PUMaC 2013 · 加试 · 第 1 题

1. ♦ Definition 2.10. Let D be a link diagram with n crossings. Number the crossings 1 , . . . , n . n Let = ( , . . . , ) ∈ { 0 , 1 } . Then D denotes the smoothing of D where crossing i is 1 n resolved via a -resolution for i = 1 , . . . , n . (Note that D is also a knot diagram.) ♦ i n Definition 2.11. Let D be a link diagram, and let = ( , , . . . , ) ∈ { 0 , 1 } be a 1 2 n smoothing of D . Define s ( ) = the number of 0-resolutions in D 0 s ( ) = the number of 1-resolutions in D 1 o ( ) = the number of circles in D . 8 (We use the letter o because it looks like a circle!) Also, define s ( ) − s ( ) 0 1 〈 D, 〉 = A 〈 D 〉 . ♦ Remark 2.12. We will omit the commas in the ( , . . . , ) notation to avoid clutter. For 1 n example, = 10011 is short for = (1 , 0 , 0 , 1 , 1). ♦ Example 2.13. If D = , and the top crossing is labeled 1 (so the bottom crossing is labeled 2), then D = , D = , D = , D = . (2.3) 00 01 10 11 Also, s (00) = 2, s (01) = s (10) = 1, s (11) = 0, s (00) = 0, s (01) = s (10) = 1, 0 0 0 0 1 1 1 s (11) = 2, o (00) = o (11) = 2, o (01) = o (10) = 1. ♦ 1 2.3.1. Problem: (5) Let D be a link diagram with n crossings. Show that ∑ 〈 D 〉 = 〈 D, 〉 , n ∈{ 0 , 1 } where the sum is over all smoothings of D . Solution: Start with 〈 D 〉 and repeatedly apply BP3 until none of the diagrams have any more crossings. The result is a sum over all the smoothings of D . 〈 〉 〈 〉 − 1 The coefficient A of and the coefficient of A of in BP3 means that the s ( ) − s ( ) 0 1 coefficient of 〈 D 〉 will be A . 2.3.2. Problem: (3) Let D be a link diagram and let be a smoothing of D . Show that 2 − 2 o ( ) − 1 〈 D 〉 = ( − A − A ) . Solution: We know D has o ( ) circles. Apply BP2 over and over again (a total of 2 − 2 o ( ) − 1 o ( ) − 1 times), then apply BP1 once to get the the polynomial ( − A − A ) . ′ 2.3.3. Problem: (2) Show that if and differ by one resolution, then ′ o ( ) = o ( ) ± 1 . Solution: Changing one resolution either merges two circles in D or separates one circle into two. 9 2.4 Invariance under Type II and Type III Moves (10 points) Because of our discussion of invariants and Reidemeister moves at the end of Section 1.2, we should study how the bracket polynomial behaves under Reidemeister moves. ′ 2.4.1. Problem: (5) Show that if link diagrams D and D are related by one application ′ of a Type II Reidemeister move, then 〈 D 〉 = 〈 D 〉 . That is, show that 〈 〉 〈 〉 = . Solution: Start with the diagram D = . Label the two crossings so that 1 is on the left and 2 is on the right. The four ways of resolving these crossings are given by D = , D = , D = , D = . 00 01 10 11 Following Example 2.6 or using Problem 2.3.1, we have 〈 〉 〈 〉 〈 〉 〈 〉 〈 〉 2 − 2 = A + + + A . (2.A) 〈 〉 〈 〉 2 − 2 Using BP2, we get = ( − A − A ) , so three of the four terms on the right hand side of (2.A) cancel out: 〈 〉 ( 〈 〉) 〈 〉 2 − 2 2 2 − 2 − 2 A 〈 〉 + 〈 〉 + A 〈 〉 = A + ( − A − A ) + A = 0 . 〈 〉 〈 〉 We are left with = . This proves that the bracket polynomial is unchanged under a Type II move. ′ 2.4.2. Problem: (5) Show that if link diagrams D and D are related by one application ′ of a Type III Reidemeister move, then 〈 D 〉 = 〈 D 〉 . That is, show 〈 〉 〈 〉 = . Solution: We could start with the diagram and relate it to the 8 diagrams that correspond to the 8 possible ways of resolving the three crossings. Then we could do the same thing with . We would indeed see that both diagrams give the same result. However, we can make a cleaner argument by noting what happens if we only resolve the crossing between the two upper strands: 〈 〉 〈 〉 〈 〉 − 1 = A + A 〈 〉 〈 〉 〈 〉 − 1 = A + A 10 To complete the proof, it suffices to show that 〈 〉 〈 〉 〈 〉 〈 〉 = and = The equality on the left holds by two applications of Problem 2.4.1. The equality on the right holds because we can slide the bottom strand. 2.5 Type I moves (3 points) In the previous section, you showed that the bracket polynomial is invariant under Type II and Type III Reidemeister moves. If it is also invariant under Type I moves, the the bracket polynomial would be a genuine link invariant. However, this is not the case. 2.5.1. Problem: (3) Show that 〈 〉 〈 〉 − 3 = − A . Solution: We have 〈 〉 〈 〉 〈 〉 − 1 = A + A 〈 〉 〈 〉 − 1 2 − 2 = A + A ( − A − A ) 〈 〉 − 3 = − A . 2.6 Writhe of an oriented link (4 points) Definition 2.14. Given an oriented link diagram we can define positive crossings and neg- ative crossings by Figure 2.3. ♦ (a) positive crossing (b) negative crossing Figure 2.3: Two types of crossings for an oriented diagram Definition 2.15. Let n ( D ) and n ( D ) be the number of positive and negative crossings,

• − respectively, of an oriented link diagram D . ♦ Definition 2.16. For an oriented link diagram D , the writhe of D is w ( D ) = n ( D ) −

n ( D ). ♦ − 11 Fact 2.17. As the following diagrams show, if we reverse the direction of all components of a link, the crossing types (positive/negative) do not change. ↔ ↔ ♦ Remark 2.18. Since a knot is a link with one component, we can define positive and negative crossings for knot diagrams without specifying an orientation on the knot. Note that this is not true for links in general. If we reverse the orientations of some (but not all) of the components of a link, then some crossing types will change. ♦ 2.6.1. Problem: (2) Show ( ) ( ) w = w − 1 ( ) ( ) w = w − 1 . Solution: On the left-hand side, there is a negative crossing, which contributes − 1 to the writhe. On the right-hand side, the negative crossing is removed, while the rest of the link diagram remains unchanged. 2.6.2. Problem: (2) Show ( ) ( ) w = w ( ) ( ) w = w ( ) ( ) w = w ( ) ( ) w = w . Solution: Take a look at the first line. The diagram on the left-hand-side has one positive crossing and one negative crossing, so this part of the knot diagram contributes nothing to the writhe. Clearly on the right-hand side, there is no contribution to the writhe either. The same observation holds for the other three lines. Fact 2.19. The writhe is also unchanged under type III moves. ♦ 2.7 The Jones Polynomial (7 points) Because of Problem 2.6.1, the writhe is not invariant under Type I moves. 12 2.7.1. Problem: (7) For an oriented link L with diagram D , show that the polynomial − 3 w ( D ) ( − A ) 〈 D 〉 is an invariant of the link L . − 3 w ( D ) Solution: Because of Remark 1.5, it suffices to show that ( − A ) 〈 D 〉 is un- changed under the three types of Reidemeister moves. Because both the writhe and the bracket polynomial are invariant under Type II and − 3 w ( D ) Type III moves, we only need to check that the quantity ( − A ) 〈 D 〉 is unchanged under Type I moves. This is immediate from (2.5.1) and (2.6.1): 〈 〉 〈 〉 − 3 w − 3 w ( )+3 − 3 ( ) ( − A ) = ( − A ) · ( − A ) 〈 〉 − 3 w ( ) = ( − A ) . Definition 2.20. Let L be an oriented link, and let D be a diagram of L . The Jones − 3 w ( D ) polynomial of L , denoted V ( t ), is obtained by taking the expression ( − A ) 〈 D 〉 and L − 1 / 4 setting A = t . ♦ Fact 2.21. Because of Problem 2.7.1, the Jones polynomial is an invariant of oriented links. For knots, recall that the writhe does not depend on the orientation. If we retrace our arguments above, we see that the Jones polynomial is also an invariant of (unoriented) knots. ♦ Fact 2.22. For the trefoil (see Figure 1.1b), we have − 4 − 3 − 1 V ( t ) = − t + t + t . K You might want to verify this yourself. ♦ Remark 2.23. Vaughan Jones received the Fields Medal for his discovery of the Jones polynomial. It is a pretty big deal! ♦ 3 Detecting chiral knots 3.1 Mirroring knots (25 points) flip Definition 3.1. For a knot K , let K denote the mirrored knot. (In other words, make K out of a piece of string, and hold it in front of a mirror! The knot you see in the mirror flip is K .) ♦ flip Definition 3.2. A knot K is amphichiral if it is equivalent to K . Otherwise, it is chiral . ♦ Consider, for example the two diagrams in Figure 3.1. We may ask if they are the same knot. 13 flip (a) diagram of K (b) diagram of K Figure 3.1: Mirror images of a trefoil 3.1.1. Problem: (20) Show that for any knot K , − 1 V flip ( t ) = V ( t ) . K K Solution: Suppose we start with a smoothing D of D . What do we get when flip flip we mirror this to ( D ) ? The answer is a smoothing of D . See the example in Figure 3.A. flip flip =011 =100 Figure 3.A: Relations between mirroring and smoothing. (In the labeling, the three crossings are numbered from top to bottom.) The example makes a general pattern clear. Every smoothing of D corresponds to flip the dual smoothing ˆ of D . The dual smoothing ˆ is obtained by reversing every resolution in . (That is, interchange 0s and 1s.) For example, if = 011, then ˆ = 100. We can write the relation as flip ( D ) = D . (3.A) ˆ Observe that s ( ) = s (ˆ ) and s ( ) = s (ˆ ). Using (3.A), it follows that 0 1 1 0 ∑ ∑ 〈 〉 〈 〉 flip s ( ) − s ( ) flip − ( s (ˆ ) − s (ˆ )) 0 1 0 1 D ( A ) = A ( D ) ( A ) = A 〈 D 〉 ( A ) . ˆ n n ∈{ 0 , 1 } ∈{ 0 , 1 } 2 Recall that the bracket polynomial of a smoothing 〈 D 〉 ( A ) is some power of ( − A − − 2 − 1 A ). Hence, 〈 D 〉 ( A ) = 〈 D 〉 ( A ), giving us ∑ 〈 〉 flip − 1 s (ˆ ) − s (ˆ ) − 1 − 1 0 1 D ( A ) = ( A ) 〈 D 〉 ( A ) = 〈 D 〉 ( A ) . ˆ n ∈{ 0 , 1 } flip Also, w ( D ) = − w ( D ), since positive crossings in D become negative crossings in flip D and vice versa. We can translate the results we have just obtained to a statement about the Jones polynomial. 14 3.1.2. Problem: (5) Is the trefoil amphichiral or chiral? Justify your answer. Solution: The trefoil is chiral. Let K be the knot represented by Figure 3.1a. By − 4 − 3 − 1 Fact 2.22, we have V ( t ) = − t + t + t . K 4 3 It follows that the Jones polynomial of the mirrored knot is V flip ( t ) = − t + t + t . K Since the two polynomials are not the same, we can conclude that the trefoil and its mirror are distinct. Thus, the trefoil is chiral. (Both knots are usually referred to as the “trefoil.” If we want to make it clear which one we are talking about, we call Figure 3.1a and Figure 3.1b the “left-handed” and “right-handed” trefoils, respectively.) Remark 3.3. Recall that the Jones polynomial is not just a knot invariant but also an oriented link invariant. Thus, Problem 3.1.1 still holds if we replace “knot” with “oriented link.” ♦ 4 Bound on crossing numbers 4.1 Crossing number (3 points) Definition 4.1. The crossing number of a knot K is the minimum number of crossings needed to draw the knot in a plane. It is denoted c ( K ). ♦ 4.1.1. Problem: (3) While the diagram given in Figure 4.1 has seven crossings, show that the crossing number of that knot is not 7. Figure 4.1 Solution: We can flip one of the two halves over to get a diagram with six crossings. (For example, flipping the right half over leaves us with Figure 4.A.) This shows that the crossing number of the knot is at most 6. Figure 4.A 15 4.2 Reduced diagrams (5 points) Definition 4.2. A knot diagram D is un-reduced is it has the form of Figure 4.2. (That is, there are exactly two strands in the region between X and Y that go from X to Y . Furthermore, these two strands cross each other exactly once.) A knot diagram D is reduced if it is not un-reduced. X Y Figure 4.2 ♦ 4.2.1. Problem: (5) Show, by drawing an example, that it is possible to have two diagrams ′ D and D of the same knot K , such that • D is reduced and ′ • D has fewer crossings than D . ′ Solution: Here is one possible answer. Let D be the unknot in its usual form: . Let D be the unknot as depicted in Figure 4.B. Figure 4.B: Reduced diagram of the unknot with three crossings ′ The diagram D is reduced and contains three crossings, whereas D contains no cross- ings. 4.3 Knots and graphs (5 points) Every knot diagram corresponds to a planar graph. (See, for example, Figure 4.3.) Definition 4.3. A graph is a set of vertices (e.g., the black dots in Figure 4.3b) that are joined together by edges (e.g., the lines between the dots in Figure 4.3b). A graph is planar if it can be drawn on the plane in such a way that its edges intersect only at their endpoints. ♦ Fact 4.4. A planar graph divides the plane into regions, called faces. (The exterior of a graph is considered a face too.) If V , E , F are the number of vertices, edges, faces (respectively) of a planar graph, then Euler’s formula says that V − E + F = 2. ♦ 16 (a) trefoil (b) corresponding graph Figure 4.3 Example 4.5. The planar graph Figure 4.3b has V = 3, E = 6, F = 5 (don’t forget to count the exterior face!). Thus, V − E + F = 2, as expected. ♦ 4.3.1. Problem: (5) Show that for a knot diagram with n crossings, the corresponding graph has n + 2 faces. Solution: Apply Euler’s formula: V − E + F = 2. Because the knot diagram has n crossings, we know V = n . Because each crossing involves two strands, each vertex of the graph has four edges coming out of it. Using this fact, we know that the set of ordered pairs { ( e, v ) : e is an edge and v is an endpoint of e } has E · 2 = V · 4 elements. Thus, E = 2 n and we have F = n + 2. 4.4 Coloring faces of knots (15 points) It is always possible to color the faces of a link in an alternating black-white pattern, as shown in Figure 4.4. (Recall that we are considering the exterior of the knot diagram to be a face as well, which explains why Figure 4.4b is a valid checkerboard coloring of the trefoil.) (a) trefoil (b) trefoil Figure 4.4: Examples of checkerboard colorings Fact 4.6. By resolving a crossing, we still get a checkerboard coloring. (See, for example Figure 4.5.) ♦ (a) checkerboard (b) 0-resolution (c) 1-resolution Figure 4.5: Resolving a crossing still gives us a checkerboard coloring. 17 Given a checkerboard coloring of a knot diagram, we can divide the crossings into two types. (See Figure 4.6.) Definition 4.7. The coloring in Figure 4.6a is called “0-separating” (because a 0-resolution separates the two black regions). The coloring in Figure 4.6b is called “1-separating.” ♦ (a) 0-separating (b) 1-separating Figure 4.6: Two coloring types at a crossing Definition 4.8. A knot diagram is alternating if the strand alternates between going over and going under at crossings. A knot is alternating if there is a alternating diagram of the knot. ♦ 4.4.1. Problem: (15) Show that a knot diagram is alternating if and only if the diagram admits a checkerboard coloring consisting of only 0-separating crossings. Solution: To prove the reverse direction, consider the portion of a knot diagram given in Figure 4.Ca. Suppose we start at point A and move rightwards along the horizontal strand. We will reach the crossing at the center of the diagram. (Observe that this is a 0-separating crossing.) In this crossing, the horizontal strand we are traveling along goes underneath the vertical strand. A A ? (a) (b) A (c) Figure 4.C We keep moving rightwards after we pass the crossing. Eventually we will reach another crossing. (See Figure 4.Cb.) There is only one way to make this crossing 0-separating. The horizontal strand must go over the vertical strand. (See Figure 4.Cc.) 18 The general pattern is clear: if we want this knot diagram to have only 0-separating crossings, then the strand must alternate over-under-over-under. This completes the proof of the reverse direction. To prove the forward direction, suppose we have an alternating knot diagram. Start at a crossing C . Color the diagram in a checkerboard pattern so that C has a 0-separating coloring. Now, consider traveling around the knot starting at C . Every time we reach a crossing, the shaded region alternates between left and right and our position at a crossing alternates between over and under. Thus, all the crossings are 0-separating. 4.5 The span of the bracket polynomial (18 points) Recall the definitions of 〈 D, 〉 and o ( ) given in section Section 2.3. Definition 4.9. For a Laurent polynomial f ( x ), we define hp( f ) to be the highest power of x that appears in f , and we define lp( f ) similarly to be the lowest power. We define span( f ) = hp( f ) − lp( f ). ♦ Definition 4.10. Let D be a link diagram. We let 0 = (0 , 0 , . . . , 0) denote the smoothing with all 0-resolutions and 1 = (1 , 1 , . . . , 1) denote the smoothing with all 1-resolutions. ♦ 4.5.1. Problem: (5) Let D and D be knot diagrams of the same knot. Show that 1 2 span 〈 D 〉 = span 〈 D 〉 . 1 2 Solution: There are two ways to approach this. First solution: we only need to show that span 〈 D 〉 is unchanged under the Reidemeister moves. Because 〈 D 〉 is itself unchanged under type II and III moves, we only need to − 3 check type I moves. Recall that a type I move multiples 〈 D 〉 by ( − A ) . This does not change the span, so we are done. Second solution: Suppose D and D are diagrams of the knot K . From the definition 1 2 of the Jones polynomial, we see that span 〈 D 〉 = 4 span V = span 〈 D 〉 . 1 K 2 4.5.2. Problem: (3) Let D be a knot diagram and let be a smoothing of D . Show that hp 〈 D, 〉 = s ( ) − s ( ) + 2 o ( ) − 2 . 0 1 Solution: s ( ) − s ( ) 2 − 2 o ( ) − 1 0 1 hp 〈 D, 〉 = hp A ( − A − A ) = s ( ) − s ( ) + 2 o ( ) − 2 0 1 4.5.3. Problem: (5) Let D be a knot diagram. Show that hp 〈 D, 0 〉 ≥ hp 〈 D, 〉 for all smoothings . 19 Solution: Let D be a knot diagram and let be a smoothing of D . Suppose we ′ change a 0-resolution of to a 1-resolution to get a new smoothing . ′ ′ As we go from to , the quantity s − s decreases by 2. Since o ( ) = o ( ) ± 1 by 0 1