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Issue 1, March 2001

Physical Sciences & Mathematics
The Circular Braid Group and Its Relationship to the Standard Braid Group

John Singler
University of North Carolina at Asheville
Advisor: David Peifer, Ph.D.
Professor of Mathematics, University of North Carolina at Asheville

Abstract

In this paper, we will define the circular braid group on n strings and examine its relationship to the standard braid group on n strings. A braid on n strings has n strings attached to two bars, where one end of each string is attached to each bar. The strings cross as they please. A circular braid on n strings has n strings attached to two circular bars and the strings may also cross around as they please. As all of the braids on n strings form a group, Bn, it will also be shown that the circular braids on n strings form a group, which we will denote n. We will then continue on to define a function from n to Bn+1 and examine it. It will be shown that the function is not an isomorphism between the two groups, but that it is likely that it is an isomorphism between n and a subgroup of Bn+1. In conclusion, further opportunities for research concerning the above function, the presentation of n and suggestions for finding an isomorphism between n and Bn+1 will be made.

Introduction

Braid theory is a tool used to study mathematical knots. One might think that braids and knots are quite different, but in fact they have a very close connection - every knot can be made into a braid. Click here for an intuitive proof and some links concerning the subject. Knot theory has direct applications to theoretical physics and DNA research. Click here for more discussion on the applications of knot theory and other links. Also, see the Geometry Center for knot pictures and other information.

Background

Braid theory originated in 1925 when Artin examined the relation between the braid group on n strings and the algebraic braid group. Braids are geometric in nature and can be represented both geometrically and algebraically. Geometrically, an element of a braid group on n strings is represented by n strings attached to two bars, one end of each string is attached to each bar. The strings twist and tangle around each other, however they may not turn back upward, cross over or go around the bars. A standard practice of representing the crossings is to never place two crossings exactly horizontal of each other. An example of a braid on four strings is given in Figure 1.


Figure 1


Algebraically, the braid groups themselves are represented by group generators. In a group, relators are sequences of generators in a group for which the product is equal to the identity. An obvious relator in any presentation is any element times its inverse. Since this is obvious in any presentation of a group, inverse pair relators are omitted from the presentation. Relators are often rewritten as relations in which a word will equal another word.

Two braids can be combined by concatenation, i.e. placing the bottom bar of one braid on top of the other. Note that only braids with the same number of strings may be combined by concatenation. Along with this operation, all the braids on n strings form a group, Bn. In 1947, Artin showed that the braid group on n strings has a relatively simple presentation:

In this presentation, s1,s2,…,sn-1 are the generators. si represents the ith string crossing over the i+1st string, while si-1 represents the i+1st string string crossing over the ith string. The equations on the si's are the relations. The first relation explains how crossings which have at least a string between them can be rotated in their order without effect on the braid (Figure 2).



Figure 2


The other relation explains a subtle change that can be made when crossings are in a certain order which will also not change the braid (Figure 3).


Figure 3


The crossings being represented by the si's is enough information to represent all of the individual elements of Bn as a word on the generators, the si's. For example, the braid in Figure 1 can be represented as the word s2s3s1-1s2-1s1s3. For more information on Bn, see Magnus, and for more on group presentations see Gallian or Magnus.

Now I will move on to define the central aspect of this paper, the circular braid group on n strings, n. The circular braid group has many similarities to the standard braid group and its definition is a derivative of Bn. The one major difference between the two is that the ends of the bars that hold the strings of a standard braid never touch, whereas the ends of the bars of a circular braid meet forming two circles. We will then move on to examine the relationship between the two groups and we will soon note that n is more closely related to Bn+1 than Bn. Defining a function from n to Bn+1 will bring some clarity, but will not yet provide an exact relationship between the two groups. To conclude, plenty of opportunities for further research on this topic will be presented.

The Circular Braid Group n

Definition: A circular braid on n strings

A circular braid on n strings has n strings attached to the outer edges of two circles which lie in parallel planes in R3. The strings may tangle around each other but they may not cross the planes in which the two circles lie nor may they pass through the center of the cylinder-like area (the void space) created by the circles and strings. Also, the strings may not turn back upwards.


Definition: Relationships in a circular braid on n strings

When a string crosses another string it is an over-crossing if the string passes out of the cylinder to cross the other string. Define ai to be the ith string passing over the i+1st (mod n) string and ai-1 to be the i+1st (mod n) string passing over the ith string. Working in mod n is necessary for the nth string to be able to cross over the 1st string and vice versa (a phenomenon impossible in the standard braid group unless accompanied by multiple other crossings). Just as in Bn, every circular braid can be written as a word on the ai's.


Definition: The circular braid group n

The circular braid group n consists of all possible circular braids on n strings. Just as in the standard braid group Bn, the operation between two elements is defined as the concatenation of their representations as words in the ai's. Geometrically, this attaches one circular braid on top of the other by removing the circle on the bottom of the one braid and on the top of the other and then attaching string i to string i. By using the same arguments that are used to prove the Bn is a group, it is easily shown that n is a group. The operation is obviously associative, the inverse of any one element is the inverse of its word in the ai's and the identity is the empty word, the braid with no crossings where string i begins and ends in the ith position. Therefore, n is a group under the operation defined above.


A Standard Method to Geometrically Represent an Element of n

I have established some guidelines to follow in order to have a standardized system of notations for easily representing an element of the circular braid group. First, draw the upper circle of the element. To the right of the center of the circle on the edge of the circle which faces out, place string one and continue until half of the strings have been placed. However, do not continue placing the strings around the back of the circle. Instead, start to the left of the center of the circle on the edge of the circle which faces out and continue to place the strings until they are exhausted. If there are an odd number of strings, place an odd number of strings to the right on the circle and an even number to the left. All of this will place a potential crossing between the nth and 1st strings right in the center of your vision. The strings that have been placed on the outer edges of the circle we will refer to as the outer strings.

Now except for the outer strings, represent the crossings between the strings just as you would in the standard braid group. As for the outer strings, we will represent the crossing between them by the outer strings passing behind all of the other strings (although no actual crossings are taking place) and then crossing where the nth and 1st strings cross in our field of vision. This crossing will take place on the rear of the cylinder. Represent the right outer string crossing over the left outer string by the left outer string crossing over the right. In actuality, this represents the right crossing over the left, for it is the right string which is passing out of the cylinder. Figure 4 gives two examples of circular braids with their respective words.


Figure 4


Conjecture: The Presentation of n

Noting that an element of n is simply an element of Bn wrapped around and connected to itself in R3, it seems plausible that the presentation of n must be quite similar to the of Bn. Obviously, the same relations hold in n that held in Bn with the addition of the relations holding over crossings from the nth string to the 1st string. For this reason, the presentation of n seems obvious:



Although it is not proven that this is the group presentation of n, we will work with this assumption for the rest of this paper.

If this is in fact the presentation of n, there is an immediate corollary. Dyck's Theorem says that given two groups, G and H, with the same number of generators, with H having all of G's relations plus finitely many more distinct relations, then H is a homomorphic image of G. Note that n has the same number of generators as Bn+1 and has all of the same relations as Bn+1 with three more (all the relations involving crossings between the nth and 1st strings). Therefore, by Dyck's Theorem, n is a homomorphic image of Bn+1. This being known, we will now devote our attention to defining a function between the two groups and examining it to see if we might obtain an isomorphism between n and Bn+1.


The Function : n —> Bn+1

Our previous observation that an element of Bn wrapped around and connected to itself in R3 yields a circular braid suggests a natural function from n to Bn - ripping apart the circles of a circualr braid and stretching it out to become a standard braid. Now of course it is possible to tear the circles apart anywhere, but in order to standardize our function, let us tear the circles between the nth and 1st strings. Quickly examining our function geometrically, however, we note that our function is not one-to-one (Figure 5).


Figure 5


This example gives us much insight into the geometrical relationship between n and Bn. In this example, stretching a crossing between the nth and 1st strings across the void space of the cylinder yields the exact same braid as does flattening out a circular braid in which the strings skirt the edge of the void space of the cylinder but did not completely tangle around the void space. This clearly shows the importance of the void space in the cylinder.

Therefore, we must alter our function so that it takes into account the void space inside the cylinder. In our new function, we will again rip apart the circles between the nth and 1st strings of a circular braid and stretch it out to become a standard braid. However, our improved function will also add an extra string upon translation - which will be represented by a double line - to account for the void space inside the cylinder. Hopefully, taking into account the void space in this matter will solve our previous problem of our function not being one-to-one.


Definition: The Function from n to Bn+1

Let n. Let be a function from n to Bn+1 such that () does the following:
  1. Every ane in the word is replaced with the word:
    (s1s2...sc-1)(sn-1-1sn-2-1 ...sn+2-1), scs c+1esc-1 ...
    (sc-1-1sc-2-1 ...s1-1)(sn+2sn+3...sn-1)
    where = 1 and c = n/2 if n is even or c = (n+1)/2 if n is odd

  2. Let abe be a crossing between the outer strings. They are replaced with sc-1sc+1esc or equivalently sc+1scesc+1-1.
  3. Any other aie is replaced with s(n+1)-ie if i < c or sn-ie if i > c.

Since it is not entirely obvious that represents pulling apart an element of n stretching it out, and adding an extra string where the void space once existed (the void string), we will look at some pictures to solidify the claim.

Property 1 of represents the crossing ane being stretched out over all of the other strings as the circular braid is being pulled apart. Pulling apart the circles where the nth and 1st strings cross will pull the nth strings across all of the strings to its left yielding s1s2…sc-1 and sc-1-1sc-2-1…s1-1 and the first string across the strings to its right yielding sn-1-1sn-2-1…sc+2-1 and sc+2-1sc+3-1…sn-1-1. The nth string also will cross the void string and the 1st string yielding scsc+1esc-1. This algorithm for determining the exact word comes from the fact that the circular braid is ordered to have a certain number of strings on each side of the nth and 1st strings (Figure 6).



Figure 6


Property 2 of deals with crossings between the outer strings and their relation to the void space. First of all, no other strings besides the outer strings and the void strings will be involved in this mapping, since the circular braid is being straightened out in the same plane in which lies the crossings between the outer strings. Since the crossings are behind the void space, the void string will cross over one outer string, the outer strings will cross, and then the void string will cross the outer string, yielding sc-1sc+1esc (Figure 7). If desired, using a relation of Bn gives us the equivalent word sc+1sc+1esc+1-1.



Figure 7


In property 3 of , the ordinary crossings on a circular braid are mapped. Consider the case where i > s. The crossings which involved the ith and i+1st strings will now involve the n-(i-1)st and n-ith strings in the new standard braid, which is sn-ie. Now consider i < s. The addition of the void string moves all of the strings to the right of the first string over, causing the crossing numbers to change. The crossing itself is not changed, but only moved over to allow for the addition of the void string. Therefore, the crossing will now involve the (n+1)-ith and (n+1)-(i-1)th, yielding s(n+1)-ie (Figure 8).




Figure 8


Now that our function is established, we will examine if it is well-defined. This seems to be clear, because it is highly unlikely that tearing apart a circular braid could in any way yield two different standard braids. However, as the interest lies in knowing the precise relationship between n and Bn+1, we will now examine if maps the relations of n to the same word in Bn+1.

Let us first examine if maps the relation aiaj = ajai, where |i-j| > 2, to the same word in Bn+1. That is, does (aiaj) = (ajai)? Well, if i and j do not equal n, it is obvious that the relation will hold upon mapping since no extra crossings will be added upon mapping, and ai and aj will still have at least one string between them. The crossings will be able to commute and the relation will yield the same word in Bn+1.

However, if i or j equals n, then a whole string of new si's will be added upon mapping. Let i n. This implies that i cannot equal 1 or n-1 since |i-j| > 2. Also, suppose that i > c. We will now use relations of Bn to manipulate the resulting word. Examine: (In order to simplify, we will only examine pieces of the word at a time.)



The case of i < c is similar. Note that using the relations of Bn in the order that they are used keeps the order of the s's intact to form (an) even though different si's and si-1's commute throughout the word. This geometric example in 5 (Figure 9) may help to clarify what is happening in the proof.



Figure 9


Now examine the relation aiaqai = aqaiaq where q=i+1 mod n, to see if yields the same word in Bn+1 upon mapping. If either i or q do not equal n or b, no extra crossings are added and the relation will obviously hold upon mapping. This is a direct consequence of the crossings being at least one string apart. Now let us consider the other two cases.


Case 1: Suppose i = b.

Note that

Since b represents either n/2 if n is even, or (n-1)/2 if n is odd, we have two cases to consider:
    If n is even, note (n-1)-b = (n-1)-(n/2) = (n-2)/2 = (n/2)-1 = c-1.
    If n is odd, note (n-1)-b = (n-1)-(n-1)/2 = (n-1)/2 = (n+1-2)/2 = ((n+1)/2)-1 = c-1.

Therefore,


Setting q = b is proved by a similar argument.

Case 2: Suppose q = n.
As the algebra behind this proof is incredibly complex and lengthy, we will examine what is happening geometrically (Figure 10).





Figure 10



First examine the braid (a5a4a5). Let string i denote the string that is in the ith position at the top of the braid. First we cancel the inverse pair at the far right of the braid. Now, note that string 6 passes everywhere under string 1. Pulling string 6 as far right as possible and string 1 as far left as possible yields the second picture in Figure 10. Finally, pulling string 2 up slightly yields the last picture in the Figure which is (a4a5a4). Clearly, all braids of the form (anan+1an) will have geometric representations of the same form which will also be able to be manipulated as above. Therefore, (anan+1an) = (an+1anan+1). Setting i = n can be proved similarly. Since holds the relations from n to Bn+1, is well-defined.

Showing to be operation preserving is not as difficult of a task. It comes from the fact that in essence separates a circular braid into separate generator elements, transforms each individually, and then uses the separate results to piece together a braid in Bn+1. The generic results for each generator combine together, inverse pairs cancel, relations are employed, and the resulting braid is exactly the same as the braid which results when the circular braid is split apart to form a standard braid. In the same way, adding circular braids together and then performing on the sum yields the same result as performing individually on the circular braids and then adding their images together. Thus is operation preserving.


The Relationship Between n+1 and Bn+1

Now that we have shown that is a homomorphism between n and Bn+1, it now remains to examine whether is an isomorphism between the two groups. Note that takes the void space inside the cylinder and translates it into a string. When this string is crossed, it is always crossed again. Therefore, the crossing will always form the relation sc-1sc+1esc or sc-1sc+1esc-1 by definition of . These crossings always take the void string back to its original position. Thus is not onto all of Bn+1 since the void string will never end up at any other position then where it started from.

However, all of the strings will move around freely as of any circular braid will dictate. Therefore, it seems possible that (n) will be onto a subgroup of Bn+1 where the c+1st string will be in the c+1st position at the beginning and end of each braid. It also seems likely that is a one-to-one function between n and Bn+1. This follows from the fact that the void space inside the cylinder has been accounted for. All of the possible words in n have a definite geometric relationship to each other and to the void space. Since is well-defined, operation preserving and a geometrically intuitive function between the two groups, it is possible that all of these relationships have been accounted for in and that it is indeed an isomorphism between n and the above mentioned subgroup of Bn+1.

Conclusions

To conclude, there are plenty of opportunities for further research on this topic. It remains to be known if we have the correct presentation for n and whether in fact an isomorphism between n and a subgroup of Bn+1. Also, one might explore adding new generators to the above given presentation of n and see if they might account for always taking the void string to the same position. This might truly yield an isomorphism between n and Bn+1 which would then allow us to examine knot invariants in the circular braid groups.

References

E. Artin. Theory of Braids. Annals of Mathematics, vol. 48, 1947, pp. 101-126.

Gallian J. Gallian. Contemporary Abstract Algebra. Houghton Mifflin Co., Boston, MA, 1998

Magnus W. Magnus, A. Karrass and D. Solitar. Combinatorial Group Theory. Dover, 1966/1976.


Journal of Young Investigators. 2001. Volume Three.
Copyright © 2001 by John Singler and JYI. All rights reserved.
 		 
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