Three-Variable Bracket Polynomial for Two-Bridge Knots

Transcription

1 Three-Variable Bracket Polynomial for Two-Brige Knots Matthew Overuin Research Experience for Unergrauates California State University, San Bernarino San Bernarino, CA August 23, 2013 Abstract In this paper, we erive recursive formulas for a twist connecte to a tangle an for a twist that connects to a tangle in two places. We use these formulas to erive the three-variable bracket polynomial for two-brige knots. This erivation allows us to calculate the number of states corresponing to the maximal exponent of in a two-brige knot with no single crossings. Introuction A knot is simply efine as a close curve in R 3. An alternating knot is a knot where the crossings alternate between going uner an going over. A reuce alternating iagram of a knot can be thought of as the simplest way to epict the knot which minimizes the number of crossings. A flype is a move that creates a twist in the knot. The Tait Flyping Theorem states that two reuce alternating iagrams of the same knot can be eforme into each other by a sequence of flypes. The crossing number is efine as the number of crossings in a knot. A twist is a portion of the knot with a set of consecutive crossings. An A-twist is a twist where the A regions are containe insie the twist. An A-region is a region above a crossing where the overcrossing goes to the left. A B-twist is a twist where the B regions are containe insie the twist. A B-region is a region above a crossing where the overcrossing goes to the right. The twist number is efine as the number of twists in a knot. A two-brige knot is a knot which has only two maxima an minima. A torus knot is a knot that can be embee onto the surface of a torus. A two-torus knot is a knot that can be embee onto the surface of a 2-torus. The three variable bracket polynomial of a reuce alternating knot is invariant uner flypes. 1

2 . A-Twist B-Twist 2-Brige Knot The three- variable bracket polynomial is one way of escribing a reuce alternating iagram of a knot uniquely. The three-variable bracket polynomial of a knot can be compute by using the following rules: 1) O = 1. 2) O U K = K. 3) = A +. Example 1. The three variable bracket polynomial of the trefoil in Figure 1 is foun in the following way, = B + A = B(A + B )+ A(A +B ) = =B(A(A + B ) + B(A + B ) + =A(A(A + B ) + B(A + B ). = A 3 + 3AB 2 + 3A 2 B + B 3 2. Figure 1. 2

3 Results Lemma 1. For a B-twist, the three variable bracket polynomial is given by, = (A+B) n. Proof. Let n=1, then we have that, = A + B = (A+B). Therefore, the formula works when n=1. Now suppose that the formula is true for n-1, then we have that, = (A + B) n 1. We also have the following recursive relation, = A + B. = (A+B). Combining the inuction hypothesis with the recursive relation, we arrive at the formula, = (A+B) n. Therefore, by inuction, the lemma must be true. Lemma 2. For a B-twist, the three-variable bracket polynomial is given by, 3

4 = (A + B)n B n + B n. Proof. Let n=1, then the recursive efinition of the three variable bracket polynomial gives, = A + B. Therefore, the statement is true for n=1. Now suppose that the statement is true for n-1, then, = (A + B)n 1 B n 1 +B n 1. We also have the following recursive relation, = A + B. Using Lemma 1 an combining the inuction hypothesis with the recursive relation above, we arrive at the following formula, = (A + B)n B n +B n. Therefore, by inuction, the lemma must be true. For A twists, we simply change the A s to B s an the B s to A s in the above formulas. The proof is left to the reaer. The following formulas apply to A- twists. = (B+A) n. 4

5 = (B + A)n A n + A n. These formulas allow us to calculate the bracket polynomial for two brige knots quickly. The example below emonstrates this. Corollary 1. For a two-torus knot, T 2,n, the three variable bracket polynomial is given by, T 2,n = (A + B)n + ( 2 1)B n. Proof. T 2,n = (A + B)n B n + B n. = (A + B)n B n + B n. = (A + B)n + ( 2 1)B n Lemma 3. Suppose we have the knot shown in the figure below, where c 1 is an A-twist an c 2 is a B-twist, the three-variable bracket polynomial is then given by, = (B+A)c 1 A c 1 ( (A+B)c 2 B c 2 + B c2 )+A c1 ((B + A) c2 ). Proof. = (B+A)c 1 A c 1 + A c1. Note that, = (B + A) c2. = (A+B)c 2 A c 2 + B c2. 5

6 Substituting the secon an thir equations into the first equation, we get the equality in the lemma. In the theorem to follow, we mean L(c 1, c 2, c 3,...c n ) to be the three variable bracket of a two-brige knot with twists. Where c 1 is the number of crossings in the first twist, c 2 is the number of crossings in the secon twist, an so on. α ci α ci = (A + B)ci B ci = (B + A)ci A ci if c i an β ci = B ci (B + A) ci+1 if c i is a B-twist. if c i an β ci = A ci (A + B) ci+1 if c i is an A-twist. Theorem 1. The three variable bracket polynomial of any two-brige knot can be recursively efine as, L(c i, c i+1, c i+2,...c n ) = α ci L(c i+1, c i+2, c i+3,...c n ) +β ci L(c i+2, c i+3, c i+4,...c n ). Proof. We will treat the two brige knot with an o number of twists first. We can assume that the top twist is an A-twist. By Lemma 2, we have that, = (B + A)ci A ci + A ci. We also know that, = (A + B) ci+1. By Lemma 1 an knowing that box i + 1 is a B-twist. Substituting, we arrive at the following equality, = (B + A)ci A ci + A ci (A + B) ci+1.. Letting L(c i, c i+1,...c n ) =, we arrive at the following result, 6

7 L(c i,...c n ) = α ci L(c i+1,...c n ) + β ci L(c i+2,...c n ). Next, we will look at the case when the two brige knot has an even number of twists. We know that in this two brige knot, the top twist must be a B-twist. By Lemma 2, we have that, = (A + B)ci B ci + B ci. By Lemma 2 an knowing that box i + 2 is an A-twist, = (B + A) ci+1. Substituting, we arrive at the following result, = (A + B)ci B ci + B ci (B + A) ci+1. Letting L(c i, c i+1,...c n ) =, we arrive at the following result, L(c i,...c n ) = α ci L(c i+1,...c n ) + β ci L(c i+2,...c n ). We notice that L(c n ) =. This is just a 2-torus knot where c n is an A-twist. Therefore, the polynomial L(c n ) = (B + + ( 2 1)A A)cn cn. Theorem 2. The initial conitions for the theorem above are that with L(c n ) = (B + A) cn + ( 2 1)A cn an L(c 2 ) = (B+A)c 1 A c 1 ( (A+B)c 2 B c 2 + B c2 )+A c1 ((B+ A) c2 ). Proof. We can untwist the knot until we get the bottom twist, an the bottom twist is just a torus knot of type (2,n). This torus knot has the equation from the lemma which happens to be the equation state above. Now the secon initial conition is erive from the fact that we have two twists in a torus knot. One happens to be an A-twist, the other happens to be a B-twist an we arrive at the formula above. This follows irectly from the lemma. From the previous theorem, we see that the three variable bracket polynomial is a summation of a bunch of sequences of α s an β s. For instance, α c1 α c2...α cn is 7

8 a term in the three variable bracket polynomial. We also notice that we can obtain other terms by replacing any two consecutive α s with one β. For instance, α c1 β c2 α c4...α cn an α c1 β c2 β c4 α c6...α cn are two such terms in the polynomial. Example 2. The three-variable bracket polynomial of the two-brige knot L(3, 4, 4) is equal to α c1 α c2 α c3 + β c1 α c3 + α c1 β c2 where α c1 = (B+A)3 +( 2 1)A 3, α c2 = (A+B) 4 B 4, α c3 = (B+A)4 A 4, β c1 = A 3 (A + B) 3, an β c2 = B 4 (B + A) 4. Theorem 3. Consier a two brige knot with all twists having more than one crossing each. The maximal egree of in the bracket polynomial of the twobrige knot is equal to the crossing number minus the twist number. Proof. Consier the all alpha term, α c1 α c2 α c3...α cn. The highest exponent of here is C T. This is obtaine by aing up all the terms, c 1, c 2,...c n an subtracting the twist number. Now suppose we replace α ck with β ck, then we elete α ck+1 as well. Since all twists have two or more crossings, the exponent of in the resulting sequence has to be less than or equal to C T. Therefore any sequence in the three variable bracket polynomial must have the highest exponent of less than or equal to C T. We can conclue from this that the maximal egree of the bracket polynomial is equal to the crossing number minus the twist number in this case. In the next theorems, we mean a block of 2-twists to be a set of consecutive twists in a two-brige knot with 2 crossings each. We mean the length of a block of 2-twists to be the number of 2-twists in the block. For example, in Figure 2, the two-brige knot has a block of 2-twists with length 4.. Figure 2 Theorem 4. Let L be a two-brige knot with no twist of single crossings an with the bottom twist having three or more crossings. The number of states corresponing to the maximal egree of is enote by, F n1+1f n2+1...f nk +1. Where n 1 is the length of the first block of 2-twists, n 2 is the length of the secon block of 2-twists, an so on. Proof. Consier the two brige knot L(c 1, c 2,...c n ), we claim that F n1+1...f nk +1 is equal to the number of sequences in the three variable bracket polynomial having a highest power of of C T. We will first prove this by inuction by consiering a two-brige knot with one block of 2-twists, an increasing this block by a 8

9 length of 1. Suppose we have a two-brige knot with only one 2-twist an without twists of single crossings. We know that one such term in the three-variable bracket polynomial is α c1 α c2 α c3...α cn. This clearly has a highest exponent of of C T. Now consier the term α c1 α c2 α c3...α ck 1 β ck α ck+2...α cn where α ck+1 is a 2-twist. Calculation shows that this term also has an exponent of of C T. If we replace any other α with a β, the exponent will go own. Hence there are only two possible terms corresponing to the highest exponent of of C T. This is inee equal to F 1+1 = F 2. Now suppose we have a two-brige knot with only one block of two-twists of length n, an suppose that the number of terms corresponing to the highest exponent of is F n+1. Block = c j, c j+1...c j+n 1. Now suppose we increase the length of the block by aing on one more 2-twist. Then we a on c j+n. Now consier the term, α c1 α c2...α cj α cj+1...α cj+n...α cl. If we replace α cj+n 1 with β cj+n 1, then we elete α cj+n, an the term becomes α c1 α c2...α cj α cj+1...β cj+n 1 α cj+n+1...α cl. Now we are ealing with the block = c j, c j+1...c j+n 2. An the number of terms corresponing to the highest power of here is F n. In aition, we can have the term, α c1 α c2...α cj α cj+1...α cj+n α cj+n+1...α cl. An the number of terms corresponing to the highest exponent of here is F n+1. In total, the number of terms corresponing to the highest exponent of in the resulting block is F n + F n+1 = F n+2. Which is precisely the Fibonacci sequence. Now suppose we have a two-brige knot with only one block of 2-twists of length n. It was shown that the amount of states corresponing to the maximal egree is F n+1. Now suppose we have k blocks of two twists of length n 1, n 2, n 3...n k respectively. An suppose that the number of terms corresponing to the maximal egree of is F n1+1f n2+1f n3+1...f nk +1. Now suppose we a on another block so we have k + 1 blocks in total. For each sequence on the k + 1 block we have F n1+1f n2+1f n3+1...f nk +1 sequences, so in total we have F n1+1f n2+1f n3+1...f nk+1 +1 sequences corresponing to the k + 1 blocks of 2-twists. In conclusion, the theorem is true. Theorem 5. Let L be a two brige knot with no twist having a single crossing or a ouble crossing an with the bottom twist having three or more crossings. The number of states corresponing the maximum egree of minus one is equal to n + c, where n is the number of 3-twists. Proof. Consier the two-brige knot, L(c 1, c 2,...c n ). Now consier the all alpha state, α c1 α c2 α c3...α cn. The exponent of here is C T. An the secon highest exponent of is C T 1. Now suppose we replace α k with a β k where α k+1 is a 3-twist. Then our new term is α c1 α c2 α c3...β ck α ck+2..α cn. Since α k+1 is not in the term, the exponent of goes own by three. But the exponent of also goes up by one since β k replaces α k. Therefore the exponent of in the above term is C (T 1) = C T 1. Therefore, this state contributes 1. If we replace any other α with a β, then the resulting exponent of must be less than C T 1. So we can only replace two alphas with one beta. But we can o this with all 3-twists, so the number of states contribute to the secon 9

10 highest exponent of must be equal to n where n is the number of 3-twists. We can also look at the all alpha term, clearly the number of states corresponing to the secon highest exponent of is equal to c. So in total, there are n + c states corresponing to the secon highest power of. Open Questions How can we exten these results to two-brige knots with single crossings? That is, what are the number of states corresponing to the highest exponent of an the secon highest exponent of in a two-brige knot if single crossings are allowe? What form o the states take on corresponing to the highest exponent of an the secon highest exponent of? Acknowlegements I woul like to thank Dr. Trapp for proviing guiance through this program an I woul also like to thank Dr. Dunn for his lectures on ifferential geometry. I woul also like to thank California State University, San Bernarino an NSF grant for jointly funing this research program. References [1] Kauffman, L. H., State moels an the Jones Polynomial, Topology, 26, pp , 1987 [2] Twist Numbers of Links from the Jones Polynomial. CSUSB REU, [3] Lee,E.,Lee,S.,Seo,M., A Recursive Formula for the Jones Polynomial of 2- Brige Links an Applications, J. Korean Math. Soc. 46 (2009), No. 5, pp