PUMaC 2013 · 加试 · 第 3 题
PUMaC 2013 — Power Round — Problem 3
题目详情
- In order to apply the third relation, we need to resolve crossings. Start with a diagram D and fix a crossing. If D and D are the 0- and 1-resolutions of this crossing, then 0 1 − 1 BP3 states that 〈 D 〉 = A 〈 D 〉 + A 〈 D 〉 . For example, 0 1 〈 〉 〈 〉 〈 〉 − 1 = A + A Example 2.6. Let’s compute the bracket polynomial of the diagram . (It is a diagram of the Hopf link.) Applying BP3 gives us 〈 〉 〈 〉 〈 〉 − 1 = A + A . (2.1) Using BP3 again, 〈 〉 〈 〉 〈 〉 − 1 = A + A 〈 〉 〈 〉 〈 〉 (2.2) − 1 = A + A . Combining (2.1) and (2.2), we see that 〈 〉 〈 〉 〈 〉 〈 〉 〈 〉 2 − 2 = A + + + A . Invoking BP1 and BP2 gives us 〈 〉 〈 〉 〈 〉 〈 〉 2 − 2 = = 1 and = = − A − A . 4 − 4 Putting everything together gives us 〈 〉 = − A − A . ♦ 7 Fact 2.7. For the trefoil, we have 〈 〉 7 3 − 5 = A − A − A . You might want to verify this yourself, to make sure you understand how to compute bracket polynomials of knot diagrams. ♦ 2.3 Smoothings (10 points) In Example 2.6, we decomposed the Hopf link into four diagrams. The four diagrams corre- spond to the four ways of resolving the two crossings of . Each of these diagrams is called a smoothing . Definition 2.8. Given a link diagram D , a smoothing of D is a diagram in which every crossing of D has been resolved (either by a 0-resolution or a 1-resolution). Note that a smoothing has no crossings. ♦ n Definition 2.9. Let { 0 , 1 } denote the set of n -tuples, where each component is either 0 or
解析
- In order to apply the third relation, we need to resolve crossings. Start with a diagram D and fix a crossing. If D and D are the 0- and 1-resolutions of this crossing, then 0 1 − 1 BP3 states that 〈 D 〉 = A 〈 D 〉 + A 〈 D 〉 . For example, 0 1 〈 〉 〈 〉 〈 〉 − 1 = A + A Example 2.6. Let’s compute the bracket polynomial of the diagram . (It is a diagram of the Hopf link.) Applying BP3 gives us 〈 〉 〈 〉 〈 〉 − 1 = A + A . (2.1) 7 Using BP3 again, 〈 〉 〈 〉 〈 〉 − 1 = A + A 〈 〉 〈 〉 〈 〉 (2.2) − 1 = A + A . Combining (2.1) and (2.2), we see that 〈 〉 〈 〉 〈 〉 〈 〉 〈 〉 2 − 2 = A + + + A . Invoking BP1 and BP2 gives us 〈 〉 〈 〉 〈 〉 〈 〉 2 − 2 = = 1 and = = − A − A . 4 − 4 Putting everything together gives us 〈 〉 = − A − A . ♦ Fact 2.7. For the trefoil, we have 〈 〉 7 3 − 5 = A − A − A . You might want to verify this yourself, to make sure you understand how to compute bracket polynomials of knot diagrams. ♦ 2.3 Smoothings (10 points) In Example 2.6, we decomposed the Hopf link into four diagrams. The four diagrams corre- spond to the four ways of resolving the two crossings of . Each of these diagrams is called a smoothing . Definition 2.8. Given a link diagram D , a smoothing of D is a diagram in which every crossing of D has been resolved (either by a 0-resolution or a 1-resolution). Note that a smoothing has no crossings. ♦ n Definition 2.9. Let { 0 , 1 } denote the set of n -tuples, where each component is either 0 or