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Math 3361-Modern Algebra — Lecture 13 10/13/14-10/15/14 1. More on the Conway Polynomial On Monday, I had you construct the Conway polynomial starting with simple knots and working up. In order to do this, you have to know where you’re going. Starting with the knot you’re interested in is more straight forward. We are given that the Conway polynomial for the unknot is 1, and it’s pretty easy to show that any unlinked link has Conway polynomial 0. You can basically just start with a knot, and keep changing crossings until you’re down to unknots and unlinked links. Let me illustrate with the figure-eight knot. ∇A − ∇B = x ·∇C Figure 1. A is an unknot, B is a figure-eight knot, and C is a simple link. The figure-eight knot is the knot B in Figure 1, and I’ve given it an orientation as shown by the arrows (a different orientation might yield a different polynomial). I’ve also chosen to work with the crossing in the middle. With the orientation I’ve chosen, this crossing is left-handed. Changing the crossing yields a knot with a right-handed crossing, the knot A, which is an unknot, and a knot with a smoothed crossing, the knot C, which is a simple link. ∇C − ∇D = x ·∇E Figure 2. C is the simple link from Figure 1, D is an unlinked link, and E is an unknot. The simple link C in Figure 1 is not an unknot nor an unlinked link, so I want to break it down further. I’ve chosen the crossing at the upper-right, which is a right-handed crossing. See Figure 2. Changing this crossing to a left-handed crossing yields the unlinked link D, and smoothing this crossing yields the unknot E. I am now down to unknots and unlinked links, and have put this information into the equations ∇A −∇B = x ·∇C (1) ∇C −∇D = x ·∇E 1 2 to get 1 −∇B = x ·∇C (2) ∇C − 0 = x · 1. The second equation tells us that ∇C = x, and it follows that 2 (3) ∇B =1 − x , so this is the Conway polynomial for the figure-eight knot. 2. Connect Sums correspond to Polynomial Multiplication One really amazing property of the Conway polynomial is the following. (4) ∇A#B = ∇A ·∇B. For example, if we were to take the connect sum of a figure-eight knot and a trefoil, then the Conway polynomial for the connect sum would be (5) (1 − x2)(x2 +1)=1 − x4. This by itself proves that the figure-eight and trefoil cannot be connect sum inverses, since the connect sum in this case is not the unknot, which would have Conway polynomial 1. Let’s go through the computation of the Conway polynomial for the connect sum of a figure-eight and a trefoil. In the process, we’ll see why the Conway polynomials multiply. Figure 3. The figure-eight knot connect sum with the trefoil. Look at Figure 3. I’ve connect summed the trefoil to the figure-eight. I made the trefoil small, because I’m not going to do anthing to it. I’m just going to repeat the computations for the figure-eight knot. The connect sum of the trefoil and figure-eight knot is knot B in Figure 4, and it has a left-handed crossing where the arrows are. Knots A and C have the right-handed and smoothed versions. ∇A − ∇B = x ·∇C Figure 4. Changing a crossing on the connect sum of the figure-eight and the trefoil. 3 Continuing on as we did before, we break C down further. The upper-right crossing is right-handed, so if we change that crossing to its left-handed and smoothed versions, we get the knots of Figure 5. ∇C − ∇D = x ·∇E Figure 5. Breaking down C a little further We the two equations ∇A −∇B = x ·∇C (6) ∇C −∇D = x ·∇E , which look the same as with the figure-eight by itself, but the knots A and E are trefoils instead of unknots. The knot D is still an unlink. Let’s let ∇T be the Conway polynomial for the trefoil, and substituting this information into the equations (6), we get ∇T −∇B = x ·∇C (7) ∇C − 0 = x ·∇T . This gives us (8) ∇C = x ·∇T , and then (9) ∇T −∇B = x · (x ·∇T ). Therefore, 2 2 (10) −∇B = x ·∇T −∇T = (x − 1) ·∇T , and 2 2 (11) ∇B = −(x − 1) ·∇T = (1 − x ) ·∇T . In this particular case, if we let F be the figure-eight knot, then we’ve just established that (12) ∇F #T = ∇F ·∇T . 3. Homework 13 For problems 1-5, assume the following. In the computations we did with the figure-eight knot connect sum with the trefoil (F #T ), suppose we connect summed with another knot K instead of T . We would have K 2 where T is in Figures 4 and 5. Let ∇K be the Conway polynomial for K. Remember that ∇F =1 − x . 1. What would the knot A be in the new Figure 4? (a) The unknot. (b) A trefoil. (c) An unlinked link. (d) The knot K. (e) none of these 2. What would the knot D be in the new Figure 5? (a) The unknot. (b) A trefoil. (c) An unlinked link. (d) The knot K. (e) none of these 4 3. What would the knot E be in the new Figure 5? (a) The unknot. (b) A trefoil. (c) An unlinked link. (d) The knot K. (e) none of these 4. What would ∇C be? 2 (a) x ·∇K (b) ∇K (c) x ∇K (d) 1 (e) none of these 5. What would ∇B be? 2 2 (a) (1 − x ) ·∇K (b) ∇T (c) (1 − x )∇T (d) ∇K (e) none of these Mystery Knot A Mystery Knot B Figure 6. A#B, the connect sum of two mystery knots. For problems 6-10, refer to Figure 6. It shows two mystery knots connect summed together with a twist (for fun). You’re going to show that the unlinked link of A and B has Conway polynomial 0. 6. Is the crossing shown left-handed or right-handed? (a) right-handed (b) left-handed (c) it depends on what knot A is (d) it depends on what knot B is (e) none of these 7. Let’s say that A#B = R. The other two knots/links will be L and S (as in Lecture 12), if we change this crossing. What is L? (a) The same knot, A#B = R (b) The mirror image of the knot A#B = R (c) Definitely an unlinked link. (d) Definitely the unknot. (e) none of these 8. What’s special about the knot/link S? (a) It’s an unlinked link (b) It’s an unknot (c) It’s the same as its mirror image. (d) It’s very flexible and can tie a trefoil into its right arm. (e) none of these 9. What can we say about ∇S ? (a) ∇S must be zero. (b) ∇S might be zero, but might not be. (c) ∇S = 1. (d) ∇S =6 0. (e) none of these 10. What can we say about the Conway polynomial of unlinked links in general? (a) They must be zero. (b) They might be zero, but don’t have to be. (c) They can’t be zero. (d) They must be 1. (e) none of these Homework continued on next page. 5 Figure 7. A knot like the figure-eight knot but with another twist in the middle. 11. Find the Conway polynomial for the knot shown in Figure 7. The polynomial will be ∇K = a + bx + cx2 + dx3 + ex4. 11a. a =? 11b. b =? 11c. c =? 11d. d =? 11e. e =?.
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