the same as the braid in (Fig. 6), the
middle strand has simply been pulled from left to right. So, in our braid word we have σ1 σ2 σ1= σ2 σ1 σ2. Also, if we encounter anything like (Fig. 7), it is obvious that we can just undo the two crossings. So if σi and σi-1 happen to be next to each other in a braid word, they just cancel. These are the basic two moves needed – any other permitted move will be a consequence of these two. So, if we want to establish that two braids are the same, we simply how – employing a sequence of these changes – the first word can be changed into the second. In particular, we show that a braid is trivial by showing that all the σis cancel.
But how do we show that two braids are different? How do we know that there is simply no way of transforming one braid into the other, no matter how hard we try?
First, it is important to note that all braids have inverses[i]. That is, we can undo any braid by adding its mirror image to it. Pulling and stretching will then reveal that the new braid is trivial. Adding the mirror image amounts to rewriting the braid word from right to left changing the
powers from -1 to +1 and vice-versa as we go along. Now, two braids are the same if and only if their inverses are the same. So, if we are comparing two braids and the inverse of the second undoes the first, then the two are the same. This is why in comparing two braids we usually check whether the first combined with the inverse of the second can be reduced to the trivial braid.
Second, two braids cannot be the same if their strands end up in different places. This is simply because no permissible move allows us to change where the strands end up (think about the rigid frame again). A braid, where the strands end up where they started – so strand 1 ends in position 1, strand 2 in position 2, and so on – is called a pure braid. Hence the only potential candidates for being trivial are pure braids.
The first step in checking whether two braids are the same is thus to check whether their strands end up in the same places. If they don't, they are different. If they do, we need to proceed.
Let us distinguish four kinds of braid words. They are either: empty (then they are trivial), sigma-positive (the sigma with the lowest index appears only to a positive power), sigma-negative (the sigma with the lowest index appears only to a negative power) or neither. In the last case, the braid word contains both a σi and a σi-1, where i is the lowest index. So, because i is the lowest index, all other crossings occur to the right of this crossing. This is called a handle, for pretty graphical reasons it seems.
Now for the nifty thing: the handles in a word can be reduced and doing so will return an equivalent braid! If a word contains any handles, we can get rid of them and reduce the word to one that is either sigma-positive, sigma-negative or empty. All that is needed is a prescription (or algorithm) how to reduce the word. Fortunately, such an algorithm does exist and it is fairly simple. Furthermore, its run-time grows "only" exponentially, making it the best available algorithm to date[ii]. The algorithm is as follows:
1. If the word is empty, sigma-positive or sigma-negative, finish. If the word is empty, the braid is trivial. If the word is not empty, the braid is not trivial.
2. If the word is neither, find the handle that ends first and reduce it. Go back to step 1.[iii]
Now, suppose you have a braid word that is neither empty, nor sigma-positive nor sigma-negative. This means that if i is the lowest appearing index in the word, there exist both σi and σi-1 in the word. We distinguish between a positive (of the form σi …σi-1) and a negative (of the form σi-1…σi) handle. The reduction works thusly: first, delete the σi and σi-1. If the handle was positive, replace any σi+1 appearing in the word with σi+1-1σi σi+1 and any σi+1-1 with σi+1-1σi-1σi+1. If the handle was negative, replace any σi+1 with σi+1σi σi+1-1 and any σi+1-1 with σi+1-1 σi-1 σi+1. Leave all other crossings as they are (See Fig. 8 and Fig. 9). This may result in new handles being introduced into the braid, but the algorithm always terminates (provided we always reduce the handle that ends first first).
We thus have a relatively simple and fast method for deciding whether two braids are the same or not and whether any given braid is trivial. Furthermore, the algorithm is such that it can be run by a computer, saving us a lot of tedious work. There is a very nice applet to be found here, which runs the algorithm and graphically illustrates the reduction as the program runs.
Braids can also be used to codify information. Most notably, if you imagine that a strands unfolds over time, it could be seen as the trajectory of a particle. A n-stranded braid would then give the relative trajectories of n particles. Similarly for dancers. In many parts of the Alps there is a tradition to dance around a big tree erected on a central place during May, the May Pole. Dancers would hold ribbons that are attached to the tree and braid them in a certain pattern. The dancers' relative movements are encoded in the final braid. It is thus possible to reconstruct the dance from the braid.
Recall again (Fig. 1). This is the braid you get after one repeat of the Ceilidh "Gypsy Shawl"[iv]. For a reference, you might find this video of the dance relevant. Braids allow for a fascinating exploration of dances through mathematics.
It is quite intriguing that such a rather simple object like braids should lead to challenging and deep mathematics. In addition to their aesthetic complexity, it is because of their mathematical beauty that braids are of such great interest. Braids allow for a connection with many a mathematical field, be it topology, group theory or geometry.
They are very curious objects, indeed.
[i] By the way, for all natural n, braids of n strands form a group! Unfortunately, the present essay does not allow to explain this beautiful concept, but the facts that braids form groups is just too curious to pass without mention.
[ii] Less fortunately, the proof for why it always converges (so why it always manages to present a reduced word and does not simply continue to reduce and introduce handles) is quite complex and beyond the scope of this essay. It relies on an ordering of braid groups and can be found in P. Dehornoy's: "Ordering Braids".
[iii] Cf. E. Dalvit: "New Proposals for the Popularisation of Braid Theory" – all the beautiful colourful drawings have been taken from this thesis.
[iv] Ceilidhs are traditional Scottish Folk Dances, which are usual held at gatherings. The intention is that no prior knowledge should be needed and every dance at a Ceilidh will be explained by a caller at the beginning. Even if you are a very dispassionate dancer, do not miss out on the chance to attend a Ceilidh!
Posted by Carl Pierer at 12:40 AM | Permalink