Math. Ann, 298. 327-348 (1994) Malhematlsche Annalen �9 Springer-Verlag 1994

SO3(R)-representation curves for two-bridge knot groups

Michael Heusener* Fachbereich Mathematik, Universit/it Siegen, Holderlinstrasse 3, D-57068 Siegen, Germany

Received May 18, 1993

1980 Mathematics Subject Classification (1985 Revision): 57M25, 57M05

Given a 2-bridge knot or link group the corresponding SO3(IR) representation space may be described by a real plane algebraic curve (see [3]). These real algebraic curves are closely related to the nab-rep curves investigated by Riley [14]. The study of the geometry of the representation curves arise in its first step the question of reducibility. Contributions to this question have been made in [3], [ 15] and [131 .

It was a conjecture of Riley that the representation curve of a 2-bridge knot group has extra algebraic components if the knot admits an unusually large group of symmetries which preserve both space and knot orientation. The next question which one might ask is what is the genus of the representation curve. It was a conjecture of Burde [3] that the representation curves splits always into rational components.

The aim of this paper is to give an affirmative answer to Riley's conjecture (see also [13]) and to present an example of an irreducible non-rational representation curve which disproves Burde's conjecture.

In fact, we obtain both results by an analysis of the symmetries of represent- ation curves. The keynote is that the representation curve which is associated to a normal form of the 2-bridge knot is not determined by the knot type - it depends also on the choice of the normal form (see [3]). However, we expect that the representation curves which are associated to different normal forms of the same knot are related in some sense. This is made precise in Corollary 2.6 where we prove that the representation curves associated to different normal forms of the same knot are birational equivalent. We use this birational equivalence to prove Riley's conjecture (Theorem 3.1) and to present a represent- ation curve of high genus (Corollary 4.9).

The last section connects (in special cases) Casson's invariant and the signature of knots with the complexification of the representation curves.

* Supported in part by "Graduiertenf6rderung des Landes Hessen"

328 M. Heusener

1 Introduction

Given two groups G and H, a representation of G into H is a homomorphism 0: G--* H. Two representations CO and co' are equivalent iff they differ by an inner automorphism of H. Given a knot group G, we call a representation co: G--+ H abelian if its image is abelian - then it must be cyclic.

2-bridge knots and links - sometimes called 4-plats (Viergeflechte) were first investigated by Bankwitz and Schumann [2] where they are shown to be alternat- ing and invertible. A 2-bridge knot or link ~ ~ S 3 may be described by two numbers ct, fl~Tl where ~ > 0, - :t < fl < ~,gcd(ct, fl) = 1 and fl = 1 rood 2. We denote the two bridge knot or link t determined by the numbers (ct, fl) by b(ct, fl). b(~, fl) is a knot (link) iff ct is odd (even). For definitions, details and more information see [4, Chap. 12].

2-bridge knots are classified by their twofold branched covering - a method due to Seifert. It is easy to prove (see [4, Chap. 12]) that the twofold branched covering space of b(e, fl) c S 3 is the lens spaces L(~, fl). Therefore, we obtain a classification ofunoriented 2-bridge knots and links by the classification of lens spaces. However, the general result is due to Schubert:

1.1 Theorem (H. Schubert) (a) b(a, fl) and b(c(, fl') are equivalent as oriented knots (or links), i f and only i f

= ct' and fl + =- fi' m o d 2~ .

(b) b(a, fl) and b(~', fl') are equivalent as unoriented knots (or links), i f and only i f

= ~ ' and fl + - fl' m o d a .

For the proof of (a) we refer to [17]. The weaker part (b) follows from the classification of lens spaces.

Given b(c~, f l ) c S 3 we denote the fundamental group of its complement by G(~, f l ) . Using the normal form of the 2-bridge knots or links b(r fl) we can get a Wirt inger presentation:

G(~, f l ) = (S, T I L s S = T L s ) , L s = S ~ ' T ~ 2 . . . S . . . . T ~~ ' ifc~ is odd

and

G(ct, f l ) = (S , T I L s T = T L s ) , L s = S " T ~ 2 . . . S ~" ~T . . . . i f cc i seven

where el = ( - 1) [i~~ (for every real x let [x] be the greatest integer n such that n ~ x ) .

We denote by S 3 the unit quaternions

(P, ~0) = cos(�89 (p) + sin(�89 q)) P

where S 2 is the 2-sphere of pure unit quaternions defined by p2 = _ 1. There is a twofold covering 6 : S a --+ SO3(~) = IRP 3, (P, (p) ~ 6(P, (p) which is a group homo- morphism with Kernel(6) = { +_ 1}. 6(P, ~o) is a rotation of angle r with axis P.

1.2. R e m a r k It is usual to identify the unit quaternions with the group

} SU2(II;) of special unitary matrices, SU2((E)= _ ~ a~i + bb = 1 . The

isomorphism ~ : S 3 ~ SU2(I~) is given by g* : ao + a~. i + a2 "j +

SO3(R)-representation curves for two-bridgc knot groups 329

~___~( aoWi'al a2+i" ~3~. Let G = G(cr f l )be given. We would like a3.k to \

- - a2 q- i'a3 ao -- i.aa J consider non-abelian representations O: G ~ SO3(IR)

~ S ~ ( P , ~o) (1) 0 : [ T~--~cS(Q, ~o) (P * Q)

which factor through SU2(~) and which assign the same angle ~o to S and 7".

1.3 Remark Every representation 0: G-~ SO3(IR) factors through SU2(~) if ~ is odd (see [3]). Moreover, S and T are conjugate in G(~, fl) (~ is odd) and so S~--~6(P, ~o) and T~--~6(Q, ~p) holds for every representation 0: G ~ SO3(~,). The equivalence class of O, given by (1), is determined by the parameters z =

~o (P, Q ) = c o s G ~b = .~(P, Q) and y = cot 2 ~-. Here (P, Q) denotes the scalar

product in IR 3 (for details see [3]}. There is a restriction of the parameters and we denote by D ~_ IR z the domain

o = {~,y)ly_>_ 0, l < r < l } .

1.4 Theorem Given G = G(0c, fl), there exists a polynomial G.a( r ,y )~E[z ,y] ,

d e g G . e = r - [ ~ - 1[, q which only depends on ~ and fl, such that a pair (ro, Yo)eD L 2 J

determines an equivalence class of S03(lR)-representations, given by (1), !f and only !f z~,~tro, yo) = O.

Proof See [3]. []

1.5 Definition We call the affine real algebraic set

'r := ~(G(~, fi)) := 'g~.a = {(r, y) ~ 1R: I G.p(r, y) = 0}

the SO3(1R)-representation curve of b(ct, fl).

1.6. Remark Let b(zr fl) be a 2-bridge knot or link. The mirror image b*(cr fl) of b(~,/~) is given by b(cr - f l ) (see [4, Chap. 12]). Moreover, we have z~._p = ( - 1) "- ~ z,., and the representation curve does not change if we replace the knot or link by its mirror image (see [3]}.

Let ~ := ~,.p be a SO3(lR)-representation curve. It is proved in [3] that the curve ~ has singularities in the points A := ( - 1, - 1) and B := (1, - 1).

1.7 Corollary Let b(~, fl), be given. The curve c~ = cg,.a has A and B as multiple points of multiplicity

/ ~ ( A ) = [ ~ ] and / z (B) - ' / 3 [ -12

Proof. See [3]. Lq

To clear up the relation between representation curves and representation spaces, let G be a finitely generated group. We consider the set

~'(G) := ~f(G, SUz(G)) := {Q: G --* S3]Q a homomorphism} .

330 M. Heusener

~(G) has the structure of a real algebraic set (see [1]): Given a finite set S = {sl . . . . . s,} of generators there is an embedding

fs: ~ ( G ) ~ SU2(~) • • SU2(IE) c IR 4"

and fs(~(G)) =: Vs, i.e.

Vs = ((A1 . . . . . A,) e (SUd~))"I (AL . . . . . A,) satisfies the equations defining G}.

Vs ~ IR 4" is an affine algebraic set and ~(G) inherits an algebraic structure which does not depend on the finite set S of generators.

Let G be a knot group. The space d ( G ) = {0: G --* SUE I lm(o) is abelian} is an algebraic component, i.e. a maximal irreducible algebraic subset of ~(G) (for a proof see [12] or ES]). We call

~(~) := ~ ( ~ ) \ d ( ~ i z~

the variety of non-abelian representations. If G is a link group the situation is different: we consider a Wirtinger presenta-

tion of G = ( s l . . . . . s, [rl . . . . . r,_ l ) and we assume ~ (G) is embedded in R*" via the system of generators {sl . . . . . s,}. We introduce the Zariski-closed set

~ : = #~(G):= {(a 1 . . . . . A,) e~(G)ltrAi = trAj, Vl < i,j < n} .

c ~r consists of all representations ~o such that ~o(mi) is conjugate to ~o(rnj) where mi is a meridian of the link. Therefore, the value of tr o(m) is well-defined for every meridian m.

Let G := G(cz, fl) be a 2-bridge link group. We have

= {(A, B) e ~(G)[ tr A = tr B} = z~7:= {(A, B) e ~(G)[ A and B commute} .

is the union of two algebraic components of o~ each isomorphic to SU2(II~). ~7 = ~ _ w ~r where d • = {(A, B ) e ~71B = A + l}. Again we define

Zar

~(G) := ~(G)\.~(~)

In the case of a 2-bridge knot group G = G(~, fl) we have ~(G) c SU2(r • SU2(~). If b(~,/~) is a link we have ~(G(c~, fl)) = SU2(~) • SU2(r There is an obvious regular map ~ : SUDS) • SU2(~}\{( + 1, _+ 1)} -~ ~2 given by

( t r 2 a 2 tr(_A/3r_) --_tr2 A ~ (A, B) ~ \ 4 - tr 2 A ' 4 - tr2A / / '

Let (A,B)e ~l(G)resp. ~(G), we rewrite A = c o s ~ E + s in2X and B = cos2E + �9 ~o

sin ~-Y where X, Ye SU2(C), t r X = tr Y = 0 and E is the identity. Let P, Q e S 2 be

given. It is easy to see that (P, Q ) = �89 We obtain ~(A, B ) =

(co t2~ , �89189 ( ~ - ~(X), 7t- ~(Y)). Therefore the equality

(2) ~ ~ = ~(~)

fiolds.

1.8 Remark The curve r and the polynomial z~.~(z,y) are not invariants of b(~, fl). Let b(~,/~) and b(~, fl') be two equivalent 2-bridge knots or links.

SO~(IRj-representation curves fur two-bridge knot groups 331

There is an isomorphism ~0 : G(~,/~')-~ G(,,/3) mapping meridians to meridians. Given a representation o : G ( " , f l ) ~ SU~(r there is a representation 0': G(a, fl') -~ SU2(~) defined by 0 ' := 0 ~ ~o. We denote by (r, y) resp. (z', y') the parameter associated to 0 resp. 0'. ~0 maps Wirtinger generators onto Wirtinger generators for that reason it is clear that y = y'. However, in general we have T + ~' (see Fig. 2) i.e. the parameter z depends on the choice of gener- ators - contrary to y.

2 T r a n s f o r m a t i o n b e t w e e n r e p r e s e n t a t i o n c u r v e s

In this section we would like to consider representation curves which are associated to equivalent 2-bridge knots or links but to different normal forms. In fact, let b(~, [3) and b(~,/3') be two equivalent 2-bridge knots or links, i.e./3ff - 1 mod 2,. The purpose of this section is to prove that the algebraic sets cg,.B\{A, B} and ~,~.~,\~{A, B} are biregular isomorphic.

In particular, let b(a,/3) (fl > 0) be given. There exists exactly one/3", 0 < fl*, 0 < ~ such that / 3 * - = l m o d 2 and f l . f l * - + l m o d 2 2 . According to Schubert [17] there is a homeomorphism h:SS~ S 3 such that h(b(~,/3*))= b(~, [J). h is orientation preserving iff/3-[3* - 1 mod 2~ and orientation preserving iff fi.[3* = - 1 m o d 2a .

We would like to prove (Corollary 2.6) that h: S 3 ~ S 3 induces a biregular isomorphism h : c~'~,tj\, {A, B} -~ cg~.tj,\{A, B} and we determine h by evaluating h at a finite set of points.

2.1 L)~domorphisms of the free group

Given a free group F 2 = (S, TL - ) and a canonical projection ~z: F 2 --~ G onto the group G = G(~,/3) there is an embedding

~(G) c~ ~ ( f 2) = S U 2 ( ~ ) x SU2([~ ) .

Since we are only interested in representations which are mapping 7r(S) and 7r(T) on conjugate elements, we have

~(G), ~ ( G ) = ~ ( F 2) := {(A, e) ~ SU2(~) • SUz(~:)I tr(A) = tr(B)}.

Let h" F 2 ~ F 2 be given such that there is an induced regular map ~,(h) : ~ ~ ~ . Given such a regular map we get a commutative square

~(F2) ' \{(_+l , +1)} ~,h, ~(F2) \{(_+ I, _+ l)}

D h ~ D

2,1 Example 1. Each inner automorphism of F 2 and v • given by S~-~ T • and T~--* S • induces the identity map on D.

2. Let b(, , /3) be an oriented 2-bridge link ( , even). By changing the orientation of one of its components we obtain the 2-bridge link b( , , 2 +/~) (see [4, Chap. 12]). The isomorphism h : G(~, ~ +/3) ~ G(~, r ) given by S ~ S - 1 and T~--~ T. Therefore, h induces ~(h)(A, B) = (A-l , B), using tr(A/3 T) + tr(AB) = tr(A). tr(B) one obtains h :ff~., ~ cg,.~ +, is given by fi:(x, y)~-~ ( - z , y); remember tr A = tr A-1 .

332 M. Heusener

For our purpose it is sufficient to study endomorphisms of the following form: given a sequence (~h . . . . . ~/k), ~/, ~ { -+ 1} we define h~? .~: F 2 ---, F 2 by

SI---~ X ~ . . . S , 2 T n I S • I T - , , S - ~ . . . X - , ~

T H T ~1 where

S i f k - 0 mod 2

X = T i f k - 1 rood 2 .

We need the following technical lemma

2.2. Lemma Given a sequence (ql . . . . . rlk), r/i~{___l} and an e n d o m o r p h i s m h(k • := ",lk~'t~} . . . . . ,1" F2 --'> F2 there is a p o l y n o m i a l F~ ~ . . . . "~)(xl, X2) ff 7~[~(1, X2], which on ly depends on the sequence ( r i b . . . , rla ) such tha t

Proof . (Readers not interested in calculations may skip the proof). Wc have to study the map ~(h~• ---, ~ given by

(~, ~ ) ~ ( ~ • ~ - 1 , ~•

where ~ = (P, ~p), ~ = (Q, ~p) and ~/B = 3E ~ . . . ~ " ~ " ~ (.~ is equal to ~ or ~ de- pending on k is even or not). Let Pg ~ S 2 be defined by Pk := ~[~e~[~- I. We have the equat ions

~8~ ]+l ~]3-I = (-+ Pk, ~P)

~• = (+ (2, ~o).

Therefore it follows that h},• y) = (r', y) where r' = <Pk, Q>,

C la im There are polynomials F~ := F~ n . . . . . . "~), Gk := G~ ~ . . . . . "~, Hk := H~ " . . . . . . n~ ~, I-x ~, x2] such that

( ' ) ( P k , P x Q > = s incpGk r , ~ + - ~

<P~, P> = H k %

(P x Q denotes the vector product in IR<)

P r o o f o f the claim. We are proving the claim by induct ion on k.

Fo(X1, x2) = Xl

k = O ~ P = Po ~ G o ( x ~ , x 2 ) = 0

Ho(x~ , x2) = 1

S03(N.)-representation curves for two-bridge knot groups

k = 1 ~ P1 = ~ , l p ~ - , ,

= cos ~ - s i n 2 P + r h s i n q 0 Q x P + 2 s i n 2 ( P , Q > Q .

It f o l lows <P1, Q> = <P, Q>

<P~,Q• ~sin~0(1-<P,Q>2)

<P1, P > = 2 c o s 2 ~ - 1 + 2 s i n 2

T h e s u b s t i t u t i o n s cos z p - Y a n d sin e q ~ - 1 r e su l t in 2 y + l 2 l + y

G ] ~ (x , , x2) = ,/1(1 - x~)

H ] ~ ( x l , xz) = (2x2 - - 1) + 2(1 - x 2 ) x ~ .

A n a n a l o g o u s c a l c u l a t i o n yie lds t he f o r m u l a s :

a n d

k = 1 m o d 2

F ~ . . . . ,,i = E~,21 .,~ ,t

Gt"'" " = (2x2 - 1)G~" . . . . , ,k , , + q k H ~ ' " , "" " - - qkX, --aFt""-,

H~' . . . . ~'~= (2X2 -- 1)H~'~ , ''~ " - - 4r/k(1 -- X z ) X 2 G~"._, . . . . " '~

+ 2(1 - x 2 ) x l F["" 1 ''J. '~

k - 0 m o d 2 ~

F " . . . . lo (2x2 I,t,, ,~ + .,~ ,) k = -- 1 )Fk--1 '"~ 4qk(l - - x 2 ) x 2 G ~ i T ' - i

+ 2(1 - X z ) X l H ~ . . . . ,k ,I k - - I

,Ok ,~

333

= - " ""~ '~ H ~ ' . . . . . . ,k . ~ _ q k F ~ , ~ , i G':'Tk . . . . l~, ( 2 x z 1)G~"'__'j + qkX~ _ j "" '~

a n d t he c l a i m fol lows. [ ]

As a d i r e c t c o n c l u s i o n of t he c l a i m we o b t a i n a p r o o f of the l e m m a . [ ]

2.3 C o r o l l a r y E a c h e n d o m o r p h i s m h~ • = b ~• �9 F 2 -~ F 2 induees a r a t i o n a l m a p " ' ~ k , �9 �9 , ~ 1

~ • ~ O w h i c h can be e x t e n d e d to ]R2\.{(z, - 1 )~ IR2 lz ~ IR}.

Proof . T h e c o r o l l a r y is a c o n s e q u e n c e of L e m m a 2.2. [ ]

2 .4 E x a m p l e C o n s i d e r t h e s e q u e n c e (q l , t/2), F2 := F(~ ~''~z).

F 2 ( x l , x2) = (2x2 -- 1)Xa + 4 t h q2(l - - x2)x2(1 -- x{)

+ 2(1 - x 2 ) x l ( ( 2 x z - - 1) + 2(1 - - x2)x~) .

334 M. Heusener

Assuming ql" r]2 = -- 1 we get

( 4(r2 - -1) (r + Y) ) h,2,,~(z,y)= z + ( y + l)2 ,3' -

2.5 Remark Using the observation ~l(h~)((P, q~), (eP, q~)) = (( + P, (p), ( _+ e,P, q))) for e,e{ + 1} we have that hk is the identity on the lines r = + 1. In general we have more fixed points.

2.2 The transformation h:C~,a~c(C,,t~.

The aim of this subsection is to study the transformation between the representa- tion curves which was induced by a change of the system of generators.

Let b(ct, fl) ~ S 3 be given, fl > 0 and [/* > 0 from the beginning of Sect. 2. We consider the homeomorphism h : S 3 ~ S 3 such that h(b(~, fl*)) = b(ct, fl). There is an isomorphism h #: G(~, f i*)~ G(~, fl) induced by h. Let G(~, fi~ be generated by {S, T} and let G(~, fl*) be generated by {S', T'}. We have

~S ,~_,S~,T~. . . T~, * , S - ~ T ~,*~. . . T - ~ S ,:~ (3) h,~: [ T,~__~T_ 1

(for details see [8]). Because of (3) and Lemma 2.2 there exists a polynomial F(x~, xz)c;g[x~, x2]

such that

t h = h # ( r , y ) = F r, ,y .

2.6 Corollary There is a biregular isomorphism

h:%,~'~{A, B} ~ %.~. ' ,{A, BI.

and the inverse of h is induced by h ,1.

Proof By Corollary 2.3 we extend h to a regular map h : ~ , p ' \ l R • ~ , , ~ . \ N x { - 1 }. But by Corollary 1.7 we obtain cg,,, c~ IR • { - 1 } = [A, B} be-

deg z , , t ~ = l ~ ) - I = # ( A ) + ~ z ( B ) . h i s a b i r e g u l a r i s o m o r p h i s m because it cause t _ - - /

has an inverse map induced by h~,~. /~

The curve ~o~,p is of degree n = I ~ 2 1] and meets the line y = 0 in exactly

t / 2rcv ) n points r~ = ~,cos-~--,0 , 1 < v < n (see [3]). These points z, correspond to

dihedral representations and it is possible to determine h(r~). With the intention of doing this we identify the point z, with the coset

+ v mod ~ which we denote by ( + v),.

2.7 Lemma Given h:c~,~--* c~,p, we have:

h ( r 0 = ~ -~ ( +_ ~')~ : ( +_ fl*- v)~

Proof We consider a dihedral representation O~: G(e, f l ) -~S03(~) and assume that the conjugacy class of O~ is determined by r~, i.e.

. f S~ ,6 (P ) o~.1. T~6(Q)

SO3(IR)-representation curves for two-bridge knot groups 335

Fig. 1. The Iotation in the plane spanned by P and Q

\\

C7,3

/ r~ / r2 ~ ~ 5 A r~.~T2

)i t ~

/ f

Fig. 2. l ransformat ion between the curves g7,3 and g7.5

27r~ P, QeS 2, ~ (P, Q) = - - ; remember P = (P, r with r = ~. We obtain that 6(P)is c~ a rotat ion of angle ~ with axis P. We get

e~~ ( T~-+6(Q)

with k - fl*-2 1 (see Fig. 1). In fact ~(PQ) is a rotat ion with angte 4~vcr

For that reason the angle ~ = ~z (Q, (pQ)kp(pQ)-k) is given by

2rcv 4zrvk 2~zv q, = - - + = / 3 " - - 0r ~ Gr

To the point ( c o s ( ~ ) , 0 ) we associate the coset ( + fl*v)~, which proves the

lemma. []

2.8 Example For the t ransformation h : r162 --* cr 5 we get:

"el F-'~ ~2, T2 F--~ T3, T3 b--~ "( 1 (see Fig. 2).

2.9 Remark cr is of degree n and meets the line y = 0 in exactly n points z~, v = 1 , . . . , n. By B6zout's intersection theorem cg~,~ intersects the line y = 0

336

Fig. 3. The knots b(15, 11), b(21, 13) and b(33, 23) (from left to right)

M, Heusener

transversal . So, each poin t r~. is a regular point of <g,.p. Now, h is y-level preserving. Consequent ly , h(x) is de termined by h(T~) for every xeU~(zd ~ cg'~.t~ where U~(rO = {xelRE[]x - z ~ l < e} and e > 0 is sufficiently small. Moreover , every alge- braic c o m p o n e n t of c~,.o contains at least one of the points z~; remember deg z,,a = n. But each a lgebraic map is de te rmined if the values on a Zar isk i -dense subset are given. But U,(z~) ~ ~ ,~ is a Zar iski -dense subset of ~G,p and as a result

is de te rmined by its act ion on the set {r~lv = 1 . . . . . n}. The map h may have fixed points.

2.10 L e m m a Given h:c~,,,_+cg~,,,.

h(v~) = v~ ,~ ~ lv -gcd l f l* + 1, ~) or cqv.gcd(f i* - 1, ~).

Proof By L e m m a 2.7 we have:

h(rv l : T~,=- ( • v) , : ( +_/~*~'),

�9 * ~ v - f l * v m o d e or v - - f l * v m o d a

< * c ~ l ( f l * - l ) - v or ~ [ ( f l * + l ) - v

< * ~ l g c d ( ~ , f l * - l ) - v or ~ [ g c d t ~ , f l * + l ) - v .

3 Self symmetric 2-bridge knots

Let b(~, fl) c S 3 be given, b(e, fl) is called self symmetr ic if fl ~ _+ 1 rood 2ct and f12 ~_ 1 m o d ct. The first three self symmetr ic 2-br idge knots are b(15, 11), b(21, 13) and b(33,23). Self symmet r ic 2-br idge kno t s or links possess an addi t iona l symmet ry which m a y be real ized by a ro ta t ion of angle 7r (see Fig. 3). We are now in a pos i t ion to prove a conjecture of Riley:

3.1 Theorem The S03(1R)-representation curve of a self symmetric 2-bridge knot or link is reducible.

Proof Let b(:q fl), f12 _ 1 m o d 2e a n d fl ~ + 1 mod �9 be given. Wi thou t loss of general i ty we assume fl > 0 ~ fl* = fl and we have an involu t ion h :cG, e-+ ~'~.o" Using L e m m a 2.10 we see tha t the points which are represented by the cosets

SO3(IR)-representation curves for two-bridge knot groups 337

Vv' l

T7,~-'-.5

,7

Cl 5,11

k r3 ( N / l ~T3

I'i~. 4. [ l ie rcplC~,r ctlr~t~ Y~ls I I OI1 Ihr lclt [l~llltl side ~md lhe component W~ on lhe right h a n d side

( ___ (/3 _ 1))~ are fixed by h because/J(/3 + 1) =/3/3 +/~ -= _+ (fl + 1) mod ~. The m a p h is y-level preserving and the po in ts r , . e ~ . a are all regular. As a result we see tha t [~ is the ident i ty m a p in a ne ighbourhood of a fixed point and so it has infinitely many fixed points. By / ] ( r l ) = r r , ( • y)~ = (• ~ ( • 1)~ we see tha t ~(r~) + r t ~ h + ld~,.B. Each ra t ional t r ans format ion on an i r reducible curve is the

identity, if it possesses an infinite set of fixed points. W e conclude that ~ , a must be reducible. []

3.2 Example We cons ider the curve ~15~, l i. The fixed points o f h are ~3 , zs, %} by Lemma 2.10. One easi ly sees the spl i t t ing of the curve r 1 t into the c o m p o n e n t s W1 and 1/[7"2 (see Fig. 4).

3.3. Corol lary Given a se!f symmetric 2-bridge knot or link b(ct,/3) the variety ~/'(G(o~, [~)) of non-abelian representations is reducible.

Proof Let X ~ lR N be an affine a lgebraic variety and q0 : X ~ IR M regular. If X is

i r reducible then ~0(X) zar is i rreducible. By Theorem 3.1 together with (2) the s ta tement of the coro l la ry follows. []

4 Representation curves of amphicheiral 2-bridge knots

Given an or iented kno t or link k c S 3, we denote its m i r ro r - image by k*. k is called amphiche i ra l iff k~k*. According to Schuber t ' s classif icat ion we have b(~,/3) is amphiche i ra l iff/32 _= -- 1 rood 2c~. But/32 = _ 1 rood 2c~ and fl o d d implies that ~ is odd. We ob ta in that 2-bridge l inks are never amphichei ra l . Therefore, we assume in the future tha t ~ is o d d if no th ing else is ment ioned . Amphiche i ra l 2-bridge kno t s are s t rongly negat ive amphiche i ra l , i.e. there is a ro t a t ion a b o u t n m a p p i n g b(c~,/3) on to b(~, - / 3 ) and invert ing its or ienta t ion .

338 M. Heusener

4.1 The Alexander polynomial of amphicheiral 2-bridge knots

Given b(e,,8) we denote the Alexander polynomial of b(e, fl) by A,,t~(t) . In the following we assume fl > 0. With the in ten t ion to s tudy the Alexander polynomial we look at the cont inued fraction

- [2bl , - 2 b 2 . . . . . ( - - 1) k-I 2bk]

. = 1

1 2hi +

-- 2b 2 +" . .

( - 1) k 22bk-i 4 ( - 1) k- 1 2bk

with even factors. F r o m [19] we get a formula for the Alexander polynomial

( t - 1)bl

t

A~.o(t) = 0

0

- -1

( t - 1)b 2

- 1

0

0 " " " 0

t ' 0

" ' " , ( t - -1)bk

We denote this de terminant also by Atb, . . . . b~}(t).

4�9 Remark If ct is odd we have k = 2 9 is even. b(.~,fl) is amphicheiral iff b i - - - b z g - i + 1 and b(~, fl) is self-symmetric iff bi = b2o -i+ 1.

Given bl . . . . . b, e7Z we define Zig(t) : = A{b . . . . . . b~)(t), l < k <- n. Use of the Laplace expansion formula leads to

Ak(t) = (t -- 1) hkAh- 1 (t) + tA k _ 2 (t)

s tar t ing with AdO = 1 and Al(t) = ( t - 1)bx. Again using the Laplace expansion formula one gets for amphicheiral 2-bridge knots

and for self-symmetric 2-bridge knots

A{b . . . . . . b~.b . . . . . . b , } ( t ) = A~2b . . . . . . bk}(t) + tA~b . . . . . . bk 1} ( / ) �9

4.2 Remark By subst i tut ing t 2 for t we have

(4) A{b . . . . . . bk,-b . . . . . . _b,}(t 2) = F(t) . F( -- t)

with F(t) = d{b . . . . . . bk}(rE) + tA{b . . . . . . bk ,}(t2) ' Therefore, the decomposi t ion of the Alexander po lynomia l proved in [7] is given�9 The decomposi t ion can also be given

SO3(N.)-representation curves for two-bridge knot groups 339

for all s trongly negat ive amphicheira l knots (see [5]). Analogous we obtain

(5) dl~,. .bk, bk. , b , l ( - - t 2) = F(t)" F ( - - t)

with F(t) = A~,. . ~ ) ( - - t2) ~- lAmb . . . . . bk ,~(-- t2) �9 Using the fact degA~ = k, we symmetr ize Ak by

k A(S)(t) := t-~ Ak(t) .

There exists a polynomia l V(z)~:g[z] called C o n w a y polynomia l such that

vk (z) = ~s)(t) 1 1

if we substi tute z = t 2 - t 2. The induct ion rule for Ak leads to

V~(z) = zbk Vk- l(z) + Vk- 2(z)

start ing with Vo(Z) = 1 and 171 = bzz. By the induct ion rule one concludes that Vk(z) is a polynomia l in z 2 for even k and Vk(z)/z is a polynomia l in z z for k odd.

By (4) we obtain

V~.~(z) = ( - 1 ) ~ f ( z ) . f ( - z)

iff i 2 -- -- l mod ~ with f ( z ) = Vtb,. ,b~)(Z) + Vib . ,bk ~I(Z)" Analogous , equa t ion (5) yields to

V~,~(z) = f ( z ) . f ( z )

if b(c(, fl) is self-synunetric where f (z) = V(h,. ,h~i(z) + i VIb~. ,b~ ,~(z). We are now ready to prove

4.3 Theorem The Alexander polynomial o f an amphicheiral 2-bridqe knot has no zeros on the complex unit-circle.

Proo f Given b(c(,fl), f i z = - 1 m o d c(, we assume f l > 0 . Let c ( - / ? _

[ 2 6 1 , - 2 b 2 . . . . . 262,2bl] be the cont inued fraction with even factors and e ~ e S ~ c (I~. We assume

A~.~(e~~ ~ V~.~(2 i s in~ ) = 0 .

We will show that V~.tj(ir ) ~ 0 for rclR. First we assume r + 0 by V,.p(0) = 1.

V~,t~ (Jr) = ( - 1 ) k f ( i r ) f ( - ir)

= ( - l )k f ( ir) .[( ir)

/ ( z ) = Vl~,, ,~ l (z ) + Vl~ . . . . ~ , l ( z ) e : ~ [ z ] .

As a result we have: 17~.~(ir) = 0 ~ f ( i r ) = 0. It follows f ( i r ) = 0 ~ , V~b . . . . . . b,i (Jr) = 0 and ~b . . . . ~ ,~(ir) = 0; r emember that Vk(z) is a po lynomia l in z z for even

k and Vk(z)/z is a po lynomia l in z 2 for k odd. Again the induct ion rule leads to 1 = Vo (Jr) = 0 which is a contradict ion. []

4.4 Corol lary The representat ion curve o f an amphicheiral 2-bridge knot never intersects the half line H + = { ( l , y ) ~ I R 2 1 y > - I } and the representation

340 M. tfeusener

curve of a self-symmetric 2-bridge knot never intersects the half line H - = {0, y ) ~ l y < - 1}.

Proof The first part of the corollary follows immediately from Proposit ion 2.3 of [3] and the fact that V~.#(ir) :# 0 for r~lR. In order to prove the second part, we have to show that V~,p(r) 4:0 for every self-symmetric 2-bridge knot b(~,/3) and every t e l ( . This can be done in the same way as in the proof of Theorem 4.3. V~

4.5 Remark An analogue to Theorem 4.3 does not hold for strongly negative amphicheiral knots in general. Compare the examples 818 and 1099- So Theorem 4.3 gives a necessary condit ion for an amphicheiral knot to be a 2-bridge knot.

4.2 About the genus o f representation curves

Let b(~,/3) be an amphicheiral 2-bridge knot. Without loss of generality we assume /3 > 0 ~ / 3 =/3*. Again we have a transformation

By Lemma 2.10 none of the points {~,lv = 1 . . . . . n} is fixed by h (use the fact /32 + 1 -- 0 rood ~ and gcd(~,/~) = 1). So we cannot prove a similar result to

Theorem 2.10. The map h:~g-,0-~ (g~,p possesses fixed points which are not on the line y = 0. The number of fixed points is connected to a lower bound of the genus of the curve cg,,0.

4.6. Remark Given an irreducible infinite algebraic set (s 1R 2. (~ is an irreducible algebraic curve, i.e. there is an irreducible polynomial F(xl , x2)eIR2[xl, xz] such that (g = {(xl, x2)elR21F(xl, xz) = 0} (see [6]). Let ~ c be the smallest complex algebraic set which contains (~. (gr is the complex Zariski closure of(~ ' c IR 2 c C 2.

Following [21] we have

% = {(~, ~ ) ~ r ~ ) = 0}

Let (g, ~ ~ IR 2 be algebraic curves and U ~ cg a Zariski open, dense subset. Every regular map (p: U-~ @ induces a rational map ~oc: (~r ~ @~ (see [21, Lemma 6]). As a result the biregular isomorphism

/t : (g~,p\{A, B} --4 (g,,o,k{A, B}

induces a birat ional map

he : (%.e)~ --* (%,~ , )~ .

The genus g(Cg) of the i~educible real algebraic curve ~ is defined to be the genus of the projective closure (go of (gr i.e. g(Cg) := g(C-gr (go, = C P 2.

The definition of ,q(Cgr can be found in every s tandard text book on algebraic curves.

4.7 Proposition Let C be an irreducible projective algebraic curve o f genus g over C and r : (g ~ (g a birational transformation which is not the identity. Setting

Fix (p = {xECglq~(x) = x} ,

it follows: # Fix(~o) __< 29 + 2,

SO3(IR)-representation curves for two-bridge knot groups 341

fixed poin:: ~ (

A

I

Fig. 5. l-he SO~(~}-representatmn curve % ~7,~3 on left hand side and an enlarged version on right hand side

Proof. We choose a point Q e ~ such that ~o(Q)4= Q and consider the divisor O := (g + I)Q. Let L(D) := {fek(~)iordQf> -- (g + 1) and o r d p f > 0 for P 4- Q}. (k(Cg) - denotes the function field of ~, see [18, Chapter III, Sect. 5.6] for defini- tionst. L(DI is a C-vector space containing all functions of k(Cs with only one pole of order - o r d o f < g + 1. Let I(D):= dime L(D).

By Riemann's theorem we have

l (D)>degD+ l - - g = 2 .

Thus we have a non-constant function feL(D) and we define a new function hck(~) by

h(P) :=.f(P)--.f(cp(P)). tp(Q) 4=Q ~ h possesses two poles of order ~)= - -ordQf__<g+ 1 and so the number of zeros ofh is bounded by 2~) _< 2g + 2. Every fixed point of (p is a zero of h ~ #Fix(q0 =< 2~) =< 2g + 2. []

We like to study the curve <KxT. t3 and the transformation h :~,~,a3 --+ ~17,a3 (see Fig. 5). It is easy to check that there are at least 3 fixed points: One point of intersection ~1~. ~3 c~ {1} x ~,, the other two fixed points are local maxima with respect to the y-level with the property there are no maxima on the same level (see Fig. 5 and remember h is y-level preserving). In order to apply Proposition 4.7 we have to prove the following lemma:

4.8 Lemma The polynomial z 17,, a(Z, y ) ~ [ r , y~ is irreducible.

Proof First we expand z~.y around the point B = (1, -- 1), i.e. we put z = ( ' + 1 and V = 17- 1 (see [3]). Furthermore, we substitute ~:= 2(' and obtain Z l 7,13(~ ~) = f8 (~, t/) + f7 (~, r/) + f6(~, r/) where

f8(~, r/) = r/8 + 4q7~ + 18r/6r 2 + 22~/5r 3 + 39q4~ '* + 50r/3~ 5 + 31q2~ 6

+ 9qd.7 + ~s

fv(~, t/) = 4(2r /+ ~.)(2~/6 + 8t/s~ + llt/'*r z + 32r/3r 3 + 34qz~ 4 + 14n~ 5 + 2r 6)

A(~, n) = 16~0/z + 3nr + ~2)(2q + ~)a.

342 M. Heusener

Thefk are homogeneous of degree k; rememberfo(�89 q) = 0 describes the tangent lines of ~1v,13 in B. If z17,13 is a product of two polynomials then this two polynomials are irreduciblc and they havc to be of degree 4. The reason for that is: cr does not contain any line, quadric or cubic (see Fig. 5). Ergo we assume

(6) z~, ,3(~, ,) = (g4(~, q) + ~q3(~., O))'(h4(~, ~) + h3(~., q))

where gk, h k ~ [ ~ , ~/] are homogeneous of degree k.

. 3 16+ +T3+ )(, + + The involution f~ has to map each component onto itself or it permutes the components, h preserves the components because of the existence of the fixed points. By the position of the tangent lines there is only one possible splitting:

g 3 = a ~ ( 2 ~ ! + ~ ) q + ~ and h 3 = b ( 2 q + ~ ) 2 q + ~

where a, b~t~ and ab = 4. For 94 = Y'. i+j=4ai~i~ j and h 4 = 2i+j-4 b iJ~ j the equation (6) yields to

16 equations for the 10 variables a~, b~, 0 <__ i , j _-< 4, i + j = 4. We obtain a contra- diction if we try to solve these equations. []

As a corollary of the discussion above we get:

4.9 Corollary There exists an amphicheiral 2-bridge knot such that the S03(IR)-representation curve is irreducible and has genus g > O.

This disproves the conjecture that the representation curves of 2-bridge knots always splits in rational components (see [3]).

5 Complex orientation, Casson's invariant and signature

Let V c cg,, e be an irreducible component. We consider ~ c ~2 the complex Zariski-closure and its projective closure Vr = ~p2. The non-singular model of Va: is denoted by Ve. V r tl2P u is a complex smooth irreducible curve and the map z : C P ~' ~ C P N which associates to every point x e C P N the point x(x) with complex conjugate coordinates induces an involution z: I7r -~ 174:. The set 17~ of real points is exactlythe fixed point set of z, it is a one-dimensional submanifold of the oriented surface Ve (see [18, Chapter VII, Sect. 4] for details).

Let T1 . . . . , Tr be the connected components of V~. By Harnack 's Theorem (see again [18]) the components are either independent or are connected by as ingle relation T1 + . . . + T~ = 0 in H~(Vr Z/2Z). In the second case V~ splits Vr i.e. 17~\. 17~ = 171 w I72 consists of two disjoint components.

5.1 Definition We call a SO3(lR)-representation curve divisible, if for every irredu- cible component V ~ W~,p the non-singular model 17r is split by V~.

5.2 Remark 1. I f~ , .p consists only of rational components, it is divisible because each curve in a genus 0 surface splits. Using this observation and the results of [3, Sect. 3] all curves cg,.o are divisible for fie{l, 2, 3}.

SO~(N.)-representation curves for two-bridge knot groups 343

2. It seems that every curve cg~.a is divisible (we can prove that only in special cases e.g. ~ 5. ~ is divisible because it splits into rational components).

3. If V~ splits 17r the curves l?a inherit two opposite orientations and so we have two opposite orientations of the curve V ~ IR 2 (see [20, 2.1]). We call these two orientations the complex orientations of V.

5.3 Remark It might be possible to distinguish one of the complex orientations. We cannot do that in general but the following approach should work if we assume that cgR:= cg~,~ is irreducible.

According to Thurston's Hyperbolization Theorem, S3\b(0q fl) admits a com- plete hyperbolic structure with finite volume if fl ~ + 1 mod a (remember: by a result of Schubert [16] 2-bridge knots are simple and b(~, fl) is a torus knot iff fl - + 1 rood ct). We have cgr ~ ~r and the points of (ge\cgr are corresponding to conjugacy classes of parabolic representations 69: G(ct, fl) ~ PSL2(~). Let Xor162 be the non-singular point which determines theholonomy of the excellent hyperbolic structure. We have X o q ~ . Let zt:~r cgr be given. We assume tha t cg~ is divisible then ~gr = V~ ~ V2 and Xo := n - ~(xo) is contained either in V~ or in I72. So, c ~ inherits an unique orientation from the component 17~ which contains the point s

5.1 Casson's invariant

Casson has defined an invariant 2'(k) for knots k c S 3 which may be computed from the symmetrized Alexander polynomial A~S~(t), A~S)(1)= 1 (see [1]). If 2(k) 4 :0 then k has property P.

For 2-bridge knots b(~, fl) there is a simple formula for ,~'(~, fl) in terms of the associated continued fraction (see [3]).

Let/5:= {(z, ?)1 - 1 < r < 1, 7~IR} c IR z. There is a twofold branched covering p :/5 ~ D given by p : (z, 7) ~ (z, ~2). The set p - l(cg~,a c~ D) might be identified with thc set of conjugacy classes of representations ~: G(~, f l )~ SU2(~) (see [3]). We define

/)+:= {(z,~,)~/Sl~, > 0 } and /5_ := {(r,~)c/5]7 < 0 } .

Let now ~ . ~ ~ IR 2 be given and assume that ~ , p is divisible. For every algebraic component V c ~g,,p we choose one of the two complex orientations. So every arc of V c~ D is oriented.

Let b be an arc of V n D. It is easy to lift b to an oriented arc ff c 15+ of the SUz(~)-representation space (see [3, Theorem 1.3]). To every point (r, y)Eb we have an equivalence class of SUE(C)-representations [Oc~.),~. Let (m, d) be a peri- pheral system of b(~, fl) and assume that

~t~.,,) (m) - (P, (p,,), 0 < ~0,, < ~z

~r = (P, qgc), 0 < tpe < 4n .

The orientation of b leads to a map

If Ob 4: ~ the two boundary points of b are dihedral representations or abelian representations and so we have Ob(c~b) = 1 ~S ~.

344 M. Heusener

Choos ing a fixed or ien ta t ion of S 1 w e have a well-defined mapp ing degree deg O~.

5.4 Remark Let - h the arc ob ta ined from b by changing the or ientat ion. Obvi- ously, we have deg O b = - deg Oh.

Let ~ ( V ) be the set of arcs of V ~ D. The number

b~v~ deg Ob

is independen t of the choice of complex or ientat ion. Let V1 w . . . w VN = ~ , ,p be a decompos i t ion of ~ , .p in to a lgebraic compo-

nents. We define

A(a, f l ) := ~ ~ d e g O b . [= 1 b~-~(G)

A(a, fl) is a kno t invariant . The reason for that fact: If f l . f l ' -= 1 rood 2~, there is a b i regular i somorph i sm h: cG, p--, cG,,, map p in g one complex or ien ta t ion on to another .

5.5 Remark It is possible to define A~.a only if (8,. B is dMsib le .

5.6 Example Let b(a, 1) be a torus knot . We have d = z .m -2~ where z is a gener- ~ - 1

a to r of the center of G(a, 1) (see [4, 3.28]). ~ , , 1 splits in n = - - ~ - lines G~ . . . . . G,.

F o r each ~: G(a, 1 ) ~ SU2(r we have ~(d) = - 0 (m -2~) because ~(z) = 1 (see [10]). Accordingly, we obta in q~e = 2zc - 2 ~ 0 , , .

b, = G~ c~ D is the s t raight line connec t ing the poin ts r~ = cos ~ - , 0 and

.cot v -j l, =ctg -- cxp( -- ~ti) and [ deg Ob~ I = v.

0~ 2 - 1 A(a, 1) = [deg Ob~l = 1 + . . . + n -

V=I 8

A(~, 1) : ),'(~, 1).

5.7 Example Let b(a, a - 2) , a = 1 rood 4 be given. ~ , , - 2 c~ D consists of n/2 =

arcs b~, i = 1 . . . . . n/2 (see [3, Sect. 31). Each arc b~ connects the points

'2~ and r ~ _ t. The curve c~,.~_2 is i r reducible and ra t iona l (see [3]). Therefore, it carr ies a complex o r ien ta t ion given by the pa rame t r i za t i on in [3] and each b~ has such an or ien ta t ion tha t its s tar t ing po in t is ~2~ and its end po in t is r2~- ~ (see Fig. 6). Let b / = / 5 + be the lift of b, c O.

The mer id ian and the longi tude of b(a, x - 2) are given by m = S and d = LTLs where Ls = (ST-~)"(S ~T)" and L~(S, T ) = Ls(T, S) (see [4, Chap. 12]). Let ~: G(a, ~ - 2) ~ SUz(~E) be a representa t ion which is associa ted to a poin t i n / ) + , i.e.

~ S ~--~ (P, ~) O < ~ 0 < n . b : ( Tw-~ (O, ~P) =

SO3(~,}-representalion curves for two-bridge knot groups 345

era" ~

4

~9,7

L , i

: La

L5 :x b x2 b2

Fig. 6. Ihe curves ~,~.~ and ~1,~.~1.

Fig. 7.

j f f

/ \

As in [3, Sect. 4] we think of P and Q as points on the front side of S -~ - see Fig. 7 - with M in the middle between P and Q on the equa tor and N resp. S as north- resp. south pole. Now O(Ls) = (R', 2) and o(Lr) = (R", 2) where 0 < 2 < 27t and the points R' and R" are on the front side of S 2.

Let (z, 7)e/5+ be the point which determines the equivalence class of ~: G(~, fl) -+ SU2(~). There is a polynomial v=,a(z, y)eZ[v, y] (see [3, Sect. 4]) such that

v~,.a(z, y 2) > 0 <=~ R' is on the nor th hemisphere

v~,~('r, 72) = 0 r162 R' = R" = M

v=.#(,, 72) < 0 ~=~ R' is on the south hemisphere . We obtain:

v=.~-2(z, 72) > 0 r ~(t ~) = (P, ~o:), 0 < q~: < 27z

v~,o_ 2(~, 7 2) = o ~ , 6 ( : ) = ( e , 2r~) = - 1

v=,:_2('r, ' f)<Oe>~(r 2~ < ~o: < 4 n .

346 M. Heusener

~ - 3 v~,~_ z (r, y) s 7Z [z, y] is of degree 2 = n - 1. @~ := { (z, y) e IR 2 [vs.~- 2 (r, y) = 0}

consists of n - 1 lines L~, 1 -< • _< n 1 where the line L~ connects the points

B a n d o ~ : = ~ c o s 0 . W e h a v e z ~ + l > o , > z ~ , l < K - < n - 1 .

Each line th rough B may intersect ~, ~,,_ 2 at most in one point different from B. That is because deg ~ . ~ - 2 = n and the mult ipl ici ty of ~ . , _ 2 in B is n - 1. Ergo, for a given arc b~, 1 < i <- n/2, we have ~ , c~ b~ = L2~-1 ~ b~ --:x~ consists of exactly one point x~ (see Fig. 6).

Lifting the s i tuat ion i n t o / ) § yields the following: on every arc /~ there is exactly one poin t 2i := P - l ( x ; ) which determines a conjugacy class of representa t ions ~: G(e, ~ - 2)-4 SU2(C) such that O(g) = - 1 . Moreover , we have (see [3]) that v~.~- 2(r2i) > 0 and v~.~- 2 ( z z i - l) < O, 1 <_ i < n/2.

Since b~ is or iented we are th inking of be as an ordered arc, i.e. b; = {X["~2i ~ X ~ "CZi-1}. Let Oh, :b, ~ S 1 be given by O b , ( x ) = exp(i~0l(x)/2}. By the above discussion we obtain:

0 < ~o/(x) < 2n e,. r2i > x > xi

q~e (x) = 2n , ~ x = xi

2 n < q ~ { x ) < 4 n . ~ xi > x > z2 i - l .

By fixing the ant iclockwise or ien ta t ion on S ~ we obta in deg Oh, = l, 1 <= i < n/2.

A ( ~ , ~ - 2 ) = ] ) / ( ~ , : ~ - 2 ) 1 = - - i f ~ - l m o d 4 . 4

5.8 R e m a r k As a consequence of these observat ions it is obvious to conjecture

for all divisible curves ~ , o .

5.9 R e m a r k The curve cg(17, 3) is divisible and 2'(17, 3 ) = 0 (see [3, Sect. 3]). However , so far we cannot compute A(17, 13) because the longi tude of b(17, 3) is to complicate.

5.2 S i g n a t u r e o f 2 -br idge k n o t s

An other observa t ion concerns the s ignature of 2-bridge knots. Let a divisible representa t ion curve cQB be given and let V be an algebraic componen t of ~ , ,a . F u r t h e r m o r e we assume tha t V carries a complex or ientat ion. By choosing an or ienta t ion of the plane and the line lit • {0} we define the intersect ion number

F

R x {0}) where each intersect ion poin t is count with its sing. ( R e m e m b e r # ( V k

in er c,s in e ac,,, ooints tran ve saHy) /

Again, let V1 w . . . w VN = ~ , r be a decompos i t ion of cQe into algebraic components . We define

N

s(~, p):= Z I#(V, ~ ~ • {0})1. i = 1

SO3(N)-representation curves for two-bridge knot groups 347

The construct ion of S(~, [3) can be explained in another way. Let V be an algebraic componen t of <s If b c V r~ D is an arc of second kind (see [3]) (i.e. b is an arc which connects two dihedral representations) then this two representat ions cancel each other by count ing them with sing. So only the arcs of first kind (i.e. arcs which are connect ing a dihedral representat ions with an abelian representations) are giving some cont r ibut ion to the intersection number # ( V c ~ R x { 0 } ) ; remember ~'~./~ ~ { - l } x IR -- {A} (see [3, 2.3]). But an arc of first kind is asso- ciated to a root of the Alexander polynomial on the uni t circle (see [3, 2.3]). For that reason the count ing of dihedral representat ions with sing is noth ing else but count ing the zeros of the Alexander polynomial on the uni t circle with sings. On the other hand, the signature of knots can be defined in a similar way (see [9, Chap. Xll]) .

The na tura l conjecture is:

2 . s I ~ , / ~ ) = I~ (b l c~ , /~ ) )1 �9

5.I0 Remark 1. The conjecture can be verified for all knots b(c~,/3) where 1~6 {l, 2, 3}. It is also lrue for the knot b(15, 11) and by Theorem 4.3 it is true for all amphicheiral knots assuming that their representat ion curve is divisible.

2. The conjecture above and a result of Lin [11] are well suited to each other. It reveals Lin's result in a different light and it explains (in the case of 2-bridge knots) why he comes up with the signature by count ing trace-free representat ions with a sign. Remember: Every trace-free representat ion (i.e. a representat ion ~): G SU2(Ir) such that tr ~o(m) = 0 for the meridian m) of a 2-bridge knot group is a b inary dihedral representat ion.

3. It is easy to replace the line R x {0} in the definition of S(~, fl) by any line q)o

R x {Yo}, Yo = cot 2 ~ , 0 < ~oo < n to ob ta in an invar iant Syo(e,/~) if z,. ,(1, Yo)

4= 0. It seems that we will come up in this case with the L e v i n ~ T r i s t r a m - S i g n a t u r e i.e. we conjecture that

2Syo(~, fl) - ]cr((l - ~o) W + (1 - &) wr)]

where o) = exp(i(Po) and W is a Seifert matr ix of the knot (see [9, Chap. XII]).

Acknowledgements. I am grateful to R. Riley for his recommendation to include self-symmetric links in the statement of Theorem 3.1. Moreover, 1 would like to thank G. Burde for his friendly support during the past years.

References

1. Akbulut, S. and McCarthy, J.: Casson's invariant for homology 3-spheres - an exposition. Math. Notes 36 (1990)

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