aa r X i v : . [ m a t h . G T ] M a r KNOT INVARIANT WITH MULTIPLE SKEIN RELATIONS
ZHIQING YANG
Abstract.
Given any oriented link diagram, one can construct knot invariants using skeinrelations. Usually such a skein relation contains three or four terms. In this paper, the authorintroduces several new ways to smooth a crossings, and uses a system of skein equations toconstruct link invariant. This invariant can also be modified by writhe to get a more powerfulinvariant. The modified invariant is a generalization of both the HOMFLYPT polynomial andthe two-variable Kauffman polynomial. Using the diamond lemma, a simplified version of themodified invariant is given. It is easy to compute and is a generalization of the two-variableKauffman polynomial. Introduction
Polynomial invariants of links have a long history. In 1928, J.W. Alexander [2] discoveredthe famous Alexander polynomial. It has many connections with other topological invariants.More than 50 years later, in 1984 Vaughan Jones [5] discovered the Jones polynomial. Soon,the HOMFLYPT polynomial [4][9] was found. It turns out to be a generalization of both theAlexander polynomial and the Jones polynomial. There are other polynomials, for example, theKauffman 2-variable polynomial. All those polynomials satisfy certain skein relations, whichare linear equations concerning several link diagrams. A natural questions is whether they canbe further generalized. In this paper, the author presents a new approach to construct linkinvariant. It is a natural generalization of both the HOMFLYPT polynomial and the 2-variableKauffman polynomial. This is a rewritten and improved version of an earlier preprint of theauthor [12].For simplicity, we use the following symbols to denote link diagrams. In Fig. 1, letters
E, S, W, N mean the east, south,west and north directions as in usual maps, + means positivecrossing, − means negative crossing. For example, S + means the middle of the two arrows issouth direction, and the crossing is of positive type. S means the middle of the two arrowsis south direction, and there is no crossing. Similarly, we have the local diagrams N + , N − , N,W + , W − , W, S + , S − , S . The diagram HC means that it is horizontal, and rotating clockwise.Similarly, V T means that it is vertical, and rotating anticlockwise.For a local crossing E + or E − of an oriented link diagram, we propose the following new skeinrelations. If the two arrows/arcs in the local diagram are from the same link component, then f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0(1)If the two arrows/arcs are from different components, then
Date : September 21, 2018.2000
Mathematics Subject Classification.
Primary 57M27; Secondary 57M25.
Key words and phrases. knot invariant and knot polynomial and skein relation and diamond lemma .The author is supported by a grant (No. 11271058) of NSFC.. E + E − E NS − N + W V CHC S HT V T
Figure 1.
Local Diagrams with old notations. f ( E + ) + b ′ f ( E − ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0 . (2)Now, let X denote the quotient ring Z [ b, b ′ , c , c , c , c , d , d , b ′ , c ′ c ′ , d ′ , d ′ , v , v , · · · ] /R ,where R = R ∪ { d ′ = d ′ , (1 + b + d + d ) v n + ( c + c + c + c ) v n +1 = 0 , for all n = 1 , , , · · · } . R is given in the next section. Here is our main theorem. Theorem 1.
For oriented link diagrams, there is a link invariant f with values in X and satisfiesthe following skein relations:(1) If the two strands are from same link component, then f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 . (2) Otherwise, f ( E + ) + b ′ f ( E − ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0 . The value for a trivial n-component link is v n .In general, replacing X by any homomorphic image of X , one will get a link invariant. This invariant can also be modified by the writhe, like the Kauffman bracket and the Kauff-man 2-variable polynomial [6]. Let A be a new variable. Let Y denote the quotient ring Z [ A, A − , b, b ′ , c , c , c , c , d , d , b ′ , c ′ c ′ , d ′ , d ′ , v , v , · · · ] /R , where R = R ∪{ d ′ = d ′ , AA − =1 , Av n + A − bv n + ( c + c + c + c ) v n +1 + ( d + d ) v n = 0 , for all n = 1 , , , · · · } . Then wehave the following theorem. Theorem 2.
There is a link invariant F with values in Y . For oriented link diagram D , F ( D ) = f ( D ) A − w where w is a the writhe of the link diagram, and f satisfies the followingskein relations.(1) If the two strands are from same link component, then f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 . (2) Otherwise, f ( E + ) + b ′ f ( E − ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0 . The value of F for a trivial n-component link is v n . NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 3
In general, replacing Y by any homomorphic image of Y , one will get a link invariant. The coefficients of each invariant come from some commutative ring. Homomorphisms andrepresentations of those rings define new link invariants. Some choices lead to knot polynomials.For example, if in the ring X we add the following relations c = c = c = d = d = c ′ = d ′ = d ′ = 0 and b = b ′ , then we get a generalized HOMFLYPT polynomial with three variables b, c , c . If we ask c = c ′ , then the invariant we get is equivalent to the usual HOMFLYPTpolynomial by some variable change.If we set c = c = c = c = − z/ , c ′ = c ′ = − z/ , d = d = d ′ = d ′ = z/
2, and b = b ′ = −
1, and modify it by writhe, then we can get the 2-variable Kauffman polynomial.Hence the modified invariant is a generalization of both HOMFLY polynomial and 2-variableKauffman polynomial.Compare with the well-known knot polynomials, there are a few differences here. (1) Theskein relation has 2 cases. (2) The coefficients now are from a commutative (or non commutative)ring, and there are some nontrivial relations among them. (3) The skein relation is not localhere. This means for a given oriented diagram D , if we use the skein relation, the diagram is notonly changed locally. The orientation change is globally. To avoid contradictions, not all kindsof diagrams are allowed, and the coefficients have to satisfy certain relations. This is why we donot have a polynomial invariant. The invariant takes value in a commutative ring.Our work was motivated by Jozef H. Przytycki and Pawel Traczyk’s paper [9], and V. O.Manturov’s proofs in his book [7]. Our construction and proof is a modification and improvementof their work. 2. Full resolution commutativity
Orientation of diagrams.
For simplicity, the symbol E + ( E − , etc.) has two meanings inthis paper. It denotes (i) the whole link diagram with the special local pattern, (ii) the value ofour invariant on the diagram E + . In this section, instead of writing f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 , we write E + + bE − + c E + c W + c HC + c HT + d V C + d V T = 0 . In later sections, we use f ( E + ) to denote the value of our invariant on the diagram E + .As mentioned before, we propose the following new skein relations. If the two arrows/arcsare from the same link component, then E + + bE − + c E + c W + c HC + c HT + d V C + d V T = 0 . If the two arrows/arcs are from different components, then E + + b ′ E − + c ′ E + c ′ W + d ′ S + d ′ N = 0 . Each diagram/term in the equations is canonically orientated as follows. Take the first equa-tion for example. We can draw a disk in each of the diagrams E + , E − , E + , W, HC, HT, V C, V T .Outside the disks, all diagrams are all the same, inside the disks are as in Fig. 1. Furthermore,inside the disks, E + , E − , E + , W, HC, HT, V C, V T are already oriented. Let’s start with E + ,suppose every component of E + is oriented. Then E − and E , outside the disk one take thesame orientation as in E + . Then E − and E are oriented. The link components of W can bedivided into two sets. One set, say A, contains components passing through the disk. Then wecan extend the orientation of the disk to the whole components. For other components, we justtake the same orientation as in E + . The same can be done for HC, HT, V C, V T . ZHIQING YANG
In other words, for the link components containing the arcs in the local diagram, their ori-entations are determined by the local diagrams. For all other components, the orientation isnot changed. Since we distinguish the same/different component cases, there is no contradictionregarding to the orientation assumption.There is no S or N terms in the first equation, because if the two strands are from samecomponent, this orientation assignment will cause contradiction in orientation. Under our as-sumption for the orientation, the two equations are the maximal. If one add other diagrams,then there will be contradiction for orientation.2.2. Resolution order independence condition f pq = f qp . If we want to calculate theinvariant of a diagram D , we can start at any crossing point p . When we apply the formula ata crossing p , there are two things to check, 1. the two arcs are from same/ different component,2. the crossing is positive or negative. We call the above information the crossing pattern of p . The crossing pattern determines which skein equation to use and how to use it. For example,if p is a negative crossing point, and the two arcs are from the same link component, thenwe get: E − = − b − { E + + c E + c W + c HC + c HT + d V C + d V T } . Hence if we havedefined the invariant for E + , E, · · · , we get the invariant for E − . This is similar to the usuallycalculation of Jones polynomial by using skein relation. This also motivates us to define theinvariant inductively. Such a procedure that write one term as a linear combination of otherterms in the equation will be referred to as resolving at p . We call − b − { E + + c E + c W + c HC + c HT + d V C + d V T } a linear sum . We denote it by f p ( D ).Given a link diagram D with crossings p , · · · p n . Pick two crossings, say p, q . We can use theskein relation to resolve the diagram at a crossing p . The output is a linear combination of manyterms. Each term involves a link diagram D j . We write f p ( D ) = P α i f ( D j ). Each D j also hasa crossing point corresponding to the crossing q . For each such diagram D j , we resolve it at thepoint q . We shall get f q ( D j ), a linear combination of many terms. Add the results up, we get alinear combination of linear combinations. We denote the result by f pq ( D ) = P α i f q ( D j ). It isthe result of completely resolving at two crossing points in the order p first, then q . Similarly,if we resolve D at q first, then q , we can get another result f qp ( D ). Now, we require that forany pair p, q , f pq ( D ) = f qp ( D ). The equation f pq ( D ) = f qp ( D ) is very important in this paper.Once this condition is satisfied, one need just a few more equations to get a link invariant. Weshall discuss this condition in full detail and consider several cases. Easy cases.
Resolve at p would not change the pattern of q and vise versa.For examples, D is a disjoint union of two planar link diagrams G and G , p ∈ G , and q ∈ G . In this case, when resolve p , we get diagrams D , · · · , D k . In all the D i ’s, q has thesame crossing pattern. In D , q also has the same crossing pattern. Subcase 1.
Suppose that both p, q are positive crossings, for p , the two arrows are from samecomponent, for q , the two arrows are not from same component. When we resolve p , we get − E + = bE − + c E + c W + c HC + c HT + d V C + d V T.
Since we are discussing two crossings here, we use ( E − , E + ) to denote the first crossing p ischanged to E − , the second crossing q is E + . For the first term bE − of right hand side the aboveequation, we resolve at q and get − b ( E − , E + ) = bb ′ ( E − , E − ) + bc ′ ( E − , E ) + bc ′ ( E − , W ) + bd ′ ( E − , S ) + bd ′ ( E − , N ) . Likewise, we have the following equations.
NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 5 ( E + , E + ) = −{ b ( E − , E + ) + c ( E, E + ) + c ( W, E + ) + c ( HC, E + ) + c ( HT, E + ) + d ( V C, E + ) + d ( V T, E + ) } − b ( E − , E + ) = bb ′ ( E − , E − ) + bc ′ ( E − , E ) + bc ′ ( E − , W ) + bd ′ ( E − , S ) + bd ′ ( E − , N ) − c ( E, E + ) = c b ′ ( E, E − ) + c c ′ ( E, E ) + c c ′ ( E, W ) + c d ′ ( E, S ) + c d ′ ( E, N ) − c ( W, E + ) = c b ′ ( W, E − ) + c c ′ ( W, E ) + c c ′ ( W, W ) + c d ′ ( W, S ) + c d ′ ( W, N ) − c ( HC, E + ) = c b ′ ( HC, E − ) + c c ′ ( HC, E ) + c c ′ ( HC, W ) + c d ′ ( HC, S ) + c d ′ ( HC, N ) − c ( HT, E + ) = c b ′ ( HT, E − ) + c c ′ ( HT, E ) + c c ′ ( HT, W ) + c d ′ ( HT, S ) + c d ′ ( HT, N ) − d ( V C, E + ) = d b ′ ( V C, E − ) + d c ′ ( V C, E ) + d c ′ ( V C, W ) + d d ′ ( V C, S ) + d d ′ ( V C, N ) − d ( V T, E + ) = d b ′ ( V T, E − ) + d c ′ ( V T, E ) + d c ′ ( V T, W ) + d d ′ ( V T, S ) + d d ′ ( V T, N )We can build a table for this result. We put the crossing type of the first crossing in the firstcolumn, the crossing type of the second crossing in the first row.
Table 1.
Trivial case, resolving p first.p \ q E − E W S N E − bb ′ bc ′ bc ′ bd ′ bd ′ E c b ′ c c ′ c c ′ c d ′ c d ′ W c b ′ c c ′ c c ′ c d ′ c d ′ HC c b ′ c c ′ c c ′ c d ′ c d ′ HT c b ′ c c ′ c c ′ c d ′ c d ′ V C d b ′ d c ′ d c ′ d d ′ d d ′ V T d b ′ d c ′ d c ′ d d ′ d d ′ Other other hand, if we resolve at q first, we shall get another table. Table 2.
Trivial case, resolving q first.p \ q E − E W S N E − b ′ b c ′ b c ′ b d ′ b d ′ bE b ′ c c ′ c c ′ c d ′ c d ′ c W b ′ c c ′ c c ′ c d ′ c d ′ c HC b ′ c c ′ c c ′ c d ′ c d ′ c HT b ′ c c ′ c c ′ c d ′ c d ′ c V C b ′ d c ′ d c ′ d d ′ d d ′ d V T b ′ d c ′ d c ′ d d ′ d d ′ d Compare the results, the easiest way to make them equal is to ask the coefficients equal eachother. Therefor, we ask any element from the set { b, c , c , c , c , d , d } commutes with anyelement from the set { b ′ , c ′ , c ′ , d ′ , d ′ } .Let b = b − , c i = b − c i for i = 1 , , , d i = b − d i for i = 1 , b ′ = b ′− , c ′ i = b ′− c ′ i for i = 1 , , , d ′ i = b ′− d ′ i for i = 1 ,
2. For any pair of such symbols x and x , we call them the conjugates of each other. This has an obvious benefit as follows. In a skein relation, for example E + + bE − + c E + c W + c HC + c HT + d V C + d V T = 0, we can get E + = −{ bE − + c E + c W + c HC + c HT + d V C + d V T } and E − = −{ bE + + c E + c W + c HC + c HT + d V C + d V T } . This means if we change E + to E − (or E − to E + ), we can simply replace each x to x . Thesymmetry between them will greatly simplify our discussion later. ZHIQING YANG
When we list all the subcases, we get the conclusion that any two elements from { b, c , c , c , c , d , d , b ′ , c ′ , c ′ , d ′ , d ′ } ∪ { b, c , c , c , c , d , d , b ′ , c ′ , c ′ , d ′ , d ′ } are mutually commutative. Convention : For convenience, later on in the second table, we exchange the order of theelements of all the terms. For example, cd is changed to dc . So for an entry xy , x always comesfrom resolving the first crossing point, y always comes from resolving the second crossing point. Nontrivial cases.
Now we are going to discuss the nontrivial cases. For simplicity, we use
A, B to denote the end of the first crossing p , and C, D to denote the end of the second crossing q . We also use them to denote the oriented strands. For example, ( ACB, D ) means that thethree arcs
A, C, B are from a same link component, and their order is A → C → B along thelink orientation. D is in another component. ( AC, B, D ) means that the arcs
A, C are from asame link component, and their order is A → C along the link orientation. B is in the secondcomponent, D is in the third component. AB The first crossing p CD
The second crossing q Figure 2.
The label of two crossings.Since there is a symmetry of positive/negative crossing both in the skein relation and thediagrams, we discuss only positive crossings cases. We shall tell how to deal with the other caseslater.For the nontrivial cases, there must be one link component passing through both the cross-ings p and q . There are four arcs in the two disks containing the two crossings. They canbelong to 1,2,3 link components. If there are three components, then all the possible cases are( AC, B, D ) , ( AD, B, C ) , ( BC, A, D ) , ( BD, A, C ). Case 1: ( AC, B, D ) If we resolve the 1st crossing point p first, we shall get the followings.( E + , E + ) = −{ ( b ′ E − + c ′ E + d ′ S, E + ) + ( c ′ W + d ′ N, N − ) }− ( b ′ E − , E + ) = b ′ { ( E − , b ′ E − + c ′ E + d ′ S ) + ( N + , c ′ W + d ′ N ) }− ( c ′ E, E + ) = c ′ { ( E, b ′ E − + c ′ E + d ′ S ) + ( W, c ′ W + d ′ N ) }− ( d ′ S, E + ) = d ′ { ( S, b ′ E − + c ′ E + d ′ S ) + ( N, c ′ W + d ′ N ) }− ( c ′ W, N − ) = c ′ { ( W, b ′ N + + c ′ N + d ′ W ) + ( E, c ′ S + d ′ E ) }− ( d ′ N, N − ) = d ′ { ( N, b ′ N + + c ′ N + d ′ W ) + ( S, c ′ S + d ′ E ) } Otherwise, we shall get the followings.( E + , E + ) = −{ ( E + , b ′ E − + c ′ E + d ′ S ) + ( N − , c ′ W + d ′ N ) }− b ′ ( E + , E − ) = b ′ { ( b ′ E − + c ′ E + d ′ S, E − ) + ( c ′ W + d ′ N, N + ) }− c ′ ( E + , E ) = c ′ { ( b ′ E − + c ′ E + d ′ S, E ) + ( c ′ W + d ′ N, W ) }− d ′ ( E + , S ) = d ′ { ( b ′ E − + c ′ E + d ′ S, S ) + ( c ′ W + d ′ N, N ) }− c ′ ( N − , W ) = c ′ { ( b ′ N + + c ′ N + d ′ W, W ) + ( c ′ S + d ′ E, E ) }− d ′ ( N − , N ) = d ′ { ( b ′ N + + c ′ N + d ′ W, N ) + ( c ′ S + d ′ E, S ) } NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 7
Table 3.
Case (AC,B,D), resolving p first.p \ q E E − N + N S W
E c ′ d ′ , c ′ c ′ c ′ b ′ c ′ c ′ , c ′ d ′ E − b ′ c ′ b ′ b ′ b ′ d ′ N + b ′ d ′ b ′ c ′ N d ′ b ′ d ′ d ′ , d ′ c ′ d ′ ′ , d ′ d ′ S d ′ d ′ , d ′ c ′ d ′ b ′ d ′ c ′ , d ′ d ′ W c ′ b ′ ′ c ′ , c ′ d ′ c ′ d ′ , c ′ c ′ Table 4.
Case (AC,B,D), resolving q first.p \ q E E − N + N S W
E d ′ c ′ , c ′ c ′ c ′ b ′ d ′ d ′ , c ′ d ′ E − b ′ c ′ b ′ b ′ b ′ d ′ N + b ′ d ′ b ′ c ′ N d ′ b ′ d ′ d ′ , c ′ d ′ c ′ c ′ , d ′ c ′ S c ′ c ′ , d ′ c ′ d ′ b ′ c ′ d ′ , d ′ d ′ W c ′ b ′ d ′ d ′ , c ′ d ′ d ′ c ′ , c ′ c ′ Recall that in the second matrix we write the products in a new form. For example, (
N, N + )has coefficient b ′ d ′ , but we write d ′ b ′ in the table. We exchange the order of every product inthis matrix so that the first symbol, for example the d ′ here, is always from resolving the firstcrossing point p .The relations here are: c ′ d ′ = d ′ c ′ , c ′ c ′ = d ′ d ′ , b ′ d ′ = b ′ d ′ , b ′ c ′ = b ′ c ′ , d ′ b ′ = d ′ b ′ , d ′ d ′ + d ′ c ′ = d ′ d ′ + c ′ d ′ , d ′ c ′ + d ′ d ′ = c ′ c ′ + d ′ c ′ , d ′ d ′ = c ′ c ′ , d ′ c ′ = c ′ d ′ , c ′ b ′ = c ′ b ′ , c ′ c ′ + c ′ d ′ = d ′ d ′ + c ′ d ′ , c ′ d ′ + c ′ c ′ = d ′ c ′ + c ′ c ′ .If the two crossings are both negative crossings, then we change all the coefficients x to x .For example, ( E + , E + ) = −{ ( b ′ E − + c ′ E + d ′ S, E + ) + ( c ′ W + d ′ N, N − ) } become ( E − , E − ) = −{ ( b ′ E − + c ′ E + d ′ S, E − ) + ( c ′ W + d ′ N, N + ) } . The relations become their conjugates. Forexample, c ′ d ′ = d ′ c ′ become c ′ d ′ = d ′ c ′ .If the first crossing p is a negative crossing, q is positive, then we change the first coefficients x to x . For example, ( E + , E + ) = −{ ( b ′ E − + c ′ E + d ′ S, E + ) + ( c ′ W + d ′ N, N − ) } become( E − , E + ) = −{ ( b ′ E − + c ′ E + d ′ S, E + ) + ( c ′ W + d ′ N, N − ) } . In the relations, we change thefirst variables to their conjugates. For example, c ′ d ′ = d ′ c ′ become c ′ d ′ = d ′ c ′ . Likewise, ifthe first crossing p is a positive crossing, q is negative, we get c ′ d ′ = d ′ c ′ . In short, if we havea relation xy = cd , we will add xy = cd , xy = cd and xy = cd . We will refer to this as completethe relation by the operation.To handle the ( BC, A, D ) case, let’s first introduce another conjugation induced by taking mir-ror image. Taking the mirror image of each term of our skein relation, E + , E − , E, W, HC, HT, V C,V T, S, N are changed to E − , E + , E, W, HC, HT, V C, V T, S, N (see Fig. 3). Let “ c = c , “ c = c , “ d = d , “ d = d , “ d ′ = d ′ , “ d ′ = d ′ . For other x , b x = x . For the link ( BC, A, D ), suppose thecrossing p is negative, q is positive. Then we can change the disk at p to its mirror image, andadd virtual crossings. Then the new link is the case ( AC, B, D ). In the new link, both crossings
ZHIQING YANG are positive. Although the new link (
AC, B, D ) contain virtual crossings, all the calculations wemade before are still valid. There is a one to one correspondence between the results of completeresolving (
BC, A, D ) and (
AC, B, D ) at p, q . From the results of (
BC, A, D ) to (
AC, B, D ), themirror takes E − to E + , V C to V T and so on. Since E − is mapped to E + , we have to map x to x . Since V C is mapped to
V T , we have to map c to c = “ c and so on. Because the mirroris only placed near the first crossing p , in a relation xy = cd we only change the first variablesto get b xy = b cd . Therefor, if we have a relation xy = cd from ( AC, B, D ), we will add b xy = b cd for ( BC, A, D ). Since we also have xy = cd , we can say that if we have a relation xy = cd , wewill add b xy = b cd . Sinilarly, for ( AD, B, C ) , ( BD, A, C ), if we have a relation xy = cd , we willadd x b y = c b d and b x b y = b c b d . We will refer to this as complete the relation by the b operation.In short, if there are three components, then all the possible cases are ( AC, B, D ) , ( AD, B, C ) , ( BC, A, D ) , ( BD, A, C ), but we only need to calculate the case (
AC, B, D ) and suppose that allcrossings are positive.
WHT N W V CHC S V T mirror
Figure 3.
Mirror symmetry.To get all the equations f pq = f qp , we shall list all the possible cases that how the two strandsof p is connected to the two strands of q . Up to the positive/negative crossing type symmetry,and mirror symmetry, there are only few nontrivial cases. If there are only two components passthe two disks, up to symmetry, we have ( AC, BD ), (
ABC, D ). If there is only one componentspass the two disks, up to symmetry, we have (
ACBD ) or (
ACDB ).So, we have the following five cases. 1. (
AC, B, D ), 2. (
AC, BD ), 3. (
ABC, D ), 4. (
ACBD ),5. (
ACDB ). Case 2, ( AC, BD ) Resolving p first, we shall get the following equations.( E + , E + ) = −{ ( b ′ E − + c ′ E, E + ) + ( c ′ W, W + ) + ( d ′ N, N − ) + ( d ′ S, S − ) }− ( b ′ E − , E + ) = b ′ { ( E − , b ′ E − + c ′ E ) + ( W − , c ′ W ) + ( N + , d ′ N ) + ( S + , d ′ S ) } − ( c ′ E, E + ) = c ′ { ( E, bE − + c E ) + ( W, c W ) + ( HT, c HC ) + ( HC, c HT ) + ( HC, d V C ) + (
HT, d V T ) } − ( c ′ W, W + ) = c ′ { ( W, bW − + c W ) + ( E, c E ) + ( HT, c HC ) + ( HC, c HT ) + ( HC, d V C ) + (
HT, d V T ) } − ( d ′ N, N − ) = d ′ { ( N, bN + + c N ) + ( S, c S ) + ( V T, c V C ) + (
V C, c V T ) + (
V C, d HC ) + ( V T, d HT ) } − ( d ′ S, S − ) = d ′ { ( S, bS + + c S ) + ( N, c N ) + ( V T, c V C ) + (
V C, c V T ) + (
V C, d HC ) + ( V T, d HT ) } Resolving q first, we shall get the following equations.( E + , E + ) = −{ ( E + , b ′ E − + c ′ E ) + ( W + , c ′ W ) + ( N − , d ′ N ) + ( S − , d ′ S ) }− b ′ ( E + , E − ) = b ′ { ( b ′ E − + c ′ E, E − ) + ( c ′ W, W − ) + ( d ′ N, N + ) + ( d ′ S, S ) } − c ′ ( E + , E ) = c ′ { ( bE − + c E, E ) + ( c W, W ) + ( c HC, HT ) + ( c HT, HC ) + ( d V C, HC ) + ( d V T, HT ) } − c ′ ( W + , W ) = c ′ { ( bW − + c W, W ) + ( c E, E ) + ( c HC, HT ) + ( c HT, HC ) + ( d V C, HC ) + ( d V T, HT ) } − d ′ ( N − , N ) = d ′ { ( bN + + c N, N ) + ( c S, S ) + ( c V C, V T ) + ( c V T, V C ) + ( d HC, V C ) + ( d HT, V T ) } − d ′ ( S − , S ) = d ′ { ( bS + + c S, S ) + ( c N, N ) + ( c V C, V T ) + ( c V T, V C ) + ( d HC, V C ) + ( d HT, V T ) } NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 9
Table 5.
Case (
AC, BD ), resolving p first. p \ q E E − W W − N N + S S + HC HT VC VT
E c ′ c , c ′ c c ′ bE − b ′ c ′ b ′ b ′ W c ′ c , c ′ c c ′ bW − b ′ c ′ N d ′ c , d ′ c d ′ bN + b ′ d ′ S d ′ c , d ′ c d ′ bS + b ′ d ′ HC c ′ c , c ′ c c ′ d , c ′ d HT c ′ c , c ′ c c ′ d , c ′ d V C d ′ d , d ′ d d ′ c , d ′ c V T d ′ d , d ′ d d ′ c , d ′ c Table 6.
Case (
AC, BD ), resolving q first. p \ q E E − W W − N N + S S + HC HT VC VT
E c c ′ , c c ′ c ′ b ′ E − bc ′ b ′ b ′ W c c ′ , c c ′ c ′ b ′ W − bc ′ N c d ′ , c d ′ d ′ b ′ N + bd ′ S c d ′ , c d ′ d ′ b ′ S + bd ′ HC c c ′ , c c ′ d d ′ , d d ′ HT c c ′ , c c ′ d d ′ , d d ′ V C d c ′ , d c ′ c d ′ , c d ′ V T d c ′ , d c ′ c d ′ , c d ′ The relations here are: b ′ c ′ = bc ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ c + c ′ c = c c ′ + c c ′ , b ′ c ′ = bc ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ b = d ′ b ′ , b ′ d ′ = bd ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ b = d ′ b ′ , b ′ d ′ = bd ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ d + c ′ d = d d ′ + d d ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ d + c ′ d = d d ′ + d d ′ , d ′ d + d ′ d = d c ′ + d c ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ d + d ′ d = d c ′ + d c ′ , d ′ c + d ′ c = c d ′ + c d ′ . Case 3, ( ABC, D ) Resolving p first, we shall get the following equations.( E + , E + ) = −{ ( bE − + c E + c HC + d V T, E + ) + ( c W + c HT + d V C, N − ) }− ( bE − , E + ) = b { ( E − , b ′ E − + c ′ E + d ′ S ) + ( W − , c ′ W + d ′ N ) }− ( c E, E + ) = c { ( E, b ′ E − + c ′ E + d ′ S ) + ( HT, c ′ W + d ′ N ) }− ( c HC, E + ) = c { ( HC, b ′ E − + c ′ E + d ′ S ) + ( W, c ′ W + d ′ N ) }− ( d V T, E + ) = d { ( V T, b ′ E − + c ′ E + d ′ S ) + ( V C, c ′ W + d ′ N ) }− ( c W, N − ) = c { ( W, b ′ N + + c ′ N + d ′ W ) + ( HC, c ′ S + d ′ E ) }− ( c HT, N − ) = c { ( HT, b ′ N + + c ′ N + d ′ W ) + ( E, c ′ S + d ′ E ) }− ( d V C, N − ) = d { ( V C, b ′ N + + c ′ N + d ′ W ) + ( V T, c ′ S + d ′ E ) } Resolving q first, we shall get the following equations.( E + , E + ) = −{ ( E + , b ′ E − + c ′ E + d ′ S ) + ( W + , c ′ W + d ′ N ) }− ( E + , b ′ E − ) = b ′ { ( bE − + c E + c HC + d V T, E − ) + ( c W + c HT + d V C, N + ) }− ( E + , c ′ E ) = c ′ { ( bE − + c E + c HC + d V T, E ) + ( c W + c HT + d V C, W ) }− ( E + , d ′ S ) = d ′ { ( bE − + c E + c HC + d V T, S ) + ( c W + c HT + d V C, N ) }− ( W + , c ′ W ) = c ′ { ( bW − + c W + c HT + d V C, W ) + ( c E + c HC + d V T, E ) } Table 7.
Case (
ABC, D ), resolving p first.p \ q E E − W N N + S E c d ′ , c c ′ c b ′ c c ′ , c d ′ E − bc ′ bb ′ bd ′ W c d ′ , c c ′ c c ′ , c d ′ c b ′ W − bc ′ bd ′ HC c d ′ , c c ′ c b ′ c c ′ , c d ′ HT c d ′ , c c ′ c c ′ , c d ′ c b ′ V C d d ′ , d c ′ d c ′ , d d ′ d b ′ V T d d ′ , d c ′ d b ′ d c ′ , d d ′ − ( W + , d ′ N ) = d ′ { ( bW − + c W + c HT + d V C, N ) + ( c E + c HC + d V T, S ) } Table 8.
Case (
ABC, D ), resolving q first.p \ q E E − W N N + S E c c ′ , c c ′ c b ′ c d ′ , c d ′ E − bc ′ bb ′ bd ′ W c c ′ , c c ′ c d ′ , c d ′ c b ′ W − bc ′ bd ′ HC c c ′ , c c ′ c b ′ c d ′ , c d ′ HT c c ′ , c c ′ c d ′ , c d ′ c b ′ V C d c ′ , d ′ d d ′ , d d ′ d b ′ V T d c ′ , d c ′ d b ′ d d ′ , d d ′ The relations here are: c d ′ = c c ′ , c c ′ = c d ′ , c d ′ + c c ′ = c c ′ + c c ′ , c c ′ + c d ′ = c d ′ + c d ′ , c b ′ = c b ′ , c d ′ = c c ′ , c c ′ = c d ′ , c d ′ + c c ′ = c c ′ + c c ′ , c c ′ + c d ′ = c d ′ + c d ′ , c b ′ = c b ′ , d d ′ + d c ′ = d c ′ + d c ′ , d c ′ + d d ′ = d d ′ + d d ′ , d b ′ = d b ′ , d d ′ = d c ′ , d c ′ = d d ′ . Case 4, ( ACBD ) Resolving p first, we shall get the following equations. ( E + , E + ) = −{ ( bE − + c E, E + ) + ( c W, W + ) + ( c HT, S − ) + ( c HC, N − ) + ( d V T, N − ) + ( d V C, S − ) } − ( bE − , E + ) = b { ( E − , bE − + c E ) + ( W − , c W ) + ( S + , c HC ) + ( N + , c HT ) + ( S + , d V T ) + ( N + , d V C ) } − ( c E, E + ) = c { ( E, b ′ E − + c ′ E ) + ( W, c ′ W ) + ( HC, d ′ N ) + ( HT, d ′ S ) }− ( c W, W + ) = c { W, b ′ W − + c ′ W ) + ( E, c ′ E ) + ( HC, d ′ N ) + ( HT, d ′ S ) }− ( c HT, S − ) = c { ( HT, b ′ S + + c ′ S ) + ( HC, c ′ N ) + ( E, d ′ E ) + ( W, d ′ W ) }− ( c HC, N − ) = c { ( HC, b ′ N + + c ′ N ) + ( HT, c ′ S ) + ( E, d ′ E ) + ( W, d ′ W ) } − ( d V T, N − ) = d { ( V T, bN + + c N ) + ( V C, c S ) + ( N, c V C ) + (
S, c V T ) + (
S, d HC ) + ( N, d HT ) } − ( d V C, S − ) = d { ( V C, bS + + c S ) + ( V T, c N ) + ( N, c V C ) + (
S, c V T ) + (
S, d HC ) + ( N, d HT ) } Resolving q first, we shall get the following equations. ( E + , E + ) = −{ ( E + , bE − + c E ) + ( W + , c W ) + ( S − , c HC ) + ( N − , c HT ) + ( S − , d V T ) + ( N − , d V C ) } − ( E + , bE − ) = b { ( bE − + c E, E − ) + ( c W, W − ) + ( c HT, S + ) + ( c HC, N + ) + ( d V T, N + ) + ( d V C, S + ) } − ( E + , c E ) = c { ( b ′ E − + c ′ E, E ) + ( c ′ W, W ) + ( d ′ N, HT ) + ( d ′ S, HC ) }− ( W + , c W ) = c { ( b ′ W − + c ′ W, W ) + ( c ′ E, E ) + ( d ′ N, HT ) + ( d ′ S, HC ) } NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 11
Table 9.
Case (
ACBD ), resolving p first. p \ q E E − W W − S S + N N + HC HT VC VT E c d ′ ,c d ′ c c ′ ,c c ′ c b ′ E − bc bbW c d ′ ,c d ′ c c ′ ,c c ′ c b ′ W − bc S d d , d d d c , d c S + bc bd N d d , d d d c , d c N + bc bd HC c c ′ ,c c ′ c d ′ ,c d ′ c b ′ HT c c ′ ,c c ′ c d ′ ,c d ′ c b ′ V C d c , d c d bV T d c , d c d b − ( S − , c HC ) = c { ( b ′ S + + c ′ S, HC ) + ( c ′ N, HT ) + ( d ′ E, E ) + ( d ′ W, W ) }− ( N − , c HT ) = c { ( b ′ N + + c ′ N, HT ) + ( c ′ S, HC ) + ( d ′ E, E ) + ( d ′ W, W ) } − ( S − , d V T ) = d { ( bS + + c S, V T ) + ( c N, V C ) + ( c V T, N ) + ( c V C, S ) + ( d HC, N ) + ( d HT, S ) } − ( N − , d V C ) = d { ( bN + + c N, V C ) + ( c S, V T ) + ( c V T, N ) + ( c V C, S ) + ( d HC, N ) + ( d HT, S ) } Table 10.
Case (
ACBD ), resolving q first. p \ q E E − W W − S S + N N + HC HT VC VT E d ′ c ,d ′ c c ′ c ,c ′ c c bE − b ′ c bbW d ′ c ,d ′ c c ′ c ,c ′ c c bW − b ′ c S c ′ c ,c ′ c d ′ c ,d ′ c c d , c d S + b ′ c bd N c ′ c ,c ′ c d ′ c ,d ′ c c d , c d N + b ′ c bd HC d d , d d c bHT d d , d d c bV C c d , c d d bV T c d , c d d b The relations here are: c d ′ + c d ′ + c c ′ + c c ′ = d ′ c + d ′ c + c ′ c + c ′ c , c b ′ = c b , bc = b ′ c , c d ′ + c d ′ + c c ′ + c c ′ = d ′ c + d ′ c + c ′ c + c ′ c , c b ′ = c b , bc = b ′ c , d d + d d = c ′ c + c ′ c + d ′ c + d ′ c , d c + d c = c d + c d , bc = b ′ c , bd = bd , d d + d d = c ′ c + c ′ c + d ′ c + d ′ c , d c + d c = c d + c d , bc = b ′ c , bd = bd , c c ′ + c c ′ + c d ′ + c d ′ = d d + d d , c b ′ = c b , d c + d c = c d + c d , d b = d b , d c + d c = c d + c d , d b = d b . Case 5, ( ACDB ) Resolving p first, we shall get the following equations.( E + , E + ) = −{ ( bE − + c E + c HT + d V C, E + ) + ( c W + c HC + d V T, W + ) }− ( bE − , E + ) = b { ( E − , bE − + c E + c HC + d V T ) + ( W − , c W + c HT + d V C ) }− ( c E, E + ) = c { ( E, bE − + c E + c HC + d V T ) + (
HC, c W + c HT + d V C ) }− ( c HT, E + ) = c { ( HT, bE − + c E + c HC + d V T ) + (
W, c W + c HT + d V C ) }− ( d V C, E + ) = d { ( V C, bE − + c E + c HC + d V T ) + (
V T, c W + c HT + d V C ) }− ( c W, W + ) = c { ( W, bW − + c W + c HT + d V C ) + (
HT, c E + c HC + d V T ) }− ( c HC, W + ) = c { ( HC, bW − + c W + c HT + d V C ) + (
E, c E + c HC + d V T ) }− ( d V T, W + ) = d { ( V T, bW − + c W + c HT + d V C ) + (
V C, c E + c HC + d V T ) } Table 11.
Case (
ACDB ), resolving p first.p \ q E − E W − W HC HT VC VT E − bb bc bc bd E c b c c , c c c c , c c c d , c d W − bc bc bd W c b c c , c c c c , c c c d , c d HC c b c c , c c c c , c c c d , c d HT c b c c , c c c c , c c c d , c d VC d b d c , d c d c , d c d d , d d VT d b d c , d c d c , d c d d , d d Resolving q first, we shall get the following equations.( E + , E + ) = −{ ( E + , bE − + c E + c HC + d V T ) + ( W + , c W + c HT + d V C ) }− ( E + , bE − ) = b { ( bE − + c E + c HT + d V C, E − ) + ( c W + c HC + d V T, W − ) }− ( E + , c E ) = c { ( bE − + c E + c HT + d V C, E ) + ( c W + c HC + d V T, HT ) }− ( E + , c HC ) = c { ( bE − + c E + c HT + d V C, HC ) + ( c W + c HC + d V T, W ) }− ( E + , d V T ) = d { ( bE − + c E + c HT + d V C, V T ) + ( c W + c HC + d V T, V C ) }− ( W + , c W ) = c { ( bW − + c W + c HC + d V T, W ) + ( c E + c HT + d V C, HC ) }− ( W + , c HT ) = c { ( bW − + c W + c HC + d V T, HT ) + ( c E + c HT + d V C, E ) }− ( W + , d V C ) = d { ( bW − + c W + c HC + d V T, V C ) + ( c E + c HT + d V C, V T ) } . Table 12.
Case (
ACDB ), resolving q first.p \ q E − E W − W HC HT VC VT E − bb bc bc bd E c b c c , c c c c , c c c d , c d W − bc bc bd W c b c c , c c c c , c c c d , c d HC c b c c , c c c c , c c c d , c d HT c b c c , c c c c , c c c d , c d VC d b d c , d c d c , d c d d , d d VT d b d c , d c d c , d c d d , d d The relations here are: c c = c c , c c = c c , c d = c d , bc = bc , c c + c c = c c + c c , c c + c c = c c + c c , c d + c d = c d + c d , c c + c c = c c + c c , c c + c c = c c + c c , c d = c d , c c = c c , c c = c c , c d = c d , d c = d c , d c = d c , d d = d d , d c + d c = d c + d c , d c + d c = d c + d c .In short, here are all the relations if the two crossings are all positive. Case 1: c ′ d ′ = d ′ c ′ , c ′ c ′ = d ′ d ′ , b ′ d ′ = b ′ d ′ , b ′ c ′ = b ′ c ′ , d ′ b ′ = d ′ b ′ , d ′ d ′ + d ′ c ′ = d ′ d ′ + c ′ d ′ , d ′ c ′ + d ′ d ′ = c ′ c ′ + d ′ c ′ , d ′ d ′ = c ′ c ′ , d ′ c ′ = c ′ d ′ , c ′ b ′ = c ′ b ′ , c ′ c ′ + c ′ d ′ = d ′ d ′ + c ′ d ′ , c ′ d ′ + c ′ c ′ = d ′ c ′ + c ′ c ′ . Case 2: b ′ c ′ = bc ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ c + c ′ c = c c ′ + c c ′ , b ′ c ′ = bc ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ b = d ′ b ′ , b ′ d ′ = bd ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ b = d ′ b ′ , b ′ d ′ = bd ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ d + c ′ d = d d ′ + d d ′ , c ′ c + c ′ c = c c ′ + c c ′ , c ′ d + c ′ d = NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 13 d d ′ + d d ′ , d ′ d + d ′ d = d c ′ + d c ′ , d ′ c + d ′ c = c d ′ + c d ′ , d ′ d + d ′ d = d c ′ + d c ′ , d ′ c + d ′ c = c d ′ + c d ′ . Case 3: c d ′ = c c ′ , c c ′ = c d ′ , c d ′ + c c ′ = c c ′ + c c ′ , c c ′ + c d ′ = c d ′ + c d ′ , c b ′ = c b ′ , c d ′ = c c ′ , c c ′ = c d ′ , c d ′ + c c ′ = c c ′ + c c ′ , c c ′ + c d ′ = c d ′ + c d ′ , c b ′ = c b ′ , d d ′ + d c ′ = d c ′ + d c ′ , d c ′ + d d ′ = d d ′ + d d ′ , d b ′ = d b ′ , d d ′ = d c ′ , d c ′ = d d ′ . Case 4: c d ′ + c d ′ + c c ′ + c c ′ = d ′ c + d ′ c + c ′ c + c ′ c , c b ′ = c b , bc = b ′ c , c d ′ + c d ′ + c c ′ + c c ′ = d ′ c + d ′ c + c ′ c + c ′ c , c b ′ = c b , bc = b ′ c , d d + d d = c ′ c + c ′ c + d ′ c + d ′ c , d c + d c = c d + c d , bc = b ′ c , bd = bd , d d + d d = c ′ c + c ′ c + d ′ c + d ′ c , d c + d c = c d + c d , bc = b ′ c , bd = bd , c c ′ + c c ′ + c d ′ + c d ′ = d d + d d , c b ′ = c b , d c + d c = c d + c d , d b = d b , d c + d c = c d + c d , d b = d b . Case 5: c c = c c , c c = c c , c d = c d , bc = bc , c c + c c = c c + c c , c c + c c = c c + c c , c d + c d = c d + c d , c c + c c = c c + c c , c c + c c = c c + c c , c d = c d , c c = c c , c c = c c , c d = c d , d c = d c , d c = d c , d d = d d , d c + d c = d c + d c , d c + d c = d c + d c . Remark . We list here the nontrivial relations when the two crossings are all positive. Theabove relations then should be completed by and b operations. Please refer to the discussionin case 1. The collection of all nontrivial relations will be denoted by R .If the variables satisfy the relations in R , then f pq = f qp .3. Proof of the main theorem
To define the invariant on any oriented link diagram D , we shall first assume/add someadditional data.(1) Suppose each link component has an orientation . This is already given.(2) Order the link components by integers: 1,2, · · · , m.(3) On each component k i , pick a base point p i .An oriented link diagram with order of link components and base points is called a markeddiagram . Now, we travel through component k from p along its orientation. When we finish k , we shall pass to k starting from p , · · · . Definition 3.1.
A crossing point is called bad if it is first passed over, otherwise, it is called good . A link diagram contains only good crossings is called a monotone or ascending diagram.Given a monotone diagram, each link component k i can be regarded as a map k i : S → R = R × R , and the S can be divided into two arcs α i ∪ β i , such that, (1) the map β i → R × R → R is an immersion, i.e., its image is the monotone diagram. (2) different points in β i has different z coordinates (the third coordinate in R × R = R ), hence β i → R × R → R is monotonouslyincreasing. (3) the image of α i is vertical, i.e. its projection on R is one single point, i.e., a basepoint. (4) any point in k i has smaller z coordinate than the points in k i +1 . The set of maps { k i } is called a geometric realization of a monotone diagram. Lemma 3.2.
A monotone diagram corresponds to a trivial link.
We do not use this lemma explicitly in this paper. It will help the readers to understand why wedefine the value for monotone diagram to be v n . The proof is easy. We leave it as an exercise.Now we are going to construct the link invariant for oriented link diagrams. For a givenmarked link diagram, we can define an ordered pair ( c, d ) of integers, called its index. Here c is the crossing number of the diagram, and d is the number of bad points of the diagram. ( c, d ) < ( c ′ , d ′ ) if c < c ′ , or c = c ′ and d < d ′ . Let S ( c, d ) denote the set of all marked linkdiagrams with indices ≤ ( c, d ). Note that S ( c,
0) contains exactly the monotone diagrams with c crossing points.Now let’s study the skein relations. Take f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 for example, each term has a link diagram corresponding toit. If the diagram E + is marked, as mentioned at beginning of section 2, all the other diagramsare canonical oriented. What’s more, E − is canonically marked using the same order, base pointsas E + . Suppose the marked link diagram E + has index ( c, d ), then E − has index ( c, d + 1) or( c, d − c −
1. As we will show later, the invariantactually does not depend on the order and base points of the link diagram. This tells us that wecan construct the invariant and prove its properties use induction on the index pair ( c, d ). Forexample, suppose that E − has smaller index ( c, d − ≤ ( c, d ), then f ( E + ) is uniquely determined by the skein relation. Weshall use this as the definition of f ( E + ).For any integer n >
0, we introduce a variable v n , and suppose that (1 + b + d + d ) v n +( c + c + c + c ) v n +1 = 0 is hold for all n . Proposition 3.3. If d ′ = d ′ , then there is a function f defined for marked link diagrams,satisfies the following properties.(1) The value for any marked link diagram is uniquely defined. For any trivial link diagram D ∈ S (0 , with n components, f ( D ) = v n .(2) Resolving at any crossing point, the invariant satisfies the skein relations.(3) It is invariant under base point changes.(4) f ( D ) is invariant under Reidemeister moves that never involve more than c crossings.(5) It is invariant under changing order of components.Proof. The construction and proofs are all using induction on the index pair ( c, d ), where c isthe crossing number of the diagram, and d is the number of bad points of the diagram. It isobvious that 0 ≤ d ≤ c . The initial Step.
For a diagram of index (0 , v n , where n is the number of components of the link.Then the statements (1)-(5) are satisfied for diagrams inside S (0 , The inductive Step.
Now suppose the statements (1)-(5) are proved for link diagrams withcrossings strictly less than c . This means that for any marked oriented link diagram withcrossings < c , the value of the invariant is uniquely defined, independent of choice of basepoints and ordering of link components. Hence we can choose base points and ordering of linkcomponents arbitrarily to define the invariant. Proof of the statement (1) :If the diagram D has index ( c, f ( D ) to be v n ,where n means that the link has n components.Suppose that f ( D ) is defined for diagrams of index ≤ ( c, d ), where d ≥
0. If the diagram D has index ( c, d + 1), then it has bad points. We resolve the diagram at its first bad point. Then,in the corresponding skein equation, all the other diagrams are of smaller indices than ( c, d ).Hence f is defined for those diagrams. So f ( D ) is uniquely determined by the skein relation.We take this as the definition of invariant for D . We shall prove later that if we resolve at othercrossing point we shall get the same result. NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 15
Remark . We can similarly define the invariant for marked diagrams on S . Given a markedlink diagram D on R , we can also regard it as a marked diagram on S . However, for a markedlink diagram D on S , we can have many marked diagrams on R , depending on where wepick the ∞ point. All those marked diagrams on R have the same value of invariant using thedefinition above. As a consequence, when we later prove the Reidemeister moves invariance,we can actually allow more ”generalized Reidemeister moves”. For example, if an outermostmonogon contains the ∞ point, we can use the Reidemeister move I to reduce it. Proof of the statement (2) .For a link diagram D , if D has one bad point, since f ( D ) is defined by the skein relation, itsatisfies the statement (1). If D has at least 2 bad points, and one resolve at a bad point q . If q isthe first bad point, then by definition, the equation is satisfied. If not, denote the first bad pointby p . If we resolve at p , we get many diagrams D , D , · · · and a linear sum f p ( D ) = P α i f ( D i )for some α i . Then by definition f ( D ) = f p ( D ).We resolve each D i at q , then we get the linear sum f q ( D i ). Each diagram D i has strictly lowerindices than ( c, b ). If D i has crossing number c −
1, then skein equation is proved for resolving atany point. If D i has crossing number c , then it has b − q is also a bad point of D i . In all cases, by induction hypothesis, f ( D i ) = f q ( D i ). Hence f ( D ) = f p ( D ) = P α i f q ( D i ).On the other hand, if we resolve D at q first, we get many diagrams D ′ , D ′ , · · · , each hasstrictly lower indices than ( c, b ). Hence the statements (0)-(4) are satisfied. We get a linearsum f q ( D ) = P β i f ( D ′ i ). We resolve each D ′ i at p , then we get the linear sum f p ( D ′ i ). By theargument before and our induction hypothesis, f ( D ′ i ) = f p ( D ′ i ). Hence f q ( D ) = P β i f p ( D ′ i ). Onthe other hand, the ring is designed such that P β i f p ( D ′ i ) = P α i f q ( D i )! (This is the equation f pq = f qp .) f ( D ) definition st bad point f p ( D ) = P α i f ( D i ) induction hypothesis f q ( D ) = P β i f ( D ′ i ) induction hypothesis P α i f q ( D i ) f pq = f qp P β i f p ( D ′ i )Therefor, f ( D ) = f p ( D ) = P α i f q ( D i ) = P β i f p ( D ′ i ) = f q ( D ). That is, if we resolve at q ,the skein equation is satisfied. Corollary 3.5.
If one resolve any point (not necessarily bad), the skein equation is satisfied.Proof. If q is a good point of D , we make a crossing change at q get a new diagram D ′ , then q is bad point of D ′ . The above proves that if we resolve D ′ at q the skein equation is satisfied.But this the same equation as D resolving at q . (cid:3) This means that one can resolve at any crossing point to calculate the invariant, not necessarilythe first bad point.To prove statement (3), we need the following lemmas.
Lemma 3.6. ( [11] Lemma 15.1) Suppose that p and q are two arcs in R meeting only at theirend points A and B , and let R be the compact region bounded by p ∪ q . Suppose that t , t , · · · , t n are arcs in R , each meeting p ∪ q at just its end points, one in p and one in q . Suppose thatevery t i ∩ t j is at most one point, that intersections of arcs are transverse and there are no triple points. The graph, with vertices all intersections of these arcs and edges comprising p ∪ q ∪ ( ∪ i t i ) ,separates R into collection of v -gons. Then amongst these v -gons there is a 3-gon with an edgein p and there is a 3-gon with an edge in q . Using the above lemma, and a modification of [7] Lemma 5.1, we can prove the followinglemma for marked link diagrams.
Lemma 3.7. (1) Each marked monotone link diagram D on S with < c crossings can betransformed to the unlink diagram in S (0 , using Reidemeister moves, and at each step, theresulting diagram has crossing number < c .(2) Let D be a marked monotone knot diagram D with c crossings. b is the base point. Startingfrom b , p is the first crossing point. Then after resolve D at p , any diagram D i with c − crossings is a diagram of unknot or unlink.(3) Let D be a marked monotone link diagram D with c crossings. b is one base point. Startingfrom b , p is the first crossing point. If the two arcs passing through p are from same linkcomponent, then after resolve D at p , for any the diagrams D i with c − crossings is a diagramof unknot or unlink.Proof. (1) The proof is a modification of the proof in [7]. In a link diagram D , a loop is a partof a component that starts and ends at the same crossing. A innermost loop is called simple if ithas no selfintersections. Two arcs bound a bigon if they have no selfintersections, have commoninitial and final points and no other intersections. A bigon is called simple if it does not containsmaller bigons and loops inside.By an easy innermost argument, one knows that if D has crossing then D has a simple bigonor a simple loop.Case 1. If there is a simple bigon, suppose the two arcs p, q forming the bigon are from asame link component l . Since the bigon is simple, it satisfies the condition of lemma 3.6. Thenthere is a 3-gon with an edge in p and there is a 3-gon with an edge in q . Then one of themdose not contain the base point of l . Since this is a monotone diagram, one can move the 3-gonoutside of the bigon by a Reidemeister 3 move. Thus the bigon is simplified. When there are noarcs in the bigon, one can remove the bigon by a Reidemeister 2 move.Case 2, if there is a simple bigon, and the two arcs p, q forming the bigon are from linkcomponent l, m . If at most one base point A or B of l, m lies in the bigon, then we can dealwith it as in case 1. If both A, B lie in the bigon, say A is on arc a , B is on arc b . a divides thebigon into two parts, one part, say X , does not contain B . The boundary of X contain threeparts, p ′ ⊂ p, q ′ ⊂ q, a . We regard p ′ ∪ a = p ′′ as one arc. The it forms a bigon with q ′ . Applyinglemma 3.6, one can use Reidemeister 3 moves to remove arcs inside this bigon since B is outsideof it. When there is no arc pass this new bigon, one can use Reidemeister 3 moves to move q to remove this bigon. Now the bigon contains at most one base point B . We can simplify it asabove.For a loop, we can regard it as a degenerate bigon and treat it similarly.(2) We can assume there is a small open disk U containing b . U contains only one crossing, p . There are two type of smoothings, horizontal ( HC, HT, E, W ) and vertical (
S, N, V C, V T ).See Fig. 4. When one starts at B and travel along D , one passes p, A , A , B , B .For a horizontal smoothing, D has only one component and D i has has two components D i , D i . If we take b , b as base points, then D ′ is a monotone diagram, hence is a diagram ofunlink.For a vertical smoothing, D, D ′′ both have only one component. Since the arc A A hassmaller z coordinate than the arc B B . The two arcs are also monotone with respect to z NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 17 coordinate, hence they can contract to the boundary of the disk U without obstruction. Hence D ′′ is a diagram of unknot.(3) Follows easily from (2). (cid:3) D A A B B b D ′ A A B B b b D ′′ A A B B Figure 4.
Resolve monotone diagram near base point.
Proof of the statement (3) :Given a diagram D with a fixed orientation and order of components, suppose that there aretwo base point sets B and B ′ . We only need to deal with the case that B and B ′ has only onepoint x and x ′ different, they are in the same component k , and between x and x ′ there is onlyone crossing point p . Using the base point sets B or B ′ , D has the same bad points except p .Let f B ( D ) and f B ′ ( D ) denote f ( D ) using base point sets B and B ′ respectively.We shall prove the equation f B ( D ) = f B ′ ( D ). If there is bad point other than p , say q , weresolve D at q to get diagrams D , D , · · · . Then those D i ’s has lower indices than D , hencebase point invariance is proved for them. As before, we get a marked diagram D correspondingto crossing change at q . When we apply skein equation to the bad point, for f B ( D ) and f B ′ ( D ),each f ( D i ) has same value, hence f B ( D ) = f B ′ ( D ) if and only if f B ( D ) = f B ′ ( D ). Hence wecan assume there are no other bad points.Now there are three cases. Case 1. p is a good point for both the two base point systems,then the values for D are both v n . Case 2. p is a bad point for both the two base point systems,then the skein equation tells the values are the same.Case 3, p is good in B , bad in B ′ . Then the two arcs passing through p are from same linkcomponent, and the diagram D with base point set B is a monotone diagram. Applying theabove lemma 3.7(3) to D , all D i are monotone diagrams. Suppose D has n components. Then f B ′ ( D ) is defined by the skein equation while all other terms f B ( D ) and f ( D i ) are known andhave value in { v n , v n +1 } .On the other hand, (1 + b + d + d ) v n + ( c + c + c + c ) v n +1 = 0 is hold for all n . Sincethe two arcs passing through p are from same link component, the V C, V T diagrams all have n components, the HC, HT, W, E diagrams all have n + 1 components. The values f B ( D ) and f ( D i ) fit the equations (1 + b + d + d ) v n + ( c + c + c + c ) v n +1 = 0. Hence the solution ofthe skein equation is f B ′ = v n . Proof of the statement (4) : Lemma 3.8. f is invariant under Reidemeister III move.Proof. Given two diagrams D and D ′ , which differ by a Reidemeister move III. Like above, wecan assume all other points are good. In the two local disks containing the Reidemeister moveIII, there is a one to one correspondence between the three arcs as follows. We can order thethree arcs by 1,2,3, (1 ′ , ′ , ′ in D ′ ) such that arc 1 (1 ′ ) is above arc 2 (2 ′ ), and arc 2 (2 ′ ) is abovearc 3 (3 ′ ). The one to one correspondence preserves the ordering. Their intersections induce aone to one correspondence between the three pairs of points in the two disks. Call them p, p ′ , q, q ′ , r, r ′ . If arc i intersects arc j at x , then arc i ′ intersects arc j ′ at x ′ . Suppose p is the intersection of arc 1 and arc 2 (or arc 2 and arc 3), then we resolve both p and p ′ , and get many new link diagrams. There is a canonical one to one correspondence betweenthose diagrams. So we can denote them by D , D , · · · , D ′ , D ′ , · · · . Here D , D ′ correspondto crossing change for D and D ′ , and all other diagrams are of smaller crossing numbers. Byinduction hypothesis, for those diagrams, we have f ( D i ) = f ( D ′ i ), i ≥
2. Therefor, by the skeinequation, f ( D ) = f ( D ′ ) if and only if f ( D ) = f ( D ′ ). So we can assume p is a good point.Similarly, we can assume the intersection of arc 2 and arc 3 is a good point.Now, the intersection of arc 1 and arc 3, say r , is also a good point. The reason is simple.Since we proved base point invariance, we can assume there is no base point on any of the 3 arcs.The intersection of arc 2 and arc 3 is good means we first travel arc 3, then arc 2. Likewise,intersection of arc 1 and arc 2 is good means we first travel arc 2, then arc 1. Hence we firsttravel arc 3, then arc 1, the intersection of arc 1 and arc 3 is good.So all the three intersections p, q, r are good. It follows that p ′ , q ′ , r ′ are good. Now we havetwo monotone diagrams, the invariance is clear. (cid:3) Lemma 3.9. f is invariant under Reidemeister I move.Proof. Given two diagrams D and D ′ , which differ by a Reidemeister one move. Say D hasindex ( c − , d ), where D ′ has index ( c, d ′ ). D ′ has one extra crossing point p . By base pointinvariance, we can choose base point such that p is a good point.As before, we can assume that if there are bad points other than p , we can resolve them andprove Reidemeister move one invariance inductively.Now, all other points are good, then D and D ′ are both monotone diagrams of trivial links.So f ( D ) = f ( D ′ ). (cid:3) Lemma 3.10. f is invariant under Reidemeister II move.Proof. Given two diagrams D and D ′ , which differs at a Reidemeister move II. D ′ has two morecrossings, p and q . Likewise, we can assume all other points are good. If the two crossings, p and q , one is good, the other is bad, one can use a base point change to make them both good.Then both the diagrams D and D ′ are monotone diagrams. There is nothing to prove.The only case needs a proof is that both the two crossing are bad, and base point changeswouldn’t change them from bad to good. However, changing both the two crossing will makethem both good points. And in this case, the two arcs are from different link components.In the following Fig. 5, we list all 3 possible cases of the intersections as X i , X ′ i , i = 1 , , X i or X ′ i is a monotone diagram. We apply theskein equation to the positive crossing of X i and X ′ i respectively. Then we have f ( X i ) + bf ( Y i ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0and f ( X ′ i ) + bf ( Y i ) + c ′ f ( E ′ ) + c ′ f ( W ′ ) + d ′ f ( S ′ ) + d ′ f ( N ′ ) = 0 . One can check that f ( E ) = f ( E ′ ) , f ( W ) = f ( W ′ ) , f ( S ) = f ( N ′ ) , f ( N ) = f ( S ′ ). Since weassumed that d ′ = d ′ , we have c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = c ′ f ( E ′ ) + c ′ f ( W ′ ) + d ′ f ( S ′ ) + d ′ f ( N ′ ). Therefor, f ( X i ) = f ( X ′ i ) for i = 1 , ,
3. Since either X i or X ′ i is a monotonediagram, invariance under Reidemeister II move is proved. (cid:3) Proof of the statement (5) : f is invariant under changing order of components. NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 19 X + − X ′ − + Y X − + X ′ + − Y X − + X ′ + − Y Figure 5.
Reidemeister move II invariance.Given two marked diagrams with different ordering of components. For simplicity, call them D and D . By lemma 3.7(1), they can be simultaneously reduced to trivial marked diagrams D and D ∈ S (0 ,
0) by crossing changes and Reidemeister moves never increasing crossings. So f ( D ) = f ( D ) if and only if f ( D ) = f ( D ). However, D and D ∈ S (0 ,
0) are trivial linkdiagrams with different ordering of link components. By definition, f ( D ) = f ( D ) = v n . Hence f ( D ) = f ( D ). (cid:3) Now, let X denote the quotient ring Z [ b, b ′ , c , c , c , c , d , d , b ′ , c ′ c ′ , d ′ , d ′ , v , v , v , · · · ] /R ,where R = R ∪ { d ′ = d ′ , (1 + b + d + d ) v n + ( c + c + c + c ) v n +1 = 0 , for all n = 1 , , , · · · } .Then we have the following theorem. Theorem 3.11.
For oriented link diagrams, there is a link invariant f with values in X andsatisfies the following skein relations:(1) If the two strands are from same link component, then f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 . (2) Otherwise, f ( E + ) + b ′ f ( E − ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0 . The value for a trivial n-component link is v n .In general, replacing X by any homomorphic image of X , one will get a link invariant. Modifying by writhe.
There is another closely related link invariant with values in an-other commutative ring Y . The idea is that the skein relations can reduce the calculation tomonotone diagrams, and we can regard the set of monotone diagrams as a basis and assignwrithe dependant values to those diagrams. The is an analogue of the Kauffman two variablepolynomial. Now, let A be a new variable. Let Y denote the quotient ring Z [ A, A − , b, b ′ , c , c , c , c , d , d , b ′ , c ′ c ′ , d ′ , d ′ , v , v , v , · · · ] /R , where R = R ∪{ d ′ = d ′ , AA − =1 , Av n + A − bv n + ( c + c + c + c ) v n +1 + ( d + d ) v n = 0 , for all n = 1 , , , · · · } . Then wehave the following theorem. Theorem 3.12.
There is a link invariant F with values in Y . For oriented link diagram D , F ( D ) = f ( D ) A − w where w is a the writhe of the link diagram, and f satisfies the followingskein relations.(1) If the two strands are from same link component, then f ( E + ) + bf ( E − ) + c f ( E ) + c f ( W ) + c f ( HC ) + c f ( HT ) + d f ( V C ) + d f ( V T ) = 0 . (2) Otherwise, f ( E + ) + b ′ f ( E − ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0 . The value of F for a trivial n-component link is v n .In general, replacing Y by any homomorphic image of Y , one will get a link invariant. The following proposition gives the proof of the theorem.
Proposition 3.13. f, F satisfy the following properties.(1) For a monotone diagram D , f ( D ) = A w v n , F ( D ) = v n , where w is a the writhe of the linkdiagram, n is the number of components.(2) For any link diagram D , F ( D ) = f ( D ) A − w .(3) For any marked link diagram D , f ( D ) and F ( D ) are uniquely defined.(4) The function f satisfies type one skein relations if we resolve at any point.(5) The functions f, F are invariant under base point change.(6) F is invariant under Reidemeister move I.(7) F and f are invariant under Reidemeister moves II and III.(8) F and f are invariant under changing order of components.Proof. As before, the proof is an induction on index ( c, d ). Proof of the statement (1)(2) : There is nothing to prove.
Proof of the statement (3)(4) : As in last section, for a link diagram D with bad point, weresolve it at the first bad point and use the skein equation to define f ( D ). Then f is definedinductively for all link diagrams. Since f pq = f qp still hold, we have (4). Proof of the statement (5) :This is similar as in last section. Suppose that (5) is true for diagrams with crossings ≤ c .Given a diagram D with a fixed orientation and order of components, suppose that there aretwo base point sets B and B ′ . We only need to deal with the case that B and B ′ has only onepoint x and x ′ different, they are in the same component k , and between x and x ′ there is onlyone crossing point p . Using the base point sets B or B ′ , D has the same bad points except p .Let f B ( D ) and f B ′ ( D ) denote f ( D ) using base point sets B and B ′ respectively.As in last section, we can assume the crossings different from p are all good points. p is goodwith respect to B , bad with respect to B ′ . Then the two arcs passing through p are from samelink component, and the diagram D with base point set B is a monotone diagram.Suppose D has n components. Let w ( p ) denote the sign of the crossing p , let w denote the sumof signs of all other crossings of D . Let w ( D ) denote the writhe of D . Then w ( D ) = w ( p ) + w . Claim : w ( E ) = w ( W ) = w ( HC ) = W ( HT ) = w ( V C ) = w ( V T ) = w . Proof of claim : For a horizontal smoothing, one get two new link components by smoothingat p . See Fig. 4. For the horizontal smoothing, there is a choice of base points, orientationand order such that the result is a monotone diagram. Hence we can move each components NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 21 such that the result is a disjoint union of knots diagrams. Denote it by D ∗ i Furthermore, we canassure that the diagram of each knot is not changed during the process. Hence it can be realizedby a sequence of Reidemeister II and III moves. Notice that Reidemeister move II and III donot change writhe. Hence w ( D ∗ i ) = w ( D i ).On the other hand, for a knot, its writhe is independent of the choice of orientation. Hencefor D ∗ i , w ( D ∗ i ) is the same for all E, W, HC, HT . Hence w ( E ) = w ( W ) = w ( HC ) = W ( HT ).What’s more, it is clear that W ( HT ) = w ( V C ) and w ( HC ) = w ( V T ).Notice that w ( HC ) = w . So the claim is proved.By induction hypothesis, F is a knot invariant defined for diagrams with crossing ≤ c − D at p , we get diagrams D , E, W, HC, HT, V C, V T . (We also refer to themas D i , i = 0 , , · · · , D , all diagrams D i are unlinks.Then F ( D i ) = v n +1 for horizontal smoothings and F ( D ′ j ) = v n for vertical smoothings. Also F ( D ) = v n , So f ( D i ) = A w v n +1 for horizontal smoothings and F ( D ′ j ) = A w v n for verticalsmoothings.Case 1. If p is a positive crossing, and if F ′ B ( D ) = v n , then f B ′ ( D ) + bf B ′ ( D ) = A w × ( Av n + A − bv n ) . Case 2. If p is a negative crossing, and if F ′ B ( D ) = v n , then f B ′ ( D ) + bf B ′ ( D ) = A w × ( Av n + A − bv n ) . On the other hand, A w × ( Av n + A − bv n + ( c + c + c + c ) v n +1 + ( d + d ) v n ) = 0 is holdfor all n . f B ′ ( D ) is defined by the skein equation while all other terms are known. The uniquesolution in each case is F ′ B ( D ) = v n . Hence F and f are invariant under change of base pointsfor diagrams having c crossings. Proof of the statement (6) : As in last section, we can move the base points such that crossingpoint in the Reidemeister move I is a good point. The proof is the same as in last section.
Proof of the statement (7) : Notice that Reidemeister move II does not change writhe. Checkthe following equations. f ( X i ) + bf ( Y i ) + c ′ f ( E ) + c ′ f ( W ) + d ′ f ( S ) + d ′ f ( N ) = 0and f ( X ′ i ) + bf ( Y i ) + c ′ f ( E ′ ) + c ′ f ( W ′ ) + d ′ f ( S ′ ) + d ′ f ( N ′ ) = 0 . After considering writhe, we still have f ( E ) = f ( E ′ ) , f ( W ) = f ( W ′ ) , f ( S ) = f ( N ′ ) , f ( N ) = f ( S ′ ). Since we assumed that d ′ = d ′ , we have c ′ f ( E )+ c ′ f ( W )+ d ′ f ( S )+ d ′ f ( N ) = c ′ f ( E ′ )+ c ′ f ( W ′ ) + d ′ f ( S ′ ) + d ′ f ( N ′ ). Therefor, f ( X i ) = f ( X ′ i ) for i = 1 , ,
3. Since either X i or X ′ i isa monotone diagram, invariance under Reidemeister II move is proved. Proof of the statement (8)(9) : Notice that Reidemeister move III does not change writhe.The proofs are the same. (cid:3)
If we let A = 1, we see the first invariant is a special case of the modified invariant. Some simplification
The modified invariant is a generalization of both HOMFLY polynomial and Kauffman two-variable polynomial. However, it is very complicated, hard to compute. There are some sym-metric simplification of it. For example, we let b = b ′ = b − , then x = x . Let c/ c = c = c = c , d/ d = d , c ′ / c ′ = c ′ , d ′ / d ′ = d ′ , then x = b x . The the relations aredramatically simplified.Furthermore, we add the new relations d ′ = bc ′ , cv n +1 + ( A + A − b + d ) v n = 0. Plug all thoseinto the relation set R , the new relation set contains the following. d ′ = bc ′ , dc ′ = bc ′ c ′ , b = 1 , dd = cc ′ , cv n +1 = − ( A + A − b + d ) v n , i ≥ . To go one step further, we need the famous diamond lemma [8]. One can also consultWikipedia.
An Equivalent version of the diamond lemma [8]: For every binary relation with nodecreasing infinite chains and satisfying the diamond property, there is a unique minimal elementin every connected component of the relation considered as a graph.Since d ′ = bc ′ , we delete the variable d ′ , use the following variables A, b, c, c ′ , d, v n , n = 1 , , · · · .Regard the above relations as a rewriting system as follows. dc ′ → bc ′ c ′ , b → , dd → cc ′ , AA − → , cv n +1 → − ( A + A − b + d ) v n , i ≥ . (3)Let deg v n +1 = 8 n, deg d = 4 , deg c = deg c ′ = 2 , deg b = 1 , deg A = deg A − = 1. Thenone can see that the rewriting system always decreases the degree, hence there does not existdecreasing infinite chains.To verify the diamond property, notice that for the value of F ( D ), every term contains exactlyone variable from { v n , n = 1 , , · · · } , hence there is only the following one case to check. d ( dc ′ ) → b ( dc ′ ) c ′ → ( bb ) c ′ c ′ c ′ → c ′ c ′ c ′ , or ( dd ) c ′ → cc ′ c ′ . Hence we add the new relation c ′ c ′ c ′ = cc ′ c ′ . Now the rewriting system is as follows. dc ′ → bc ′ c ′ , b → , dd → cc ′ , , AA − → , c ′ c ′ c ′ → cc ′ c ′ , cv n +1 → − ( A + A − b + d ) v n , i ≥ . (4)Now, verify the diamond property again, there are two new cases to check.( dc ′ ) c ′ c ′ → b ( c ′ c ′ c ′ ) c ′ → bc ( c ′ c ′ c ′ ) → bccc ′ c ′ , or d ( c ′ c ′ c ′ ) → c ( dc ′ ) c ′ → cb ( c ′ c ′ c ′ ) → bccc ′ c ′ . Here is another case. d ( dc ′ c ′ c ′ ) → dbccc ′ c ′ = bcc ( dc ′ c ′ ) → bccb ( c ′ c ′ c ′ ) → bbcccc ′ c ′ → cccc ′ c ′ , or ( dd ) c ′ c ′ c ′ → cc ′ c ′ c ′ c ′ → ccc ′ c ′ c ′ → cccc ′ c ′ . Hence this rewriting system satisfies the condition of the diamond lemma, any result F ( D )(or f ( D )) has a unique normal form. NOT INVARIANT WITH MULTIPLE SKEIN RELATIONS 23
Now, let Z denote the quotient ring Z [ A, A − , b, c, c ′ , d, v , v , v , · · · ] /R , where R = { dc ′ = bc ′ c ′ , b = 1 , dd = cc ′ , AA − = 1 , cv n +1 + ( A + A − b + d ) v n = 0 , i ≥ } . Then we have thefollowing theorem. Theorem 4.1.
There is a link invariant F with values in Z . For oriented link diagram D , F ( D ) = f ( D ) A − w where w is a the writhe of the link diagram, and f satisfies the followingskein relations.(1) If the two strands are from same link component, then f ( E + ) + bf ( E − ) + c ( f ( E ) + f ( W ) + f ( HC ) + f ( HT )) / d ( f ( V C ) + f ( V T )) / . (2) Otherwise, f ( E + ) + bf ( E − ) + c ′ ( f ( E ) + f ( W )) / bc ′ ( f ( S ) + f ( N )) / . The value of F for a trivial n-component link is v n .And the rewriting rules (4) given a unique nomal form for the invariant F . Example: The right hand trefoil.
Let H denote the minimal diagram of Hopf link, H ∗ denote its mirror image, D denote aminimal diagram of the right hand trefoil. Apply the skein equation to any crossing point thenwe get f ( H ) = − bv − Ac ′ v − A − bc ′ v = − bv − c ′ ( A + bA − ) v ,f ( H ∗ ) = − bv − bA − c ′ v − bAbc ′ v = − bv − c ′ ( A + bA − ) v . Then f ( D ) = − bAv − c f ( H ) + f ( H ∗ )) − dA − v = − bAv + cbv + cc ′ ( A + bA − ) v − dA − v = − bAv − b ( A + A − b + d ) v + cc ′ ( A + bA − ) v − dA − v = (( cc ′ − b ) A − bd + ( bcc ′ − A − − dA − ) v The invariant F ( D ) = A − f ( D ).In the above calculation we use the rewriting rules (4), and we use the = sign instead of → ,since they are equal in the ring. The rewriting rules (4) gives the unique normal form. In thefinal result of any F ( D ), the highest powers of d, b are ≤
1, the highest power of c ′ is ≤
2. Thisis a generalization of the 2-variable Kauffman polynomial.
Acknowledgements
The author would like to thank Ruifeng Qiu, Jiajun Wang, Ying Zhang, Xuezhi Zhao, HaoZheng, Teruhisa Kadokami for helpful discussions. The author also thanks all the organizers of2009 summer school of knot theory in ICTP, Italy. He really had a great time there and gotmany inspirations. The author also thanks NSF of China for providing traveling fee for the trip,and all the efforts made by ICTP, Italy.
References [1] C.C. Adams,
The Knot Book , W.H. Freeman and Company (1999)[2] J.W. Alexander,
Topological Invariants of Knots and Links , Transactions of the American MathematicalSociety, Volume 30, Issue 2 (April 1928), 275-306.[3] G. Burde, and H. Zieschang,
Knots , de Gruyter Studies in Mathematics, 5, Walter de Gruyter, Berlin (1985).[4] P. Freyd, D. Yetter, J. Hoste, W. B. R. Lickorish, K. Millett, and A. Ocneanu,
A new polynomial invariantof knots and links , Bull. Amer. Math. Soc. 12 (1985), 239-249.[5] V. F. R. Jones,
Hecke algebra representations of braid groups and link polynomials , Ann. of Math. (2) 126(1987), 335-388. [6] L. Kauffman,
State models and the Jones polynomial , Topology 26 (1987), 395-407.[7] V. O. Manturov,
Knot Theory . CRC Press, 2004.[8] M. H. A. Newman.
On theories with a combinatorial definition of ”equivalence” , Ann. of Math. (2) 43(1942), 223C243.[9] J. H. Przytycki and P. Traczyk,
Invariants of links of Conway type , Kobe J. Math. 4 (1987), 115-139.[10] D. Rolfsen,
Knots and links , Publish or Perish Inc., Berkeley, (1976), 160-197.[11] W.B.R. Lickorish,
An introduction to knot theory . Springer-Verlag, New York, 1997.[12] Z. Yang,
New link invariants and Polynomials (I), oriented case . arXiv:1004.2085.
Department of Mathematics, Dalian University of Technology, China
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