Combing Braids

Some braids might simplify to an elementary braid, but most will not be that nice. This begs the question -- how do we simplify more complicated braids?

On the one hand, one could attempt to simply reduce the number of crossings to a minimum. However, there is another form for a braid that focuses on simplifying the organization of crossings (even at the cost of a longer braid word). The process for getting to this useful form (called the braid's normal form) is called "combing" a braid.

We will consider this process for an important subset of braids called pure braids where the order of the strands at the beginning of the braid and at the end of the braid is the same. For example, the braid below is a pure braid, as the order of colors on the far left and far right are both: red, green, blue, and then orange (from bottom to top).

The combed version of the braid above is shown below. While it is a bit longer in terms of number of crossings, is organizationally simpler.

Notice how all crossings in the first (pink) region below involve the red strand and there are no crossings involving the red strand after this pink region. Then, all crossings in the second (light green) region involve the green strand and there are no crossings involving the green strand after this light green region. Finally, all crossings in the third (light blue) region involve the blue strand and there are no crossings involving the blue strand after this light blue region (or any other strand since we are now at the end).

To see how to turn the first (uncombed) braid into the second (combed) one, consider the following sequence of actions...

First, we color all strands of the uncombed braid black, with the exception of the red strand, and add vertical lines to make finding the algebraic expression for this braid easier to identify.

Notice there are two crossings that don't involve the red strand (see the green and blue rectangles) that occur before the red strand is done interacting with other strands. Our goal is to eliminate these.

$$x_2 \, x_1 \, x_2 \, x_2 \, x_3^{-1} \, x_1 \, x_1 \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1}$$

The blue crossing is the easiest to remove, as the commutivity of distant braids (applied 3 times) allows us to move the blue crossing toward the yellow region (where the red strand no longer crosses anything). Note in doing this, we appeal multiple times to the associativity of braids which allows us to group and regroup things (with parentheses) without altering the overall braid.

$$\begin{array}{rcl} x_2 \, x_1 \, x_2 \, x_2 \, x_3^{-1} \, x_1 \, x_1 \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_2 \, x_1 \, x_2 \, x_2 \, (x_3^{-1} \, x_1) \, x_1 \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, (x_1 \, x_3^{-1}) \, x_1 \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, (x_3^{-1} \, x_1) \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, (x_1 \, x_3^{-1}) \, x_1 \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, x_1 \, (x_3^{-1} \, x_1) \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, x_1 \, (x_1 \, x_3^{-1}) \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ \end{array}$$

Artin's relation can be applied to let us slide the green crossing to the right -- initially under the red strand..

$$\begin{array}{rcl} x_2 \, x_1 \, x_2 \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & (x_2 \, x_1 \, x_2) \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & (x_1 \, x_2 \, x_1) \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_1 \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} \end{array}$$

And then over the red strand a bit farther to the right..

$$\begin{array}{rcl} x_1 \, x_2 \, x_1 \, x_2 \, x_1 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_1 \, x_2 \, (x_1 \, x_2 \, x_1) \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, (x_2 \, x_1 \, x_2) \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_2 \, x_1 \, x_2 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ \end{array}$$

Even this last leg of the journey for the green crossing as it moves to the right can be accomplished with two more applications of Artin's relation:

$$\begin{array}{rcl} x_1 \, x_2 \, x_2 \, x_1 \, x_2 \, x_1 \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_1 \, x_2 \, x_2 \, (x_1 \, x_2 \, x_1) \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_2 \, (x_2 \, x_1 \, x_2) \, x_1 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_2 \, x_2 \, (x_1 \, x_2 \, x_1) \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_2 \, x_2 \, (x_2 \, x_1 \, x_2) \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ & = & x_1 \, x_2 \, x_2 \, x_2 \, x_2 \, x_1 \, x_2 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1}\\ \end{array}$$

With the red strands interactions with other strands now exhausted, we can ignore the red strand and focus entirely on the crossings that remain. These have been lumped together and highlighted in purple below. Notice how if we now color the green strand (i.e., the strand that started in position 2), we can essentially start the same process all over again. This time (in the purple region) we want to see only crossings involving the green strand, until those are exhausted.

To understand how to algebraically eliminate the two "black-on-black" crossings above, let us first recall braid theorems 1-3, as previously proven:

Braid Theorem 1: For all positive integers $i$, we have $x_i \, x_{i+1} \, x_i^{-1} = x_{i+1}^{-1} \, x_i \, x_{i+1}$
Braid Theorem 2: For all positive integers $i$, we have $x_i^{-1} \, x_{i+1} \, x_i = x_{i+1} \, x_i \, x_{i+1}^{-1}$
Braid Theorem 3: For all positive integers $i$, we have $x_i x_{i+1}^{-1} x_i^{-1} = x_{i+1}^{-1} \, x_i^{-1} \, x_{i+1}$

Armed with these theorems, let us now consider one algebraic way to manipulate the purple crossings seen above (shown again below, on the left) into an equivalent form (on the right) where all and only the crossings involving the green strand happen first.

One should note that while each step provided should be easy to follow from the previous step, deciding which steps to apply and their order takes some very careful thinking indeed!

$$\begin{array}{rcl} x_2 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_2 \, x_3^{-1} \, x_2 \, (x_2 \, x_3 \, x_2^{-1})\\ & = & x_2 \, x_3^{-1} \, x_2 \, (x_3^{-1} \, x_2 \, x_3)\\ & = & x_2 \, x_3^{-1} \, I \, x_2 \, x_3^{-1} \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, (x_3^{-1} \, x_3) \, x_2 \, x_3^{-1} \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, (x_3 \, x_2 \, x_3^{-1}) \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, (x_2^{-1} \, x_3 \, x_2) \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, I \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, (x_3^{-1} \, x_3) \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, x_3^{-1} \, I \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, x_3^{-1} \, (x_2^{-1} \, x_2) \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, (x_2 \, x_3^{-1} \, x_2^{-1}) \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, (x_3^{-1} \, x_2^{-1} \, x_3) \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, (x_3 \, x_2 \, x_3^{-1}) \, x_2^{-1} \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, (x_2^{-1} \, x_3 \, x_2) \, x_2^{-1} \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, (x_2 \, x_2^{-1}) \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, I \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, x_3 \, x_2 \, x_3 \, x_3\\ \end{array}$$

Substituting the second (purple) crossings for those seen in the first, we now have our original braid equivalent to the following

At this point, notice that all of the crossings involving the green strand and only the crossings involving the green strand now occur together, which was what we desired.

With the red and green strands interactions now exhausted, we ignore both of these and again focus on the crossings that remain. These have again been highlighted in a (new) purple region, with the next thread to be considered colored blue.

Had there been crossings in this new purple region that did not involve the blue strand, we would have played a similar game again to move these to the right. However, currently there are no such black-on-black crossings left. So there is nothing to do for blue.

The last strand of course has nothing else to cross with, so it is done as well. We only color it orange to match the braid diagram colors seen before this combing process began.

Artin proved that for any braid on $n$ strands, one can always find an equivalent braid in this "normal form" (again where all the crossings involving strand 1 happen first, then crossings involving strand 2, then the same for strand 3, etc.)

That said, and as the calculations above demonstrate, the braid manipulations required can be numerous and non-obvious. Artin probably put it best when he humorously wrote:

"Although it has been proved that every braid can be deformed into a similar normal form the writer is convinced that any attempt to carry this out on a living person would only lead to violent protests and discrimination against mathematics. He would therefore discourage such an experiment." -- Artin 1946