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Alexander invariants of links Bailey, James Leonard 1977

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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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