Temperley Lieb Algebra I

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1 Temperley Lieb Algebra I Uwe Kaiser Boise State University REU Lecture series on Topological Quantum Computing, Talk 3 June 9, 2011

2 Kauffman bracket Given an oriented link diagram K we define K Z[A, B, δ] satisfying the following conditions: 1. If K is regularly isotopic (i. e. they differ by planar isotopy and Reidemeister II and III moves) to K then K = K. 2. = 1, where is the unknot. 3. K = δ K 4. = A + B It is important to note that 2. is a normalization that is not always taken. Sometimes, in particular in applications to physics, the bracket of the empty diagram is normalized to 1 and thus = δ.

3 Note that the distinction between A, B is well-defined by what region is swept over if the over-crossing arc turns counter-clock wise but requires a chosen orientation of the plane. A B B A Of course 4. above implies also: = B + A and several applications of 3. imply n = δ n 1 where n is a disjoint union of n trivial circles.

4 Invariance under Reidemeister II This implies and is implied by: B = A 1 and A 2 + A 2 + δ = 0 and thus the loop value is A 2 A 2.

5 Reidemeister III The invariance under Reidemeister III now follows from the invariance under Reidemeister II:

6 Bracket and kinks There are the two kink relations: = A 3 and = A 3

7 Proof Similarly for the negative kink.

8 Examples

9 State summation Define a state of an unoriented diagram K a choice of smoothing (A or B = A 1 ) for each crossing of K. The choice for a specific crossing is called a vertex weight of the state. Because there are two choices of smoothing at each crossing there are 2 N states for each diagram with N crossings. For each state S let K S denote the product of its vertex weights. Let S denote the number of loops resulting from smoothing all crossings according to the state. Define the state summation K = S K S δ S 1. It follows that K satisfies the properties for the definition of the bracket. Thus the notation is justified, and the bracket can be calculated as a state sum.

10 Properties of the bracket For a given state of a diagram K with N crossings, if there are i A-crossings then there are always (N i) B-crossings. Thus K S = A i A (N i) = A 2i N. Since δ S 1 does only contain even powers of A it follows that the power all monomials in K is congruent to N mod-2. If K is the diagram with all crossings switched then states S of K correspond to states S of K such that S = S and if K S = A i B N i = A 2i N then K S = B i A N i = A N 2i. Thus K (A) = K (A 1 ).

11 Jones polynomial Define for each oriented diagram K f K (A) = ( A 3 ) w(k) < K > (A) where w(k) is the writhe of the diagram. Then f K (A) is invariant under all Reidemeister moves, and thus defines a link invariant (Proof below!). Definition The Jones polynomial of a link is defined by V K (t) = f K (t 1/4 ) Z[t ±1/2 ]

12 Writhe Note that the writhe of a knot diagram does not depend on the orientation of the knot: w( ) = w( ) Example w( ) = 3 w( ) = 3 Note that w(k) = w(k) if K is the oriented diagram defined by switching all crossings of the oriented diagram K.

13 Invariance of the Jones polynomial The writhe is obviously invariant under Reidemeister II and III moves. We have proved the same for K. Thus f K (A) (and also V K (t)) are invariant under Reidemeister II and III moves. Invariance under Reidemeister I follows from = A 3 = A 3 K and w( ) = w( ) + 1 = w(k) + 1, thus f (A) = ( A 3 ) w( ) (A) = ( A 3 ) w(k) ( A 3 ) 1 ( A 3 ) K = f K (A)

14 Example For the right-handed trefoil K = we have w(k) = 3 and thus f K (A) = ( A 3 ) 3 K = A 9 ( A 5 A 3 +A 7 ) = A 4 +A 12 A 16 and for the left handed trefoil K = we have w(k) = 3 and thus f K = ( A 3 ) 3 K = A 9 ( A 5 A 3 +A 7 ) = A 4 +A 12 A 16 f K (A) and thus the left and right handed trefoil are not equivalent knots.

15 Properties of the Jones polynomial If K! is the link K with the orientations of all components switched then w(k! ) = w(k) and thus V K! = V K. Because w(k) = w(k) and K (A) = K (A 1 ) it follows that f K (A) = f K (A 1 ) and thus V K (t) = V K (t 1 ). Conjecture If V K (t) = V (t) = 1 then K is the trivial knot.

Recall that a diagram is alternating if starting on any circle traversing the crossings along that circle over- and under-crossings alternate.

16 Recall that a diagram is alternating if starting on any circle traversing the crossings along that circle over- and under-crossings alternate. The Jones polynomial has been the essential too in proving the so called Tait Conjectures in the 19th century. Definition A diagram is called reduced if it has no isthmus. An isthmus of a diagram can always be eliminated as follows

17 Tait Conjectures I Theorem Any reduced diagram of an alternating knot has fewest possible crossings. (proved 1987 by Thistethwaite, Kauffman and Murasugi) Theorem A reduced alternating link diagram with writhe 0 is chiral, i. e. the underlink link K is equivalent to K. (proved 1987 by Murasugi)

18 Tait Conjectures II Theorem Given any two diagrams D 1, D 2 of a prime oriented alternating link, then D 1 can be transformed into D 2 by a finite sequence of flypes: (proved 1991 by Thistlethwaite, Menasco)

19 Literature and Exercises Literature 1. L. H. Kauffman and S. Lomonaco: Topological Quantum Information Theory 2. L. H. Kauffman: State models and the Jones polynomial, Topology 26 (1987), Exercises 1. Calculate the Kauffman bracket of projections given in the knot table, Talk 1 page Prove the switching formula A A 1 = (A 2 A 2 ) 3. Prove that alternating amphichiral knots (i. e. K is not equivalent to K) have even crossing number.

= A + A 1. = ( A 2 A 2 ) 2 n 2. n = ( A 2 A 2 ) n 1. = ( A 2 A 2 ) n 1. We start with the skein relation for one crossing of the trefoil, which gives:

= A + A 1. = ( A 2 A 2 ) 2 n 2. n = ( A 2 A 2 ) n 1. = ( A 2 A 2 ) n 1. We start with the skein relation for one crossing of the trefoil, which gives: Solutions to sheet 4 Solution to exercise 1: We have seen in the lecture that the Kauffman bracket is invariant under Reidemeister move 2. In particular, we have chosen the values in the skein relation