# The figure eight knot complement in $S^3$.

Recently I have been going through the book Hyperbolic Knot Theory by Jessica Purcell. In this book, there is an exercise in chapter 5, section 5.6, and exercise number 5.4 (Page - 101). This exercise gives us a presentation of the fundamental group of figure-eight knot complements in $$S^3$$ the presentation is as follows $$\pi_1(S^3-K)=\langle a,b : yay^{-1}=b\rangle,$$ where $$y=a^{-1}bab^{-1}$$ ($$K=$$fig-eight knot) and also the representation in $$PSL(2,\mathbb{C})$$ is given by the matrices

$$A =\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$$,

$$B=\begin{bmatrix} 1 & 0\\ -\omega & 1 \end{bmatrix}$$,

where $$\omega ^3=1$$,

and $$\Gamma =\langle A,B\rangle$$, and $$\Gamma$$ act on $$\mathbb{H}^3$$ by isometry (Here $$\mathbb{H}^3$$ upper half space model ) Now my question is does there exist two elements in $$\Gamma$$ such that one of them is hyperbolic say $$\alpha$$ ( $$trace^2$$ is real and $$>4$$) and another on loxodromic say $$\beta$$ ($$trace^2$$ not in the interval $$[0, \infty)$$ )but not hyperbolic such that fixed points of $$\alpha$$ and $$\beta$$ are in the same line and the geodesics passing through the fixed points are not intersecting each other? (More elaborately I can say geodesics $$g_{\alpha}$$ passing through the fixed points of $$\alpha$$ and $$g_{\beta}$$ the geodesic passing through the fixed points of $$\beta$$, $$g_{\alpha}$$ and $$g_{\beta}$$ lie in the same plane but they do not intersect.) NB="lie in the same plane" I mean that it is a hyperbolic plane in upper half-space model

• The matrix $B= \begin{bmatrix} 1 & 0\\ -\omega & 1 \end{bmatrix}$ May 20 at 10:23