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On Jones knot invariants and Vassiliev invariants Zhu, Jun 1995

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ON JONES KNOT INVARIANTS AND VASSILIEV INVARIANTSByJun ZhuB. Sc. (Mathematics) Suzhou University , 1982A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinTHE FACULTY OF GRADUATE STUDIESTHE DEPARTMENT OF MATHEMATICSWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAMay 1995© Jun Zhu, 1995In presenting this thesis in partial fulfilment of the requirements for an advanceddegree at the University of British Columbia, I agree that the Library shall make itfreely available for reference and study. I further agree that permission for extensivecopying of this thesis for scholarly purposes may be granted by the head of mydepartment or by his or her representatives. It is understood that copying orpublication of this thesis for financial gain shall not be allowed without my writtenpermission.(Signature)____________________Department of sThe University of British ColumbiaVancouver, CanadaDate A-pr. ],DE-6 (2/88)AbstractThe main objective of this thesis is to study invariants of knots and links. First, aminimal system is associated to each link diagram D. This minimal system is an “antichain” in the partially ordered set of all possible functions from the crossings of D to{—1, 1}. Such a system is then shown to be an effective tool for determining the highestdegree as well as the span of the Kauffman bracket (D). When applied to a diagramD with n crossings which is “dealternator connected” and “rn-alternating”, the upperbound spanKD) 4(n—m) is obtained. This is a best possible result providing a negativeanswer to the conjecture posed by C.Adams et al. Next, some examples are presented toshow that a semi-alternating diagram need not be minimal. This disproves a conjectureposed by K.Murasugi.For knots and links, if J(t) is a so-called generalized Jones invariant, then J()(i),the n-th derivative of J(t) evaluated at t = 1, is a Vassiliev invariant. On the otherhand, while the coefficients of the classical Conway polynomial are Vassiliev invariants,the coefficients of the Jones polynomial are now shown not to be Vassiliev invariants.An interesting property of Vassiliev invariants is then given to prove many known resultsin a uniform manner. It is known that all the Vassiliev invariants with order less thanor equal to n form a vector space called the n-th Vassiliev space. It turns out that twolens spaces have the same Vassiliev spaces if and only if their fundamental groups areisomorphic. Consequently, Vassiliev spaces do not distinguish manifolds.iiTable of ContentsAbstract iiList of Figures vAcknowledgements vi1 Introduction 12 Kauffman bracket 52.1 Introduction 52.2 Minimal System 62.3 Degree of Kauffman bracket 102.4 The Kauffman bracket and binomial coefficients 162.5 rn-almost alternating diagrams 202.6 Semi-alternating diagrams 253 Jones knot invariants and Vassiliev invariants 303.1 Introduction 303.2 Relationship between quantum group invariants and Vassiliev invariants 313.3 Some known invariants which are not Vassiliev invariants 343.4 Numerical invariants which are approximable by Vassiliev invariants . 403.5 Knot connected sum in finite type spaces 443.6 Some properties of Vassiliev invariants 484 Finite type invariants for lens spaces 514.1 Introduction 514.2 An operation on finite type spaces 521114.3 Similarity classes 574.4 Finite type spaces for lens spaces 62Bibliography 66ivList of Figures2.1 Two different smoothings 62.2 A non-adequate knot diagram . . 102.3 A diagram for the unknot 162.4 Link 97 192.5 A counterexample 232.6 An alternating diagram for the counterexample 242.7 Signed crossings 262.8 A link diagram D and related graphs G, F0(G) and F(G) 272.9 An equivalent diagram 283.10 Related knot diagrams 313.11 Torus knot of type (2, 2n + 1) 343.12 A singular torus knot 353.13 A singular torus knot with 2n double points 363.14 A singular torus knot with 2n + 1 double points . . . 394.15 Connected sum of two knots in solid torus 554.16 Arc winding numbers 60vAcknowledgementIt is a great pleasure for me to thank my supervisors, Dr. Dale Rolfsen and Dr Kee Lam,who spent lots of valuable time to guide me through my research and to read a draftof this thesis. Special thanks also go to professors Joan Birman and Xiao-Song Lin ofColumbia University for their useful comments. Finally, I would like to thank my wifeYunfang for her devotion and support.viChapter 1IntroductionA knot (link) K is an embedding of an (several copies of) oriented circle 5’ in 3-space 1R3or the 3-dimensional sphere S3. The knot type of K is the topological type of the pair(S3,K), under homeomorphisms which preserve orientations on both S3 and K. Whenwe say “two different knots”, we usually mean “two different knot types”.The theory of classical knots in 3-dimensional space iii has played a very important,if not crucial, role in the theory of 3-dimensional manifolds. The most fundamentalproblem in knot theory is the classification problem of knot types. The first systematiccontribution to this problem is by P.G.Tait and his coworkers [67] a century ago. Theycompiled a table of knots with relatively few crossings. Now knots have, in theory, beenclassified by Haken [24] and Hemion [25]. But the classification is by means of an algorithm that is too complex to use in practice. On the other hand, the complementaryspace of a knot determines the knot type [79] and [22]. However, the topological equivalence classes of the complementary spaces of knots is in general not easy to determine.Thus one is led to seek simple invariants for knots which classify large classes of specificknots. The Alexander polynomial [3], /jc(t), a Laurent polynomial in variable t, is onesuch knot invariant which was and is still a most useful one.Let us denote by K, K_ and K0 the knots which are defined by diagrams that areidentical outside a neighborhood of a particular double point, and differing in the mannerindicated in Figure 3.10.Then the Alexander polynomial is characterized by the following crossing change1Chapter 1. Introduction 2formula— = (—together with zXu = 1, where U is the unknot.Modifying the formula tor1J(K+) — tJ(K_) = (v’ —together with J(U) = 1, one can obtain another very different polynomial invariant,nowadays called the Jones polynomial. This is a remarkable discovery by V. Jones[30]. Although there are infinitely many knots sharing the same Jones polynomial, see[33], whether the Jones polynomial can distinguish nontrivial knots from the unknotis still an important open problem. See [39] and [64]. It is well-known that the Jonespolynomial is retrievable from the Kauffman bracket. Now the problem can be restated asto whether there is a knotted link diagram whose Kauffman bracket is a monomial. Basedon knowledge on the bracket, Kauffman [36], K. Murasugi [56] and M. Thistlethwaite[69] settled some hundred-year-old conjectures of P.O. Tait which imply that the aboveproblem on Kauffman bracket can be answered affirmatively within the family of reducedalternating link diagrams. Later, Lickorish and Thistlethwaite [50] generalized this tothe wider family of adequate link diagrams. In Chapter 2 of this thesis, we generalizethis further to subadequate link diagrams. Based on our techniques, we are able to solvea conjecture of C. Adams and his coworkers [1] and a conjecture of K.Murasugi [58].Motivated by the discovery of the Jones polynomial, more knot polynomials weresubsequently found. Notably, the HOMFLY polynomial and the Kauffman polynomial.A unified setting for finding such knot invariants is available now, see [32] and [74]. Allsuch invariants are called generalized Jones invariants. Since generalized Jones invariantsdo not distinguish a knot from its reverse (a knot obtained from the original knot byChapter 1. Introduction 3reversing orientation) and since a knot may not be equivalent to its reverse [73], thegeneralized Jones invariants are not enough to classify knots. Around 1990, Vassiliev [75],using an approach pioneered by Arnold, introduced a set of numerical knot invariantsvia a totally different method. Instead of studying one single knot at a time, he studiesthe space of all knots and constructed a sequence of subspaces in the 0-th cohomologygroup of the space of all knots. These spaces have become known as Vassiliev spaces andthe elements in Vassiliev spaces are called Vassiliev invariants.Although the generalized Jones invariants and the Vassiliev invariants come fromdifferent viewpoints, they are closely related. For example, Bar-Natan [7] proved thatevery coefficient of the Alexander Conway polynomial is a Vassiliev invariant. Actually,a deep relationship between generalized Jones invariants and Vassiliev invariants hasbeen found in [13] and [44]. There the authors show that, after a suitable change ofvariables, each coefficient in the Taylor series expansion of a generalized Jones invariantis a Vassiliev invariant. The first result in Chapter 3 is that every coefficient in the Taylorseries expansion at 1 of a generalized Jones invariant is a Vassiliev invariant. In otherwords, the n-th derivative of a generalized Jones invariant evaluated at 1 is a Vassilievinvariant.In contrast with the Alexander Conway polynomial, we show that every coefficient ofthe Jones polynomial is not a Vassiliev invariant. Hence the classical Alexander Conwaypolynomial and the Jones polynomial are, in this sense, quite different.Whether every numerical invariant can be approximated by Vassiliev invariants is animportant open problem, see [13]. We prove in Chapter 3 that the space of numericalinvariants which can be approximated by Vassiliev invariants is a dual space of somevector space. Finally, we give an interesting property of Vassiliev invariants from whichwe can retrieve, in a uniform manner, many known results.In Chapter 4, we study Vassiliev spaces for 3-manifolds, in particular for the lensChapter 1. Introduction 4spaces. We show that two lens spaces have isomorphic Vassiliev spaces if and only iftheir fundamental groups are isomorphic. Consequently, we see that Vassiliev spaces donot distinguish manifolds.Chapter 2Kauffman bracket2.1 IntroductionSince V. Jones invented his knot invariants [30], many applications have been found. Oneof the most important applications is a solution to a hundred-year-old conjecture of P.G. Tait. See [36], [56] and [69]. The key method Kauffman used to settle this conjectureis to determine the span of the Kauffman bracket for reduced alternating link diagrams;i.e., to determine the highest and the lowest degree of the Kauffman bracket for suchlink diagrams. W. B. R. Lickorish and M. B. Thistlethwaite generalized this method toadequate link diagrams [50]. In this chapter, we generalize the method to a larger classof link diagrams by associating a minimal system to each link diagram (for the definitionsee the following section), which is closely related to the highest or the lowest degree ofthe Kauffman bracket. Our approach is easy to apply and allows us to generalize someresults of [36] and [50]. It is also interesting to note that our method can be used toprove some identities on binomial coefficients via Kauffman bracket. See section 2.4.In their work [1], the authors defined rn-almost alternating projections of n crossingsfor a link. (We may call such a projection an rn-almost alternating diagram). They provedthat for a dealternator reduced and dealternator connected rn-alternating diagram D, thespan of its Kauffman bracket is less than or equal to 4(n — m — 2), and conjectured thatthe span is less than or equal to 4(n — m — 1) if the condition “dealternator reduced” isremoved. As an application of our methods, we show that the conjecture is true if one5Chapter 2. Kauffman bracket 6/\ /\A-smoothing A’-smoothingFigure 2.1: Two different smoothingsreplaces 4(n — m — 1) by 4(n — m). An example is then provided to show that this upperbound is best possible. The conjecture is hence completely solved.K.Murasugi conjectured in [58] that every semi-alternating link diagram is minimal,namely that it realizes the minimal crossing number for the link. We construct a sequenceof semi-alternating link diagrams which are not minimal. Hence Murasugi’s conjectureis false. In [58] K. Murasugi also raised the question whether a semi-alternating linkdiagram is adequate. Our semi-alternating link diagrams turn out to be not adequate.The answer is thus negative as well.In what follows, by a diagram we always mean a link diagram which is connected asa graph.2.2 Minimal Systemlet D be a diagram for some link, with crossings c1,c2,.. , c. A state for D is a function{cI1 i n} —+ {1, —1}. Let sD be the diagram with its crossings nullifiedaccording to an A-smoothing if .s(c) = 1, and anA1-smoothing if .s(c) = —1, see figure2.1.Chapter 2. Kauffman bracket 7We abbreviate s(c) by s(i), and denote the number of components of sD by IsDI.Let S {ss is a state for D}. Then we can impose a natural partial order on 5,namely, s <s’ if and only if s(i) s’(i), for every i with 1 i <n. This turns S into aposet, or partially ordered set.DenoteSq = {s E Ss(i) = n — 2q},O q n,then S = {1}, where 1 is the constant map taking value 1, and similarly, Si-, = {—1}.Lemma 1 Let s, s’€S with s s’, thens’(i)+2s’Dand equality holds if and only if sD = s’D — r, where r satisfies2r = s(i) — s’(i).Proof. Let s E Sq. Since s s’, there are s si > 5r = s’ such thatZs(i) = si(i) + 2 for 0 p r. We then have 5p E 5q+p for every q.Note that sDI = s1D + 1 and s(i) = 2 +s1(i), hences(i) + 2IsDI = s÷1(i)+ 2 + 2(Is+iD + 1)=s1(i) + 2Is+1DI + 2 ± 2 s1(i) + 2s+iDIand equality holds if and only if IsDI = s1D — 1.It is then easy to sees(i)+2IsD > s’(i)+2Is’Dand equality holds if and only if jsD = Is’D — r, where r satisfies 2r = s(i) — s’(i).0Chapter 2. Kauffman bracket 8Remark 1 If sD = s1D + 1, thens(i) + 2sD —4 =s1(i) +Recall that the Kauffman bracket for diagram D is defined as follows:(D) = (Ds)sESwhere(Djs) = AE5(i)(_A_2—A2)IsDI_1.From the definition of (Dis), it is easy to see that the highest degree of (Ds) is lessthan or equal to s(i) + 2sDI — 2 and that the lowest degree of Ds) is greater than orequal to s(i) — 2sD + 2. Therefore, by Lemma 1, we have the following well knownresult:Corollary 1 The highest degree of (D) does not exceed n + 21D — 2 and the lowestdegree of KD) is at least—n — 2 — 1D +2.Proposition 1 Let D be a diagram with crossings c1, , c,-. Then for every integerq 0, there are unique elements s ,5j ES (up to permutation) such that(a) {s,. . , sk} is an antichain in the poset 5, i.e. for every pair Si, 5j with i j,Si Si,(b) For every 8r E {Si, . ,5r() + 2ISrDI 1(i) + 21D — 4q = n + 21D — 4q.and(c) {s,. is minimal, i.e. if {s’,... ,4} C S satisfies (a) and (b), then forevery s, there exists an 5r E {si,. . such that s,Chapter 2. Kauffman bracket 9Proof. Let= {s E S s(i) + 2sD n + 21D — 4q}.Clearly, 1 e B B 0.Let {Si,• , .Sk} be the set of all minimal elements in B. i.e. for every 3r there isno s e with s r such that s < r• It is clear that {s,. . . , sj} satisfies (a) and(b). To show (c), it suffices to show that for any s $ satisfying (b) there exists anr {si,. . , s,} such that .s r. But this clearly follows from the definition of Bc,hence the proposition.Definition 1 The antichain {s,. . ,s,} satisfying (a), (b) and (c) in Proposition 1 iscalled the q-th minimal system of the diagram D. The O-th minimal system is called theminimal system.Remark 2 is, in fact, equal to U1 3j, where j = {s $s s}. Let us callthe closure of the q-th minimal system. By Lemma 1, B C for q q’. If q is thesmallest integer such that B S, then 11D1 + n — 2q = I — 1DI, and — 1DI n + 1 — q.If q is the smallest integer such that BD) = S, where is the mirror image of D, thenq+=n and 1D<n+1—q.Remark 3 Recall that the diagram D is +adequate if 1DI > sDI for every s E Siand is adequate if it is +adequate and the mirror image is also +adequate. Hence foradequate or +adequate diagrams, their minimal systems are always {l}.Example 1. Let D be a diagram of the knot 101.5.6. We draw a diagram of 1D bynullifying its crossings with respect to state 1. See figure 2.2.From the diagram of 1D, one can easily find its minimal system{S125, 3345, 367, 81267}Chapter 2. Kauffman bracket 10Figure 2.2: A non-adequate knot diagramwhere 5ijk is a state taking value —1 only at {i,j,. , k}. By the above remark, D isnot +adequate.Let us finish this section with the following lemma which can be easily checked.Actually it is a special case of Proposition 2 in section 2.4.Lemma 2=0 r = nk=O ( (_1)r() 0 r <n2.3 Degree of Kauffman bracketLet us denote by DegD) the highest degree of a Kauffman bracket, hereafter simplyreferred to as the degree of the Kauffman bracket for D. If D denotes the mirror imageof the diagram D, then the lowest degree of (D) is equal to —Deg(D). Hence, to knowthe lowest degree of (D), we only need to know how to determine the degree of Kauffmanbracket.Knot 10156 1DChapter 2. Kauffman bracket 11Lemma 3 Suppose {Si, 2,••• , sk} is an antichain in S satisfying fl1 {1}, where= {s 6 Ss > s}. Let B = that is, B = B° in the notation of Remark S above.Then(_1)iBnSjI = 0,where X denotes the number of elements in the set X.Proof. We prove the lemma by induction on k. For k = 1, note that the assumptionimplies s1 1, hence we may assume s 6 8r for some r > 0, and since lB fl S =i fl Si I= (), it follows from Lemma 2 thatn S (_1) (j =0.i=o i=o JNow assume k > 1, let B1 = ., B2 = and B’ = B1 fl B2. Define s1 A sj to be thefunction given by i A s(i) = max{si(i),s(i)}, 1 i n. Notice that for every s e B’,s s1 and s > 8r for some 5r, hence the set of all minimal elements of B’ is containedin {Si A 2, s1 A 33, , A sk}. This set is clearly an antichain satisfying the conditionsin the present lemma. Since the number of minimal elements in B’ is less than k, byinduction assumption we have(1)’lB’nS3l= 0,and with the same reason,(_1)ilBi n S = 0and(_1)iIB2nSl =0.Hencen Sl = (_1)iIBi n Sl + (—1)IB2n S1 — (_1)uIB n Sl = 0.Chapter 2. Kauffman bracket 120Corollary 2 For an antichain s, we have(—1)IBflSl 0 if and only if s = 1.We may generalize Lemma 3 to the followingLemma 4 Suppose , 4},. , {s,. , s} are antichains in S. If for every rwith 0 < r < p, there exists a partition {Fr_i, Gr_i} of {1, 2,•• , kr_i} such that {s’li EFr_i} and {sli E Gr_i} satisfy the conditions in Lemma Sand ‘r} is theset of all minimal elements in (UjEFr_ij)fl(UjEGr_ij) and iffurther{s,... , s} satisfiesthe conditions in Lemma 3, then(1)ilB n SI = 0whereB =Proof. The case p = 0 is proved in the above lemma. Assuming the lemma true up to<p—i, we let B1 = UEFO.i and B2 = UEGOi so that B1 flB2 = and {s,...,{sç,. , s,,} satisfy the conditions in the present lemma. By induction assumptionwe have(—1)lBi n B2 n S1 = 0,and by Lemma 3(_1)ilB nSl = 0, i = 1,2.Hencen s1 = (_1)uIBi n sl + (—1)IB2n Sl — (—1)lBi fl B2 fl SI =0.Chapter 2. Kauffman bracket 13Theorem 1 Suppose a link diagram D with n crossings has minimal system {Si, 2, , sk}playing the role of {$,. , 4} in Lemma . ThenDeg(D) <n + 2I1D — 2.Proof. Denote >Z by B as before. Then by Lemma 1 and the definition ofKauffman bracket, it is not difficult to see that the term of degree n + 2I1D — 2 for (D)is(_l)ISDI_1A8(i)+2ISDI_2 = (_l)18D1_1AEs(i)+218D1_2sEB j=OsEBnS= (_l)I1DI+i_1AE1()+2I1DI_2j0 sEBflS3= (_l)11D1+i_1Afl+211D1_2j0 sEBflS3={(_1)1lDI_l(_1)iB nTherefore, Deg(D) < n + 21D — 2. DRemark 4 The proof of the theorem indicates that the term of degree n+21D —2—4qis related only to B, B1 .. , For example, if we know B, B1, we can determine theterm of degree n + 21DI — 6.Now we try to decide when Deg(D) is equal to n + 21D — 2.Lemma 5 Suppose {$,. . , $},... , {s,. . . , s} are antichains in S satisfying all theconditions in Lemma 4 except that the last condition is replaced by {sç,. . , s} = {1}.Thenn S = (-i)where B =Chapter 2. Kauffman bracket 14Proof. Just mimic the proof of Lemma 4, noting that oniy the last term in the sum isnonzero and equal to (_1)1’. We leave the details to the reader. 0Definition 2 A link diagram D with a minimal system {si,. . ., Sk} is called +subadequateif there are antichains {s,. . . ,{5,. . . ,s} with {s,. ,s} =such that these antichains satisfy the conditions in Lemma 5. D is called subadequate ifboth D and the mirror image Ti of D are +subadequate.Remark 5 Clearly + adequate diagrams are +subadequate diagrams. Hence adequatediagrams are subadequate.Example 1 in section 2.2 is not +adequate but is +subadequate. In fact, {3125,5345}and {5367, 51267} form a partition for the minimal system. Let B1 = U andB2 = U Then {512, 53} is an antichain with B1 fl B2 = U . Clearly{512} and {53} form a partition satisfying fl = {1}. It is therefore +subadequate.Moreover, by the following theorem, Deg(D) = 10 + 4 — 2 = 12. It is also easy to find asubadequate diagram which is not adequate. For example, a diagram with two crossingsfor the trivial link of two components is subadequate, but not adequate.Theorem 2 If D is a +subadequate diagram, thenDeg(D) = n + 21D —2.Proof. As in the proof of Theorem 1, the term of degree n + 21D — 2 for (D) is(_l)I8DI_1AES(i)+2ISDI_2 = (l)(SDI_1AES(i)+213D1_1séB j=O seBflS,= (_l)I1DI_1((_l)iIB n Sj)A2IDI_2.Chapter 2. Kauffman bracket 15By Lemma 5, the number inside the parenthesis is nonzero, therefore, by Corollary 1,Deg(D) = n + 21D — 2. oCorollary 3 Suppose D have minimal system {s,. ., Sk} (k < 2). ThenDegD) = n + 21D — 2if and only if= {1}The proof of this corollary is trivial.Using subadequate diagrams, we can generalize many results in [361 and [50]. Wewill not go over these one by one, but only mention the following corollary. Recall that.span(D) Deg(D) + Deg(D), then we haveCorollary 4 If D is subadequate, then span(D) 2n + 2(1D + — 1D) — 4. Inparticular, if n> 1, then span(D) > 0 and the Jones polynomial VD(t) 1.Lemma 6 Let {5i,• , sk} be an antichain in S. Suppose there is a partition {F1,. . . , Fi}with 1 2 of{1,...,k} such that fl{ i e F} {1}forl p 1, and that forp q,(U{i E F}) fl (U{i E Fq}) = {1}.Then(—1)BnSI= 1—1where B =Proof. Denote B = U{i E F}. By the assumption we haveChapter 2. Kauffman bracket 16n crossingsFigure 2.3: A diagram for the unknotfor j > 0, hence= BflsoI+(—1)BnsI=j=1 p=ln=j=Oin=p=l j=O= 1—i.The last equality follows from Lemma 4. 0By the above lemma, we can generalize Theorem 2 to some other cases. We leave thisto the reader.2.4 The Kauffman bracket and binomial coefficientsIn this section we prove some identities on binomial coefficients via Kauffman bracket. Infact we only use a very simple diagram; the same idea can be applied to other diagrams.We do not know whether one can find some new or really interesting binomial coefficientidentities in this way.Proposition 2Z0(—i)() () = 0 for 0 k < n, where () = 0 if i <k.Chapter 2. Kauffman bracket 17Proof. Let D be the diagram shown in figure 2.3. Then the minimal system is clearly{—1} and 1D = 1, hence B = S and sD = 1 +j for every .s e S,. By the definitionof The Kauffman bracket we have(D) = A2(A — A2)i0 sES,= A2(—A—A2)S= (_1)i[()An + ()An-4 + ... + ()An-4k + ... + ()Am-4i1 ()-r() (+ ()(i+ [ () (;)n + () (;)n + () (;)-+(1)u[() (jAn+ () (j- +... + () (j-+(1)n[()(:)Afl+(j (jAn-4 + ... + (:) (:)A4iTherefore the coefficient of A’4’’, summed along the k-th column, is() (j = E(1)i () (j.On the other hand, (D) = A3, hence for k < n we have() (;) = 0. DRemark 6 Some special cases, for example, 0(_i)i() = 0 when k = 0, are wellknown.Chapter 2. Kauffman bracket 18As an application, we prove the following theorem which allows us to compute thehighest degree of Kauffman bracket in some non +subadequate cases. See Example 2below.Before stating the theorem, we need the fact that if {Sj,. , .s} is an antichain in S,then fl1 = for some s E S. For instance, sj fl.= Si A .s.Theorem 3 Suppose D is a link diagram with minimal system {s,... , Sk}. If =for some s Sr then the highest degree ofAE)(A2—A2)I°I’sEBis less than n + 2I1D —2— 4r. Here B =Proof. The case k = 1 follows from the above proposition. Mimic the proofs ofLemma 3 and Theorem 1 for the general case. 0Example 2.It is not difficult to see that the diagram D in Figure 2.4 has minimal system {S12},hence D is not +subadequate, and that its first minimal system is{ 127, 128, 129,S1234 1256, S3456}.Choose a partition as follows {{S127, 128, 129, 51256}, {31234, 53456}}, thenB1 fl B2 = 12 U 56where B1 = 127 U 8128 U 129 U 1256 and B2 = 81234 U 83456.Since 12 fl 56 = {1} the conditions in Lemma 5 are satisfied. Therefore, we have(—1)jB’nS= = 1,Chapter 2. Kauffman bracket 1964)5Link 97 1DFigure 2.4: LinkwhereB’ = 8127 U 8128 U 129 U 1234 U 1256 U 3456•Hence the highest degree ofAES(i)(_A2— A2)I13I1 + AE8()(_A2 — A2)I9D1_3sEB’—B sEBis n + 21Dj —6Denote 12 by B. Applying the above theorem to SEB A()(—A2—A2)15D1_l, weconclude that the highest degree of SEB AE5(i)(_A2—A2)I11 and the highest degreeof SEB A()(—A2—A2)I’I3 are both less than n + 21D — 2 — 8 = n + 211D1 — 10.Hence, the highest degree of (D) is equal to the highest degree ofAE5()(_A_2— A2)15D1_ + AE5(z)(_A_2— A2)15D1_3sEW—B sEBwhich is n + 21DI — 6Chapter 2. Kauffman bracket 202.5 rn-almost alternating diagramsFirst of all, let us recall some concepts from [1]. A link diagram is said to be alternatingif the pattern of over and undercrossings alternates as one traverses a component. A linkdiagram is called rn-almost alternating if m is the smallest integer such that m crossingchanges produce an alternating diagram. Hence a 0-almost alternating diagram is analternating diagram, and we call 1-almost alternating diagram simply almost alternating.Let D be an rn-almost alternating diagram with crossings c1,c2,• ,c,-. Then wemay assume that changing the first m crossings of D produces an alternating diagram.Let D’ be a diagram obtained from D by nullifying the first m crossings. Clearly, wehave 2m such diagrams corresponding to different smoothings. A crossing of D is callednugatory if some nullyfying of the crossing disconnects D. D is reduced if D has nonugatory crossings. D is called “dealternator reduced” if every D’ is reduced. D is called“dealternator connected” if every D’ is connected. A loop ir is called a “dealternatorsevering path” if ir intersects D only at some of the first m crossings. It is clear that Dis dealternator connected if and only if D has no dealternator severing paths.Lemma 7 Let D be a diagram with minimal system {s, S2, ,Sk}. Ifs E B Ft Sm thenDegc(D) n — 2m + 2sDI — 2where B =Proof. By Corollary 1, we haveDeg(D) n + 21DI — 2.It follows from the definition of B that 1DI = sD — m. It is then not difficult tosee that the lemma is true. oChapter 2. Kauffman bracket 21Let i be the mirror image of D. By the obvious one-one correspondence betweenthe crossings of D and the crossings of D, We can think of a state s of D as being alsoa state of D. So we can denote the set of states for tY by S as well. Denote by B andthe closure of the minimal systems for D and 7 respectively. Recall that the span of (D)is the highest degree of (D) minus the lowest degree of (D), that is, Deg(D) + DegD).We have the followingCorollary 5 Suppose .s Sm is a state for a diagram D. If s e B fl B, thenspan(D) 4(77 — m).Proof. By the above lemma we havespan(D) 2(n — 2m) + 2{sD + IsDI} — 4.Following Kauffman [36] we havesD+sD<n+2.Hence the inequality. DProposition 3 Let D be a link diagram and s be a state of D in Sm. If each diagramD’ obtained from D by nullifying the crossings at which s takes value —1 is +adequate,then the closure of the minimal system associated to D is contained in the closure of s;that is, B C ..Proof. Let s’ , then s’ A s . and there exists an s” with s’ < s” < s’ A s suchthat s” and s’ A s take different values only at one crossing, say c. Notice that ifs” isnot in B then neither is s’. Hence we may assume s’ s” without loss of generality.We claim s’D < s’ A sD so that s’ is not contained in B by the definition of B. ToChapter 2. Kauffman bracket 22justify our claim, we may regard s’ and s’ A s as states for D’, where D’ is a diagramobtained from D by nullifying all crossings at which s takes value —1 in A’-direction ifs’ A s takes value +1. Clearly, as states for D’, s’ S1(D’) and s’ A s = 1 in S(D’). Byassumption, D’ is adequate. Hence s’Dl = Is’D’l < 1D9 = sDj. DRemark 7 The above proposition is also true if we replace s by an antichain in S.Corollary 6 If D is a dealternator reduced rn-almost alternating diagram, then the closure of its minimal system is contained in , where s is a state of D satisfying s(i) = —1if and only ifi m.Proof. This follows from proposition 3 and the fact that each D’ is a reduced alternating diagram, hence each D’ is +adequate.In what follows we always assume D is reduced, namely, has no nugatory crossings.Proposition 4 Suppose D is a dealternator connected and rn-almost alternating diagramand .s is a state of D satisfying s(i) = —1 if and only if i rn. Then B contains .Proof. The proof is essentially contained in the proof of Lemma 4.3 in [1], we do notrepeat it. DBy Corollary 6 and Proposition 4 we haveCorollary 7 If D is a dealternator reduced and dealternator connected rn-almost alternating diagram, then D has minimal system {s}, where s is defined as in proposition4.Proposition 5 If D is dealternator reduced, or dealternator connected, or rn-almostalternating, then so is . Moreover, if changing the first rn crossings in D produces analternating diagram, then the same is true for D.Chapter 2. Kauffman bracket 238(D 1D -1DFigure 2.5: A coiinterexampleTheorem 4 If D is a dealternator connected rn-almost alternating diagram, then span(D)4(n-m).Proof. By Proposition 4, B contains . By Proposition 5, B also contains . Hences e B fl . By Corollary 5, span(D) 4(n — rn). 0Note that Deg(D) < n + 2I1D — 2 for rn 1 implies that Deg(D) n + 211D1 — 6by the definition of Kauffman bracket. If D is a dealternator reduced and dealternatorconnected rn-almost alternating diagram, then, by Corollary 7 and Theorem 1, Deg(D) <n + 21DI — 2. Thus Deg(D) n + 2I1D — 2 — 4. Similarly we have Deg(D)n + 21DI —2 — 4. Thereforespan(D) 2(n — 2rn) + 2{sD + s} — 4 — 8 4(n — m — 2)and we have proved the following theoremChapter 2. Kauffman bracket 24Theorem 5 (Theorem 4 in [1]). If D is a dealternator reduced and dealternator connected rn-almost alternating diagram, then span(D) 4(n — rn — 2).In the paper [1], the authors conjectured that removal of the condition “dealternatorreduced” in the above theorem would lead to span(D) 4(n— rn — 1). It turns out thatTheorem 4 is the optimal result (see example 3).Example 3. Let D be the diagram in Figure 1.5. Clearly, D is a dealternatorconnected 2-almost alternating diagram. From diagrams 1D and —1D, we see that Dhas minimal system {512, 845} and has minimal system {512, 33}, where 5j denotes thestate taking value —1 only at i and j. It is then easy to see that D is subadequate. ByTheorem 2 we havespan(D) = 2n + 2(I1D + ) —4 32 4(10 — 2).which is exactly 4(n — m).Remark 8 We can compute the span by isotoping the diagram D into an alternatingdiagram, (Figure 2.6), because the span of D and that of the alternating diagram are theFigure 2.6: An alternating diagram for the counterexampleChapter 2. Kauffman bracket 25same.2.6 Semi-alternating diagramsIn this section, we will resolve some problems in [58]. To this end, we first fix notationsand definitions. We refer to [58] for more details.Let G be a graph. Let V(G) and E(G) be the (finite) sets of vertices and edges of C,respectively.A graph C is said to be signed if either +1 or —1, called sign, is assigned to eachedge. More precisely, C is a signed graph if G is a graph equipped with a sign functionfG : E(G) —+ {1, —1}. We could call an edge e positive if fG(e) = 1 and negativeotherwise.A graph C is said to be separable if there are two subgraphs H and K such thatC = H U K and H fl K = {v0}, where both H and K have at least one edge and v0 is avertex.Let D be a link diagram of a link L. D divides R2 into a finite number of domains.We shade the domains in checkerboard fashion with the infinite domain shaded. Take apoint v from each unshaded domain R. These points form a set of vertices of a graphG. For any two unshaded domains R and 1?3 meeting at a crossing c of D, we assign anedge e to C to join vertices v and vj and sign it according to whether the crossing c ispositive or negative. See Figure 2.7.The signed plane graph C obtained above is called the graph of the link diagram D.Conversely, given a signed graph C, one can construct uniquely the link diagram D of alink having G as its signed graph.Let C be a signed plane graph. For a vertex v of G, let e1,e2,• e be an enumeration of all edges emerging from v in the counter-clockwise order. Set (v) =Chapter 2. Kauffman bracket 26//\sign +1 sign=—lFigure 2.7: Signed crossings=i sign(ej)sign(eji), e1 = e1. Then y(v) = {n + (v)} 0 is called the alternation index at v. -y(G) = minvy(v) is called the alternation index of G, where theminimum is taken over all vertices of G. Now we construct two graphs P0(G) and F(G)from C as follows.First we subdivide G by adding one vertex w to each edge e of G so that e,is divided into two edges e, e’. The resulting graph is denoted by G’. By definingsign(e) = sign(e7) = sign(ej), C’ becomes a signed plane graph. Define F0(v) =F(v) = star(v) if all the edges of G’ emerging from v have the same sign. If not, supposee2,1• , e2,2,••• ,.e2k are the edges of G’ emerging from v, where.signej,j = (_l)i+1,i = . ,2k. Let wjj be the other end of Then 1’0(v) is definedto be the plane (unsigned) graph consisting of ni—k edges, e1,• , e,e2,••.,e2,2,e3,1 , e3,fl_1, e3,e4,2 ,e4,, , e2kl, e2k,2,• , e2k,k, and their ends. F(v),on the other hand, consists of edges, e1,2 , ei,,, e2,1 , e2,2_1e2,e3,2 , e3,3e4,1e4,4_1 e4, , e2k,1,• , e2k,flk_1, e2k, and their ends. Here, ëj(l = 1,2,.•• ,2k) is the edgejoining two vertices wi,,1 and wii,i, where W2k+1,1 = wi,1. Finally, F0(G) and P(G) areobtained from G’ by replacing each star(v) by Po(v) and I’(v) respectively. F0(G) andI’(G) are called the over-graph and under-graph of C, respectively.Example 4.Chapter 2. Kauffman bracket 27ACF0(G) P(G)Figure 2.8: A link diagram D and related graphs G, P0(G) and F(G)Chapter 2. Kauffman bracket 28In the link diagram D of Figure 2.8, each box represents some positive twists ofcrossing number two or more. If we shade the domains of the diagram, only threedomains A, B and C are white domains. So its graph G has three vertices, labeled alsoby A, B and C respectively. Clearly, the graph has only negative edges except the topedge. Hence, from the definition, we get the graphs P0(G) and P(G) as shown in Figure2.8.A signed graph G is said to be semi-alternating if (1) 7(G) 2 and (2) both F0(G) andP(G) are connected and non-separable. A link diagram D is said to be semi-alternatingif the signed graph GD associated with D is semi-alternating. A link L is semi-alternatingif L admits a semi-alternating diagram.K. Murasugi posed the following conjecture in [58]Conjecture. A semi-alternating diagram D is a minimal diagram of the link represented by D. In other words, the link diagram D realizes the minimal crossing numberfor the link.It is not difficult to check that the diagram in Example 4 is semi-alternating. However,Figure 2.9: An equivalent diagramChapter 2. Kauffman bracket 29the diagram is not minimal because it is obviously equivalent to the diagram in Figure2.9. Hence the above conjecture is false.It is also in [58] that K. Murasugi raised the problem — whether a semi-alternatingdiagram is adequate.Let D be the diagram in Example 4. It is not difficult to see that D has minimalsystem {s}, where s takes value —1 only at the top vertex. By Corollary 3, D is not+subadequate. Heilce a semi-alternating diagram need not to be +subadequate, and weget a negative answer as well.Chapter 3Jones knot invariants and Vassiliev invariants3.1 IntroductionThis Chapter is divided into two parts. In the first part we investigate which knot invariant is Vassiliev invariant. In [13], J.S.Birman and X.S.Lin established a fundamentalrelationship between general Jones invariants and the Vassiliev invariants via a substitution t &. We show here that this relation can be rearranged into a different form(Theorem 6), probably more natural, from which one can prove that every derivative ofa general Jones invariant, evaluated at 1, is a Vassiliev invariant.While Bar-Natan observed that every coefficient of the Conway polynomial is a Vassiliev invariant, it is a little bit surprising that every coefficient of the Jones polynomialis not a Vassiliev invariant. This new result is proved in Theorem 7 below. On the otherhand we also prove that the Jones polynomial evaluated at any nonzero complex number,except the cubic roots of unity, is not of finite type.In the second part of this chapter, we study the question: which numerical knotinvariant can be approximated by Vassiliev invariants? We prove that the space of numerical invariants which can be approximated by Vassiliev invariants is a dual space ofsome vector space. Finally, we give an interesting property of Vassiliev invariants fromwhich we can retrieve, in a uniform manner, many known results about these invariants.For definitions and notations in this chapter, we refer to [7], [12] or [13].30Chapter 3. Jones knot invariants and Vassiliev invariants 31K K_ K0Figure 3.10: Related knot diagrams3.2 Relationship between quantum group invariants and Vassiliev invariantsLet S’ be the unit circle in the complex plane with a given orientation. A singular knotof order n is a piecewise linear immersion L: S1 —+ 1R3 which has exactly n transversedouble points. Two singular knots L and L’ are equivalent if there exists an isotopy—+ R, t [0, 1] such that h0 = id, h1L = L’ and the double points of hL areall transverse for every t E [0, 1].We denote K, K and K_ the singular knots identical outside a small ball arounda crossing and different inside a ball as shown in the Figure 3.10.Let IC, be the set of equivalence classes of singular knots with exactly i double points.In particular, IC0 = K, the set of all knot types.Let’s abbreviate U3 K3 by K>1 and denote the Q-vector space generated by a set Aby Q(A). Then the n-th finite-type space F is defined as the vector space generated bythe set K>0 subject to the following relations(1) K = K — K_ for K K>1Chapter 3. Jones knot invariants and Vassiliev invariants 32(2) K = 0 for K K>1.Any element in the dual space Hom(F, Q) of F, but not in Hom(F_1,Q) is calleda Vassiliev invariant of order n or finite type invariant of order n.Recall that a quantum group invariant or a generalized Jones invariant is a knot orlink invariant obtained from a trace function on a”R-matrix representation” of the familyof braid groups {Bn = 1,2,3••.}; refer to [12] for more details.Theorem 6 Let J= 00 ct be a quantum group invariant and00 j(m)(fl(t_1)mmO m.be the Taylor series of J at 1, where J(m)(l) denotes the m-th derivative evaluated at 1.Then the constant term is 1 and the coefficient ‘‘‘) of (t — 1) is a Vassiliev invariantof order m form 1.Proof. First we note that a Vassiliev invariant of order m times a nonzero constant isalso a Vassiliev invariant of order m and that the sum of a Vassiliev invariant of order m,and a Vassiliev invariant of order m’ is a Vassiliev invariant of order less than or equalto max{m,m’}.It is easy to see that the constant term is J(0)(1) = J(1) = 1, so we assume m 1.By the definition of derivative, we haveJ(m)(1) cn(n—1)...(n—(m—1)).fl —00Sincen(n — 1)... (n — (m— 1)) =— ( j)m_l + ( +1<i<m—1Chapter 3. Jones knot invariants and Vassiliev invariants 33we haveJm(1) =— ( cn1 + + (_1)m_l(m — 1)! can.n=—00 1zm—1 n=—oo n=—c.On the other hand, substituting t ex in J = c,-t and expanding cx in Taylorseries, we have00 00 00 m 00 00 mJ = c,e” = = ( >Zfl=—oo n=—co m=O m=O n=—ooBy a theorem of Birman and Lin, see [13, Theorem 1], the coefficient of Xm00 m 1 00,, -.--=is a Vassiliev invariant of order m for every m 1. Therefore, by the remark at thebeginning of this proof, J(m)(1), as a linear combination of Vassiliev invariants of orderm, is a Vassiliev invariant of order less than or equal to m. Since there is only oneterm in the summation having order m, is a Vassiliev invariant of order m. Theproof is completed. DCorollary 8 The m-th derivative of a quantum group invariant evaluated at 1 is a Vassiliev invariant of order m. In particular, the m-th derivative of the Jones polynomial,evaluated at 1, is a Vassiliev invariant of order m.Proof. Clear from the proof of the above theorem.Corollary 9 For every quantum group invariant J, J’(l) = 0 and J”(l) = av2, where ais a constant and v2 is the first nontrivial Vassiliev invariant.Proof. The first equation follows from the fact that there are no nontrivial Vassilievinvariants of order 1. The second one follows from that there is only one nontrivialVassiliev invariant v2 of order less than or equal to 2, up to constant multiplication. DChapter 3. Jones knot invariants and Vassiliev invariants 34Figure 3.11: Torus knot of type (2, 2n + 1)Remark 9 For the Jones polynomial J, since J”(l)(trefoil)= —6, we get J”(l) = —6v2.It is known that the second coefficient of the Conway polynomial = L”(1) = —J”(1),hence we reprove the fact that v2 is the second coefficient of the Conway polynomial.3.3 Some known invariants which are not Vassiliev invariantsBar-Natan observed that every coefficient of the Conway polynomial is a Vassiliev invariant. In contrast, we haveTheorem 7 Every coefficient of the Jones polynomial is not of finite type.Proof. First we prove the constant term c0 of the Jones polynomial J = ct’is not of finite type.Notice that if the minimal degree of the Jones polynomial J(K) for a knot K isgreater than zero, then co(K) = 0. Denote the torus knot of type (2, 2n + 1) by K21,see Figure 3.11.According to V.Jones [31], we havetnii’ rz —_______f-i 42n+2 43 I 42n+3‘ 112n+1)— 1 — t2 “ — —&—j— &2n + 1Chapter 3. Jones knot invariants and Vassiliev invariants 352n + 1Hence0(K21)= 0 for ii 1 and co(Ki) = 1.Now extend c0 to singular knots as usual, i.e.,2nc0(K) = co(K+) — co(K_).Let K2 be the singular knot as shown in Figure 3.12,By the above observation, we haveco(K2m) = (_1)P(2jco(K2fl_ )+l) = (_1)2n= 1 0.pzO PSince n is arbitrary, we see that c0 is not of finite type.Now we turn to Cr for r> 0. It is well known that for the trefoil knot K we haveJ(K)=t-i-t3—t4Let K’#K2+i be the connected sum of K2+i and r copies of K. By a property of theJones polynomial, we have J(Kr#K2+1)= J(KyJ(K21). It follows that the minimaldegree of J(K#K2+1) r + 1 for n 1, hence we haveCr(Kr#K2n+i) = 0 for n 1.As before we extend Cr to singular knots. Since Cr(K’#Ki) = Cr(K’) = 1, we haveCr(KT#K2Th) (_1)P(2jCr(Kr#K2n_p)+l) = (_1)2n 0.p=o PFigure 3.12: A singular torus knotChapter 3. Jones knot invariants and Vassiliev invariants 362n - 1 2nFigure 3.13: A singular torus knot with 2n double pointsHence Cr is not of finite type.Applying the above argument to the mirror images of knots we can prove that Cr isalso not of finite type for r 0. This completes the proof.Remark 10 Theorem 7 implies that there are infinitely many integer-valued invariantswhich are not Vassiliev invariants. We can have a even stronger result, see Corollary 13and Corollary L below.Remark 11 Theorem 7 is also true for the HOMFLY polynomials of one variable.Theorem 8 The genus, the minimal (maximal) degree of the Jones polynomial and thespan of the Jones polynomial are not of finite type.Proof . Let K2 be as in Figure 3.13, and K21 be the torus knot of type (2,2n + 1).Denote the genus of knot K by g(K), then we know that g(K21)= minimal degree ofJ(i)=nforn0andg(K_i)=0.Now extend g to singular knots byg(Kx) = g(K) — g(K).Chapter 3. Jones knot invariants and Vassiliev invariants 37Then we haveg(K2) = (_l)P (2fl)g(Kl)+l)pO P2n—1=pz0 P2n—1 \ 2n—1= (2n-1) (_i)fl)- (_i)fl)p= (2n_1)(_1)P()_(2fl_1)(_1)2fl(:n)-(-1) (2;) + (_i)2n (j 2n= -(2n - l)(_l)2n(j + (_1)2n()2n=(l)2n(jLO.The second equality holds because g(K1)= 0, the fifth equality follows from Proposition 2 for k = 1 and 0.Since n is arbitrary, we have proved the theorem for genus and the minimal degree ofthe Jones polynomial.For the span of the Jones polynomial, we know that span(K2i)= n + 3, hence asimilar argument can be applied. This completes the proof. DRemark 12 For K2+1 where n 1, the unknotting number is n, the minimal crossingnumber is 2n + 1 and the braid index is 2. Hence the above proof also shows that theseinvariants are not of finite type. These facts are already known. See J.S.Birman [12].Remark 13 Theorem 8 is also obtained by J. Dean [15], Rolland Trapp [72] and others.Chapter 3. Jones knot invariants and Vassiliev invariants 38In connection with Theorem 6, we may ask if Theorem 6 holds for Taylor seriesexpansion in powers of (t — c) with c 1. The answer is no because of the followingtheorem.Theorem 9 The Jones polynomial evaluated at any nonzero complex number, except1, and w2, is not of finite type, where w is the primitive root of x3 = 1.Proof. As before we extend the Jones polynomial to singular knots. Let K2 denote thesingular knot as in the proof of Theorem 7. Since= 1 _2(1 — — t3 +t23)we haveJ(K2) = (—1) (2n)J(K_)+i)p=o P2n=1—2—— t3 + t2(H3)p=o P= 1 -t2{(-1) ()t2 (1)(2j— (i)P( jt2n_P+3 + E(_1)P( n)t32n_P+3p=o P p={(l — t)2 —t2(1 —t3)2 —t3(1 — t)2 +t3(1 —t3)2}= 1 — t2——— t2(1 + t +t2)(1—= 1 2(1 — t)2(1 — t)(1 —i2(1 + t +t2)1.It follows that if t is not a root of the above term, then J(K2) is nonzero. Hence, toprove the theorem we only need to deal with these roots.To this end we consider some other singular knots. Let K2’ be the knot as shownin Figure 3.14.Chapter 3. Jones knot invariants and Vassiliev invariants 392n+2 2n+1Figure 3.14: A singular torus knot with 2n + 1 double pointsSimilar to the above computation, we obtainJ(K2’)= 1— t)2’(1 —t3)(1 t2(1 + t +t2).It is easy to see that —1 is the only common root of (1 —t2(1+t+t2)’)and (1 _t2(1 +t +t2)). Since t = —1 is a zero of order 1 for (1 — t2(1 + t +t2)—’), J(K2)(—1) 0.Therefore, given any to which is not a cubic root of unity, for every positive integer nthere exists a singular knot K with more than n double points such that J(K)(to) 0.Hence the Jones polynomial evaluated at to is not of finite type. The proof is completed.DRemark 14 We do not know whether every numerical invariant can be approximatedby Vassiliev invariants. However Theorem 9 together with Theorem 6 imply that thereare infinitely many non-finite-type invariants which can be approximated by finite-typeinvariants.Remark 15 The Jones polynomial evaluated at any cubic root of unity is known to bethe constant 1. Hence we have completely determined whether the Jones polynomial,evaluated at a complex number, is of finite type.Chapter 3. Jones knot invariants and Vassiliev invariants 40Remark 16 A similar proof shows that the one-variable HOMFLY polynomials evaluatedat a complex number outside the unit circle are not of finite type.3.4 Numerical invariants which are approximable by Vassiliev invariantsRecall that a numerical knot invariant is a map v : IC —+ Q. Let us restate the followingimportant problem described in [13].Problem 1. Given any numerical invariant v IC—f Q, does there exist a sequenceof Vassiliev invariants {v : K;—+ Q i = 1,2,. . .} such thatlim v(K) = v(K)for every K€IC?This problem can be posed in the following way: is the Vassiliev invariant space densein the 0-th cohomology of the knot type space? Another closely related problem isProblem 2. Do Vassiliev invariants distinguish knots?We are going to study the closure of the Vassiliev invariants in the 0-th cohomologyof the space of knot types. Some equivalent conditions will then be given which lead toa better understanding on Vassiliev invariants.Recall that the n-th finite type space F is defined as the vector space generated bythe set IC>0 subject to the following relations(1) K=K+—K_ for KeIC>1(2) K = 0 for K E K;>,iChapter 3. Jones knot invariants and Vassiliev invariants 41Note that we could allow n = in the definition. More precisely, F is the Q-vectorspace generated by the set K>0 subject only to relation (1). Since Q(K>0) = Q(K0)Q(K>1), and relation (1) simply kills the second summand, we haveProposition 6 F = Q(K0).It is clear that there is a sequence of finite type spaces and mapsF1 F ... ,‘ F0where each map is a natural and epimorphism. If we regard the sequence as an inversesystem, We have a projective limit denoted bylimFClearly, for every finite-type space F, we have a natural epimorphism F —+ Fsuch that the diagramcommutes. Passing to limit, we have a homomorphismlifl1F.Now the problem arises whether Tt is injective. More generally, what is the dimensionof the kernel of 7r? What kind of elements are in this kernel? Analogously, one can askF_1Chapter 3. Jones knot invariants and Vassiliev invariants 42whether it is surjective. If not, what is the dimension of its cokernel? we will answersome of these questions, which, as we can see, are very closely related to the Vassilievinvariants.Let us denote the Vassiliev space of order n by V = Hom(F, Q). Then it is clearthatV0 C V1C •••C V C V+1C •••C VC Hom(M,Q) C Hwhere V= U V, M is the image of F under it and H = Hom(F, Q) is the O-thcohomology group of the knot type space. Note that the inclusions are in the followingsense: if v e V,1, then v E Vi-, if and only if v factors through F, that is, if and only ifthere exists a v’ V, such that the following diagram commutesHence V is the set of all Vassiliev invariants of order less than or equal to n.Let A C H. The closure of A in H, denoted by A, is defined to be the set of allelements f H such that there exists a sequence {f} of A which pointwisely convergesto f, namelylimf(K)=f(K) for every Ke)C.Then we haveF+1FTheorem 10 V = Hom(M, Q), where M is the image of F under it.Chapter 3. Jones knot invariants and Vassiliev invariants 43Proof. Let us first proveVc Hom(M,Q).ConsiderF—M—-*•••-—F.and denote the composite of the above homomorphisms by d. Then we clearly haveKer(ir) C Ker(d) for every n.Suppose now f E V, then there exists a sequence {f} of V such thatlimf(K)=f(K) for every KEIC.Without loss of generality, we may assume that f,-, V,,, for every n. It follows thatf(Ker(d)) = 0, and hence f(Ker(7r)) = 0. Therefore, f(Ker(ir)) = 0. This meansthatf E Hom(M,Q).Now we prove the other inclusion. Supposef:M—Qis a nontrivial fullctional. Then there exists an x0 M such that f(xo) 0. It followsthat f is surjective, hence we haveM = Ker(f) e Q(xo).Choose x1,x2, , xj, to be a basis for Ker(f), then x0,x1,x2, ,x,••• is a basis forM. Since x0 0, there exists an n such that the image of x0 is not zero in F. Since dis surjective and F is finite dimensional, we can choose {x0,x,1,• , x,} such that theimage of this set is a basis for F and n1 <m2 < n, with r such that for any x notappearing in the set, d(x) is a linear combination of d(x)’s with j < i. Definef:F—+QChapter 3. Jones knot invariants and Vassiliev invariants 44by taking f(d(x0))= f(xo) and f(x) = 0 for every i.Note that dm(XO) 0 in Fm for every m n, and therefore we can define fm EVm in a similar way for every m n. We claim that the sequence {f} pointwiselyconverges to f. In fact, any finite set {x0,x1,•• , x} must, by construction, appearin some ,x,}. But then it appears in {XO,Xmi, ,Xmr} for every m n.Hence we have1imf(x) f(x).This proves our claim, in other words, f V. DCorollary 10 The answer to Problem 1 is positive if and only if Ker(ir) is trivial.Corollary 11 The answer to Problem 2 is negative if and only if there are two differentknots K and K’ such that K — K’ Ker(ir).Proof. If the Vassiliev invariants do not distinguish two knots K and K’, then for everyVassiliev invariant v, we have v(K) = v(K’), and hence, K K’ in F for every n. Bythe definition of projective limit, K — K’ e Ker(ir).Conversely, if K—K’ E Ker(r), we clearly have that v(K) = v(K’) for every Vassilievinvariant v. This completes the proof. D3.5 Knot connected sum in finite type spacesLet K be a knot in R3. DefineK#:C—Kby taking K#(K’) = K’#K. It induces a homomorphismQ() Q() ,‘ Q(io) ,‘ F.Chapter 3. Jones knot invariants and Vassiliev invariants 45Extend K# to singular knots as followsK#(K) = K#(K) — K#(K).It is not difficult to see that K induces a homomorphism, still denoted by K,F —+ F.Proposition 7 K is an isomorphismProof. This is a corollary of Gussarov’s result, see [23] and [60]. Or one can also argueas in the proof of Proposition 10 below.It is easy to see that the diagramcommutes, and that K# induces a homomorphism, still denoted by K#,lim F —+ lim F.It is again an isomorphism.Notice that the diagramChapter 3. Jones knot invariants and Vassiliev invariants 46K#Fc,D F00commutes, and by passing to limit we get the following commutative diagramK#FM MIt follows that K#(Ker(7r)) C Ker(’r). This property is useful in the proof of thefollowing theorem.We say that K and K’ are distinguishable by Vassiliev invariants if there exists aVassiliev invariant v such that v(K) v(K’). K and K’ are weakly distinguishableby Vassiliev invariants if there exist a Vassiliev invariant v and a knot L such thatv(K#L) v(K’#L). Based on the same observation preceding Theorem 10, we may nowprove a property on Vassiliev invariants similar to that of the quantum group invariants.Theorem 11 K and K’ are distinguishable by Vassiliev invariants if and only if theyare weakly distinguishable by Vassiliev invariants.F1Chapter 3. Jones knot invariants and Vassiliev invariants 47Proof. One direction is clear. Suppose there exists a Vassiliev invariant v and a knotL such that v(K#L) v(K’#L). If v(K) = v(K’) for every Vassiliev invariant v, byTheorem 10, K — K’ E Ker(ir), hence L#(K — K’) = K#L — K’#L € Ker(ir) forevery knot L. By Theorem 10 again, v(K#L) = v(K’#L) for every knot L. This is acontradiction. The proof is completed. 0K# is always injective because it is an isomorphism on the projective limit. Thefollowing theorem shows that this homomorphism reflects some important properties ofVassiliev invariants.Theorem 12 K—K’€ Ker(ir) if and only if K#=K :M—4MProof. Suppose K# = K, then K = K’ in F for every n, therefore, K—K’ E Ker(7r).Conversely, let K — K’ e Ker(ir), then K = K’ in F for every n. For fixed n, let L bean arbitrary knot, then L#(K) = L#(K’). HenceK#(L) = L#K = L#(K) = L#(K’) = L#K’ = K(L).Therefore, K#(L) = K(L). 0Conjecture. K# = K : M —+ M if and only if K#ir(1() Kir(K) if and only ifK#() = K(M).The basis for proposing this conjecture is the following proposition.Proposition 8 If Problem 1 has a positive answer, then the above conjecture is alsotrue. If problem 2 has a positive answer, then the first part of the conjecture is true.Proof. Clearly we haveK# = K K&7r(AC) = Kir(K) K(M) = K(M).Chapter 3. Jones knot invariants and Vassiliev invariants 48Now suppose that there are Kand K’ with K# K ( hence K K’ ) such thatK#(M) = K(M). Since K#(K) I(PC), we may, without loss of generality, choose aknot L e K#(K) but not in K(K). Note that the images of Q(K#(K)) and Q(K4(K))under ir are the same because K#(M) = K(M). Hence there exists an element x EQ(K(K)) such that r(x) ir(L). Clearly, x L in F, therefore 7t is not one to one,hence Ker(ir) {0}. The proposition now follows from Corollary 10.The second conclusion of this proposition is shown in the same way. 03.6 Some properties of Vassiliev invariantsThe following theorem is an interesting observation which allows us to retrieve, in auniform manner, some known interesting results about Vassiliev invariants.Theorem 13 Let K be a knot which is equal to the unknot in F. Then for every knotL, K#L = L in F, where K denotes the connected sum of i copies of K.Proof. Since K is equal to the unknot in F, K# = id on F. Therefore, K1#L =K(L) = L for every positive integer i. 0According to Gusarov and Yamamoto (see [60], for example), there exists a nontrivialknot K which is equal to the unknot in F. By Theorem 13, K#L’s have the sameVassiliev invariants. Since K#L’s are clearly different knots, we conclude thatCorollary 12 (Lin and Stanford) There are infinitely many knots sharing the same Vassiliev invariants of order < n.Corollary 13 Let v be a Vassiliev invariant of order n. Then for every x E Q,1v(x) fl K = 0 or oc.Chapter 3. Jones knot invariants and Vassiliev invariants 49Proof. If v’(x) fl K 0, choose a knot L such that v(L) = x. Let K be the knotwhich is equal to the unknot in F. By Theorem 13, v(K#L) = v(L) for every i. Hencethe corollary. DCorollary 14 Numerical invariants such as the minimum crossing number, the unknotting number, the braid index, the numerical invariants with finite support, etc are notVassiliev invariants.Proof. The minimum crossing number of a knot is zero if and only if the knot isthe unknot. Therefore, by the above corollary, the minimum crossing number is not aVassiliev invariant. Similarly, we can prove the other cases. DCorollary 15 (Trapp) The complement of the Vassiliev invariants is dense in the 0-thcohomology of the knot type space.Proof. Clearly, every numerical invariant can be approximated by numerical invariantswith finite support. By the above corollary, these invariants are in the complement ofthe Vassiliev invariants, hence the corollary.Theorem 14 If one of the numerical invariants listed in Corollary 14 can be approximated by Vassiliev invariants, then the Vassiliev invariants distinguish nontrivial knotsfrom the unknot, in other words, there exists no nontrivial knot K such that K — U EKer(ir) for the unknot U.Proof. If there is a nontrivial knot K such that K — U E Ker(Tr), then using a proofsimilar to the proof of Corollary 13, we can show that every numerical invariant v E Vhas the property described in Corollary 13. But invariants listed in Corollary 14 do notChapter 3. Jones knot invariants and Vassiliev invariants 50have this property, thus they are uot approximable by Vassiliev invariants, contradictingthe hypothesis.Finally let us make a few comments on r. While we do not know whether KerQir) = 0,it is not difficult to see that if the kernel is nontrivial then it is infinite dimensional. Thecokernel of it, on the other hand, has uncountable infinite dimension. So far we do notknow what role is played by the cokernel of it in knot theory, but its interest cannot beruled out.Chapter 4Finite type invariants for lens spaces4.1 IntroductionLet M be a connected and oriented 3-manifold, S’ be the unit circle in the complexplane with a given orientation. A singular knot of order n is a piecewise linear mapL : S’ —+ M which has exactly n transverse double points. Two singular knots Land L’ are equivalent if there exists an isotopy h : M —+ M, t E [0, 1] such that= id, h1L L’ and the double points of hL are all transverse for every t E [0, 1].We denote by the set of equivalence classes of singular knots with exactly ndouble points. Clearly, £ = £° is the set of all knot types.Since L is oriented, for each double point we have two resolutions, denoted by L+ andsee Figure 3.10. For more details, see [46].Let C be the set of homotopy classes of loops in M, i.e. the set of conjugacy classesof iri(M). Then it is easy to see that every singular knot L belongs to a unique loophomotopy class; moreover, if L is equivalent to L’, then L and L’ belong to the sameloop homotopy class.Let R be a ring with unit, and R(A) be the free module over R generated by a set A.DefineF(M) =where is the submodule generated by the following relations(1) L = L — L_, L E £ for some i51Chapter 4. Finite type invariants for lens spaces 52n+1(2) L(x = 0, L (n+1)F(M) is called the n-th finite type space of M. The dual space of F(M), denotedby V,,(M), is called the Vassiliev space of M, and elements of V(M) are called finitetype invariants or Vassiliev invariants.For M = R3, we already know much about V(IR3). Refer to [7], [12], [13] and [46]for more details. But we know little about V(M) for a general 3-manifold M. X.S. Linproved in [46] that V(M) = V(IR3) when 7ri(M) = ir2(M) = 0. We are still lookingforward to more developments along this direction. Since the solid torus plays a veryimportant role in the study of 3-manifolds, it is our goal to understand more about V(M)for a solid torus. In this chapter, we first study the finite type spaces for a solid torus,then show that the Vassiliev spaces of lens spaces L(p, q) and L(p’, q’) are isomophic ifp = p’. Consequently, we see that Vassiliev spaces do not distinguish manifolds.4.2 An operation on finite type spacesLet c C, and £)(M) = {L E E c}, then defineF,(M) =>0R())/where .— is generated by the following relations(1) = L — L_, L E £ for some in+1(2) LcE’Ex = 0, L EProposition 9 F(M)= eCEC F(M).Proof. It is easy to see that if L c then L and L_ are also in c. Hence the propositionfollows. DChapter 4. Finite type invariants for lens spaces 53Corollary 16 V(M) = fJcV,(M), where V(M) is the dnal space of F(M).It follows that the study of F(M) is reduced to the study of each F,(M) which isstill a complicated task. Let M = R3 and K be a knot in F. DefineK#C—÷Lby taking K(L) = K#L. This induces a homomorphismR(C) - R(L) —÷ —÷ F,(R3).Extending K# to singular knots byK#(L) = K#(L+) — K#(L_),we can define a homomorphism, still denoted by K#,K# : F(13)—+ F(IR3).By Proposition 7, K# is an isomorphism.Let us now study the case when the manifold M can be embedded in 1R3, for example,handle bodies, Whitehead manifolds, etc. LetM—-R3be an embedding, and c E C be a loop homotopy class. Given K c, choose a 3-ballinside of M, denoted by B, such that K fl B is a simple line segment. We try to constructa mapK# : £(B) —kFor each L £(B), let K#(L) be a “connected sum” of K and L. Here the “connectedsum” is performed inside the ball B. Strictly speaking, we understand the “connectedChapter 4. Finite type invariants for lens spaces 54sum” as follows: for each knot type, choose a representative knot L and a band connectingone point on L with a point on B fl K, then do the connected sum along the band. SinceB — K U L is path connected, we always assume the band is contained in B. In this waywe get a map from £(B) to £(M), hence a map to F(M), also denoted by K#. Nowextend K# to F(B) via the formulaK#(L) = K(L) — K#(L_),and obtain homomorphismsF(B) F,(M) -- F(R3),where i. is induced by inclusion map i. It follows from Proposition 7 that the compositionof the above homomorphisms is an isomorphism.Theorem 15 F,(M) = F(R3) ker(i).Proof. By the above remark, we have an exact sequence0 —* Ker(i) —+ F(M) -- F(R3).This sequence splits because we have a homomorphism K#(i*K#)1 : F(IR3)—+ Fc(M)such that iI(iK#)’ = id. Therefore we haveF,c(M)= F(R3) ker(i).Let us study the special case when M is the solid torus T. Since C = iri(M) = Z inthis case, we haveF(T) = CEZF(T).andChapter 4. Finite type invariants for lens spaces 55Two knots in a solid torus Connected sum of two knotsFigure 4.15: Connected sum of two knots in solid torusCorollary 17 For every c € Z, F,c(T) = F(IR3)e ker(i).Similar to the “connected sum” described before, we have a “connected sum” forknots in T. We describe it by diagrams. See Figure 4.15.Note that the orientations should coincide with each other. Note also that the “connected sum” is dependent on the band on which the surgery is performed, and hencemight be noncommutative and nonassociative.Now that each pair of knots K, K’ can be assigned a “connected sum”, denoted asusual by K#K’, we can extend such “connected sum” to singular knots in T simply byLX#L’ L+#L’ — L#L’.Note that L #L’ may be represented by a singular knot. With this operation one caneasily prove the followingTheorem 16 For n 1, F(T) CEzF,(T) is a graded algebra (possibly nonassociative) with the unknot serving as unit element.Remark 17 A similar “connected sum” can be introduced for any manifold M.Chapter 4. Finite type invariants for lens spaces 56Choose a “connected sum” as above, and let K be a knot. DefineK#:L—4Cby taking K#(L) = K#L. This induces a homomorphismR(C) R() , , F(T).Extend K# to singular knots as follows:K#(L) = K#(L+) — K#(L_).It is not difficult to see that I induces a homomorphism, still denoted by K#,K# : F(T) —f F(T).Proposition 10 K# is an isomorphism.Proof. First we assume that the knot K has winding number zero. Then K# takesF(T) into F(T), and we only need to show that K# restricted to F is an isomorphismfor every c E Z.For n = 0, F(T) is a space of dimension one generated by a knot, say L, which is inc. K# carries L to K#L which is equal to L in F°(T) because K can be changed intothe unknot via crossing changes. Hence the proposition is true in this case.Assume the proposition true for values < n and consider the following commutativediagramChapter 4. Finite type invariants for lens spaces 570K IF,(T) . F(T)F1(T) K F1(T)The bottom K# is an isomorphism by induction assumption. The top K# is anidentity map because K#L can be changed into L via crossing changes. Therefore, bythe five lemma, the middle K3 is also an isomorphism.Now we assume that K has winding number r. Choose a knot K’ such that K’ haswinding number —r. For example, we may choose K’ to be the knot obtained from K byreversing the orientation of K. Then K#K’ has winding number zero, hence (K#K’)# isan isomorphism by the previous argument. Moreover, this is indeed true for any chosen“connected sum”. Hence the composite IK is also an isomorphism. Likewise forKIi. Therefore K# is an isomorphism. The proof is completed. 04.3 Similarity classesLet us recall the definition of similarity classes. Following [46], let L : S1 —+ M bea singular knot. The pre-image of a double point consists of two points in S. Identifying these two points into one for each double point, we get a 1-dimensional compactChapter 4. Finite type invariants for lens spaces 58polyhedron, denoted by FL. Let L and L’ be singular knots of order n in M. Wesay that L is similar to L’ if (1) there is a homeomorphism FL —+ FL’ which liftsto an orientation preserving homeomorphism 31 S’, and (2) there is a homotopyS’ —+ M, t E [0, 1], with 4o = L, 4 = L’, and such that there are finitely manyparametric values 0 < t1 < <tj,, < 1 so that• e £‘), i = 1,... ,k;• for different t’s in the same connected componemt of [0, 1] — {t1,. tk}, the ‘sare equivalent singular knots in and• when t passes through t, q changes from one resolution of to the other.The equivalence classes of singular knots so defined are called similarity classes.Choose a representative from each similarity class to form a set.Proposition 11 F,(M) is generated by where Q = {L E 1ZL E c}.Proof. For n = 0, all the knots in c are similar. We can choose a knot as the representative so that F(M) is generated by the chosen knot, hence Proposition 11 is true in thiscase.To prove that it is also true for n, under the induction assumption that it is true for< n— 1, we consider the exact sequenceR(Q) -- F,(M) --*F1(M) —f 0Let x E F(M) and x’=?r(x). By the induction assumption x’ can be written in theform= >a3L L3 E U<_1with oniy finitely many as’s nonzero.Chapter 4. Finite type invariants for lens spaces 59Regarding ZaL as an element in F(M), we have x— ZaL = g(y) for somey E R(Qj. Since y is a linear combination of elements of f, so is g(y). Hence x is alinear combination of elements of U0 Q. The proof is completed. DIt is clear that similarity of L and L’ in T implies their similarity in R3. Let’s fix asimilarity class L in R and ask how many similarity classes in T correspond to L. Ifthere were only finitely many similarity classes for each L, then F,(T) would be finitelygenerated. Unfortunately, this isn’t quite the case. This is because one can always letone arc go clockwise around T p times and let another arc go counterclockwise aroundT q times so that p — q = c, while different values of p and q may correspond to differentsimilarity classes. In other words one can easily construct infinitely many similarityclasses for every c. On the other hand, in a slightly different way, we will define a finitelygenerated module which is important to the study of the finite type space of lens spaces.Given a knot L, assume that all the double points are in the upper part of the solidtorus, or equivalently in a given 3-ball in T so that outside of the ball, the knot consistsof parallel strings. Such a singular knot embedding is called a special embedding. It isclear that every singular knot in T has a special embedding. Now delete all double pointsof a special embedding for the singular knot to get several arcs, labelled as l(i = 1, . . k).For each i, we choose an oriented arc lj in the upper part of the solid torus so that thebeginning point 1 is the end point of and the end point of i is the beginning point of Iand so that ij has no other intersection with L. Then the composite I * i, is a loop in thesolid torus which represents an element in the homology group H1(T) = Z. This element,an integer, is called the “winding number” of I with respect to the special embedding.For every sequence a1, , ak of integers, one can construct a knot L’ such that thecorresponding winding numbers are a1, ak and L is similar to L’ in R3, as Figure4.16 shows.Chapter 4. Finite type invariants for lens spaces 60Figure 4.16: Arc winding numbersIt follows from the following proposition that any singular knot which is similar toL in R3 can be constructed in this way. As noted before there are infinitely many suchknots that are not similar to each other in T.Proposition 12 Let L be a singular knot contained in a 3-ball in T. Then any singularknot K in T which is similar in 1R3 to L is similar in T to a singular knot constructedfrom L via letting l(1 i k) go around T a1 times , where a2 is the winding numberof an arc of K which corresponds to i for some special embedding of K.3 times -1 timesChapter 4. Finite type invariants for lens spaces 61Proof. We may assume that all the double points of L and K are in the upper partof the solid torus. Since L and K are similar in R3, we can move the double pointsof L via an isotopy of T to match the double points of K according to the one to onecorrespondence determined by the similarity property. Hence we may assume that L isalready in this position. Now we let each arc l go around the solid torus a times , wherea is the winding number of the corresponding arc of K. We denote the resulting singularknot by L’. We need to prove that K is similar to L’ in T.The corresponding arcs of K and L’ constitute a loop which is homotopically trivialbecause their winding numbers are the same. Therefore the arcs of K are homotopicto the corresponding arcs of L’ relative to the end points. By Denn’s lemma, everycorresponding arc pair bounds a disk. Perturbing this disk if neccesary, we can ensurethat it avoids the double points and intersects with knots K and L’ transversely. Withsuch disks, one can easily construct the required homotopy. The proof is completed. DFor a given integer p 0, one can introduce a relation between singular knots in T.Namely, L and L’ are said to be p-related if one of them can be obtained from the otherby taking connected sum with some knot of winding number +p. The relation generatedby p-relation and similarity relation is called p-similarity.Modulo the p-relation in the finite type space F(T), we haveProposition 13 F(T)/ = Z{F(T)/‘-}, where is the p-relation. Moreover,this module is finitely generated.Proof. Since F(T) = zF(T), it is clear that if two singular knots are p-similar, thedifference of their grades in F(T) must be a multiple of p and that any element of F(T)is p-similar to some element of F,0(T) (simply let one string go ±p times around thesolid torus T). Also notice that there are some identifications within the same summandsof F(T). Therefore the first part of the proposition is true.Chapter 4. Finite type invariants for lens spaces 62For the second part, we need to note that F(T) is generated by the set of representatives from similarity classes of order less than or equal to n , hence F(T)/ ‘-‘.‘ isgenerated by the set of representatives from p-similarity classes in T of order less thanor equal to n. Since the set of similarity classes in R3 of order less than or equal to n isa finite set, it follows from Proposition 5 that the set of p- similarity classes is a finiteset too ( of cardinality not excceding the number of similarity classes in R3 times p2n )Hence the second part of the proposition. 04.4 Finite type spaces for lens spacesNow we turn to study lens spaces and show that their finite type spaces can be describedby the finite type space of the solid torus.Let L(p, q) be a lens space. Then it can be obtained by attaching a 2-handle to thesolid torus T along a torus knot of type (p, q) followed by attaching a 3-handle. Fromthis point of view, it is not difficult to see that every singular knot is isotopic in L(p, q)to a singular knot contained in the solid torus. Therefore we can define F(L(p, q)) usingonly those singular knots contained in T. Notice that adding a 2-handle is to add somerelation to the module F(T) = eF(T) and adding a 3-handle does not affect themodule. Hence, to determine the finite type space of L(p, q), we only need to figure outthe relation corresponding to the 2-handle. The relation can be described as follows. LetL and L’ be two singular knots such that L’ can be obtained from L by sliding a stringof L over the attached disk. We say that L’ is a sliding of L in this case. The relation issimply the one generated by slidings.Lemma 8 Sliding a string over the attached disk is equivalent to taking connected sumwith an oriented torus knot of type (p, q), where the orientations may be different forChapter 4. Finite type invariants for lens spaces 63different slidings. Sliding a double point over the attached disk is equivalent to takingtwice connected sum with an oriented torus knot of type (p,q).Proof. Since the boundary of the attached disk is the torus knot of type (p, q), slidinga string over the attached disk is equivalent to taking connected sum with an orientedtorus knot of type (p, q). Sliding a double point over the attached disk can be realized byfirst “parallelly” moving two strings near the double point over the attached disk, thenmoving the double point along the band bounded by the two strings to where the doublepoint is located after the double point sliding. Since the second moving is clearly anisotopy of the solid torus T, it is easy to see that sliding a double pint over the attacheddisk is equivalent to taking twice connected sum with an oriented torus knot of type(p,q). DTheorem 17 Two singular knots in T are similar in L(p, q) if and only if they arep-similar.Proof. Let L and L’ be two singular knots in T which are similar in L(p, q). Then L’can be obtained from L via crossing changes, slidings and isotopies which are performedoutside of the 3-handle. Since the torus knot of type (p, q) has winding number +p, byLemma 8, slidings do not change p-similarity. Crossing changes and isotopies clearly donot change p—similarity. Hence, L and L’ are p-similar. This proves one implication. Forthe reversed implication, assume that L and L’ are p-similar. Notice that any knot ofwinding number +p is similar to the (p, q)-type torus knot in T. Hence a singular knotconnected sum with a knot of winding number +p is similar to the same knot connectedsum with the (p, q)-type torus knot. Hence L is related to L’ by crossing changes, isotopiesand slidings. In other words, L and L’ are similar in L(p, q). The proof is completed. DNow we are going to prove the main theorem of this chapter.Chapter 4. Finite type invariants for lens spaces 64Theorem 18 F(L(p,q)) where is the p-relation. Moreover, thefinite type spaces of lens spaces are finitely generated.Proof. We knowF(T) EEZF(T).Since “sliding” is equivalent to “connected sum”, for any singular knot K e n E Z, thereexists a singular knot K’, obtained from K by taking iterated connected sum with a(p, q)-type torus knot, such that K’ E i with 0 < i <p — 1. Hence we haveF(L(p,q)) Jqwhere the relation q is generated by slidings with respect to the torus knot of type(p, q). Since the relation q is contained in the relation ‘-‘-‘, we have a quotient mapi: F,(T)/ F,!(T)/ for every i.Now we prove the theorem by induction on n. Since every F1j(T) is one dimensional,the theorem is clear for n 0. Assume it is true for n — 1, and consider the followingdiagram:F(T)/q__II IChapter 4. Finite type invariants for lens spaces 65Here i denotes the quotient map, and R(L4) is a submodule of F,(T) generated bysimilarity classes of order n in i.By Theorem 17, the relation‘q, restricted to similarity classes, is the same as thep-relation. Hence the top is actually an identity map. Since the bottom ii is anisomorphism by induction assumption, it follows from the five lemma that the middle iis an isomorphism too. This completes the proof of the first part of the theorem. Thesecond part follows from Proposition 13. 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