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I have been working through Rolfsen's "Knots and Links" and have found myself frustrated by exercise 4 on page 58. It concerns the Wirtinger Presentation of the figure eight knot, where the (unsimplified) knot group can be found with the four generators $x_1,\ldots,x_4$ and relations: $$ x_1 x_3 = x_3 x_2 $$ $$ x_4 x_2 = x_3 x_4 $$ $$ x_3 x_1 = x_1 x_4 $$ $$ x_2 x_4 = x_1 x_2. $$ The exercise says "Show combinatorially that the fourth relation is a consequence of the other three." Originally, I assumed this was just a simple algebraic manipulation, but after a while of no success, I am assuming that I am going about this the wrong way. If I just am missing the simple algebra solution, I would be satisfied and move on. Otherwise, what kind of "combinatorial" argue is intended here? I know that you are always suppose to be able to eliminate one of your relations, but cannot see how in this case. Thank you in advance for any response.

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    $\begingroup$ This question will be moved to math.stackexchange soon. FYI, any one of the relations is a consequence of the remaining relations. And this is something that holds for any Wirthinger presentation of any knot. It's probably best if you write the relations in the form $x_i = x_j^{\pm} x_k x_j^{\mp}$ and write them in the order where the $x_i$'s follow the cyclic ordering of the knot. $\endgroup$ Aug 23, 2013 at 17:06
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    $\begingroup$ But there is a nice geometric reason for this fact, as well as a "you can just compute it" reason. The above comment leads you to the computation. To see it geometrically, think of the 3-sphere as the 1-point compactification of $\mathbb R^3$, and think of a closed knot in $S^3$ as a 1-point compactification of a "long knot", meaning an embedding of $\mathbb R$ in $\mathbb R^3$ that "goes to infinity" in a controlled (linear) way. You then argue that the complement of the long knot in $\mathbb R^3$ is homeomorphic to the complement of the closed knot in $S^3$, then compare the $\pi_1$ pres $\endgroup$ Aug 23, 2013 at 17:20
  • $\begingroup$ @RyanBudney Thank you for the responses. Now I am wondering what what constitutes a question for math.stackexchange as opposed to mathoverflow? $\endgroup$
    – N. Owad
    Aug 23, 2013 at 17:37
  • $\begingroup$ math.stackexchange is meant for any kind of mathematics. MathOverflow is oriented more towards research-related problems. $\endgroup$ Aug 23, 2013 at 17:40

1 Answer 1

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We do algebraic manipulation on the first three equations to derive the fourth one. equation (1) is equivalent to the following $x_{1} = x_{3}x_{2}x_{3}^{-1}$.\ equation (2) is equivalent to $x_{4}x_{2}x_{4}^{-1} = x_{3}$\

Now put eqn(1) into eqn(3) to get: \begin{align*} & x_{4}x_{2}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{4}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow\;\;\;\; & x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow\;\;\;\; & x_{4}x_{2}x_{4}^{-1}x_{2}x_{4}x_{2}^{-1}x_{4}^{-1} = x_{2}x_{4}x_{2}^{-1}\\ \end{align*} But, $x_{4}x_{2}x_{4}^{-1} = x_{3}, x_{4}x_{2}^{-1}x_{4}^{-1} = x_{3}^{-1}$, so we have

\begin{align*} & x_{3}x_{2}x_{3}^{-1} = x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow \;\;\;& x_{1} = x_{2}x_{4}x_{2}^{-1}\\ \Longleftrightarrow \;\;\;& x_{1}x_{2} = x_{2}x_{4} \end{align*}

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