Exercises - Braids

  1. Draw braids on $5$ strands that match the given expressions:

    1.   $\displaystyle{x_1^{\phantom{1}} x_2^{-1} x_1^{-1} x_4^{\phantom{1}} x_3^{\phantom{1}} x_1^{-1} x_1^{-1}}$
    2.   $\displaystyle{x_3^{\phantom{1}} x_2^{-3} x_1^{\phantom{1}} x_4^{\phantom{1}} x_2^{\phantom{1}}}$
    3.   $\displaystyle{x_4^{3} x_1^{-3}}$
    4.   $\displaystyle{x_3^{\phantom{1}} x_2^{-1} x_1^{\phantom{1}} I x_1^{-1} x_2^{\phantom{1}} x_3^{-1}}$

    a) b)
    c) d)

  2. Write an expression to match each braid shown below:

    a) b)
    c) d)

    1.   $x_3^{-1} x_2^{\phantom{1}} x_4^{-1} x_3^{\phantom{1}} x_4^{-3} x_1^{-1}$
    2.   $x_3^{\phantom{1}} x_2^{-2} I I x_2^{2} x_3^{-1}$
    3.   $x_1^{-1} x_3^4 x_2^{-2}$
    4.   $x_1^{\phantom{1}} x_2^{-1} x_3^{\phantom{1}} x_4^{-1} x_1^{-1} I x_2^{\phantom{1}}$
  3. 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.

  4. 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.

    1. $x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1} \quad \textrm{ and } \quad x_3^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}}$

    2. $x_4^{\phantom{1}} x_2^{-2} x_2^{2} x_4^{-1} \quad \textrm{ and } \quad x_4^{\phantom{1}} x_2^{-2} x_2^{\phantom{1}} x_4^{-1} x_2^{\phantom{1}}$

    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.

    2. 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.

  5. Each expression below can be simplified to a single elementary braid -- find it.
    1.   $x_1^{\phantom{1}} x_1^{-1} x_2^{\phantom{1}} x_3^{\phantom{1}} x_3^{-1}$

    2.   $x_2^{-1} x_4^{\phantom{1}} x_3^{-1} x_3^{\phantom{1}} x_4^{-1} x_2^{\phantom{1}} x_1^{\phantom{1}}$

    3.   $x_3^{-2} x_3^{4} x_1^{\phantom{1}} x_3^{-2}$     (Hint: use the commutativity of distant braids)

    4.   $(x_2^{5})^{-3} (x_2^{2})^{7}$

    5.   $x_2^{\phantom{1}} x_3^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1} x_2^{-1}$     (Hint: use Artin's Relation)

    1.   $x_2$, after appealing to the inverse property twice

    2.   $x_1$, after appealing to the inverse property three times

    3.   $x_1$, after using commutativity of distant braids twice to get the $x_1$ on the far right end

    4.   $x_2^{-1}$, after expanding both powers of powers, and applying the inverse property repeatedly

    5.   $x_3$, 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

  6. Use what you know to algebraically manipulate each of the following into the form $x_2^{a} x_4^{b} x_6^{c} x_8^{d}$ for some suitable $a$, $b$, $c$, and $d$, presuming both represent braids on $10$ strands (Hint: do all the elementary braids present below commute with one another?):
    1.   $\displaystyle{(x_2^3 x_4^{-1} x_6^2 x_8^{-5})^{3}}$

    2.   $\displaystyle{\frac{x_2^{\phantom{1}} x_6^{3} x_8^{-1}}{x_4^{-3} x_6^{5}}}$

    3.   $\displaystyle{\frac{x_2^3 (x_4^5 x_6^{-2})^3}{x_6^{\phantom{1}} (x_8^4 x_4^{-3})^{-2}}}$

    4.   $\displaystyle{\left(\frac{x_2^{\phantom{1}} x_4^3}{x_8^0 x_4^2}\right)^2 \div \frac{x_4^{\phantom{1}}}{x_2^{-1}x_6}}$

    1.   $x_2^9 x_4^{-3} x_6^6 x_8^{-15}$

    2.   $x_2^{\phantom{1}} x_4^3 x_6^{-2} x_8^{-1}$

    3.   $x_2^3 x_4^9 x_6^{-7} x_8^8$

    4.   $x_2^1 x_4^1 x_6^1 x_8^0$   (which is more simply written as $x_2 x_4^{\phantom{1}} x_6^{\phantom{1}}$, of course)

  7. 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.

  8. Which of the following represents a combed braid? $$x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1} \quad \textrm{ vs. } \quad x_1^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1} x_3^{\phantom{1}} x_1^{\phantom{1}} x_4^{-1}$$

    $x_3^{\phantom{1}} x_2^{-2} x_2^{2} x_3^{-1}$
    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 a combed pure braid

    Looking at the first, we check to see if all the first crossings involve the first strand (red). As red never crosses anything, we move onto the green strand. However, the first crossing does not involve the green strand -- it involves the blue and orange strands. As such, the first braid is not combed.

    Looking at the second, we check to see if all the first crossings involve the first strand (red). They do -- and as is also required -- there are no more crossings involving the red strand after these initial ones. Then, we check to see if all the crossings with the second strand that remain happen next. They do. Noting that the blue strand is not involved in any crossings past that point, we can ignore this third strand. The fourth (orange) strand is next. Sure enough, the next group of crossings (there is actually only one crossing left) involves this orange strand. The last strand (which is black here) never plays into whether the braid is combed, as if it is involved in any crossings, they have already been accounted for.

    Despite all the second braid matching so many criteria for a combed braid, we have only defined what we mean by "combed" in the context of pure braids -- i.e., braids whose order of colored strands at the beginning and order of colored strands at the end match. Consider the order we start with in this braid (red, green, blue, orange, and black) versus the order with which the braid ends (blue, green, red, black, orange). As these differ, this is not a pure braid. Without a definition for what combed non-pure braids should look like, we play it safe and say "this is not a combed pure braid".

  9. Prove using the algebraic properties of braids we've developed that the following pairs of braids on 4 strands are equivalent:

    1. $x_2^{-1} x_1^{\phantom{1}} x_3^{\phantom{1}} x_3^{-1} x_2^{\phantom{1}} x_1^{\phantom{1}}$   and   $x_1 x_2$

    2. $x_1 x_2 x_1 x_2 x_1 x_1 x_1$   and   $x_1 x_2 x_2 x_2 x_2 x_1 x_2$

    3. $x_3^{\phantom{1}} x_2^{\phantom{1}} x_3^{-1}$   and   $x_2^{-1} x_3^{\phantom{1}} x_2^{\phantom{1}}$

    1. Use inverses and then Artin's relation.

    2. Use Artin's relation repeatedly.

    3. 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.)

  10. 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".

  11. 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.