PUMaC 2013 · 加试 · 第 4 题
PUMaC 2013 — Power Round — Problem 4
题目详情
- Using calculators and Mathematica (or similar programs), is allowed. Print and online sources are not allowed. No communication with humans outside your team about the content of these problems is allowed.
解析
- PUMaC competitors: for taking the Power Round! –Alan Chang ^ ¨ A Fun fact: You can make a happy face in L TEX with . This is because gives you ¨ x (notation for second derivative) and gives you ^ . 1 Contents 1 Preliminaries 3 1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Reidemeister moves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2 The Jones Polynomial 5 2.1 Resolving a crossing (4 points) . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 The Bracket Polynomial (0 points) . . . . . . . . . . . . . . . . . . . . . . . 6 2.3 Smoothings (10 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.4 Invariance under Type II and Type III Moves (10 points) . . . . . . . . . . . 10 2.5 Type I moves (3 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.6 Writhe of an oriented link (4 points) . . . . . . . . . . . . . . . . . . . . . . 11 2.7 The Jones Polynomial (7 points) . . . . . . . . . . . . . . . . . . . . . . . . . 12 3 Detecting chiral knots 13 3.1 Mirroring knots (25 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4 Bound on crossing numbers 15 4.1 Crossing number (3 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 4.2 Reduced diagrams (5 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.3 Knots and graphs (5 points) . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.4 Coloring faces of knots (15 points) . . . . . . . . . . . . . . . . . . . . . . . . 17 4.5 The span of the bracket polynomial (18 points) . . . . . . . . . . . . . . . . 19 4.6 Connected link diagrams (17 points) . . . . . . . . . . . . . . . . . . . . . . 20 4.7 Reduced alternating knot diagrams (42 points) . . . . . . . . . . . . . . . . . 21 4.8 Back to the Jones polynomial (13 points) . . . . . . . . . . . . . . . . . . . . 23 2 1 Preliminaries 1.1 Definitions Although there are more technical definitions, for this Power Round, it is enough to think of a knot as something made physically by attaching the two ends of a string together. Since knots exist in three dimensions, when we need to draw them on paper, we often use knot diagrams . Figure 1.1 contains examples of knot diagrams. (a) unknot (b) trefoil (c) trefoil (again) (d) figure-8 knot Figure 1.1: Examples of knot diagrams, with the names of the knots they represent. As we can see from Figure 1.1b and Figure 1.1c, different knot diagrams can represent the same knot. To see that these two are really the same knot, we could make Figure 1.1b out of a piece of string and move the string around in space (without cutting it) so that it looks like Figure 1.1c. Tip 1.1. Make sure you understand the distinction between “knot” and “knot diagram”! Do not use these terms interchangeably in your solutions. If you use one when you mean the other, the proof will be incorrect (logically) and this will lead to a significantly lower score. ♦ There are some restrictions on knot diagrams: (1) each crossing must involve exactly two segments of the string and (2) those segments must cross transversely. (See Figure 1.2.) (a) triple crossing (b) non-transverse crossing Figure 1.2: Examples of invalid knot diagrams There are two ways to travel around a knot; these correspond to the orientations of the knot. An oriented knot is a knot with a specified orientation. On a knot diagram, we can indicate an orientation via an arrow. (See Figure 1.3.) Sometimes we’ll use more than one piece of string, so we define a link to be a generaliza- tion of a knot: links can be made by multiple pieces of string. For each string, we attach the two ends together. (Note that we do not attach the ends of two different strings together.) The number of components of a link is the number of strings used. (Observe that every knot is a link with one component.) A link diagram is a straightforward generalization 3 (a) one orientation (b) the other orientation Figure 1.3: Two orientations of the trefoil of a knot diagram, and an oriented link is a link where all the components have specified orientations. Figure 1.4 contains examples of two-component links. (a) unlink (b) Hopf link (c) Whitehead link Figure 1.4: Examples of links with two components Tip 1.2. Pay close attention to the problem statements in this Power Round. If a problem asks you to prove something for links, it is not enough to prove it for knots! ♦ A knot invariant is something (such as number, matrix, or polynomial) associated to a knot. A link invariant is defined similarly for links. An example of a knot invariant is the crossing number , which we will now define. Suppose for a knot K , we take a diagram D of K and count the number of crossings in D . This number is not an invariant of K because K has many different diagrams that differ in number of crossings. For example, in Figure 1.5, we see two different diagrams of the unknot. (a) 0 crossings (b) 3 crossings Figure 1.5: Two diagrams of the unknot, with different number of crossings. Thus, we have not yet successfully defined a knot invariant. However, if we consider all diagrams of K and take the minimum number of crossings over all diagrams, then we do have an invariant of K . This is called the crossing number of a knot. 4 1.2 Reidemeister moves Consider the kinds of moves in Figure 1.6, which you can perform on a knot diagram. These are called Reidemeister moves . We can think of Type I as adding/removing a twist, Type II as crossing/uncrossing two strands, and Type III as sliding a strand past a crossing. ↔ ↔ ↔ (a) Type I (b) Type II (c) Type III Figure 1.6: The three types of Reidemeister moves Fact 1.3. Suppose we start with a knot diagram D and perform one of the Reidemeister 1 moves on it so that we end up with a knot diagram D . The knot diagrams D and D 2 1 2 might be different, but they will represent the same knot. (This is clear from the diagrams in Figure 1.6.) ♦ Fact 1.4. In 1926, Kurt Reidemeister proved that given two diagrams D and D of the 1 2 same link, it is always possible to get from D to D via a finite sequence of Reidemeister 1 2 moves. This is a remarkable fact! ♦ Remark 1.5. Suppose there is a quantity that we are trying to show is a knot invariant, but it is defined in terms of a knot diagram. There is a possibility that the quantity is different for different diagrams of the same knot, in which case our quantity would not be a knot invariant. However, if we can show the quantity is unchanged when we alter the diagram via any Reidemeister move, then we know it is an invariant, because of Fact 1.4. ♦ 2 The Jones Polynomial 2.1 Resolving a crossing (4 points) Definition 2.1. Suppose we start with a crossing of the form . The 0-resolution of this crossing is and the 1-resolution is . (For example, see Figure 2.1.) ♦ (a) trefoil (b) 0-resolution (c) 1-resolution Figure 2.1: Resolving a crossing 5 Remark 2.2. A diagram may need to be rotated so that the crossing in concern appears as ◦ . For example, after a 90 rotation, we see that the 0- and 1-resolutions of are and , respectively. ♦ Remark 2.3. Here is one way to think of a 0-resolution: if we are traveling along a knot and reach a crossing in which we are on the upper strand, then we turn left onto the lower strand. (For a 1-resolution, we would turn right instead.) ♦ 2.1.1. Problem: (2) Start with the diagram of the figure 8 knot given in Figure 2.2. Observe that the crossings are labeled A , B , C , D . Draw the diagram in which crossing B has been resolved with a 0-resolution and identify the resulting knot/link. A B C D Figure 2.2 Solution: The diagram becomes . It is a Hopf link. 2.1.2. Problem: (2) Once again, start with the diagram of the figure 8 knot given in Figure 2.2. It is possible to resolve one of the crossings so that the resulting diagram is a trefoil. Which crossing can we resolve ( A , B , C , D ) and how do we resolve it (0- or 1-resolution)? Simply state an answer. Solution: There are four possible solutions. Any one is enough. • Resolve crossing A or B with a 1-resolution. • Resolve crossing C or D with a 0-resolution. 2.2 The Bracket Polynomial (0 points) Definition 2.4. As you all know, a polynomial in x is something of the form n n − 1 a x + a x + · · · + a x + a . n n − 1 1 0 6 A Laurent polynomial in x is like a polynomial except you can use negative powers. In other words a Laurent polynomial is something of the form n n − 1 m a x + a x + · · · + a x , n n − 1 m 3 − 1 − 4 where m and n are integers with m ≤ n . (For example, x + 2 x + 4 + x − 5 x is a Laurent polynomial.) Quite confusingly, a Laurent polynomial is not necessarily a polynomial. ♦ Definition 2.5. The bracket polynomial of a link diagram D is a Laurent polynomial in the variable A and is denoted 〈 D 〉 . It is completely determined by three rules: 〈 〉 = 1 (BP1) 〈 〉 2 − 2 D t = ( − A − A ) 〈 D 〉 (BP2) 〈 〉 〈 〉 〈 〉 − 1 = A + A (BP3) (The BP stands for “bracket polynomial.”) ♦ Let’s go through what these rules mean, one by one.