Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

migrated from Aug 23 '13 at 19:09

This question came from our site for professional mathematicians.

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. – Ryan Budney Aug 23 '13 at 17:06
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 – Ryan Budney Aug 23 '13 at 17:20
@RyanBudney Thank you for the responses. Now I am wondering what what constitutes a question for math.stackexchange as opposed to mathoverflow? – N. Owad Aug 23 '13 at 17:37
math.stackexchange is meant for any kind of mathematics. MathOverflow is oriented more towards research-related problems. – Ryan Budney Aug 23 '13 at 17:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.