PUMaC 2013 · 加试 · 第 2 题

Problem 2.3.3, the quantity 2 o can increase by 2 or decrease by 2. In either case, we ′ have 2 o ( ) ≥ 2 o ( ) − 2, so ′ ′ ′ s ( ) − s ( ) + 2 o ( ) − 2 ≥ s ( ) − s ( ) + 2 o ( ) − 2 . 0 1 0 1 ′ By Problem 4.5.2, it follows that hp 〈 D, 〉 ≥ hp 〈 D, 〉 . Thus, suppose we are given a smoothing ˜ . If we start at 0 and changing the 0- resolutions to 1-resolutions one by one to get to , the value hp 〈 D, 〉 does not increase in any of the steps. 4.5.4. Problem: (2) Let D be a knot diagram with n crossings. Show that hp 〈 D 〉 ≤ n + 2 o ( 0 ) − 2 . Solution: hp 〈 D 〉 ≤ max hp 〈 D, 〉 = hp 〈 D, 0 〉 = n + 2 o ( 0 ) − 2. 4.5.5. Problem: (3) Let D be a knot diagram with n crossings. Show that span 〈 D 〉 ≤ 2 n + 2( o ( 0 ) + o ( 1 )) − 4 . Solution: Analogous to Problem 4.5.4, we have that lp 〈 D 〉 ≥ lp 〈 D, 1 〉 = − n − 2 o ( 1 ) + 2 . Combine this with Problem 4.5.4 to get the desired inequality. 4.6 Connected link diagrams (17 points) Definition 4.11. We say that a link diagram is connected if its corresponding graph is connected. For example, the usual diagram for the unlink is not connected. However, if the two components overlap in the diagram (as in ), then the diagram is connected. (Note that knot diagrams are always connected.) ♦ 4.6.1. Problem: (15) Let D be a connected link diagram. Show that if D has n crossings, then o ( 0 ) + o ( 1 ) ≤ n + 2 . Solution: We proceed by induction on n . For the base case ( n = 0), we have the usual diagram for the unknot ( ). Since there are no crossings to resolve, we have 1 o ( D ) = o ( D ) = 1, which completes the base case. 0 1 1 For this proof, we write o ( D ) instead of o ( 0 ) to emphasize that D is the knot we are taking the 0 0 -smoothing of. 20 For the inductive step, suppose that the statement is true for all connected link dia- grams with n crossings. Take a connected link diagram D with n + 1 crossings. Pick one of the crossings of D (call it x ). We can resolve x via a 0- or a 1-resolution; observe that at least one of the two resulting diagrams is connected. Suppose, without loss of generality, that applying the 0-resolution on x leaves us with a connected link diagram. Let this diagram be E . (See Figure 4.D.) (a) D (b) D (c) D 0 1 (d) E (e) E (f) E 0 1 Figure 4.D: Various smoothings of D and E ( x is the crossing on the bottom of D ) Note that E and E are also smoothings of D . In fact: 0 1 • E = D . 0 0 • E and D differ at exactly one crossing (namely, x ). 1 1 Thus, o ( E ) = o ( D ) and o ( E ) = o ( D ) ± 1. Furthermore, E is a connected link 0 0 1 1 diagram with n crossings, so by the inductive hypothesis, o ( E ) + o ( E ) ≤ n + 2. 0 1 Putting the results together gives us o ( D ) + o ( D ) ≤ n + 3, which completes the 0 1 induction. 4.6.2. Problem: (2) Let D be a knot diagram with n crossings. Show that span 〈 D 〉 ≤ 4 n. Solution: Combine Problem 4.5.5 and Problem 4.6.1. 4.7 Reduced alternating knot diagrams (42 points) The goal of this section is to show that alternating knots are nice. Definition 4.12. A knot diagram D is reduced alternating if it is both reduced and alter- nating. ♦ Fact 4.13. Given an alternating knot K , there is a diagram D of K that is reduced alter- nating. (This is not hard to show.) ♦ 21 4.7.1. Problem: (7) Show that if D is a reduced alternating diagram and is a smoothing with exactly one 1-resolution, then o ( 0 ) = o ( ) + 1 . Solution: Because the diagram is alternating, we can give the diagram a checkerboard coloring so that all the crossings are 0-separating. See for example, Figure 4.Ea. (a) D (b) D (c) D 0 Figure 4.E Then when we look at D , all the circles bound the shaded faces in the coloring. (See 0 Figure 4.Eb.) When we change one of the 0-resolutions to a 1-resolution, we add a “bridge” across a white region. (See Figure 4.Ec.) Because the diagram D is reduced, the bridge connects two separate black regions. Thus, the number of circles decreases by one, so o ( 0 ) = o ( ) + 1. 4.7.2. Problem: (5) Show that the assumption that the diagram D is reduced is necessary for Problem 4.7.1. Give an example where D is alternating but o ( 0 ) 6 = o ( ) + 1. Solution: Here’s one possible answer. (a) D (b) D (c) D 0 Figure 4.F 4.7.3. Problem: (30) Let D be a reduced alternating knot diagram with n crossings. Show that span 〈 D 〉 = 4 n. Solution: The key is to observe that most of the problems in the previous section have a nicer analogue for reduced alternating knot diagrams. Lemma 1 (analogue of Problem 4.5.3) . If D is a reduced alternating diagram and is a smoothing with exactly one 1-resolution, then hp 〈 D, 0 〉 > hp 〈 D, 〉 . Proof. Since, o ( 0 ) = o ( ) + 1, we can modify the proof of Problem 4.5.3 to get a strict inequality. 22 Lemma 2 (analogue of Problem 4.5.4) . If D is a reduced alternating diagram, then hp 〈 D 〉 = n + 2 o ( 0 ) − 2. ∑ Proof. We have 〈 D 〉 = 〈 D, 0 〉 + 〈 D, 〉 . By Problem 4.5.3 and Lemma 1, we know 6 = 0 ∑ that the highest power in 〈 D, 〉 is strictly lower than the highest power in 〈 D, 0 〉 . 6 = 0 Thus, hp 〈 D 〉 = hp 〈 D, 0 〉 = n + 2 o ( 0 ) − 2. Lemma 3 (analogue of Problem 4.5.5) . If D is a reduced alternating diagram with n crossings, then span 〈 D 〉 = 2 n + 2( o ( 0 ) + o ( 1 )) − 4 . Proof. Analogous to Lemma 2, we have lp 〈 D 〉 = lp 〈 D, 1 〉 = − n − 2 o ( 1 ) + 2 .. Using this result together with Lemma 2 gives us the lemma. Lemma 4 (analogue of Problem 4.6.1) . Let D be an alternating knot. Then o ( 0 ) + o ( 1 ) = n + 2 . Proof. Give D a checkerboard coloring with only 0-separating crossings. The key observation is that o ( 0 ) counts the black faces and o ( 1 ) counts the white faces. Thus, o ( 0 ) + o ( 1 ) equals the total number of faces. By Problem 4.3.1, we are done. The equality span 〈 D 〉 = 4 n follows by combining Lemma 3 and Lemma 4. 4.8 Back to the Jones polynomial (13 points) We are almost there! 4.8.1. Problem: (3) Let K be a knot. Show that c ( K ) ≥ span V . K Solution: By definition of c ( K ), there exists a knot diagram D of K with exactly c ( K ) crossings. Then Problem 4.6.2 gives us span 〈 D 〉 ≤ 4 c ( K ). 4.8.2. Problem: (5) Let K be an alternating knot. Show that c ( K ) = span V . K Solution: Note: this proof is short, but it’s a little tricky. All the steps must be in the proper order! 23 Let K be an alternating knot. There exists a reduced alternating knot diagram D of K . Let n be the number of crossings of D . By the definition of c ( K ), we know c ( K ) ≤ n . By Problem 4.7.3, we know span 〈 D 〉 = 4 n , which gives span V = n. K By Problem 4.8.1, we have span V ≤ c ( K ) ≤ n. K Putting span V = n and span V ≤ c ( K ) ≤ n together shows that span V = c ( K ). K K K (Incidentally, we have also shown that c ( K ) = n , which will be useful for the next problem.) 4.8.3. Problem: (5) Determine the crossing number c ( K ) of the knot K depicted by the diagram D in Figure 4.7. Justify your answer. Figure 4.7: D , crazy knot diagram Solution: The answer is 11. Observe that D is reduced alternating and has n = 11 crossings. By Problem 4.7.3, we know span 〈 D 〉 = 4 n , which gives span V = n. K By Problem 4.8.1, we have span V ≤ c ( K ) ≤ n. K Putting span V = n and span V ≤ c ( K ) ≤ n together shows that c ( K ) = n = 11. K K 24 Index 0-resolution, 5 link, 3 0-separating, 18 link diagram, 3 1-resolution, 5 link invariant, 4 1-separating, 18 negative crossing, 11 D , 8 flip K , 13 oriented knot, 3 V ( t ), 13 L oriented link, 4 〈 D, 〉 , 8 planar graph, 16 〈 D 〉 , 7 positive crossing, 11 hp( f ), 19 lp( f ), 19 reduced, 16 span( f ), 19 reduced alternating, 21 0 , 19 Reidemeister moves, 5 1 , 19 n { 0 , 1 } , 8 smoothing, 8 n ( D ), 11 + writhe, 11 n ( D ), 11 − o ( ), 8 s ( ), 8 0 s ( ), 8 1 alternating knot, 18 alternating knot diagram, 18 amphichiral, 13 bracket polynomial, 7 chiral, 13 component of a link, 3 connected link diagram, 20 crossing number, 4, 15 graph, 16 Jones polynomial, 13 knot, 3 knot diagram, 3 knot invariant, 4 Laurent polynomial, 7 25