How to see isometries of figure 8 knot complement The figure 8 knot complement $M$ is the orientable double cover of the Gieseking manifold, which implies that $M$ has a fixed-point free involution. If we think of $M$ with its hyperbolic metric, this involution is an isometry. Is there some way of visualizing this isometry?
I know how to produce one relatively easy to see isometry of $M$. The fundamental group of $M$ is a two generator group. Lift geodesic representatives of a pair of generators to $\mathbb{H}^3$. These geodesics have a mutual perpendicular, and $180^{\circ}$ rotation about that geodesic descends to an involution of $M$. However, this map has fixed points, and I'd like to "see" one that doesn't.
 A: In terms of 'seeing' the isometry $\rho$ needed to realize the Gieseking manifold as a quotient of the figure 8 knot complement $S^3-K$, we require a slight of hand, namely thinking of another name for the manifold. The reason for this is simple, the Gieseking manifold is non-orientable so $\rho$ must reverse orientation. However, there are no non-orientable manifolds covered by $S^3$, so we will not be able to 'see' $\rho$ acting on $S^3$.
Instead, we can think of $S^3-K$ by another name as follows. The figure 8 knot complement is fibered. In fact, it can be thought of as a once-punctured torus bundle with monodromy given by $\pmatrix{2 & 1\\ 1 & 1}$. It turns out $\rho$ acts on once-punctured torus bundle by a glide reflection (loosely translate 'half-way' up and reflect).     
A: If you're interested in the involution only defined on the complement, Igor's answer does a fine job. 
But the involution extends to an involution of $S^3$ and perhaps you'd like to see that? 
I think the symmetry is a little tricky to traditionally visualize, because as a map of $S^3$, thought of as the unit sphere $S^3 \subset \mathbb C^2$ it's of the form $(z_1,z_2) \longmapsto (\overline{z_1}, -z_2)$.  If you stereographically project at one of these fixed points, this becomes the involution of $\mathbb R^3$ given by $(x,y,z) \longmapsto (-x,-y,-z)$.  Both the fixed points have to be on the knot, so this means we have to find a "long" embedding of the figure-8 knot in $\mathbb R^3$ which is invariant under the antipodal map.   That's easy.   Here are two such (approximate) positions:


In the latter picture the fixed point is easier to see -- the blue straight line intersects the knot in 5 points, one at the "center" and four other points indicated by blue balls. 
Looking through the code I used to generate the latter picture, the parametrization I use of the figure-8 knot is:
$$t \longmapsto (5(t^3-3t), 0.25(t^7-42t), t^5-5t^3+4t)$$
which is easy to verify is equivariant with respect to the above involution, as all the terms in the polynomial are odd. 
A: The figure 8 complement decomposes into 2 regular ideal tetrahedra (see Thurston's notes, here for example.) This gives the involution quite explicitely.
