Chapter 1, Decomposition of the Figure-8 Knot Video Solutions, Hyperbolic Knot Theory

Problem 1

A polyhedron is a closed 3-ball whose boundary is labeled with a finite graph, containing a finite number vertices and edges, so that complementary regions, which are called faces, are simply connected.

An ideal polyhedron is a polyhedron with all vertices removed. That is, to form an ideal polyhedron, start with a regular polyhedron and remove the points corresponding to vertices.

Vikash Ranjan

Numerade Educator

Problem 2

Orientations on the edges can be chosen to run in either direction; that is, arrows on the edges can run from overcrossing to undercrossing or vice versa, as long as we are consistent with orientations corresponding to the same edge. We have chosen the orientations in Figure 1.6 to simplify a later step, and to match a figure in Chapter 4. The opposite choice for any edge is also fine.

Check back soon!

Problem 3

As a warm-up exercise, determine the polyhedral decomposition for one (or more) of the knots shown in Figure 1.11. Sketch both top and bottom polyhedra.

Your solution should consist of two ideal polyhedra, i.e. marked graphs on the surface of a ball, with faces and edges marked according to the gluing pattern. For example, the complete diagrams in Figures 1.8 and 1.10 form the solution for the figure- 8 knot.

Check back soon!

Problem 4

The examples of knots we have encountered so far are all alternating, as in Definition 0.14. The diagram of the knot $8_{19}$ in Figure 1.12 is not alternating. In fact, the knot $8_{19}$ has no alternating diagram.
Determine the polyhedral decomposition for the given diagram of the knot $8_{19}$. Note: as above, many ideal vertices are obtained by shrinking overstrands to a point. However, you will have to use, for example, Figure 1.3 to determine what happens between two understrands.

Check back soon!

Problem 5

Recall that the valence of a vertex in a graph is the number of edges that meet that vertex. The valence of an ideal vertex is defined similarly.
(a) If a knot diagram is alternating, we obtain a very special ideal polyhedron. In particular, all ideal vertices will have the same valence. What is it? Show that the ideal vertices for an alternating knot all have this valence.
(b) What are the possible valences of ideal vertices in general, i.e. for nonalternating knots? For which $n \geq 0 \in \mathbb{Z}$ is there a knot diagram whose polyhedral decomposition yields an ideal vertex of valence $n$ ? Explain your answer, with (portions of) knot diagrams.

Check back soon!

Problem 6

In the polyhedral decomposition for alternating knots, the polyhedra are given by simply labeling each ball with the projection graph of the knot and declaring each vertex to be ideal.
(a) Prove this statement for any alternating knot. That is, prove that the decomposition gives polyhedra whose edges match the projection graph of the diagram.
(b) Show that for nonalternating knots, this is false. That is, the decomposition does not give polyhedra whose edges match the projection graph of the diagram.

Check back soon!

Problem 7

A graph admits a checkerboard coloring if all the complementary regions can be colored either white or shaded, with white faces meeting shaded faces across the edges. Any 4 -valent graph can be checkerboard colored, particularly projection graphs of knot diagrams.

In the case of an alternating knot, faces are identified from the top polyhedron to the identical face on the bottom polyhedron, and the identification is by a gear rotation: white faces on the top are rotated once counter-clockwise and then glued to the corresponding face on the bottom; shaded faces on the top are rotated once clockwise and then glued. This is shown for the figure-8 knot in Figure 1.13. Prove that for the decomposition of any alternating knot, faces are identified by a gear rotation.

Check back soon!

Problem 8

The diagrams we have encountered so far are all reduced, as in Definition 0.6, but we can follow the above procedure for nonreduced diagrams. For example, we can obtain a polyhedral decomposition for diagrams which contain a nugatory crossing.

Show that the polyhedral decomposition of a knot diagram will contain a monogon, i.e. a face whose boundary is a single edge and a single vertex, if and only if the diagram has a simple nugatory crossing.

Check back soon!

Problem 9

Recall that a bigon is a region of a graph bounded by exactly two edges and exactly two vertices. Note that when a bigon appears in our polyhedral decomposition, the two edges of the bigon must be isotopic to each other. Hence, we sometimes will remove bigon faces from the polyhedral decomposition, identifying their two edges.

Let bigons be bygone. - William Menasco
For the figure- 8 knot, sketch the two polyhedra that we get when bigon faces are removed. How many edges are there in this new, bigon-free decomposition? The resulting polyhedra are well-known solids in this case. What are they?

For each of the polyhedra obtained in Exercise 1.3, sketch the resulting polyhedra with bigons removed.

Check back soon!

Problem 10

Suppose we start with an alternating knot diagram with at least two crossings, and do the polyhedral decomposition above, collapsing bigons at the last step. What are possible valences of vertices? Sketch the diagram of a single alternating knot that has all possible valences of ideal vertices in its polyhedral decomposition.

What valences of vertices can you get if you don't require the diagram to be alternating but collapse bigons? Can you find 1 -valent vertices? For any $n>4 \in \mathbb{Z}$, can you find $n$-valent vertices?

Check back soon!