...,
8L2 (.1q)REPRESENTAnONS
OF TORUS KNOTS
By
UANGXU
Bachelor of Science
Hubei University
Wuhan, Hubei
People's Republic of China
1995
Submitted to the Faculty of the
Graduate College of
Oklahoma State University
in partial fulfillment of
the requirement for
the Degree of
MASTER OF SCIENCE
August, 2001
Oklahoma State' University L.ibrary
SL2 (.1q)REPRESENTAnONS
OF TORUS KNOTS
Thesis Approved:
~j:ir J ~iS w3 Advisor
~';leoe
o
11
,I
JI
i
I
ACKNOWLEDGEMENTS
I wish to express my sincere appreciate to my advisor, Dr. Weiping Li for inspiration,
encouragement, and constructive guidance about the thesis, because without him this
would not be possibk
Special thanks are extended to Dr. Jim Cogdell and Dr. Roger Zierau for kindly
agreeing to serve on my committee.
I would also like to express my thanks to my friends and parents, for their support and
love during these years.
Finally, I would like to thank the Department of Mathematics in OSU for supporting
during these one and a half years.
ill
TABLE OF CONTENTS
Chapter Page
1. INTRODUCTION ]
2. FIRST METHOD FOR (n, 2) TORUS KNOT 1
Conjugacy Classes in SL2(!fi;) 2
Preliminary Results " 3
Counting the number of solutions , 7
3. COUNTEREXAMPLE FOR THE RESULT IN [7] 17
4. FINDING THE ZETA FUNCTION .19
lV
Table
LIST OF TABLES
Page
Table I 2
Table 11. " 3
Table III 16
Table IV 17
v
1. 'INTRODUCTION
"
The Casson invariant defined in [1] was originally 8:Il invariant of integral homology
3spheres, and it was extended to an invariant of knots in integral horp.olGgy
3spheres through surgery relations. The intrinsic exten.sion of the Casson invariant
for knots was developed in 15] and generalized to the sympelctic Floer homology in
[3]. The CassonLin invariant defined in [3, 51 was an invariant of knots as roughly
counting irreducible SU(2)representations of the knot groups with tracezero along
all meridians.
There is an attempt to define a characteristicp Casson invariant in [7]. One would
like to have the characteristicp Casson invariant preserving all properties of. the
Casson invariant in [1, 3, 5]. Thus one can count representations of the knot group
into SL2(:Fq ), where Fq is a finite field and q = pS. Note that the finite field version
of SU(2) does not have the some nice property as 8U(2), and the Lie algebras of
8U(2) and SL2 (0) are pretty much the same as isomorphism. In [4], the 8L2(0)
knot invariant was studied and the relation with 8U(2) knot invariant is discovered.
For finite field case, instead of representation varieties, there is no nice intersection
pairing since the representation spaces are not projective. Thus one simply count the
representations of knot groups into SL2 (:Fq ). Our work is motivated by the study of
Sink [7]. We use some explicit methods to count the representations of (n, m) torus
knot. Our main results can be stated in the following.
Theorem 1.1. There is a formula to compute the number of representations of (n, 2)
torus knot into SL2 (:Fq) up to conjugacy (see Table III).
Our formula in Table III has a discrepancy with the result in [7]. We compute some
explieit solutions to show that there are errors in the counting of !7] in §3.
Using the numbers of representations we counted in §2, we formulate a zeta function
of the numbers of representations of SL2 (:Fq ) where q = pS for sEN. The zeta
function of the (n,2) torus knot can be computed in an explicit manner. Then we
can verify the modified zeta function ),.1/ is a polynomial for n prime, or Ii product
of two primes, or a square of a prime. We expect this phenomena true for general n
as it was claimed in Theorem 3 of [7].
The thesis is organized as follows: we first review the conjugacy classes and stabilizers
in SL2 (:Fq ) in §2.1. Few elementary number theoretic countings and cyclic
group properties are presented in §2.2. In §2.3, we start to counting the number of
representations of (n,2) torus knot group into SL2 (:Fq ) up to conjugacy and prove
Theorem 1.1. We give two counterexamples to statements in [7] in §3. In §4, we study
the Zeta function of (n, 2) torus knot and verify that the modified Zeta function ),1/:
is a polynomial for some special cases.
2. FIRST METHOD FOR (n,2) TORUS KNOT
We win count representations SL2 (:Fq ) of the fundamental group G of the (n,2)
torus knot up to conjugacy, where :Fq is a finite field with q = pS for a prime number
p> 2. As we know, the fundamental group of (n, 2) torus knot can be presented by
1
< a, blan = b2 >. Any representation p : G + SL2(Fq) can be onetoone identified
with (p(a), p(b)) E SL2 (Fq) x SLz(Fq) such that p(at = p(b)2. Hence we have the
correspondence between a representation and a pair (A, B) of SL2(Fq ) matrices with
An = B 2 . Since we count the representations up to conjugacy' and' we can always
conjugate on.e of A and B into a standard representative of one of the conjugacy
classes, we may choose A to be the element and find the number of B satisfying
An = B 2 .
2.1. Conjugacy Classes in 8L2 (:Fq). We have to find the conjugacy classes in
8L2 (:Fq ), and this is wellknown in [2], and see Table 1. Let € be an element of :Fq
with JE ¢:. :Fq.
Table I
Case No. Representative No. of Elts in Class No. Classes
1. 1_(10)  0 1 1 1
2.
3.
4.
5.
6.
7.
8.
(
1 1. 0
(~ ~)
(~ ~)
(~1 ~l)
(~1 ~1)
(~ ~l )
(€~ ;)
1
,i 1
2
q(q + 1)
q(q  1)
1
1
1
1
1
Note that we have certain redundancies in the above table; namely in Case 7, the
matrices (~ ~l ) and (X~l ~) are in the same conjugacy class. Also in that
same Case 7, we do not allow x to be ±l since they are covered in Case 1 and Case
2

2 respectively. Also, in Case 8, the matrix (x y) is conjug.ate to the matrix Ey x
( EXy xY) and further cannot let y = 0, otherwise it reduces to' Case 1 or 2.
The following table gives the group of stabilizers for each case.
Table II
Case No.. The stabilizer of each representative in SL'};(:Fq)
1.
2.
3.
4.
5.
6.
7.
8.
SL2 (:Fq )
.SL2(:Fq )
{( ~1 ~l) IxErq }
{( ~,1 ~1) IxE:Fq }
{( ~1 :1) IxE:Fq }
{( ~1 ~1) IxE:Fq }
{( ~ ~1 ) lyE :Fq  O}
{( E~ ~) la2
 fb2 = I}
We list the characteristic polynomials for each case. For Case 1, Case 3 and Case
4, the characteristic polynomial is ~?  2), + 1 = O. For Case 2, Case 5 and Case
6, the characteristic polynomial is ).2 + 2), + 1 = O. For Case 7, the characteristic
polynomial is A'};  (x + XI)A+ 1 = O. For Case 8, the characteristic polynomial is
).2 _ 2XA + 1 = O.
In the following paper, we use CA to be the set of the elements conjugate to A in
SL2 (:Fq ) and use r A to be the stabilizer of A.
2.2. Preliminary Results. Before we begin to discuss solutions of each case, let us
give elementary propositions which we will use later.
Proposition 2.1. Let a, band m be integers such that m > 0 and (a, m) = d, where
(a, m) is the greatest common divisor of a and m. If d f b, then ax =b (mod m)
3
has no solutions. If d I b, then ax == b (mod m) has exactly d incongruent solutions
modulo m.
The proposition is important to prove the fonowing propositions and it is in Section
4.9 of (6].
Proposition 2.2. Given a finite cyclic group G, IGI = m, where 1·1 is the cardinality.
Let Gn ={xnIx E G}. For a yE Gn, the number of solutions for xn =y is (rn, n).
Proof. Since G is a cyclic group, let g be the generator of G. Then there exists at
with 0 ~ t < m  1,. such that gt = y. Write x = g5 for some s. Then the equation
xn = y is reduced to the equation ns _ t (mod m). Clearly there exists a solution for
ns  t (mod m) since y E Gn. By Proposition 2.1, it has (n, m) solutions. 0
Proposition 2.3. For a finite cyclic group G, let IGI = m, then IGnl = (':::n).
Proof. It is clear that jGnl * I{x E Glxn = y with y E Gn}1 = IGI. By Proposition
2.2, IGnl = ( m ). 0 n,m
Proposition 2.4. For a finite field :Fq , G is cyclic subgroup of:Fq with respect to
multiplication with 1 E G. Let m be even. Then there is a solution for the equation
xm = 1 in G if and only if (m, IGDIJ¥l.
Proof. Let 9 be the generator ofthe cyclic group G. Clearly glGI = 1 and (I? = 1.
Let gt = 1 with 0 < t < IGI. Then g2t = 1 which means 2t = IGI, i.e., ,q~ = 1.
Like the discussion in the proof of Proposition 2.2, to solve the equation xm = 1
is equivalent to solve the equation 8m =~ (mod IGI). By Proposition 2.1, there is
a solution for equation xm = 1 in Gif and only if (m, IGDI¥. 0
Proposition 2.5. For a finite field :Fq, then if q =1 (mod 4), then there are two
solutions for the equation x2 = 1 in the :Fq. If q == 3 (mod 4), then there is no
solution for the equation x 2 = 1 in the :Fq .
Proof. To determine whether there is a solution for x 2 = lis equivalent to
find a solution in :Fq  0, which is a cyclic group with respect to multiplication. By
Proposition 2.4, there exists a solution for x2 = 1 in:Fq if and only if (2, q  1)1?
Therefore if q=1 (mod 4), then there are two solutions for the equation x2 = 1 in
the :Fq. If q=3 (mod 4), then there is no solution for the equation x 2 = 1 in the
~. 0
Proposition 2.6. The set rl = {( x y) I(x y) E SL2 (:Fq )} is a cyclic
f.y x Ey x
group. Let:F' be the vector space over :Fq with the basis 1 and."fE. Let C = {( E
(:FI)*I(q+1 = I}. Let ¢.: rl + C be ¢(( x y)) = x + ."fEy. Then ¢ is a isomorEy
x
phism, and ~nl = ICI = q + 1.
4
1
The statement is shown in Chapter 5 of {2].
For the convenience of the discussion, we let a1 = a  {±I}.
Propos.ition 2.7. For the cyclic group nas we defined in Proposition 2.6, if q =1
(mod 4), there is no solution for the equation A2 = 1. If q=3 (mod 4). there are
two solutions for the equation A2 = 1. where A E n.
Proof. Clearly IE n. By Proposition 2.6, IQI = q+ 1. By Proposition 2.4, there
is a solution for A2 = 1 if and only if (2, q + 1)19;1. Therefore if q ~ 1 (mod 4)1
there is no solution for the equation A2 = 1. Ifq == 3 (mod 4), there are two
solutions fOT the equation A2 = I, where AE n. 0
Proposition 2.8. Let p, q be coprim.e natural numbers. Suppose that the equation rr  1 (mod q) has solutions. Let mo be the smallest positive solution of the equation.
If m is another solution for r =1 (mod q), then m = kmo for some positive
odd integer k.
Pruof. Note by Fermat's little theorem, there exists a solution for pX =1 (mod q).
Let ml he the smallest positive solution for r  1 (mod q). First we show ml > "/.0.
Otherwise there exist a and b such that mo = ami + b with a 2::: 1 and 0 < b <
ml < mo. Therefore pb =1 (mod q) which is contradictory to the assumption of
rna is the smallest positive solution of pX == 1 (mod q). Clearly Ifmo= q (mod q)
for some even number l. So if mo!m, then m = kmo for some odd integer k.
Assume mo f m, there exist positive integersc and d such that m = emu +d with
a < d < mo. If c is even, then we get pd = 1 (mod q), which is impossible. If
c is odd, then we get pd == 1 (mod 1), but d < mo < mIl so it is false. Therefore
molm. 0
Proposition 2.9. Fora finite field Fq , if x is in Fq with yX r;. Fql then .;x:::r fJ. Fq •
If the square root of both x and yare not in f'q, then JXY E Fq•
Proof. Let 9 be the generator of the cyclic group Fq  0 with respect to multipllcation.
Since VI r;. :Fq , x = 98 for some odd number s. Clearly XI = 9q I  8
, then
.;x:::r ¢. Fq• Let y = gl. By the same reason, t is odd too. We get xy = gs+t, so
vfXY E Fq .
Proposition 2.10. Given A,. B be representatives of different eonju.qate classes such
that r A nrB = {±I}. For aGE SL2 (:Fq), let B1 = GBGJ. Then r A nrRJ = {±I}.
Proof The pair of (A, B) has the following possibilities:
1) A in Case 3, B in Case 7;
2) A in Case 3, B in Case 8;
3) A in Case 4, B in Case 7;
4) A in Case 4, B in Case 8;
5) A in Case 5, B in Case 7;
5
6) A in Case 5, B in Case 8;
7) A in Case 6~ B in Case 7;
8) A in Case 6, B in Case 8;
9) A in Case 7, B in Case 8;
Clearly fBI = GfBGi . Given C E fA, D E fBI such that C and D =1= ±1. By
Table II, C and D have different characteristic polynomials. Clearly {±I} c fAnf13) .
Therefore fAn f Bt = {±I}. 0
Proposition 2.11. (1) Let Bo= (~o ~Ol) with Xo =I ±1 and C = (~ :) E
SL2 (:Fq). lfCBoC1 is diagonal, then C must be (~ ~) or (~ ~) and CBOC1 =
B±o1 .
(2) IfCBoCl is not diagonal, then fCBoCt n f so = {±1}.
Proof. (1) By a simple calculation
C I( X 0_1 )1 C1 = ( adxo  bc~r ab(xo  Xj}l) )
o xo cd(xo  xo) bcxo + adxo1
.
Since xo =1= ±1, C (~o ~OI ) Cl is diagonal if and only if a = 0, d = 0 or
b = 0, e = O. If C = (~ ~), then CBoC 1 = Bo. If C = (~ ~), then
CBOC1 = Bol
•
(2) Since CBOC1 is not diagonal, C cannot be in the form of (1). Note that
fcBoct = crBocl. For any (6 ~1 ) E f Bo with Y =1= ±1, C (5 ~l) C·1 is
not diagonal. Hence f CBoC) n f 130 = ±I. 0
I ,
Proposition 2.12. (1) Let Eo = (~ ;) with y =1= 0, C = (~ ~) E SL2 (:Fq).
If B = CBOC1 E D, then C E n or C = ( a
b
b ), and B = Btl.
E a
(2) lIB rI D, then f B n fBo = {±1}.
Proof. (1) By a simple calculation, we get
C (x y) C1 = ( X + bd€y  acy
Ey x E~Y  e2y
If C (x y) C1 E 0, then x+ bay  acy = x+ aey  Ebdy~ which implies
Ey x
(ac  €bd)y = O. Since y =1= 0, we get ae  €bd = O. Therefore a2  b2E = ±1.
Combining with ad  be = 1 and ac  Ebd = 0, we get a = d, C = Eb, i.e, C E D;
6
or a= d, C = €b. If C = (~ ~), then CBOCl = B. If C = (:b ~a) t
CBoC1 = BOI .
(2) If B ¢:. n, then C cannot be in form of (1). Note that r B =CfBoC1• For any
A = (u v) E f Bo with v=1= 0, CACl ¢:. n. Hence f B n f Bo = {±I}. 0
€v U
2.3. Counting the Number of Solutions. As we have said, we take the matrix A
to be in the standard form of each case shown in Table I, and count the solutions B
of the equation
(1)
up to conjugacy in each case. It is obvious that Equation (1) is equivalent to the
equation
(2)
In the first 7 cases, we count solutions of Equation (2) instead of solving Equation
(1) directly. In this section, we write B = (~ ~) with
(3) adbc=1
in :Fq• Clearly B1 = ( d ab).
c
Notice that (n,2) = 1, so n is always odd in this section. Now we discuss the
number of solutions of Equation (1) case by case.
In order to get the n~mber of representations up to conjugacy, we need to modulo
more from the stabilizers of A and B. Let r A = {G E SL2 (:Fq)IGAGl = 4} be the
stabilizer of A. Hence any element G E fA acts on (A, B) as a single conjugacy orbit
(A, GBGl
) with fixed conjugacy representative of A. This only affects solutions
(A, Bd and (A, B2 ) with B] and B2 in the same conjugacy class under r A.
Case 1. Let A = (~ ~). Then An = (~ ~), Equation(2) is reduced to'
B I = B, I. .e.,
(4) (
a b) = ( d b). c d c a
Therefore d = a, b = b, and c = c. Then b = c = O. Combining with equation (3),
we get a2 = ad = 1 which has solutions a = ±1. Therefore we have two solutions for
B = (~ ~) and (~1 ~ 1 ).
Note that for both two solutions of (A, B), we have r A = r B = 3L2 (:Fq ). Hence
there are two distinct solutions of I = B2 up to conjugacy.
7
Case 2. Let A ~ (~1 ~l)' Then An =' (~1 ~1). Equation (2) becomes
(5) ( d b ) = (a b )
c a cd'
which is equivalent to a = d. Combining with equation (3), we get d'}; +be = l.
There are two sllbcases: (a) be = 0, (b) be =I O. We count the number of solutions for
each subcase and add them up.
2.a: For be = 0, it is dear that
l{(b, c, d)ld2 + be = 1, be = O}I = I{(b, e)lbe = OJ! * l{dld2 =I}!.
There are 2qI solutions in Fq for be = O. For the equation cP = 1, by
Proposition 2.5, there are two solutions for d? = 1 if q =1 (mod 4), and there
is no solution for d2 = 1 if q =3 (mod 4). Therefore the number of solutions
for cP + be = 1 with be = 0 in Fq is
{
2(2q  1) if q =1 (mod 4)}
o if q = 3 (mod 4).
2.b: For bc =I 0, we get the following relation
I{(b,. e, d) Id2 + be = 1) be =I O} I= I{dld2 =I I}I* I{bib =I O} I·
According to Proposition 2.5, we have l{dld2 =I I}I = q  2 if q _ 1 (mod 4),
and l{dld2 =I I}I = q if q=3 (mod 4). Obviously I{blb =f O}I = q  1. Hence
the number of solutions for cP  be = 1 with be =I 0 is
{
(q. 2)(q  1) if q=1 (mod 4),
q(q  1) if q == 3 (mod 4).
Summarizing the previous discussions, the number of solutions of Equation (1) in
Case 2 is
{
q(q + 1) if q =1 (mod 4)}
q(q  1) if q  3 (mod 4).
For all the solutions B satisfying B 2 =  I, B = (~d ~) with cP + be = 1)
then its characteristic polynomial is ,x.2 = 1. By the characteristic polynomial of
8L2(Fq), the solution must be in Case 7 or Case 8 for x + XI = aor 2x = O.
If q=1 (mod 4), then v=r E SL2(Fq ) by Proposition 2.5. Hence x +x 1 = 0 has
a solution, all the solutions are conjugate Bo = (~o ~o J ) with the fixed Xo and
x~ = l.
Let CBo = {CBoCIIC E 8L2(Fq )}. Since Bo is a solution for B 2 = I, any
element in CBo is also a solution. Therefore the collection of the solutions of B2 = I
is equal to CBo' We can write all the solutions in this case as (I, CBoCI
) for some
8
G E SL2 (:Fq). On the other hand the stabilizer of 1 is S~(:Fq). Therefore up to
conjugacy, we only have one solution (I, (~o ~Ol ).
If q =3 (mod 4), then .J=I rt. :Fq by Proposition 2.5. There is no solution for
x+x1 = O. So all the solutions must be in Case 8. By Proposition 2.9, Jc1 E :Fq •
Therefore all the solutions are conjugate to the matrix B~ = (EJ~Cl tC1
).
Let eBb = {GBb,GIIGE SL2 (:Fq )}. We can write all the solutions in this case as
(I, GBbG') for G E SL2(:Fq). For r 1 , SL2(rq ), up to conjugacy we only have
one solution (I, ( EJ 0c 1 tc
1
)).
Therefore in Case 2, up to conjugacy the number of solutions is l.
Case 3. Let A = (~ ~). Then An = (~ ~). Equation (2) becomes
(6) AnB1 = (d=;C ~b+an) = (~ ~).
One has the equivalent relations:
{
~;:~: a b
c= c
a=d.
By c = c, we get c = O. Combining equation (3) and a = d, we have a'l: = 1 which
has solutions a= ±l. Thus b +an = b is reduced to b = ±21n. Therefore we have
two solutions in this case, i.e.,
B = (~l ~~ln).
(
1 21n) (1 21n) Clearly 0 1 cannot be conjugate to 0 1 in·SL2 (.rq) for they
have different eigenvalues. Therefore we have two distinct solutions in this case.·
Case 4. Let A = (~ ~). Then An = (~ ~E). Equation (2) is equivalent to
(7) AnB1 = (d __~CE ~b+anf) = (~ ~).
One has the equivalent relations:
Id  nCE = a
b+ anE = b
c=c
a=d.
Similarly as Case 3, we get C = 0 from c = c. Combining equation (3) and a = d, we
have a2 = 1 which has solutions a = ±l. Thus b+anE =bis reduced to b = ±21n€.
9
In this case the solutions are
= (±1 ±2lnE) B 0 ±1 .
(
1 21nE). . (1 2ln€). . Clearly· 0 1 : IS not conjugate to 0 1 ,m SL2 (:Fq ) SInce they
have different eigenvalues. Therefore we have two distinct solutions in this case.
Case 5. Let A = (01 11 ). Then An = (10 n1 ) . Equation (2) becomes
(8)
One has the equivalent relations:
{
dnc~ a
b+an = b
c=c
a= d.
From b+ an = b we get an = O. If p t n, a must be O. Combining d  nc = a and
a = d, we get c = 0, which contradicts to the assumption ad  be = 1. Therefore
there is no solution.
(
Ifpin, then An = 01 01 ) .
If q =1 (mod 4), by Case 2, all the solutions can be written as (A, GEoGI) where
Bo = (~o ~Ol ) and G E SL2 (:Fq ).
We define an action r Aon CBo : fAXCB ~ CB by (H,B) ~ HBHl where H E fA
and B E CB. Now we just need to count the orbits under the action of r A. Clearly
IfA(B)I = IfAlfA n fBI·
Accordm. g to Table II, fA = { (±01 x±1 ) Ix E:Fq}, and. f Bo = { (y0 0yl ) Iy E.
:Fq  OJ, and fA n f Bo = {±I}. For any B = GBOG1
, by Proposition 2.10, fA n
r B = {±I}. Therefore IfA(B)! = IfAlfA n fBI = q for every B E CBo ' Since
ICBol =q(q + 1), the number of orbits under the action of r A is q + 1. Therefore up
to conjugacy the number of solutions is q + 1.
If q _ 3 (mod 4), by Case 2, an the solutions for B 2 = 1 are conjugate to
B~ = ( eJ0c vC2). (±1 x ) l 0 . According to Table II, r A = { 0 ±1 Ix E :Fq },
fB~ = {( ~ ;) Ix2 Ey2 = I}, rAn fBh = {±I}. By the same discussion as
above, we get the number of solutions up to conjugacy is q  1.
10
Thus the nurriber of solutions of equation (1) is
{
q + 1 if q  1 (mod 4) and pin,
q  1 if q=3 (mod 4) and pin,
o Otherwise.
Case 6. Let A = (~1 ~ 1 ). Then An = (~1 ~€1)' Equation (2) becomes
(9) AnB1 = ( d ~ ne€ ~~ an€ ) = (~ ~).
One has the equivalent relations:
Id  neE = a
b+ anf = b
c=c
a=d.
From b+ anE = b we get anE = O. With the same method as in Case 5, if p f n,
then there is no solution. If pin, then An = (~l ~1)' By the same discussion
of Case 5, the number of solutions of equation (1) is
{
q + 1 if q = 1 (mod 4) and pIn,
q  1 if q 3 (mod 4) and pin,
o Otherwise.
Case 7. Let A = (~ ~l)' x1= ±l. Then An = (x; ~n). Equation (2) is
reduced to
(10) AnB1 = ( dX~n b~:) = (a db).
ex ax c
One has the equivalent relations
{
dxn = a
bxn = b
cxn = c
axn = d.
From (l+xn )e = (l+xn )b = 0, Equation (10) is split to two subcases, (a) l+xn = 0,
(b)l + x n =1= O.
7.a: For 1 + x n = 0, we have An = 1. It is clear that
1{(A,B)IAn= B2,An = I}I = I{AIAn= I}I * I{BIB2 = I}j.
Because n is odd, 1 E (Fq  o)n. By Proposition 2.2, there are (q  1, n) many
solutions for xn = 1. Since we do not allow x = 1, there are (q  1, n)  1
many solutions satisfying An = 1 and A =1= ±1. We know (~ ~l ) and
11
(X~1 ~) are conjugate to each other. On the other hand~or two representatives
A and A' of Case 7, A' is conjugate to A if and only if A' = A±. Up to
conjugacy we get (n,q;I)1 many solutions for An = I , where A = (~ ~l )
for some x with x =I ±l.
Fix Al to be (~l ~ll ) such that xf = 1 and Xl =I. ±l.
If q  1 (mod 4), by Case 2, the set of collection of solutions for A~ = B2
is CHo where Bo = (~o ~OI ). Like in Case 5, we define an action' f A.I on
fBo:f Al X CBo t CBo by (H, B) t HBR i . Let NrA ] be the number of orbits.
Clearly f AI = f Bo = fB,j}1. Then IrA1(Bo)1 = IfAj(Bol)1 = IfA,/fAlnfBol =
1.
Given a B = GBOGl, and B =i Btl, By Proposition 2.11, B is not diagonal,
f Al n f B = {±/}. Therefore IfA1 (B)I = IfAI/rAl n fBI = q;l.
Hence
ql
2(NrA\  2) + 2 = q(q + 1).
We get Nr Al = 2(q + 3). Therefore up to conjugacy the number of solutions
for A~ = B 2 is 2(q + 3).
If q =3 (mod 4), by case 2, the set of solutions for A~ = B 2 is CBo where
B~ = (fJ~Cl tC1
). By Table II, fAn f Bo = {±I}. By the same
argument of Case 5, up to conjugacy, the number of solution for Ar = B2 is
ICBol = q(q 1) = 2q
IfAJfAl n f 80 1 (q  1)/2 .
Recall up to conjugacy the number of solutions for An = lis (n,q;1)1,
therefore the number of solutions for An = B2 with An = 1 is
{
((n,q 1) 1)(q + 3) if q 1 (mod 4),
((n, q  1)  l)q if q  3 (mod 4).
7.b: If x n + 1 =I 0, then b = c = Q. Combining a = dxn and equation (3), we get
d2xn = 1, which is equivalent to the equation xn = d2 . The number of solutions
for x n = d2 is equal to
L I{xlxn = g}1 * l{dld2 = g}l,
gEG
where G = (Fq  0)2 n (Fq  Qt. Since (n,2) = 1, G = (F  Q)2n. By Proposition
2.2, the number of solutions for x n = 9 and (d1)2 = 9 with 9 E G are
(n, q  1) and (n,2) respectively. By Proposition 2.3, IGI = (q~~,;n)' Therefore
we totally have q  1 solutions for the equation rfxn = 1.
12
If q == 1 (mod 4), then 1 E (Fq  0)2. So we must throw the solutions of
xn = d2 = 1 out.. The number of solutions of xn = d 2 = 1 is
l{xlx1t = 1}1 * !{dld2 = 1}1 = 2(n,q  1).
Obviously 1 E (Fq  0)2. So we have to throw out 2 solutions out again. So
the number of solutions is q  3  2(q  1, n) for q == 1 (mod 4).
If q=3 (mod 4), then 1 ~ (Fq  0)2. We just throw out 2 solutions if q=3
(mod 4), that is to say the number of solutions is q 3 for q =3 (mod 4).
Clearly in this subcase both A and B are in the standard form of Case 7.
Clearly (A, B) is not conjugate to (A, B). We know that if (A, B) is a solution,
then (A1, B1) is a solution and conjugate to (A, B). On the other hand, (A, B)
and (A', B' ) are conjugate to each other if and only if A' = A±l, in this situation
B' = B±l. Therefore up to conjugacy, we have
{
q;3 _(n, q 1) if q _ 1 (mod 4),
~ if q == 3 (mod 4),
many solutions for An :/= I and An = B2 ,
Add the number of solutions in case 7.a and case 7.b up, we get the number of
solutions for the Equation (1) in case 7 is
{
((n, q  1)  l)(q + 3) + q;3  (n, q  1) if q=1 (mod 4),
((n, q  1)  l)q + ~ if q == 3 (mod 4).
Let E(s) = HI + (1)8),0(8) = Hl (1)8). As we know if p =3 (mod 4), then
p8  1 (mod 4) for seven, p8 =3 (mod 4) for somes odd. The previous formula is
equivalent to
{
(n,p8  1)  l)(p8 + 3) + p';3  (n,pS  1) if p =1 (mod 4),
((n,pS  1)  1)(pS + 3E(s» + p";3  E(s)(n,p8  1) if p=3 (mod 4).
Case 8. Let A= (x Y), where y :/= O. By Proposition 2.6, An = (u v) for Ey x €V u
some u, v E :Fq . So Equation (1) is sphtted into 3 subcases:(a) An = I, (b) An = I,
and (c) An 1= ±J.
8.a: For An = I, we get
{(A, B)IAn = B2
, An = I}l = I{AIAn = I}! * I{BIB2 = I}I·
By Case 1, we know I{BIB2 = I}I = 2. By Proposition 2.2, the number of
solutions of An = I is equal to (n, q + 1). Since A is not allowed to be I, we
throw one solution out. Clearly (A, I) and (A, I) are not conjugate to each
other. On the other hand A is conjugate to At in n if and only if A' = A1
for A E n. Hence we totally have 2(n,q~1)1 = (n,q + 1)  1 solutions for the
equation An = B 2 with An = I.
8.b: For An = I. It is dear that
I{(A, B)IAn= B2
, An = I}I = I{AIAn = I}I * I{BIB2 = I}I·
13
First we count the solutions for the equation An = I . By Proposition 2.2,
the equation An =  1 has (n, q + 1) many solutions in the set n. We have to
throw one solution out for A i= 1. With the similar discussion of Case B.a, up
to conjugacy we have (n,q~l)1 solutions for An = 1.
Fix Al to be (Xf.Yll YI) such that Af = 1 and Al =1= 1. XL
If q == 1 (mod 4), then by Case 2, the set of collection of all solutions B is
C80 where Bo = ( ~o . ~Ol ) as we have defined as in Case 2. By Table II,
f Ai n f Bo = {±I}. By the same argument of Case 5, the number of solutions
for A? = B 2 up to conjugacy is
If q 3 (mod 4), the set of collection of Af B2 is CSo where Bb
( f.J af.I tC1
) as defined in Case 2. Clearly r AI = f Bo '
Similarly we define the action f Al on CBo:fAl X CBo t CBo by (H, B) t
H BH1
. We use Nr Al to be the number of orbits.
Clearly f AI = f Bo = f(Bo).1. Then IrAJ(Bb)/ = IfAl((B~)I)1 = IfA1/fA1 n
fBbl = 1.
Given a B = GBoG1, and B =1= Btl. Then B ¢ n. By Proposition 2.12,
f AJ n f B = {±1}. Therefore IfAl (B)I = IfAJfAI n rHI = ~. Since ICBol =
q(q  1), then
q+1
2(NrAl  2) + 2 = q(q  1).
We get ICBo / rv I = 2(q  1).
Since up to conjugacy the number of solutions for An = 1 is (n,q~I)J , we
get the number of solutions for An = H2 with An = 1 is
{
((n,q + 1) l)q if q _ 1 (mod 4),
((n,q + 1)  l)(q  1) if q _ 3 (mod 4).
8.c: For An f. ±1, let An = (u v) with some u,vand v=1= O. By computation,
EV U .
(
a2 + be ab + bd ) .
B 2 = ae + de be + ,CP . Hence EquatIOn (1) becomes
.1
(11) (f.: ~) = ( :: ~ ~~ :;:~~ ) .
14
r
One has the equivalent relations
Iu = a2 + be
: v = ab + bd
, EV = ae +bc
! u=bc+d2
•
From (a + d)b = v and v =1= 0, we get a + d =i O. From u = a2 + be and
u = ~. + be, we get a2 = cP. Therefore a = d. From v = ab + bd, €V = ac + de
and a= d, we get 2abc = 2ae, i.e., a(bc  c) = O. Since a = d and a + d =f:. 0,
we get c = tb. This means B E 0 too. So the problem is reduced to solve the
equation An = B2 in the group D. With the same discussion as we used in Case
7, we get [HA, B)IAn = B2 , A, B E O}I = q + 1. Since An =i I, we must throw
out 2(n, q + 1) solutions. By Proposition 2.7. the square root of I is in n
if and only if q = 3 (mod 4), By Proposition 2.7. In this condition, we throw
2(n, q + 1) many solutions out again.
Cle.arly in this subcase both A and B are representatives of Case 8, then they
have the same stabilizer.. On the other hand, we know that if (A, B) is a solution,
then (A1, B1
) is a solution and conjugate to (A, B). we also know that in Case
8, (A, B) and (A', B' ) are conjugate to each other if and only if A' = A±l. So up
to conjugacy the number of solutions for the equation An = B2 with An =i ±1
are
{
~(n,q+1) ifq=l (mod 4),
~  2(n,q+ 1) if q  3 (mod 4).
Summarize above we totally have
{
((n, q + 1)  l)q + ~  1 if q =1
((n, q + 1)  l)(q  1) + ~  (n, q + 1)  1 if q =3
By the same discussion as in Case 7, the above is equivalent to
(mod 4),
(mod 4).
j
{
((n,p~ + 1)  l)pS + P~;1 if p=1 (mod 4),
((n,pS + 1)  l)(pB  O(s)) + P~;l  O(s)(n,pS +1) if p=3 (mod 4).
Now we can give Table III for the number of solutions for each case which completes
the proof of Theorem 1.1.
15
Table III
Case No. No. of Solutions Up To Conjugacy
l. 2
2. 1
3. 2
4. 2r+1 if q =1 (mod 4) and pin,
5. ql if q  3 (mod 4) and pin,
0 Otherwise. r+1 if q 1 (mod 4) and pin,
6. ql if q =3 (mod 4) and pin,
a Otherwise.
7. { ((n, q  1)  1) (q + 2) + 9 if q _ 1 (mod 4),
((n,ql)l)Q+? if q=3 (mod 4).
8. { ((n,q+ 1) l)q+ ~ if q=1 (mod 4),
'((n,q + 1)  l)(q  2) + ~ ifq=3 (mod 4).
16
.3.. COUNTEREXAMPLE FOR THE RESULT IN, [7J
In [7], the number of solutions of An = B2 is computed case by case up to conjugacy.
The following table is a result of (7]. '
Table IV
Case No. No. ,ofSolutions
1. 2
2.
3.
4.
5.
6.
7.
8.
{
q(q + 1) if q 1 (mod 4),
q(q  1) if q 3 (mod 4)
2
2
{
Coase 2 pin
pfn
{
Coase 2 pin
pin
{
(n,q;1)lq(q+1)+~_(q_l,n) ifq=l (mod 4),
(n,q;l)lq(q _ 1) + ~ if q=3 (mod 4).
~ 1  ~(1 + (l)~).
The table IV of [7J is completely different from our Table III except Case 1, Case
3 and Case 4. The main difference is in the result of Case 8. The formula given in [7]
in Case 8 is ~  1  HI + (l)Y). While in this paper, by our computation, the
number of solutions of the same case is
{
((n, q + 1)  l)q + ~ if q=1 (mod 4),
((n, q 1)  l)(q  1) + ~  (n, q + 1) if q _ 3 (mod 4).
They are not equivalent to each other. There is an error in [7] by the following
examples.
Example 1. Let q = 5, n = 3. Then Fq = {O, 1,2, 3,4}. Since .j2 ~ Fq , we choose
E = 2. Write A = (: :) as the pair (a, b),. where a, b E Fq and a2
 €b2 = 1.
( u v) (a b)n . If €V u €b a ,wewnte(u,v) = (a,b)n. Letn={(a,b)la2 Eb2 =
1 in Fq }. We determine a by a2 = 1 + Eb2
• Thus
n = {(I, 0), (4,0), (2,2), (2,3), (3,2), (3, 3)}.
17
i
'II!,
II
Let 112 ~ {(u,v)I{u,v).= (a,b)Z tor som'e (a,b) E OJ, . Clearly if (u~v).....:; (a,bJ~7
then u ~ 0,2 +ebz, 11 = Zab. Thus
0.2 ;;:; ((l,O'), (2,2), (2, 3)}.
Let na ~ ({u, v)j(u, v) ;; (a, b);;} for some (a~ b) En}. Obviously if (tt,V) := (a;u)'\
then u =a3'+3ab2e,v = 3a?b+ tri. Thus
US ={(I, 0), (1, O)}.
. .
So for any A E 0, A3 = ±I. The number of solutions of both A3 = HZ with
A E n {±l} is equal to
I{BIB2 = I}I *' HAIA3
:= I,A E nIl! + ·1{BIB2 = I}I * I{AIA3 = I,A E {!dl.
. .
By computation, Ha,b)j{a,b)3=,I} = {(l,O),(2,3),(2,2)}. The solutionsfof
B2 .~ I is ±I. Since (2,3)1 = (2,2), the matrix (2,3) are conjugate to the matrix
(2,2), so up to conjugacy we have 2 pairs solutions for A3 = B2 = I in Case 8:
l.A= (t~ ;),B = (~ ~).
2.A = (t~ ;), B = ( ~1 ~1 ) .
We also get {(a,b)l(a,b)3 = I} = {(I, 0), (3,2), (3, 3)}. Up to conjugacy weordy
have one solution A = (3,2) for A3 = 1 in Case 8. We list the solution of B2 = I
below, where the matrices in the same row are the orbits under the action of r (3,2)'
1.. (~~)t(~ i),(~ ~),
2. (~~),(; ~),(; :),
3. (20) (3 0)"(02) ·13 ' 1 2 ' 20'
4. (30) (2 0) (03) 42' 43' 3O'
5. (2 0) (21) (13) 23' 03' 14'
6. (~ ~),(~ ~),(~ ;),
7. (; ~),(~ ~),(~ ~),
8. (~ ~),(~ i),(~ ~),
9. (~ ;),(~ ;),(~ ~),
18

10. (~ ~)", (~ ~), (~ ~).
Therefore up to conjugacy we have 12 solutions for A3 = B2 in Case 8.
By the formula given in {7], the number of solutions is
q + 1 _ 1 _ ~ (1 + (1) ~) = 5 + 1 _, 1 _ ~ (1 + (1) &;1) = 1.
22 2 2
By the formula given in table II, the number of solutions is
ql ,51
«n.,q + 1)  l)q +2 = «3,5+ 1)  1)5 + 2  1 = 12.
Example 2. Let q = 7, n = 3. ,By simple calculation,'VS ¢ :Fr , let € = 3. Then
n = {(I, 0), (6,0)., (2, 1), (2, 6); (0,3), (0,4), (5, 1), (5, 6)},
n2 = HI, 0), (6',0), (0, 3), (0, 4)},
n3 =n.
Clearly there is no matrix A E 0 1 satisfying A3 = ±I. So
{(A,B)tA3 = B2,A End = {(A,B)IA3 = B2,A, BE nIl,
which is listed below
{(2, 1), (0, 3», (2,6); (0, 4», «(5, 1), (0,4)), (5,6), (0, 3»)}.
Clearly «(2,1), (0,3» and «2,6), (01 4» are conjugate to each other, and ((5,1), (0, 4))
and «5,6), (0,3» ar~ conjugate to each other. Up to the conjuagacy, the number of
solutions is 2. By the formula given by [7J, the number of solutions is
q+ 1  1 !I..::..!.
2  1  2(1 + (1) 2 ) = 5.
By the formula given in table II, up the conjugacy the number of solutions is
q1 71
«n,q+ 1) l)(ql) + 2'  (n,q+ 1) = ((3,8) 1)(7 1) + 2  (3,8) = 2.
4. FINDING THE ZETA FUNCTION
What we are going to do now is to use the previous information on the counting of
the representations to evaluate the Zeta function of a torus knot. First we give the
definition.
Definition 4.1. Let K be a Torus knot in 3space. Fix a prime number p. Let Ns
be the number of representations of the knot group into SL(:Fp')' up to conjugacy.
The the Zeta function is simply the formal power series
(12)
00 NTs
Z(P,T) = exp(E _8_)
s
8:=::1
19
We will keep p fixed, once and for all. Now notice that if N; is the number of
solution that come from Case i above then: Ns = L:~=1 N;,
Z(p, T) =
where Zi = exp(I::l Nr:~), so basically we can just find the Zeta function for each
individual case, then multiply the answers together when all done.
Case 1. Recall that in this situation we have 2 solutions, no matter what q = pS
was. So we can say Ns
1 =2, so
00 NITs 00 T8
exp(L _8_) = exp(L 2)
s s 3=1 5=1
00 TS
exp(2L ;) = exp(2log(1 T»
8=1
1
Case 2. From table III, we have 1 solution up to conjugacy. So
1
Z2=
IT
Similarly we get
Z _ 1
3  (lTP'
Case 5. From table III, we have no solution if p f n. If pin, the number of solutions
is equal to
{
q + 1 if q =1 (mod 4),
q  1 if q=3 (mod 4),
If p =1 (mod 4), then plJ _ 1 (mod 4) for all s. If p 3 {mod 4), then p8 =3
(mod 4) for s is odd, p8 =1 (mod 4) for s is even. Therefore the previous formula is
equivalent to
{
ps + 1 if P=1
pS+(_l)S ifp=3.
20
(mod 4),
(mod 4).
Therefore for p =1.(mod 4)' and p. j n, we get
Z5 =
00 N 5Ts
exp(" _S_)
LJ s
s=l
(~ (1 +pll)Ts)
exp LJ
s
8=1
1
(1  pT)(l  T)'
For p=3 (mod 4) and pin, we get
. 00 N5Ts
Z5 = exp(L _s)
8
0=1
00 (pS + (_1)8)TS
exp(L )
s
8=1
1
(1  pT)(l +T)'
In general,
Similarly
Z5 = { 1 1
1'+!
(lpT)(H(l) rT)
{
1.
Z6 = 1
(lpT)(l+(l)~T)
if p f n,
if pin.
if p tn,
if pin.
To calculate Z7, we need to deal with L:(pS  1,n)~~. So we first discuss how to
calculateL::I«(ps  1,n)  l)~·. By Fermat's little Theorem, for each din, there
exists a 'ljJ(d), such that p1/J(d) _ 1 (mod d) if (d,p) = 1. Clearly if (n,p) =I 1, then
(n,pS 1) = 1 for all s. In this situation, we obtain
Z1= {
V(1T)6
lpT
(IT)4(t+T)2
IpT
if p =1 (mod 4)
if p =3 (mod 4).
Suppose (n,p) = 1.
Given a natural number p, we can write n = pi! p~2 ... p;r for some distinguish prime
number Pi. So we calculate three different cases i)n = PI, ii)n = pi, iii)n = PIPZ
where PI,P2 are different prime number, with P1,P2 > 2. Then the other cases can be
computed by the same algorithm. Write F(s) = L:~I(pS  1,PI)  1)~~.
21
i: Let n = Pi: Let ml be the smaHestpasitive integer satisfying pS =1 (mod PI)'
If pm =1 (mod pd, we can write m = kmI for some integer k. Therefore
00 TS
F(s) = '.L"..'<«(pS  1,PI)  1) S
s=1
ii: Let n = pi. Let ml as above. Let m2 be the smallest positive integer satisfying
pS =1 (mod pf). Clearly mllm2. Therefore
00 Tm,k 00 Tm2k . 00 rm2k
= (PI  1)L m1k  (PI  1)L m2k + (vi  1)L m2k
k=1 k=1 k=i
= _PI  1 In(l rynl) + PI : 1 1n(1 _ Tm2) _ pi  1 1n(1 _ rm2)
ml m2 m2
= _PI :1 1n(1 Tm.) _ Pi(Pl  1) In(l _ Tm2).
ml m2
iii: For n = PIP2 Choose ml as above.. Let m3, m4 be the smallest positive integers
satisfying pS =1 (mod P2) and pS = 1 (mod PIP2) respectively. Clearly ml ImJj
and m3 I m4. Then
00 . Ttl
F(s) = L«(ps l,PIP2)  1)
8=1 S
00 Tmlk .Tm4k 00 Tmak
= (PIl)L  (pl1)L + (P2 l)L·
~I~k ~k ~J~k
00 Tm4 k 00 Tm4 k
 (P2 1)L+(PIP2 1)2:
k=1 m4 k k=1 m4k
= _Pi  1 1n(1 _ Tml) + PI 1 1n(1 T m4)
ml m4
_ P3  1 In(1 _ yma) + P2  1 In (1 _ Tm4) _ PIP2  1 In (1 _ T m 4)
m3 m4 m4
= _PI  1 1n(1 _ Tm]) _ P2  1 1n(1 yma) _ (PI  1)(P2  1) In(l _ Tm4).
ml m3 m4
Tosimplifytheformula,letj(pl) = PIi,!(P2) = Pt(Pll),andj(PJP2) = (PlI)(P2 i).
ml m2 ma
So for n = PI, we get,
22
n \ I P
Dror n = P'2i' we get Ii
I. .'
.. p =1 (mod 4),
Z.., =
1 1
(Ipml Tml )HpJ) (1_Tmq2f(Pl) (1_pm2T'R2 )/(pl)
if P=1 (mod 4),
Zr =
1 I 1.
(l(pT)m1)/(pJl * (lTm.t)t<Pl) * (I_(T)ml)f(Pl)
ifp= 3 (mod 4).
For n = PIP2, we get
1 I
1 * I
(1rm3 2f(P3) (1_pm 4Tm 4 )!(P3)(lTm 4 )2/(P31
(lT)5
.jlpT if P=1 (mod 4),
v.(1T)4(1+T)
lpT if P=3 (mod 4).
To compute the Zeta function of case 8, we use the same method as we have shown
in the computation of Zeta function of Ca.'5e 7. And by the same reason as above,
we only give the Zeta function for i)n = PI, ii)n = pi, iii)n = PIP2 for some PI,P2
are prime number with Pl,P2 > 2 and PI =1= P2· To simplify the discussion, we always
suppose there is a solution for the equation pS == 1 (mod n).
23
i)For n =Pt. Let m~ be the smallest positive solution for pS =1 (mod ph. By
Proposition 4.1, we get if pm = 1 (mod Pt), then m = (2k + l)mt for some integer
k ~ O. Clearly mt = 2m1
t . We compute the subcase of P=1 (mod 4). So
S 1
N~ = ((Pl'PS + 1)  l)pS + P ;
We compute In Zs first.
co pS _ 1 TS
In(Zs) = I)((n,pS + 1) _l)pS + 2 )s
s=1
PI  1 00 (pT) (2k+l)m~ '1 00 pSYs _ TS
= ml L 2k + 1 + 2L S
k=l s=1
= PI  1 In 1 + ,(pT)m: + ~ In 1  T
2m~ 1  (pT)m) 2 1  pT
=f(pdln(l+(pT)~)+~ln(IT).
1  (pT) 2 2 1  pT
In general we get
if p =1 (mod 4),
if p =3 (mod 4).
I I
, (l+(pT)mj
l )!(PI) (~)!(Pl)
.l_(pT)mJ l+Tml
I ,...,=:;:
(
1+(_T)ffll )f(PI) (IT)2 * l(T)ffi'l (JpT)(l+T)
ii)For n = pro Choose m~ as above, let m~ be ~he smallest positive solution for
pS =1 (mod pD. Clearly mllm2. And m2 = 2m~. By the similarly discussion as
above, we get Zs equal to
if P =1 (mod 4),
iii)For n = PlP2' Choose m~ as above, let m~ be the smallest positive solution
for pS =1 (mod ph, m~ be the smallest positive solution for pI! _ 1 (mod PIP2).
Clearly m~lm~, m~lm~, and m~, m;are coprime. We also get m3 = 2m; and m4 = m~.
Therefore Zs equal to
24
if p =1 (mod 4),
In [7], a modified Zetafunction of a knot is defined as
(13)
00 N'Ts
),K(p, T) = exp(~(_S_)), L.J s
s=]
where N~ is the number of nondiagonal representations of the fundamental group
of the torus knot into SL(Fps), counted up to conjugacy.
Let p =1 (mod 4). We show that for n = PI, Xil(p, T) is a polynomial. Here we
have two subcases 1) p t n. 2)p I n. Denote expeL: (N;;'T") by Ai where (N;)' is the
number of nondiagonal representations in Case i.
1) Let Ptn.
i: Both representative is Case 1 are diagonal, so
ii: Both representative is Case 2 are diagonal, so
),2 = 1
iii: Obviously no diagonal representations since A is not diagonal. So ),3 = (1_\.)2'
iv: Obviously no diagonal representations since A is not diagonal. SO),4 = (1_\.)Z.
v: By table III, we get Ng = 0, so ),5 = 1.
vi: By table III, we get Nt = 0, so ),6 = 1.
vii: The representations of subcase 7.b are diagonal. In subcase 7.a, for the solution
(A, B), it is diagonal if and only if B is diagonal. For each fixed AI,
there are two diagonal solutions satisfying A~ = B2 . So (N;)' = ~((n,plJ  1)1)(
2(pS + 3)  2) = ((n,pS  1)  1)(ps + 2). We get
1
),7 = (1 _ pml Tml )J(Pr) (1  Tml )2J(PJ)'
viii: Obviously all the presentations in Case 8 are not diagonal. So
), = 1 + (pT)m'] )f(pI) ~
8 1 (pT)m'] V~'
25
Therefore
A2 = 1 * 1 * (1  1T)4 * (I 1T)4 * 1 * 1
1 1+ (pT)m'l 2f(PI) 1  T
* (1 pm1TmJ)2f (pd(1_ Tm1 )4f (pJ) *.(1 (pT)m'l) 1 pT
1 1 1
(1  T)7 ((1  (pT)m~ )4f (p.) (1  Tml )4f (pJl 1 pT'
Hence A2 is a polynomial.
2)Let pin.
i: Both representative is Case 1 are diagonal, so
ii: Both representative is Case 2 are diagonal, so
iii: Obviously no diagonal representations siIlce A is not diagonaL So A3 = (1~T)2'
iv: Obviously no diagonal representations since A is not diagonaL So A4 = (1!T)2'
v: Clearly no diagonal representations since A is not diagonal, So A5 = (1PT~(lT)'
vi: Clearly no diagonal representations since A is not diagonal. So A6 = (I pT)(1T)'
vii: The representations of subcase 7.b are diagonal. In subcase 7.a, since pin,
(n, pS  1)  1 = 0, we get
viii: Obviously all the presentations in Case 8 are not diagonal. So _[glT A8  1pT'
Therefore
2 111
A = 1 * 1 * (1 _ T)4 (1  T)4 (1  pT):.!(l  T)2
1 1 T
,.c::;=:,.;: * 1 * (I  pT)2(1  T)2 1  pT
1
(1  T)l1 (1  pT)5'
Clearly A2 is a polynomial.
26
Let n = Pt. Then
1 I( 1 + (pT)m~ 2f(pJl 1 + (pT)m'2 2/CPr) r  T
* (1  Tm2)4f (PD 1  (pT)m;) (1 ~ (pT)m~) 1  pT
111
 (lT)7 ((1 (pT)m'I)4f (Pl)(1 Tm1 )4f (Pd ((1.. (pT)m~)4f(pr)(1 Tm2 )4f (pn
1
* . 1pT
Clearly ,\2 is a polynomial.
Let n = PIP2. Then
,\2 = 1 1 1
(1  T)7 ((1  (pT)m'l )4f (ptJ (1  Tml )4f (PtJ ((1  (pT)mj )Af(P2 (1  Tm3 )4f (P2)
1 1
*((1  (pT)mI4 )4f (P1P2 ) (1 Tm4 )4!(PIP2)*1 p.T
Obviously ,\2 is a polynomial.
27
REFERENCES
[1] S. Akbulut, J. MaCarthy, Casson's invariant for oriented homology 3spheres, an exposition,
Math. Notes 36, (1990).
[2] W. Fulton and J. Harris, Representation Theory: a first course, SpringerVerlag, (1950).
[3] W. Li, CassonLin's invariant and Floer homology, J. Knot Theory and its Ramifications,
Vol 6, No.6 (1997), 851877.
(4] W. Li, Knot and Link Invariants and Moduli Spaces of Parabolic Bundles, to appear in
Communications in Contemporary Mathematics.
{5] X. S. Lin, A knot invariant via represent.ation spaces, J. Diff. GeOID., 35, 337  357 (1992).
[6] Kenneth H. Rosen, Elementary Number Theory And Its Applications,AddisonWesley,
(2000).
[7] JeHery M. Sink, A ZetaFunction For A Knot Using SL2(:Fp') Representations, Knots in
Bellas' 98, Proceedings of The International Conference on Knot Theory and Its Ramincation,
452470.
28
~
VITA
Liang Xu
Candidate for the Degree of
Master of Science
Thesis: SL2 (.7q)REPRESENTAnONS OF TORUS KNOTS
Major Field: Mathematics
Biographical:
Education: Received Bachelor of Science degree in Mathematics in Hubei
University, Wuhan, Hubei, China in July, 1995. Attended Graduate
School in Wuhan University after that. Completed the requirements
for the Master of Science degree with a major in Pure Mathematics at
Oklahoma State University in August 2001.
Experience: Employed by Wuhan University, Department of Mathematics as a
graduate teaching assistant from 19971998; employed by Central
China Normal University, Department of Mathematics as a lecturer
from 19981999; employed by Oklahoma State University,
Department of Mathematics as a graduate teaching assistant from
2000 to 200] .