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.
[Math] Wirtinger Presentation for the Figure eight knot, Rolfsen exercise
knot-theory
Related Solutions
Here is a picture taken from Out of Line "Paths and Knot Spaces"
There is more discussion at Topology and Groupoids p,350, and this simple crossing diagram in some sense assumes one is using the fundamental groupoid: insistence on one base points is not natural to the knot situation.
I have demonstrated the crossing relation to children using a copper pentoil and rope, and ended up with this string wrapping on the pentoil:
and asking one of the children to show how the loop comes off the knot!
First, welcome to MSE!
And yes, the connect sum of two fibered knots is fibered.
This can be seen relatively easily by considering the topological definition of connect sum. Place your two separate knots inside an $S^2$ each. Then pull a small arc with no knotting outside of the sphere. Then we create the connect sum by taking the interior of the two spheres and gluing them together along their boundary so that the two points where the knots intersect the sphere line up.
Fibered knots are knots which have a Seifert Surface that can rotate around the knot. Since the knots originally had this property, then there was a small part of that Seifert Surface that would have been on the outside of those spheres which also did this. This was a disk with part of its boundary on the sphere's boundary. We just need to line these boundaries up on each sphere, which is always possible since two curves with common endpoints on a sphere are isotopic.
Hope this helps!
Best Answer
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*}