Whether in the context of hairstyles, friendship bracelets, or even parachute cords -- most will be familiar with the notion of a braid.

As can be seen in the images above, each braid starts with some number of strands which are repeatedly crossed under/over each other in some way. Note that we typically don't allow the strands of a braid to "turn back up".

We can represent the particular crossings of a braid with a braid diagram like the one shown below. Note the diagram shown describes a braid similar to (but longer than) the hair braid above.

Of course, other braids will have a different number of strands and/or a different sequence of crossings. Some may even include sequences of crossings that don't even repeat, such as the one shown below:

Taking inspiration from braids in the real world, "tugging" on the strands in one direction or another -- even when new crossings result (as long as we don't allow any one strand to pass through another) -- can lead to different representations of what is essentially the same braid. As an example, consider the following two diagrams which actually represent the same braid.

While the two braid diagrams above represent the same braid, certainly the one on the left seems "simpler" in some capacity. This raises the question: "*How does one simplify a given braid diagram?*" Remember this question -- we'll come back to it in a bit.

Admittedly, drawing braid diagrams like the ones previously seen -- and especially when they are not fully simplified -- can be tedious. However, there is a much easier way to represent braids!

Towards this end, observe that if we "tug" in just the right places, we can always jiggle any particular crossing a little bit to the left or right, as desired. In this way, we can arrange any braid (with a finite number of crossings, anyways) so that no two crossings happen simultaneously as we scan the braid from left to right.

As an example, consider the braid diagram involving 5 strands presented earlier, which is shown again below on the left. Numbers and vertical lines have been added to help make the positions of the crossings easier to identify.

In the diagram below on the left, multiple crossings sometimes happen simultaneously between consecutive pairs of vertical lines. For example, between the first pair of vertical lines, the strands at positions $1$ and $2$ cross (red and green) and the strands at positions $3$ and $4$ cross (blue and orange). Similarly, between the second pair of vertical lines, the strands at positions $1$ and $2$ again cross (green and red) and the strands at positions $4$ and $5$ cross (blue and black).

However, with a bit of tugging on the strands we can ensure only one crossing happens at a time. Notice how in the diagram on the right the initial red/green crossing has been jiggled a bit to the left and the initial blue/orange crossing has been jiggled a bit to the right. In this way, the red/green crossing now happens first, and the blue/orange crossing now happens second.

Indeed, once things have been "jiggled" in this way, what we see happening between pairs of consecutive lines reduces down to just a few simple possibilities for $5$ strands (there would of course be more if there were more strands involved):

Importantly, if we have names for these possibilities (above we have used $a$ through $h$), then we can describe the braid in question with a simple sequence of letters. So for example, using the above we might identify the braid we've been discussing with the following sequence of letters (also known as a "word"): $$aeahchchedh$$

As much as this can help reduce the tedium of describing a braid from drawing a complicated picture to just writing down a sequence of letters -- implicit in the above is an even greater revelation. Notice it suggests a natural way to combine two braids together to produce a new (longer) braid -- through *concatenation*!

Consider the two small braids below, which are combined by concatenating the second to the first to form a longer braid. Note, we'll use the "$*$" operator to indicate the action of concatenation:

One should note that combining two braids (that share the same number of strands) always results in another braid, again with the same number of strands. In general, when combining two things of the same "type" (here, both braids on $n$ strands) via some operation (here, concatenation) and the result is always of the same type, we say the operation is **closed**.

That might seem like an unusual term to use, but consider the following bizarre, but hopefully illustrative example:

Suppose you live on LEGO world. You and everything around you are made of Lego bricks. If you combine (by taking apart and reconnecting) any two things in LEGO world, the result is still made of Lego bricks. As much as you might try, there is no way you can combine two things made of Lego bricks and get a Play-Doh sculpture, for example. Play-Doh sculptures are in this way "inaccessible" to you -- they are "closed off" from you.

Closed operations will become very important to us later, but just to mention a couple of specific examples to reinforce the idea: Note that addition is closed with respect to even integers, but not odd ones. Similarly multiplication is closed with respect to integers, but division is not.

Turning attention back to braids -- note that denoting the result of concatenating braids $B_1$ and $B_2$ with $B_1 * B_2$ subtly suggests this operation of concatenation behaves in a way similar to "multiplication". The use of an asterisk "*" after all is a frequent way to denote a product.

Let's think about that for a moment -- can that even make sense in the context of braids? Beyond being a closed operation like multiplication (note the product of any two real numbers is a real number), in what other ways could braid concatenation behave like a "multiplication"? Some natural questions come to mind:

*Is braid concatenation associative?*

Recall, this is certainly a property of real-number multiplication: $(ab)c = a(bc)$*Is there an identity braid?*

That is to say, is there something that functions like the value $1$, in that for any real number $x$, we have $x \cdot 1 = 1 \cdot x = x$? (i.e., some special value that when we multiply some other value by it, that other value's "identity" is preserved)*Do braids have inverses?*

We certainly have multiplicative inverses for real numbers (provided they aren't zero). That is, there is a real-number $x^{-1}$ for every real-number $x$ (namely $1/x$) where the products $x \cdot x^{-1}$ and $x^{-1} \cdot x$ both equal the multiplicative identity, $1$.

Let us consider each of these in turn. For convenience, for the second and third questions, we'll assume the number of strands involved is $4$, but generalizations to some other number of strands should be both natural and (hopefully) obvious.

**Q: Is braid concatenation associative?**

That is to say, are $(B_1 * B_2) * B_3$ and $B_1 * (B_2 * B_3)$ the same braid?

Absolutely! This is an immediate result of how concatenation works. We don't even need to consider any specific braids to see this. Just let the yellow, pink, and green rectangles below represent arbitrary braids $B_1$, $B_2$, and $B_3$, respectively.

**Q: Is there an identity braid?**

Again, recall the multiplicative identity for real numbers is the value $1$ as we can multiply any real value by $1$ and leave it unchanged. Notice this works in both directions -- that is to say, for any real value $x$, it is true that $x \cdot 1 = 1 \cdot x = x$.

Similarly, the additive identity for real numbers is the value $0$ as we can add $0$ to any real value $x$ and leave it unchanged. (Again, reversing the order from $x+0$ to $0+x$ has no impact -- both result in $x$.)

If we seek a braid identity, then we seek a braid that could be concatenated to any other braid and leave its sequence of crossings unchanged.

Consider that unique braid on some number of strands, $I$, that has no crossings at all!

Clearly, as the below suggests, concatenating such a braid $I$ to any other braid $B$ (with the same number of strands, of course) leaves $B$ essentially unchanged.

The reverse is easily shown to hold as well (i.e., $I * B = B$ for any braid $B$).

As such, the braid $I$ on $n$ strands with no crossings serves as an identity for the concatenation of braids on $n$ strands.

**Q: Do braids have inverses?**

Here again, let us restrict our attention to "braids on $n$ strands" for a particular $n$.

Following the pattern of multiplicative inverses discussed earlier, we then seek for any such braid $B$ an inverse $B^{-1}$ where $B * B^{-1} = I$ and $B^{-1} * B = I$ (assuming $I$ denotes the braid identity)

Remember the simple braids that we previously used to identify a braid of $5$ strands with a sequence of letters? Here's a similar set of braids for braids of $4$ strands:

Regardless of the number of strands involved, notice that these always occur in pairs where the same two strands cross -- with one with one strand on top and the other where the other strand is on top. Indeed, each of these pairs is an *inverse pair*, as suggested by the names given to the six simple braids immediately above. After concatenating each such pair, only a couple of tugs on the strands are needed to simplify the result to the identity braid $I$ (on $n$ strands), as the below demonstrates for one such pair:

Just to be explicit about the naming convention adopted above, note that for any $i = 1,2,3,\ldots$, we let $x_i$ denote the braid where strands at positions $i$ and $i+1$ cross, with the strand at position $i$ going "over" the strand at position $i+1$. We denote the inverse of $x_i$ by $x_i^{-1}$, where the strand at position $i$ goes "under" the strand at position $i+1$. As a matter of verbiage, we call the set of all such $x_i$ and their inverses the **elementary braids** on $n$ strands.

Armed now with these inverse pairs of elementary braids, we can build inverses for more complicated braids.

We can think of the individual crossings as actions taken on the strands that change their state, much like the individual actions of putting on one's socks, shoes, and rain boots (which go over one's shoes) each change the state of your feet. The inverse action to some combination of these can be found by "undoing" each individual action, but in reverse.

Suppose one puts on one's socks, and then shoes, and then rain boots, in that order. *We could consider other orders, but are likely to over-stretch our socks in doing so.* 😆 To undo this combination of three individual actions (returning one's feet to their bare state), one removes the rain boots, then removes one's socks, then removes one's socks. (Note the reverse order!)

Likewise, if we apply elementary braids $x_1$, $x_3^{-1}$, and $x_2$ in that order, we can undo them by applying their inverses $x_2^{-1}$, $x_3$, and then $x_1^{-1}$, in that order.

Below are the braid diagrams for concatenation and simplification of the example just described. Note that, given how far this similarity between braid concatenation and real number multiplication seems to be going, we'll go ahead and adopt some additional notational conventions often used for products.

Specifically -- just as we often abbreviate $a \cdot b$ with $ab$ when dealing with products of real numbers $a$ and $b$ -- we'll often omit the "$*$" operator between variables representing braids (elementary or otherwise), leaving their concatenation assumed by their adjacency. We may also now start referring to such concatenations "braid products", or simply "products" when the context is clear.

As you consider the braid product being simplified below, note how we take advantage of the associativity of braid concatenation to evaluate the product of the center-most two elementary braids at each step -- which, being an inverse pair, results in the identity braid $I$ which can then be removed as it has no effect (except the last $I$, of course).

We don't actually need the pictures, now though. We can proceed to simplify things in a completely algebraic way, as shown below.

One will notice in the following calculations that we have added yet another way to write the concatenations (again mirroring the familiar ways we often write products of real numbers). Specifically, we've allowed the braid product $B_1 * B_2$ to also be written as $(B_1)(B_2)$. $$\begin{array}{rcl} (x_1 x_3^{-1} x_2)(x_2^{-1} x_3 x_1^{-1}) & = & x_1 x_3^{-1} x_2 x_2^{-1} x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} (x_2 x_2^{-1}) x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} I x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} (I x_3) x_1^{-1}\\ & = & x_1 x_3^{-1} x_3 x_1^{-1}\\ & = & x_1 (x_3^{-1} x_3) x_1^{-1}\\ & = & x_1 I x_1^{-1}\\ & = & x_1 (I x_1^{-1})\\ & = & x_1 x_1^{-1}\\ & = & I \end{array}$$

In truth, the above is a bit verbose -- showing all the intermediate steps each time an inverse pair produces an identity braid $I$, which then combines with whatever comes next to leave only whatever comes next.

In practice, this combination of steps is so common we often omit this level of detail when writing the steps taken to simplify a braid -- writing only something similar to the below (which one will notice mirrors the "braid words" given with the pictures above):

$$\begin{array}{rcl} (x_1 x_3^{-1} x_2)(x_2^{-1} x_3 x_1^{-1}) & = & x_1 x_3^{-1} x_2 x_2^{-1} x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} x_3 x_1^{-1}\\ & = & x_1 x_1^{-1}\\ & = & I \end{array}$$There is precedence for this. Consider what happens when you simplify the following fraction (where $b \neq 0$). In doing so, remember that $b/b = 1$, the multiplicative "identity". $$\frac{ab}{b} = a \cdot \frac{b}{b} = a \cdot 1 = a$$ Something like this happens every time one cancels a common factor in the numerator and denominator of a fraction -- but we often skip all that detail, writing only $$\frac{ab}{b} = a$$

The above clearly establishes there is some sort of "multiplicative arithmetic" we can apply to braids, but we must be careful to not let our analogy go too far. One significant difference between braid multiplication and the multiplication of numerical values with which we are well familiar is that braid multiplication is not generally commutative.

That is to say, we don't always have $B_1 B_2 = B_2 B_1$ for any braids $B_1$ and $B_2$.

As a simple example of this, consider the following two braids. Notice the first "product" on the right can't be simplified to the second. For example, the strand initially at position $1$ ends up in position $4$ in the first product, but not in the second.

The lack of general commutivity for braid products certainly decreases the ease with which we may manipulate braids, but as Alexander Graham Bell once said: "When one door closes, another opens."

Braids do actually enjoy a commutativity of "distant" elementary braids. That is to say, adjacent elementary braids $x_i$ and $x_j$ will commute if they are far enough apart that they don't involve a common strand position. Noting that this only happens when $i$ and $j$ are at least two positions apart, we can equivalently say for adjacent elementary braids $x_i$ and $x_j$: $$x_i x_j = x_j x_i \textrm{ when } |i-j| \ge 2$$

This is perhaps easier to understand with an example. Note in the diagram below, we can change the order of the pink elementary braids without effectively changing the overall braid. However, if we change the order of the yellow elementary braids, the overall braid is a different braid.

Named after Emil Artin, one of the leading mathematicians of the twentieth century and who developed the theory of braids as a branch of an area in mathematics known as algebraic topology, Artin's relation provides the last piece of the puzzle when establishing the strange arithmetic of braids.

With a little mental "tugging" on the strands below, one should easily be able to convince oneself that this relation holds for elementary braids $x_i$ and $x_{i+1}$ multiplied (i.e., concatenated) in the given way.

What is important, however, is that this special braid relation will allow us to manipulate braids now in a completely algebraic way -- never having to draw pictures like those above, if desired.

We have seen that braids on $n$ strands can be represented by algebraic expressions/words consisting of concatenations of elementary braids $x_1,x_2,x_3,\ldots,x_{n-1}$ and/or their inverses $x_1^{-1},x_2^{-1},x_3^{-1},\ldots,x_{n-1}^{-1}$.

These braid expressions are not unique to a given braid, however. We can show two braid words represent the same braid by algebraically manipulating one into the other in accordance with the following rules:

Assuming $i$ and $j$ are taken from $1,2,\ldots,n-1$ as appropriate, $B_i$ represents an arbitrary braid word, and $I$ represents the identify braid of no crossings)

Braid Associativity | $(B_1 B_2) B_3 = B_1 (B_2 B_3)$ |

Multiplication by the Identity | $I \, B_i = B_i \, I = B_i$ |

Inverse Relations | $x_i \, x_i^{-1} = I = x_i^{-1} \, x_i$ |

Commutativity of Distant Braids | $x_i \, x_j = x_j \, x_i$ when $|i-j| \ge 2$ |

Artin's Relation | $x_i \, x_{i+1} \, x_i = x_{i+1} \, x_i \, x_{i+1}$ |

Now let's put these to use! Consider the following way to simplify a braid $B$ on 3 strands where $$B = x_3^{-1} \, x_2 \, \, x_3 \, x_2 \, x_3^{-1}$$ Note that with only three strands, we won't be able to take advantage of the commutativity of distant braids. Further, we have no inverse pairs adjacent to one another, so we can't use any inverse relations yet.

However, there is an opportunity to apply Artin's relation. Notice, once we take advantage of this, we see two inverse pairs that can then be eliminated -- greatly simplifying the resulting expression!

$$\begin{array}{rcl} B & = & x_3^{-1} \, x_2 \, \, x_3 \, x_2 \, x_3^{-1}\\ & = & x_3^{-1} \, (x_2 \, \, x_3 \, x_2) \, x_3^{-1}\\ & = & x_3^{-1} \, (x_3 \, x_2 \, x_3) \, x_3^{-1}\\ & = & (x_3^{-1} \, x_3) \, x_2 \, (x_3 \, x_3^{-1})\\ & = & I \, x_2 \, I\\ & = & (I \, x_2) \, I\\ & = & x_2 \, I\\ & = & x_2 \end{array}$$Knowing that intermixed elementary inverse braids $x_i$ and $x_i^{-1}$ can result in cancellations (see example below) leaving a product/concatenation of $x_i$ with itself some number of times, we might find it useful and more compact to abbreviate a braid $b$ (elementary or otherwise) multiplied/concatenated by itself $p$ times by $b^p$. $$\begin{array}{rcl} x_i \, x_i \, x_i \, x_i^{-1} \, x_i \, x_i^{-1} \, x_i \, x_i & = & x_i \, x_i \, (x_i \, x_i^{-1}) \, (x_i \, x_i^{-1}) \, x_i \, x_i\\ & = & x_i \, x_i \, I \, I \, x_i \, x_i\\ & = & x_i \, x_i \, x_i \, x_i\\ & = & x_i^4 \end{array}$$

Other times, such products simplify to a concatenation of $x_i^{-1}$ with itself some number of times (see example below). In these cases, abbreviating $(b^{-1})^p$ with $b^{-p}$ also seems natural and more compact.

$$\begin{array}{rcl} x_i \, x_i^{-1} \, x_i^{-1} \, x_i^{-1} \, x_i \, x_i^{-1} \, x_i \, x_i^{-1} & = & (x_i \, x_i^{-1}) \, x_i^{-1} \, (x_i^{-1} \, x_i) \, (x_i^{-1} \, x_i) \, x_i^{-1}\\ & = & I \, x_i^{-1} \, I \, I \, x_i^{-1}\\ & = & x_i^{-1} \, x_i^{-1}\\ & = & x_i^{-2} \end{array}$$Of course, there is a third option for such products. It could be that pairing off the $x_i$ and $x_i^{-1}$ elementary braids and eliminating them leaves nothing but $I$ in the end, as the following suggests: $$\begin{array}{rcl} x_i \, x_i^{-1} \, x_i^{-1} \, x_i^{-1} \, x_i \, x_i \, x_i^{-1} \, x_i & = & (x_i \, x_i^{-1}) \, x_i^{-1} \, (x_i^{-1} \, x_i) \, (x_i \, x_i^{-1}) \, x_i\\ & = & I \, x_i^{-1} \, I \, I \, x_i\\ & = & x_i^{-1} \, x_i\\ & = & I \end{array}$$

Given all of the above -- and for consistency with the rules to come -- let us make the following two additional definitions:

- $x_i^0 = I$
- $x_i^1 = x_i$

As a consequence, we can now quickly argue the following must hold for ANY integers $p$ and $q$:

$x_i^p \, x_i^q = x_i^{p+q}$

*(add exponents when multiplying powers)*$(x_i^p)^q = x_i^{pq}$

*(multiply exponents when finding a power of a power)*

As an immediate application of the first of these, we also know

- $x_i^p \, x_i^{-q} = x_i^{p-q} \quad \textrm{ for ALL integers } p \textrm{ and } q$

Seeing the above, and remembering that the division of one real number by another is equivalent to multiplying the first by the reciprical (i.e., the multiplicative inverse) of the second, one might be inclined to define and denote the "division of one braid by another" (and the equivalent "fraction of braids") in the following way: $$b_1 \div b_2 = \frac{b_1}{b_2} = b_1 b_2^{-1}$$

Doing this leads to even more (familiar?) results for elementary braids:

$\displaystyle{x_i^{-p} = \frac{I}{x_i^p}}$

*(negative exponents are connected to recipricals of powers)*$\displaystyle{\frac{x_i^p}{x_i^q} = x_i^{p-q}}$

*(subtract exponents when dividing powers)*

For commutative $x_i$ and $x_j$ (i.e., "distant" elementary braids $x_i$ and $x_j$, where $|i-j| \ge 2$), we can say even more! In particular, for all integers $p$:

$(x_i x_j)^p = x_i^p x_j^p$

*(exponents distribute over products)*$\displaystyle{\left( \frac{x_i}{x_j} \right)^p = \frac{x_i^p}{x_j^p}}$

*(exponents distribute over quotients)*

**Be careful!** Don't use these last two rules when $x_i$ and $x_j$ fail to commute!