Grade 9 & 10 Introduction to Group Theory Part 3

Today we are going to learn about braid groups! But before we jump in, let’s do a quick recap of what we have covered so far. Recall that a group is a set

G

, together with a binary operation

∙

, such that the following group axioms hold:

So far we have seen two main examples of groups. The first being symmetry groups, and the second being symmetric groups. Today we are going to look at another class of groups called braid groups!

Braid Groups

Like most groups, the study of braid groups has applications all over mathematics in areas such as topology, knot theory, physics, fluid mechanics, and algebraic geometry. Besides it’s far reaching applications, braid groups are also visually nice and fun to play with because the elements in the underlying set of a braid group actually look like braids! Recall again that a group is a set together with a binary operation. Together, we will learn about the underlying set of a braid group. And throughout various exercises, you will figure out what the binary operation is!

The underlying set of a braid group

The underlying set of a braid group is a certain collection of objects called braids. As hinted at earlier, these objects do in fact look like the braids you see in your everyday life. Here are some examples of braids out in the wild:

The braids that we are going to learn about today are just like the ones in Figure 1. Since we are doing mathematics though, we need to precisely define what we mean by a braid. For us, a braid is an object that is constructed via the step by step guide below.

Step by step guide for how to construct a braid

Six examples of braids. In each example, eight dots are arranged into a left column of four dots and a right column of four dots, and four strings (curved or straight lines) connect each dot on the left to a dot on the right as described below.

First braid: Four horizontal strings connect each dot on the left to the corresponding dot on the right.
Second braid:
- A string connects dot 2 on the left to dot 1 on the right.
- A string, placed on top of the previously described string, connects dot 1 on the left to dot 2 on the right.
- A string connects dot 3 on the left to dot 3 on the right.
- A string connects dot 4 on the left to dot 4 on the right.
Third braid: The same connections as the second braid except the string connecting left dot 2 to right dot 1 lies on top of the string connecting left dot 1 to right dot 2.
Fourth braid:
- A horizontal string connects left dot 1 to right dot 1.
- Strings connect left dot 3 to right dot 2 and left dot 4 to right dot 3.
- A string, placed on top of the two previously described strings, connects left dot 2 to right dot 4.
Fifth braid:
- Horizontal strings connect left dot 1 to right dot 1 and left dot 2 to right dot 2.
- A string connects left dot 3 to right dot 4.
- A string connects left dot 4 to right dot 3 as follows: It starts at left 4, then goes over the string connecting left 3 to right 4, then continues on and goes over the string connecting left 2 to right 2, then changes direction to go under the same string connecting 2 to 2, and then ends at right 3.
Sixth braid:
- A string connects left 4 to right 1.
- A string that goes over the previous string connects left 3 to right 2.
- A string that goes over the previous two strings connects left 2 to right 3.
- A string that goes over the previous three strings connects left 1 to right 4.

All of the diagrams in Figure 2 are examples of braids. In some of the diagrams, you may have noticed that there are "spaces" or "breaks" in some of the strings. Let’s explain this by first looking at the second braid:

In this diagram, we see that there are spaces in the string connecting the "B" dots. These spaces are used to indicate that the string connecting the "A" dots lies on top of the string connecting the "B" dots at that spot. In other words, the string connecting the "A" dots crosses over, or goes over, the string connecting the "B" dots. Let’s look at another braid:

The spaces here tell us where the string connecting the "A" dots crosses over the string connecting the "B" dots and the string connecting the "C" dots. For another example, consider the following braid:

The spaces here are a bit more involved than in the braids above. First off, the string connecting the "C" dots crosses over the string connecting the B" dots. We also see that the string connecting the "C" dots goes over the string connecting the "A" dots, and then goes underneath it. This is not a knot, and so is allowed. Let’s now look at some non-examples:

Three diagrams. In each, eight dots are arranged into a left column of four dots and a right column of four dots, and four strings connect dots on the left to dots on the right as described below.

First diagram: Three horizontal strings connect left dots 2,3,4 to corresponding right dots 2,3,4. A fourth string connects left dot 1 to right dot 1 as follows: It starts at left dot 1, then goes over the string connecting left 2 to right 2, then loops back around and goes under that same string, then crosses over itself on the way to right dot 1.
Second diagram: Strings connect left dot 1 to right dot 1, left dot 2 to right dot 1, left dot 3 to right dot 3, and left dot 4 to right dot 4. No strings go over or under any others.
Third diagram: Strings connect left dot 1 to right dot 1, left dot 3 to right dot 2, left dot 3 to right dot 3, and left dot 4 to right dot 4. No strings go over or under any others.

All of the diagrams in Figure 3 are not braids. In the last two diagrams, we can see that not every dot is connected to a string (and as a result, there are dots connected to more than one string), and so they are not braids. Now consider the first diagram in Figure 3:

Here, every dot is connected to exactly one string, however, there is a knot! The string connecting the "A" dots makes a knot around the string connecting the "B" dots. So, this is not a braid.

Definition 1

The objects formed using the above "step by step guide for how to construct a braid" are called braids.

Definition 2

Let $n \in {1, 2, 3, \dots}$ . A braid formed using $n$ strings is called an $n$ -braid. We use $B_{n}$ to denote the set of all $n$ -braids.

For each

n

, the set

B_{n}

together with a certain binary operation forms a group. This group is called the braid group on $n$ strings. We will come back to this soon!

Exercise 1

Identify which of the below diagrams are $4$ -braids.

A description of the diagram follows.

Six diagrams. In each, eight dots are arranged into a left column of four dots and a right column of four dots, and four strings connect dots on the left to dots on the right as described below.

First diagram:
- Two horizontal strings connect left dots 2 and 3 to corresponding right dots 2 and 3.
- A third string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes under the string connecting dot 2 to dot 2, then changes direction to go under the same string again to end at right dot 1.
- A fourth string connects left dot 4 to right dot 4 as follows: Starting at left dot 4, it goes over the string connecting dot 3 to dot 3, then changes direction to go over the same string again to end at right dot 4.
Second diagram:
- Three horizontal strings connect left dots 2, 3, and 4 to corresponding right dots 2, 3, and 4.
- A fourth string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes over each of the previous three strings, then changes direction to go over the same three strings again to end at right dot 1.
Third diagram:
- Two horizontal strings connect left dots 2 and 3 to corresponding right dots 2 and 3.
- A third string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes over the string connecting dot 2 to dot 2, then changes direction to go under the same string and then end at right dot 1.
- A fourth string connects left dot 4 to right dot 4 as follows: Starting at left dot 4, it goes under the string connecting dot 3 to dot 3, then changes direction to go over the same string and then end at right dot 4.
Fourth diagram:
- A horizontal string connects left dot 1 to right dot 1.
- A second string connects left dot 2 to right dot 2 as follows: Starting at left dot 2, it goes under the string connecting dot 1 to dot 1, then loops back around and goes over the same string, then crosses over itself on the way to right dot 2.
A third string connects left dot 3 to right dot 4.
A fourth string, that goes under the third string, connects left dot 4 to right dot 3 follows.
Fifth diagram:
- Two horizontal strings connect left dots 2 and 3 to corresponding right dots 2 and 3.
- A third string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes under the string connecting dot 2 to dot 2, then loops back around and goes over the same string, then crosses under itself on the way to right dot 1.
- A fourth string connects left dot 4 to right dot 4 as follows: Starting at left dot 4, it goes under the string connecting dot 3 to dot 3, then loops back around and goes over the same string, then crosses under itself on the way to right dot 4.
Sixth diagram: No strings go over or under any others.
- A horizontal string connects left dot 1 to right dot 1.
- A second string connects left dot 2 to right dot 3.
- A third string connects left dot 3 to right dot 3.
- A fourth string connects left dot 4 to right dot 3.

Exercise 2

Choose a small $n$ , say $1 \leq n \leq 6$ , and make $n$ -braids. Feel free to use a few different $n$ values.

Sometimes two or more

n

-braids appear to be different, but they are actually the same braid. Let’s look at some examples. Consider the following two braids:

These two braids are different or, in other words, distinct. Note that the bottom two string connections in each braid are identical, so we only need to focus on the top two. In the first braid, the string connecting the "A" dots crosses over the string connecting the "B" dots. In the second braid, the opposite happens; the string connecting the "B" dots crosses over the string connecting the "A" dots. Given this, these two braids are different. Now, consider these two braids:

In each braid, eight dots are arranged into a left column of four dots and a right column of four dots, and four strings connect dots on the left to dots on the right as described below.

First braid:
- Two horizontal strings connect left dots 3 and 4 to corresponding right dots 3 and 4.
- A third string connects left dot 1 to right dot 2, both marked as A.
- A fourth string connects left dot 2 to right dot 1, both marked as B. This string goes over the third string.
Second braid:
- A first string connects left dot 2 to right dot 1, both marked as B.
- A second string connects left dot 3 to right dot 3, both marked as C.
- A third string connects left dot 4 to right dot 4, both marked as D.
- A fourth string connects left dot 1 to right dot 2, both marked as A, as follows: Starting at left dot 1, it goes over the string connecting the "B" dots, then under the one connecting the "C" dots, then over the one connecting the "D" dots, then changes direction to go over the one connecting the "D" dots again, then under the one connecting the "C" dots again, ending at right dot 2.

These two braids look quite different from each other, but they are actually the same braid! We can see this by "pulling" and "stretching" the strings. Let’s start by looking at the strings that connect the "A" dots and the "B" dots. Our goal is to make these strings look the same on each braid by pulling and stretching them. To do this, let’s make the second braid look like the first. Note that the string connecting the "A" dots in the second braid lies on top of the string connecting the "D" dots, and lies below the string connecting the "C" dots. Because there is no intertwining or crossings between these strings, we can pull the string connecting the "A" dots towards the top of the braid; it will slide over the string connecting the "D" dots and slide underneath the string connecting the "C" dots. After doing this, we can see that the strings connecting the "A" dots and "B" dots are the same on each braid. Also, after doing this, we can see that the string connecting the "C" dots is the same on both braids, and similarly, the string connecting the "D" dots is the same on each braid. Given this, the two braids are considered to be the same! Note that for the example on page 8, there is no way to pull and stretch the strings to make them the same on each braid; this is because pulling and stretching strings cannot undo or create crossings.

To conclude, if two

n

-braids can be made to look the same by "pulling or stretching the strings", then we say that the two

n

-braids are the same. In other words, if you can change, or transform, one braid into another (by pulling or stretching the strings) without moving the starting and end points of the strings, then the two braids are the same. If two

n

-braids are not the same, then they are different. Note that the length of strings does not matter when in comes to determining if two braids are the same or different.

Exercise 3

Consider the six $3$ -braids below. How many different $3$ -braids are there?

A description of the diagram follows.

In each of the six braids, six dots are arranged into a left column of three dots and a right column of three dots, and three strings connect dots on the left to dots on the right as described below.

First braid: Three horizontal strings connect left dots 1, 2, and 3 to corresponding right dots 1, 2, and 3. No strings go over or under any others.
Second braid:
- A first string connects left dot 1 to right dot 2.
- A second string connects left dot 2 to right dot 3.
- A third string connects left dot 3 to right dot 1 as follows: Starting at left dot 3, it goes over the second string, then under the first string, ending at right dot 1.
Third braid:
- Two horizontal strings connect left dots 2 and 3 to corresponding right dots 2 and 3.
- A third string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes under the string connecting dot 2 to dot 2, then changes direction to go under this same string and then end at right dot 1.
Fourth braid:
- A horizontal string connects left dot 2 to right dot 2.
- A second string connects left dot 3 to right dot 3 as follows: Starting at left dot 3, it goes under the string connecting dot 2 to dot 2, then changes direction to go under this same string again and end at right dot 3.
- A third string connects left dot 1 to right dot 1 as follows: Starting at left dot 1, it goes over the string connecting dot 3 to dot 3, then over the string connecting dot 2 to dot 2, then changes direction to over the same two strings again and end at right dot 1.
Fifth braid:
- A bottom horizontal string connects left dot 3 to right dot 3.
- A second string connects left dot 2 to right dot 1.
- A third string connects left dot 1 to right dot 2, going over the second string.
Sixth braid:
- A bottom horizontal string connects left dot 3 to right dot 3.
- A second string connects left dot 2 to right dot 1.
- A third string connects left dot 1 to right dot 2 as follow: Starting at left dot 1, it goes over the second string, then over the bottom string, then changes direction to go over the bottom string again and end at right dot 2.

Binary operation on braids

So far we defined the underlying set for braids groups. Specifically, for each non-zero

n \in N

, there is a group called the braid group on

n

strings, whose underlying set is

B_{n}

. We still need a binary operation on

B_{n}

to make it into a group! Now that you have some practice working with braids, you are going to try to figure out what the binary operation on

B_{n}

is.

Exercise 4 **$* $ Spoilers follow $ *$**

Try to define a binary operation $∙$ on $B_{n}$ so that $(B_{n}, ∙)$ is a group. Here are some suggestions and comments to help you get started:

Asking for a binary operation on $B_{n}$ is the same as asking "how can we combine two $n$ -braids to make another $n$ -braid?".
Feel free to use a small $n$ , say $n = 3$ or $n = 4$ , to figure out the binary operation.
Don’t worry too much about trying to formalize the binary operation. Playing around and trying to combine explicit $n$ -braids in a natural way is a great way to get you on the right track!
Keep in mind that we want this binary operation and $B_{n}$ to form a group. Once you have a guess of what the binary operation on $B_{n}$ is, try to see if the group axioms hold for $B_{n}$ and your binary operation.

Consider the set

B_{n}

consisting of all

n

-braids. We can combine two

n

-braids in

B_{n}

by a binary operation called concatenation. We use the symbol

*

for concatenation.

Definition 3

Let $b_{1}$ and $b_{2}$ be two $n$ -braids. Then $b_{1} * b_{2}$ is a new $n$ -braid which is formed by connecting the right set of dots in $b_{1}$ to the left set of dots in $b_{2}$ and then erasing these dots so that the strings from the two braids form a connection. Specifically, if we label each set of $n$ dots 1 through $n$ from top to bottom, then we connect the $k$ th dot from the right set of dots in $b_{1}$ to the $k$ th dot in the left set of dots in $b_{2}$ . This is a binary operation on $B_{n}$ called concatenation.

Note: In general, given two

n

-braids

b_{1}

and

b_{2}

b_{1} * b_{2}

is not necessarily equal to

b_{2} * b_{1}

Two braids, each with two columns of four dots, with an $*$ between them.

First braid (to the left of $*$ ):
- A string connects left dot 1 to right dot 2.
- A string, going under the previous string, connects left dot 2 to right dot 1.
- Horizontal strings connect left dots 3 and 4 to corresponding right dots 3 and 4.
Second braid (to the right of $*$ ): Same as the first braid except the string connecting left dot 2 to right dot 1 goes over the string connecting left dot 1 to left dot 2.

In the first braid, the dots in the right column are marked as 1, 2, 3, and 4, from top to bottom. In the second braid, the dots in the left column are marked in the same way.

As suggested in Definition 3, you can see that we labelled two sets of dots 1 through 4. This is completely optional, and is just used as a way to help understand how concatenation works. To compute this concatenation, we connect dot 1 in the left braid with dot 1 in the second braid, dot 2 in the first braid with dot 2 in the second braid, and so on. The result is as follows:

To make this into a

4

-braid, we simply erase the middle set of dots. The result is as follows:

We can also shorten the strings to make the diagram tidier. The result is as follows:

In short, to compute

b_{1} * b_{2}

, you can imagine sliding the two braids together so that the right set of dots in

b_{1}

lines up with the left set of dots in

b_{2}

. Next, we will convince ourselves that

B_{n}

under concatenation

*

forms a group. In other words, we will convince ourselves that the group axioms hold for

B_{n}

with

*

Exercise 5

Convince yourself that $(B_{3}, *)$ is a group.

$* *$ See solutions to Exercise 5 before reading on $* *$

The explanation for Exercise 5 can be generalized to show that

(B_{n}, *)

is a group for all non-zero

n \in N

. Let’s roughly see why

(B_{n}, *)

is a group for any

n

. To do this, we need to convince ourselves that the group axioms hold for

B_{n}

with concatenation.

Axiom 1: For associativity to hold, we need

(b_{1} * b_{2}) * b_{3}

to be the same

n

-braid as

b_{1} * (b_{2} * b_{3})

, for all

b_{1}, b_{2}, b_{3} \in B_{3}

. This axiom is a bit hard to argue rigorously without introducing a lot of extra notation. However, for us, it is plenty good enough to understand why it works by looking at concrete examples. See the solutions to Exercise 5 for a concrete example worked out.

This braid consists of

n

parallel horizontal strings. If you concatenate this braid with any other

n

-braid

b

, the strings of

b

just get longer. Since the length of strings does not matter, the result is just

b

. So, this braid is the identity element and we call it

{id}_{B_{n}}

Axiom 3: Lastly, we want to show that every

n

-braid has an inverse. The inverse of an

n

-braid is obtained by flipping it over horizontally. For a concrete example, consider the following photo:

Call the braid on the left

b

. Flipping

b

over horizontally is the same as reflecting

b

in the red vertical line. To obtain the inverse of

b

, all we have to do is reflect

b

in this line. The braid on the right is what we get after reflecting

b

in this line. We will soon see that the braid on the right is the inverse of

b

, so let’s call it

b^{- 1}

. Indeed, we compute that

b * b^{- 1}

By stretching and pulling on the strands of

b * b^{- 1}

, we see that it is equal to

{id}_{B_{n}}

. In the same way, one can show that

b^{- 1} * b = {id}_{B_{n}}

. So

b^{- 1}

is the inverse of

b

. And in general, the inverse of any

n

-braid can be obtained by using the method illustrated in this example. If you think of an

n

-braid as a certain "tangling" of

{id}_{B_{n}}

, then it’s inverse is the

n

-braid that undoes, or is the reverse, of this tangling.

Grade 9/10 Math Circles
An Introduction to Group Theory Part 3

Review from last time

Braid Groups

The underlying set of a braid group

Step by step guide for how to construct a braid

Definition 1

Definition 2

Exercise 1

Exercise 2

Exercise 3

Binary operation on braids

Exercise 4 **$* $ Spoilers follow $ *$**

Definition 3

Exercise 5

Definition 4

Grade 9/10 Math Circles An Introduction to Group Theory Part 3

Review from last time

Braid Groups

The underlying set of a braid group

Step by step guide for how to construct a braid

Definition 1

Definition 2

Exercise 1

Exercise 2

Exercise 3

Binary operation on braids

Exercise 4 ∗∗ Spoilers follow ∗∗

Definition 3

Exercise 5

Definition 4

Grade 9/10 Math Circles
An Introduction to Group Theory Part 3

Exercise 4 **$* $ Spoilers follow $ *$**