The Culler-Shalen seminorms of pretzel knots
Thomas W. Mattman
Department of Mathematics and Statistics
McGill University, Montréal
May 2000
A Thesis submitted to the Faculty of Graduate Studies and Research
in partial fulfillment of the requirements of the degree of Ph.D.
@ Thomas Mattman 2000
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Abstract We use the theory of Culler-Shalen seminorms to investigate the SL2(@)-character
variety X ( K ) and PSL2(@)-character variety .Y(K) of a Montesinos knot K. When Ii
has three tangles, Bi /a i , b2/n2, B1/a3, the PSL2(C)-character variety of the triangle
group A(ai , a2> ag) includes into .T(K). We apply this observation to make several
deductions about (ÿ, q, r ) pretzel knots.
For hyperbolic (-2,3, n) pretzel knots, ive calculate the Culler-Shalen seminorms
esplicitly and deduce that X ( K ) consists of esactly two (respectit-ely three) algebraic
curves when 3 t n (respectively 3 1 n). This leads to a classification of the finite and
cyclic surgeries of these knots. \Ne obtain similar results for some (-3,3, n) pretzel
knots.
CVe classify cyclic surgeries on (-2, p, q) pretzel knots (p, p odd and positive). The
(-2; 3,7) is the only non-toms knot in this family admitting non-trivial cyclic surg-
eries. We show that a (p, q, 2m) pretzel h o t admits at most one non-trivial h i t e 1 1 surgery so long as i, + + < 1. Moreowr, for m < -1, these knots have no
non-trivial finite surgeries.
Combining these results with work of Delman we classify cyclic surgeries on Mon-
tesinos knots. If a non-toms Montesinos knot I\' admits a non-trivial cyclic surgerx
then it is the (-2,3,7) pretzel knot and the surgery is 18 or 19. Further, if h- has
a non-trivia1 finite surgerK then it is either a ( - Z p , q ) pretzel knot with 5 5 p 5 q
odd, or else it is a (-2.3' q ) pretzel knot with q = 7 or 9.
Résumé On utilise la théorie des seminormes de Culler-Shalen pour examiner S ( K ) (la
SL2(C)-variété de caractères) ,et .T(K) (la PSL2(C)-variété de caractères) d'un noeud
de Montesinos K. Lorsque K a trois tangles f l l /crl , ,&/a2. P3/a3, alors la PSL2(@)-
variété de caractères du groupe de triangle A(al , L Y ~ : a3) est incluse dans S ( K ) . Cet te observation permet de tirer plusieurs déductions concernant les (p, q; r) noeuds
bretzel .
On calcule explicitement les seminormes de Culler-Shalen pour les (-2,3, n) noeuds
bretzel hyperboliques. On déduit que S (K) est constitué de deus courbes algébriques
quand 3 ( n et de trois courbes algébriques quand 3 1 n. Ceci donne la classification
des chirurgies cycliques et finies de ce type de noeuds. Des résultats similaires sont
obtenus pour certains (-3,3, n) noeuds bretzel.
On classifie les chirurgies cycliques pour les (-2, p, q ) noeuds bretzel (p, q étant
impairs e t positifs). On prouve que le (-2.3: 7) noeud bretzel est le seul noeud
non-torique dans cette famille admettant des chirurgies cyclicpes non-triviales. On 1 démontre aussi que le (p, q, 2m) noeud bretzel avec ipi + & + < 1' admet au plus
une chirurgie finie non-triviale. Si de plus, rn < -1, alors ce noeud n'admet aucune
chirurgie finie non-triviale.
En combinant ces résultats avec le travail de Delman, on classifie les chirurgies
cycliqiies des noeuds de h4ontesinos. Si Lin noeud non-torique de Montesinos A-
admet une chirurgie cyclique non-triviale, alors il est le (-2' 3,7) noeud bretzel et la
chirurgie est 18 ou 19. De plus, si K admet une chirurgie finie non-triviale, soit il est
un (-2: p, q ) noeud bretzel avec 5 5 p 5 q, soit il est un (-2, 3. q ) noeud bretzel avec
q = 7 ou 9.
Acknowledgment s
1 gratefully acknowledge the assistance and encouragement of my supervisor Steve
Boyer tlirough my many years of doctoral study His unflagging good humour and
patient responses to even the most inane questions were a tvonderful resource for
me- 1 have very much appreciated and benefitted from his friendlÿ and approachable
manner. -4t several critical junctures he has provided suggestions and advice without
which 1 could not have completed this thesis. His love of mathematics and his knack
for clear exposition, both as a teacher and a writer, remain an inspiration.
1 am indebted to Jacques Hurtubise for his advice, particularly at an early stage in
my studies. and for happily signing al1 forms that came his \ v a - John hlcKay, Olga
Kharlampovich, and Peter Russell generously took the time to Iisten to my questions
in their fields of espertise and provided answers and suggestions.
1 thank Paul Libbrecht for his energ- in organizing innumerable seminars and
- reading projects: and for his efforts to drag me: kicking and screaming, tomards some
semblance of mathematical culture. On a more mundane level: he showed me how to
produce the many figures which appear in the following pages. On a more persona1
Ievel, he has been a great friend.
Yassir Miladi: Luc Patry, and Nabil Sayari have been our fellotvs i r i many of these
seminars and 1 thank them for many useful conversations ancl cliscussions. 1 am
particularly gratefül to Nabil for suggesting the link between cyclic and toroidal
surgeries. He also provided a translation of the abstract. Steve New sent helpful
suggestions al1 the way from Thailand.
-4 substantial part of this thesis was written during a visit to Nihon University in
Tokyo and I am grateful to Kimihiko Motegi and the Department of Mathematics
for their hospitality during my stay.
With the aid of the Internet, 1 have been fortunate to profit from the espertise
of colleagues whom 1 have not even met (at Ieast not without electronic mediation).
Edwin Clark, Mourad Ismail, and especially Richard Stong tielped me to understand
Tchebyshev polynomials as part of my work on p-representûtions. Dave Rusin and
Kurt Foster suggested techniques for finding cyclotomic roots of -4lexander polyno-
mials. Felipe Voloch and Torsten Sillke told me about Mann's Theorem, and Andreas
Caranti and Rick Thomas pointed me torvards the work of Edjvet in group t h e o l
1 am indebted to Mark Ballora for bringing the quote of . Jer l Garcia to m y atten-
tion.
To my friends and family, and especially Shigemi: 1 s q thanks for your tremendous
love and support through the years. This accomplishment is as much yours as mine.
This research has been supported by grants and scholarships from NSERC, FCARt
and the ISM
For me? playing is like unraveling a whole bunch of little knots. Some
shows I never get past the knots. I spend the whole night trying to
untie them, and rny consciousness never breaks throiigh to the first
person out there.
-J . Garcia (1 942-1998)
List of Figures
1. Introduction
1.1. Conjectures of Poincaré and Thurston
1.2. Contributions of this thesis
1.3. Outline
2. Character Varieties and the Culler-Shalen Seminorm
2.1. Character varieties
2.2. Culler-Shalen seminorrns
2.3. Polygons of the twist knots
2.4- Preliminary lemmas
2.5. Riley's p-representations
3. Montesinos Knots
3.1. Seifert spaces and Montesinos knots
3.2. Montesinos' t heorem
3.3. Extending representations
4. Cyclic and Finite Surgeries on Montesinos Knots
4.1. Infinite fillings of (p, q, 2m) pretzel knots
4.2. Son-integral bounday slapes of pretzel knots
4.3. Cyclic surgeries of (-% p, q) pretzel knots
4.4. Finite surgeries on pretzel knots
4.5. Surgeries on hlontesinos knots
5 . Character Variet ies of Pretzel Iinots
5.1. The total norm of (2,p,q) pretzel knots
5.2. (-2,3, n) pretzel knots
5.3. (-3: 3, n) pretzel b o t s
6. Conclusions and Questions
-Appendix -4. Zeroes of -Alexander Polynomials
References
The twist knot K,,.
Fundamental polygon of the twist knot K n (n > 1).
Fundamental polygon of the twist knot hi, (n < 0).
Newton polygon of the twist knot K,, (n > 1).
Newton polygon of the twist b o t h', (n < 0).
The (-2,3: n) pretzel knot.
The Montesinos link m(e; @i/crl : &/a2; . . . y Or/a,).
- . n annulus preserved by T . 35
The quotient of T by r, 36
The double of the quotient of T. 36
Illustration of the disks Di used to construct V. 37
Illustration of the balls Bi in S3. 31
The trivial link transforms to the Montesinos link. 38
-A lift of the meridian p. 40
The reflection of S2 inducecl by T. 41
The diagram V. 48
The tree of a l/ï tangle. 49
The trees for the ( -2 ,3 ,7) pretzel knot. .5 O
The (2, pl q ) pretzel h o t . 65
A spanning surface of IL. 87
The handlebody H. 87
T h e regular fibres a2 and (XTJ("-~)/~)~ intersect once inside the circle. 89
T h e meridian p represented by a c u m y.
The fundamental polygon of Kg.
The fundamental polygon of KAT.
The Newton polygon of 1 ' (n 2 7 and 3 1 n).
2'7 The Newton polygon of K, (n 5 -5 and 3 in).
28 The (-3,3; 4) pretzel knot.
29 The fundamental polygon of the (-3,3,4) pretzel knot.
30 The Newton polygon of the (-3,3,4) pretael knot.
1.1. Conjectures of Poincaré and Thurstcjn. The Poincaré Conjecture states
t hat any closed, simply-connected %manifold is homeomorphic to 3" That t his
conjecture has remained open for close to a century points to the subtlety of 3-
dimensional topolog-. -4lthough S3 can be regarded as the simplest of %manifolds:
we still do not know that there are no clever imitations or "fake" S3?s out there which
share most of its important properties.
The work of Lickorish and Wallace suggests one approach to this conunclrum. They
showed, independently? that any 3-manifold can be constructed through Dehn surgery
on a knot or link in S3. In particular any fake S3 counteresample to Poincaré%
conjecture could be so constructed. Of course, starting with such a knot or link, one
can always recover S3 through "trivial" surgew We say that a knot has property P
(for "Poincaré") if this is the only Dehn surgery which results in a simpl-connected
manifold. The property P conjecture states that every non-trivial knot in S3 has
property P. -1lthough property P is knoïvn to hold for large classes of knots, this
conjecture also remains open.
The work of Culler and Shalen [CSl, CS?' CGLS], and Boyer and Zhang [BZl, BZ2,
BZ-L] shows the feasibility of taking a slightly larger perspective. Rather than surgeries
leading to a simpiy-connected manifold they investigate those resulting in a rnanifold
with cyclic or finite fundamental group. Essentiallyv, they show that surgery on a
hyperbolic knot in S3 c m produce a t most three manifolds witli cyclic fundamental
group and a t most five with finite group (see Theorems 2.2.1 and 2 - 2 2 for more
precise statemeiits) . For esample, since trivial surgery produces S3, and as simply-
connected manifolds have cyclic fundaniental group, a given hyperbolic knot could
have a t rnost two surgeries which violate the Poincaré Conjecture. Thus one could
imagine a program to prove the Poincaré Conjecture by systematically going through
al1 the hyperbolic knots in S3, finding the cyclic or finite surgeries (of which there
are at most five for a given knot): and checking if any of them is a counteresample
to the conjecture. Of course, one would still be left to esamine non-hyperbolic knots,
not to mention links.
X more realistic appraisal cornes in the context of another of the great open prob-
lems of 3-dimensional topology, Thurston's hyperbolization conjecture. Roughly
speaking, this conjecture contends that "rnost %manifolds are hyperbolic." For es-
ample, Thurston h'as proved that al1 but a finite number of Dehn surgeries on a
hyperbolic knot result in manifolds which are again hyperbolic. Moreover, the con-
jecture implies t hat the non-hyperbolic ones eit her contain an incompressible sphere
or torus, have cyclic or finite fundamental group, or else are Seifert fibred manifolds
(these outcornes are not disjoint). -4long with Gordon and Luecke's studies of incom-
pressible surfaces, the investigation of cyclic and finite surgeries by Culler, Shalen,
Boyer, and Zhang forms half of a two-pronged attack on Thurston's conjecture.
Culler-Shalen seminorms have played a central role in these investigations. They
were first introduced by Culler and Shalen [CGLS] as part of the proof of the Cyclic
Surgery Theorem. Later, Boyer and Zhang [BZl] estended the idea to the s tud- of
finite surgeries eventually leading to a proof of the Finite Filling Conjecture [BZ4].
1.2. Contributions of this thesis. By focusing on a specific class of knots, the
pretzel knots, we can use the methods of Culler, Shalen, Boyer, and Zhang to obtain
stronger results. Rather than just showing that there are veq- few cyclic or finite
surgeries, Ive can show that in many cases there are none (aside from the trivial
surgery) and give a listing of the knots which admit interesting fillings. By further
restricting to the (-2: 3, n) pretzel knots, we can explicitly calculate the Culler-Shalen
seminorms and use t hat information to give a complete description of the character
variety of these knots. Therefore this thesis rmkes contributions to the understand-
ing of finite and cyclic surgeries of pretzel knots and illustrates how Ciiller-Shalen
serninorms can be used to describe the character variety of a knot.
Delman [Del] showed that if a hyperbolic Montesinos knot admits a finite surgery.
then it is a pretzel knot of the form (2k + 1,21 + lo -2m), k , l , m being positive
integers. Our contribution is a thorough understariding of cyclic and finite surgeries
on these pretzel knots.
In particular, we prove that there are no non-trivial cyclic or finite surgeries on a
(-2,3, n) pretzel knot K unless one of the folloming holds.
0 K is torus, in which case n = 1,3, or 5.
6
n = 7, in which case 18 and 19 are cyclic fillings while 17 is a finite, non-cyclic
filling.
a n- = 9, in which case 22 and 23 are finite, non-cyclic fillings.
A coiisequence is that the ( - 3 , 3 , T ) is the only non-torus (-2. p, q) pretzel knot
(p, q odd, positive integers) which admits a non-trivial cyclic surgery- We show that
a (p, q, 2m) pretzel knot admits a t most one non-trivial finite surgery so long as 1 + + & < 1. For rn < - 1, we show further that (p, q, 2m) pretzels admit no
non- t rivial finit e surgeries.
This c m be combineci with the work of Delman to deduce:
Theorem 4.5.1. The only non- toms Montesinos h o t awhich admits a non-trivial
cyclic surgery zs the (-2,3: 7 ) pretzel knot. The non-trivial cyclic surgeries on thzs
h o t are of slope 18 and 19.
Theorem 4.5.2. If a non- toms Montesinos knot I< adrnits a non-trivial finite surgenj:
then one of the following holds.
K is n (-2? p, q) pretzel knot with 5 < p 5 q odd and the filling is not cyclic.
a A- is the (-2,3: 7 ) pretzel h o t and the filling is along slope 1T,18, or 19.
IC is the (-2,3,9) pretzel h o t and the Jilling is along slope 22 or 23.
For certain pretzel knots we are able, not only to classify finite and cyclic surgeries.
but also to determine the structure of the character variety. \'e begin by ciemon-
strating the inclusion of PSL2(@)-charaçter varieties .y(il(n l . a?: a3)) C -y(fC) mhen
I< = I i ($ i /a i , ,&/a2, f i3 /%) is B three-tangle Montesinos knot and il(crl, a*, a3) the
associatecl triangle group. The (p, q, r) pretzel knots, for mhich 1' = K ( l / p , l / q , l l r ) ,
are important esamples. These knots are small, so the SL2 (C)-character variety -1- ( K )
will consist of algebraic curves. Generally, if K also admits a Seifert surges., we can
proceed to calculate the Culler-Shah seniinorms esplicitly ancl t hereby enunlerate
the curves in S ( K ) . For hyperbolic (-2,3, n) pretzel knots (n an odd integer), ive
deduce that S(K) consists of esactly twvo (respectively three) curves when 3 1 n (re-
spectively 3 1 n). We obtain similar rcsiilts for some (-3.3. n) pretzel knots. (This
is an espanded form of the papers [BXIZ, hiatl; Mat?].)
It should be emphasized that such a detailed description of the character variety of
a knot is rare. It is true that the character varieties of torus knots are well understood
and Burde [Bu] has shown that X ( K ) consists of exactly two curves for certain 2-
bridge knots. However, our description of the character varieties and Culler-Shalen
seminorms of a family of pretzel knots marks substantial progress in t his area.
These seminorm calculations also provide a link wit h the A-polynomial invariant
of [CCGLS]. For example, ive determine the Newton polygon for the A-polynomial
of the (-2,3, n) pretzel knots. This is significant, as it rernains difficult to compute
the A-polynornial. For esample, the (-2,3: 7 ) is the only one of these knots whose
polynomial is included in the table of [CCGLS].
In addition to such concrete contributions, we would like to think of this thesis
as a handbook of computational techniques for use in the understanding of Culler-
Shalen seminorms. For example' we illustrate how Ohtsuki's [Oht] work on ideal
points can be used to deduce the Culler-Shalen seminorms of the twist knots. (This
again leads to informat ion about cyclic and finite surgeries and the- Newton polygon
for the -4-polynomial of these knots.) We explain how to show Z,( fp) = 1 when
x is the character of a p-representation. We discuss in some detail Hatcher and
Oertel's algorithrn for calculating the boundary slopes of Montesinos knots using
pretzel knots as examples. Finally, in the -Appendis, we present calculations of the
zeroes of Alexander polynomials for certain pretzel knots.
1.3. Outline. Following the introductory first chapter, the second chapter presents
a brief overview of the theory of character varieties of knots arid the Ciiller-Shalen
seminorm. To illustrate the power of this theory, we esplicitly calculate the Culler-
Shalen seminorms of the twist knots and show how such a calculation can be used
to make deductions about finite surgeries on those knots. We also construct the
Newton polygon for the A-polynomial of the twist knots. We nest state some prelim-
inary lemmas about commutativity of representations and representat ions of triangle
groups. We complete the second chapter with an argument tliat Z,(f,) = 1 when x
is the character of a prepresentation of a twist knot or a ( -2 ,3 , n) pretzel h o t .
The third chapter introduces Montesinos knots. -4fter outlining the tlieory of
Seifert fibrecl spaces, we prove a theoreni of hIotitesinos wtiich illustrates the close
connection between Seifert spaces and Montesinos knots- This chapter closes with
a proposition showing that representations of the double cover of a knot group can
often he extended to the mhole knot group. It follows that the character variety of
a triangle group includes into the character variety of the corresponding three-tangle
kf ontesinos knot.
This is exploited in the last two chapters where we investigate a specific kind of
three-tangle Montesinos knot, the (p, q, r) pretzel knot. In Ckapter 4 we look at cyclic
and finite surgeries of such knots. We first show that if a (p, q, 2 m ) pretzel knot adrnits
a finite surgert., that surgery is odd integral and near a non-integral boundary dope.
We then present Hatcher and Oertel's algorithm for calculating boundary slopes in
order to refine Our results. We next classify cyclic surgeries on (-2, p, q) pretzel knots
and finite surgeries on ( p , q, -2772) pretzel knots (rn > 1). Combined with Delman's
work: this gives a complete classification ofcyclic surgeries on Montesinos knots (there
are only two non-trivial cyclic surgeries amongst the non-torus Montesinos knots) and
a near complete analysis of finite surgeries.
In Chapter 5; we make explicit calculations of the Culler-Shalen seminorms of
several knots. We first determine the minimum of the total seminorm for (2. pl q )
pretzel knots. CVe theri calculate the Culler-Shalen seminorms of (-2,3, n) pretzel
knots and thereby decluce the structure of the SL2(@)-character variety of those
knots. We also investigate to what estent we can make analogous computütions for
the (-3? 3, n) pretzel knots.
Finally, in Chapter 6 we suggest some ways in which this work coiild be estended
and discuss questions arising from our research. The thesis concludes with an .Ap-
pendis concerning zeroes of the Alexander polynomials of (-2,3, n) pretzel knots.
2.1. Character varieties. Character varieties wi11 play a fundamental role in what
follorvs. CVe give here a brief outline of the important facts for our purposes. The
standard references for SL2(C)-character varieties are the two articles of Culler and
Shalen [CSl, CS21 as well as their part (Chapter 1) of [CGLS]. Boyer and Zhang's
[BZl, Section 21 is also a good reference and [BZ2] develops the theory of PSL2(C)-
character varieties. For a more leisurely introduction, we recommencl Tanguay's
thesis [Ta].
For a finitely generated group ~r; let R = Hom(rr, SL2(@)) denote the set of SL2(@)-
representations. Then R is an affine algebraic set referred to as the representation
uariety. In particular, if 5~ is generated by gi , g*, . . . g,, then the entries of the matrices
p(g*) E SL2(C) can be taken as coordinates. The algebraiç set R is then determined
by the equations det(p(gi)) = 1 along with those arising from any relations amongst
the gi in n. Thus R c Cn. A different choice of generators will result in a set R'
isomorphic to R.
The character variety ,Y is the set of characters of representations in R. It is also
algebraic and generated by
(see [GhcI, Corollary 4-12] and [VI). If we let t : R + Clcl denote the function which
takes p to ( ~ ~ ( g ) ) ~ ~ ~ ? then the image t (R) is isomorphic to S. Note that IGI; the
cardinality of G, is n(n2 + 5 ) / 6 .
The focus of this thesis is the case where T = ï r L ( M ) is the fundamental group of
the complement M of a h o t K in S3. That is, Ii is the image of a smooth enibedding
of S' in 9 and M = S3 \ !V(IC), N ( K ) being an open tubular neighbourhood of the
h o t . We will be especiallÿ interested in the case mhere 1 ' is hyperbolic. This means
&, the interior of -41, admits a hyperbolic metric. In other words, ;{f= w/T where
is hyperbolic 3-space and r C PSL2(C) = Isom+(m), the group of orientation
preserving isometries of .
In this context, r "- rr so that ï provides a natural PSL2(C)-representation, the
holonomy representation, Po : n + PSL2(@). Thurston has shown (see [CS1 , Proposi-
tion 3.1. l] for a proof) that Po lifts to an irreducible SL2(C)-representation po. More-
over, the irreducible component -Yo c -': containing X, is a curue, Le.. di-(&) = 1
[CGLS, Proposition 1-1-11. FVe refer to -Yo as the canonical cume.
The set of characters of reducible representations d so forms a cuwe n-hich is iso-
morphic to @. (See for esample [Ta, Proposition 2-5-51. This is t m e tvheneyer 111
is the complement of a knot in S3, hyperbolic or not.) Moreover; when the knot is
small, any irreducible component of .Y containing the character of an irreducible r e p
resentation is again a curve. (A knot is small provided there are no closed essential
surfaces in M. -4 surface is essential if it is incompressible, orientable, and properiy
embedded in 1b.I such that no component is parallel into &CI and no S2 cornponent
bounds a ball.)
For such a curve Xi c X, let dc denote the smooth projective variety birationally
equivalent to Xi. The birational equivalence is regular a t al1 but a finite number of
points of Si called ideal points. -4s in [CGLS, Section 1.~51~ the complernent of the
ideal points in ,Ti rnay be identified with .Y," mhere u : -t -Yi is norrnalization [Shf.
Chapter II $51.
We will have occasion to use PSL2 (@)-character varieties: part icularly when me
investigate r-curves (see Section 2.2.3). We mil1 generally use a bar to denote PSL2(C)
versions of familiar objects. Thus R is a PSL2(C)-representation 1-ariety ancl .y a
PSL2(@)-character variety The theory of these objects is similar to the SL2(@)
theory and ive refer the reader to [BZ2, Sections 3 and 41 for details.
Each ideal point of .ci can be used to construct a sirnplicial tree on whicti ir acts
non-trivially (see [CGLS, Section 1.21). This gives a splitting of the group ii as the
fundamental group of a graph of groups. This in turn yields an incompressible surface
in the knot exterior M. In this svay ideal points are associated with incompressible
surfaces in 1;C.I ancl vice versa.
2.2. Culler-Shalen seminorms. Ciiller-Shalen seminornis were introduced in the
first chapter of [CGLS]. The authors show that the canonical cilme So c S (sce
Section 2.1) can be used to constriict a norni on surgery space b- = H,(ai\l$) and
11
this norm plays a central role in their proof of the Cyclic Surgery Theorem- Boyer
and Zhang [BZl; BZ2] recognized that this construction can be extended to any curve
in S containing the character of an irreducible representation. However, in general
the result is not a norm, but rather a seminorm. These seminornis, which we will
cal1 Culler-Shalen seminorms, again figure prominently in Boyer and Zliang7s proof
of the Finite Filling Conjecture [BZ4]. Moreover, one of the themes of this thesis is
that these seminorms are also of great use in understanding Seifert surgeries-
2.2.1. Dehn Surgery. Before coming to the seminorm construction, ive briefly intro-
duce Dehn s u r g e . The classical reference here is Rolfsen's book [Roll. We also
recommend Boyer's survey [B2].
Let AI = S3 \ N ( K ) mhere iV(K) is an open tubular neighbourhood of a knot
K in S3. A Dehn surgery on I< (or a Dehn filling of M) is the closed manifold
M ( a ) = M u, S1 x D2 obtained by attaching a solid torus, S1 x D2, and M dong
their boundary ton, T2. This glueing is determined by the image a c aM of a
meridional disc {pt) x aD2.
To be precise, the surgery is determined by the isotopy class of a in d k f . There
is a standard choice of basis {p, A } of H1(aiLf; Z) where p is the class of a meridian
of the knot (the boundary of a disc transverse t,o the h o t ) ancl X is the class of a
preferred longitude (a simple closed curve in dlL1 which intersects a meridian once
transversally, and whose class in homology HL (d l ) is zero) [Rol, Section TE]. Wïth
respect to this basis. a. represents the classes * (ap + bX) (depending on orientation).
W e will sometimes (by abuse of notation) write CI = (a, 6 ) . The ambigiiity in sign
can be removed by taking the ratio 9. Indeed, since the isotopy class of a can be
identified (canonically, see [B2, Proposition 2-41) with E QU {i}, ive will frequently
confuse a with its isotopy class and f . Using the basis {p, X}, lines of slope in
the universal cover, W x R of aM = T2 = SL x S L , project ont0 curves of class
a in Bhf . For this reason, ive will often refer to a (and other primitive classes in
HI ( B M ; Z)) as a "slope." Finally. taking advantage of the Hurewicz isomorphism,
HI (alif; Z) i HL ( a M ) S ZeZ, we will have occasion to think of a or % as a tiomotopy
class although this identification is only valid up to conjugation.
If srl(M(a)) is cyclic, we wîll say a is a cyclic surgery. The simplest esample is
trivial surgery: A I ( $ ) = h.1 ( p ) = S3. On the other hand, the only way to obtain
r l ( M ( a ) ) - Z is through surgery on a trivial h o t , Le., the standard, unknotted,
embedding of S1 in S3 [Gall. So cyclic surgeries will generally be examples of finite
surgeries, Le., surgeries for which rrl(M(û.)) is finite. A third important category,
from our point of view, is the case where M ( a ) is a Seifert fibred space. (We twill
discuss this case in more detail in Section 2.2.4 below.)
2-3-2. Construction of Culler-Shalen seminorms. Let us now describe the construc-
tion of a Culler-Shalen seminorm given a curve Xi c .Y. For E s = iri (AI), clefine
the regular function 1, : S + @ by I, (x,) = ~ ~ ( 7 ) = By the Hurewicz
isomorphism, a class ni E L = Hi (aM; Z) determines an element of (BAI): and
therefore an element of r; well-defined up t o conjugacy The function f, = 1: - 4
is again regular and so can be pulled back to ,fi, the smooth projective variety bi-
rationally equivalent to -Xi- For y E L, l l ~ l l ~ is the degree of f , : -Yi + C'PL. The
seminorm is extended to V = XL(Bh.I; R) by linearity.
The power of Culler-Shalen seminorms is perhaps best illustrated by the fol-
lowing two theorems which relate them t o cyclic and finite surgeries. Here si =
min{llnflli ; E L. llrlli > O} denotes the minimal n o m and a boundarg dope ,O E
Wl(dM; Z) is the class of the boundary of an essential surface properly embedded in
M.
Theorem 2.2.1 (Corollary 1 - 1 4 [CGLS]). If a is not a bovndary slope and xi (M ( a ) )
zs C ~ / C ~ Z C , then llalli = Si.
Theorern 2.2.2 (Theorem 2.3 [BZl]). I f a is not a boundary slope and al(Al(a)) is
finite, then llalli < TTMZX(%~, Si + 8).
Indeed, boundary slopes play a fundamental role in the theory of Culler-Shalen
seminorms. Let A(?, /3) denote the minimal geometric iritersection number of curves
represeiiting the classes y and ,fi so that A(%; 2) = lad - bcl. In the context of a knot
in S3 for which p is not a boundary slope, Lemma 6.2 of [BZl] can be rearranged to
Say:
Lemma 2.2.3 (Lemma 6.2 [BZl]).
where the ai are non-negative integers and the s u n is over the set of bovndary dopes
Roughly speaking the ai- count ideal points OF Avi associated to Bj (see Section 2.1).
In particular, if there are no such, then ai = 0.
22.3. N o m curves and r-curves. Lernma 3.23, provides a practical way to describe
the clifference between norm curves and r-curves. .A n o m curve Si is one on which
no f, is constant, (1 # y E r I ( B M ) ) - In this case [ I - I l i is a norm (rather than just a
seminorrn). In terms of Lemma 2.2.3, this means a t least two of the a; are non-zero.
An important example of a norm curve is the canonical curve, ,Yo, in the case of a
hyperbolic knot.
If Si c S is noi a norm curve (and X is the character variety of the complement
of a small knot in S3 [BZ2; Section 5 ] ) , then there is a boundary slope r such that f,
is constant on Xi and we will refer to as an r-curve. In terms of Lernma 2.2.3, this
rneans that a; = O if fij # r and [ ( A / ( ( ~ = siA(-/7 r). In particular, since 111 ($) = S3, ive
have I l I l i = si (Theorem 2.2.1. Note that % is not a boundary slope by Lemma 2.3.1.)
This means r is distance 1 from 6 and is therefore an integer. If is srnaIl; then any
r-curve ,\; includes into the PSL2(C)-character variety .T(M(T)) (see [BZ2. Esarnple
5 . IO]).
-4s ive mentioned in Section 2.1, K small aIso implies eaçh component Si of S is a
curve. Therefore each component leads t o a Culler-Shalen seminorm. In this contest,
wve will often want to look at the total norm 11 I l T = Ci I I I l i and its corresponcling
minimal value S = Ci si, the sums being taken over the components Si containing
irreducible characters. If M is hyperbolic, this total norm is a norm (and not just a
seminorm) since one of the components is the canonical c u n e So (for which I I - I l o is
a norm).
2.2.4. Seifert szlrgery. If M(o) is a Seifert fihred space, we will say that a is a Seifert
surgery slope. Seifert fibred spaces will be described in more detail in Section 3.1.1.
14
Briefly, when the base orbifold is SZ, such a space is a surgecy on S2 x S1 which
involves the removal of n solid tori Ti which are then reattached using a homeomor-
phism of mi. Just as in the case of Dehn surgery, we can describe the attaching
homeomorphisms by fractions 2 E Q U {$). In case n < 3, we will say that the
resulting manifold is a small Seifert space.
-4ccorcling to Thurston's hyperbolization conjecture, Dehn surgery on a hyperbolic
knot in S3 \vil1 result in a manifold which is either hyperbolic, has finite or cyclic
fundamental group, contains an essential toms or sphere: or is small Seifert (these
outcomes are not disjoint). Moreover, there are only a finite number of surgeries
in Q U {$ } which can result in a non-hyperbolic manifold [Thu]. These are called
exceptional surgeries.
There are two important approaches to the study of exceptional surgeries. Surg-
eries producing essential surfaces are amenable to investigation via the t heory of
intersection graphs developed by Gordon and Luecke (see [Gol] for a survey). On
the other hand, Our discussion of Culler-S halen seminorrns showed Iiow the techniques
of Culler. Shalen, Boyer, and Zhang allow us to understand cyclic and finite surgeries.
-An important theme of this thesis is tha t Seifert surgeries can also be studied using
Culler-Shalen seminorms. In particular, when LW (a) is a small Seifert surgex-y we can
often show ]laIli 5 Si + C where C is a constant depending only on the surgery
coefficients b l . b2, b3 (compare Theorems 2.2.1 and 2-22). This idea mil1 be developed
more explicitly in Section 5.2 (see also [BB]) where ive calculate the Culler-Shah
seminorms of small Seifert fillings of the (-2,3, n) pretzel h o t .
2.2.5. Fundamental and Newton polygons. Given a Culler-Shalen seminorm II - I I aris-
ing fmm a curve in the character variety of a knot K , rtre cal1 B7 the norrn-disc of
radius s in I' = RI ( â M ; IR), a fundamental polygon for K. When ( 1 . I I is a norm
(rather than just a seminorm), the polygon B is compact, conves. and finite-sided
with vertices which are rational multiples of boundary slopes in L = H l ( d M ; Z). It
is syrnmetric (-B = B) and centred a t (O: O). I t also provides a nice way to visualize
finite and cyclic surgeries. Any cyclic surgeries will lie on i3B while finite surgeries
must lie within '2B (assuming s > 8 and neglecting cyclic or finite siirgeries along
boundary slopes, see Theorems 2.2.1 and 2 . 2 2 ) . Recall that the only Z surgery of a
15
knot in S3 is on the trivial knot ([Gall). so the cyclic surgeries on aB d l generally
be evamples of finite surgeries.
The fundamental polygon also provides a direct connect ion wit h the -4-polynomial
invariant of [CCGLS]. This is a polynomial in Z[l,m]: -4 = '&j, bï,limj. The
Newton polggon of such a polynornial in two variables is ~ h e conves hull in IR2 of
{ ( 2 , A lbi7j # 0)- Let N denote the Newton polygon of the A-polynomial for a hyperbolic knot
K. Let Bo be the fundamental polygon of the Culler-Shalen seminorm associated
to the canonical curve ,Yo C S (i.e. Xo contains the character of the holonorny
representation). Boyer and Zhang have shown that these polygons are dual in the
following sense-
Theorem 2.2.4 (Theorern 1.4 of [BZ4]). The line through any pair of antipodal uer-
tices of Bo is parallel to a side of AT. Conversely, the line through any pair of antzpodal
vertices of AT is paralle1 to a side of Bo.
Thus given the fundamental polygon Bo, one can deduce iV, at least up to scaling
and translation. As it remains difficult to calculate the -4-polynomial, this is morth-
while and ive include diagrams of several Newton polygons in this thesis (Figures 4?
5, 26, 27 and 30). The conventions Ive use are that iV meets the l and m axes but
lies in the first quadrant. The scale is provided by Shanahan's width function:
Definition 2.2.5 (Definition 1.2 of [Shn]). The p / q width w(p/q) of iV is one less
than the number of lines of dope p/q which intersect !V and contain a point of the
integer lattice.
We require t hat Shanahan's widt h correspond to the canonical Culler-S halen semi-
norm: w ( p / q ) = IIq/pllo. Given an expression for I I - I l o as in Lemma 2.2.3. we can
quickly find fV as is illustrated by the esample of the twist knots which follows.
2.3. Polygons of the twist knots. In Chapter 5 ive will niake a rather detaileci
calculation of the Culler-Shalen seminorms of some pretzel kiiots. -1s a prelude, and to
illustrate the power of the serninorm approach, we investigate twist knots. .1lthough
these can be treated in much the same fashion as the pretzel knots (see [BMZ]), they
have some special features mhich allow a more direct approach, which we aciopt here.
FIGURE 1. The twist h o t Kn.
Figure 1 shows the twist knot h' = Kn. Since the trivial knot & and the trefoil
KI are not hyperbolic, we will assume n # 0, 1. Burde [Bu? Section 31 has shown that
within the character varïety -1- of a twist knot there is a unique cume dyo containing
irreducible characters (which is t herefore the canonical curve) .
This allows us to determine s directly for this curve as it is simply related to the
number of dihedral representations. One can show (see [Ta, Section 5-31) that the
class (2? 0) E Hl(diCI; Z) has norm s+2d where d is the nurnber of dihedral characters
of K . Hence s = 11(1,O)II = f11(2,0)II = ?s + d and thus s = 2d.
The number d is equal to (H - 1)/2 (see for esample [Kl, Theorem 101) where
H = card(Hl(CÎ)), C2 being the second branched cyclic cover of the h o t ([Rol,
Section 10.C]). Since card(H&)) = IA(-l)I ([Rol, Corollary 8.D.31): we see that s
may be derived from the Alexander polynomial, whicli is A(t) = nt2 f (1 - 2n)t + IL in the case of the twist knot k<n ([Rol, Exercise 7.B.71):
41n1, if n 5 -1
4n - 2 : i f n > 2
Twist knots are exarnples of '>-bridge knots. These kiiots are sometimes callecl
rational knots as each has an associatecl rational nitmber. For Kn, the associatecl
number is =- Hatcher and Thurston [HT] have shown that the incompressible
surfaces of such a knot are given by continued fraction expansions of the associatecl
rational number,
We require Ib,l 2 2 and denote such a continued fraction by [b,, - . . : b , ~ ] or simply
[b, 1 - There are three expansions of -&- The first two we can write imniediately:
I - - = [Zn - 1,2]. 2 n - 1 + ~
The third depends on the sign of n. If n is positive we have
1-2: 2: -2,2,. . . - -2.2. , , -31 \ #
I
n-L pairs
with n - 1 pairs -2,2, while for n negatit-e,
12, -2, 2. -2'2,. . . ' 2, -2,2, -31 *
~ n l - L pairs
with In1 - 1 pairs 2, -2. For each continued fraction: let b f = caïd({(-l)j+'bj > 0))
and b- = card({(-l)jilbj < O}).
There is a unique continued fraction n-hose entries bj are al1 even. It corresponds
to the longitude of the h o t . Let 6: and b; be the bC. b- of this surface. For the
twist knot, it is the first continued fraction [2n, -21 which has al1 entries even, so
b,f = 2, b; = O when n > O and b i = 1 = 6; when n < 0.
The boundary slope for the surface associated to [b,] can non- be found through
comparison with the longitude. It's given as
Thus, wlien n > 0, the second continued fraction has b7 = 1 = b- and the bounctary
slope is 2[(1- 1) - (2 - O)] = -4. The third boundary slope is 2[(0 - (Zn - 1)) - 21 =
-(4n + 2). When n < O, we have the boundary slopes -4 and -472.
Ohtsuki [Oht] shows how to determine the Culler-Shalen seminorm given these
continued fraction espansions. His strategy is to determine esplicitly the number of
ideal points for each incompressible surface. He shows that there are $ I I j (lbj 1 - 1)
18
' This means S U C ~ unless al1 the nj are even in which case there are f nj(lbjl - 1) - 5-
that, up to a constant k-, the Culler-Shalen seminorm is given by
the sum being taken over the possible continued fraction espansions of the rational
nurnber associated to the h o t . (This is almost the formula of [Oht. Proposition 3-21.
Unfortunatelx after correctly calculating the number of ideal points on the previous
page, Ohtsuki has neglected to subtract $ J p l in stating the Proposition.)
Remark: Note that there is no ideal point associated to a surface whose contin-
ued fraction has al1 Ibj/ = 2. These are preciseiy the surfaces which are the fibres
of a fibration of the knot over S1 (see the remark following the Corollary to [HI.
Proposition 11). In other words, for a fibred Zbridge h o t , tliere are no icleal points
associated to a surface which is a leaf in the fibration. This is true more generally
(see for esample [CL]). However, such a boundary slope may be represented by other
surfaces which are not leaves in a fibration. An esample of this is the (-3: 3: -3)
pretzel knot, which admits a fibration with a Seifert surface IeaF. In spite of
this, as we shall see in Section 5.2, there are ideal points associated to the Longitude
because there are other surfaces which are not leaves of a fibration. but nonetheless
have the longitude as boundary slope.
Using Ohtsuki's formula when n > O, ive have
while for n < 0,
Given that ]11/011 = -5 (since M(i/O) = S3 has cyclic fundamental group and 1/0 is
not a boundary slope) we can use the calculation of s above (Eqiiation 2.1) to see
that k = 2-
FIGURE 2. Fundamental polygon of the twist knot ICn (n > 1).
FIGURE 3. Fundamental polygon of the twist knot IL (n < 0).
The fundamental polygon B OF the twist knot & is illustrated in Figures 2 ( r i > 1)
and 3 (n < O ) . In particular, note that the fundamental polygon lies below the line
9 = 1/2. This shows that thcse knots admit no rion-trivial cyclic or finite surgeries
20
as such surgeries would either lie within 2B (assuming n # 2 so that s 2 8, [BZl,
Theorem 2-31) or else occur at a boundary slope. Since 2B includes no integer lattice
points (besides (a, O), -2 5 a 5 2), the first possibility does not arise. On the other
hand, since twist knots are small, the boundary slopes are also not cyclic or finite:
Lemma 2.3.1. If kf is srnall and a! is a boundary slope, then M(a) zs not cyclic o r
finite.
Proof: By [CGLS, Theorem 2.0.31, M ( a ) is not finite, and it is cyclic only if M ( a )
S1 x S2. However: Gabai [Gall has shown that, amongst knots in S3, only slope O
s u r g e - on the trivial h o t can produce S1 x S2. O
For h;, s = 6 so that max(2.s: s + 8) = s + 8 = 14 (see Theorem 2.2.2). Here, the
highest point of B has y = s/4(s - 1) = 7/26. So, in this case as well, $B includes no
integer lattice points other than (a, O)? and therefore K2 also admits no non-trivial
finite surgeries.
Given the Culler-Shalen serninorm calculations, as above, we can also immediately
deduce the Newton polygon. For esample: for n > 0; we see that the edges of dope
O and 5 rnust have Yength" n - 1. while the edge of dope has only length
one. In other words, the vertical segments corresponding to the bounclary slope O
have length n - 1 while a segment of dope -114, like that from (O. n) to (4(n - 1): 1 )
is (n - 1) times the vector (4, -1):
( O . n) + (n - 1)(4: -1) = (O, n) + (4(n - 1 - n) = (4(n - 1); 1 ) -
Figures 4 (n > 1) and 5. (n < O ) illustrate the Newton polygons for these knots.
2-4. Preliminary lemmas. Our goal is to develop the theory of SL2(C)-character
varieties of Montesinos knots, particularly (p, q, r ) pretzel knots. In this section we
present several useful lemmas about the structure of SL2(@)- and PSL-(C)-character
variet ies.
2.4.1-
tivity.
Commutativity of representations. The first set of lemmas
Recall that -4 E SI,;?(@) is parabolic if it is conjiigate to a
with a # O. *1
relate to commuta-
matris of the form
FIGURE 4. Xewton polygon of the twist knot fi-, (n > 1).
FIGURE 5 . xewton polygon of the twist h o t & (n < 0 ) .
Lemma 2.4.1. -4; B E SL2(@) \ {f I } commute zff they are both diagonal or both
pambolic (afler a n appropriate conjugation).
Proof: Conjugate so that -4 = ( O ,;a ) Let B = ( 1 : ) . Thrn
Suppose A is diagonal. Then fl = O and, since -4 # f 1' a' # 1. The lower left
entries of AB and B A imply c = O- Similarllv, since f i = 0, the upper right entries of
-4B ancl B A imply 6 = O ancl so B is diagonal.
22
Suppose that -4 is parabolic, Le., a = +1 and P # O. The Iomer right entry of AB
and BA then implies c = O while the upper right entry obliges a = d. It follorvs that
B is also parabolic. I I
Remark: -4n irreducible SL2(@)-representation is not abelian. However: a reducible
- SL2(@)-representation could be non-abelian. For esample p : 2 / 3 * L/3 -+ SL2(@)
defined by taking the generators to ( O ) m d ( ' l ) where < is a prirn- 0 c- O c-l
itive third root of unity. On the other hand, the chara& x,' OC a reducible SL2(C)-
representation p is always the character X, of a diagonal (hence abelian) character
po. Simply replace each mat ri^ p(g) = O l /n O l / a
notice that any relations which the p(g) satisfy will&o hold for
Definition 2.4.2. We ,will say of a rnatrix .il = d= ( 1 d ) t PSL2(C) thnt il
is diagonal (or parabolic, etc.) if the corresponding rnatiix ( r d ) E SL2(C) is
diagonal (parabolic etc.). If A E PSL2(@) zs of the fonn f -;) we will
say that -4 is antidiagonal. We denote by N the set of diagonal and antidiapnal
matrices in PSL2 (C) :
Lemma 2.4.3. Suppose -4, B E PSL2(C) \ {f l} cornmute. Then, afier an appropri-
ate conjugation, one of the following vrill obtain.
1. -4 and B are both diagonal.
2. -4 and B are both parabolic.
3- -4 E iV and B = E (or vice versa).
Proof: Similar to proof of Lemma 2.4.1.
23
Definition 2.4.4 (18221). A representation fi : G + PS&(@) is called irreducible if it is not conjuyate to a representation whose image lies in
Otheriuise i t zs called reducible.
Lemma 2.4.5. Let p : G + PS&(C) be an irreducible representation. I'J p(G) is
abelian, then P(G) 2 2 / 2 @ 2 / 2 .
Proof: Since p is irreducible, 39 E G 3 p(g) # f 1. Conjugate so that ~ ( g ) =
4 = 3~ ( ' ) . If u = tl and ,B + O . the only matricer rhich cornmute with O l/cu
A are ot-her parabo~ics and P is reducible. If a # f 1. we can conjugate so that
~ ( g ) is diagonal. So without loss of generality? we can assume ,8 = O. If cr # f i .
the only matrices which commute with -4 are again diagonal and P(G) is reducible.
Therefore a = *i and A = E. By the previous lemma, the remaining elements of
P(G) are in N . Since p is irreducible. there is a t l e s t one antidiagonal element:
So: if B and C commute: then b2 = kt?. Thus' either B = C : or else C = B.4 = BE,
and these are the only antidiagonal elements of P(G). Since id and E are the onIy
diagonal elements which commute with B: we see that P(G) = { k I , E: B' B E )
2 / 2 @ 2/2. O
Lemma 2.4.6. Let p : G -, PSL2(C) 6e an irreducible o r non-abelian representation.
Suppose A is in the center ZpSL,(q (fi(G)) = {B E PS&(C)IBC = CB V C E p(G)}-
Then, either
Proof: -4pply Lemma 2.4.3.
24
Lemma 2.4.7. Let p : G + PS& (C) be an irreducible o r non-abelzan representation.
Let A E &(C) and suppose A = Ad&) - -4 = ~ ( ~ ) - ' . @ ( g ) for each g E G . Then
-4 = o.
Proof: -4s p is irreducible or non-abelian, ive can find elements g1,fi E G so that,
after a conjugation, P(gL) # I I is upper triangular while P(g2) has non-zero lower
e t e t 4s 4 = ( : b ) cornmutes with p(g,); we ree t h r t c = O B u t then,
in order for -4 to commute with p(g2) as well, a and 6 must also be zero. O
2-42. The triangle group. The triangle groups A ( p . q , r ) are intimately related to the
(p, q, r) pretzel knots which are the main focus of Chapters 4 and 5-
Definition 2.4.8. The triangle group A ( p , q: r) has presentatzon
9 ( p , q, r ) = (x, 9, zlxP, gq. z'? XYZ)
= (x, 7JlxP, yq, ( x ~ J ) ~ ) .
In particular, we will need t o count the characters of A ( p , q, r ) on several occasions.
By [BB, Proposition DI, the number of PSL2(@)-characters of A ( p , q, r) is
where 1x1 denotes the largest integer less than or equal to x. This count inclucles
chüracters of reducible representations. The character of a reducible representation
is also the cliaracter of a diagonal (hence abelian) representation (see the Remark of
Section '2.4.1). So, to count the characters of reducible representatioris we can look
a t representations of H, (A(p ,q , r ) ) . Let a = gcd(p.q,r), b = gcd(pq.pr,qr). Then
H l (A(p, q, r ) ) = Z / n @ Z/(6/a) while IHi (h(p, q, T ) ) 1 = 6 . Consequently. the number
of characters of H l ( A ( p , q, r ) ) is
+ 1 , if a = 1 (mod 2),
~ f j + 2 , i f a ~ 0 (niod2).
Lemma 2.4.9. Let G = A ( p , q, r) with p and q odd. If p zs an irreduci6le PSL2(C)-
representation of G and -4 E ZpSL,(c)(jj(G)), then A = +I.
25
Proof: Suppose instead that A is conjugate to E (see Lemma 2.4.6). Conjugate
so that p(G) c N . Since the set of diagonal matrices is closed under multiplication
and p is irreducible, a t least one of the generators is taken to an antidiagonal matris.
However, antidiagonal matrices are of order 2 in PSL2(C), so ive have a contradiction
if r is odd. If r is even, the corresponding gerierator is mapped to an antidiagonal
matrix while the other two generators are mapped to diagonal matrices. This is a
contradiction since the product of the three generators is the identit . O.
Lemma 2.4.10. If p and q are odd, then no irreducible PSL2(@)-representation of
A ( p , q, r ) has jnite dihedral image.
Proof: Suppose instead that P : A ( p , ql r ) -+ PSL2(C) has dihedral image D. If D - 2/2@2/2: then p(x2) = f I and ~(y') = f 1: x and y being the generators of A ( p , g , r)
(Defini tion 2.4.8). Since these also have odd order p and q respectivel_v, it follows that
~ ( x ) = f 1 and p(y) = f 1 and the representations is trivial, a contradiction.
Thus, we can assume the indes two cyclic subgroup C of D has order at least 3. It
bllows that, after an appropriate conjugation, C consists of diagonal matrices (use
Lemma 2.4.3 and the infinite order of non-identity parabolics). -4s P is irreducible,
p ( x ) . Say: is not diagonal. Then ~ ( x ) E D \ C has order two. -4s above. this implies
~ ( x ) = H, a contradiction. O
2.5. Riley's p-representations. -4s we shall see. an effective way of calculating the
Culler-Shalen seminorm of a slope cu is through cornparison with the meridian p and
we will encounter ecluations of the form
where Z,(f,) denotes the zero off, a t a point x in the character variety (for esample,
see Equation 5.12). Boyer and Ben Abdelghani [BB, Theorem A] have recently shown
that, subject to some mild conditions, the "jump" Z,(f,) - Z,(fp) is 2.
Prior to this work, jump calculations broke clorvn into two cases. Let p : sr,(dl) + SL2(@) he a representation with character x. If p ( s i i ( B M ) ) is not parabolic, argu-
ments like those in [BZl , Section 41 can often be applied to show Z,( f,,) = O and
Z,( f a ) = 2 resiilting in a jump of two. IF p ( ~ ~ ( t 3 M ) ) is parabolic (see Section '2.4.1
for a definition), then following Riley [Ri], we Say tha t p is a prepresentation. We
were able to show Z,(f') = 1 for the character of a prepresentation in two rat her dis-
parate situations: the twist knots and the (-2,3, n) pretzel knots. Given Z,(f,) = 1
one can again apply the techniques of [BZl, Section 41 to see that Zr( fa) = 3 and
the jump is two.
It may well be that Z,(f,) is always one a t the character x of a prepresentation.
as we know of no evidence to the contrary. W e will therefore include Our calculations
for t hese two families of knots as eviclence supporting t his conjectiire.
2-5.1. The twist h o t s . Vie introduced the twist knots Kn in Section 2.3. -+pin-
we assume n # 0.1 so that hi, is hyperbolic. Let il1 = S3 \ iV(Kn) denote the
cornplement of such a knot. Burde [Bu; Section 31 has shown that within the SLz(@)-
character variety of a twist knot there is a unique curve Xo containing the characters
of irreducible representations.
Proposition 2.5.1. T h e characters of the irreducible p-representations of a hyper-
bolic twist knot I\n are smooth points of So. Furthemore Z,( fp) = 1 at the character
x of any such representation.
Proof: The minimal norm s on is twice the nurnber of dihedral characters d
(Equation 2.1). On the other hand: Ritey [Ri] shows that irreducible paraboliç char-
acters x 114th x(p) = 2 correspond to the roots of a polynomial of degree (1. Moreover,
he argues ([Ri, Theorem 31) that the polynomial has no repeated roots. Le., there are
d such points in -Yo. Similarly there are d points with x(p ) = -3 and s in all. Let us
label them x i , . . . .x, and let ü l ( x i ) = { y i l , . . . , giki)? i = 1.. . S' mhere Y : .Y; + -yo is normalization [Shf. Chapter II $51. Then IIpII is equal to the sum of clegrees a t
these points:
Since Zy,,(fp) 2 1, w e must have ki = 1, (i = 1 .... 9) and.sett ing yi = yi,:
Z,, ( f,l) = 1- If Xi were sinpular. then Z,, (f,) > 2 ([Ta? Lemme 5-42]) . We conclude
t herefore tliat each ri is a smooth point of .Yo and 2,- I , , , ~ ( fp) = 1. O
1 n crossings
FIGURE 6 . The (-2,3, n ) pretzel h o t .
2 - 5 2 . (-2; 3, n ) pretzel knots. Figure 6 shows the (-2; 3: n) pretzel knot hn. We will
look a t this knot in detail in Section 5-2, If n is even, K, is a link. Also. ICI: Ii-3,
and K5 are torus knots and therefore not hyperbolic. So, we will assume that n is
an odd integer, n # 1,3? 5. Let M = S3 \ N ( I i , ) clenote the knot exterior and S its
SL2 (@)-charac ter varie ty.
Proposition 2.5.2. The characters of the irreducible p-representations of the com-
plement M of a hgperbolïc (-2,3, n ) pretzel knot are simple points of -Y. Furthermore
Zr( f,) = 1 at the character x of any such representation.
Proofi Starting from the Wirtinger presentation [Rol, Section 3.D], we can show
generators f: g and h are as indicated in Figure 6 ( c ~ m p a r e [Tr]). Let x = X, he the - character of an irreclucible prepresentation with x(p) = 2. Following Riley, ive c m
assume
Then the relation h f hg = f hg f implies u - = 311. hloreover a representation
with u - u = -1 can be replaced by one with u - 'L' = 1 simply by chang-ing the
signs of both ,u and v. -4s these two representations will have the same character.
we can assume u - ,u = 1. -ifter the substitution u = u + 1, the upper left entry of
p(hfhg)-p(fhgf) becomes u 2 ( u w - ( o + 1 ) ( ~ + 2 ) ) . So u = O or w = (v+ l)(u+Z)/u.
If u = 0; the secorid relation of T implies ttiat the characters of prepresentations
correspond to the distinct roots of a polynomial of degree (In1 - 1)/2 (compare [Ri.
Theorem 31).
If .v # O the same relation implies that u is the root of a polynomial p,(v) of degree
In - 31 - 1. These are the polynomials described in Lemmas 2.5.3 and 2.5.4 below
where it is shown that each p, has distinct roots. Thus. there are In - 31 - 1 distinct
characters with v # 0.
Moreover, the characters mith v = O differ from those with u # O as may be seen
by examining trace(p( f g ) ) = 2 - (u + l)?. If v = 0' the trace is 1. To have the
same trace with v # O, we must choose v = -2. This irnplies w = O and p(g) and
p ( h ) cornmute. The second relation of srl(Kn) then simplifies to g f = f h and it is
easy to verify that the choice u = -2, w = O and IL = o' + I = -1 is not consistent
with p(g f ) = p( f h). Thus there is no chûracter mith v # O equal to a character with
.t. = o. So? in al1 there are 3(1n - 21 - 1 ) / 2 irreducibie parabolic characters x with x(p) = 2.
Similarly there are 3(ln - 21 - 1)/2 points with x ( p ) = -2 and S in all. (We will
see in Section 5.2 that S = 3(ln - 21 - l ) , where S is the total minimal norrn.) Let
us label the points XI:.. . :xs and let Uu ,iE, u,-'(xi) = {w~,. - . : y ikz} : i = 1 - - . S>
where vj : Xy + Xj is normalization. Shen llpilT is equal to the sum of degrees a t
t hese points:
Since Z,,, (f,) 2 1, ive must have ki = 1, (i = 1 . . . S) and, setting yi = Y~/, I :
Z,,(f,) = 1. If xi were singular. then Z,,(f,) 2 2 [Ta. Lemme 5.4.21. \.Ve conclude
t herefore t hat each xi is a simple point of .Y and 2,- 1 ( f , ) = 1. O
Lemma 2.5.3. Let p, be the polgnornia1.s of degree 2 - n dejined recursively 6 3 the
equations
where n 5 -1 zs odd. T h e n p, has distinct roots.
Proof: (1 a m indebted to Richard Stong [St] for this argument.) First note that p,
has leacling term (- l)("+ ')'2v2-7' and constant term pn (O) = (-2)-("+ ')/-. Therefore
the roots of p, are al1 algebraic integers ancl .u = O is never a root. Since v = O
29
is not a root: the roots of p, are the same as the roots of the Laurent polynomial
L( 1 - n ) / 2 = u ( ~ - ~ ) / ~ ~ ~ - Notice that for n 2 0,
Lo(.u) = -u - 1 + I /v .
L l ( v ) = u2 + 2 u + l + 1/u7 and
O = Ln + ( u t 1 +2/2.)L,-, +Ln+
Letting y = -(c + 1 + 2/u) /2 we see that Ln satisfies the Tchebyshev recursion:
Ln - 2yLn-i + Ln-2 = 0-
T h e standard solutions to this recursion are the Tchebyshev polynomials T,(y) = sin(nt) cos(nt) and P,(y) = where cost = y. It is well-known that T.(y) and P,&)
are polynornials in IJ? To(y) = 1 , Tl(./) = g' Po(y) = 0' P L ( y ) = 1' and both s a t i s -
the same recursion as L. Therefore,
where cost = y = -(v + 1 + 2/v)/2,
47.7) = ( L O ( Û . ) ~ + (Li (o.) - y ~ ( l ( u ) ) 2 / sin2 t ) L/2
and B ( v ) is given by the formulas
cos B ( v ) = Lo(v) /A(.u)
s inB(v ) = - ( L L ( v ) - y L o ( t ~ ) ) / ( . 4 ( ~ ) s i n t ) .
Note that this formula requires special interpretation i f 9 = 1 or - 1 . In this
case sin t = O and Pn requires a limit to define. However, in our case: y = 1 means
c2+3u+2 = Oso v = - 1 or -2, and y = -1 means v 2 - v + 2 = O so u = ( l f i&) /2 .
One easily checks that:
Ln(+) = 1/2: and
Thus none of these four values of ,u is ever a root of any of the p, or Ln and we need
not w o r l about this special case.
Returning to the discussion above, for al1 other v we have Ln(.ü) = 4 v ) cos(nt(v) + B(v)) and A(E) is nonzero. Therefore roots of Ln occur esactly for zeroes of the cosine.
Now we look for a double root. We have
L',(u) = Af(o) cos(nt(u) + B(u)) + 44 (nt' ( u ) + B1(z ) ) sin(nt(u) + B(v))
= (-d'/-A) Ln (v) + 4 ( v ) (nt' (v) + Bt(.u)) sin(nt (u) + B(u)).
Since at any root of Ln me have sin(nt(v) + B(v) ) = f 1, a double root can only
occur if ive simultaneously have a root of ntr(,u) + B'(2;) = 0.
Since cos t = y,
t'(v) = -y1/ sin t = (u2 - 2)/ ( 2 2 sin t ) ,
and since cos B = Lo/A,
B'(zJ) = (Lo.4' - Lo-4) /(A2 sin B) .
Now as -4sin B = ( y L o - Li)/ sin t we find
The only zeroes of f ( u ) occur at u = f.& at mhich point B'(u) # O. Thus the
only possible double root left to consider is when
This is equivslent to
(272 + 5)u2 + 4 u + (2 - 4n) = 0,
If this u were a root of Ln, then it would be a double root and these are the only
values of ,u with this property. However the p,, (and therefore the Ln) have roots
which are algebraic integers. The cpantity whose square root we take in the formula
for v is congruent t o 2 (mod 4) and hence is not a square. Therefore u is not an
algebraic integer unless 272 + 5 clivides 4 (sec: for esample, Corollary 2 in Chapter 2
of [hlrc]). This only occurs for n = -2 or -3. Thus these trvo bad values of v are
never roots of Ln for n > O (and therefore not roots of the corresponding pn). Thus
p, has no double root. Cl
Lemma 2.5.4. Let p, be the polynomials of degree n - 4 deJined recursively bg the
equations
a n d pn-,(u) = - ( ( a u 2 + v + 2)pn(v) + vZPn+2(li))
where n 2 T i s odd. Then p, has distinct roots.
Proof: Similar to tha t of the previous lemma
3.1. Seifert spaces and Montesinos knots. Montesinos knots are knots whose
2-fold branched cyclic cover C2 is a Seifert fibred spaçe having as base orbifold S2
with a certain number of cone points. Since 7r1(&) = %/(p2) : ive can think of C2 as
a kind of Seifert surgery of the knot, and we will take advantage of this similaritu.
To better understand these knots, we begiri with a brief introclirction to the theory
of Seifert spaces.
3.1.1. Seifert jbred spaces. This presentation of Seifert spaces is taken from Chapter
4 of Hatcher's [Ha] notes (see slso [Sco, Section 31).
Definition 3.1.1. A model Seifert fibring of SL x D2 i s a foliation of S I x D2 bg
circles, called fibres constructed as fo1low.s. Start ing with [O? 11 x D2 foliated by the
segments [O: 11 x { p t ) , ident i fy the àïsks { O ) x D2 and {l) x D2 by a %îrD/o rotation,
for P b f Q-
Definition 3.1.2. A Seifert fibred space is a three-manifold with a foliation by cir-
cles such that each fibre has a neighbozlrhood d$eomorphic. preserving fibres, to a
neighbourhood of a fibre in some model Seifert fibring of S1 x D2.
Definition 3.1.3. The multiplicity o j a fibre circle C in a Seifert fibred space 2s the
number of t imes a smal l disk transverse to C rneets each nearby fibre. The fibres of
multiplicity I are regular fibres and the other fibres are singular.
Definition 3.1.4. T h e base orbifold of a Seifert fibrtd space i.s the two dimensional
orbijold obtained by identifying each Jibre to a point. T h e images of the s i n g d a r fibres
are called cone points, the order of a cone point beiny the mz~ltipficity of its singular
fibre.
We will be most interested in the case where the base orbifold is S2 dong with
some cone points. (For an introduction to the theory of orbifolds, see [Sco? Sec-
tion 21.) Such a Seifert fibred space is completely cleterminecl by a listing of the
model fibrings of neighbourhoods of the singdar fibres ancl we wiil dcnote it as
V(,Bi/crl. ,82/a2, . . - &/a,). In this caseo the ai are the cone point indices for the
base orbifold which w e mil1 denote by S'(al 0 2 , . . . , a,).
FIGURE 5. The Montesinos link m(e; ,&/cul' , & / ~ t ~ . . . . . &/a,).
3-12. Montesinos knots. A Iink as illustrated in Figure 7 is callecl a Montesinos link
and will be denoted by m(e; &/aL, fi2/a2,. . . , /3,/cr,). Here ,B/a describes a rational
tangle and we assume that a and ,LI are relatively prime integers and cu > 1. The a;
give a continued fraction espansion of p/n,
Generally, ive can arrange e = O by combining those twists with one of t h e tangles
,&/cri. In this case, wve will write simply m(& / a l , ,32/a2, . . . : $/a,).
As the rest of the thesis will be concerned exclusively with Montesinos knots, ive
here introduce some notation for this situation. Let i\( = S3 \ X(m) denote the h o t
cornplement, .z the 2-fold cyclic cover of A$, and C2 the 2-fold branched cyclic cover
(sec [Rol, Chapter 61 and [Rd, Section 10.C]). Theorem 3.2.1, ivhich we prove in the
next section, shows that C2 = V ( ~ O / ~ O , ,BL/a17 &/a2, . . . Bk/an) with ,&/cuO = -e.
Miich of mhat follows depends on the relationships arnongst the f~inclümental groups
of these manifolds. - First, ii = ni (134) is an indes two subgroup of .ir = 7i3(91). The group of E2 can be
understood in two ways. On the one hand, 7iL(C2) = ?/(p2) = i i / ( j i ) ivhere Ir E ri LI
is the class of a meridian of m and f i E .ii the class of the loop in Ad n-hich (cloiibie)
covers that meridian. On the other hand, due t o the Seifert structure, we have the
esact sequence (for example, see [Sco, Section 31)
where F 2 Z is the center of x1(C2), generated by the class h of a regular fibre,
and irprb(l3) is the fundamental group of the base orbifold L3 = S2 ( a l , . . . a,). In
particular, we have the presentations ([Sco, Section 21)
and ([BuZ, Equation 12.311)
Note that when r = 3, rve recover the triangle group (Section 2-42): ryrb(f3) =
A(al : a2, a3). We conclude this section with a lemma pertinent to this case.
Lemma 3.1.5. Suppose 1k.I is Seijert fibred over S2(p, q, r ) wzth py q? r 2 2 und that
P : ni (ilf) + PSL2 (@) is a n irreducible or n o n - u b e h n representation. Then P factors
through A ( p , q: r ) .
Proof: -4s above, t e have the exact sequence
where F = (h) S Z. We need to argue that ~ ( h ) = f 1.
Lemma 2.4.6 says that if P(h) # f 1, then we can corijugate so that p(h) = E ancl
P(nl ( ! I l ) ) c iV. If p, q, r are al1 ocld, rve can argue as in the proof of Lemma '2.4.9.
So we can assume p is even and that p(x ) is antidiagonal where x is t he generator
of r i , (M) satisfying xP = ha (see Ecluation 3.5). -As the iinticliagonal elements have
order two, p(x2) = H. Using the relation XP = ha we see that p(h) has orcler dividing
a where (a, p) = 1. However, p(h) = E has order tmo. This contradiction shows thüt
p ( h ) = f I in the case of even p as rvell, CI
32. Montesinos' theorem. In this section me present Théorème 1 of monte sin os^
Orsay notes and its proof closely folloming those notes to which we refer the reader
for more details (see also [BuZ, Section 12.D]).
FIGURE 8. ,An annulus preserved by T .
Theorem 3.2.1 (Théorème 1 of [Mo]). Let V = V(,&/cro PL /al. &/a2. . . . . ar/a,)
be a Sezfert fibred space with (oo7 Po) = (1: -e), ai, ,Bi relatively prime, and ai > 1
(1 < i 5 r). There i s an involution r of V such that the quotient CZ-/r is homeomor-
phic to S3 with the Montesinos link rn = m(e: pi /ai, &/a2: . . . ,&/ar) as branching
set, i.e. m is the image of the jixed points under the quotient map C* --+ I,*/r. More-
over, T i.s fibre and orientation preseruing and acts on the base orbifold S' us a
refiection in a circle passing through the cone points.
Remark: It follows that V is C2, the two-fold branched cyclic cover of m.
The proof recpires the following elernentan; lemma mhich ive state wit hout proof.
Lemma 3.2.2. Let S be a manifold containing a 3-bail B h n d let f be an auto-
rnorphism of S2 = aB3. Suppose F is an extension of f to an outornorphisrn of
B3, the cone of S2 ( for exarnple, F = cone( f)). Then there is a homeornorphism
Proof: (of Montesinos' Théorème 1) W e will define the involution r in several steps.
Consicler the solid toms T = B2 x SL equipped mith a n angle ïr rotation T :
( z 2 ) + ( 2 ) . In Figure 8 we've shaded an annulus { r l E R) preserved by r.
The quotient T/T of T by this rotation is a bal1 B3 and the image of the fised
points consists of two unlinkecl arcs. To the teft of Figure 9 ive see two choices of
fundamental dornain of T with the induced identifications. On the right. w-e see a n
equatorial disc of the quotient B3. The lightly shaded bancl at right represents the
quotient by T of the anniilus of Figure 8.
FIGURE 9- The quotient of T by T .
Fixed points
F~GURE 10. The double of the quotient of T.
The rotation T of T defiries an invol~ltion r on the clouble torus D ( T ) = S" SSL.
The double of the quotient is then D(B3) = S%nd the image of the fised points is
two disjoint, unknottecl circles (Figure 10). On the base S2 = D ( B ~ ) of the trivial
fibration S' x SL = D(T) : T acts as reflection in the circle D ( B 1 ) .
Now, the Seifert space V = V(@,-,/crO, ,Cl1 , 01, . . . , &/a,) can be constructecl starting
mith S2 x SL by removing the interiors of r + 1 disjoint solid tori = Di x SL
(i = O : . . . , T ) ancl then replacing tliem using an oriented automorphisrn di of mi which takes the niericlian mi to the curve aimi + Dil i , li being the longitude.
We ccan assume ttiat the disks Di used in constructing 1.- are centerecl on B L ( c
B2 C D(B2)) as iri Figure 11. So we can dioose the autoniorphisni Qi of fl* so that
FIGURE I l . Illustration of the disks Di used to construct b-.
FIGURE 12. Illustration of the balls Bi in S3.
it cornmutes with rlfln. Thus ive can construct an involution on V r\.hich we will
again cal1 T.
The quotients T i / r are 3-balls Bi. LVe obtain V/T by removing the interiors of the
r + 1 balls from (S2 x SL ) / r = S3 and replacing them osing the automorphism Gi of
aBi = B ( ~ / T ) induced by Qi. In Figure 12, the balls Bi = (Di x SL)/r are lined itp
along the band (BL x S ' ) /T just as the Di were lined up dong B1 in Figure 11.
.-\pplying Lemma 3 . 2 2 with S = S3 shows that V/T remains an S3. Moreover: the
lemma shows that the image of the fixed points under the quotient map L- -+ V/T,
is obtained from the trivial link (of S' x SL see Figure 10) by replacing the image of
this trivial link in eacli Bi with its image under an estension Fi of 4i to Bi.
Case where V = S2 x SL General case
FIGURE 13- The trivial link transforms to the ?vlontesinos linkt
Claim 3.2.3. There is an oriented automorphism c$i of a which sends mi to aimi+
piXi, cornmutes with rlmi and induces o n dBi a n automorphism Ji which extends to
an autornorphism F, of Bi such that the image under F, of the trivial tangle i n B3 zs
the rational tangle fi/ni.
We ref'er the reader to [Mo] b r a proof of this claim.
The claim completes the proof since i t shows that the extensions Fi transform the
trivial link to the Montesinos link m(e; Dl /ai ; /?*/a2, . . . Br /a,.) (see Figure 13). 0
3.3. Extending representations. When a Montesinos h o t m has r = 3 tan-
gles, the group of the base orbifold i r p r b ( ~ ) is a triangle group, I ( p , q, r ) (see Sec-
tion 2.4.2). We can assume that p , q , r are positive and since m is a knot, p and q are
odd. W e show that PSL2(C)-representations of this triangle group extend to become
representations of the knot group n = rrl(S3 \ N(m)) .
Proposition 3.3.1. Let f i be an irreducible PSL2(@)-representntion of ;i whzch fac-
tor .~ through A(p, q , r ) . Then Po has a unique extension to 7.
Proof: Suppose and f3' were tmo extensions. Let a E T \ 5. For any $5 E
39
So A = p(o)-'$(n) cornmutes with po(fl) and consequentiy with each element of
Po(+). By Lemma 2-49. -4 = f 1: as P is irreduçible. Thus p and 3 agree on a. hence
on 7r and there is at most one extension of Po.
Claim 3.3.2. There wzll be an extension P i f and only if there is a matrix A E
PSL2(C) such that
Proof: (of daim) Suppose p is an extension and let -4 = ~ ( p ) . CVe knom that po
factors t hrough A ( p , q' r ) c al (E2) = %/ {p2) = % / ( f i ) : so A* = p0(p2) = II . Clearly
--L~o(,D)-A-L = po(p,8b-i) as well.
Conversely, suppose ive have such an A. Note that 7;- = .ii i~ p?. Define p by
Let a E ii. By hvpot hesis,
Then,
whence:
So, va, ,B E ii,
FIGURE 14. .A lift of the rneridian p.
Thus p is a homomorphism whicli estends f i . U(c1aim)
To see that Po estends, ive must demonstrate the esistence of such an -4 corre-
sponcling to conjugation by p. Let T (Theorem 3.2.1) be the involution of the 2-fold hr
branched cyclic cover C2 and choose base points 5 f d M and Zo E Fis( r ) c E2.
Then conjugation by p corresponds to
mhere p~ is the lift of p beginning a t 5 (see Figure 14).
So we have the folloming commutative cliagram
From Theorern 3.2.2: we have the diagram
where 7 corresponds to reflection in the equator of S' (see Figure 15). If we choose
paths as illustrated, then 7 has the effect of taking the generators of A ( p . q , r) =
(a, b 1 aP: bq, (ab)') to their inverses. Combining these ideas with the representation
of h ( p , q, r ) incluced by f i , ive have the diagram:
xo = base point
FIGURE 15. The reflection of S2 induced by r .
ir -+ al (C2) --+ A(p. q, r) PSL2(@) Since ;Tj takes a and b of A ( p , q 7 r) to their inverses, and a b to n-'b-' which is
conjugate to (ab)- ' , we see that tr(& O Yj) = tr(&). On the other han& since Po is
irreducible, q50 is as well and we deduce (see [CSl, Proposition 1-52] ) that there is
an A E PSL2(@) with O i2 = -4qioA-'- In other words, we have found an -4 such
According to the claim. ive can complete the proof by shoiving that -4' = & I . Since
T is an involution: G 1 and A* cornmutes with every element of the irreducible
representation qjo(3(p, q7 r ) ) . By Lemma 2.4.9, -4' = 3~1. O
Scholium 3.3.3. An9 representation P which extends Go is such that P(p) h.as order
two.
We have been assuming that m = m(flL/al, ,82/cr2: ,3:$/%) is a knot with three
tangles. Then Z2 is Seifert fibred over S2 with cone points p = lal 1, q = 10-1 and
r = In3[. This means T ~ ~ ( S * ( ~ . q: r ) ) = A@, q , r). In other words, mhen there are
three tangles? we can use the proposition to see that the PSLÎ(C)-charactcr wriety
of A ( p , q , r) includes into that of m: Av(ll(p, q. r ) ) c ,Y(rn).
This suggests that for other Montesinos knots as tvell, ~ ( i r ~ ' ~ ( l 3 ) ) C S ( m ) where
B is the base orbifold of XP, the two-fold branched cyclic cover of m. Irideecl. it
is difficult a t first glance to see how the proof depends on rn having three tangles.
Essentially the argument breaks down when aprb (B) has more generators because.
although tvill still take generators to their inverses, it mil1 no longer do sa for al1
products of generators. So there is no guarantee that T2 will preserve characters.
This means we cannot use [CSl, Proposition 1-5-21 to show the existence of -4.
On the other hand: it may well be that 7, preserves the characters of many of
the T P ~ ~ ( B ) representations. For esample when r = 4, rprb(f3) has three generators
a, b:c so tha t characters are deterrnined by their value on the tvords a: b, c: ab. ac:
bc and abc (see Section 2.1). Of these, al1 but one: ac: is taken by 5 to a word
having the sarne trace. Since the SL2(R)-character variety of ~ f " ~ ( L 3 ) has (real)
dimension two, we know that there are an infinite number of characters. It seems
plausible tha t some of these may in fact be presemed by %. It would certainly he
interesting t o investigate this further. Horvever. in this thesis, we will instead apply
the observation X(A q, r)) c .Y (m) to deduce some consequences for pretzel knots.
This is the subject of the nest two chapters.
4. CYCLIC AND FINITE SURGERIES ON MONTESINOS KNOTS
-4s Ive have seen, three-tangle Montesinos knots are closely related to the triangle
groups. In this and the following chapter, we exploit t his connection in a study of the
(p, ql r ) pretzel knots. These form an important subset of the t hree-tangle &lontesinos
knots. The ( - 2 , 3 , n ) pretzel knots of Section 3-52 are examples.
Pretzel knots with cyclic permutations of the indices p' q, r are obviously isotopic
and will be considered equivalent. On the other hand, there is also an evident isotopy
between the (p, q, r) and the (r , q , p ) pretzel knot. (For esamplel given a cliagram of
the h o t such as that of Figure 6: rotate by 180" about a vertical line to the right
of the diagram.) This means that any permutation of the indices pl q. r mil1 result
in an isotopic (hence equivalent) knot. Moreover, taking the mirror reflection of a
pretzel knot corresponds to chaaging the signs of a11 the indices. -4s this reduces to
an isomorphisrn of the fundamental group ;r: we will ordinarily consider the knots
( p . q, r) and (-p, -q, -r) equivalent. Note hoivever that such a mirror reflection d l
change the signs of boundary and surgery slopes. For esample, it d l result in a
reflection of the fundamental and Xewton polygons in a vertical line.
In this chapter we use the methods of Ciiller, Slialen: Boyer, and Zhang to corne
to a thorough understanding of cyclic and finite surgeries on pretzel knots. This
complements the io rk of Delman [Del] who provided a classification of cyclic and
finite surgeries on al1 other Montesinos knots.
4.1. Infinite fillings of ( p . q, 2m) pretzel knots. Let K = KP,,,, be a pretzel knot
wliere p = 2k + 1, q = 21 + 1 and r = 2m. We ivill be assuming that l/lpl + l/lql + 1/17nl 5 1: and. by [ K a y Theorem III], this ensures that fi is hyperbolic.
Lemma 4.1.1. If lllpl + l/lql + lllrnl 5 1 then euerg filling (2nlb) of the knot
complernent hl is infinite.
Proof: To simplify the notation. we will assume p, q, r > O . For the general case,
one need only take absolute values.
-4s A ( p , q, m) is infinite, our strategy is to construct a represeritation of AI (2a /b )
with image A ( p : q, m).
Let Po be a faithful, non-abelian PSL2(@)-representation of A ( p , q , m). Using the
obvious homomorphism, Po is also a representation of l ( p , q, r ) , the group of the
base orbifold of the twofold branched cyclic cover of the h o t . Then, as in Propo-
sition 3.3.1, po 'Lextends" to a PSL2(@)-representation P of the knot group .ir which
in turn lifts to an SL2(C)-representation p. (The obstruction to such a lift is in
H2(ri:Z/2) [BZ2, Section 31. For a knot in S", the second coliomology is trivial
and there is no obstruction.) Moreover, Scholium 3.3.3 shows 2, (/,-) > 2, (f,) for
x = X p . It follows that ~ ( g ) = f 1-
On the other hand, we can determine the image of X in A ( p : q, r) = (f, g, h 1 fr: gP, h g , fgh) to be X = gk/mgk+lhL f mh' fL - (This derivation is esplained in more
detail in Lemma 5.12.) Now, as Po is a representation of l ( p , q, m), we have Po( f ") =
*I and consequently p o ( i ) = II. Then P(X) = f 1 as well.
So, for any filling of the from a = 2n/b, we have p ( a ) = f I wherice P factors
through q (M(a)). Since po (A@, q, m)) is infinite, we see that iil (M(û)) must also
be infinite. 1 I t remains to construct such a representation Po. Note that as + + 5 1,
A(p, q, m) is infinite. Moreover, either {p, q, m) = (3, 3,3} , and A ( p , q, rn) is a set of
isometries of the Euclidean plane F , or else + $ + & < 1 and A ( p , q, rn) represents
isornetries of m . Since both E2 and EX2 imbed isometrically into HP. in either caseo
A ( p , q, m) is contained in PSL2 (C) t-he set of orientation preserving isomet ries of 8.
This provides the required fait hfd , non-abelian representat ion. O
So under the hypothesis of the lemrna, every 3alb filling of K is infinite. This
meûns that any finite surgeries would have to be of the form (?a + l)/b and therefore
would have norm 11 (2a + l)/bllT 5 S + 8 [BZl, Theorem 2-31. On the other hancl- the
2alb fillings will have norm superior to S + 8. (See Section 2.2.3 for the definition of
S and I I - IlT, the total norm.)
Proof: ,\gain, ive assume p, q , r > 0. We first observe that there are a t least 3 irreclucible PSL2(@) characters of A ( p , q , rn)
using Equations 2.3 and 2.3. This can be verified directly if nias (p, q;m) < 11. Let
us assume then that mau (p, q, n ~ ) > 11.
As p and q are both odcl, we can simplify the two equations sornewhat to see that
the number of irreducible characters of A ( p , q: m) is
where b = gcd (pq, pm, qm) .
Without loss of generalits p 5 q; and there are two cases: m = m a . ( p : q: m), and
q = mau (p, q , m). If rn = max ( p , q , m), then b 5 pq and the number of irreducible
characters is a t least
(p - l ) (q - l ) ( m - 1)/4 + 1 - ( p q + 1)/2
If p or q is greater than 3, this is a t least 2(m - 3) + 1 - m = m -5. If p = q = 3,
we again have the bound m - 3 + 1 - 3 = rn - 5 . So either way there will be a t least - r irreducible characters since we're assuming m = max ( p , q , m) > 11.
Suppose next that q = max (p, q, m). As above ive see that there are at least
( p - l ) (m - l ) (q - 3)/4 + 1 - ( p + m)/2 irreducible characters of A(p, q, n). If p 5
or rn > 3 there are a t least 3(q - 3)/2 + 1 - q = (q - 7)/2 > 2 characters. On the
other hand, the remaining possibilities (p = 3 or 5 and m = 2 or 3) also yield a t least
3 irreducible characters.
Therefore, t here are a t least 3 irreduci ble PSL2 (@)-characters of A ( p . q . nz) . As we
saw in the previous lemma, these al1 factor through to becorne characters of M(a)
d i e n a is of the from 2a /b . None of these charactcrs are dihedral (Lemma 2.4.10),
so each is covered twice in SL2(@). -4s they are the characters of irreducible repre-
sentations of a triangle group, they are smooth points of S ( M ) (see Lemma 5.1 -3).
hloreover, they are zeroes or f, which are not zeroes off,. (-As in the previous lemma;
these are characters of representations which take p to an element of order two-) I t
follows from [BB, Theorem A] tha t 2, (fa) = Z,( f,) + 2? and since ive have sis such
z, we see that 112u/bllT = llcrllT 3 IIpII + 12 = S + 12. O
Theorem 4.1.3. If Ii = I(,,q,, is a pretzel h o t with po q odd, r euen und 1 /lpl + l/lql + ;?/Ir1 < 1: then K admits at rnost one non-trivial jînite suryen/. Moreover
svch a surgery slope u zs odd antegral and there is a non-integral boundnry slope in
(u - 1: u + 1).
Proof: The conditions on p: q: r ensure that K is hyperbolic [Kaw, Theorem III] (see
also Lemma 5.2.1).
Let a be a finite stirgery of such a knot. We have already obsen-ed (Lemma 4.1.1)
that cr = (2a + l)/b. Since meridional s u r g e - is cyclic, ive can apply [BZl. Theorem
1-11 to see that b 5 2.
If a = (2a + 1)/2 were a finite filling, then, by [BZl, Theorem 2-31: [lallT 5 S + S.
At the same time, llpllT = I I - pllT = S. The line joining n = (2a + 1' 2) ancl
p = (1,O) in surgery space V r H I (âM; W) R2 passes tlirough (a + 1: 1) while the
line through a and -p passes through (a: 1). It follows that Ils + and llallT are
both Iess than S + 4 Since one of them is even, t his contradicts Lemma 4.1.2.
So any non-trivial finite fillings must be odd integral. Suppose there were two such.
Each would have norm at most S + 8. The line joining them woulcl necessarily pass
through some even integral surgeries tvhich would therefore also have norm a t rnost
S + S. This again contradicts Lemma 4.1.2.
'iom suppose that 2a + 1 is a non-trivial finite filling. Then 112n + 1: lllT 5 S + S
while 112a, IllT 2 S + 12 by Lemma 4.1.2. Let P c 1- clenote the norm-bal1 of
radius S + 8. By [CGLS? Proposition 1.1.2], P is a firiite-sicled çonves polygon whose
vertices are multiples of boundary slopes. In particular: (-a + 1, 1) is not a vertes of
P (Lemma 2.3.1).
Sinçe (2a + 1: 1) is inside P and (2a, 1) is not. there is a segment of d P which
intersects the line y = 1 between them. Let k(c, cl) be the vertes of this scg~rient
which lies on or above y = 1; i.e., II E Q and c / d is a boundary slope. Consider
the segment from the origin to k(c , d) . As both endpoints are in Pt this segment is
also. It crosses y = 1 at ( c / d . 1) which must lie between (2a: 1) and (2(a + l ) , 1) .
(Otherwise, the segment joining (2a + 1, 1) ancl (c/d, 1) passes through (-a, l ) , say.
Since both enclpoints are in P , this segment is in P ancl in particular (2n . 1) is in P:
a contradiction.)
Thus 12n + 1 - cldl < 1, as recpired.
Corollary 4.1.4. If a knot satisfies the conditions of the theorem and has no non-
zntegral boundary slopes, then i t admits no non-trivial finite svrgeries-
Corollary 4.1.5. Alternating knots which satisfîj the conditions of the theorem admit
no non-trivial finite surgeries.
Proof: This follows since alternating Montesinos knots have no non-integral bound-
ary slopes (see [HO, p.4621). 0
Remark: .\ (p l q: r ) pretzel knot is alternating iff p: q: r are al1 of the same sign.
More genera l l~ an alternating Montesinos knot m can he written in the Form
m(e: ,B l /a i ' , B 2 / a 2 ' . . . , &/a,) with -e and al1 , 3 i / ~ i ha\-ing the same sign. Incleed, it
is clear that such a knot is alternating. O n the other hand, a recluced cliagram of an
alternating knot will realize the knot's crossing number iff the cliagram is alternating
(see [Kau, LIU: Thi]). By [LT, Theorem 101, the crossing number of m is realized by a
diagram like Figure 7. If rn is alternating, this diagram will therefore be alternating
whence -e and al1 the &/cri tvill have the same sign.
Note that the second Corollary also follows from Delman ancl Robert's [DR] proof
that alternating knots satisfy strong property P.
4.2. Non-integral boundary slopes of pretzel knots. -4s ive have seen (Theo-
rem 4.1.3): finite surgeries on ( p t q , 2m) pretzel knots are intirnatelu related to non-
integral bounclaq- slopes. In this section Ive will use the methods of Hatcher and
Oertel [HO] to calculate these slopes. \Ve first illustrate the method using the esam-
ple of the (-2: 3 ,7) pretzei knot. We mil1 assume familiarity with the notation and
conventions of [HO].
Boundary slopes are found using "edgepaths" in the Diagrani 23 (Figure 16). Points
of 2) are labeled by triples (a: 6, c) and have vertical coordiiiate Wope" c / ( a + b) and
horizontal coordinate b / ( a + b ) . Thus horizontal lines in 27 represent points which ail
have the same slope. Let @ / q ) denote the vertes (1: q - 1' p ) and @ / q , ris) the edge
joining @/q) and (r ls) . Such a n edge esists only if the slopes are of distance one:
4 ( p / q 7 r / s ) = 1p.s - rq1 = 1. The diagrani also includes horizontal segments from
O>/q) to ( O . q , p) on the right-hand edge. \Ve will denote such horizontal segments by
@/% plcl) -
FIGURE 16. The diagram D.
An edgepath of V is a piecewise linear path in the 1-skeleton of V. Boundary
slopes of a Montesinos knot m(pl/ql, p2/q2,.. . pn/qn) are determined by a sequence
of edgepat hs -fi (i = 1 . . . n) satisfying the following conditions:
El: The starting point of Ti lies on the edge (pi/qi7 pi/qi). If the starting point is
not bi/qi), then the edgepath is constant. Le., it never leaves (pi/qi, pi/qi).
E2: -{i is minimal, Le., it never stops and retraces itself, nor does it ever go along
two sides of the same triangle of V in succession.
E3: The ending points of the yi7s are rational points of V which all lie on one
vertical line and whose vertical coordinates add up to zero.
E4: n/i proceeds monotonically from right to left, "monotonically" in the weak
sense that motion a i m g vertical edges is permitted.
Thus each tangle pi/qi gives rise to a tree in V corresponding to potential edge
paths *fi. For esample, the tree for 1/7 is illustrated in Figure 17.
Non-integral boundary slopes will be given only by edgepaths wliich have no vertical
edges and end a t a vertical line u = uo E 0 before reaching the left edge of D (the
"Type I" edgepaths of [HO]). That is, the -/i end at (uo, ~ i ) = (b/(a + b) c i / ( a + b))
with C vi = O.
-4s ive have seen (Figure 17), the trees are cluite simple in the case of the tangles
l/pi of a pretzel h o t . This means finding points where vi = O is not so difficult.
For esample, let's look a t the (-2,3: 7) pretzel h o t . The trees corresponcling t o
FIGURE 17. The tree of a 1/T tangle.
the tangles -112' 113' and 117 are aligned above one another in Figure 18. Let
,UT = Cui. AS ui is a linear function of L between vertices: it ivill be helpful to
calculate the value of z!i at each u which is a vertes on one of the trees. \Ve have
included this data in Figure 18.
Since each tree has two branches, there are a t most eight different edgepaths for a
given value of u. For esample when u = 0, the possible values of UT are 2 (+ + +),
1 (+ + -. + - +, or - + +), O (+ - -, - + -, - - +), and -1 (- - -), where
+ + -: for esample. is meant t o indicate that rve've chosen the upper branch of the
- 112 tree. the upper brandi of the 113 tree and the lower branch of the 117 tree.
When u = 112, the possible values of v~ are 112 (+ + + or - + +), 11.1 (+ - + or - - +): 1/12 (+ + - or - + -), and -116 (+ - - or - - -). Since the value of
2 ; ~ varies linearly between vertices, we see that it c m have no zero for O < .u 5 1/2.
For esample, for the + + + edgepaths, ,UT = 2 at u = O and o ' ~ = 112 a t u = 11'2.
Therefore it is positive throughout the interval O < u 5 1/2.
When u = 213, there are only two possible values of UT, namely 116 (* * +) and ..
-1/18 (* * -), where we've used * to mean "+ or -," i.e.. the two choices lead t o
the same result here. This shows that o-~. has a zero on the * + - paths as u~ = 1/16
when a = 112 and UT = -1/18 when u = 213.
Finally, when 2~ = 611, there is only one choice for UT: nanlely UT = -112 + 113 + 1/7 = - 1/42. So there is a n additional zero of UT on t h e * * + edgepüths. Indeetl,
when u = 5 / û , we have UT = -112 + 113 + 116 = O on * * +. Thus there are csactly
FIGURE 18. The trees for the (-2; 3: 7) pretzel h o t .
two Type 1 edgepaths for the (-2' 3? 7) pretzel knot: * + - with 1/2 < uo < 2/3 and
* * + with uo = 5 / 6 . Thesc arc the only candidates which could yielcl a non-integral
bounclary slope.
-4s explained in [HO] an edgepath is a way of describing a surface S in the knot
complement. The twist associatecl to this surface is r ( S ) = 2(e- - e+) where e+ ( e - )
is the number of edges in the edge paths which increase (decrease) slope.
For esample the twist of the edgepath system which terminates at cto = 5 /6 is -2
since the -1/2 and 1/3 edgepaths are constant and the l/Ï ectgepath increases on
one edge from 1/7 to L/6.
The * + - systern ending with uo between 1/3 and 2/3 is more subtle as it involves
fractions of edges. The value of UT for u in this intemal is
So = O ivhen u = uo = 3/5. This is half way along (1/2,1/3) in the 1/3 tangle
and 31.1 of the way along (0/1'1/7) in the 1/7 tangle. Thus the -1/2 tangle has no
increasing or clecreasing edges, the 1/3 tangle contributes 112 to e+ ancl the 1/T tangle
contributes 314 to e - . The twist of this surface is then r (S ) = 2(e- - e,) = 1/2.
Finally, the boiindary dope of a surface is given by r ( S ) - r(So) where So is a
Seifert surface for the knot. For a (p, q, r ) pretzel knot, a Seifert surface c m be
found by taking the + + + system of edgepaths and estending to ~x a t the left of
V. For the (-2.3: 4) pretzel knot, the three tangles contribute 1.2, ancl 6 upward
edges respectively so t hat r(.so) = - 18. The two surfaces constructed above t herefore
have bounda- slope 16 and 18 1/2. In particular, this includes the one non-integral
boundary slope of this h o t : 18 1/2.
We now prove several temmas about non-integral boundary slopes on pretzel knots
tvhich will prove iiseful in the following sections.
Lemma 4.2.1. Let K be a ( - 2 : p , q ) pretzel knot zuith p , q odd and 3 5 p < q . if
p 3 7 (respectivelg q 3 7) then -
p 2 - p - a p-3
( resp . q 2 - q - 5 ) g-3
i.5 a non-integral boundary slope of K . Moreouer, these are the onlg non-integral
boundanj slopes of ri-.
Proof: Let us assume p 3 5. The case p = 3 is analogous, but involves small
deviations from the general case.
We follow the procedure outlined above and look for Type 1 edgepaths. As before, 1 I at u = 0, UT is 2,1, 0, or -1. At u = $: there are four values for v ~ : (t + +)
1 1 (* - +), (* + -), and '(A + - - 1) (* - -) and so there are no edgepath 2 p I q-1
s>*stems terminating at uo E (O; $).
At u = y, there are orily two values of UT depending on whether n-e take the
upper or lower branch in the l/q tangle. For * * +' we have UT = -$ + + a = 7- Since p > 5 , this is negative and there will be zeroes of 2 ; ~ on both the * - + and
* + + edgepaths. -At * * -, UT = -2 + $+fi. AS 5 5 p 5 q, this is again negative
and UT has an additional zero in the * + - system of edgepaths. 1 Now, *UT = -- + $ + < O at u = !i2 and there are no further zeroes of q for 2 '7 '7
u 2 e. P
Thus there are three candidates surfaces wit h edgepath systems terminating at 1 p-L
uo E (?, ) Note that , as each of these systems is constant in the -1/2 tangle.
t hese will al1 be incompressible surfaces (see [HO: Proposition 2-11).
However, the * + ; system will not contribute a non-integral boundary slope since 1 it in fact terminates at uo = $ where UT = -- + ! + f and therefore involves no 2
fractional edges.
For the * - + system
mhich is zero when u = -uo = &. In the l /p tangle, uo corresponds to a point on the edge (0/1, l/p):
( O ) + (1 - ) ( p ) = a(1,O;O) + (1 - a ) ( l ; p - 1J)
Taking the u coordinate from the last iine, ive have
which implies cr = 5- The l l p tangle therefore contributes to the total of e - .
and the l / q tangle has q - 3 increasing edges. plus a certain fraction of the (1/2; 1/3)
edge:
- 2 - p -
1 -) in (u: v) coordinates. ( 3 - 8 ' 2 - 8
That is,
mhence ,8 = 5 and the l/q tangle contributes q - 3 + to e+.
Finally, since the edgepath in the -1/2 tangle is constant, we find that the twist
corresponding to the ~c - + system is
On the other hand, the twist of a Seifert surface (+ + + estendecl to oo) is r(&) =
2(0 - (1 + p - 1 + q - 1)). Putting it al1 together, the boundary dope of the surface
S given by the * - + system is
1 - ~ ( S O ) = 3 3 - q + - ) - 2 ( 0 - ( l + p - l + q - 1 ) ) p - 3
Analogous arguments show that the * + - surface has boundary dope W. 1
We turti now to the case of a ( p , q , -r) pretzel knot K with 4 5 r = 2m, even and
p = 2k + 1 and q = 21 + 1 both odd. Further, we assume 3 5 p < 4. iF7e n i i l ggie ttwo
lemmas describing the non-integral boundary slopes of K depending on the relative
values of p and r .
In each case we mil1 list fractions which are candidates for the non-integral bound-
ary slopes of K with the understanding that for certain choices of p, q. ancl r: these
fractions will turn out t o be integers. Moreover, ive will only verify that these can-
didates are truly boundary slopes when we can easily use [HO' Proposition 2-11 to
do so? that is, when one of the edgepaths is constant. In general, the fractions are
the boundary dope of some surface in hl, but when we cannot apply Proposition
2.1, it is tedious to ve r ib whether or not that surface is essential. (Also. there is a
small error in [HO] ivhen they discuss testing whether or not candidate surfaces are
essential. See DU^].) Since a list of candidates will be adequate for ils. we will not
endeavour to investigate further.
In short, if K admits a non-integral boundary slope, then that slope will appear in
the list of slopes in the appropriate lemma (depending on the values of p and r ) .
Lemma 4.2.2. If p 2 2 r + 1, then
and
are the non-integral boundary .dopes of K.
Remark: Substituting r = 2, we see that Lemma 4.2.2 generalizes Lemma 4.2.1.
Proof: As in Lemma 4.2.1, we need to look a t edgepath systems à la Hatcher and
Oertel [HO].
When u = O, UT is 2, 1,O: or -1. At u = 9. there are four values for ET: ! (*++), r -1 (*+-), and ;(-+%- 1 r-1 1) (*--) . XS the first three values - (* - +)O r ( p - 1) 9 1
are positive and the last is negative, there are no edgepath systems terminating a t
.O E ( O , 9). p-l When IL = e, the two values of vT are -$ + = (* * +) and -j + + p ( , - , , P
(* * -), so the * - +, * + + and * + - edgepaths are al1 Type 1 each terminating a t
a uo E (5: y)- -4s these edgepaths are constant in the -l/r tangle, eacli results
in an incompressible surface [HO; Proposition 2.11- Since - + + < O, there are P
no other Type 1 systems for this knot. 1 1 1 No,, the * + + edgepath will terminate at uo = - where UT = -; + a + = 0.
This system will not leacl to a non-integral bounciary slope as it involves no fractional
edges.
As in Lemma 4.2.1, the * - + and * + - edgepaths will give surfaces having
boundary slope
p(p - 1) + 1 - 3r q(q - 1) + 1 -3r P+
and 9-L-r - 2
respect it-ely.
Lemma 4.2.3. Ij p < r , the only potential nomintegral boundary dope of K is
Proof: As usual, a t u = 0: the values of are 2. lr O. and -1. When u = m. we P .
have Z.T = 2 - = 'p(rPl) Zr- -' (+ * +), UT = (- * +)' = --EL + ' f P p(r-1) P p(q-1)
(+ * -), and 2 > ~ = JE!-- (- * -). NOW, these are ail positive and. when IL = 9. p(q- 1)
l - is positive as wvell. So the only Type 1 system is in the - - - UT=;+; edgepat h
and it terminates a t a cl0 E (O, y). The associated boundary slope is as given in
Equation 4.7. Note that we cannot use [HOo Proposition 2-11 so it may be that the
fraction of Equation 4.7 is not a boundary slope. In that case, Ii' would have no
non-integral boundary slopes. O
4.3. Cyclic surgeries of ( - Z p , q ) pretzel knots. In this section. let K be a
(-2, P, q) pmtzel knot (p, q odd ancl positive) and M = Sv N(IC) .
Lemma 4.3.1. M ( 2 ( p + q)) contains an zncompressible toms.
Proof: The obvious (see Figure 20) spanning surface of such a knot is a once-
purictured Klein bottle and it's double cover, T1 is a twice-punctured toms which
meets dhf in two parallel curves of dope 2 ( p + q ) . We can use the methods of Hatçher
and Oertel [HO] to see that 2 ( p +CI) is in fact a boundary slope of K . Folloming the
proof of [CGLS, Theorem 2-03] we see that T is a minimal surface realizing this
boundary siope. (Since 2 ( p + qj is not the longitude slope. such a minimal surface
is connected, separating and has at least two boundary cornponents.) As T is non-
planar, we apply [CGLS, Proposition 22-11 (along with the fact that I\' is a srnall h o t
[Oe, Corollary 41) to see that the filled toms T is incompressible in M ( 2 ( p + g ) ) . Cl
Proposition 4.3.2. Suppose hm, a (-2, p. q ) pretzel knot ( p < q odd and positive),
admits a non-trivial cyclic surgeq. Then one of the follovling holds.
1. h- is toms and therefore admits an infinite number of cyclic surgeries- In this
case either p = 1, q = 1, or else {p, q } = {3,3) or {3: 5 ) .
2. h' is the (-3.3,7) pretzel knot and the surgery is 18 o r 19.
ProoE Theorem III of [Kaw] shows that is toms iff it is as characterized in 1 (see
a h Lemma 5.2.1).
Since I< is small ([Oe, Corollary 41). we can assume that I( is hyperbolic. The cyclic
surgeries 18 and 19 of the (-253, 7) pretzel knot were first observed by Fintushel and
Stern (see [FS. Section 41). Our task is to show that there is no other choice for p
and q leading to a cyclic surgery.
The case p = 3 is the subject of Section 5.2 where we will see that there are no
non-trivial cyclic surgeries when q 2 9 and that the cydic surgeries of the (-a,& 7 )
pretzel knot are as stated.
If p = 5, the boundary slopes [HO] are 0,14, l5,9-,, 2q + 10; and 2q + 12. By %-
[Dul , Theorem 4-11, a nori-trivial cyclic surgey could occur only at 2q + 4 or 2q + 5 .
However, as we explain below, a cyclic surge- would have to be within distance 5
of the toroidal surgery 2q + 10 (see Lemma 4.3.1). So the only candidate is 2q + 5 .
Xow the (-2,s; 5) pretzel has no non-integral boundary slopes so ([Dul. Theorem
4-11) it has no non-trivial cyclic surgeries. -4s for (-2.5, ï), SnapPca [Wee] shows
that 2q + 5 = 19 surgery on this knot is hyperbolic. So we can assunie q 2 9.
Suppose (for a contradiction) that 2q + 5 is indeed a cyclic surgery. By [BZl?
Lemma 6.2], the (total) norm can be written
If 2q + 5 is cyclic it has minimal norm s, as does the meridian sorgery p ([CGLS.
Corollary 1-1-41 and Lemma 2.3.1). The norm of 2q + 4 will also be of interest, and
it will be bounded by the minimal norm S.
Subtracting the first two equations, Ive have
while subtracting the second from the t hird leaves
2 where q = 5 1. 9 '
Combining this with Equation 4.8, ~ v e have
Since a; 3 O: this shows al = a2 = a3 = a5 = a6 = O. On the other han& for a norm,
a t least two of the ai must be non-zero. This contracIiction show that there can be
no non-trivial cyclic surgery when p = 5.
So let us assume 7 5 p 5 q. Dunfield[Dul, Theorem 4-11 has shomn that any
non-trivial cyclic surgery on a knot such as K must lie near a non-integral surgery.
Combining this with Lemrna 4.2.1, the only candidates for a non-trivial cyclic surgery
are 2 p + 4, 2p + JI 2q + 4, and 2q + 5. Suppose that u is one of these candidates slopes
and hl (u) is a cyclic filling. Since I< is strongly invertible, the Orbifold Theorem
implies that M ( u ) admits a geometric decomposition (see [CHK: Corollary 1.71).
Now, as A(u, 2 (p + q ) ) > 5, 1b.l (u) is irreducible [Oh, \Vu] and atoroidal [Go21 and
therefore has a geornetric structure.
Note that srt (iCI(u)) 2 ([Gall): ço ;rl (i\.I(u)) is finite. The geornetry is therefore
S" and as ST~(I\~(U)) is finite -clic, me deduce that M ( u ) is a lens space. However.
this contradicts [Go3? Theorem 1.11 which states that the distance between a lens
space surgery such as u and a toroidal surgery such as 2 ( p + q) is a t most 5. We
conclude that there are also no non-trivial cyclic surgeries in this case. O
4.4. Finite surgeries on pretzel knots. We turn now to the case of a (p,q, -r)
pretzel knot K wtiere 4 5 r = 2m is even and p = 2k + I and q = 21 + 1 are both
odd. We will assume 3 5 p 5 q.
Lemma 4.4.1. M(2(p + q)) contains a 7 ~ incompressible toms.
Proof: Identical to that of Lemma 4-3.1.
Proposition 4.4.2. If p > 2r + 1 , then K admits n o non-trivial finite su~geries.
Proof: By Theorem 4-1.3, a non-trivial finite surgery would be close to one of the
non-integral boiindary slopes of Lemma 4-22. However,
L
and similarly for the other slope of Lcmma 4.2.2. Therefore, any non-trivial finite
surgery would be of distance (in the sense of minimal geometric intersection) greater
than 10 from the toroidal surgery 2 ( p + g ) . However this contradicts work of Ag01 [.-\g]
and Lackenby [LI showing that the distance between exceptional surgeries is 5 10. O
Proposition 4.4.3. If p 5 r - 5? then h' admits no non-trivial finite surgeries.
Proof: -4s in the previous proposition, we observe that
Thus the lone non-integral boundary dope of Lemma 4.2.3 is too far from the toroiclal
bounclary slope 2 ( p + q ) (by Theorem 4.1.3 a finite filling coulcl only occur a t an ocid-
integral slope, which would therefore have to be within clistance 9 of the even number
2 ( p + q) 1. O
Combining the results of the last few sections, we see that we now have a fairly
precise description of what a finite filling s on a (p, q, -r) pretzel knot would look like.
By Theorem 4.1.3, s would have to be odd-integrai and near a non-integral bounclary
dope and by Propositions 4.4.2 and 4.4.3 ive would have to have p+3 3 r 3 ( p - 1)/2.
We now propose to explicitly calculate the fundamental group of such a filling. We
will then project ont0 a smaller group G aiid observe that G is generically infinite.
59
The Wirtinger presentation [Rol, Section 3.D] of a (p, q, -r) pretzel knot is (com-
pare [Sr. Equation 11):
The longitude being
filling along an odd integral dope s results in
We can obtain a more manageable factor group G by adding the relators
yx- ', and (zx)* :
wliere w = ( z ~ ) ( P - ~ ) / ? . This is an esample of a group which Coxeter [Cl lias called
(2, a, b; c ) = ( R , S 1 Ra. sb, (RS)~ , ( ~ ~ 3 ~ ) ' ) .
Thus G = (2: p. 1.9 - 2pl; r /2) . kloreover. iil (M(s)) mil1 be infinite mhtinever G is.
And indeed: these groups are usuallp infinite as Edjvet has shown:
Theorem 4.4.4 (Main Theorem of [El). If 2 5 a 5 b, 2 5 c and (2:a,b;c) # (2,3.13; 4), then the group ( 2 , a, 6; c) is finite i j and onlv if it is one of the folloving:
(2) (2,2. b; c) (2 5 b , 2 5 c ) ;
(iQ (23,b:c) (3 5 b 5 6 4 5 c);
(zii) ( 2 , 3 , 7 ; c ) ( 4 5 c 5 8 ) ;
(zv) (2: 3: b: c ) (8 5 b 5 9,4 5 c 5 5);
(v) (2:3:6;4) (10 5 b 5 11);
(ui) ( 2 , 4 b;2) (4 5 b);
(ix) (274,7;3);
(x) (2,5: b; 2) (5 5 6 5 9):
Since p is odd, if p 5 1s - 2p1, then G is infinite unless p = 3 or 5 .
Similarly, if 1s - 2pl 5 p, we see that G is infinite unless 1s - 2pl = 3 or 5 (1s - 2pl
is also odd) whence s 4 2 p + 5. On the other hand. by [;\go LI' the finite filling s
and the toroidal filling 2(p + q) (Lemma 4.4.1) are separated by a t most 10. Since s
is odd and 2(p + q) is even, they in fact differ by a t most 9. Thus. 9 3 2 ( p + q ) - s > 2(p + g ) - (2p + 5) = 2q - 5. It followvs that 2q 5 14, whence 3 5 p 5 q 5 7.
So ive can assume that 3 5 p 5 7. That is, if p 3 9, the knot admits no non-trivial
finite surgeries.
Now, eorlier work shows that there are no non-trivial surgeries unless 9 5 T < -
p + 3. So, given r 2 4, and assuming 3 5 p 5 7, we see that ive are left to consider
4 5 r 5 10. (Since Theorem 4.1.3 does not apply to the knots ( - 4 3 , 3). (-4: 3 , s )
and (-6,3: 3) we will consider those separately below.)
m r = 4
- p = 3. When q = 7, there are no non-integral boundary slopes ancl therefore
no non-trivial finite surgeries (Theorem 4.1.3). For q 2 9: the boundary
slope of Lemma 4.2.3 is 2q + 12 - 4(q - l ) / (q + 2) E (2q + 8:2q + 9) so
that the only candidate for a finite filling is s = 2q + 9. But then s - 2 p =
9+2(q-p) 2 21 and G is again infinite (Theorem 44.4). Therefore (-4,3, q)
pretzel knots admit no finite surgeries (q 2 7).
- p = 5. Suppose 1s- 2pl 5 3. Then s 5 2p+3 and since a finite filling must be
witithin distance 10 of the toroidal slope 2(p + q) ([.-\g: L] and Lemma 4.4. l) ,
10 3 2(p+q) - (2p+3) 2 2q- 3 =+ q 5 5 (q being ocld). Since the (-4. 5 , s )
pretzel knot admits no non-integral boundary slopes, we cannot have a. non-
trivial finite surgery in this case. Therefore wve can assume 5 < 1.5 - 2pl. In
order for G to be finite, we must have 1s - 2pl 5 9 (Theorem 4.4.4). Since s
61
would also have to be within distance 10 of 2(p + q) we deduce q 5 9. The
(-4,s: 7) pretzel knot has 12615 as its only non-integral boundary slope
and therefore s = 25 is the only candidate for a finite filling. But then
s - 2 p = 15 contradicting an earlier assurnption. Similady: b~ esarnining
the non-integral boundary slopes, we find that the only candidate for a finite
filling on (-4,s: 9) is s = 29 with s - 2 p = 19 > 9. SO the (-4,s: q ) pretzel
knots admit no finite surgeries.
- p = 7. Suppose 1s - 2p( 5 5. Then q 5 7. Hoivever, since the (-4,7, 7)
admits no non-integral b o u n d a l slopes, it cannot lead to a non-trivial finite
surgery. So we can assume 1s - 2pl 2 7. Since G is then infinite. ive see that
(-cl, 7; q ) pretzel knots admit no non-trivial fini te surgeries.
Escept for possibly ( -43: 3) and (-4,3: 5 ) : which will be considereci separately
below, (-4, p, q) pretzel knots admit no non-trivial finite surgeries.
r=6
- p = 3. When q 2 9, the non-integral boundary slope of Lemma 4.2.3 is in
(2q + 12,2q + 13) and the only candidate for a finite filling is s = 2q + 13.
Then s - 2 p = 13 + 2(q - p) 2 25 and G is infinite (Theorem 4.4.4). When
q = 5, using non-integral boundary slopes, we find that the only candidate
for a non-integral boundary slope is s = 23, whence .s - Zp = 17 and G
is infinite. When q = 7. there are no non-integral boundary slopes. So
(-6: 3, q ) pretzel knots admit no non-trivial finite surgeries ( q > 5 ) .
- p = 5. When q 2 15: the only non-integral boundary slope is in the interval
(2q + 12,2q + 14) (Lemrna 42 .3 ) and the only candidate for a finite filling
is s = 2q + 13. Then s - 2 p = 13 + 2(q - p) > 33 and G is infinite. When
q = 5 or 13, there is no non-integral boundary slope. For 7 < q < 11: by
non-integral surgeries, the only candidate For finite filling is s = 2q + 15 + s - 2 p = 15 + 2(q - p) 2 19 and G is infinite. So ( - 6 , 5 , q) pretzel knots
admit no non-trivial finite surgeries.
- p = 7. We can assume 1s - 2pl 5 7 as otherwise G is infinite. Then q < 7.
Since the (-6,7,7) pretzel knot admits rio non-integral boundary slopes, ive
conclucle that (-6,T, q ) pretzel knots have rio non-integral finite surgeries.
62
Thus, aside from possibly (-6,3: 3): the ( - 6 , p , q ) pretzel knots admit no non-
trivial finite surgeries.
0 r=8
- p = 3 . By Proposition 4-43 (-8,3, q ) pretzel knots admit no non-trivial
finite surgeries.
- p = 5 . FVhen q 2 15, by Theorern 4.1.3 and Lemma 4.2.3. the only candidate
for a finite filling is s = 2q + 17. Then s - 2p = 17 + 2(q - p) 2 37 and G is
infinite. When q = 5 . or 13: there are no non-integrai bounclary slopes and
for 7 5 q 5 11: the only candidate is s = 2q + 19 which would again lead to
G infinite.
- p = 7. We c m assume 1s - 2pl 5 7 whence q 5 7. Since (-S,7' 7) has no
non-integral boundary slopes, we are done in this case as well.
Therefore (-8, p, q) pretzel knots admit no non-trivial finite surgeries.
0 r=10
- p = '7. Again 1s - 2p( 5 7 q 5 7. Since (-10:7,7) has no non-integral
boundary slopes, we7re done-
The (-10, p7 q ) pretzel knots aIso have no non-trivial finite surgeries.
It remains t o esamine the knots (-4: 3 , 3 ) , (-43,s) and ( - 6 : 3,3) which are not
coverecl by Theorem 41 .3 . We can follom the procedure outiined above to see that
the group ? i l ( M ( s / t ) ) of a s / t filling on a (3, q, -r) pretzel knot projects ont0 G =
(2: 3' 1s -2pt l ; r / 2 ) . Since r / 2 is 2 or 3: this mil1 be an infinite filling unless 1s -2tpl =
1s - 6tl 5 6 and 1s - 6tl # 2 (Theorem 4.4.4). B y [ B Z l , Corollary 1-31, a finite
filling s / t has denominator 1 or 2. On the other hand, by the work of Ag01 [.Ag]
and Lackenby [LI, such a filling must also be close to the toroidal filling 2 ( p + q )
(Lemma 44.1) : 10 2 A ( s / t 7 2 ( p + q ) ) = 1s - 2 t ( p + q ) ( = 1.5 - I t ( 3 + q ) I . These
considerations leme only a few candidates for finite surgery.
0 (-4,3,3) Since 6 2 1s-6t17 1s-6tl # 2. 10 2 I.s-2t(3+q)l = 1s- 12tl: and t = 1
or 2. the only candidates for a finite surgey on this knot are integral surgeries
2 , 3 , 5 , 6 : i,9,10,11,12 and half-integral surgeries .s/2 with s ocld, 15 5 s 5 17.
Moreover, Lemma 23.1 allows us to eliminate the boundary dope s = 12. One
can verify directly, using SnapPea [Wee], that none of the remaining candidates
are finite surgeries.
(-4,3,5) Here. 6 2 1s - 6t1, 1s - 6tl # 2: ancl 10 > 1s - 2t(3 + q)l iniply
s = 6,7,9.11,12. We can ûlso eliminate the boundary dope s = 12 ancl again
verify directly that these fillings d o not yield manifoids with finite fiindamental
P 'UP - e (-6,3,3) Since q = 3 and s = 12 is again a boundary slope. rve are left mith the
same candidates for a finite filling as t e had in the case of the (-4.3: 3) pretzel
knot. FVe can verif? directly that none of these are finite fillings.
In summary then, Theorems 4.1.3 and Theorem 4.4.4 combine to show that a
( p . q , -r) admits no non-trivial finite surgery unless 4 5 r 5 10 and 3 5 p 5 7. IVe
then investigated those cases directly t o observe that they also admit no non-trivial
finite surgeries. We have therefore provecl the following.
Theorem 4.4.5. A (p, q, -r) pretzel knot, with 4 5 r euen and 3 5 p 5 q odd admits
no non-trivial finite svrgeries.
4.5. Surgeries on Montesinos knots. CVe can combine t h e results of the last t r o
sections with the work of Delman to classify cyclic surgeries on 54ontesirios knots.
Theorem 4.5.1. The only non-toms Montesinos knot which admits a non-tr-iuial
cyclic surgery is the (-2.3.7) pretzel knot. The non-trivial cyclic surgeries o n this
knot are of slope 18 and 19.
Remark: A torus knot admits an infinite riumber of cyclic fillings.
Proof: Delman [Del] has shown that if such a knot admits a cycliç filling? then it is
a pretzel knot of the form ( p , q , - r ) : with 2 5 r cven and 3 5 p 5 q otld. -4s only
the trivial knot admits a Z filling ([Gall) Theorem 4.45 implies fiirther that r must
be 2. Proposition 4 - 3 2 completes the proof. O
For finite surgeries. ive have:
Theorem 4.5.2. If a non-toms ikiontesznos knot K adrnits a non-t~izrial finite surgeq,
then one of the followiny holds.
O K is a ( - 2 , p , q ) pretzel knot vrith 5 5 p 5 q odd and the fillzng is not cyclic.
O K is the (-3: 3, 7) pretzei knot a n d the f i lhg zS aiong dope 17,18, o r 19.
0 Ir- is the ( -2 ,3 ,9) pretzel h o t and the fillzng zs along dope 22 or 23.
Proof: Again' Delman [Del] allons us to reduce to the case of a (p : q. -r) pretzel
knot and Theorem 4 . 4 5 further shows that r = 2. The finite surgeries on (-2,3, n)
pretzel knots are classified in Section 5.2. That a non-trivial finite filling of (-2: p. q )
with p 2 5 is not cyclic follows from the previous theorem. O
Remark: Although the surgeries listed for the (-2: 3' 7) and (-2.3: 9) knots are
finite surgeries, ive know of no instances of a finite surgery on a knot ( - 2 , p , q) with
.5 5 p 5 q- Indeed, we expect. that there are none.
We continue our investigation of pretzel knots this time restricting to knots of the
form ( - 2 , 3 , n) and (-3,3, n) . The Seifert fillings of these knots allom us to make an
explicit calculation of their Culler-Shalen seminorms. This in turn leads to a precise
description of t heir SL2 (@)-character c-arieties. Moreover, it alloms us to construct the
Newton polygon For the A-polynomial of these knots (much as rve did for the twist
knots in Section 2.3). The study of Seifert fillings of these knots gires a concrete
demonstration of the power of Culler-Shalen seminorrns, not only in the study of
cyclic and finite surgeries, as we sarv in the previous chapter, but also in the study of
Seifert fillings.
This chapter. which is largely an expanded form of the papers [BA4 Z: Matl: blat2],
begins with a calculation of the total minimal seminorm of the ( 2 , p. q ) pretzel knots.
5.1. The total norm of (2,p,q) pretzel knots. Let Ii = K ( 2 , p : q) be a pretzel
knot as illustrated in Figure 19. Since K is a knot, p = 2k + 1 and q = 21 + 1 are
crossings
FIGURE 19. The (2, p, q) pretzel kiiot.
odcl. Let :\I = Sv \iV(K), li = (M). The two-fold branched cyclic cover E2 is a
Seifert manifold with base B = S2(2, Ipl, Iql) and x ~ ' ~ ( l 3 ) = A(2, Ipl, Iq ( ) .
For each irreducible component R, of the representation variety R = R(M) rvhich
contains an irreducible character, let ,Y, = t (R) . By [CSl, Proposition 1.4.4]: Si is
an affine variety and a closed sub-variety of the character variety S = S(J1). We
will denote the smooth projective cornpletion of Si by ,Ti. Since Ii is srnall ([Oe.
Corollary JI), the .C, are curves ([CCGLS, Proposition 2.41).
Let z E Si mith Z,(f,z) > Z,(f,). We cm find S, the suni of the minima of
the Culler-Shalen seminorms over the components -Yi of S, by enumerating such
"jumping points" and then showing that each contributes two to S. Thtrs S will
simply be twice the number of jumping points-
-1s a first step, ive show that x is not an ideal point using a modificz.tion of the
argument of [CGLS, Proposition 1-6-11.
Lemma 5.1.1. For each cr E Hl(t3M; Z), f,~ = (/J2 + 4 fa.
Proof: Let ,yp be a point of X. Let p(a) = ( z d ) E SL2(@) Thin
= (a' + d2 + 2bc)* - 4 (since ad - bc = 1 ):
cl Let x be an ideal point with Z,( f,z) > Z,( 1,). Then f,?(x) = O so that x is not a
pole of fp?. By Lemma 5.1.1, x is also not a pole of f, and therefore I , (x) # ai as
well. It follows that either LW admits a closed essential surface, or else p is a boundary
slope (see [CGLS. Proposition 1-3-91). However, since is a Montesinos h o t with
less than four tangles, neither is true (see [Oe. Section 1 and Corolla- 41).
Thus we csn assume x E .Y: and wite ~ ( x ) = X, with p E Ri. (U : Sr +- Si is
normalization. See Section 2.1 or [Shf. Chapter II, S3].) We will argue that x is the
character of an irreclucible representation. Suppose instead that p were reducible.
Then, bu conjugating, we can take p to be a representation into the iipper triangular
matrices. Xow replace each matris p(g) = O a-'
obtain a diagonal representation po with the same character v ( x ) .
Since (1" O ) ( a b ) ( l j n O)-'= ( b h 2 )
O n O a-[ O n O ü1
ancl R, is closed under conjugation [CSl, Proposition 1-1-11, ive cari find representa-
tions on arbitrarily close to po. But, as Ri is closed, po € &. So wittiout loss of
generality, ive c m assume p is diagonal.
The Zariski tangent space a t p may be identified with a subspace of the space
of 1-cocycles Z L ( q . ~ l ~ ( @ ) ~ ~ ~ ) (see [GI, Section 1-21 or [Wei, Section 31). We can
see that l& is four dimensional since, as tve have meritioned, ,\ii is one dimensional
(the knot being small) and by [CSI? Corollary 15-31, dim = dirn -Y, + 3. Tlius.
clirn (Z1 ( x ; sZ2 (C) 2 4. New: by Lemma 5.1.1, zeroes of f, are also zeroes of fp2 ancl moreover, the orcler
of zero agrees a t such points. So Zr( fp2) > Zl ( fp) implies 2, (j,) = O and t herefore
trace(&)) # f 2. On the other hand, trace(^(^*)) is I 2 : so p ( $ ) = AI (we're
assuming that p is a diagonal representation). It follows that p(p) =
Since sr is normally generated by 11, p(sr) = 2 / 4 is cyclic.
( y :J- Given this, Ive can calculate dirn (Z1 (T ; S I ~ ( @ ) ~ ~ ~ ~ ) ) clirectly. Using [BN? Theorem
1.1 (i)17 dirn ( H L (H; s Z ~ ( @ ) ~ ~ ~ ) ) = bl (x; 812(@).4dp) = 1. (This argument is esplainecl in
more detail and in a more general context in Section 5.2-1.) We can also determine
dirn B1 (7i; sZ2(@) rldp) as we haïe the surjection
Since p(p) = the kernel is the one-climensional set {A E s12(@) 1 -4 =
(: 0.) } rvhile s12 (@) has dimension 3. Therefore. dirn ( B L (7: s12(@) ..id,)) = 2
This contradiction with our earlier estirnate of the dimension of the cocÿcles shows
that there can be no jiirnp a t the character of a reducible representation.
Thus we can assume that p is irreducible. Then, by [CGLS, Proposition 1-52] ,
p ( $ ) = fI and f i , the inducecl PSL2(@)-representation of T r : will factor tliroiigb
nl(&). If p is non-abelian, Lemma 3.1.5 shows that P factors through to give an
irreducible representation of A(2, lpl, lql). O n the other hanci: if P is abelian, it
factors through the finite group Hl (&). In this case f i (+) is cyclic and extending to
sr and lifting we see tha t p has binary dihedral image in SL2(@).
Furthemore, any such dihedral representation will result in a jumping point. For
let u(z) = X , be the character of the binary dihedral SL2(@)-representation p. The
corresponcling PSL2(C) representation p has as image a diheclral group normally
generated by p(p). Therefore P(p) is of order twvo and consecpently p(p) # &I mhile
p(p2) = + I - This implies trace(p(p)) = O so that Z,( f,?) > Z,( fp).
The nitmber of such dihedral characters ci can be related to the ,Alesander poly-
nomial AK( t ) - Indeed, d is equal to (carcI(Hl(C2)) - 1)/2 ([KI. Theorem 101) while
card(Hl(E2)) = lAK(-1)/ ([Rol, Corollary 8.D.31). So d = (lAn(-l)l - 1)/2.
By Equations 2.2 and 2.3, there are 121 LY] - irreducible PSL2(C)-characters of
A(2, lpl, IqI). The corresponding representations each estend to an irreducible repre-
sentation of si. These in turn can be estended to sr (Proposition 3.3.1). Moreover,
(Scholium 3.3.3) any representation p which extends Po is such that p ( p ) has order
twvo. Thus, as waç the case for the dihedral represeiitations, the irreducible represen-
tations of A(2, IpI. 1q1) al1 lead to jumping points where Z,(f,z) > Z,( f,).
There are LY] - L ~ J irreducible PSL2(@)-characters of ti mhich factor through the
triangle group A(-: Ipl, Iql) and ([Ah.(-1) 1 - 1)/2 = (Ipq + 2 ( p + q) 1 - l ) / Z irreducible
dihedral characters. (Note that by [Mru, Proposition 141 or [Hi: Theorem 1.21:
By Lemma 2-4-10. none of the dihedral characters go through A(2, Ipl, Iq/). Since
PSL2(@)-dihedrai characters are covered once in SL2(@) and otlier diaracters are
covered twice ([BZl: Lemma 5-51); we see that there are (Ipql - (Ipl + 141) + Ipq + Z(p + q)1)/2 jumping points where Z,(fp2) > Z,( f,). This allows us to calculate S
once \w7e have shown that Z,(f,z) - Z,(f,) = 2 a t each of these points.
The idea is to follow the argument of [BZl, Section 41 (see also [BB, Tlieorcm
A]). The essential requirements are that p(srl(%1)) P {fI} and that v(x) is a
smooth point of ,\'i. Suppose first that p is a representation which factors through
A(2. lpl, Jql). \.Ve proceed by showing that the corresponding point y = X , in Y, the
character variety of ii. also satisfics these recliiirements.
69
Lemma 5.1.2. Let Po be the PSL2(@) representation indvced by po. If p 0 ( r L (a$)) c { f 1) then p zs an octahedral representation.
Proof: hctually we \ d l argue that if the preferred longitude X has trivial image in
A(2, Ipl, lql). then p is octahedral.
Let 1 E ir denote the class of a lift of a representative of A. Trotter [Tr] has
esplained how to find the image of ;\ in A ( p ? qt r ) in the case of a (p, q, r) pretzel knot
with p, q, r al1 odd. We will folloiv the same procedure for the (2: p, q) pretzel.
Starting wit h the Wirtinger presentation [Rol. Section 3.D] for sr wi t h generators
f: g and h as indicated in Figure 19; ive look a t the index two subgroiip consisting
of words of even length and quotient out by the relations f = g' = h' = 1. The
resulting group is al (Z2) - K/{p2). Quotienting again by the center: brings us to
h ( 2 , Ipl, ]pl) = (a , b. c 1 a*, b l p l , c1q1, abc) mhere a = g f b = fh and c = hg. Beginning
with the arc labeled h and tracing out the knot, we find the longitude X E sr:
We can take
Suppose that p o ( X ) = il. Then P-l A P = &.A where P = ~ ~ ( b ~ + k ~ + l ) . It follows
that P is of the form
Let po(b) = B. Note that Bk is not diagonal. If i t mere: then B = B-2k \vould
be as well and would therefore commute with -4. In this case, po woukl be reducible
which is a contraclict ion. Therefore, after conjugat ing by diagonal matrices. ive can
write Bk in one of the following two rvays:
where f r is the trace of Bk and .LL E @ (compare [T322' Esample 3.21).
Taking C = &(c) , ive see that C-' = B ~ P . So the relator abc implies
1 = ABC
- _ 4 ~ - 2 k ~ - 2 1
= A B - ~ P B ~ P .
In other words, Bk=L = P B ~ P . If
then no matter which of the two forms we choose for Bk, the equation B k A = PB"P
will not be satisfied. Therefore
For the time being, Ive will assume
and return to the other case later.
Equating BkA and PBkP (in PSL2(C)) gives rise to the equations
Using these, we may simplify
On the other hand,
Therefore (trace(C-'))2 + ( t r a ~ e ( B - ~ ) ) * = 2. Since C and B are of finite order'
trace&-') = &(< + <-') and trace(l3-9 = &(C + C-') mhere 5 and C are roots of
unity and we have
where ai = 1 (i = 1 . . - 4 ) and as = 2.
Following Mann [Man? Definition 21, an equation OF this form is called irrediicible,
provided there is no equation of the form
where bi = ai or bi = O with a t least one but not al1 the bi zero. It's clear that a t least
two of the bi would have to be zero in Equation 5.10- (Otherwise the complimentary
equation woultl have only one terrn.) So there are basically two possibilities.
If Equation 5-10 is of the form <*+c2 + 2 = O, then we see that c2 = c2 = -1. But
then <-2 = Ç - ~ = -1 as well in contracliction to Equation 5.9. The other possibility
for Equation 5-10 is of the form <* + < - 2 + 2 = O. Again this implies <' = <-' = -1.
Now, since trace(C-') = &(< + we can diagonalize C-' to put it in the form
i ( i!l ) i h e n C = C-" cliagonalizes to II. In other worcls C ir the identity.
This contraclicts irreducibility of po.
We conclude that Equation 5.9 is irreclucible. By Mann's Thcorem [Man, Theorern
11, the solutions to the equation are 30th roots of unity. Checking amongst the 30th
roots of uni- ive see that the only solution is c2 +<-2 = c2 +<-- = - 1 so tliat <* and
c2 are both third roots of unity- -4s: before w e can argue that B and C diagonalize to
z t O ) and t ('. O ) respectii-el) \k conclude that B and C both O C-' O c-?
have order 3. .
If ive take
then again we find that B and C have order 3.
NTe began nvith the assurnption that po(X) = f I and deduced that B and C are of
order 3. Thus po (+) = A(3: 3: 3) witli generators A, B and C. In ot her words Po(?)
is =I.I, the tetrahedral group. As [ ~ i : 51 = 2, the lift to ~ ( a ) is either A4 or $4. Nom,
p(p) has order two and normally generates ~ ( x ) . Since A4 has no such order two
generator, we conclude that P ( K ) = S4, the octahedral group. Thus p(ir) is a (binary)
octahedral representation into SL2(@).
This completes the proof of the claim. III - So, as long as p is not octahedral, po(iri (BM)) # { f 1)- Then, po (r1 (a-c)) Q {f 1)
and p(~~(t9d.I)) {f 1) as well.
We now turn to the srnoothness of v(x). -&gain we will first show y is smooth. -4s
the Zariski tangent space at po can be identified with a subspace of the cocycles. \ire
proceed by investigating the group cohomology.
-4s a first step, we observe that Z L (iil(C2); s l2 (c ) 2 ZL(A(2, Ipl, IqI): .d2(C)
Indeed, the Seifert structure of C3 gives the exact sequence (Eqilation 3.4)
where F = (h) n Z is the group of a regular fibre. The projection 4 induces a
hornornorphisrn <P : Z1(il(2, Ip(, lql); s12(Q)) -f Z1("1 (C2); &(@) i,dp,).
To construct the inverse, rve show that u(h) = 0 for each u E Z1 ( ï ï I (C2); .d2 (c) -4dPO)-
Incleed, for al1 g E ni (&). u(hg) = ~ ( g h ) . On the other iiand. po(h) conirnutes mith
fi(A(2, lpl, 141)). Sinçe f i is irreducible. this implies po(h) = f I by Lcmnia 2.4.9-
Putting it together:
Thus u ( h ) E s 1 2 ( @ ) cornmutes with p o ( 7 r l ( C 2 ) ) . By Lemma 2-47. u(h) = O. Nom
define 9 : 2' ( T I (x2) ; s12 (@) ldpo ) -f ZL ( h ( 2 ; ( P I 7 141) 1 s ~ ? ( @ ) b~ *(SU) ( O ( g ) ) =
u ( g ) . Since u(h) = 0: @ is well defined. Moreover, it7s an inverse of <P and we have the
required isomorphism. This isomorphism also descends to the level of cohomology:
H 1 ( a l ( x 2 ) ; s l 2 ( @ ) AdPo) H L (A(& lp(, IqI); SI*(@) , d p o ) . o n the other hand, we can
argue that the cohomology of the triangle group is trivial directly.
Lemma 5.1.3. If p : A ( p ; q: r ) + PSLî(C) is an irreducible or non-abelian repre-
sentation, then d imc(H ( A ( p , q' r ) ; s l z ( C ) = 0.
Proof: We begin with B 1 ( A ( p , q: r ) ; .&(@) Recall the surjection
By Lemma 2-47, the kernel is empty and
Thus i t sufficcs to argile that dimê(ZL q, r ) ; sh ( C ) ,dp)) = 3 and, as the coboond-
aries are contained in the cocyclcs, it will in fact be enough to argue that the climen-
sion is at niost 3.
Let A ( p l q- r ) = (a, b 1 a* = bq = (ab)'). We can assume
cornmute) and similarly, For B = f (b: :: ) , either b1b2 # 0 or ii3i,PO.
Recall tha t
mhere g - u(gt) denotes the adjoint action of ~ ( g ) on u(gt). Then a cocycle u E
Z 1 ( A ( p , q, r ) : .s12(C) .4dp) is determined b ~ * its value a t a and b:
SO Z 1 ( A ( p : q' r ) ; sL(@) .4d0) c s12(@) @ S I - (C) P c. W e can therefore establish the
inequality dimc(Z1 ( A ( p , q_ r ) ; .s12(@) .4dp)) 5 3 by finding three independent equations
relating the Xi and Yi-
Now,
we have
whence x 1 = O. This is our first equation.
Similarly, if we let ( y : -::1 ) = A%(b)A; then
so that .wl = O as well.
Since 7u1 = (a la4 + anas)yl + a3a4y2 - alazCI.3 and either ala- # O or a3a.[ # O' this
is a second, independent equation in the xi : Yi coordinates.
Finally, the relation (ab)' = 1 allows us to deduce a third equation
which is again non-trivial since either b l b or b3b.* is non-zero.
76
Thus ZL(A(p, q, r); sl2(C),,& is an algebraic set in @6 cut out by a t least three
independent linear equations and so of dimension at most three. O
Now, using the remarks which precede the lemma, and the observation that the
PSL2(@) representation f i and the SL2(C) representation po result in esactly the
same adjoint action on &(@) , we see that d i q ( H L (sil (&): &(@) = O. SO we
can proceed as in [BZl; Section 41 to show that d iqH1(5r : . s / ~ ( @ ) , \ ~ ~ ~ ) = 1 ancl y
is simple in Y. (Note that we will make the distinction between smooth and simple
points of a character variety ..A simple point is a smooth point which lies on a unique
irreclucible component of the variety. See [Shf. Chapter 2 521.)
Proposition 5.1.4. Let p be an SL2(C) -epresen ta ton of a jînitely generated group
sr and po the restriction to a normal sub~roup of finite index i?. T h e n
Proof: The Lyndon - Hochschilcl - Serre spectral sequence gives us the exact sequence
(see [Rot. Theorem 11.51)
mhere --IC = {a E A 1 g - a = g,Vg E G} denotes the set of fised points of the motlule
A ~tncier the group action G. Now, HL (TI?; SI^(@)..^^^)^) = O since si/* is finite and
( - ~ / ~ ( @ ) , \ ~ p ) ~ is a cornples vector space. so we have
cl In our case, the proposition shows that d i m H L (T ; s lZ(Qldp) 5 1 whence ~ ( x ) is
a smooth point of Xi (and in fact a simple point of -Y).
Remark: We have been iising the ideas of [BZl, Section 41 mhereby, under appro-
priate conditions7 x = X , is smooth in X(T) esactly wlien dimc H1(; i ; ~1~(@)..\dp) = 1.
Interpreted in this contest: the proposition says "simpIe points of S ( 5 ) lift to simple
points of X(R) ."
n
Thus if p factors through A(2, Ip[, Iql) and is not octahedral. then v ( x ) is a smooth
point of Si and p(x l ( a M ) ) $? {II}. Following the reasoning of [BZl: Section -11: we
conclude that ZZ(fp?) - Zz (fp) = 2.
Ive can see that the jump a t a dihedral character is also two by adapting Tan-
guay's [Ta, Propostion 5.3.31 arguments for dihedral characters of two-bridge knots to
the present situation. Again, the argument cornes down to showing H ( T : .d2 (C) )
has dimension 1 when p is a dihedral representation. Let p ( ~ ) = D4,, the binary
dihetlral group of order 4m. Then --Ldp(r) C .-\ut(SL2(C)) is isomorphic to D2m9 the
dihedral group of order 2772.
Tanguay shows that the Betti number bl ( K ; s ~ ~ ( @ ) . ~ ~ ~ ) can be related to the Betti
numbers of several covers of M:
where the r d are the kernels of the maps
and 4 and p are the Euler and Mobius Functions respectively Xow, bl (i;; @) = 1 [Rol,
Esercise 2.E.61 and b l (ir; @) = 1 [Rol, Section 8.D]. So shorving that b l ( T ; ~ [ ~ ( @ ) . ~ ~ p ) =
1 and diheclral characters are smootli reduces to arguing that b l ( r d ; @) = d . - - Let be the covering of d l corresponding to r d . Then also covers hf and
this covering may be estended to an orbifold covering Ed -+ S 2 . We mil1 argue that - bl (&) = O . Then, since Cd is obtained from by filling dong cl tori, O = b l ( C d ) > b l ( - z d ) - d whence b l ( l F d ) 5 d. On the other han& since -cd has d t o r d boundary
cornponents, Lefschetz duality allows us to argue that b l ( z d ) = dirn ~ ~ ( . . \ . l d ; @) 2 d.
Therefore bl (rd) = d, as required.
It remains to show that b l ( X d ) = 0: and this is where we must introduce some new
ideas beyond those used by Tanguay. We have the diagram
where the horizontal rows are the exact sequences arising from the Seifert structure
of Cd and Ca: E r F E Z represent regular fibres, and Bd is the base orbifold of
C d . NOW. Im(P) is normal since P is a regular cover. This implies Im(P") is normal.
Since E and F are abelian, Im(Pf) is ais0 normal. Thus, the cokernels will be groups
and ive can use the Snake Lemma to obtain the exact sequence
Norv. ker(PU) = O since Bd + A(2: lpl, IqI) is an orbifold covering space. Thus n is Ly
injective. Since P comes from the dihedral covering + iCI + M: we see that
coker(P) 2 Z / d . By the injectivity of a, coker(Pf) v a where a 1 cl. On the other
hand. as the degree d of the Seifert cover Cd +- E2 is the product of the degree a in the
fibres and the degree of the orbifold cover c, ive see that card(coker(P")) = c = d/a.
Howvever, since Im(a) = ker(p), Im(,8) also has cardinality c:
The projection A(2, Ipl, lql) + coker(PM) i Z/c is an abelian representation, and
as such factors through Hl (A (2, Ipl, Iq 1)) = Z/b: where b = gcd(pq, 'Zp, 2 q ) = gcd(p, q ) .
So the covering Bd -+ S2(2, Ipl, Iq[) is either trivial: or else of degree c > 1 dividing 6.
In partic~ilar~ c is odd. So. Bd is either S2(2, [pl, Iql) or else S"('1.2. ...? 2, Ipl/c, Iql/c),
i.e., c cone-points of orcler 2. Given Bd, we have an esplicit formula (Ecluation 3.5)
for r i r l ( C d ) involving the orders of the cone points. Die can then show that Ht(Cc1) is
torsion by examining its order ideal (see [Rol. Section 8.B]). Tlierefore bl(Xd) = O as
recjuirecl.
Octahedral representations can be treated in a similar fashion. In this case,
where 1i = pA1(D6) with Ds a clihedral subgroup of index four in the octahedral group
S.I. Of course bl (T; @) = 1 as before. so t e will want to argue that bl (a: @) = 2 .
Let fi be the covering of M corresponding to DG. Then ~$1 is an irregiilar covering h -
of degree 4 which also covers M. As before, we estend thc coveririg .il -+ JI to
a degree two map between Seifert spaces: X + C2. This leads to a diagrani quite
similar to t hat for the diheclral representation:
In this case, fi, the base orbifold of 2, is either a 1-1 or a 2- 1 cover of S2(3: IpI, IqJ).
In particular it is a regular cover and coker(Prr) is either trivial or cyclic of order
2. Since coker(Ptr) is abelian, it is again a factor of Hi(A(2; lpl, Iql)) S Z / b where
b = gcd(p, q) is an odd nurnber. Thus B = S2(2. Ipl, lql) as well and, as in the dihedral
case- we find bi (2) = O. Since ~$1 has two b o u n d a - components, we may now argue
that bl (a; @) = 2. Thus b, (ri; s~~(C)&&, ) = 1 and v(z) is again a smooth point of Si
yielding a jump of two.
So each of the jumping points v(x) E Xi is a simple point of .Y (in particiilar,
u - ' ( ~ ( x ) ) = x) and results in a jump of two: ( f ) - 2 ) = 2. This allows us
to calculate S once we have observed that 2si = 21 1p11i = 1 lp-1 li and that Zx( f p z ) 2
Z,( f,), tJz E S ([CGLS, Proposition 1.1.3]).
2.2. (-2,3, n) pretzel knots. Let Ka denote the ( - 2 . 3 : n ) pretzel knot mhicli we
introcluced in Section 2.5. If n is even, ICn is a link, so ive witl take n odd. Since Kr,, is
a ?v\ontesinos knot with three tüngles. it is small [Oe. Corollary 41 and conseqiientl~
not a satellite knot. Therefore II-,, is either a toms knot or liyperbolic.
Lemma 5.2.1, h', is a toms knot zff n = 1: 3 . 5 .
80
Proof: Kawauchi [Kaw, Theorem III] shows more generally t hat a ( p , q, r) pretzel
knot with Ipl; lql? Ir1 > 1 is toms only if {p, q, r ) = {-2,3- 3) or {-2,3: 5). We
present here a direct argument specific to the (-2,3, n) pretzel knots.
W e will show that if n # 1,3,5, then the Alexander polynomial AK, ( t ) is not
the Alexander polynomial of a torus knot. Recall ([blru. Proposition 141 or [Hi,
Theoreml.2]) that
while the polynomial for the p, q torus knot is [Rol, Section T.D.81
In the discussion that follows, we will use representatives for these polynomiczls which
have a positive constant term and no negative porvers of t.
Suppose first that n 2 7. Then AK, (t) terminates with the three terms t" t + 1
mhile AK P 4 (t) terminates with t P - t + 1 (ive will assume O < p < q ) . Thus, if Kn is a
pl q torus h o t , then p = 3. Since deg AKn (t) = n + 3 and deg AKP, (t) = pq+ 1 - p - q .
we see that q = (n + 5)/2 > 6. Actually q 2 7 since it's also relatively prime to 3.
However, although AK,(t) includes the term t5 (when n 2 T ) , AA-p,, ( t ) has no t5 term
(when p = 3 and q 3 7.) This contradiction shows that Kn is not a torus knot when
n 2 7.
If n < O, AKn(t ) includes the term -2t (or -3t when n = -1) while ilh;.,(t)
always has t coefficient -1. Therefore h', ( n < O) is not a torus knot either. O
Since ive are interested in hyperbolic knots, tve d l assume tha t n is an odd integer,
n # 1,3,5. In addition, we will generally assume n # -1 as that case corresponds to
the twist knot K2 treated in Section 2.3. However, we d l vcrify that our coriclusions
also hold true for this knot.
As in Equation 5.1 1, (p = -3, q = -n.)
= 3(ln - 21 - 1) (since n 6 [O, 61 ).
5.2.1. Boundzng the Seifert slopes. Bleiler and Hoclgson [BH, Propositions 16 & 171
have shown that 2n + 4 (respectively 2n + 5) surgery on K, results in a manifold
which is Seifert fibred over S2(2,4, In - 61) (respectiveiy S2(3,5: In - 51/2)). (Note
that there is a small error in [BH, Proposition 1'71 which refers to '4n + 14 surgery on
the (-2: 3,2n+7) pretzel knot." It shoulcl read "4n+ 18 surgery on the ( - 2 : 3. 2n + 7)
pretzel knot ." )
We can bound the Culler-Shalen seminorms of these slopes in much the same way
we calculated S above. We will count the irreducible characters of the surgered
manifold and then show that each such character contributes 2 to the seminorms of
the corresponding slope.
2n + 4 By Equations '2.2 and 2.3, there are In - 61 - 1 irreducible PSL2 (@)-characters
of A(2,4, In - 61). Since d = (IlK(-1)1 - 1)/2 = (In - 61 - 1)/2: half of them are
dihedral. This gives $(ln - 61 - 1) irreducible SL2(@)-characters.
Recall t hat
XE ,Tl where Zx(f) denotes the order of zero of f at x. Since the meridian p of K,, is
not a boundary slope, Z,(jJ 5 Z,(f2n+.l) for each x ([CGLS. Proposition 1-1-31).
This suggests that we approach the calculation of the total norm (see Section 2.3.3)
112n + 41IT by cornparison with IlpllT = S :
Since il1 is small and 2n + 4 is not a strict boundary class. we may apply [CGLS,
Proposition 1.6.11 to see that Z,(/2,+i) = Zl(fp) at ideal points. Thiis the sum of
Eyuation 3.12 may be restrictecl to x E -Y:.
Reducible characters do not contribute to the sum of Ecluation 5.12 either. For
suppose v(x) = X, were the character of a reducible representation p E R, with
Zl (h+4) > Zx( f p ) - Since f i is closed and invariant under conjugation, i re can
assume that p is diagonal. Then, as in Section 5.1, we can argue that the dimension
of ZL (T ; is at Ieast 4.
On the other hancl, Zr( fin+.l) > Zz( fp) implies p(2n+4) = f 1 ([CGLS, Proposition
1-5-41). So, if ive take P as the PSL2(@) representation corresponding to p, ttien P
factors through Hl ( M ( 2 n + 4)) P Z/(2n + 4). Since p(?r) is normally generated by
We can use [BN, Theorem 1.1 (i)] to find b; (ir; ~ l ~ ( @ ) ; , ~ , ) . Indeed,
where ,8 = 112 is a (2nf4) th root of unity. Of course [Roi? Exercise 2.E.61, bl (sr: C) = 1.
Now @B is @ with the L-action induced by t - c = c,L? where t is a generator of Z.
Since ri surjects ont0 H 1 ( M ) Zr this gives a cr-action on C.
We can also think of Hl (M) as acting on ~$1: the infinite cyclic cover of M' and de-
fine a C[t. t-l]-moduIe structure on Ifi ( ~ f l ; C) ((see [Rol: Section i.;\]). In this contest
H1(ir: @, ) = c o k e r ( ~ ' ( & l ; C) 3 H' (LG; C)) rvhere t - represents multiplication
by t - p. Since the Alexander polynomial AK,,(t) is the generator of ~ ~ ( 0 : @) as a
@[t: t-']-module, rve can argue that coker(t - 8) = O if AKn (P) # O .
In -1ppendix -4 we show tha t AKn( t ) admits no roots which are 2n + 4th roots
of unity. Thus H1(a ; CB) = O and b1(r: = 1. FVe can then argue as
in Section 5.1 that clim BL ( K ; . ~ l ~ ( @ ) . ~ ~ ~ ) = 2 and dim Z L ( a : . ~ l ~ ( @ ) . ~ ~ ~ ) = 3. This
contradicts our earlier estimate for the dimension of the cocycles ancl Ive conclude
that there can be no jump a t the character of a rectucible representation p.
It remains to examine the x E satisfying Z,(J2,+4) > Z,(f,) and such that
v ( x ) = xp is the character of an irreducible representation p. .As meiitioned above'
Zl(f2n+LL) > Zz ( fp) implies the corresponding PSL2 (C) representation factors
through srl ( M ( 2 n + 4)) .
-As M(2n + 4) is Seifert fibred over S2(2, 4. In - 61): P factors through an irreducible
representation p' : A(2,4, (n - 61) + PSL2(C) (see Lemma 3.1.5.) Nonr: as in Sec-
tion 5.1, H1(7rI(M(2n + 4 ) ) ; ~ 1 ~ ( @ ) . ~ ~ p ) % H1(A(2,41 ln - 61); S L ~ ( ~ ) . . \ ~ ~ ) is trivial.
Thus, arguing as in [BZl; Section 41, we can cleduce that u(x) is a smooth point of
.Yi (and in fact a simple point of X) so that V-l(u(x)) = Z.
Therefore the jumping points u(x) where Z,(fz,+,r) > Z,(/') are simple points of .Y
and correspond to irreducible PSL2(@) characters p which Factor through A(2: 4: In - 61). Conwrsely, any stich representation induces a jumping point. This is immediate
if the representation is diagonalizable on 7rl (t3M) since then p ( p ) is of finite orcler,
but not fI. On the other hand, p(Zn + 4 ) = fI. Thus Z,(f2,+.1) > O = Z,(f,). If
p(irL(aM)) is parabolic, we can appeal to [BB. Theorem A]. In any case' as the jiimp
Zl(f2nc4) - Zr( fp) will be two (see [BB? Theorern A]) a t each of the $(ln - 61 - 1)
SL2(@) characters: we can evaluate the sum of Equation 5.12 to fincl that 112n +41Ir =
S + 3(ln - 61 - 1).
Zn + 5 Since M ( k + 5) has odd orcler first homology, irl (M(2n + 5)) has no dihedral
characters and its irreducible characters correspond to those of A(3, 5: In - 51/2).
When n $ 5 (mod 30), the Alexander polynomial AK, (t) admits no zeroes which
are 272 + -5th roots of uiiity (Lemma -4.1.1). So: as with 2n + 4, there is no jump a t
the character of a reducible representation. Equations 2.2 and 2.3 show that there
are In - 41 - 1 irreducible PSL2(@)-characters of A(3- 5: In - 5112) each of which
is double covered in SL2(@). Thus srl ( M ( 2 . n + 5)) has %(In - 41 - 1) irreducible
SL2(ê)-characters. As was the case for 2n + 4: each of these contribute two to the
Culler-Shalen seminorms of 2n + 5 so that 112n + JllT = S + 4(ln - 41 - 1).
This equation also holds for n = 5 (mod 30) although by a different argument. In
t his case, t here are In - 4 1 - 5 irreducible PSL2 (@)-characters of 11 (3.3: In - 5 1/2) and
consequently 2(ln -41 - 5) irreducible SL2(@)-characters of RI (M(2nf5)) . Four of the
reducible characters of A(3,5, ln-31/2) correspond to reducible representations which
project ont0 Z/ l5. -4s shomn in -4ppendis .A7 the Alexander polynomial admits 15th
roots of unity among its zeroes when 15 1 (n - 5): so ive cannot neglect these reclucible
characters. Indeecl, as we mil1 now show, each one will be covered twice in SL2(C) by
characters contributing two so that we again have 112n + 511~ = S + ~ ( I B - 41 - 1).
If n G -5 mod 30, then In - 5(/2 = l5k and Hl(A(2.3; lnl)) 2 Z/l5. On the other
hand, by Lemma X.1.2, AK, (t) admits primitive 15th roots of unity ancl t h e - are
simple zeroes of AI(, (t) .
Let < = e2*ji/l5 be a primitive 15th root of unity and let p be the reducible SL2(C)
representation of ir induced by
84
Shen p(p)15 = f 1 and p is a cover of one of the reducible PSL2(@) representations of
A(2,3, Inl) projecting ont0 2/15. In other wordso we can think of p as a representation
of sr which factors through 1b1(2n + 5 ) . Corresponding to the eight primitive 15th
roots of unity, ive have eight SL2(@) characters. ÇVe mil1 show that the jump a t each
of these characters is 2.
Frohman and Klassen [FK, Theorem 1-11 show that such a representation p is the
endpoint of an arc of irreducible representations. So p E &, a component of the
SL2(@)-representation variety containing an irreducible representation. The corre-
sponding character x = X, lies on the curve Si = t (&).
Since < = e2"i/1.5 is a primitive 13th root of unitu; ~ , ( p ) = 2 cos(irj/15) # I 2 .
So Z,(f,) = O' rc is a non-trivial character, and. moreover, x( i i l (BM)) # {H). (-1
character is trivial if X ( T ) C ( I 2 ) . See [P, Section 3-21.) On the other hand. since p
factors through M ( 2 n + 5): p(2n + 5 ) = I and Zz (f~,,+j) > O- SO Zz( An+5) > Zz (fp)
and there is a jump a t x.
Now Proposition 1.5.2 of [CGLS] shows that there is a non-abelian representation
p' E R, with character x and p'(2n + 5 ) = f 1. Since x(iri(dM)) # (33): we see that
p'(srl(aM)) {f 1). Finally, as in Section 5.1: ive can argue that HL (si1(;\,I(2n + 5 ) ) ; s12(C) = O. This allows us to apply [BB. Theorem A] and conclude that the
jump, z,(f~,+,) - &(fp), is 2-
5.2.2. An application of Lemma 6.2. Lemma 6.2 of [BZl] relates the Culler-Shalen
norm on a norm cilnie to the boundary siopes. In the case of a h o t . such as Ii",: for
which p is not a boundary slope ive c m write
tvhere the ai are non-negat ive integers and the pi are boundary siopes.
The boundary slopes of IC, can be fourid using the methods of [HO] (see also
Section 4.2). \Ve have = 0, 82 = 2nf6, a = 16 (respectively 10) and = "'-J~ n-
(respectively 2 ( n + 1)'/n) when n 2 7 (respectively n 5 - 1).
85
The calculations of the 1s t two sections imply the following inequalities for any
Culler-Shalen seminorm:
(5.14) llpll = s 5 3(ln - 21 - 1):
(5.15) s < 11272 + -111 5 s + 3(ln - 61 - 1) and
(5.16) s 5 112n + 511 5 s +4( ln - 51 - 2).
These strongly restrict the possible values of the coefficients ai.
For esample,
(5.17)
(5.15) s <
suppose n 2 7. Then the equations above become
2[(2n + 4 ) a l + 2a2 + (2n - 12)a3 + a4] s + 3 ( n - 7 ) and
It will be useful to subtract s from each of the last two equations:
Since ai 3 O, Equation 5-17 implies a . ~ 5 3. In fact, in order to have a norm
(rather than a seminorm), we would need a t least two of the ai > O. This condition
further restricts a . ~ < 2. (Seminorms which are not norms will be disciissecl further
in Section 5.2.4.)
Given a.! 5 2, Equation 5.21 implies (Zn + 3)a1 5 2(n - 6) so that a l = O. Theno
the same equation implies a3 5 1. We will argue that, in facto a., = 2 and a3 = 1.
Suppose instead that a4 5 1. Since al = O, Equation 5.21 becomes (272 - l')cia 5
2n - 13 so that a3 = O. This is a contradiction since if al and a3 are both zero. then
Equation 5:21 in fact says a.[ = O as wvell, and only a2 is non-zero. This n-ould mean
that I I - 1 1 is not a norm.
Therefore a 4 = 2. Since al = O? Equation 5-31 implies that n3 > O. Thus a3 = 1.
F i n a l l ~ given these values, Equation 5.20 can be rearranged to see that O 5 a? 5
(n - 5) /2 . This implies s = 2n - 4 + 2a2! 112n + 411 = s + 2(n - 8) + 2a2 and
1 1 % + 511 = s + 4(n - 7).
For n < -3 tliere are four possible solutions.
1. ai = O o a3 = a 4 = 1 and 0 5 a2 5 ( 1 - n ) / 2 . Then
and Il2n+ 511 = 2(7 - 3n) +?a2 = s + 4(3 - n).
2. ai = a4 = 1, a3 = O and O 5 a2 5 (1 - n ) / 2 . Then
and 112n + 511 = -6(n + 1) + na2 = s - 4 ( n + 2) .
3. If n 2 -23, ai = 1, a3 = a4 = O and O < a2 5 (n + 25) /2 . Then
and 11272 + 511 = - 2 ( 2 n + 5 ) +2a2 = s - 2 ( 2 n + 6 ) .
4. If n = -3, al = a4 = O and a2 =a3 = 1. Then
5.2.3. N o m Cumes. For n > 7: we see immediately that there is a t most one norm
cuwe in the character variety. Indeed. if there were two, each woulcl have s > 2n - 4
mhich would force S > 3 ( l n - 21 - 1) = 3(n - 3). Similarly, for negative n_ we see
that there can be at most one norm curve of type 1 or 2. In orcler to see how the
type 3 solution interacts with the other two: ive use the Seifert slopes.
For a type 1 norm curve, we have 112n + 511 = s + 4(3 - n) which irnplies al1 the
jumping points for 2 n + 5 surgery lie on that curve. W'e've shown that the jumping
points are simple, so they cannot lie on any other curve. Thus any other norm cuwe
would have 112n+Sjl = S. However, examining the other solutions we see thnt 112n+5ll
is greater than s (unless n = -3). So for n 5 -5, if there is a curve corresponding to
a solution of type 1, then there are no other norm curves. Using similar arguments,
we can show that for n 5 -11: there is at most one norm curve. For n = -9, there
are either two norm curves each corresponding to a type 3 solution or else there is
only orle norm curve.
In the following, ive will assume that there is only one norm curve So and we n-il1
denote its CulIer-Shalen norm by I I I l o . The cases -9 < n 5 -1 will be trcatecl
separately Iater.
FIGURE 20. -4 spanning surface of hl', .
FIGURE 21. The handlebociy H .
5.2.4. ?--Cumes. We mil1 argue that h', admits an r-curve only if r = 2n+6. For this.
we'll use the graph manifold structure of M(2n+6): M(2n+6) = .\.I1uAl2: is the union
of two Seifert fibred manifolds MI and !CI?_ along a torus. We can construct Ml by
thicktlning the spanning surface of Figure 20. The surface is a punctiired Klein bottle
so 114' is a twisted 1-bundle over the Klein bottle. We denote the complementary
manifold S3 \ iV(Jl l) by .LI2. \Ve can get a better handle on the Seifert structure of
the Mi (i = 1,2) using the ideas of Dean [Des].
A key observation [Dea, Lemma 2.2.11 is that an irreducible Haken manifold, like
l l f i i , with fundamental group G,,, = (x' y 1 xmyn) is Seifert fibred, the base orbifold
being a disc with cone points of order m and n. Since Ml is obtained from the
obvious genus 7 handlehody H in Figure 21 by adding a 2-handle along ttie knot,
we can c o m p t e it's fundamental group. Indeed, with respect to the generators
a, b of al(H). the knot represents the relstor b-'ab-'a-' . Making the change of
basis, b-'a + c: a + d - l , the relator becorries c2&. Thus Mi is Seifert fibred
88
over D2(2, 2) with ( b - l ~ ) ~ or a2 representing a regular fibre and fundamental group
ri(ii.r,) = (c, d 1 7d2) .
.A similar argument allows us to identify M2 using the generators x and y of the
complementary handlebody H f (see Figure 21). In this contest. the knot represents
the word ZIxY(n-1)~2xy(n-1)/2x. After the change of basis y(n-1)/2r + tu. y-' + ;: the word becomes z("-~)/'Iu~. Thus & is Seifert fibred over D2(3, ln - 31/2)
and has fundamental group n l (M2) = (w: r 1 ~ ' ~ ~ - ~ ) / ~ w ~ ) . Lforeover: a regiilar fibre
corresponds to ( x ~ ( " - ' ) ' * ) ~ .
We can now argue that the fibres intersect once on the common boiindary of ML
ancl hl2. Indeed, Figure 22 shows how a fibre of ML representing a' and a fibre of i&
representing ( x ~ ( " - ' ) / ~ ) ~ have intersection nurnber one. Note also that the :IIL fibre
a2 becomes y - 2 ~ - ' = Y (n-5)1?(zy(n-L)i2)-1 = Y ,(j-n)/2w-l in whereas the
fibre goes t o b-lab-lab-la = ( b - ' ~ ) ~ a - l = Zd.
Proposition 5.2.2. The PSL2(C)-character varietg .T(i\1(2n + 6)) contains exactlg
one cvrue when 3 1 n. Othemise clim X ( ~ ( 2 n + 6 ) ) = 0.
Proof: We will argue that irreclucible PSL2(C)-characters of M(2n + 6) are either
isolatecl or else factor through 2 / 2 +2/3. The result then follows froni [BZ2: Example
3.21.
-An irreducible representation p : M(2n + 6) + PSL2(C) will be non-abelian or
else have image 2 / 2 @ 2 / 2 (Lemma 2-43). O n the other hand. if it's abelian. it
also factors through H L ( M ( 2 n + 6)) E Z / ( 2 n + 6) and there is no cyclic groiip
which contains 2 / 2 @ 2 /2 . Therefore if P is irreducible, it's also non-abelian. Let
pi : K ~ ( M ~ ) + PSL2(C) (i = 1' 2) be the inducecl representations. If one of these is
non-abelian, we can show that it kills the corresponding fibre. Let hi E *,(dl,) be
the claçs of a regular fibre.
Claim 5.2.3. If Pi is non-abelian, then pi(hi) = 311.
Proof: (of Claim) Suppose fi is non-abelian.
Since hl generates the center of iil(iCIL); by Lemma 2 -43? if P l (hL) # I I then
we cün conjugate so that p i (h i ) = E and p i ( ~ i ( M l ) ) c iV (see Definition 2 -42 for
notation).
- - . .. .. - = (-2,3, n) pretzel h o t ----- = a* --- = (q( n- I ) /2 )3
FIGURE 22. The regular fibres a2 and ( x $ " - ' ) / ? ) ~ intersect once
inside the circle.
Since pl is non-abelian, a t least one of the generators, say c: of al (.\,Il) = (c, d 1 c2d2)
is sent to an antidiagonal element and is t herefore of order two. But since k2 generates
the center of (Ml) , this means that the image of the center is trivial, ancl, in
particular, pl (h l ) = a contracliction.
Sinrilarly, if f i is non-abelian, then we can assume (for a contradiction) tliat
Im(fi) c N and &(h2) = E. AS before, one of the generators of nl(1W2) = (w, z 1
90
w ~ z ( " - ~ ) / ~ ) must be antidiagonal. If pt(ui) is aotidiagonal, then pl(w" is as well.
This is a contradiction since both w%nd h2 generate the center Z(7il(M2)) P Z and
we started out by assuming fi(h2) = E. This argument also shows that f i ( z ) cannot
be antidiagonal in case (n - 3)/2 is odd. If (n - 3)/2 is even, we can repeat the
argument we used for p l - So we also get a contradiction in this case. O(C1aim)
Let us assume p ( r 1 ( T ) ) @ {f 1 ) . We wish to show that X , is then isolated
in .Y(M(2n + 6)). Since regular fibres intersect once on T' their images generate
P(?rl(T)). The Claim therefore shows that in order to satisf!- p(;rl (T)) $ {f I } . a t
least one pi is abelian with p*(hi) # -ti. So, suppose p- is abelian and pl is not. As above, Pl ( h l ) = f 1. Since the glueing
torus T contains regular fibres? wve can assume hl E .ni (T). +As the Pi's agree on the
intersection rl (T), f i ( h l ) = I I as well. However, we've seen earlier that a regular
fibre hl represents the word in r1 (Ah) - Since tliis word is killed. & factors
t hrough
which is cyclic
This means
n-6 # O ) . We
since pl ( h i ) =
of order In - 61.
that fi(h3) is of finite odd order (remember that n is odd. so that
can conjugate so that &(h3) = f ( O 1y6 ) with q f *i, + i Noiv,
f 1; Pi Factors through the orbifold group TP'~(LI,) = (c, d 1 c2, 8)
where 0, is the base orbifold of M l . -Again, pl (h2) = f i (h2) is of finite orcler clividing
In - 61 and represents the word c3d. Thus pl factors t hrougli (c. d 1 c2, 8, 1 \
which is diheclral of order 21n - 6 1. Also, (h,) = f i (h2) = z t ( OV ) is in the
image of the cyclic subgroup which is therefore diagonal ( ~ e m m a 2.4.3). .
We have now given a rather specific description of p. Restricted to nl (h l2 ) : it is
cyclic of order dividing (n - 61 and diagonal. Restricted to srl(i\.ll) it factors through
D21n-61 with the cyclic subgroup having image in the diagonal matrices. Morcover,
A= (u ;,) mith 71 # f 1, &i is common to the images of ni ( M l ) ancl (A&)-
There are only a finite number of characters consistent with such a representation.
Thiis characters of this form are isolated in the sense that they cannot forrn a cun-e.
Next, assume pl is abelian and f i is not. Xgain fi(h2) = &I which implies
goes through ?rl (M1)/(h2) = (c, d 1 c?62,c3d) = 2/4. Then P(hl) is of order dividing
4 ancl since it is non-trivial: it has order 2 or 4. O n the other hand, as the fibre ha
is killed: & goes t hrough the orbifold group 2/3 * Z/(ln - 31/2). Moreover, since
Pl(hl) = &(h2) has order 2 or 4 we deduce that f i factors through A(3, In - 31/2,2)
or A(3, In - 31/2:4). So we are again in a rather restrictive situation. The number
of characters of such a triangle group is finite (see Ecluation 2.2) and: since the third
generator of the triangle group is of order 2 or 4 and generates fSl(Ml), there are also
only a finite number of characters for such a representation p. So they are isolated.
Finally ive turn to the case were both Pl and & are abelian. This means tha t
p(.lrl(T)) must commute with everything in the image of the non-abelian represen-
tation p. FVe are assuming o(nl(T)) # {H), so, by Lemma 2 - 4 3 and after a n
appropriate conjugation, P(IT~ (T)) = {f 1, E} and P(nI (M(2n + 6))) C N .
Now, as pl and fi are abelian, they will factor through 2@Z/2 and Z @ L / g , (mhere
g = gcd(3, In - 31/2)) respectivel. Since p is non-abelian, there is an antidiagonal
matris in its image. On the other hand: since E E p(.i-rl (T)), at least one of the
piYs has image including both diagonal ancl antidiczgonal elements. The only way this
could happen, while the pi in question remains abelian woulcl be for it to have image
2 / 2 8 2 /2 (see Lernma 2-45). We see that this is feasible only on the Ml sicle and
conclucle that pL(nl(ll.ll)) 212 @ 212.
In particular, PI (h l ) has orcler one or two. So &(nl (-1.6)) either factors through
Z/ln - 61 (as above) or else through (w, z 1 ~ ~ r ( ~ - ~ ) / ~ , (i(a-n)/2w-1)27 [w, 21) whicli
is cyclic of order 2172 - 61. Moreover, we can assume that Im(fi) consists only of
diagonal matrices. Again there are only a finite number of characters correspondirig
to representations of this form so they are isolated.
92
In summary then, the only way t o construct a curve in d v ( ~ l (2n + 6)) is by making
use of representations jj which kill the glueing toms T and therefore factor through
irl ( M ( 2 n + 6))/.ir1 (T ) = (ni (Ml) *-(TI ~ L ( J [ ~ ) ) / ~ L ( T )
= rl (M1)/7ïl (T) * TL ( ~ 1 1 2 ) / ~ 1 ( T )
= 2 / 2 * z / g .
where g = gcd(3, In - 3112). If g = 1, this is an abelian representation, contradicting
an earlier assiimption, and there is no cun-e in X(i\i(2n + 6). If g = 3. (Le. if 3 1 n).
we see tha t we are looking a t representations of 212 * 2/3. Since S(Z/.~ + 2/3)
contains exactly one curve (see [BZ'Z: Esample 3.21): ive conclude that this is also the
case for X ( ~ ( 2 n + 6)). O
Thus if 3 1 n, there is a unique ciinTe in x(M (2n + 6)). Moreover, since the
representation (see [BZl. Esample 3-21 or Equation 5.22 belon?) is diheclral. this
curve contains the character of a clihedral representation- It follo~vs that the curve is
covered by a unique ciirve in the SL2(@)-character variety .Y(M(2n + 6)) (see [BZl,
Lemma 5-51]). Thus there is exactly one r-curve, cal1 it SI. mith r = 2n + 6 in this
case.
Moreover. we can show si = 2 for this curve. Recall that sl = IlplIl is the degree of
f,. So ive rieed to understand the image of p under the composition ir -+ rl ( A I (2n i-
6)) + 2 1 2 + 2/3 = (c: d 1 c?,d3). We can construct p in terms of a curve n/ in the
genus two surface which connects points on opposite sides of the h o t : see Figure 23.
The idea is that we can break up the mericlian as the sum of a loop in :\IL ancl a
loop in M2. We will then show tha t those project to c ancl d respectively. Tlie first
loop is -! plus a small arc joining the two endpoints of y in the interior of H. the
obvious genus two handlebod. The second loop is y plus a small arc joining the two
endpoints of y and passing through the c o m p l e m e n t a ~ genus two handlebody H'.
In ;il ( M L ) , y represents ab-l which is conjugate to b-la and therefore projects
ont0 the generator of 2 / 2 = ?ri (Ml)/?rl (T). In 7r1 (1Li2)[2)< -y represents x ~ ( " - ~ ) " which
projects to the generator of 2/3. Thus p is rriapped to cd in 212 * 2 1 3 .
94
of two (three) curves when 3 j n (3 1 n) and n 2 7 or n 5 - 11. This observation also
holds true for -9 5 n 5 -1 as we d l now verif .
n = -9 Here 3 / n' so there is an r - c u m ,Yl with r = 272 + 6 = -12 and .sl = 2.
We can show that tbere is no r-cume with r = O or 10 as we did before and we've
already mentioned that if there were two norm curves, they would both correspond
to a type 3 solution. Let us verify that this cannot happen. Suppose then that there
were two norm curves ;Yo and J&- Since S = 30 and si = 2' we see that
where we have given the a2's (see Eqiiation 5.13) superscripts showing which cun-e
they corne from. This implies a: + a; = 12. But then
which contradicts. the equation 112n + 41Ir = S + 3(jn - 61 - 1) = S + 42. Thiis we
see that there is esactly one norm cunre and one r-cun7e when rz = -9.
n = -7 In this case there is no r-curve for r = Ln + 6 = -S. By esamining the
norm of the 2n + 5 = -9 dope, ive see that if there is a norm curve of type 1: it is
the only curve in X(1L7) containing an irreducible character. Sirnilarly, if there is
a norm curve of type 2, then there is no r-curt-e with r = 10. Howec-er: we cannot
immediately eliminate the possibility of an r-curve for r. = 0.
Incleed, ive know that for the type 2 norm c u m -Yo
and 112n + 5110 = 1 1 - 9110 = so - 4(n + 2) = so + 20.
On the other hand, if there were an r-curve Si with r = 0, then 112n+411 = I I - 1011 5 sl + 3(5 - n) - 8 = .sl + 28 and also 112n + 4111 = I I - 101li = slil(-l0,O) = los1.
This implies
95
Since .sl is an even integer, we see that sl = 2- Similarl~r, an esamination of the
-9 slope also leads us to the conclusion that SI = 2. So we cannot eliminate the
possibility of an r-curve ivith r = O directly as we did earlier. We need to examine
possible combinations of curves.
For example, suppose S(Kh'ï), had a type 2 norm cirrve So and one r-cun-e -Y1
for r = O and no other norm or r-curves. Then II - 9110 = so + 20 and
Thus, if we assume that these are the only two curves, ive see that we cannot
accoiint for al1 the jumping points associated with the -9 s1ope. Thercfore this is
not a possible configuration for iY(K-7). By analyzing the possible combinations of
norm curves and r-curves in this way, ive see that the only possibility is that there is
esactly one norm curve of type 1 and no r-cuwes.
n = -5 .A similar analysis shows that ,Y(IC,) contains exactly one norm curve and
it's of type 1.
n = -3 Since 3 1 n, ive know that there is an r-curve ,E-l for r = 272 + 6 = 0:
96
If Ive follow the same strategy as in the previous cases we find that there are two
possible configurations:
1. In addition to the r-curve there is one type 1 norm curve ,Yo with
II. Here there is a type 2 norm cun-e -Yo:
as well as an additional r-curve with T = 10:
Both configurations are consistent ivith S = 3(ln - 21 - 1) = 1%' I I - % I l T =
S + 3(ln - 61 - 1) = S + 24 and I I - l(lT = S + 4(ln - 41 - 1) = S + 24.
In order to show that the second configuration does not arise, recall t hat the
PSL2(@) analogue of .\i, would include into . T ( ~ ( l O ) ) (see Section 7.2.3 or [BZ2,
Erample 5-10]). Now, hl(- 1) is Seifert fibred over S2(3. 4,s) and the jumping points
for [I - 111 come from the six irreducible PSL2(@)-characters of A(3: 4.3). If the
second configuration is valid, five of these characters are on -x-? and therefore come
from representations lying in ~ ( ~ ( 1 0 ) ) . We will argue that at least two of them do
not.
Indeed two of the characters correspond to representations diich factor E hrough
A(2,3,5) which has order 60. On the other hand, if such a representation P is also
in ~ ( M ( 1 0 ) ) ~ then it annihilates both the 10 and the -1 slopes. In other words,
the kernel of p contains an index eleven subgroup of ~ ~ ( & \ l ) . Therefore P(T~(BIV))
is either Z / l l or else trivial. On the other hando p(xL(dM)) also Factors through
2 3 5 ) Thus p(r i ( B i l l ) ) is trivial and since n1 (61) is normall- geiierated by the
peripheral group, p(ïr,(i\-I)) = {f I } as well. This contradicts the fact that p is an
irreducibte representation. Therefore, the irrediicible representations which factor
throiigh A(2,3,5) are riot in ~ ( ~ ( 1 0 ) ) . This shows that the second configuration is
not possible.
n = -1 This knot was treated using different methods in Section 2.3 (where it is
identified as the twist knot 16). W e saw tha t there is one norm curve of type 1 in
the character variet?;.
Thus for any hyperbolic pretzel knot AmnI the character variety contains one norm
cume ,Yo and one curve of reducible characters. If 3 1 n t h e is an additional r-curve
XI for the slope r = 2n + 6 with sl = 2-
If 3 f n, then so = 3(In - 21 - 1) and the Culler-Shaler~ norm is given by
when n 2 7 and
when n < -1.
If 3 1 n, then so = 3 / n - 2 1 - 5 and the Culler-Shalen norm is
when n 2 7 and
when n 5 -1.
'lote that the longitude has an associated ideal point only when n = -1 or n = -3.
That is. for the boundary slope pi = 0, the associated ai is zero unless n = -1 or
n = -3. -4s Ive mentioned in the Remark of Section 2.3. this corresponds to the
fact that these knots fibre over SL with a Seifert surface as fibre unless rt = -1
(see [Gaz, Section 61). Altliough 1c4 admits such a fibration there is norietheless
a t least one ideal point associated to the longitude. This is because there are other
essential surfaces which realize the longitude as a boundary dope but are not fibres
in a fibration of the knot. Essentiall_v, when .n = -3, the boundary slope 2 n + 6 tirrris
out to be zero, and corresponding to the surfaces which realizes 2n+G in al1 ttic other
knots, t here is a surface realizing the longitude dope of O for this h o t . However, this
surface is not the leaf of a fibration of K4, just as the corresponding 2n + 6 surfaces
do not play that role for any other Kn-
FIGURE 24. The fundamental polygon of Kg.
Fundamental Polygon a n d Newton Polygon. Figure 24 shows the fundamental poly-
gon of 6 üssociated t o the norm curve Xo. For Kg, ma'c(2so. so + S) = 2so so
any finite surgery slopes must lie in 2B (see Section 2.2.5). Hoivever. as we see in
Figure 24, the only slopes inside 2B are 21: 22, 23 and p = I /O. -4ccording to Snap-
Pea [Wee], M ( 2 1 ) is llyperbolic and so (-2: 3.9) admits exactly two non-trivial finite
surgeries: 22 arid 23.
Xs n increases, the fundamental polygon for the norm curve niaintains the same
aspect but becomes smaller. For I I 5 n 5 19. the only slopes inside 2 B are 272 + 4.
2n+5 and p and once n 2 21, onIy 2n+4 and p rernain. Now M(2n+-4) ancl d i ( 2 n + 5 )
are Seifert fibred over 1 1 ( 2 , 4 In - 61) and A ( 3 : 5 , In - 51/2) [BH, Propositions 16 91
171. Since. for n 2 11, 1 / 2 + 1/4 + l / ( ln - 61) < 1 and 1/3 + 1/3 + 2 / ( l n - 51) < 1,
these are hyperbolic orbifolds and consequently M ( 3 n + 4) and M(2n + 5 ) are not
fini te siirgeries. Thus the (-2: 3: n) pretzel knots admit no non-trivial finite surgeries
when n 2 I l .
Figure 23 gives the fundamental polygon of the (-2: 3, -7) pretzel knot and illus-
trates the situation for n 5 -1. \Vhen n < 1: s o < 8. Since 2 B lies below the line
y = 1 there are no non-trivial finite surgeries liere. When n = - 1: we have a twist
FIGURE 25. The fundamental polygon of
knot, and ive have already obsemed that they have no non-trivial finite surgeries
(Section 2.3).
Thus there are esactly five non-trivial finite surgeries on the (-2,3: n) pretzel
knots. As we have mentioned, IC9 admits two. We can use the same methods to see
that I(7 has three (see [BMZ, BZl]), two cyclic fillings 18 and 19 and one non-cyclic
fioite filling of dope 17. These fillings were already known ([FS? Section 41 and [BH.
Propositions 16 and 171). The content here is that these are the only esamples of non-
trivial finite surgeries. As none of these fillings are simply connected this constitutes
a proof of Property P for these knots. However, Property P was aIready known as
these knots arc strongl- inwrtible [BS].
We are also in a. position to determine the Newton Polygons for these knots (see
Section 2.2.5).
The vertices of the Newton polygon are
(0, O), (16.1). (n2 - 2n - 15, (n - 5)/2), (2(n2 - n + 3), n - 2):
mhen n 3 7 ancl 3 f n (see Figure 26 which ma? be comparecl with [Shn? Figure 51);
FIGURE 26. The Newton poiygon of 1cn (n 3 7 and 3 j n).
b = (lO,3(l- n)/2) c = (2(n2 + 2n + 6 ) , (3 - n)/2) a
d = (3(n2 +2n+3), 1)
a = (O, (1 - 3 n ) / 2 )
f = (n2 + 2n - 3, -n) e = (3n2 + 6n - 1,O) e
- - - - - - - - - - - - - - - - - - - - - - - - -
F~GURE 27. The Newton polygon of I(, (n -5 a n d 3 rnr (3n2 - 6n - 31,(3n - 13) /2 ) , (3(n" 272 - 5) . (3n - 1 1 ) / 2 )
when n = 3k, k 2 3;
(0, (1 - 3n)/2) , (10'3(1 - n)/2), (n' + 272 - 3; -n),
(2(n2 + Sn + 6), (3 - n)/Z) , (3n2 + 6n - 1, O ) , (3(n2 + 2n + 3). 1)
when n 5 -5 and 3 f n (see Figure 27);
(0 , -(3n + 1 ) / 2 ) , (10, ( 1 - 3n) /2 ) , (n2 + -In t . 3 , -n):
(2(n2 + 2n + 6 ) , (1 - n)/2), (3nZ + Sn + 5: O), (3n2 + 8n + 15.1)
FIGURE 28. The (-3,3:4) pretzel knot.
when n = 3k, k 5 -1; and
(O: O ) : (O: l), (4,2), (IO, 11, (14: 2), (14: 3)
when n = -1.
5.3. (-3, 3, n ) pretzel knots. We now turn to the (-3,3, n) pretzel knots. We begin
by examining the (-3,3,4) pretzel knot K as illustrated in Figure 25. First note that
K is hyperbolic. Since it's small [Oe, Corollary 41, it's not a satellite h o t . The only
other possibility is tha t it is a torus knot. However, as it has -Alexander polynomial
([Mru, Proposition 141 or [Hi: Theorem 1-21) AK( t ) = ( t2 - t + 1)' this is not possible
(see the argument of Lemma 5.2.1 or [Kaw, Theorem III]).
Note that lAK(-l)l = 9 so that there are 4 dihedral characters contributing to
S. On the other hand, Equations 2.2 and 2.3 stiow that there are three irreducible
PSL2(@) characters of 4(3 ,3 ,4) . Using the methods of Section 5.1 ive see that there
are 10 ( = 4 + 2(3)) jumping points in the SL2 (@)-character variety. each contributing
a jump of 2. Thus S = 20.
Cising the Montesinos trick (for esample, see [BH]), me can see ttiat Jf(1) is Seifert
fibred over S2(2, 5 , ï ) . By Equations 2.2 and 2.3, there are 6 irreducible PSL2(@)-
representations of A(2,5, 7) and none of these are dihedral representations since we
are looking a t a surgery r = 1 with odd numerator. So each of these is covered by two
SI+(@)-characters each of which in turn contributes ttvo to the seminorms of r = 1
surgery. That is, 11 1 I l T = S + 24 (see Section 5.2.1 .)
By [HO]. the boundary slopes of I< are -14, O and 8 / 5 so that [BZl, Lemnia 6-21
gives us
11 y11 = 2[a1A(7, -14) + a2A(7, O) + n3A(7. 8/ii)].
In particular,
Subtracting, Ive find
Now Equation 5.23 shows that a3 < Z on a norm curve. On the other hanci, from
Equation 5-25 we see that a3 = O implies al = O wliich is not possible for a norm
curve. (Recall that for a norm curve at least two of the ai's are non-zero.) Therefore
n3 = 1, Shen Equation 5-35 implies a l = 1. Finally: Ecpation 5.23 allows us to
bound a?: O 5 a* 5 4. In particular, on a norm ciirve, we have 11111 = s + 24- -411 the
jumping points for r = 1 surgery lie on the norm curve. As they are simple points,
this shows that there is a t most one norm cume. Cal1 it .Xo.
Since an r curve, S1; would again have none of the r = 1 surgery jumping points:
Ive see that llllli = SI. But? as the norm is given by ,ildll = s1A(-{, r) : the only
candidate for r among the boundary slopes is r = O. Moreover, a s was the case for
2n +6 surgery on the (-2,3' n) pretzel knot, M ( 0 ) = Mi u M2 with MI Seifert fibred
over D2(2, 2) and M2 Seifert over D2(3: 3 ) . So as in the previous section: t here is a
unique r-curve for r = O and sl = 2.
In sumrnary then, there is one norm curve: one r-curve with r = O and one curve of
reducible characters in the character variety of 1;. The minimal norni on the r-curve
is .sl = 2. -4s for the norm curve. so = 18 and the norm is given by
Figures 29 and 30 show the fundamental and Newton polygons of this h o t . Note
that as the fundamental polygon lies belorv the line IJ = 1/2 (and 2.so > so + 8):
I< admits no non-trivial cyclic or finite surgeries. (Delman [Del] has already shown
that this knot admits a persistent lamination and ttierefore has no non-trivial finite
or cyclic surgeries.)
FIGURE 29. The fundamentai polygon of the (-3: 3: 4) pretael h o t .
FIGURE 30- The Yewton polygon of the (-3. 3. 4) pretzel knot.
5.3.1. Other n. F\-e nonr generalize to the (-3.3, n) pretzel knot which we mil1
denote by IL. Note that I L , is the mirror reflection of K,, so we will assume
n > O. This family includes some knots we have already looked at: is a twist
knot and 1(2 is the reflection of the (-2,3, -3) pretzel knot. Since K0 is not prime.
it's not hyperbolic and therefore not amenable to the techniques we developed in the
last tivo sections. On the otlier handl when 3 5 n 5 6 : 1(, is small. hyperbolic. and
moreover admits a Seifert surgery a t slope r = 1. (We have verified this for n = 3: 4.6
iising the Montesinos trick. For n = 5 we have only the evidence of SnapPea [\.Ireel.)
This means we can again apply the machine- of the last section to work out the
Culler-Shah seminorms of the knot.
1 O4
But what of n 2 77 Why stop a t n = 6? \Ve are obliged to stop since ive have
no evidence of h> admitting a Seifert filling for n 2 7. Indeed, the Seifert surgenes
occur according to a nice pattern. By [HO]. the boundary slopes of I<;, are -(2n + 6 ) .
O ancl 8/(n t 1).
O and 8/(n + 1)
For 1 < n 5 6: the Seifert surgeries lie between the boundary slopes
as the following table illustrates.
-AS the boundary slope 8/(n + 1) moves across the integers toward 0: those integers
cease to be available for Seifert surgeries. For example, when n > 7: the boundary
slope is 5 1 and there are no more Seifert surgeries. 1 should emphasize that this
is based on experimental evidence. These knots may admit other Seifert surgeries
beyond those I've listed in the table. In addition, although 1 (or others) have shown
that al1 the other surgeries in the table are Seifert, the only evidence I have in the
n = 6 case comes from SnapPea [Wee]. Nonetheless, it is a curious pattern and it
would be nice to understand this phenomenon.
Thus we can onIy hope to apply our machines- to & when 1 5 n 5 6. The first
two cases are treated elsewhere and n = 4 \vas discussed in detail above. For I ï3
Our method breaks down as the equations correspondirig to Equations 5.23,.5.2-4 and
5.25 above don't result in a unique solution for the ai's. Since K5 is not strongly
invertible, we canriot use the Montesinos trick to work out the indices for the Seifert
surgery of slope 1. Without that information, we cannot complete the analysis of
that knot.
Hoivever. Ks is tractable. Using the filling M ( l ) , which is Seifert over S2(2, 3,13),
we arrive a t the same conclusions as for there7s one norm curve with so = 22 and
and one r-cume with r = O and sl = 2. This means that K; also admits no non-
trivial cyclic or finite surgeries. (Again, Delman [Del] had shown this previously using
ctifferent methods.)
6- C O N C L U ~ I O N ~ AND QUESTIONS
In this thesis we have shown how the techniques developed by Culler, Shalen:
Boyer, and Zhang can be fruitfully applied to the study of Montesinos knots and in
particular pretzel knots. These techniques allow for a classification of cyclic surgeries
as well as a good understanding of the finite surgeries. By taking advantage of the
Seifert fillings of the (-2,3' n) and (-3: 3, n) pretzel knots, we were able to esplicitly
calculate the Culler-Shalen seminorms for those knots and parlay that information
into a precise description of the character varieties. These knots therefore serve as a
concrete esample of the usefulness of the Culler-Shalen scminorms not only for the
study of cÿclic and finite surgeries, but also for the study of Seifert fillings.
Along with some progress in understanding pretzel knots. our research raises many
questions. We list sorne of these cluestions and suggest w q s in which Our work coulcl
be estencled.
0 In Section 2.3 we used Ohtsuki's [Oht] work to directly compute the Culler-
Stialen seminorms of the twist knots. Ohtsuki shows how to esplicitly construct
the trees associated with icleal points of the character variety of a 2-bridge
knot. He can t hen count t hese ideal points by eniimerating the associated trees.
This is a beaiitiful construction and it is evidently of great use as it alloivs
rapicl calculation of the Culler-Shalen seminorrns. It would be advantageous to
estend his ideas to other classes of knots. Given the good understancling of
the Culer-Shalen seminorrns of ( p : qt r ) pretzel knots presented in t his thesis it
seems natural to esperiment with estending Ohtsuki's ideas to pretzel knots as
a first step towards eventually looking at larger classes of knots.
0 CVe have presented calculations showing that Z,(i,) = 1 at the character x of a
p-representation of a twist knot or (-2,3, n) knot. These calculations coulcl be
estended to other knots. This would be interesting as there is still little known
about the zeroes of the f, functions. For esample it may w l l be that Z,( f,) = 1
at the character of a p-representation on anÿ knot as we know of no evidence to
the contrary. Calculations for other knots might be useci to support or disprove
t his conjecture.
a In Section 3.3 we observed that, when m = m(,&/crl 1 3 2 / ~ t 2 ~ P 3 / a 3 ) is a three-
tangle Montesinos knot, there is an inclusion of the PSL2(@)-character varieties:
- q ( h ( a i , a,, as)) c ly(m). CVe then went on to use this observation to make
many deductions about the pretzel knots m(l/p, l / q , l / r ) . An obtious question
is, to what extent do our results extend to other three-tangle Montesinos knots'?
On the other hand, as we rnentioned in Section 3.3, it seems platisible that
there is also an overlap of character varieties when the Montesinos knot has
more than three tangles. To see hou- this might work in a more general contest,
recall (Section 3-12) that X2, the two-fold branched cyclic cover of a Montesinos
knot, is a Seifert fibred space with base orbifold B. In the case of a three-tangle
Mont esinos kno t , the triangle group is zfrb (B) .
Now, the SLÎ (R)-character variety of ?ryrb(l3) includes the Teichmüller space
of B which is homeomorphic to t being the number of tangles of the
Montesinos knot. I t follows t hat t h e PSL2 (@)-character variety -x-(rprb (B) ) also
has dimension at least 2(r - 3). It seems too much t o hope that . Y ( I F ~ ~ ~ ( B ) ) c S(m) when there are more than r = 3 tangles. Nonetheless. given the large
dimension of , Y ( R P ~ ~ ( B ) ) i t seerns likely that there is a t least sorne overlap of
t hese character variet ies. bIoreover, seeing how the dimension of ~ ( i r ? ' ~ (B) )
grows linearly with the number of tarigles r , perhaps this is also true of the
.y(m) character varieties. In otlier words, Montesinos knots are likely examples
of knots having character varieties of arbitrarily large dimension. The solicl
understanding of t hree-tangle knots presented in t his thesis would be an excellent
platform from which to launch an esploration of the rich structiire such higlier
climensionâl character varieties likely present.
0 The conclusion of the proof of Theorem 4.4.5 relies on Theorem 4.4.4 (the main
theorem of [El), a ratlier powerful result in group t h e o - It is clear that this
group theoretic result could be of great use in understanding infinite fillings of
many other Montesinos knots since fillings of these knots d l often admit factors
of the type (2, a, b; c) clescribed in Theorem 4-44. Indeedo the original t heorem
of Edjvet's [El paper açtually refers to a more general type of group (d? a, 6; c) .
On the other hand, it might also be possible to go the other way. Perhaps
conclusions about fini te fillings derived from Culler-Shalen t heory or ot her topo-
logical arguments could be used to make sorne deductions in group theory. For
esample, the finiteness of (2,3,13: 4) remains an open question. If this group
could be recognized as a factor of a finite filling of some knot that xould prove
the group finite.
Clearly it woulcl be nice to complete the analysis of finite surgeries on Montesinos
knots (Theorem 4-52) by understanding finite surgeries on (-?-y, q ) pretzel
knots with 5 5 p 5 q. For example, [Dol. Theorem 4-11 shows that çyclic
surgeries are near non-integral b o u n d a l slopes mhile Theorem 4.1.3 shows that
the same is true of finite surgeries, a t least on ( p , q , -r) pretzel knots with
r 2 4. It seems plausible that the same is true for (-2,p,q) pretzels and likelv
for other knots as well. Proving this would be a powerful first step in completing
the classification of finite surgeries on Montesinos knots.
The detailed calculations of Culler-Shalen seminorms in Chapter 3 depend largely
on the existence of Seifert fillings of the (-2,3, n) and (-3,3, n) pretzel knots.
Such calculations could likely be carried out for other Montesinos knots admit-
ting such Seifert fillings. Therefore, one way to estend this research is to use it
as motivation for a more thorough investigation of Seifert surgery on 3lontesinos
knots. -4 particularly provocative point in this regard is the pattern of Seifert
surgeries of (-3: 3, n) pretzel knots ihstratecl by the table in Section 5.3.1.
-Aside from providing information about finite and cyclic surgeries, cietailed cal-
culations of the Culler-Shalen seminorm provicie a lot of information about the
-4-polynomial invariant of a knot. Converselx the -4-pol-nomial can be usecl
to construct Culler-Shalen seminorms. Recently, David Boyd has proposed
techniques for efficient calculation of -4-polynomials. This is an exciting de-
velopment since, as difficult as the determination of the -4-polynomial is, it is
nonetheless even more difficult to get at the Ciiller-Shalen seminorm? particii-
larly if one wants to consider knots other than the 2-bridge knots (for which
Ohtsuki's methods can be employecl) or knots admitting Seifert fillings. For
example, given -4-polynornials for the (-'Zlp, q ) pretzel knots ( 5 5 p 5 q ) one
could complete the classification of finite surgerïes on Montesinos knots begun
in Theorem 4.5.2.
Lemma A.1.1. Let h ( t ) be the Alexander polynornial of the (-2: 3.72) pretzel knot
K . (In particular, n i s odd.) Suppose A(C) = O where C 2s a primitive mth root O/
unity. Then one of the following i s tme.
a 31n and rn = 6.
a lOI(n - 1) and m = 10-
9 121(n - 3) and rn = 12.
9 151(n-5) a n d m = 15.
Proof: The Alexander polynomial never admits zeroes which are prime powers of
u n i t . Indeed, by [BuZ, Theorem 8-21]: Hl&) is finite iff no root of the Alexander
polynomial is an rnth root of unity. Here Cm denotes the m-fold branched cyclic
cover of the knot (see [Rol, Section 10.C]). Using the Milnor [Mi] sequence 1.e can
show that bL(C,) = O whenever rn is a prime power of unity.
We next show that C = e2"'~m is not a root when m 2 18 by looking at a few cases.
By [Mru: Proposition 141 or [Hi, Theorem 1-21:
- l. - - - D- ( t ) .
t + 1
Suppose first that m > 2(2 - n) > O. Then,
(2-n T (2-n)n So; Re(D-(C) = 2 c o ~ ( - & ) [ c o s ( ~ ) - cos(-=) + cos(-("+')')]. nt Since rn >
2(2 - n); this is positive as long as n < -4. In fact, D-(C) is also positive when
n = -1: -3, so we see that C is not a root of the Alexander polynomial when n < O
and 2(2 - n) < m.
Nest suppose m > 2 ( n + 4 ) and n > O. Then.
and C is not a root of the -Alexander polynomial.
Since n = n' mod rn implies Cn = y': we can assume -m/2 < n - 2 5 ml?. We
have already looked a t the case .m > 2(2 - n) w -m/2 < n - 2, as well as the
case m > 2(n + 4) w m/2 > n + 4. SO it remains to investigate n - 2 5 m/2 and
m/2 5 n + 4, Le., 2n - 4 5 m 5 2n + 8. Let us now assume that m 2 18. Since
m < 2n + 8, this implies n 2 5.
If m = 272 + 8, then in fact ReA(c) = O, so we work instead with the imaginary
part:
(n-2)-r Thus C is a root of the Alexander polynomial only if cos(?) = cos(+). But since
these two values of cosine are distinct, ImD+(<) # 0, and < is not a root of the
Alexander polynomial in this case either.
If m = 2n + j, with -4 5 j 5 7, ive can use the real value:
112
Since the t e m in square brackets is never zero (given m 2 18), we see that C is not
a root of the Alexander polynomial when rn > 18-
Nest, let's look at the case where ,m = 14. Again,
ancl <as we've mentioned this value depends only on n mod m. By running through
the odd integers 1: 3,5 , . . . : 13 with .m = 14 in Equation -1.26 we see that the value is
never zero. It follows that the Alexander polynomial can never have a root < which
is a fourteenth root of unity.
Thus far: we have shown that the Alexander polynornial has no zeroes which are
mth roots of unity with m a prime power, nor with rn 2 18, nor with m = 14. The
only remaining candidates are m = 6: 20,12, and 15. To find out which values of
n yield a root. we again use Equation A 2 6 and strbstitute the odd integers mod m
to see which gives zero. It turns out that n = 3 (mod 6): n m 1 (rnod 10): n -= 3
(mod 12): and n = 5 (mod 15) respectively are the only candidates. Verifying that
these values also yield zero for ImD+(C) completes the proof. Q
Kurt Foster [FI has proposed a slightly different proof of this lemma. Here is a
sketch of his argument.
Proof: (of Lemma A. 1.1) Multiplying A(t) by t + 1 gives the eqiiation
Plugging in t = e2ri'm. and doing a bit of algebra yields
This may be recast (assuming (n + l ) /m isn't a half-integer) as
Direct verification rules out m = 3 4 , and 5 as possibilities, and turns up the
solution m = 6 when n = 3 (mod 6). The left sicle of Ecl~iation -4.29 is positive for
m > 6: and is > 1 for m > 9. Inspection rules out any solutions in the case m = 9.
We assume henceforth that m > 6. We clearly m l restrict rz to O 5 n < m.
113
Taking logs (assuming m isn't 9) and applying the Mean \-due Theorem tells us
that
r / m tan(x) = ln(2 cos(3x/m)), nn /m < x < (n + l)sr/m.
If ,m is sufficiently large, we may then write
which will: for each m, force the value of n- The limiting value of the expression
being subtracted From m/2 is l/ ln(2) as m increases without bound, so i t will be
between 1 and 1-5 for sufficiently large m. I t follows t hatl if m is large enough, the
integrality of m forces n = m/2 - 2 if m is even, and n = (m - 1)/2 - 1 if m is odd.
Direct substitution of these values then shows the relation of Equation -\29 fails.
-411 that rernains is to mop things u p for small values of m. -1 simple computer rou-
tine checked the values of n = 7 to 10 (escept n = 9) , to see whet her Equation -1.29
held to within 2-30. The program Elagged the pairs
m = 8, n = 6; rn = 10, n = 1; rn = 12, n = 3; and m = 15, n = 5 .
The first value of n is even, so doesn7t fit the original problem - Equation -1.2'7
isn't divisible by ( t + 1) if n is even - but it does give a solution of Equation -4.29,
and yields the primitive 8th roots of unity as solutions to Equation -4.27 when n = 6
(mod 8). The other pairs do give solutions (direct wrification).
The program also indicates that m 3 40 is "sufficientIÿ large" for the "large n"
argument to a p p l . Cl
Remark: -4s a consequence of Lernma Al. 1, A( t ) has no zero which is a (2n + 4) th
root of unity. For, suppose C were such a zero. Then ( n-oulci also be a 6th, lOth,
12th or 15th root of unity Hoivever: if it were a 6th root, then 6 1 2n + 4 which
precludes 3 1 n. Similar arguments apply for 10th: 12th, and 15th roots. Analogous
reasoning shows that h(t) admits a 2n + 5th root of unity zero iff n = 5 mod 30.
Lemma A.1.2. Let A, (t) be the Alexander polynornzal o j the (-2,3: IL) pretzel knot
K When n - 5 niod 30, &(t) admits primitive 15th roots of unit?/ UR zeroes and
moreouer: they are simple zeroes of A,.
114
Proof: &(t ) = (F4 - tn+2 + tn+' + t3 - t2 + l ) / ( t + 1 ) . Let &(t ) = t8 - t7 + t5 -
t" + t3 - t + 1 be the polynomial whose roots are the primitive 15th roots of uni t . We
will show that &5(t ) is a Factor of I l , ( t ) by induction. We iilustrate the induction
for n > O. The argument For n < O is similar. When n = 33. @15 is incleed a factor of
An ( t ) .
Suppose +15 is a factor For some n with n E 5 mod 30.
Since q5i5 is a factor of both t30 - 1 and i l n ( t ) . we see that it's a factor of An+aa(t)
To see that these are simple roots, it suffices to show that they are not also roots
of the derivative AR(t). -\gain, we d l assume that n > O. The case where n < O is
analogous.
AL(t) =
Suppose < is a common root of and 3;. Since < is a 15th root of unity and
n 5 mod 30. we have
Thus < is a root of
This implies that the irreducible polynomial 415 divides / ( t ) . Howevcr. it's easy to
verify that ol5 does not in fact divide f ( t ) . O
115
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