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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.

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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:

alt text

alt text

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.

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Nice! What did you use to generate the picture? –  auniket Nov 19 '11 at 0:45
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The latter image was generated in two steps. The knot and the line were computed in some C++ code that I wrote. Then the 3-dimensional data was output into a PoVRay format, which was then rendered in PoVray: povray.org The top image I just grabbed in a Google search. –  Ryan Budney Nov 19 '11 at 1:49
    
Maybe I'm confused; isn't the OP trying to visualize a fixed-point free involution? –  j.c. Nov 19 '11 at 6:14
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The only fixed points are on the knot -- it's free on the knot complement. –  Ryan Budney Nov 19 '11 at 6:24
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The figure 8 complement decomposes into 2 regular ideal tetrahedra (see Thurston's notes, here for example.) This gives the involution quite explicitely.

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