| | |
Draw braids on $5$ strands that match the given expressions:
a) |
b) |
c) |
d) |
Write an expression to match each braid shown below:
a) |
b) |
c) |
d) |
Do the first two elementary braids in the expression $x_3^{-1} x_2^{\phantom{1}} x_4^{-1} x_3^{\phantom{1}} x_4^{-1} x_1^{-1}$ commute? Do the last two commute? Explain.
The first two, $x_3^{-1}$ and $x_2^{\phantom{1}}$, are not "distant" elementary braids (i.e., $|3-2| \not \ge 2$) and thus fail to commute. The last two, $x_4^{-1}$ and $x_1^{-1}$, are "distant" elementary braids (i.e., $|4-1| \ge 2$) and thus do commute.
For each pair of braid words given (where both represent braids on $5$ strands), decide if they both can be simplified to the same expression. If they can, show how to accomplish this algebraically. If they can't, make an argument as to why.
$x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1} \quad \textrm{ and } \quad x_3^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}}$
$x_4^{\phantom{1}} x_2^{-2} x_2^{2} x_4^{-1} \quad \textrm{ and } \quad x_4^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_4^{-1} x_2^{\phantom{1}}$
These two braid words can not be simplified to the same expression. To see this, draw the braids and note that the strands end up in different places. For example, the green strand below ends up in the same position as it started in the first braid, but where the orange strand started in the second braid.
Make sure you don't read too much into this argument, however. A different ordering of colored strands on the right is a sufficient but not necessary condition for two braids to be different.
That is to say, just because two braids have the same ordering of colors on the right does NOT mean they are the same braid.
Examples abound -- consider $I$ vs. $x_1^2$ on just two strands, for instance. There is no way to "untwist" the second into the first, despite the fact that when drawn with each strand a different color -- the order of colors on the far right is the same for both.
These two braid words can be simplified to the same expression -- as (verbosely) justified by the following. (Note: you need not show all the steps below to effectively justify this claim, but they are included here for clarity.) $$\begin{array}{rcll} x_4^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_4^{-1} x_2^{\phantom{1}} &=& x_4^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_2^{\phantom{1}} x_4^{-1} & \textrm{by commutativity of distant braids}\\ &=& x_4^{\phantom{1}} (x_2^{-1} x_2^{-1}) x_2^{\phantom{1}} x_2^{\phantom{1}} x_4^{-1}&\textrm{"powers" of braids}\\ &=& x_4^{\phantom{1}} x_2^{-1} (x_2^{-1} x_2^{\phantom{1}}) x_2^{\phantom{1}} x_4^{-1}& \textrm{associativity}\\ &=& x_4^{\phantom{1}} x_2^{-1} I x_2^{\phantom{1}} x_4^{-1}& \textrm{inverses combine to give the identity}\\ &=& x_4^{\phantom{1}} x_2^{-1} x_2^{\phantom{1}} x_4^{-1}& \textrm{property of identity ($I x_2 = x_2$)}\\ &=& x_4^{\phantom{1}} (x_2^{-1} x_2^{\phantom{1}}) x_4^{-1}& \textrm{associativity}\\ &=& x_4^{\phantom{1}} I x_4^{-1}&\textrm{inverses again} \\ &=& x_4^{\phantom{1}} x_4^{-1}&\textrm{property of identity ($I x_4 = x_4$)}\\ &=& I&\textrm{inverses again}\\\\\hline\\ x_4^{\phantom{1}} x_2^{-2} x_2^{2} x_4^{-1} &=& x_4^{\phantom{1}} (x_2^{-1} x_2^{-1}) x_2^{2} x_4^{-1}&\textrm{"powers" of braids}\\ &=& x_4^{\phantom{1}} x_2^{-1} x_2^{-1} (x_2^{\phantom{1}} x_2^{\phantom{1}}) x_4^{-1}&\textrm{"powers" of braids}\\ &=& x_4^{\phantom{1}} x_2^{-1} (x_2^{-1} x_2^{\phantom{1}}) x_2^{\phantom{1}} x_4^{-1}&\textrm{associativity} \end{array}$$ Note this last expression (after applying associativity) was identical to the third step in the previous calculation. As such, from here we can simplify in the same way to get $I$ for this braid word as well.
$x_1^{\phantom{1}} x_1^{-1} x_2^{\phantom{1}} x_3^{\phantom{1}} x_3^{-1}$
$x_2^{-1} x_4^{\phantom{1}} x_3^{-1} x_3^{\phantom{1}} x_4^{-1} x_2^{\phantom{1}} x_1^{\phantom{1}}$
$x_3^{-2} x_3^{4} x_1^{\phantom{1}} x_3^{-2}$ (Hint: use the commutativity of distant braids)
$(x_2^{5})^{-3} (x_2^{2})^{7}$
$x_2^{\phantom{1}} x_3^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1} x_2^{-1}$ (Hint: use Artin's Relation)
$x_2$, after appealing to the inverse property twice
$x_1$, after appealing to the inverse property three times
$x_1$, after using commutativity of distant braids twice to get the $x_1$ on the far right end
$x_2^{-1}$, after expanding both powers of powers, and applying the inverse property repeatedly
$x_3$, after replacing $x_2 x_3 x_2$ with $x_3 x_2 x_3$ via Artin's relation, and then applying the inverse property twice
$\displaystyle{(x_2^3 x_4^{-1} x_6^2 x_8^{-5})^{3}}$
$\displaystyle{\frac{x_2^{\phantom{1}} x_6^{3} x_8^{-1}}{x_4^{-3} x_6^{5}}}$
$\displaystyle{\frac{x_2^3 (x_4^5 x_6^{-2})^3}{x_6^{\phantom{1}} (x_8^4 x_4^{-3})^{-2}}}$
$\displaystyle{\left(\frac{x_2^{\phantom{1}} x_4^3}{x_8^0 x_4^2}\right)^2 \div \frac{x_4^{\phantom{1}}}{x_2^{-1}x_6}}$
$x_2^9 x_4^{-3} x_6^6 x_8^{-15}$
$x_2^{\phantom{1}} x_4^3 x_6^{-2} x_8^{-1}$
$x_2^3 x_4^9 x_6^{-7} x_8^8$
$x_2^1 x_4^1 x_6^1 x_8^0$ (which is more simply written as $x_2 x_4^{\phantom{1}} x_6^{\phantom{1}}$, of course)
Prove using the algebraic properties of braids we've developed that the following pairs of braids on 4 strands are equivalent:
$x_2^{-1} x_1^{\phantom{1}} x_3^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}} x_1^{\phantom{1}}$ and $x_1 x_2$
$x_1 x_2 x_1 x_2 x_1 x_1 x_1$ and $x_1 x_2 x_2 x_2 x_2 x_1 x_2$
$x_3^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1}$ and $x_2^{-1} x_3^{\phantom{1}} x_2^{\phantom{1}}$
Use inverses and then Artin's relation.
Use Artin's relation repeatedly.
Try concatenating what is given to a "well-chosen-value of $I$" on the left so that Artin's relation can be used. (Note, this is not disimilar to how we often add a well-chosen value of zero to an expression, or multiply it by a well-chosen value of $1$, as a trick to simplifying some expressions in algebra.)
Prove the braid property that $x_i^{-1} x_{i+1}^{-1} x_i^{\phantom{1}} = x_{i+1}^{\phantom{1}} x_i^{-1} x_{i+1}^{-1}$.
Draw both braids above and think about what you do with the strings to turn one into the other. Let these manipulations inform what you do algebraically in your proof, and model your argument in a similar way to the proofs of "Braid Theorem 1", "Braid Theorem 2" and "Braid Theorem 3" presented in the Combing Braids section.
Below is a sequence of elementary braid operations and/or applications of braid theorems proved in the section that have been made during one part of the combing a complicated braid. Identify the sequence of braid operations/theorems applied in this process. As an example, the first step is an appeal to associativity, the second is an application of Braid Theorem 2, the third uses the property of an identity, and so on..
$$\begin{array}{rcl} x_2 \, x_3^{-1} \, x_2 \, x_2 \, x_3 \, x_2^{-1} & = & x_2 \, x_3^{-1} \, x_2 \, (x_2 \, x_3 \, x_2^{-1})\\ & = & x_2 \, x_3^{-1} \, x_2 \, (x_3^{-1} \, x_2 \, x_3)\\ & = & x_2 \, x_3^{-1} \, I \, x_2 \, x_3^{-1} \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, (x_3^{-1} \, x_3) \, x_2 \, x_3^{-1} \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, (x_3 \, x_2 \, x_3^{-1}) \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, (x_2^{-1} \, x_3 \, x_2) \, x_2 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, I \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, (x_3^{-1} \, x_3) \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, x_3^{-1} \, I \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, x_2 \, x_3^{-1} \, (x_2^{-1} \, x_2) \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, (x_2 \, x_3^{-1} \, x_2^{-1}) \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_3 \, x_2 \, (x_3^{-1} \, x_2^{-1} \, x_3) \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, (x_3 \, x_2 \, x_3^{-1}) \, x_2^{-1} \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, (x_2^{-1} \, x_3 \, x_2) \, x_2^{-1} \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, (x_2 \, x_2^{-1}) \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, I \, x_3 \, x_2 \, x_3 \, x_3\\ & = & x_2 \, x_3^{-1} \, x_3^{-1} \, x_2^{-1} \, x_2^{-1} \, x_3 \, x_3 \, x_2 \, x_3 \, x_3\\ \end{array}$$The fourth step involves using the properties of inverses (backwards). Note, this can be interpreted in the same way as "multiplying by a well-chosen value of one" -- except here, we would say "concatenating by a well-chosen representation of the identity braid".
Which of the following represents a pure braid? $$x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1} \quad \textrm{ vs. } \quad x_3^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}}$$
 |
 |
| $x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1}$ pure |
$x_3^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}}$ not pure |
The first is a pure braid as the order of the colored strands at the beginning and end of the braid agree, while for the second braid this is not true (the green, blue, and orange strands change order) -- so the second braid is not a pure braid.
Which of the following represents a combed braid? $$x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1} \quad \textrm{ vs. } \quad x_1^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1} x_3^{\phantom{1}} x_1^{\phantom{1}} x_4^{-1} \quad \textrm{ vs. } \quad x_1 x_2^{-1} x_2^{-1} x_1 x_2^{-1} x_2^{-1} x_3 x_3 x_4 x_4$$
|
 |
 |
| $x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1}$ not combed |
$x_1^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1} x_3^{\phantom{1}} x_1^{\phantom{1}} x_4^{-1}$ not combed |
$x_1 x_2^{-1} x_2^{-1} x_1 x_2^{-1} x_2^{-1} x_3 x_3 x_4 x_4$ a combed braid! |
With regard to the first braid, we should check to see if it is a pure braid first. We start with this as doing so is easy, and if we don't have a pure braid it can't possibly be combed braid (as we haven't defined what combed means for non-pure braids). Noting the order of the colored strands at the beginning and end of the braid match (from bottom to top: red, green, blue, orange, black), we deduce it is indeed pure.
However, we need more to establish this braid as combed. We look at the first strand (red), and check to see if all the first crossings (up to some point) involve the red strand, and then after that point no crossings involving red occur. Here, red never crosses anything -- which (vacuously) satisfies this requirement. Then we move onto the second (green) strand, starting where we left off with red one (i.e., still at the beginning), and do the same thing. Unfortunately, the first crossing does not involve the green strand -- it involves the blue and orange strands -- but later ones do. As such, the first braid is not combed.
As for the second braid -- we again first check if it is a pure braid. Consider the order of colored strands with which we start the braid (from bottom to top: red, green, blue, orange, and black) as opposed to the order of the same when braid ends (from bottom to top: blue, green, red, black, orange). As these differ, this is not a pure braid -- and consequently not a combed braid.
Finally, in the third braid, we notice it is indeed a pure braid as the order of colored strands at the beginning and end match (i.e., from bottom to top: red, green, blue, orange, black). Further, the first 3 crossings all involve the first (red) strand with no red crossings seen past that, the next two involve the second (green strand) with no green crossings past that, the two after that both involve the third (blue) strand with no blue crossings after that, and finally the last two crossings involve the orange strand. Thus, this is a combed braid!