## Archive for April 21st, 2010

April 21, 2010

I’ve been going through Thurston’s book ‘The Geometry and Topology of Three-Manifolds‘ in a reading course with Amie Wilkinson. In Chapter 3, p32, when he’s constructing a hyperbolic structure on the Whitehead link complement, there is a picture on how to glue the 2-cells to the knot, to quite Thurston, ‘the attaching map for the two-cells are indicated by the dotted lines.’ However, for me it’s impossible to see where are the dotted lines going. So I reconstruct it here with some more clear pictures. The construction itself was a cool reading that I wish to share.

First, we have the Whitehead link, looking like the first figure below:

We attach three 1-cells (line segments) as in the second figure, note that the ‘x’ in the middle represents a line segment orthogonal to the screen, connecting the top and bottom line in the figure ‘8’ loop.

Now we will start to attach four 2-cells to the 1-complex above: First, we attach a 2-cell spanning the top part of the figure ‘8’ loop, spanning one side of the middle segment and two sides of the top segment (denote this by cell A):

Do the same with the bottom half (cell B). Note that each cell is attached to three edges, hence they are triangles without vertices in the knot complement with three one-cells attached.

For the other two cells, we attach as follows (cells C and D):

Combining the four 2-cells, we get something like the figure showed below. Note that at the top, cell A is under cell C in the left, intersecting the surface spanned by cells C and D at the edge, and comes above cell D to the right of the edge.

It’s easy to see that the complement of the above 2-complex does not separate $\mathbb{R}^3$, hence it’s a 3-cell with eight faces (i.e. it has to go through both sides of each 2-cell in order to fill the 3-space) each of its face has three edges. Hence we may glue an octahedron to the 2-complex after the gluing, pairs of faces of the octahedron will be identified groups of four edges will be identified to single edges. Hence to put a hyperbolic structure on the link complement, it suffice to put an hyperbolic structure to the octahedron with vertices deleted.

Since each edge is glued up by four edges of the octahedron, it suffice to find an octahedron (without vertices) in the hyperbolic 3-space that has all adjacent faces intersect in dihedral angle $2 \pi / 4$ i.e. all adjecent faces are orthogonal in the hyperbolic space. But this is achieved if we inscribe the regular octahedron into the Klein model (also called projective model of hyperbolic 3-space.

The gluing map for the faces are merely rotations and reflections of the ball which are certainly hyperbolic isometries. Hence this gives a hyperbolic structure to the link complement.