ALEXANDER INVARIANTS OF LINKS ': • by James Leonard B a i l e y B.Sc.(Hons) Ca r l e t o n U n i v e r s i t y (Ottawa), 1969 M.Sc. Ca r l e t o n U n i v e r s i t y (Ottawa), 1970 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY i n the Department of Mathematics We accept t h i s t h e s i s as conforming to the req u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA June, 1977 © JAMES LEONARD BAILEY In presenting th is thes is in p a r t i a l fu l f i lment of the r e q u i r e m e n t s f o r an advanced degree at the Un ivers i ty of B r i t i s h C o l u m b i a , I ag ree that the L ibrary sha l l make it f ree ly ava i l ab le for r e f e r e n c e and s t u d y . I fur ther agree that permission for extensive copying o f th is t h e s i s f o r scho la r ly purposes may be granted by the Head o f my Department o r by h is representat ives . It is understood that c o p y i n g o r p u b l i c a t i o n o f th is thes is f o r f inanc ia l gain sha l l not be allowed without my wri t ten permission. Department of Mathematics The Univers i ty of B r i t i s h Columbia 2075 Wesbrook P l a c e Vancouver, Canada V6T 1W5 June 22, 1977. i i A b s t r a c t . Supervisor: Dr. Dale R o l f s e n . In the three main s e c t i o n s of t h i s t h e s i s ( chapters I I , I I I , and IV; chapter I c o n s i s t s of d e f i n i t i o n s ) .we explore three methods of studying Alexander polynomials of l i n k s which are a l t e r n a t i v e s to Fox' f r e e d i f f e r e n t i a l c a l c u l u s . In chapter I I we work d i r e c t l y w i t h a p r e s e n t a t i o n of the l i n k group and show how to o b t a i n a p r e s e n t a t i o n f o r the Alexander i n v a r i a n t . From t h i s we deduce that the order i d e a l of the Alexander i n v a r i a n t i s p r i n c i p a l f o r l i n k s of two or three components ( the case of one component i s w e l l known ) but n o n p r i n c i p a l i n general f o r l i n k s of four or more components. In any event we show that only one determinant i s needed to o b t a i n the Alexander polynomial. In chapter I I I we use surgery techniques to c h a r a c t e r i z e Alexander i n v a r i a n t s of l i n k s of two components i n terms of t h e i r p r e s e n t a t i o n m a t r i c e s . We then use t h i s to show that the Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials when the l i n k i n g number of the two components i s zero or both components are unknotted and the l i n k i n g number i s two. Chapter IV uses S e i f e r t surfaces to prove a g e n e r a l i z a t i o n of a theorem of K i d w e l l which r e l a t e s the i n d i v i d u a l degrees of the Alexander polynomial to the l i n k i n g complexity, to present an a l g o r i t h m f o r c a l c u l a t i n g the Alexander polynomial of a two-bridge l i n k from a two-bridge diagram and to prove a conjecture of K i d w e l l i n the s p e c i a l case of two-bridge l i n k s . These r e s u l t s are then used to generate l i n k polynomials from allowable p a i r s ( a concept introduced i n chapter I I I ) and these r e s u l t s i n t u r n are used to produce methods of generating a l l o w a b l e p a i r s . i i i Table of Contents. I n t r o d u c t i o n 1 Chapter I : D e f i n i t i o n s . 4 Table 1. A l i s t of standard symbols. 6 Chapter I I : Computing the Alexander polynomial from the l i n k group. . 7 S e c t i o n A: L i n k s of two components. 8 S e c t i o n B: L i n k s of three components. 14 S e c t i o n C: L i n k s of more than three components 15 Chapter I I I : Surgery techniques. 24 S e c t i o n A: Using s u r g e r i e s to unknot knots. 25 S e c t i o n B: An example. 29 Se c t i o n C: The c h a r a c t e r i z a t i o n theorem. 32 S e c t i o n D: The Torres c o n d i t i o n s . 37 S e c t i o n E: C h a r a c t e r i z i n g l i n k polynomials. 40 Chapter IV: S e i f e r t surfaces. 44 S e c t i o n A: A theorem of K i d w e l l . 45 S e c t i o n B: Two-bridge l i n k s . 48 Table 1. 58 Table 2. 59 Table 3. 60 S e c t i o n C: Generating l i n k polynomials from al l o w a b l e p a i r s . 61 S e c t i o n D: Generating a l l o w a b l e p a i r s . 68 B i b l i o g r a p h y 70 Appendix: The general case of K i d w e l l ' s theorem. 72 1 I n t r o d u c t i o n . In t h i s t h e s i s I had hoped to decide whether or not the Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials, and i f not to f i n d added c o n d i t i o n s which would c h a r a c t e r i z e them. My p l a n of a t t a c k on t h i s problem seemed q u i t e reasonable: s t a r t w i t h l i n k s of two components; f i n d a c h a r a c t e r i z a t i o n of the Alexander i n v a r i a n t ; use t h i s to c h a r a c t e r i z e l i n k polynomials; g e n e r a l i z e the r e s u l t s to l i n k s of more than two components. My general f e e l i n g was that " what's good f o r two components must be good f o r more than two components." I knew that the order i d e a l of the Alexander i n v a r i a n t f o r l i n k s of two components i s p r i n c i p a l , so the Alexander i n v a r i a n t seemed simpler to work w i t h than Fox' f r e e d i f f e r e n t i a l c a l c u l u s . Research proceeded smoothly to the c h a r a c t e r i z a t i o n of the Alexander i n v a r i a n t , but the step from the i n v a r i a n t to the polynomial l e d to the problem of c h a r a c t e r i z i n g a l l o w a b l e p a i r s ( chapter I I I s e c t i o n E ) which has proved to be impossible ( at l e a s t f o r me. ) Consequently the problem of c h a r a c t e r i z i n g l i n k polynomials remains unsolved. Chapter I contains the d e f i n i t i o n s which are c e n t r a l to the t h e s i s . Concepts which are important but standard ( such as r e g u l a r p r o j e c t i o n , W i r t i n g e r p r e s e n t a t i o n , l i n k i n g number e t c . ) are omitted; the reader who. i s u n f a m i l i a r w i t h knot theory i s urged to r e f e r to R o l f s e n 2, e s p e c i a l l y chapters three, f i v e , seven, and e i g h t . Chapter I I presents an a l g o r i t h m f o r c a l c u l a t i n g the Alexander polynomial from a p r e s e n t a t i o n of the l i n k group using the Alexander i n v a r i a n t as an a l t e r n a t i v e to Fox' f r e e d i f f e r e n t i a l c a l c u l u s . In s e c t i o n s A and B we present the theory f o r two and three components r e s p e c t i v e l y and show that the order i d e a l i n these cases i s p r i n c i p a l . In c o n t r a s t to t h i s i s s e c t i o n C which deals w i t h more than three components and i s taken up w i t h 2 a proof of the s t a r t l i n g r e s u l t that the order i d e a l i n t h i s case i s not p r i n c i p a l when i t i s n o n t r i v i a l . This came as q u i t e a s u r p r i s e to me because I had always thought of " knots " and " l i n k s " a s being the proper d i v i s i o n w h i l e here the break occurs between l i n k s of three and four components. I do not know i f t h i s i s r e l a t e d i n any way to the f a c t that Burau and Gassner r e p r e s e n t a t i o n s are known to be f a i t h f u l f o r n < 3 w h i l e f a i t h f u l n e s s i s unknown i f n > 3 ( see Birman. ) Chapter I I I deals w i t h surgery techniques f o r l i n k s of two components. Sect i o n A b r i e f l y reviews surgery techniques and contains a lemma which w i l l be needed i n s e c t i o n C. S e c t i o n B contains an example. The main r e s u l t of the chapter i s i n s e c t i o n C; here Alexander i n v a r i a n t s of l i n k s of two components are c h a r a c t e r i z e d . This i s used i n s e c t i o n D to reprove the Torres c o n d i t i o n s i n the case of two components and to show that i f there are r e s t r i c t i o n s on l i n k polynomials other than the Torres c o n d i t i o n s these w i l l have to come from a study of all o w a b l e p a i r s . The concept of a l l o w a b l e p a i r s i s introduced i n s e c t i o n E where i t i s shown that the Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials i f the l i n k i n g number of the two components zero or i f both components are unknotted and the l i n k i n g number i s two. Chapter IV explores the technique of S e i f e r t surfaces i n the study of l i n k polynomials. In s e c t i o n A we prove a g e n e r a l i z a t i o n of a r e s u l t of K i d w e l l which r e l a t e s the i n d i v i d u a l degrees of the Alexander polynomial to the l i n k i n g complexity. S e c t i o n B turns to the s p e c i a l case of two-bridge l i n k s which have long been known to be p a r t i c u l a r l y s u i t a b l e f o r a n a l y s i s . In t h i s s e c t i o n we prove that two bridge l i n k s are interchangable and present an a l g o r i t h m f o r c a l c u l a t i n g the Alexander polynomial from a two-bridge p r e s e n t a t i o n . As a c o r o l l a r y to t h i s we prove a conjecture of K i d w e l l i n the s p e c i a l case of two-bridge l i n k s . The work i n s e c t i o n s C and 3 D was done i n the hope of f i n d i n g a way to c h a r a c t e r i z e a l l o w a b l e p a i r s without working d i r e c t l y w i t h the matri x i n the c h a r a c t e r i z a t i o n theorem. Unfortunately the r e s u l t s found are too complicated to do t h i s . S e c t i o n C gives methods of generating l i n k polynomials from a l l o w a b l e p a i r s and s e c t i o n D uses these r e s u l t s as w e l l as others i n the t h e s i s to compile a l i s t of methods f o r generating a l l o w a b l e p a i r s without r e s o r t i n g to ma t r i c e s . I would l i k e to thank s e v e r a l people f o r t h e i r help w h i l e I was working on t h i s t h e s i s . F i r s t my s i n c e r e s t thanks to Dale R o l f s e n who, besides being my supervisor has a l s o become a c l o s e f r i e n d ; to Kee Lam and Denis Sjerve f o r s e r v i n g on my committee; to Roy Westwick and Ben Moyls f o r t h e i r h e l p f u l encouragement when I was f e e l i n g that matrices must be the most damnable things ever invented; to Mark K i d w e l l f o r a s t i m u l a t i n g correspondence; and f i n a l l y , s p e c i a l thanks to A l i Roth whose diagrams help to r e l i e v e the monotony of the t e x t . 4 CHAPTER I: D e f i n i t i o n s . A link i s a homeomorphic image of y ( f i n i t e l y many ) d i s j o i n t 3 3 o r i e n t e d c i r c l e s i n S ; we f u r t h e r assume that S i s o r i e n t e d and the components of the l i n k are indexed. We w r i t e L = ^ ( j ^ Q ' *** U where each SL i s homeomorphic to and y i s the number of components or m u l t i p l i c i t y of the l i n k . Two l i n k s K and L are equivalent i f f the components have the same index set ( so K = k^ \j k^ \j ••• \jk^ and L = SL^\j SL2\I U &u ) a n ^ there i s an o r i e n t a t i o n p r e s e r v i n g homeomorphism 3 3 S ->S which r e s t r i c t s to o r i e n t a t i o n p r e s e r v i n g homeomorphisms k^ 1^ on each component. A l i n k i s tame i f i t i s equivalent to a polygonal ( that i s piecewise l i n e a r ) l i n k . A l l l i n k s w i l l be t a c i t l y assumed to be tame. In studying l i n k s , l i n k diagrams are o f t e n used. Roughly speaking, a link diagram i s what you would get i f you were to take a photograph of a l i n k . We w i l l f u r t h e r assume that there are at most double p o i n t s and that l i n e segments i n t e r s e c t t r a n s v e r s a l l y ( i n the l i t e r a t u r e t h i s i s c a l l e d a r e g u l a r p r o j e c t i o n ; see R o l f s e n 2 f o r d e t a i l s . F i g ure II.A.1 3 should make the concept c l e a r enough. ) I f L i s a l i n k then X = S -L i s a link complement and ir^(X) i s a link group. The Hurewicz homomorphism h:ir^(X) -»• H^ (X) def i n e s a r e g u l a r covering space p:X X c a l l e d the universal abelian covering space of X, namely the covering space so that p ^ ( T r^(X)) = ker h. As an immediate consequence of t h i s d e f i n i t i o n we see that a loop i n X l i f t s to a loop i n X i f f i t s l i n k i n g number w i t h each component of the l i n k i s zero. This i s u s u a l l y a convenient way to check when a given cover i s the u n i v e r s a l a b e l i a n cover. Since X i s a r e g u l a r covering space i t s group of covering automorphisms i s H^(X). But the components of the l i n k are indexed and o r i e n t e d , so by Alexander d u a l i t y there i s a ca n o n i c a l isomorphism H^(X) - 2Z^ . Let x^,X2»... 5x^ be the generators f o r the group of covering automorphisms of X corresponding to the ca n o n i c a l generators of TL*1 and take a e H^(X) . We de f i n e ± 1 i„ i i i - i X l X2 ••• X y a = X l * ° X 2 * X y * ( a ) where x^: H^(X) —>- H^(X) i s the homomorphism induced by the covering automorphism x: X —> X . Since we can a l s o m u l t i p l y by i n t e g e r s and , add t h i s allows us to de f i n e the a c t i o n of a polynomial on an element of H., (X) . In other words we have a 7L 72^ = A -module s t r u c t u r e on 1 y H^(X); t h i s module w i l l be c a l l e d the Alexander invariant of the link. A module w i l l be c a l l e d a link module provided i t i s isomorphic to the Alexander i n v a r i a n t of some l i n k of m u l t i p l i c i t y y • Given a p r e s e n t a t i o n of a A module i n terms of generators and r e l a t i o n s A = (a.. , a „ , . . . , a : r , r „ ,. . . ,r ) 1 z n 1 l m the corresponding presentation matrix i s (a„) where a „ e A i s n k defined by r . = E. ,a.,a.. Given a p r e s e n t a t i o n matrix the i d e a l E. l j = l 13 3 v A generated by the determinants of a l l (n-k)x(n-k) submatrices i s c a l l e d the k-th elementary ideal and depends only on A (see Zassenhaus page 90.) In case k=0, E^ i s c a l l e d the order ideal of A. I f A i s a unique f a c t o r i z a t i o n domain we l e t A be a generator of the minimal p r i n c i p a l k i d e a l c o n t a i n i n g E ; i t i s defined up to u n i t s i n A. Again the case k = 0 i s s p e c i a l : A^ i s c a l l e d the Alexander polynomial of A; A^ i s c a l l e d the Alexander polynomial of the l i n k ; a link polynomial i s a polynomial which i s equal to the Alexander polynomial of some l i n k . S n the u n i t sphere i n lRn+''" or anything homeomorphic to i t . I t i s sometimes thought of as the one p o i n t c o m p a c t i f i c a t i o n of ]R n. 8 boundary, e i t h e r i n homology or topology. T the i n t e r i o r of T. = equivalence i n the category i n question, i . e . homeomorphism f o r t o p o l o g i c a l spaces, isomorphism f o r groups, modules, e t c . 2Z the group of i n t e g e r s , w r i t t e n m u l t i p l i c a t i v e l y . K y TL<& 7L® • • • © 7L ( y copies of Z ) 5ZG the i n t e g r a l group r i n g of G. ZZZS^.'It i s thought of as f i n i t e Laurent polynomials i n y commuting v a r i a b l e s x ^ j x^, , x^ w i t h i n t e g r a l c o e f f i c i e n t s . £a,gj a8a 3^ ^ when a and 3 are elements of a group. [a,b| the closed i n t e r v a l between a and b when a and b are r e a l numbers. ±3 ±1 -1 a 6a 3 . a and 3 are elements of a group. (a,B,Y»***) the i d e a l generated by a, 3, y, ••• . r a a! b b!(a-b)! I I : .-a. a -a. a i = l l 1 2 n y ? .a. a. + a. 4- • • • + a ^1=1 I 1 2 n t r A the transpose of the ma t r i x A. diag(a,b,•••) the diagonal m a t r i x w i t h a, b, ••• down the diagonal. l k ( a , b ) the l i n k i n g number of the two c y c l e s a and b. sgn a +1 i f a i s an even permutation and -1 i f o i s an odd permutation. (2m,2) torus l i n k has 2m cr o s s i n g s Table 1. A l i s t of standard symbols. 7 CHAPTER I I : computing the Alexander polynomial from the l i n k group. There are three s e c t i o n s i n t h i s chapter, one f o r each of two, three, and more than three components. In each of these i s given an a l g o r i t h m f o r c a l c u l a t i o n of the Alexander i n v a r i a n t as a A module given a p r e s e n t a t i o n of the fundamental group of the l i n k , as w e l l as the Alexander polynomial. We use t h i s a l g o r i t h m to show that the order i d e a l i s p r i n c i p a l i n the case of l i n k s of two and three components ( the case of one component i s w e l l known ) but i s n o n p r i n c i p a l i n general f o r l i n k s of more than three components. 8 CHAPTER I I : Computing the Alexander polynomial from the l i n k group. The Alexander polynomial of a l i n k i s u s u a l l y defined using Fox' f r e e d i f f e r e n t i a l c a l c u l u s . What he c a l l s the Alexander matr i x i s a A -module p r e s e n t a t i o n matrix f o r the r e l a t i v e Alexander i n v a r i a n t y H^(X,XQ) where x^ e X i s a poi n t and x^ = p ^ ( * Q ) ( the n o t a t i o n i s the same as i n chapter I . ) The Alexander polynomial i s then defined to 1 k k+1 be A ,~ Although E f E ,~ _ * i n general, Levine 2 shows H^(.A,XQ,) HJ^A; H ^ V A , X Q ; that A^ ,~. = A„ +l~ _ v and so the two d e f i n i t i o n s of Alexander polynomial H^ (.A; H ^ ^ A J X Q ; are equivalent. In t h i s chapter we present a method of c a l c u l a t i n g the Alexander polynomial given a p r e s e n t a t i o n of the l i n k group as an a l t e r n a t i v e to Fox' f r e e c a l c u l u s ; i t i s a g e n e r a l i z a t i o n of the method used i n Ro l f s e n 2 f o r c a l c u l a t i n g knot polynomials. S e c t i o n A: Lin k s of two components. We s t a r t t h i s s e c t i o n w i t h an example. Example 1: - 1 - l Figure 1. The l i n k i n f i g u r e 1 has fundamental group 2 I f h : T r^(X) -»• H^(X) = TL i s the Hurewicz homomorphism we have h(£) = x, h(£) = h(n) = y where x and y are the f i r s t and second generators of 2 TL r e s p e c t i v e l y . I f we set a = n ? - 1 so that h(a) = 1, t h i s can be r e w r i t t e n as TT^X) = ( £, C, a : S a C ^ " 1 ^ " 1 ? " 1 ^ - 1 ? " 1 ^ ; ^ C C " 1 ^ " 1 ^ " ^ - 1 ? " 1 ) This can be f u r t h e r r e w r i t t e n as (*) TT^X) = ( K, 5, « : a ? £ , "^"V1-*5 J [5, £J _ S a ^ t"] c f ? ) [ T -1 -1 ±t ±1 -1 s, t l = s t s t and s = t s t . To b e t t e r v i s u a l i z e what i s happening we repl a c e X by a c e l l complex Y w i t h the same fundamental group: Y i s the one poin t union of three c i r c l e s w i t h two d i s c s attached, the boundaries of the d i s c s being attached to loops which represent the r e l a t o r s . We w i l l continue 2 to c a l l the loops Z, a and h:7r^(Y) -»• TL w i l l again be given by h(^) = x, h(5) = y, and h(a) = 1; Y w i l l be the covering space corresponding to ker h. For the sake of c l a r i t y l e t Z and Z be the 1-skeleton of Y and Y r e s p e c t i v e l y . Then i t i s c l e a r that • ^ ( Z ) •= ( Y 1 2 » A ~ ) where y^ 2 = 0>> ^} ' ^ p r e s e n t a t i o n f o r TT^(Y) and hence f o r TT^(X) i s obtained by adding the l i f t s of words r e p r e s e n t i n g the r e l a t o r s i n TT^(Z) to TT^(Z) as the r e l a t o r s of TT^(Y). This e x p l a i n s why we chose the p r e s e n t a t i o n (*); i n t h i s form i t i s p a r t i c u l a r l y easy to read o f f the new r e l a t o r s . For example a^y ^a~^a** becomes ya-yc^2~ a +y '''a, where the Hurewicz homomorphism h : i r^(Z)- -> H^(Z) i s given by h(Y-^2) = c ^ a n d h(a) = a. Thus we see that HjCX) = ( a, c 1 2 : ( y - l + y - 1 ) a - y c 1 2 , - ( l - x ) y a + ( l - x ) c 1 2 ) = ( a : ( l - x ) ( l - y ) ( l + y " 2 ) a ) 10 presented as a module. Hence we have 4 (x) = ( d - ^ d - y X i + y " 2 ) ) , A H X ( X ) = d - x X i ^ C i + y " 2 ) ~ 2 \ < x ) - 1 i f k>0 i f k>0 Notice that i n t h i s case the Alexander i n v a r i a n t i s c y c l i c ( t h i s i s not t r u e i n general ) and the order i d e a l i s p r i n c i p a l ( t h i s i s always true f o r two component l i n k s , as w i l l be shown below.) 0 o 0 0 a 0 a Z. Y c o n s i s t s of Z to which d i s c s have been e q u i v a r i a n t l y attached to loops r e p r e s e n t i n g the l i f t s of the r e l a t o r s . ZV Y c o n s i s t s of Z w i t h d i s c s attached to loops r e p r e s e n t i n g the r e l a t o r s . F i g ure 2. 11 We summarize the above example as an algorithm for computing a a presentation of the Alexander invariant as a module from a presentation of a link group in the case of two component links. 1. find a presentation for TT^(X) in the form ( £, a± l<i<n : r. l<j<n+l ) 2 where the Hurewicz homomorphism h : i r ^ ( X ) -> (X) = 2Z takes h(?) = x, h(£) = y, and h(a.) = 1 i f l f i f n where x and y are the two generators of 2 2Z . Notice that there is one more generator than relator. Using the Wirtinger presentation of a link i t is possible to show that the defect of any link group ( that i s , the number of generators minus the number of relators ) is at least one ( see Rolfsen 2 for a more complete discussion ). Hence we can always assume that we have a presentation with defect one by adding the t r i v i a l relator wherever necessary. 2. write the relators in terms of l<i<n, y ^ = ^ ,£j and their conjugates. In the example this was done by inspection, but here we present an algorithm for reducing a relator to this form. Since we w i l l need this result later we prove i t for any number of components. Lemma 1: Suppose that r i s a word in the symbols E,^ a n <^ a j ^=J=n' Suppose further that the sum of the exponents of £ is zero for l<i<y- Then r can be written as a word in y = ^ , z,7^ l _ i < j < V j l<i_ n> a n a their conjugates. Proof: Notice that wa = wa.w~lw that i s (1 . ) Wa = a w. We can apply (1 . ) whenever r contains an preceded by a £ to rewrite r as a product of conjugates of the a_^ 's followed by a word W in the 5 j ' s -12 Next consider W. We know that (2.) = ±1 ±i ±i where C^ . = 5^ • In the event that we can find a £^ followed by a i r 1, in ¥ we can apply (2.) to i t ; that is i f W = C^ ^2 w e c a n w r ' x t e WmWlh*lW2 " ^ i t V ' J W z " t l ' ^ i f * ^ i ^ 2 ' ±1 Now replace W by ^•^•^=-^2 a n d c o n t i n u e a s above u n t i l no £^ is followed ±1 i by a ^ i f 1. Then we have written W as a product of symbols of the C^ >?.jJ followed by where f/^ i s W with the 's omitted ( a l l the 5~^'s have been moved to the end, and since their exponent sum is zero they ±1 ±1 cancel each other out. ) Now replace W by and £^ by £^ and so on un t i l r has been written as a product of conjugates of the a^'s followed by a product of conjugates of |j^>£jj where l<i<jfv. The proof of the lemma is completed by noticing that .-1 „-l , , b.c cb and (a ) = a . 11 3. We can now write down a presentation matrix for H^(X). It is of the form ( a. l<i<n, c.. : R. l<i<n+l ) x = = I J 3 = = where R^. is obtained by writing r_. additively and replacing B by ±h(zd)h(B) where h:iT (X) H^X) and h-ir^X)--> H^X) are the Hurewicz homomorphisms ( h was defined on the previous page and fi is defined by h(a.) = a. and h(y..) = c...) As an immediate consequence of this algorithm we have 13 Proposition 2: For two component l i n k s the order i d e a l E^ i s p r i n c i p a l . Proof: (X) has a square p r e s e n t a t i o n matrix ( n+1 generators and n+1 r e l a t o r s . ) 11 Remark: For l i n k s of two or more components the i d e a l E,^ ,~r „ . i s the Hj^X,' x Q ) product of the n o n p r i n c i p a l i d e a l generated by the elements f o r l<i£y where y i s the m u l t i p l i c i t y of the l i n k and a p r i n c i p a l i d e a l generated by the Alexander polynomial. Hence i n the case of two component l i n k s i t i s s l i g h t l y simpler to work w i t h the Alexander i n v a r i a n t r a t h e r than the Alexander matrix. 14 Sectio n B: L i n k s of three components. The case of three components i s not s u b s t a n t i a l l y d i f f e r e n t from the case of two . components. Again we present an a l g o r i t h m f o r c a l c u l a t i n g a pr e s e n t a t i o n f o r the Alexander i n v a r i a n t as a module. 1. f i n d a p r e s e n t a t i o n f o r i r^(X) of the form T T 1 ( X ) = ( £ L F -£ 2, £ 3 > a± l<i<n : r l<j<n+2 ) 3 where h:ir^(X) -* (X) = ZZ i s the Hurewicz homomorphism which takes 3 h(£„) = x^ f o r i = 1,2,3 where x _ / i s the i - t h generator of ZZ and h(a^) = 1 fo r l<i<n. 2. use lemma 1 to w r i t e the r e l a t o r s i n terms of the a.'s, the Y..'S where ' i j = fs^sSjJ l l 1 < J ^ 3 , and t h e i r conjugates. 3. i n t h i s case Z i s a wedge of n+3 c i r c l e s and Z i s homeomorphic to the set 3 { (x,y,z) e TR : at most one of x, y, z i s not an int e g e r } 3 w i t h a wedge of n c i r c l e s attached at each v e r t e x . Again Z has a ZZ a c t i o n on i t and H^(Z) has a p r e s e n t a t i o n ( a. l<i<n, c 1 2 , c 1 3 , c 2 3 : ( l - x ^ c ^ + C l - x ^ c ^ + U - x ^ c ^ ) as a A 3 module. Here again we have the Hurewicz map h : i r^(Z) H^ (Z) defined by h(a.) = a. and h(y..) = c . where l<i<j<3. Hence a p r e s e n t a t i o n i i i l i l = = f o r H^(X) as a A 3 module i s ( a i 1= 1= n' c i 2 ' c 1 3 ' c23 : ( 1 - X 3 ) C 1 2 + ^ 1 _ X 2 ^ C 1 3 + ^ 1 ~ X 1 ^ C 2 3 ' R i 1 = i = n + 2 ) where i s obtained by w r i t i n g r ^ a d d i t i v e l y and r e p l a c i n g (3 by ±h(w)h(3) Since we again have an equal number of generators and r e l a t o r s we again have Proposition 3: For three component l i n k s the order i d e a l i s H^ (.A; p r i n c i p a l . 11 15 S e c t i o n C: L i n k s of more than three components. The method of c a l c u l a t i n g a p r e s e n t a t i o n m a t r i x f o r the Alexander i n v a r i a n t from a p r e s e n t a t i o n f o r the fundamental group i n the general case should now be c l e a r , although i t holds a b i t of a s u r p r i s e . 1. f i n d a p r e s e n t a t i o n f o r T T^(X) of the form T T ^ X ) = ( ^ I V * a j 1<j<n : r k l<k<n+y-l ) where h:ir (X) -> H ^ X ) = ZZ y i s the Hur ewicz homomorphism which takes hCS^) = x^ f o r i = 1, 2, ••• , y where x^ i s the i - t h generator of and h(a^) = 1 f o r i = 1, 2, ••• , n. 2. use lemma 1 to w r i t e the r e l a t o r s i n terms of the a.'s, the Y..'S where Y^J = i [ ^ i ' ^ j ] ^ = 1 <i= l J» a n c* t h e i r conjugates. 3. i n t h i s case Z i s a wedge of n+y c i r c l e s and Z i s homeomorphic to the set { p o i n t s i n TR^ : at most one coordinant i s not an i n t e g e r } w i t h a wedge of n c i r c l e s attached at each v e r t e x . Again Z has a ZZ^ a c t i o n on i t and H^(Z) has a p r e s e n t a t i o n ( a i l<i<n, c „ l<i<j<y : l<i<j<k<y ) as a A module. Here R = ( l - x . ) c . . - ( l - x . ) c . , + (l-x.)c„ and the y l j k k I J 3 i k i j k Hurewicz homomorphism h r i r ^ Z ) -* H., (Z) i s defined by h(a.) = &. and hCy, . , . ) = where l l i < j < y . Hence a p r e s e n t a t i o n f o r H^(X) as a module i s (t) ( a. l<i<n, c.. l<i<j<y : R.., l<i<i<k<y, R. l<i<n+y-l ) l = = - l j k = J = l = = where R^ i s obtained by w r i t i n g r ^ a d d i t i v e l y and r e p l a c i n g (3 ± W by ±h(w)h(g) Notice that we have a new w r i n k l e : we have n + generators and n + y - 1 + r e l a t o r s and n + y - 1 + - n - = C^" 1 > 0 i f y > 4. Hence the corresponding p r e s e n t a t i o n m a t r i x i s not square. Theorem 4: For l i n k s of more than three components the order i d e a l i s the product of the n o n p r i n c i p a l i d e a l I generated by elements of the form 16 .(1-x.) •*• where n. > 0 and .n. = , and a p r i n c i p a l i d e a l (generated i=l I I = ^ i = l l 2 by the Alexander polynomial) and hence i s not p r i n c i p a l i n general. Proof: Let P be the presentation matrix corresponding to the presentation (+) for H 1 ( X ) ; i s generated by the determinants of a l l submatrices 1 H^(.A ; obtained by deleting ^ rows from P. Since submatrices with zero determinant do not e f f e c t t h i s generating set, i t i s of value to notice that the rows of P axe not independent; i n f a c t , for any 1 < i < j < k < m < u we have (*) R. ., : (1-x )R... - (l-x.)R.. + (l-x.)R., - (l-x.)R.. = 0 ljkm m l j k k ljm j lkm l jkm We can c a l c u l a t e how many rows must be deleted to obtain a l i n e a r l y independent set by using vector spaces over the f i e l d of quotients of A . Let (respectively V^) be the vector space of dimension (respectively cV) with basis the set { R . : l<i<i<k<u } (respectively { R . : l<i<j<k<m<u}) 4 l j k = ijkm = Then we have a map -»- determined by (*)• and the number we want i s dim(imV,). 4 Lemma 5: dim(imV.) = ch" 4 3 Proof: The set { imR.., } i s not independent. To see what i s happening i n l j km general notice that we have a ZZV a c t i o n on MV which r e s t r i c t s to the Section on Y with fundamental region the hypercube nV_^ I^ where I = ,1^ for a l l i . The elements c „ can be thought of either as the squares I.j_Xl\j o r t h e i r boundaries; the r e l a t o r s R. ... correspond to the cubes I.xi.xT or t h e i r l j k r I j k boundaries; the r e l a t i o n s R. ... correspond to the hypercubes I . x i . x T x i or ijkm l j k m the i r boundaries, and so on. If V i s the dimensional vector space with n n basis { R.. : l<i<i<•••<m<y,n subscripts } there i s a homomorphism i j • • -m = = V V , defined by l n -> 3 i n . Then n n-1 dim(imV.) = dim(V,/imV,_) 4 4 5 = dimV, - dim(V r/imV^) 4 5 6 But so 17 = dimV, - dimV r + dimV, - ••• +(-l) ydimV 4 5 6 y = . c y - + ^ + ( - D v c y . 4 5 6 y o = ( i - i ) y = c£ - c\ + - • • • + ( - i ) V dim(imV 4) = -( • c[J - C'J + - ) = d f 1 1 Hence we must d e l e t e from P at l e a s t ^ of the rows corresponding to the R. ., 's to o b t a i n a submatrix w i t h nonzero determinant. Since t h i s i s e x a c t l y i l k the number of e x t r a rows we see that we can never d e l e t e a row corresponding to a R^ i n (t) and get a maximal submatrix of P w i t h nonzero determinant. We now need some way to compare the determinants of an a r b i t r a r y submatrix M of P w i t h independent rows w i t h a f i x e d submatrix M . Definition: i s the submatrix of P obtained by d e l e t i n g the rows corresponding to R.., where n £ { i , j , k }. The remaining rows are independent s i n c e c i £ n ^ j has nonzero c o e f f i c i e n t e x a c t l y once. In the next lemma there i s a problem w i t h keeping t r a c k of the order of the s u b s c r i p t s ; i n order to get around t h i s , l e t R . . . -. = R , where { i , j , k } = { a, b, c } 11 j J j K j 3.DC and a < b < c. S i m i l a r l y f o r Rr. . n -, • {t,i,k,m} c y ~ 2 2 Lemma 6: detM = ± ^ 1" x") detW D a y-2 3 2 d - x B ) 2 Proof: The r e l a t i o n s R r. . n 1 a l l o w us to w r i t e (1-x )R r. . i n terms of ( l - x . ) R r . . n, ( l - x . ) R r . „,, and ( l - x . ) R r . w i t h ±1 c o e f f i c i e n t s . Put the r e l a t i o n s R r . . i n t o some order, say R T. . i x , j , a , 3 } { I ^ J ,a,B}' R r. . •-. and so on. Let be the m a t r i x obtained from M by r e p l a c i n g { i 2 , i 2 , a , 3 > a 18 the row R r. . , by R,. . . Because of R r . . ' „, we see that {•ij.-Jj_.ct} J { l j . j j . g } { i j . j j . a . 3 } det M = ± rj-—=- det AT . a ( 1 - x g ) Now consider A / ^ and R r . . and proceed as before to o b t a i n A/^2^ w i t h { i 2 , J 2 , a , B > d e t M ( 1 ) ='±4 det A f ( 2 ) ( 1"V that i s ( 1 _ x a ) 2 (2) det M = ± det AT ' a v 2 ( l - x e ) A f t e r 2 steps we o b t a i n a matrix which d i f f e r s from M only i n the order of i t s rows, and hence ,y-2 det A? = ± a 1-x ' C a 2 det A T cr2 Corollary 7: (1-x ) I det Af IT Lemma 8: the polynomial i n c o r o l l a r y 7 i s maximal i n the sense that i f p e A and p Idet M f o r a l l l i n k s and c a l c u l a t i o n s of M then p I(1-x ) ct y or ct a or a Proof: By symmetry we need only consider the case a = y. Since we have p det M f o r a l l l i n k s , i n p a r t i c u l a r p must be i n v a r i a n t up to u n i t s i n y' y y A^ under permutations of { 1, 2, ••• , y-1 }. We complete the proof by E x h i b i t i n g a f a m i l y of l i n k s so that i f p det Af and p i s i n v a r i a n t under y 1 y y permutations of { 1. 2. ••• , y-1 } then p (1-x ) y y Example: The d a i s y chain of l e n g t h y. 19 5 1? 2? 1 ^ 3 ^ 2 5 y - 2 5 y - l 5 y - 2 Figure 3. The fundamental group of t h i s l i n k i s ( KV e2, ••• , s y : [ c ^ g , [e2,e3], - , [VrS] } and so i t s Alexander i n v a r i a n t has a p r e s e n t a t i o n ( c.. l<i<j<y : R 1 5 R 2 , ••• , Ry_15 R. j k l<i<j<k<y ) where R. = c. .,,. M i s the m a t r i x corresponding to the rows R.. and R.. x x,x+l y r xjy x Notice that the rows R. have a 1 i n the c. . , .. column and 0 elsewhere, so x x,x+l we can omit the rows corresponding to the R^'s and the columns corresponding to the c. .,,'s and o b t a i n a matri x whose determinant d i f f e r s from that of x, x+1 M by a f a c t o r of ±1. Of the remaining columns, those corresponding to c. . l<i<i<y have a (1-x ) i n the row R.. and zeroes elsewhere, so we can xj = v xjy e matri x below „y-2 omit these C 2 1 - (y-2) = ry-2 C 2 rows and columns to o b t a i n whose determinant d i f f e r s from that of M by a f a c t o r of C l y C2y C3y c, 4y °y-2,y E12P ' - d - x 2 ) ( i - X l ) 0 0 0 R23, o - d - x 3 ) ( l - x 0 ) 0 0 34y 0 0 • a • • -d-x 4) ( l - x 3 ) 0 • • 0 0 0 0 -(1-x . y - l 20 The determinant of t h i s m a t r i x i s i ^ l - x ^ ) ( 1 - x ^ ) ( 1 - x ^ ) ••• (1-x _ j ) a n d so det M ' = ±(1-x.)(1-x.) ••• (1-x ,)(l-x") . Since the f a c t o r (1-x,) i s y 2 3 y-1 y 1 missing i t i s c l e a r that these l i n k s have the re q u i r e d p r o p e r t i e s . 11 We are now i n a p o s i t i o n to compare the determinant of an a r b i t r a r y submatrix M of P w i t h independent rows w i t h that of M . The c o n s t r u c t i o n used g e n e r a l i z e s that of lemma 6. Put the rows R.., which are rows of M and x j k l<i<j<k<y i n t o some order, say R. . . , R. . , , ••• , R. . , . The row 1 l J l k l V 2 k 2 Vnn R. . , can be w r i t t e n as a l i n e a r combination of the rows of M w i t h V i k i u c o e f f i c i e n t s i n A . At l e a s t one row which takes part i n t h i s l i n e a r Vi combination i s not a row of M, say R , . Let be the matrix obtained a l V l from M by r e p l a c i n g t h i s row w i t h R. . . . By the c o n s t r u c t i o n i t i s c l e a r V! x 1 J 1 k 1 that c o e f f i c i e n t of R. . , X 1 J i R l (1) detW- = —'• 1—4= P detAT : y coeffxcxent of E , a i V i We can re p l a c e M by to o b t a i n and so on. A f t e r n stages we a r r i v e V at a matrix which d i f f e r s from M by at most a permutation of the rows. Hence , product of c o e f f i c i e n t s of rows added , ' „„ det M = ± det M. y product of c o e f f i c i e n t s of rows deleted Now we need a b e t t e r i d e a of what these c o e f f i c i e n t s look l i k e . To s t a r t out w i t h we have the set of r e l a t i o n s s = { n ,. . , ( l - x )R. n ,. . , (l - x . )R. . . } ^ l ^ l ' k l ^ a V l V a*VV kn' y > VnV which has the f o l l o w i n g p r o p e r t i e s : 1. each row of M which i s not a row of i s contained i n a r e l a t i o n y which i s an element of S which can be used to express i t i n terms of the rows 21 which are i n . 2. the coefficient of R. ., is always an integral multiple of II ,. . , (1-x ). xjk ' 6 r a/x,j,k v a I want to modify this set at each stage so that at the m-th stage 1. and 2. continue to be true with replaced by M^.This is done as follows: the relation which was used to go from ^ to M^m^ (which must contain the row R. . , ) i s removed from the set. Some of the relations which i J fe-rn m m remain may contain R. . , , but by hypothesis 2. these can be multiplied by m m m integers and added to an integral multiple of the newly removed relation to to eliminate the R. . , term; these linear combinations form my new set. x J k J m m m In view of 2., after cancelling like terms from the numerator and denominator we have m. AHV (i-x.) 1 det M = ± — — — det M y n. B n J = 1 ( l - x ) 1 m. where A, B e ZZ , n.>0, m.>0, and YV ,n. = 7^ Since AIlV ,(1-x.) 1-|det M , i= i= Lx=l x ^1=1 x i=l x 1 y ! by lemma 8 we have m. C„ -n AJlJ = 1(l-x ) 1 = ±(l-x ) P for some 0 < n < 2. Hence = y = 2 det M det M ^ J l . _ ± = A C„ n. (1-x ) 1 BIlV ,(l-x.) 1 U 1=1 X is a polynomial where B e 2 Z , n . >0, TV , n. =C y . I do not know i f the x = ^1=1 I 2 case B f ±1 can occur. The d i f f i c u l t part of the proof w i l l be finished i f , given n. > 0 l 1 < i < y with I^_2 n£ = 2 we can find a matrix M satisfying 22 det .Af det M y _ ry-2 n. ( i ^ ) 2 " i - i ^ - i ) 1 F o r t u n a t e l y t h i s i s r e l a t i v e l y easy. Look at lemma 6 and apply the proof w i t h a = y, [3=1, but stop a f t e r the n^-st stage. This gives us a matrix A f ^ w i t h _ n l d e t M = + C 1 xn> d e t M C D y n d - X j ) 1 Now look at the set of cl! 2 r e l a t i o n s { R r . . „ -, }.At l e a s t 2 - n, > n„ 2 {x,j,2,y} 2 1 = 2 of these do not con t a i n rows which have been e l i m i n a t e d , so use these to get M ( 2 ) w i t h n l + n 2 ( 1"V (2) det Af = ± det A T ' y n n ( l - x j ) ( l - x 2 ) I t i s c l e a r that we can continue i n t h i s way; at the m-th stage there are at l e a s t c l ! 2 - ( n 1 + n „ + * " + n n ) > n r e l a t i o n s which do not co n t a i n 2 1 2 m-1 = m eli m i n a t e d rows. We e v e n t u a l l y a r r i v e at M w i t h n +n + ••• +n (1-x ) 1 2 det M = ± det Af y i n . y-1. . l n i = 1 ( i - x . ) which i s equivalent to the re q u i r e d statement. A l l that remains to complete the proof i s toprove that I i s not p r i n c i p a l . We do t h i s by c o n t r a d i c t i o n . We can de f i n e an epimorphism A^ -*• TL by s e t t i n g x^ = -1 f o r a l l i . Under t h i s homomorphism the generators of I y-2 y-2 2 2 are sent to 2 so I i s sent to the i d e a l generated by 2 .On the other hand, i f I were p r i n c i p a l i t would have to be a l l of Ay s i n c e 23 rV-2 _y-2 2 2 (1-x^) and (1-x..) are r e l a t i v e l y prime. As a c o r o l l a r y to the proof we have Corollary 9: The Alexander polynomial of a l i n k i s given by det M. c y _ 2 2 ( l - x y ) Bemark: From t h i s we see that the Alexander polynomial of the l i n k i n the example i s ( l - x 2 ) ( 1 - x ^ ) ( 1 - x ^ ) ••• (1-x 24 CHAPTER I I I : Surgery techniques. This chapter deals almost e x c l u s i v e l y w i t h l i n k s of two components. Sectio n A b r i e f l y reviews surgery techniques and contains a lemma which w i l l be needed i n s e c t i o n C. Secti o n B contains an example. The main r e s u l t of the chapter i s i n s e c t i o n C; here the Alexander i n v a r i a n t s of l i n k s of two components are c h a r a c t e r i z e d i n terms of t h e i r p r e s e n t a t i o n m a t r i c e s . This i s used i n s e c t i o n D to reprove the Torres c o n d i t i o n s i n the case of two components and to show that i f there are r e s t r i c t i o n s on l i n k polynomials other than the Torres c o n d i t i o n s these w i l l have to come from a study of allo w a b l e p a i r s . The concept of a l l o w a b l e p a i r s i s introduced i n s e c t i o n E where i t i s shown that the Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials i f the l i n k i n g number of the two components i s zero or i f both components are unknotted and the l i n k i n g number i s two. 25 Chapter I I I : Surgery techniques. S e c t i o n A: Using s u r g e r i e s to unknot knots. Although surgery techniques had been used f o r some time i n the study of 3-manifolds and even to u n t i e knots ( see f o r example Hempel ), Levine was the f i r s t to use them to analyse knot and l i n k complements. The idea has s i n c e proved u s e f u l to s e v e r a l authors i n c l u d i n g R o l f s e n , Goldsmith, Shaneson, B a i l e y - R o l f s e n , and L i c k o r i s h . We begin t h i s s e c t i o n by q u i c k l y reviewing the ideas which w i l l be used. I f we are using a l i n k diagram to study a l i n k L, we can change a c r o s s i n g at the expense of adding a surgery torus as f o l l o w s : e n c i r c l e the c r o s s i n g w i t h a s o l i d torus T so that ( i . ) T i s unknotted, ( i i . ) T l i e s i n a b a l l which i s d i s j o i n t from any other surgery t o r i which may be present, and ( i i i . ) I k ( c, JL ) = 0 f o r i = 1, 2, ••• , u where c i s the ( o r i e n t e d ) c e n t r e l i n e of T and L = \j SL^V '" U Then there i s an autohomeomorphism 3 3 f:S -T -»• S -T so that the p r o j e c t i o n of the r e s u l t i n g l i n k f (L) i s the same as the o r i g i n a l one except that the e n c i r c l e d c r o s s i n g i s reversed. See f i g u r e 1. The autohomeomorphism which reverses the c r o s s i n g may be v i s u a l i z e d 3 as f o l l o w s : S -T i s again a s o l i d t o r u s ; cut along a m e r i d i o n a l d i s c , give a r i g h t or l e f t hand t w i s t as re q u i r e d and r e a t t a c h along the m e r i d i o n a l d i s c . We w i l l o r i e n t the c e n t r e l i n e c, the lo n g i t u d e 1, and the meridean m of T so that c, I, and f(m) are a l l homotopic i n T. L a b e l l i n g f(2") by l k ( f(m), c ) = ±1 gives us enough in f o r m a t i o n to r e c o n s t r u c t f up to iso t o p y and hence i s enough to recover the o r i g i n a l l i n k . More g e n e r a l l y , an i n t e g e r n which l a b e l s a surgery torus ( as i n s e c t i o n B ) w i l l i n d i c a t e 26 that a m e r i d i o n a l d i s c f(m) which r e a l i z e s the surgery t r a v e l s once l o n g i t u d i n a l l y around the torus i n the d i r e c t i o n of the c e n t r e l i n e c and l k ( f(m), c ) = n. See Ro l f s e n 2 Figure 1. Lemma 7.-Given a two component l i n k L = there are f i n i t e l y many d i s j o i n t s o l i d t o r i T. i = 1 , 2, ••• w i t h c e n t r e l i n e s c. and an 1 x autohomeomorphism f: S3-(9" y i j •••)->• S3-(9'i:u ^ i j ••• )so that ( i . ) the t o r i are unknotted and un l i n k e d , ( i i . ) l k ( c ± , £ ) = l k ( c±, f(A ) ) = 0 i = 1, 2, ••• , j = 1, 2. ( i i i . ) f(o"!Z\) = 32\. Each torus i s l a b e l e d by a ±1 as above. ( i v . ) f(L) i s a ( 2, 2m )-torus l i n k p i c t u r e d ' i n the t a b l e of standard symbols. Proof: Take a l i n k diagram of L. Since we can change some of the c r o s s i n g s 27 of &2 t o unknot i t , there are s o l i d t o r i T i = 1, 2, ••• n^ as i n the 3 n 3 n previous d i s c u s s i o n and a homeomorphism f^:S " U ^ i i ^ ^ _ ^ i - l " ^ i S O t^ i a t 3 3 f ( ^ 2 ^ ^ s u nk n°tted. We can now take a homeomorphism h^:S -> S which i s i s o t o p i c to the i d e n t i t y and so that there i s a r e g u l a r p r o j e c t i o n of h^of^(L) i n which h^of^C^) 1 S the standard p r o j e c t i o n of the unknot. We f u r t h e r i n s i s t that there be a short d i s t a n c e where the two components are cl o s e to and p a r a l l e l to each other which we engulf i n a b a l l B ( see f i g u r e 2. ) Figure 2. Next, f o l l o w h^of^(Jl^) from where i t leaves B to i t s f i r s t c r o s s i n g w i t h h ^ o f ^ ( f l ^ ) • We want to change cr o s s i n g s so that t h i s p a r t of h^of^(£^) l i e s e n t i r e l y over / under the r e s t of h ^ o f ^ ( ^ ) depending on whether that f i r s t c r o s s i n g has h^of^(Jt^) over / under h^of^(1^) r e s p e c t i v e l y . To do t h i s we place s o l i d t o r i h^(2\) i = n^+1, ri^+2j••"* , n^ around the cr o s s i n g s to be reversed and l e t f 0 : S 3 - , , ? 2 ,,hAT.) -+ S 3-.,? 2 ,,b.AT.) which 2 u i=n^+l 1 I u i = n T + l 1 l simultaneously reverses the c r o s s i n g s as r e q u i r e d . We are now i n \a p o s i t i o n 3 to f i n d an autohomeomorphism h^ of S which i s i s o t o p i c to the i d e n t i t y , leaves B f i x e d as a s e t , and so that there i s one fewer c r o s s i n g between the components of the image of L o u t s i d e of B and one more i n s i d e B. We can 28 f u r t h e r i n s i s t that the only part of the p r o j e c t i o n of the l i n k which changes l i e s on the image of from j u s t i n s i d e B to j u s t past the f i r s t c r o s s i n g . Figure 3 should help c l a r i f y these ideas. ( the surgery t o r i have been omitted from the middle and r i g h t diagrams. ) Figure 3 This l a s t step can be repeated u n t i l the images of and have no more crossi n g s outside B, say a f t e r m steps. may s t i l l be knotted, but we can e n c i r c l e c r o s s i n g s w i t h surgery t o r i h o ••• oh,(T.) i = n +1, ••• m i l m 3 and apply an autohomeomorphism f of S -ii. h o ••• oh,(T.) so that J m+l u i>n m 1 l m f ,,oh of o ••• oh 1of,(£ 1) i s unknotted. The re q u i r e d homeomorphism i s then m+l m m 1 1 1 f = h71o o h _ 1 o f ,,oh of o ••• oh,of, :S n-u .T. -> S 3-n .T.. U 1 m m+l m m 1 1 3 ^ x x u x x Remark: the proof of t h i s lemma g e n e r a l i z e s to more than two components, the r e s u l t being a pure b r a i d r a t h e r than a ( 2 , 2m ) torus l i n k . ( The ( 2 , 2m ) torus l i n k s are the pure 2—braids. ) 29 S e c t i o n B: An example. Surgery techniques are u s u a l l y too cumbersome f o r r o u t i n e c a l c u l a t i o n s of Alexander i n v a r i a n t s ; the f o l l o w i n g example i s to i l l u s t r a t e the techniques used i n the next s e c t i o n . The reader should be f a m i l i a r w i t h the examples presented i n the chapter on surgery c a l c u l a t i o n s of Alexander polynomials of knots found i n R o l f s e n 2. In order to o b t a i n an i n t e r e s t i n g p r e s e n t a t i o n m a t r i x we take the l i n k L obtained by performing the s u r g e r i e s as i n s t r u c t e d i n f i g u r e 4 and sewing a thickened d i s c across B to a r e g u l a r neighbourhood of y^ ( t h i s produces a l i n k w i t h l i n k i n g number two between the components.) Figure 5 3 shows the u n i v e r s a l a b e l i a n cover Y of Y = S - ( ^ i j B ) together w i t h the l i f t s of the surgery t o r i and one l i f t of y^, l a b e l l e d y^. The f u l l l i f t of y Q i s obtained by t a k i n g the t r a n s l a t e s of y^. The surgery i n s t r u c t i o n s ( that i s -1 on !Tj and +1 on T ) l i f t to surgery i n s t r u c t i o n s -3 on and +1 on T^. I f Z = Y-( the l i f t s of the surgery t o r i ) then i t i s c l e a r that (Z) i s a f r e e module of rank 3, the generators being ( t h i s i s the l i f t of the commutator of the two most obvious generators of ^ ( Y ) ) , and ct^ which are Alexander dual to the c e n t r e l i n e s of T and r e s p e c t i v e l y . Sewing m e r i d i o n a l d i s c s to the surgery t o r i as i n s t r u c t e d gives Rj_: -XOQ + ( y - 3 + y " 1 ) ^ + (x~ly~l-y~l) c*2 = 0 R 2: yo^ + (xy-y) + = 0 w h i l e sewing a thickened d i s c to a neighbourhood of y^ gives RQ• (l+xy)oo + ( l - x ) ( l - y ) x _ 1 a i - ( l - x ) ( l - y ) y " 1 a 2 = 0. Hence a p r e s e n t a t i o n m a t r i x f o r H.,(X) i s 30 ' 1+xy ( l - x ) ( l - y ) x _ 1 - ( l - x ) ( l - y ) y _ 1 ' . o . " I -1 " I " I -x y-3+y x y -y y xy-y 1 Notice that i f we ignore the upper l e f t hand entry and the f a c t o r - ( l - x ) ( l - y ) i n the f i r s t row, we get a matri x w i t h an Hermitian symmetry. This w i l l be explained i n the next s e c t i o n . -1 T L Figure 4 mil ^ i " H I ii 32 S e c t i o n C: The c h a r a c t e r i z a t i o n theorem. We now r e s t r i c t our a t t e n t i o n to l i n k s of two components. This s e c t i o n i s devoted to a theorem which c h a r a c t e r i z e s l i n k modules as modules which have a p r e s e n t a t i o n m a t r i x w i t h a symmetry c o n d i t i o n . Theorem 2: A module i s a l i n k module i f f i t has a p r e s e n t a t i o n m a t r i x of the form - ( l - x K l - y ) ^ ^ ^ ) ' 1-xy 1-xy BCx'^y" 1)^ 4(x,y) where A(x,y) i s a square ma t r i x , B(x,y) i s a row mat r i x , both w i t h e n t r i e s i n A2, s a t i s f y i n g 4(x,y) = A(x_1,y_1)tr and A(l,l) = diag(±l,±l, ••• ,±1). Further , 4(x,l) ( r e s p e c t i v e l y A(l,y) ) i s a p r e s e n t a t i o n m a t r i x f o r the f i r s t ( r e s p e c t i v e l y second ) component of the l i n k and I i s the l i n k i n g number of the two components. Here " t r " means transpose and diag(±l, ±1, ••• ,±1) i s a diagonal m a t r i x w i t h ±l's down the diagonal. Proof: Suppose L = i s a l i n k . By lemma 1 there are surgery 3 t o r i T i = 1, 2, ••• , n and an autohomeomorphism f of S -^2". so that f ( L ) i s a ( 2 , 2m ) torus l i n k . Let B be as i n the proof of lemma 1, and 3 l e t Y = S - (B ij f ( L ) ) . Then Y i s an open genus 2 handlebody and there i s a ca n o n i c a l isomorphism H^(Y) = the two generators being Alexander dual to f(&^) and fCJ^) r e s p e c t i v e l y . As i n the case of l i n k s , the u n i v e r s a l a b e l i a n cover Y of Y has as i t s group of covering automorphisms, and we see th a t H (Y) a l s o has a A module s t r u c t u r e on i t . In f a c t IT (Y) = ZZ * 7L i z \ O so i t i s c l e a r that H^(Y) i s the f r e e A9 module generated by h(££5 1) = fi where h:7r^(Y) -> H^ (Y) i s the Hurewicz homomorphism, and £ and £ are elements of ir^(Y) which are sent to the f i r s t and second generators of H^(Y) - ZZ&ZZ 33 r e s p e c t i v e l y by the Hurewicz homomorphism hrir^CY) •+ H^(Y). Here I am t h i n k i n g of 7Tj(Y) as a subgroup of TTJ(Y), SO E,t,E, ^ e TTJ(Y). We can v i s u a l i z e Y as 3 f o l l o w s : s i t u a t e f(Z-))-B i n S -B so as to have apparent l i n k i n g number +1. Next, imagine f (L)-B being completed to a ( 2, 2 ) torus l i n k by adding l i n e segments i n B. The complement of a ( 2, 2 ) t o r i s l i n k i s S ^ S ^ l 3 so i t s u n i v e r s a l a b e l i a n cover i s M x i R x M = ]R ( i n the diagrams we have drawn I R x l R x H as i f i t were I R * » x (0,1) .) I f we remove the l i f t of B 3 from t h i s cover we o b t a i n Y. This means that Y i s embedded i n S and we can 2 use a l i n k i n g number argument. Noti c e that the TL a c t i o n on Y extends to a 2 3 2 3 ZZ a c t i o n on TR and hence to a TL a c t i o n on S w i t h 0 0 as a f i x e d p o i n t . 3 ~ The c y c l e C q i n H^(S -Y) which i s Alexander dual to a runs from 0 0 up through the p i l l a r which ct Q i s s i t u a t e d on ( l i n k i n g once ) and c o n t i n u i n g up to «>. Let Z = Y - I J ? ,T. and Z the l i f t of Z to Y. Since the l i n k i n g x = l 1 6 number of each surgery torus w i t h the components of the l i n k i s zero, the preimage of a surgery torus c o n s i s t s of a torus which covers i t 2 homeomorphically plus a l l the t r a n s l a t e s of i t under the TL a c t i o n . From t h i s we see that H^(Z) i s a f r e e A^ module of rank n+1. We are now i n a p o s i t i o n to recover the u n i v e r s a l a b e l i a n cover of the l i n k . F i r s t , n o t i c e that each time a m e r i d i o n a l d i s c i s sewn to a torus i n Z a f a m i l y of m e r i d i o n a l d i s c s i s sewn e q u i v a r i a n t l y to the corresponding f a m i l y of t o r i i n Z, g i v i n g one r e l a t o r f o r H j ( X ) . These r e l a t o r s can be c a l c u l a t e d e x p l i c i t l y as f o l l o w s : l e t c. be the c e n t r e l i n e of the i'-th 1 surgery torus T., y. = f(m.) where m. i s a merideanof T. ( o r i e n t e d as i n 1 1 1 1 1 s e c t i o n A ) and c\ and y\ some l i f t of c^ and y_^ r e s p e c t i v e l y to X. An e x p l i c i t set of generators f o r Hj(Z) i s the set { where i s Alexander dual to c\ ( that i s , 34 l k ( a., x\ma ) = { * " i=J.and k=m=0 } 1 m 0 otherwise )• The r e l a t o r R^ obtained by sewing a m e r i d i o n a l d i s c to p. i = 1, 2, ••• , n i s R. = £ . , r . ., x-'y^a l i,k,m i i k m J m i k where r... = l k ( y., x J y c "ijkm ~ 1 K~ ( y., x J y c. ). There i s another r e l a t o r , R , obtained by l m o sewing a d i s c across B, the c y c l e \iQ i n 9B to which i t i s sewn being determined by the l i n k i n g number of the two components of the l i n k , i s the image of under a homeomorphism g:85 W which c o n s i s t s of t-1 L i c k o r i s h t w i s t s about t (• see f i g u r e 6 ) . The l i f t y of y can be described o o as f o l l o w s : s t a r t at the basepoint and f o l l o w a u n t i l t i s reached; f o l l o w t u n t i l you have covered t t-1 times; continue along the l i f t of ct^ which 7_1 2.-1-you are at ( namely x y ) u n t i l t i s reached again; f o l l o w t u n t i l t has been covered -(£-1) times; continue along a to the basepoint. From ° 7 t h i s we see that the c o e f f i c i e n t of a i n R w i l l be l+xy+*•* + (xy) =-:— •jo o J J 1-xy Since y i s homotopic to -(1-x) ( l - y ) ^ r - ^ ^ — £ i n S 3-Y ( see f i g u r e 6 f o r o 1-xy o the case I - 3 ) we see t h a t the c o e f f i c i e n t of a i n R w i l l be m o v / i \ / i \ l - ( x y ) ^ * 1 k -S. . (1-x) (1-y) , — r_.. x J y j , k J 1-xy Oikm J 1 k~ where r = l k ( cn, x y c ). ojkm 0 J m We can now w r i t e down a p r e s e n t a t i o n m a t r i x f o r (X): the entry i n the i - t h row j - t h column i s the c o e f f i c i e n t of oL i n R^ i,j=0,1,2, ••• ,n. This i s the m a t r i x promised by the theorem\A(x,y) i s the m a t r i x obtained by o m i t t i n g the row and column corresponding to R and a r e s p e c t i v e l y , and o o -1 -1 t r B(x ,y ) i s the column matrix corresponding to a o m i t t i n g the R^ entry. We must show i t has the r e q u i r e d p r o p e r t i e s . F i r s t , Figure 6 35 i k~ g k. l k ( u., x y c ) = l k ( c., x y c ) unless i=m and i=k=0 x J m x J m J - j _k_ = l k ( x y c. , c ) J x m = l k ( c , x ^y . ) m x -1 -1 t r and hence 4(x,y) = 4(x ,y ) and the top row ( the row corresponding to R ) must be o 1-xy v / v " 1-xy Once we have shown that 4 ( x , l ) i s a p r e s e n t a t i o n m a t r i x f o r the Alexander module of the f i r s t component corresponding to the given s u r g e r i e s as described i n R o l f s e n 1, i t w i l l a l s o f o l l o w that 4(1,1) = diag (±1,±1, ••• ) F i r s t , n o t i c e that 4(x,y) i s a p r e s e n t a t i o n matrix f o r the l i n k L' obtained by sewing a d i s c across and preforming the given s u r g e r i e s ( so the l i n k i n g number of the two components i s 1 ). Sewing a d i s c across gives a ( 2, 2 ) torus l i n k , and we can co n s t r u c t i t s u n i v e r s a l a b e l i a n cover i n two stages: the u n i v e r s a l a b e l i a n cover of the f i r s t component i s ]RxiRx]R and the l i f t of the second component together w i t h the p o i n t a t 0 0 i s the 3 3 unknot i n ]R + 0 0 = S . The u n i v e r s a l a b e l i a n cover of t h i s knot then gives the u n i v e r s a l a b e l i a n cover of £' as the composite cover. This plus the lemma i n R o l f s e n .3 makes i t c l e a r that 4 ( x , l ) i s a p r e s e n t a t i o n matrix f o r the Alexander i n v a r i a n t of the f i r s t component of L.The second component i s symmetrical w i t h the f i r s t , so 4(1,y) i s a p r e s e n t a t i o n m a t r i x f o r the Alexander i n v a r i a n t of the second component of L. To prove the converse we need only s i t u a t e unknotted u n l i n k e d surgery t o r i i n Y together w i t h surgery c o e f f i c i e n t s which produce the r e l a t o r s R. i = l , 2, • • • , n . Here we f o l l o w Levine 2. Let x D = [o, n+lj|x[o, l ] x [ 0 , l] and l e t D ± = [ i - 1 , i ] x [ o , l j x | j ) , ij , i = 1,2, • • • ,n+l. Let T.C D . be an unknotted s o l i d torus f o r i = 1, 2, ••• , n and l e t x x ' ' ' 36 3 ° i:D + S -V be an embedding where aC i (D) H B C i (D ,,) ( so i (D-D ,,)CY ) n+1 n+1 Follow the i n s t r u c t i o n s given i n Levine 2 pages 78 to 80 to modify T to give the c o e f f i c i e n t of i n R^. In order to avoid making the n o t a t i o n too cumbersome we continue to c a l l the modified torus T^; s i m i l a r l y f o r the other t o r i . But now we must f u r t h e r modify T as f o l l o w s : i f the c o e f f i c i e n t of a i n R. i s E,. . x ,c..x 1y^ then s i t u a t e d i s j o i n t 2-discs D. . o 1 ( x , j ) e l x j J xj i n i(D so that 3D.. = S.. i s homotopic to a i n i(D ,,)-B and use these n+1 xj xj o n+1 and use these f o r f u r t h e r m o d i f i c a t i o n s of as above. Next look at R^ and proceed as f o r R^. In a d d i t i o n we must take care of the c o e f f i c i e n t of which we do as f o l l o w s : l e t the c o e f f i c i e n t of oL be E-.. T ( x ^ - l ) c . where ( 0, 0 ) i I. S i t u a t e s"l". and sT. i n 1 ( i , j ) e l x j xj x j a small neighbourhood If of ( N i s chosen small enough that 717(1 ? \ = 0 f o r a l l i and N^\(D.) = N ) where S?. and S.. bound d i s j o i n t d i s c s i n Y 1 xj xj J and l k ( S +., 2\ ) = - l k ( sT., T, ) = c... Let a'., be an arc i n N and l e t i j 1 i J 1 13 13 u.. represent x^y^ i n H,(X) = ZZ . Take the boundary connected sum of a!. 13 1 J xj and u „ along an arc between them to o b t a i n a „ , and l e t S „ be the boundary connected sum of S~l". and S.. along a.., being c a r e f u l not to introduce t w i s t s xj xj x j ' ( so S „ and 2^ are u n l i n k e d ). F i n a l l y , a boundary connected sum of and S „ gives the r e q u i r e d t o r u s . Notice that and are s t i l l u n l i n k e d . The way to proceed should now be c l e a r . 11 37 S e c t i o n D: The Torres c o n d i t i o n s . As a c o r o l l a r y to the c h a r a c t e r i z a t i o n theorem we have Corollary 3: The order i d e a l i s generated by a polynomial of the Hj(.XJ form 1 i-i (*) A(x,y) = A(x,y) - (1-x) (1-y) B(x,y) where A(x,y) = A(x 1,y l ) , B(x,y) = B(x 1,y l ) , and A ( x , l ) = A j ( x ) , A ( l , y ) = A 2 ( y ) . A j ( x ) and A 2(y) are Alexander polynomials f o r the f i r s t and second components r e s p e c t i v e l y of the l i n k and 1 i s the l i n k i n g number of the two components. Proof: I f we c a l c u l a t e the determinant of the matrix given i n the c h a r a c t e r i z a t i o n theorem by expanding along the f i r s t row we get the polynomial (*) where A(x,y) = det4(x,y) and B(x,y) = det 0 S(x,y) B f r ' ^ y " 1 ) " 4(x,y) The symmetry c o n d i t i o n s on A(x,y) and B(x,y) f o l l o w from the Hermitian character of A(x,y) . 11 Corollary 4 ( Torres ): The Alexander polynomial A(x,y) of a two component l i n k s a t i s f i e s (Tl) A ( x , l ) = f^- A j ( x ) A ( l , y ) = A 2 ( y ) (T2) A(x,y) = x l _ 1 y Z " 1 A ( x " 1 , y " 1 ) U Because of c o r o l l a r y 4 i t i s n a t u r a l to ask i f (*) puts more r e s t r i c t i o n s on p o s s i b l e l i n k polynomials than j u s t ( Tl) and (T2). I t turns out that i t does not. Proposition 5: A polynomial i s i n the form (*) i f f i t s a t i s f i e s (Tl) and (T2) 38 Proof: "only i f " i s c o r o l l a r y 4. For the " i f " p a r t , l e t A ^ x ) and A (y) defined by (Tl) i f I $ 0 and l e t A ^ x ) = A (y) = 1 i f I = 0. (T2) be forc e s A j ( x ) = A ^ x l) and A 2 ( y ) = A ^ y - 1 ) . (Tl) insures that A ( l , l ) = Z A j d ) = ZA 2(1) so A (1) = A (1) = ±1. Then P(x,y) = A(x,y) - A ^ x ) A ^ A ^ l ) s a t i s f i e s P ( x , l ) = P ( l , y ) = 0 and consequently has the form P(x,y) = ( l - x ) ( l - y ) Q ( x , y ) that i s I A(x,y) = A 1 ( x ) A 2 ( y ) A 1 ( l ) + (1-x) (l-y)Q(x.y) where by (T2) we know that Q(x,y) = (xy) Q(x ,y ). I t i s easy to check that R ( x > y ) , J ^ a l i MH21 ( x y ) i - z . i - ( x v ) z - 1 i - ( x v ) I + 1 i - i 1-xy 1-xy 1-xy 1-xy v y j = 1 so A(x,y) = A 1 ( x ) A 2 ( y ) A 1 ( l ) + (1-x) (l-y)Q(x,y)R(x,y) I I = CA.(x)A,(y)A 1(l) + ( l - x X l - y ) 1 : ^ (xy) 1 - Z Q ( x , y ) } 1-xy 1 2 1 1-xy - ( l - x ) ( l - y ) 1 : ( x y ) Z " 1 { 1 : ( x y ) H l ( x y ) 1 - Z Q ( x , y ) } 1-xy 1-xy Let A(x,y) = A j ( x ) A £ ( y ) ^ ( 1 ) + ( 1 - x ) ( 1 - y ) ( X y J _ ^ y ( x y ) Q(x,y) and B(x,y) = ^ y ) 1 1 Q ( x y ) . T h e n 1-xy xy A ( x " 1 , y - 1 ) = A . ( x _ 1 ) A . ( y ' ^ A r i ) + (1-x" 1) (1-y' 1) ( X Y ) "^^"QCX"^"1) l - ( x y ) " 1 l-l -1 = A 1 ( x ) A 2 ( y ) A 1 ( l ) + ( x - l ) ( y - l ) ( x ^ (xy) 2"*Q(x,y) xy--( = A 1(x)>A 2(y)A 1(l) + ( x - l ) ( y - l ) x y _ W Q(x,y) 39 = A(x,y). The symmetry condition on B(x,y) is handled similarly. The other conditions are obvious. 11 40 Se c t i o n E: C h a r a c t e r i z i n g l i n k polynomials. In c o r o l l a r y 4 we c h a r a c t e r i z e d Alexander polynomials of two component l i n k s as those polynomials of the form (*) where A(x,y) and B(x,y) were as i n the proof of the c o r o l l a r y . U nfortunately t h i s does not a l l o w us to decide when a given polynomial i s a l i n k polynomial. We now examine t h i s question. Definition: An ordered p a i r ( A ( x , y ) , B(x,y) ) of elements of A^ i s allowable i f there i s a square matrix 4(x,y) and a row matrix B(x,y) both -1 -1 t r w i t h e n t r i e s i n A^ s a t i s f y i n g 4(x,y) = 4(x ,y ) and 4(1,1) = dia g ( + l , * so that A(x,y) = det4(x,y) 0 B(x,y) 4(x,y) We w i l l use t h i s n o t a t i o n ( p o s s i b l y w i t h s u b s c r i p t s ) i n the remainder of t h i s s e c t i o n . B(x,y) = det R , -1 - K t r #(x ,y ) ) The problem i s now to c h a r a c t e r i z e a l l o w a b l e p a i r s . I t i s c l e a r that we must have A(x,y) = A(x \ y A ( l , l ) = ±1, and B(x,y) = B(x *,y *) I do not know i f these c o n d i t i o n s are s u f f i c i e n t . In t h i s s e c t i o n we f i n d a method f o r generating enough a l l o w a b l e p a i r s to prove that the Torres c o n d i t i o n s c h a r a c t e r i z e two component l i n k polynomials provided the l i n k i n g number of the two components i s zero or both components are unknotted and the l i n k i n g number i s two. Notation: i f C. = I 0 (x ,y ) B ±(x,y) 4.(x,y) 1, 2. then 41 C\°2 0 . -1 - l . t r ^ 1 ( x ,y ) n , -1 - l . t r S 2 ( x ,y ) BAx,y) B 2(x,y) 0 ^ ( x . y ) 0 4 2 ( x , y ) Lemma 6: det C,*^ = + A 2 B i Proof:Suppose that AA*,y) i s nxn and 4 2 ( x , y ) i s mxm and l e t C^C^ ~ c ^ where i, j = 1, 2, ••• , n+m. Then det C *C_ = E sgnoII I! 1 +^c. 1 2 a 6 i=0 x a ( i ) Let X = { a | a leaves 0, 1, ••• , n f i x e d as a set }, Y = { o | a leaves 1, 2, ••• , n f i x e d as a set }, Z = { a | there i s an i where l£i<n so that o(i)>n }. XrtY. n o t i c e that XnY = { a I a(0) = 0 } and hence I T ^ c . = 0 i f o e 1 i=0 i a ( i ) I f l<i<n and a(i)>n then c. ,.s = 0 and so n. _c. ,.s = 0 i f a e Z. A l s o , = = i o ( i ) i=0 i a ( i ) every permutation i s i n at l e a s t one of the three s e t s . Hence det C *C_ = E „sgna irT^c. ... + E ,.sgna n ^ ^ c . .. . 1 2 oeX & i=0 i a ( i ) aeY 6 i=0 i a ( i ) I f a e X we can t h i n k of a = a^oa^ where i s a permutation of the set { 0 , 1, 2, ••• , n } and i s a permutation of { n+1, „m+n , n+m }. Then m+n Z a E X S g n a n ' i = 0 c i a ( i ) = E a 1 Z a 2 S g n a l S g n a 2 n i = 0 c i a 1 ( i ) n i = n + l C i a 2 ( i ) n „ T T H + T I - ^ a ^ l n i = 0 c i a 1 ( i ) ' Z a 2 S g n a 2 I I i = n + l c i a 2 ( i ) = A 2B r S i m i l a r l y S ^ g n a n ^ c ± a ( i ) = A 1 B 2 IF Theorem 7: ( 1, B(x,y) ) i s an allo w a b l e p a i r f o r any B(x,y) s a t i s f y i n g B(x,y) = B(x 1 , y V Proof ."Write B(x,y) = a + Y ,. . * , n . a . . ( x y + x y ) w i t h a s u i t a b l e UU L (,i, j )f (.1), U; i j r e s t r i c t i o n on the index set to i n s u r e that not both of ( i , j ) and ( - i , - j ) are present . Let 42 C. _ 1 ~ J x y -a 1 i J x y -a.. 1 0 • 0 0 -1 0 1 0 0 -1 J i f (i,j)?KO,0) c 00 = 0 1 U i j 0 1 1 -1) 0 1 U - i j 0 1 0 1 U - I J 0 1 .1 -1. - a 0 0 t i m e s i f a Q 0<0 a^g times i f aQQ >0 and even 0 0 0 -1 Then A..(x,y) = 1 and B i ; J(x,y) = \ a. . ( xN^ + x \ J ) aoo so lemma 6 gives det ..C.. = B(x,y) and so C ( 1 , 3 ) i j a l l o x a b l e p a i r ( 1, B(x,y) ). 11 a 0 0 + l , terms i f aQ>>0 and odd ( i , j ) ^ ( 0 , 0 ) ( i , j ) = ( 0 , 0 ) , . * . \ C . . r e a l i z e s the Remark: Looking at C we see that 4(x,y) = diag(±l,±l, ••• ,±1). Ret r a c i n g the steps i n the c h a r a c t e r i z a t i o n theorem we see that the surgery t o r i corresponding to C can be chosen to be unlinked from L'; hence the l i n k so obtained has both components unknotted. Corollary 8: 1. The Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials when the components have l i n k i n g number zero; i n f a c t any such polynomial can be r e a l i z e d by a l i n k w i t h two unknotted components. 2. The Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials when the components have l i n k i n g number two and both components are unknotted. Proof: 1. A polynomial s a t i s f i e s the Torres c o n d i t i o n s w i t h l i n k i n g number zero i f f i t i s of the form ( 1 - x ) ( l - y ) B ( x , y ) where B(x,y) = B(x \ y ^ ) . But f o r any such B(x,y) we know by theorem 7 that ( 1, B(x,y) ) i s all o w a b l e and the r e s u l t f o l l o w s from the above remark. 43 2. The Torres c o n d i t i o n s f o r l i n k i n g number two and both components unknotted f o r c e the polynomial to have the form ( 1+xy ) - ( 1 - x ) ( l - y ) B ( x , y ) where B(x,y) = B(x \ y . Again ( 1, B(x,y) ) i s a l l o w a b l e f o r any such B(x,y) and the c o n c l u s i o n f o l l o w s from the above remark. IF Remark: Levine 2 shows that the Torres c o n d i t i o n s c h a r a c t e r i z e l i n k polynomials when the l i n k i n g number of the two components i s one. This a l s o f o l l o w s from our r e s u l t s s i n c e i t i s c l e a r that ( A ( x , y ) , 0 ) i s a l l o w a b l e f o r any A(x,y) = A(x \.y *) and A ( l , l ) = ±1 ( t a k e 4(x,y) = A(x,y) and B(x,y) = 0 ). S i m i l a r l y Levine's theorem 2 i n the case of two component l i n k s f o l l o w s from our work by t a k i n g the d i r e c t sum of the given p r e s e n t a t i o n m a t r i x and the m a t r i x w i t h a s i n g l e entry as the new p r e s e n t a t i o n matrix. We have a l s o answered the question f o l l o w i n g theorem 2 a f f i r m a t i v e l y i n the case of two component l i n k s : the polynomial of a l i n k of two components wi t h s p l i t t a b l e l i n k i n g matrix ( which f o r l i n k s of two components simply means that the l i n k i n g number of the components i s zero ) may be m u l t i p l i e d by any polynomial s a t i s f y i n g P(x,y) = P(x \ y *) and the product r e a l i z e d as the Alexander polynomial of a l i n k . 44 CHAPTER IV: S e i f e r t surfaces. In t h i s chapter we explore the technique of S e i f e r t surfaces i n the study of l i n k polynomials. In s e c t i o n A we prove a g e n e r a l i z a t i o n of a r e s u l t of K i d w e l l which r e l a t e s the i n d i v i d u a l degrees of the Alexander polynomial to the l i n k i n g complexity. Section B turns to the s p e c i a l case of 2-bridge l i n k s , which have long been known to be p a r t i c u l a r l y s u i t a b l e f o r a n a l y s i s . In t h i s s e c t i o n we prove that 2-bridge l i n k s are interchangable and present an a l g o r i t h m f o r c a l c u l a t i n g the Alexander polynomial from a 2-bridge p r e s e n t a t i o n . As a c o r o l l a r y to t h i s we prove a conjecture of K i d w e l l i n the s p e c i a l case of 2-bridge l i n k s . The work i n s e c t i o n s C and D was done i n the hope of f i n d i n g a way to c h a r a c t e r i z e a l l o w a b l e p a i r s without working d i r e c t l y w i t h the matri x i n the c h a r a c t e r i z a t i o n theorem. Unfortunately the r e s u l t s found are too complicated to do t h i s but may be of use i n the f u t u r e machine c a l c u l a t i o n s of l i n k s . S e c t i o n C gives methods of generating l i n k polynomials from a l l o w a b l e p a i r s and s e c t i o n D uses these r e s u l t s as w e l l as others i n the t h e s i s to compile a l i s t of methods f o r generating a l l o w a b l e p a i r s without r e s o r t i n g to matrices. 45 Chapter IV. S e i f e r t surfaces. S e i f e r t surfaces have been used e x t e n s i v e l y to study the Alexander i n v a r i a n t s ( and, more g e n e r a l l y , the a b e l i a n i n v a r i a n t s ) of knots, but so f a r t h e i r use i n studying l i n k i n v a r i a n t s has been r e s t r i c t e d to boundary l i n k s ( that i s , l i n k s which bound d i s j o i n t S e i f e r t surfaces ) as i n G u t i e r r e z and reduced Alexander polynomials ( see Torres and K i d w e l l . ) The c o n s t r u c t i o n f o r knots can be i t e r a t e d to con s t r u c t the u n i v e r s a l a b e l i a n cover of a l i n k as f o l l o w s : l e t L = J l ^ U ^ U *"* U & be a l i n k . Then £^ bounds a S e i f e r t surface which can be used to con s t r u c t where 3 Xj = S -I ( see Ro l f s e n 2 f o r d e t a i l s . ) a l s o bounds a S e i f e r t surface 3 ~ i n S , say S^. Let S be the preimage of ^ " ^ l ^ n i " ^ e c a n c u t a ± o n S t h i s surface and assemble countably many copies of X^-S i n t o a covering space °f x j s o that a loop a i n X^-fc^ l i f t s to a loop i n X ^ i f f a - i S = 0 ( that i s , the a l g e b r a i c i n t e r s e c t i o n of a and S i s zero.) From t h i s i t 3 i s c l e a r that X ^ i s the u n i v e r s a l a b e l i a n cover of X 2 = S SL^.: I f t h i s c o n s t r u c t i o n i s continued i t e v e n t u a l l y y i e l d s the u n i v e r s a l a b e l i a n cover of X = S 3-L. Section A: A theorem of K i d w e l l . Definition: Let L = A^yJl^ be a l i n k and S be a S e i f e r t surface f o r £^ w i t h S and &2 i n general p o s i t i o n . I f Og = 2(genus of S) + the number of times i n t e r s e c t s S then cij = min^ i s the linking complexity of w i t h £^. The linking complexity of the link L i s the ordered p a i r ( a^, )• Remark: I f £ ^ i s unknotted and we i n s i s t that S be a d i s c then we get the d e f i n i t i o n of order as found i n K i d w e l l . In general the order i s greater than the l i n k i n g complexity; f o r example the n-th i t e r a t e d double of the Whitehead l i n k has order ( 2n, 2 n ) ( see K i d w e l l ) w h i l e i t i s c l e a r that 46 they have l i n k i n g complexity ( 2 , 2 ). Consequently the f o l l o w i n g theorem sharpens as w e l l as g e n e r a l i z e s the main r e s u l t i n K i d w e l l . F i r s t we need D e f i n i t i o n : " I f A(x,y) = E?_ p . ( y ) x 1 where p.(y) i s a polynomial i n y and i=n* l P (y) i- 0 / p (y) then deg A(x,y) = m-n. n m x Theorem 1 ( Kidwell ): I f A(x,y) i s the Alexander polynomial of a l i n k L = £^ij&2 w i t h l i n k i n g complexity ( a^, ) then - 1 > deg^A(x,y). 3 Proof: Let = S -J^s X^ i t s u n i v e r s a l a b e l i a n cover, and S the l i f t of S-SL^ t o X 2 w n e r e ^ i s a S e i f e r t surface f o r Jt^. I f Y = X^-S we can c o n s t r u c t X by i d e n t i f y i n g { Y.=Y } and { N.=5x(-l,l) }. F 7 7 a p p r o p r i a t e l y . From the Mayer - V i e t o r i s sequence ••• -> H^CjN.) t H ^ y Y . ) -> HjCX) -> HQCU N.) £ ^ ( y Y . ) -We o b t a i n the short exact sequence 0 -* cokerijj -»- H^X) -* kercfi -»• 0 By lemma 5 of Levine 2 we know th a t A(x,y) = A . -A. . so we now J J cokerij; ker<j> examine the l a t t e r two. 1. A , , . A p r e s e n t a t i o n f o r cokeriL can be obtained from one f o r I L O J Y . ) cokerijj c 1 w l by adding the r e l a t i o n s ijj>(a\) where { a A i s a set of generators f o r H^(tjN^) These r e l a t i o n s are of the form ^_(a\) - i|>+(a^)x= 0; s i n c e the r e l a t i o n s f o r H , ( i i Y . ) do not i n v o l v e x we see that deg A ' , < the number of generators 1 U I x cokeri); = of H ^ ( i j N ^ ) . TAAS-ly) i s a f r e e ZZ module on 2g + h generators where g = genus of S and h = number of times £^ i n t e r s e c t s S. Let t h i s set of generators be represented by c y c l e s { a^ ^ n S-l^viheve we can assume that l k ( a , H ) = I > 0 and l k ( a., ^ ) = 0 i f i ^ 1. When I $ 0 the l i f t s a^ of a^ f o r i > 2 w i l l generate H^(5) and H^(ylSL) w i l l be a f r e e module on 2g + h - 1 generators, g i v i n g ^ e g ^ A ^ ^ ^ - ai ~ • To f i n i s h 2 l-l the proof of the theorem we show that A, , = 1 + x + x +•••"+ x i f l>0 ker<f> and A. = 0 i f I = 0. kercj) 47 2. A^ e r^. Notice that S has I components i f I > 0 and i n f i n i t e l y many components i f I = 0. In either case, V u V = ( a ; ( y1-1 )a = o ) as a A module. It is easy to see that b = (y-l)a w i l l generate ker<|> so that 1- 1 I-1 we must have kertj> = ( b ; _ ^ b = 0 ). Hence A, , = , y . H 1-y ker<}> 1-y Corollary 2: If L i s a boundary link then A(x,y) = 0. Proof: Boundary links have I = 0 in the above proof. U -The proof of this theorem generalizes to links of more than two components. This is found in the appendix. 48 S e c t i o n B: Two-bridge l i n k s . We now turn our a t t e n t i o n to 2-bridge l i n k s and prove that order = l i n k i n g complexity = deg xA(x,y) f o r t h i s s p e c i a l c l a s s of l i n k s . We s t a r t by proving that = a^. Proposition 4: 1. 2-bridge l i n k s are interchangeable. More p r e c i s e l y , there 3 i s an i s o t o p y of S which interchanges the two components, p o s s i b l y a l s o r e v e r s i n g the o r i e n t a t i o n s of the two components. 3 2. There i s an is o t o p y of S which reverses the o r i e n t a t i o n s of the two components. Proof:Every 2-bridge l i n k can be put i n the form of f i g u r e 1. Figure 1. I f we remove B^ ( a 3 - b a l l s i t u a t e d as i n f i g u r e 1 ) then o 3 o i . A^ = L - Bj C. s -Bj = B^ i s a p a i r of d i s j o i n t t r i v i a l spanning arcs ( as i n f i g u r e 2 ) and i i . L can be e f f e c t i v e l y recovered by sewing a thickened d i s c t o a r e g u l a r neighbourhood of c i n f i g u r e 1, that i s L i s recovered by sewing a thickened Figure 2. disc to a regular neighbourhood of hCc^) in figure 2 where H:!^ i s the orientation preserving homeomorphism from figure 1 to figure 2 and h = HlBB^. Up to isotopy leaving 9A^ fixed h is a composite of the homeomorphisms which are described in figure 3. Figure 3. 50 .3 More p r e c i s e l y , i f 3B 2 i s the u n i t sphere i n TR , p o i n t s on 3B 2 are determined by the two angles ( 6, cf> ) i n s p h e r i c a l coordinates where £ <f> < |. I f 3A„ = { ( 0, ±} ) , ( TT, ±£ ) } then * 2 >• v - 4 v ^ h ( e, * ) = J ( 6+TT, <f> ) <j> > | ( e+2<j,+^ , <(, ) ( e, <j> ) cj, < ~ 3 h 2 i s obtained by r i g i d l y r o t a t i n g TR through TT about the l i n e 9 = 0, IT <)> = £, app l y i n g h^ and r o t a t i n g back again. We a l s o have a ZZ^B ZZ^ a c t i o n on ( B 2 > A 2 ); one generator, x^ ( r o t a t i o n through TT about a v e r t i c a l a x i s ) interchanges the two components, w h i l e the other x ( r o t a t i o n through TT about a h o r i z o n t a l a x i s ) reverses the o r i e n t a t i o n of the two components. ( On B , 9, cj> ) = ( 6+TT, <j> ) and x2( Q, cj> ) = ( -6, -<|> ).) Since \^oh^ = h ^ o i ^ and x^oh^ = h ^ o \ ^ o i 2 we see tha t i f C i s any set i n v a r i a n t under the ZZ^ $ ZZ 2 a c t i o n then so i s h^(C); by symmytry the same i s tru e of h 2 ( C ) . The proof i s completed by n o t i c i n g that c 2 i s i n v a r i a n t under the ZZ^ZZ^ a c t i o n . 11 I f you look at f i g u r e 1 you w i l l see tha t any 2-bridge l i n k can be b u i l t up i n stages as i s represented schematicly i n f i g u r e 4. We use t h i s o b servation to c a l c u l a t e the Alexander polynomial of a 2-bridge l i n k . Theorem 4: The Alexander polynomial of a 2-bridge l i n k i s the determinant of an mxm matrix ( where m i s the number of stages i n f i g u r e 4 ) w i t h diagonal e n t r i e s f n^x-lKy-l) n ± + 0 '-n^x-lMy-l) - (1+xy) n ± > 0 ai±(x,y) = n i ( x - l ) ( y - l ) - (x+y) n > 0 superdiagonal e n t r i e s a i , i - l ( x ' y ) = x y 51 0-th stage i - t h stage (m+1)st stage Stage T b e -*-in^ a t l e f t has ( see c o r o l l a r y 5 ) ° PjCx.y) = ( x - l ) ( y - l ) - (x+y) = 1 A 1(x,y) = ( x - l ) ( y - l ) - (x+y) = l-2x-2y+xy The l i n k 222 ( see Conway f o r n o t a t i o n ) has t h i s as l i n k polynomial. P 2(x,y) = ( x - l ) ( y - l ) e 2 = -1 A 2(x,y) = ( x - l ) ( y - l ) ( l - 2 x - 2 y + x y ) + xy The l i n k 221112 has t h i s as l i n k polynomial. P 3(x,y) = - ( x - l ) ( y - l ) - (xy+1) e 3 = 1 A 3(x,y) = ( - ( x - l ) ( y - l ) - ( x y + l ) ) ( ( x - l ) ( y - l ) ( l - 2 x - 2 y + x y ) + x y ) -xy(l-2x-2y+xy) The l i n k 221111122 has t h i s as l i n k polynomial. Figure 4. 52 subdiagonal e n t r i e s i - l , l 1 i f a '(l',D = "2 -1 i f a (l'.l) = 0 and a l l other e n t r i e s zero. Proof: Let the l i n k be L = l^^SL^ > X^ be the u n i v e r s a l a b e l i a n cover of 3 Xj = S - & j , and D be the l i f t of the d i s c spanned by i. as i n f i g u r e 4. We 3 construct the u n i v e r s a l a b e l i a n cover of X = S - L by c u t t i n g along D. Each stage i n f i g u r e 4 adds a generator to Hj(D) as a module. There are 16 cases to consider. The f i r s t four are contained i n f i g u r e 5; the corresponding r e l a t o r s are contained as the f i r s t four e n t r i e s of t a b l e 1. The r e l a t o r s are obtained from the diagram as f o l l o w s ( the n o t a t i o n i s the same as i n Rol f s e n 2, chapter 7, s e c t i o n B.) Case 1. x3. + n. (x- l ) 3 . - xg. , . b. * 3. + n. (x- l ) 3 . - 3. n. > 0 i i l l + l l l - l l i i I = V y 3 i - l + ^(x-lXy-l) - (x+y)}3 ± + x 3 ± + 1 Case 2. xg. + n. ( x - l ) 3 . - xg.,, *- b. % x3. , + n. (x-1)3 . - 3. n. > 0 i i I I + I . i . l - l l , l I I = R ±: x y B ^ + { n ± (x-1) (y-1) - (x+y)}3 ± + x g ± + 1 Case 3. 3. , + n . ( x - l ) g . - x g . j . *• b. $ n . ( x - l ) g . n. > 1 l - l I I i+I i i i l = V " B i - 1 + ^(x-DCy-DBi + x 3 i + 1 Case 4. xg. . + n.(x-1)3. - xg.,, ^ b. i n . ( x - 1 ) 3 . n. > 1 l - l l l i+I i i i l = R i : " x e i - l + n 1 ( x - l ) ( y - l ) 3 i + x 3 ± + 1 -The next four cases are obtained from the previous four by t a k i n g the m i r r o r image of the diagram and r e p l a c i n g x by x \ 3.. by -g , and by This gives r i s e to the next four e n t r i e s i n the t a b l e . The l a s t e i g h t cases are obtained from the f i r s t eight by r e v e r s i n g the o r i e n t a t i o n of D. This has the e f f e c t of r e p l a c i n g y by y ^ and accounts f o r the l a s t e i g h t e n t r i e s i n the t a b l e . We can now b u i l d a p r e s e n t a t i o n m a t r i x f o r (X). The entry i n the 55 i - t h row, j - t h column i s the c o e f f i c i e n t of B.. i n R^. We immediately see that a „ = 0 i f | i - j | > 1. Noti c e that case 1 can be preceded by cases 1, 2, 3, and 4 and can be followed by cases 1, 2^, 3^, and 4^.This plus the r e l e v a n t observations f o r the other cases allows us to conclude t h a t , around the i - t h diagonal entry, the p r e s e n t a t i o n matrix looks as i n t a b l e 2. By simultaneously m u l t i p l y i n g the rows and columns by x m y n as we move down the diagonal we can make every entry on the subdiagonal ±1 and every entry on the superdiagonal xy. Table 3 l i s t s t h i s set of p o s s i b i l i t i e s . But t h i s i s the form promised i n the theorem. 1T As a c o r o l l a r y to the proof we have Corollary 5: Let I be a l i n k as i n f i g u r e 4 where and hence D have been o r i e n t e d . Let P^Cx^) and e i = 1, 2, ••• , m be as i n f i g u r e 6 and i n d u c t i v e l y d e f i n e A 1(x,y) as f o l l o w s : A 1(x,y) = P 1(x,y) 2 1 A (x,y) = P 2(x,y)A (x,y) - e 2xy A 1(x,y) = P i ( x , y ) A 1 ^x.y) - e ^ x y A 1 - 2 ( x , y ) i > 2 Ao Then L has Alexander i n v a r i a n t isomorphic to ( A m(x,y) ) Proof: I t i s c l e a r that the Alexander i n v a r i a n t of a 2-bridge l i n k i s c y c l i c ; to see that the polynomial i s c o r r e c t , take the p r e s e n t a t i o n m a t r i x i n the above theorem and expand by the l a s t row and column to o b t a i n i t s determinant. IT K i d w e l l has conjectured that f o r a l t e r n a t i n g l i n k s the i n e q u a l i t y i n theorem 1 ( w i t h l i n k i n g complexity replaced by order ) i s an e q u a l i t y . We now prove t h i s f o r 2-bridge l i n k s . Definition: I f A( x , y ) = £™_ nP^(y) x l where p^(y) i s a polynomial i n y and P (y) ^ 0 ^ p (y) then min A ( X » Y ) = n a n (i m a x A(x,y) = m. Hence n m x x Figure 6 57 deg A(x,y) = max A(x,y) - min A(x,y). X X X Proposition 6: deg xA(x,y) = . - 1 f o r 2-bridge l i n k s . Proof: by i n d u c t i o n on the number of stages i n f i g u r e 4. I f f i g u r e 4 has m stages i t i s c l e a r that ct^ < m+l so by theorem 1 i t s u f f i c e s to show that deg A m(x,y) = m. I f m = 1 or 2 we see that min A m(x,y) = 0 and X X max xA m(x,y) = m; assume that t h i s i s t r u e f o r m < i - 1 . Then we have m i n x P i ( x , y ) A 1 ^(x.y) = 0 max^P^ (x,y)A"*" ^(x,y) = i min^xyA"*" 2 ( x , y ) = 1 max^xyA^ 2 ( x , y ) = i - 1 Hence min A ^ X j y ) = 0 and max A^x.y) = i and the r e s u l t i s proved. H X X 58 Case Relator 1 y B i - l + t n ^ x - D C y - l ) - (x+y)>3± + x 3 ± + 1 n > 0 2 x y e i - l + {n± (x-D (y-D - (x+y)}3± + x3 ±- + 1 n > 0 3 -6. + n . (x- l ) (y- l )B. +x3 . . , n. > 1 1-1 X X 1+1 X = 4 _ x p i - l + n i ( x - l ) ( y - l)3 1 + x B ± + 1 n > 1 *1 X y e i - 1 " ^ ( x - l H y - l ) + (l+xy)}31 + B i + 1 n > 0 21 y B i - l " tn (x-1)(y-1) + (l+xy)}3± + B n > 0 3. -x3. - n . (x- l ) (y- l )B . + p n. > 1 1 1-1 1 x i+I l = 41 " B i - 1 " n ± ( x - l ) ( y - l ) B i + 3 ± + 1 n. > 1 1. 3 ±_ 1 - {n (x-1)(y-1) + (l+xy)}3± + x y B ± + 1 n > 0 2 1 x B ^ j - {n 1 (x-l)(y-l) + (l+xy)}3± + x y B ± + 1 n ± > 0 3 1 - y 8 i _ i " n 1 ( x - l ) (y - l )3 ± + x y B . + 1 n. > 1 4 -xyB. . - n.(x-1)(y-l)B. + xyB _ n. > 1 l - l x J 1 i+I i = *1 x 6 i - l + tn 1 (x- l ) (y- l ) - (x+y)}3± + y B i + 1 n > 0 l\ B i _ 1 + in.(x-1)(y-1) - (x+y)}Bi + y B ± + 1 n > 0 3} " x y B i - l + n i ( x - l ) ( y - l ) 3 i + y B i + 1 n ± > 1 * \ " y B i - l + n i ( x - l ) ( y - l ) 3 ± + y B i + 1 n ± > 1 Table 1. Table 2 is on the next page. 59 r * Case 1 x y n.(x-l)(y-l)-(x+y) 0 1 Case 3 -1 0 xy n.(x-1)(y-1) Case 2 xy n^x-1) (y-l)-(x+y) Case 4 -x n ±(x-1)(y-1) ±y Case 1, xy 0 0 1 -n1(x-l)(y-l)-(l+xy) 1 ±xy Case 3, -n.(x-1)(y-1) txy x 01 Case 2, | y -n.(x-l)(y-l)-(l+xy) 1 0 ±xy * Case 4 xy -1 -n.(x-l)(y-l) 0 ±xy xy Case T| 1 -n (x-1)(y-l)-(l+xy) xy 0 ±1 * Case 3 -y o -n.(x-1)(y-1) ±1 Case 2^ 1 x -n.(x-1)(y-l)-(l+xy) xy ±1 Case 4 -xy -n i(x-1)(y-1) 0 ±1 Case 1 1 x (x-1)(y-l)-(x+y) y 0 ±x * r * Case 3, -xy n.(x-l)(y-l) ±x Case 2 T * 1 0 xy n1(x-l)(y-l)-(x+y) ±x 0 1 y * r * Case 4 -y n^x-lKy-l) 0 ±x Table 2. Cases 1, 2, l | , 2^ xy 1 ^(x-lXy-D-Cx+y) 0 ±1 0 xy Cases 1^, 2^, l \ 2^ xy 0 1 1 -iuCx-1) (y-l)-(l+xy) xy 0 ±1 * 1 1 Cases 3, 4, 3^, 4^ xy -1 n.(x-l)(y-l) 0 ±1 0 1 xy Cases 3^, 4^, 3^, 4^" xy -1 -n^x-DCy-l) xy 0 ±1 * Table 3 Section C: Generating l i n k polynomials from a l l o w a b l e p a i r s . Theorem 7: I f ( A(x,y ) , B(x,y) ) i s an a l l o w a b l e p a i r and the polymomials A 1(x,y) are as i n c o r o l l a r y 5, then A 1(x,y)A(x,y) ± ( 1 - x ) ( 1 - y ) A 1 " 1 ( x , y ) B ( x , y ) and A 1(x,y)A(x,y) ± ( 1 - x " 1 ) ( l - y " 1 ) A 1 + 1 ( x , y ) B ( x , y ) are l i n k polynomials. Proof: The a l l o w a b l e p a i r i s generated by a p a i r of matrices 4(x,y) and B(x,y) which i n t u r n determine a set of surgery i n s t r u c t i o n s f o r a l i n k . The r e l e v a n t o b servation i s that the c y c l e i n dB i s r e s t r i c t e d only by U Q = 0 e H^(Y) and has no s e l f i n t e r s e c t i o n s . Hence can generate any 2-bridge l i n k . The c y c l e U Q i n 9$ determinesthe r e l a t o r and i t i s c l e a r that the c o e f f i c i e n t of ct^ i n RQ may be taken to be the Alexander polynomial of any 2-bridge l i n k , say A 1(x,y) ( the c o e f f i c i e n t of i n R^ i s the Alexander polynomial of the l i n k obtained by i g n o r i n g a l l the surgery t o r i . ) I f T i s as i n f i g u r e 7 then the c o e f f i c i e n t which m u l t i p l i e s B(x,y) to complete the p r e s e n t a t i o n m a t r i x i s 1. £ i , j l k ( v xiyJT )x±yj The lower h a l f of f i g u r e 7 i s j u s t a rearranging of the top h a l f so that B i s emphasized. H^(B) i s a f r e e module on the generator g which i s dual to the c y c l e b which s t a r t s at runs through the g r i d l i n k i n g (3 once and continues to +°°; hence T = ( l - x ) ( l - y ) b . I f we sew the d i s c to U Q i n B ( r a t h e r than Y ) the r e s u l t i n g polynomial i s 2. I . } . l k ( y Q , xVb )xV This i s the Alexander polynomial of l i n k obtained by sewing a d i s c to B along u^. Figure 8 shows the r e l a t i o n s h i p between and L^; we see that i f a = 0 then ^ has one fewer stage than L so we l e t the Alexander L—> o V Figure 7 63 polynomial of L^ be A ^ + 1 ( x , y ) , w h i l e i n the other cases L^ has one more stage than L so we l e t the Alexander polynomial of L^ be A'*" ^ ( x , y ) . Comparing 1., 2. and knowing that T = ( l - x ) ( l - y ) b we see that the r e q u i r e d c o e f f i c i e n t i s ( l - x ) ( i - y ) i i " 1 ( x , y ) or • ( l - x ~ 1 ) ( l - y " " 1 ) A i + 1 ( x , y ) . S t r i c t l y speaking we have determined A^(x,y) f o r j = i - 1 , i , i+1 only up to a f a c t o r of i x y . I f we i n s i s t that X V A ^ X ,y ) = AJ(x,y) ( which we r e q u i r e by the Torres c o n d i t i o n s ) we have determined them up to a f a c t o r of ±1. That t h i s i s unimportant i s seen i n lemma 8 which f o l l o w s . The p r e s e n t a t i o n m a t r i x i s A1 (x,y) ±(1-x) (1-y) A±_ 1 (x,y)B(x,y) or f A i(x,y) ±(l-x ] ; ) ( l - y 1 ) A 1 + 1 ( x , y ) B ( x , y ) 4(x,y) B , -1 - l s t r B(x ,y ) whose determinants are those promised by the theorem. 11 Lemma 8: I f ( A(x,y ) , B(x,y) ) i s an a l l o w a b l e . p a i r , then so i s ( A(x,y),-B(x,y) ) Proof: Let 4(x,y) and B(x,y) be the matrices which generate the al l o w a b l e p a i r . We can assume that 4(x,y) i s nxn where n i s even ( otherwise r e p l a c e i t by 4(x,y)$(;l] .) Then det(-4(x,y)) = A(x,y) and 0 -B(x,y) -4(x,y) p/ -1 - l x t r -B(x ,y ) = -B(x,y) 11 Remark: This lemma can be r e a l i z e d g e o m e t r i c a l l y by r e v e r s i n g the o r i e n t a t i o n 3 of S . The f o l l o w i n g g e n e r a l i z e s theorem 7. Theorem 9: Let ( A^(x , y ) , B^.('x,y) ) be allowable p a i r s f o r i = 1, 2. Let L and L are the same below the line Figure 8. on the boundary Figure 9. 65 I l-l A i ( x ' Y ) = JrSy1 A i ( x ' y ) - (1-x)(1-y)IlSy1 V X' y ) i = 1, 2 Then A* + 1(x,y)A2(x,y) - A*(x,y)A^ 4" 1 (x,y) i s a l i n k polynomial. Proof: F i g u r e 9 shows the c o n s t r u c t i o n which r e a l i z e s the promised polynomial. Let (x,y) and B_^(x,y) be the ma t r i c e s , the t a n g l e s , the surgery complements, and t h e i r covers as i n the c h a r a c t e r i z a t i o n theorem which correspond to the allo w a b l e p a i r ( A^(x , y ) , B^(x,y) ) f o r i = 1, 2. A p r e s e n t a t i o n m a t r i x f o r the constructed l i n k can be c a l c u l a t e d from the Mayer - V i e t o r i s sequence f o r Z = Z UZ ; i t w i l l have a l l the r e l a t o r s corresponding to the s u r g e r i e s i n Z^ and Z^ as w e l l as the r e l a t o r s obtained when 3Z^ i s i d e n t i f i e d w i t h SZ^. H^(3Z\) has two generators: the l i f t s of u. and v ( see f i g u r e 10 ) 0 o Figure 10. i 1> I f RQ' i s the r e l a t o r obtained by sewing a d i s c across 3Z^ to produce a ( 2£, 2 ) torus l i n k as i n the c h a r a c t e r i z a t i o n theorem, then the r e l a t i o n s obtained by i d e n t i f y i n g 3Z. w i t h 8Z 0 are .*}.'l+1=R**k¥l and R\ ' Z=R 2' k. In J & 1 2 0 0 0 0 short, a p r e s e n t a t i o n m a t r i x i s ^ ( x . y ) B ^ x " 1 ^ " 1 ) ^ 0 0 P Z ( x , y ) V x , y ) I i 5 r - P,(x,y)B 2(x,y) P l_ 1(x,y)B 1(x,y) P ^ ( x , y ) ^ ( x . y ) 0 0 B 2 ( x " 1 , y ~ 1 ) t r 4 2(x,y) where P^x.y) = - 1 ^ X y ^ (1-x) (1-y). I f ^ 2 ( x , y ) i s mxm then I have m u l t i p l i e d I1 -xy the l a s t m+1 columns and the l a s t m rows by -1. We want to c a l c u l a t e the determinant of t h i s m a trix. As u s u a l , l e t the entry i n the i - t h row, j - t h column be a.. so the re q u i r e d determinant i s Y n.a. ... sgn a Lo l i a ( i ) I f 4 1(x,y) i s ( n - l ) x ( n - l ) l e t T 1 = { a : a{l,2,---,n} = {l,2,---,n} } T 2 = { a : a { l , 2 , " ' , n - l , n + l } =' {1,2, • • • ,n} } = { a : e i t h e r there i s a j<n-l so that a(j)>n „ or there i s a j>n+l so that a(j)<n } Then T^, T^, and are p a i r w i s e d i s j o i n t and exhaust a l l permutations. A l s o , i f a E r „ then I I .a. ...sgn a = 0. I f a e r, then i t i s of the form a.a_ 3 x i a ( x ) & 1 1 2 where i s a permutation of { 1, 2, • • • , n } and o"2 i s a permutation of { n+1, n+2, ••• , n+m+1 } and hence 7 _ n.a. ... = 7 n.a. ...sgn o,) n.a. ...sgn a„ LoeYl i i a ( i ) La^ x io^ U ) B lLa2 I i a 2 ( i ) > 6 2 = A^ + 1 ( x , y ) A 2 . ( x , y ) . 67 Similarly ^ £ r 2 n i a i a ( i ) = "4(x'y)A2+1(x'y) • 11 There i s another possibility for the construction in this theorem, namely that suggested by figure 11. If this is carried out we see that the resulting polynomial i s A* + 1(x,y)A 2(x,y) - A*(x,y)A 2 ^x.y). I believe that the results in this and the preceeding section w i l l be useful i n the machine calculation of Alexander polynomials of links of two components. 2. glue on 9B. Figure 11. 68 Sectio n D: Generating a l l o w a b l e p a i r s . The work i n t h i s chapter was done because to Was found that the pr e s e n t a t i o n matrix i n the c h a r a c t e r i z a t i o n theorem was too d i f f i c u l t to work w i t h ; i t was hoped that a method would be found which would generate enough all o w a b l e p a i r s to show that the r e s t r i c t i o n s pointed out i n chapter I I I s e c t i o n E were a l s o s u f f i c i e n t f o r a p a i r to be al l o w a b l e . Unfortunately the methods discovered seem too complicated to a l l o w t h i s . For the sake of completeness we here present a l i s t of methods f o r generating a l l o w a b l e p a i r s . 1. ( A^~^(x,y), Q ^ X W J _ \ ) 1 S a l l o w a b l e where A^(x,y) i s as i n c o r o l l a r y 5 and A ^ l , ! ) = 0. 2. ( 1, B(x,y) ) i s all o w a b l e whenever B(x,y) = B(x *,y ( theorem III.E.7 ) 3. ( A(x,y ) , P(x,y)P(x \ y ) i s al l o w a b l e whenever A(x,y) = A(x \ y *) and A ( l , l ) = 1. The f o l l o w i n g are methods f o r generating a l l o w a b l e p a i r s from known all o w a b l e p a i r s . 4. ( A(x,y),-B(x,y) ) i s al l o w a b l e i f ( A( x , y ) , B(x,y) ) i s . 5. ( A 1 ( x , y ) A 2 ( x , y ) , (x,y)B 2(x,y) + A 2 ( x , y ) B 1 ( x , y ) ) i s all o w a b l e ( lemma III . E . 6 ) 6. Theorem 9 and the remark a f t e r i t give four ways of generating a l l o w a b l e p a i r s : ( A°(x,y)A2(x,y) - A ~ ^ ( x , y ) A 2 ( x , y ) , -2 0 1 1 A ( A 1 ( x , y ) A 2 ( x , y ) - A (x,y)A ( x , y ) , .0, ,.0, s - 1 , s -1 A (x,y)A 2(x,y) - A^(x,y)A*(x,y) (x-1)(y-1) (x,y)A 2(x,y) - A°(x,y)A2(x,y) (x-1)(y-1) (x,y)A°(x,y) - A°(x,y)A^(x,y) ( A x(x,y) A 2(x,y) - ^ (x,y)A 2 (x,y) , (x-1) (y-1) ) 69 2 0 1 - 1 AJ(x,y)A2(x,y) - A°(x,y)A^(x.y) ( A 1 ( x , y ) A 2 ( x , y ) - A 1 ( x , y ) A 2 ( x , y ) , (x-1) (y-1) ) For anyone who may want to t r y h i s hand at c h a r a c t e r i z i n g a l l o w a b l e p a i r s , ( x-l+x \ 2 ) i s the simplest p a i r which I have not been able to show i s al l o w a b l e . A l s o , K i d w e l l has a f a m i l y of polynomials which he proves cannot be r e a l i z e d by l i n k s of l i n k i n g number 3, order ( 3 , 3 ). A f a m i l y of p a i r s which r e a l i z e s these polynomials i s ( 1 - nx~ 1y" 1(l-x) 2(l-y) 2» 1 " n ( l - x ) ( l - y ) ( l + x " 1 y ~ 1 ) ) n > 0 I t would be i n t e r e s t i n g to know i f any of these are al l o w a b l e ( I have worked on the case n = 1.) 70 B i b l i o g r a p h y Alexander, J.W., " T o p o l o g i c a l i n v a r i a n t of knots and l i n k s , " Trans. A.M.S., v o l . 30, 1928, pp. 275-306. B a i l e y - R o l f s e n , "An unexpected surgery c o n s t r u c t i o n of a lens space" to appear, Pac. J . Math. Birman, J.S., Braids, links, and mapping class groups. P r i n c e t o n U n i v e r s i t y Press, P r i n c e t o n , 1975. Conway, J.H., "An enumeration ,of knots and l i n k s , and some of t h e i r a l g e b r a i c p r o p e r t i e s , " i n Computational problems in abstract algebra, John Leech, ed., Pergamon Press, Oxford, 1970, pp. 329-358. Fox, R.H., "A quick t r i p through knot theory," i n Topology of 3-manifolds and related topics, (Proceedings of the Univ. of Ga. I n s t . , 1961) M.K. F o r t , ed., P r e n t i c e H a l l , Englewood C l i f f s , N.J., New York, 1962. Goldsmith, D. "Symmetric f i b r e d l i n k s " i n Knots, Groups, and 3-manifolds, L.P. Neuwirth, ed., P r i n c e t o n U n i v e r s i t y P ress, 1975, pp. 3-23. G u t i e r r e z , M.A., "Polynomial i n v a r i a n t s of boundary l i n k s , " R e v i s t a Columbiana de Matematicas V I I I , 1974, pp. 97-109. Hempel, J . , "Construction of o r i e n t a b l e 3-manifolds," i n Topology of 3-manifolds and related topics, M.K. F o r t , ed., New York, 1962, pp. 207-212. K i d w e l l , M.E., Orders of links and the Alexander polynomial, t h e s i s , Yale U n i v e r s i t y , 1976. Levine, J . , "A c h a r a c t e r i z a t i o n of knot polynomials," Topology v o l . 4, pp. 135-141. 2 "A method f o r generating l i n k polynomials," Am. J . of Math., 89, 1976, pp. 69-84. L i c k o r i s h , W.B.R., "Surgery on knots," p r e p r i n t . Massey, W., Algebraic topology an introduction. Harcourt, Brace & World, Inc. 1967. R o l f s e n , D., 1. " L o c a l i z e d Alexander i n v a r i a n t s and isotopy of l i n k s , " Annals of Mathematics, v o l . 101, 1975, pp. 1-19. 2 Knots and links. P u b l i s h or P e r i s h , Inc., Berkeley, 1976. 3 "A s u r g i c a l view of Alexander's polynomial," i n Geometric Topology, (Proceedings of the Geometric Topology Conference 71 he l d at Park C i t y , Utah, 1974 ) A. Dold and B. Eckmann, ed., Springer-Verlag, New York, 1975. Schubert, H., " Knotten und V o l l r i n g e , " Acta Mathematica, v o l . 90, 1953, pp. 131 - 286. S e i f e r t , H., " Uber das Geschlecht von Knoten," Mathematische Annalen, v o l . 110, 1934 - 35, pp. 571 - 592. Shaneson, J . , " On c h a r a c t e r i z i n g l i n k polynomials f o r two components," unpublished manuscript, March, 1971. Torres, G., " On the Alexander polynomial," Annals of Math., v o l . 57, 1953, pp 5 7 - 8 9 . Zassenhaus, H., The theory of groups. Chelsea P u b l i s h i n g Co., New York, 1949. 72 Appendix. The general case of-Kidwell's theorem. Since we have already introduced most of the concepts needed to generalize theorem IV.A.1 to more than two components i t seems worthwhile to show how to modify the proof to obtain the more general result. D e f i n i t i o n : Let L = £^ y u ^ a link and S be a Seifert surface for £^ with S and J^U^U "*" U ^ u i n general porition. If a c = 2(genus of S) + the number of times £„y £.y ... u £ intersects S then a 1 = min a is the linking complexity of £„ y £„ y *•* y£ with £, . The linking complexity of the link L is the ordered y-tuple (a^o^»• " • ,a ) . D e f i n i t i o n : If A(x^,X2, * * • ,x^) = I ™ = n P X x 2 ' X 3 ' ' " ' > xy) xj where the coefficients of the x^'s are polynomials in the other variables and p (x„,#**,x ) £ 0 1 n 2 y p (x„,---,x ) ^ 0 then deg,A(x,,x ,•••,x ) = m-n. m I y 1 1 / y Theorem ( Kidwell ): If A(x^,X2,•••,x^) i s the Alexander polynomial of a link L = £ ^ y £ 2 y ••• y £ y with linking complexity ( a^, a^, ••• , ) then a l ~ 1 = deg 1A(x 1,x 2,•••,x^). 3 Proof: Let X„„ = S - ( £„ M £ „ M • • • M £ ), X_„ be i t s universal 23-'-y 2KJ 3 U u y 23-'-y abelian cover, and S the l i f t of S - ( £ „ i i £ „ i i • • • M £ ) to X where 2 3 V Zj*••y S i s a Seifert surface for £,. If Y = X„„ - S we can construct X by 1 23*•*y identifying { Y. = Y }. •_ and { N. = Sx(-l,l) }. _ appropriately. From the Mayer - Vietoris sequence ••• -y H^yN.) t H ^ Y . ) -> H^X) -> H^yN.) i H 0(yY.) We obtain the short exact sequence 0 cokerijj -* H^X) -> kercj) -> 0 As before A(x,,x„,•••,x ) = A -A . 1 2 y cokerijj ker<J> 1. A , ,. A presentation for cokerib can be obtained from one for H , ( I I Y . ) cokerijj 1 v x by adding the relations ty(a/) where { a, } i s a set of generators for H^(yN^) These relations are of the form ty (a^) "~ ^ + ( a ^ ) x = 0; since the relations for , ^ , ,. < the number of generators 1 cokerijj = H^(yY^) do not i n v o l v e x^ we see that deg^A of H^(ijlSL) as a A y module. (S-iSL^ U 3^ U • • • U ^,,)) ^ s a f r e e ^ module on 2g+h generators where g = genus of 5 and h = number of times IJ • • • \j I z y i n t e r s e c t s S. Let t h i s set of generators be represented by c y c l e s { a^ i n ^ - ( ^ u " " " U • Tf we form, a m a t r i x of l i n k i n g numbers where the entry i n the i - t h row j - t h column i s lkCa^jJL) we can assume that the m a t r i x has the f o l l o w i n g form: 0 • • 0 * . * . 0 • • 0 0 • • 0 * . * . 0 • • 0 0 • • 0 0 • • 0 •k • • * 0 • 0 0 • • 0 0 • • 0 0 • • 0 This m a t r i x of l i n k i n g numbers determines the l i f t s of the a^' s to 5; the l i f t of the loop a^ to S i s a path from the basepoint to the basepoint s h i f t e d by b i l b i 2 X l X 2 " - \ b. i y I f t h i s m a t r i x c o n s i s t s of zeroes then a l l the a.'s l i f t to c y c l e s and we see that deg,A , , < a,: as before we w i l l show that i n t h i s & 1 cokerij; = 1 case A, , = 0. I f there i s at l e a s t one non-zero entry i n the m a t r i x then kercb there i s at l e a s t one of these c y c l e s which l i f t s to a path which i s not a loop. We can f u r t h e r assume that such paths do not share both endpoints, s i n c e t h i s would imply that two rows of the m a t r i x were the same ;' we can m u l t i p l y one of them by -1 i f need be and add i t to the other to o b t a i n a row of zeroes which we move to the bottom of the matrix. I f there i s only one non-zero row we see immediately that H^((jN^) w i l l have 2g+h-l generators. I f there i s n > 1 non-zero rows then we must i n c l u d e commutators i n the l i f t s of the generators corresponding to these rows, and hence i n t h i s case we have 2g+h-n+C2 generators f o r H^(yN.). As i n theorem II.C.4 these are not independent; i n f a c t the techniques of lemma II.C.5 can be used to show that 74 at l e a s t 1 must b e omitted.to obtain a l i n e a r l y independent set. Hence deg1A(x1,x2,---,xtj[) < 2g+h-n+C2-c!J"1 = 2g+h-l. 2. A, . . I t i s c l e a r that A, , cannot involve x, :' H _ ( i i N . ) i s c y c l i c as a kercb ker<j> 1 0 u i J Ay module and the r e l a t o r s do not involve x^. Kercb i s generated b y the set (1.) b 2 = ( x 2 - l ) a , b 3 = (x3-.l)a, • • • , b y = (x -l)a where a i s the generator of H Q ( I J N ^ ) with r e l a t o r s (2.) ( x . - D b . - ( x . - l ) b . 2<i<j<u i j j i = J = and when we rewrite the r e l a t o r s f o r H Q ( ( J N ) to obtain r e l a t o r s for kercb we can avoid using x^. In case the matrix i n the previous section consists e n t i r e l y of zeroes, each component of 5 covers S-^l^ ij & 3 I J • * • I J H ) homeomorphically and consequently H 0 ( u N . ) = ( a ; - ) as a Ay module. Then kercb has a presentation with generators (1.) and rel a t o r s (2.). Again i t i s c l e a r that the r e l a t o r s are not independent and the techniques of lemma II.C.5 can b e used to show that ^ must be omitted before the set can b e independent. Since we started with 1 t h i s leaves u-2 r e l a t o r s . Since we have u-1 generators t h i s proves that =0. IT
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Alexander invariants of links Bailey, James Leonard 1977
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Title | Alexander invariants of links |
Creator |
Bailey, James Leonard |
Publisher | University of British Columbia |
Date Issued | 1977 |
Description | In the three main sections of this thesis (chapters II, III, and IV; chapter I consists of definitions) we explore three methods of studying Alexander polynomials of links which are alternatives to Fox' free differential calculus. In chapter II we work directly with a presentation of the link group and show how to obtain a presentation for the Alexander invariant. From this we deduce that the order ideal of the Alexander invariant is principal for links of two or three components (the case of one component is well known) but nonprincipal in general for links of four or more components. In any event we show that only one determinant is needed to obtain the Alexander polynomial. In chapter III we use surgery techniques to characterize Alexander invariants of links of two components in terms of their presentation matrices. We then use this to show that the Torres conditions characterize link polynomials when the linking number of the two components is zero or both components are unknotted and the linking number is two. Chapter IV uses Seifert surfaces to prove a generalization of a theorem of Kidwell which relates the individual degrees of the Alexander polynomial to the linking complexity, to present an algorithm for calculating the Alexander polynomial of a two-bridge link from a two-bridge diagram and to prove a conjecture of Kidwell in the special case of two-bridge links. These results are then used to generate link polynomials from allowable pairs (a concept introduced in chapter III) and these results in turn are used to produce methods of generating allowable pairs. |
Subject |
Links and link-motion Polynomials Invariants |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2010-02-19 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0079539 |
URI | http://hdl.handle.net/2429/20523 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
AggregatedSourceRepository | DSpace |
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