## How to see isometries of figure 8 knot complement

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

## 2 Answers

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.

-

1
Nice! What did you use to generate the picture? – auniket Nov 19 2011 at 0:45

1
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 2011 at 1:49
Maybe I'm confused; isn't the OP trying to visualize a fixed-point free involution? – jc Nov 19 2011 at 6:14

1
The only fixed points are on the knot -- it's free on the knot complement. – Ryan Budney Nov 19 2011 at 6:24

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

The figure 8 complement decomposes into 2 regular ideal tetrahedra (see Thurston's notes, here for example.) This gives the involution quite explicitely.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9518728256225586, "perplexity_flag": "head"}