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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 as we move from left to right along the braid. 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 for two braids on the same number of strands, we can always combine them in this way to get another braid (again with the same number of strands).
In general, when combining two things of the same "type" (e.g., two braids on $n$ strands) via some operation (e.g., concatenation) and the result is always of the same type, we say the "things" of this type under the associated operation are closed (or equivalently, that they satisfy the property of closure with respect to that operation). To understand the word choice here, note that the combination of any two braids by concatenation will never be anything besides another braid -- a flamingo, for example. Combinations of braids must stay in the "braid universe". We can't get to the universe of flamingos if concatenating is all we can do. The flamingo universe is inaccessible -- it is "closed off" from us.
Closure will become very important to us later, but just to mention a couple of specific examples to reinforce the idea: Note that even integers are closed with respect to addition, but odd integers are not. Similarly integers are closed with respect to multiplication, but not with respect to division.
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 real number multiplication. The use of an asterisk "*" after all is a frequent way to denote a product (especially in programming).
Let's think about that for a moment -- what exactly do we mean by "behaves in a way similar to real number multiplication"? The real numbers are certainly closed under multiplication, but surely we must mean more than that! As we mull over the various properties we know multiplication of real numbers enjoy -- ones we hope braids under concatenation will also enjoy -- we might find it promising to ask the following questions to this end:
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 real number $1$ with respect to multiplication, 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 (or vice-versa), 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 non-zero real number $x$ (namely $x^{-1} = 1/x$) 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, for any three braids $B_1$, $B_2$, and $B_3$, are $(B_1 * B_2) * B_3$ and $B_1 * (B_2 * B_3)$ always 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 $1$ (or vice-versa) 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 an identity braid with respect to concatenation, then we seek a braid that could be concatenated to any other braid (on either side) and leave its sequence of crossings unchanged.
Consider that unique braid on some number of strands, $I$, that has no crossings at all!
As the below clearly suggests, concatenating such a braid $I$ to any other braid $B$ (with the same number of strands, of course) leaves $B$ essentially unchanged (i.e., the strands might be a tad longer, but the crossings are all still intact, ant that's what's important).
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 this reversed 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 as "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).
Of course, we could imagine untangling the initial concatenation by visualizing tugging some of the strands up or down at key moments to reduce the number of crossing too. However, the advantage of using the properties of inverses and identities to get from one step to the next is that we no longer really need the pictures (which were cumbersome to draw in the first place) -- we can proceed to simplify things in a completely algebraic way, as shown next.
As you consider the concatenation of braids $x_1 x_3^{-1} x_2$ and $x_2^{-1} x_3 x_1^{-1}$ below and its (algebraic) simplification, notice that we have initially added parentheses around each, so that we can more easily see what is being concatenated at the moment. Of course, using parentheses in this way again mirrors yet another familiar way we often write products of real numbers. For example, $6 = (2)(3)$. $$\begin{array}{rcl} (x_1 x_3^{-1} x_2)(x_2^{-1} x_3 x_1^{-1}) & = & x_1 x_3^{-1} x_2 x_2^{-1} x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} (x_2 x_2^{-1}) x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} I x_3 x_1^{-1}\\ & = & x_1 x_3^{-1} (I x_3) x_1^{-1}\\ & = & x_1 x_3^{-1} x_3 x_1^{-1}\\ & = & x_1 (x_3^{-1} x_3) x_1^{-1}\\ & = & x_1 I x_1^{-1}\\ & = & x_1 (I x_1^{-1})\\ & = & x_1 x_1^{-1}\\ & = & I \end{array}$$
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 the steps taken as one simplifies the fraction $\frac{ab}{b}$ (where $b \neq 0$), which are shown below. (Remember that $\frac{b}{b}$ equals the value $1$, 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.
$\displaystyle{{\large \textrm{Artin's Relation}: x_i \, x_{i+1} \, x_i = x_{i+1} \, x_i \, x_{i+1}}}$
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 certainly show two braid words represent the same braid by drawing pictures of each and "tugging" the strands this way and that until these pictures are identical. However, the equality of two braid words can also be shown by applying algebraic manipulations to one to produce 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 identity 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 $b$ and $b^{-1}$ can result in cancellations leaving a product/concatenation of $x_i$ with itself some number of times (as demonstrated in the example below), we might find it useful and more compact to abbreviate a braid $b$ (elementary or otherwise) multiplied/concatenated by itself $p$ times by $b^p$. $$\begin{array}{rcl} b \, b \, b \, b^{-1} \, b \, b^{-1} \, b \, b & = & b \, b \, (b \, b^{-1}) \, (b \, b^{-1}) \, b \, b\\ & = & b \, b \, I \, I \, b \, b\\ & = & b \, b \, b \, b\\ & = & b^4 \end{array}$$
However, if the number of elementary braids of the form $b^{-1}$ exceed those of $b$ form, such products will simplify to a concatenation of $b^{-1}$ 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.
$$\begin{array}{rcl} b \, b^{-1} \, b^{-1} \, b^{-1} \, b \, b^{-1} \, b \, b^{-1} & = & (b \, b^{-1}) \, b^{-1} \, (b^{-1} \, b) \, (b^{-1} \, b) \, b^{-1}\\ & = & I \, b^{-1} \, I \, I \, b^{-1}\\ & = & b^{-1} \, b^{-1}\\ & = & b^{-2} \end{array}$$Of course, there is a third possibility. It could be that pairing off the $b$ and $b^{-1}$ elementary braids and eliminating them leaves nothing but $I$ in the end, as the following suggests: $$\begin{array}{rcl} b \, b^{-1} \, b^{-1} \, b^{-1} \, b \, b \, b^{-1} \, b & = & (b \, b^{-1}) \, b^{-1} \, (b^{-1} \, b) \, (b \, b^{-1}) \, b\\ & = & I \, b^{-1} \, I \, I \, b\\ & = & b^{-1} \, b\\ & = & I \end{array}$$
Given the above, Let us now make the following two additional definitions:
The reason for making the last two definitions above stems from the following observation: With these two definitions, we can safely say that the following simplification rules will now always hold, regardless of what integers $p$ and $q$ are involved! (Note that without these definitions, products like $x_i^3 x_i^{-3}$ or $x_i^3 x_i^{-2}$ can't be simplified with these rules.)
$b^p \, b^q = b^{p+q}$ (add exponents when multiplying powers)
$(b^p)^q = b^{pq}$ (multiply exponents when finding a power of a power)
Just for clarity, please know that the above examples and rules hold for any braid $b$, including elementary braids of the form $b = x_i$ or $b = x_i^{-1}$. As examples, it must be the case by the last two rules that $x_4^3 x_4^5 = x_4^8$ and $(x_2^7)^3 = x_2^{21}$
Noting that the first rule (the one about adding exponents) forces $b^p \, b^{-q} = b^{p-q}$ to also be true for any integers $p$ and $q$.
Taking things one step further, recall 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. In a parallel way, we can 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 results for general braids which mirror the standard exponent rules with which the reader might already be familiar.
$\displaystyle{\frac{b^p}{b^q} = b^{p-q}}$ (subtract exponents when dividing powers)
$\displaystyle{b^{-p} = \frac{I}{b^p}}$ (negative exponents produce recipricals of powers) To see why, consider replacing $I$ with $b^0$.
$\displaystyle{\frac{I}{b_i b_j} = b_j^{-1} b_i^{-1}}$ (to invert a product, invert each and combine in reverse order) Consider $x_i x_j x_j^{-1} x_i^{-1}$.
We'll have more to say about the aforementioned standard exponent rules soon -- and why there is such a strong parallel between these and the rules we are developing for braids (and other things we will learn about shortly). For now however, let us push things even farther...
Note that we can say even more for commutative braids $b_i$ and $b_j$ or their inverses (e.g., "distant" elementary braids $x_i$ and $x_j$, where $|i-j| \ge 2$ or inverse pairs). In particular, for all integers $p$ the following simplification rules also must apply:
$(b_i b_j)^p = b_i^p b_j^p$ (exponents distribute over products of commutative elementary braids)
$\displaystyle{\left( \frac{b_i}{b_j} \right)^p = \frac{b_i^p}{b_j^p}}$ (exponents distribute over quotients of commutative elementary braids)
If $b_1$, $b_2$, $b_3$, and $b_4$ are all pairwise commutative, then
$\displaystyle{\frac{b_1 b_2}{b_3 b_4} = \frac{b_1}{b_3} \cdot \frac{b_2}{b_4}}$ (quotients of products can be expressed as products of quotients)To see how commutativity plays a role in these two rules, note that to establish the first (i.e., exponents distribute over products) we can use commutivity to get all the $x_i$ expressions together to consolidate them into a single power, as shown in the following example (look carefully between the 3rd and 4th expressions):: $$(b_i b_j)^2 = (b_i b_j)(b_i b_j) = b_i b_j b_i b_j = b_i b_i b_j b_j = b_i^2 b_j^2$$ Then, we can use the first rule to prove the second (i.e., where exponents are seen to distribute over quotients): $$\left(\frac{b_i}{b_j}\right)^2 = (b_i b_j^{-1})^2 = b_i^2 b_j^{-2} = \frac{b_i^2}{b_j^2}$$
As for the last rule, note how commutativity allows us to get from the third expression below to the fourth: $$\frac{b_1 b_2}{b_3 b_4} = b_1 b_2 (b_3 b_4)^{-1} = b_1 b_2 b_4^{-1} b_3^{-1} = b_1 b_3^{-1} b_2 b_4^{-1} = \frac{b_1}{b_3} \cdot \frac{b_2}{b_4}$$
Just be careful -- don't use any of these or other rules that rely on commutativity holding when it fails to hold!
While the results and definitions in the last section will let us prove certain pairs of braid expressions describe the same braid, sometimes doing so by direct application of these results and definitions can be quite long and tedius -- especially when we find ourselves applying the same sequence of manipulations over and over again. This of course is one reason why we prove theorems in mathematics -- to let us jump over all that repetition and make a useful (read that as "usable in many different contexts") conclusion from some given knowledge. We only need to prove the theorem in question holds.
As an example, for any given positive integer $i$, note how we can justify the following sequence of manipulations with what we know already: $$\begin{array}{rcll} x_i \, x_{i+1} \, x_i^{-1} & = & I \, x_i \, x_{i+1} \, x_i^{-1} & \scriptsize{\textrm{a property of the identity braid}}\\ & = & (x_{i+1}^{-1} \, x_{i+1}) \, x_i \, x_{i+1} \, x_i^{-1} & \scriptsize{\textrm{using inverse braids to form a "well-chosen value of "} I}\\ & = & x_{i+1}^{-1} \, (x_{i+1} \, x_i \, x_{i+1}) \, x_i^{-1} & \scriptsize{\textrm{associativity of braid concatenation}}\\ & = & x_{i+1}^{-1} \, (x_i \, x_{i+1} \, x_i) \, x_i^{-1} & \scriptsize{\textrm{Artin's Relation}}\\ & = & x_{i+1}^{-1} \, x_i \, x_{i+1} \, (x_i \, x_i^{-1}) & \scriptsize{\textrm{associativity of braid concatenation}}\\ & = & x_{i+1}^{-1} \, x_i \, x_{i+1} \, I & \scriptsize{\textrm{cancellation of inverse braids}}\\ & = & x_{i+1}^{-1} \, x_i \, x_{i+1} & \scriptsize{\textrm{a property of the identity braid}}\\ \end{array}$$
For lack of a better name, let us call this "Braid Theorem 1". That is to say,
|
Braid Theorem 1 |
Having now proven this theorem, we can employ it in future manipulations. For example, suppose we were trying to decide if the following two braid words represent the same braid: $$x_2 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} \quad \stackrel{\text{?}}{=} \quad x_2 \, x_3^{-1} \, (x_3^{-1} \, x_3) \, x_2 \, x_3^{-1} \, x_2 \, x_3$$
With the theorem we just proved, we can prove they are equivalent in just 4 steps instead of the 10 steps it would require without it! $$\begin{array}{rcll} x_2 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_2 \, x_3^{-1} \, x_2 \, (x_2 \, x_3 \, x_2^{-1}) & \scriptsize{\textrm{associativity of braid concatenation}}\\ & = & x_2 \, x_3^{-1} \, x_2 \, (x_3^{-1} \, x_2 \, x_3) & \scriptsize{\textrm{braid theorem 1}}\\ & = & x_2 \, x_3^{-1} \, I \, x_2 \, x_3^{-1} \, x_2 \, x_3 & \scriptsize{\textrm{a property of the identity braid}}\\ & = & x_2 \, x_3^{-1} \, (x_3^{-1} \, x_3) \, x_2 \, x_3^{-1} \, x_2 \, x_3 & \scriptsize{\textrm{using inverse braids to form a "well-chosen value of "} I}\\ \end{array}$$
Of course, proving some theorems can lead to wondering about others. For example, the expressions involved in the statement of Braid Theorem 1 are reminiscent of those in Artin's Relation -- except for the presence of an inverse elementary braid. What happens if the inverse elementary braid is in a different location -- say, on the first $x_i$ instead of the second? Because it would be "pretty" (due to the inherent symmetry involved), we might even hope that the following turns out to be true -- that for every positive integer $i$, we have $x_i^{-1} \, x_{i+1} \, x_i = x_{i+1} \, x_i \, x_{i+1}^{-1}$.
If this were true, we could remember both results as essentially an "Artin-like manipulation with a move of the inverse from one side to the other".
Of course, we don't know if this result holds yet -- we are only hopeful. We must prove it works for any positive integer $i$ before we can use it.
Let's try to argue similarly to how the last one was argued: $$\begin{array}{rcll} x_i^{-1} \, x_{i+1} \, x_i & = & x_i^{-1} \, x_{i+1} \, x_i \, I & \scriptsize{\textrm{a property of the identity braid}}\\ & = & x_i^{-1} \, x_{i+1} \, x_i \, (x_{i+1} \, x_{i+1}^{-1}) & \scriptsize{\textrm{using inverse braids to form a "well-chosen value of "} I}\\ & = & x_i^{-1} \, (x_{i+1} \, x_i \, x_{i+1}) \, x_{i+1}^{-1} & \scriptsize{\textrm{associativity of braid concatenation}}\\ & = & x_i^{-1} \, (x_i \, x_{i+1} \, x_i) \, x_{i+1}^{-1} & \scriptsize{\textrm{Artin's Relation}}\\ & = & (x_i^{-1} \, x_i) \, x_{i+1} \, x_i \, x_{i+1}^{-1} & \scriptsize{\textrm{associativity of braid concatenation}}\\ & = & I \, x_{i+1} \, x_i \, x_{i+1}^{-1} & \scriptsize{\textrm{cancellation of inverse braids}}\\ & = & x_{i+1} \, x_i \, x_{i+1}^{-1} & \scriptsize{\textrm{a property of the identity braid}}\\ \end{array}$$ Voila! We have shown that which we hoped to demonstrate -- we have proven our result!
By the way, in math textbooks one often sees the letters "QED." written at the end of a proof. This is merely an acronym for the Latin phrase "quod erat demonstrandum" which means "which was to be demonstrated".
Let us refer to this new theorem with an equally imaginative name -- say, "Braid Theorem 2":
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Braid Theorem 2 |
This is essentially how theorems in mathematics (involving braids or otherwise) get developed. First, somebody makes a good guess as to what they think should hold. Sometimes, this stems from some desire for the world to be "pretty" or symmetric. Other times, one makes some conjecture based on patterns they see hold in specific examples. Then, they try to form an argument that connects what they know to what they hope to show -- one that can be justified at every step and turn by things already accepted as true (e.g., definitions, postulates, other previously-proven theorems, etc).
One often tries to base these arguments on tricks or techniques that have worked in the past, much as we used the proof of Braid Theorem 1 as a guide to proving Braid Theorem 2. In this sense, mathematicians can build both on their own experience and on that of others. Interestingly however, the real fun for mathematicians begins when things they or others have used in the past fail to work! Working hard to find a new "wrinkle" on an old technique or creating a brand new way to argue something altogether -- that's where things get exciting!
Sadly, the presentation of theorems and their proofs in textbooks often leave out any discussion of the "blood, sweat, and tears" that went into a proof's initial construction. If you are lucky, they might briefly discuss the inspiration for the creative argument, but almost never explicitly address the author's excitement upon discovering that their novel approach bore fruit. Instead, the proof is often simply written as efficiently as possible -- often even omitting things deemed "easy enough" for the reader to fill in.
Part of this is to draw the reader's attention to the interesting (perhaps novel?) steps.
Sgt. Joe Friday
Perhaps the greatest motivation for this habit however, lies in the very thing that drove us to develop braid words and the above algebra they afford in the first place. In this, and in many, many more examples in the future, we will see that mathematicians seem to work very hard in everything they do to save themselves effort in the future. This includes being as brief and efficient as possible with anything they write.
As an example of this less verbose way of introducing a theorem and its proof, consider the following:
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Braid Theorem 3 Proof: $$\begin{array}{rcll} x_i \, x_{i+1}^{-1} \, x_i^{-1} & = & (x_{i+1}^{-1} \, x_{i+1}) \, x_i \, x_{i+1}^{-1} \, x_i^{-1} & \\ & = & x_{i+1}^{-1} \, (x_{i+1} \, x_i \, x_{i+1}^{-1}) \, x_i^{-1} & \\ & = & x_{i+1}^{-1} \, (x_i^{-1} \, x_{i+1} \, x_i) \, x_i^{-1} & \scriptsize{\textrm{ by braid theorem 2}}\\ & = & x_{i+1}^{-1} \, x_i^{-1} \, x_{i+1} \, (x_i \, x_i^{-1}) \\ & = & x_{i+1}^{-1} \, x_i^{-1} \, x_{i+1}\\ \end{array}$$ |
Notice in the above how the steps involving concatenation with $I$ were left out, much like one rarely writes a "multiplication by 1". Also, the more common applications of associativity or properties of inverses were left implicit, while the use of braid theorem 2 (i.e., the interesting step) was highlighted.
Just remember, when reading proofs presented in a "just the facts" style like the one above, it is your job as the reader to make sure you can justify each new step from the previous ones. To help you do this, you should always keep a pencil nearby when reading a math book!
Before continuing, you might try to conjecture and then prove results similar to braid theorems 1-3 that involve the following braids: $x_i^{-1} \, x_{i+1}^{-1} \, x_i$ and $x_i^{-1} \, x_{i+1}^{-1} \, x_i^{-1}$.