A polynomial defined by the $\mathit{SL}(2;\mathbb{C})$-Reidemeister torsion for a homology 3-sphere obtained by Dehn-surgery along a torus knot

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by  Teruaki Kitano

A POLYNOMIAL DEFINED BY THE SL(2;C)-REIDEMEISTER
TORSION FOR A HOMOLOGY 3-SPHERE OBTAINED BY
DEHN-SURGERY ALONG A TORUS KNOT
6
1
0 TERUAKIKITANO
2

n
Abstract. LetM beahomology3-sphereobtainedby 1-Dehnsurgery
a n n
J along a (p,q)-torus knot. We consider a polynomial σ (t) whose
(p,q,n)
4 zeros are the inverses of the Reideimeister torsion of Mn for SL(2;C)-
irreduciblerepresentations. We giveanexplicitformulaofthispolyno-

] mialbyusingTchebychevpolynomialsofthefirstkind.Furtherwealso
T
givea3-termrelationsofthesepolynomials.
G
.
h
t
a
m 1. Introduction
[
Let T(p,q) be a (p,q)-torus knot in S3. Here p,q are coprime and posi-

1
tive integers. Let M be a homology 3-sphere obtained by 1-Dehn surgery
v n n
8 along T(p,q). It is well known that M is a Brieskorn homology 3-sphere
n
4 Σ(p,q,N)wherewewrite N for pqn+1. HereΣ(p,q,N) isdefined as
4 | |
0 (z ,z ,z ) C3 zp +zq +zN = 0, z 2 + z 2 + z 2 = 1 .
0 { 1 2 3 ∈ | 1 2 3 | 1| | 2| | 3| }
.
1 In this paper we consider the Reidemeister torsion τ (M ) of M for an
ρ n n
0 irreduciblerepresentationρ : π (M ) SL(2;C).
6 1 n →
1 In the 1980’s Johnson [1] gave an explicit formula for any non-trivial
: value of τ (M ). Furthermore, he proposed to consider the polynomial
v ρ n
i whosezerosetcoincideswiththesetofallnon-trivialvalues 1 ,which
X {τρ(Mn)}
isdenotedbyσ (t). Undersomenormalizationofσ (t),hegavea3-
r (2,3,n) (2,3,n)
a
term relation among σ (t),σ (t) and σ (t) by using Tcheby-
(2,3,n+1) (2,3,n) (2,3,n 1)
−
chevpolynomialsofthefirst kind.
Recently in [5] we gave one generalization of the Johnson’s formula for
a (2p ,q)-torus knot. Here p ,q are coprimeodd integers. In this paper, we
′ ′
showtheformulaforany torusknotT(p,q).
Acknowledgements. TheauthorwasstayinginAix-MarseilleUniversity
whenhewrotethisarticle. Hethanksfortheirhospitality. Thisresearch
waspartiallysupportedby JSPS KAKENHI25400101.
2010MathematicsSubjectClassification.57M27.
Key words and phrases. Reidemeister torsion, a torus knot, Brieskorn homology 3-
sphere,SL(2;C)-representation.
1
2 TERUAKIKITANO
2. Definition ofReidemeister torsion
Firstlet usdescribedefinitionsandproperties oftheReidemeistertorsion
forSL(2;C)-representations. See Johnson[1], Kitano [2,3]and Porti [7]
fordetails.
Letb = (b , ,b )and c = (c , ,c )betwobases forad-dimensional
1 d 1 d
··· ···
d
vectorspaceW overC. Settingb = p c , weobtainanonsingular
i ji j
Xj=1
matrix P = (p ) GL(d;C). Let [b/ c]denotethedeterminantof P.
ij
∈
Suppose
C : 0 Ck ∂k Ck 1 ∂k−1 ∂2 C1 ∂1 C0 0
∗ → → − → ··· → → →
isan acyclicchain complexoffinitedimensionalvectorspaces overC. We
assumethatapreferred basisc forC is givenforeach i. Thatis,C is a
i i
basedacyclicchain complexoverC. ∗
Chooseany basisb for B = Im(∂ ) andtakealiftofit inC ,which is
i i i+1 i+1
denotedbyb˜ . Since B = Z = Ker∂,thebasisb can serveas abasisfor
i i i i i
Z. Furthermoresincethesequence
i
0 Z C ∂i B 0
i i i 1
→ → → − →
isexact,thevectors(b,b˜ ) formabasisforC . Here b˜ is aliftofb
i i 1 i i 1 i 1
inC. It iseasilyshownth−at[b,b˜ /c]does notdepend−onachoiceof−a
i i i 1 i
lift ˜b . Hencewecan simplyden−oteit by[b,b /c].
i 1 i i 1 i
− −
Definition2.1. The torsionτ(C )of abased chaincomplexC with c is
∗ ∗ { ∗}
givenbythealternatingproduct
k
τ(C ) = [b,b /c]( 1)i+1.
i i 1 i −
∗ Yi=0 −
Remark2.2. It iseasyto seethatτ(C )doesnot dependon choicesofthe
∗
bases b , ,b .
0 k
{ ··· }
Nowweapplythistorsioninvariantofchain complexestogeometric
situationsas follows. Let X beafiniteCW-complexand X˜ auniversal
coveringof X withthelifted CW-complexstructure. Thefundamental
groupπ X acts on X˜ from theright-handsideas deck transformations. We
1
mayassumethat thisaction isfree and cellularby takingasubdivisionif
weneed. Thenthechain complexC (X˜;Z)has thestructureofachain
complexoffree Z[π X]-modules. ∗
1
Letρ : π X SL(2;C)bearepresentation. Wedenotethe2-dimensional
1
vectorspace→C2 byV. Usingtherepresentation ρ,V admitsthestructureof
aZ[π X]-moduleand thenwe denoteitby V .
1 ρ
A POLYNOMIALDEFINEDBYREIDEMEISTERTORSION 3
DefinethechaincomplexC (X;V )byC (X˜;Z) V and choosea
∗ ρ ∗ ⊗Z[π1X] ρ
preferred basis
(u˜ e ,u˜ e , ,u˜ e ,u˜ e )
1 1 1 2 d 1 d 2
⊗ ⊗ ··· ⊗ ⊗
ofC (X;V )where e ,e isacanonical basisofV = C2, u , ,u are
i ρ 1 2 1 d
{ } { ··· }
thei-cells givingabasisofC(X;Z)and u˜ , ,u˜ are liftsofthemon X˜.
i 1 d
{ ··· }
NowwesupposethatC (X;V )isacyclic, namelyall homologygroups
ρ
∗
H (X;V ) arevanishing. In thiscaseρis called an acyclicrepresentation.
ρ
∗
Definition2.3. Let ρ : π (X) SL(2;C)bean acyclicrepresentation.
1
Then theReidemeistertorsion→τ (X) C 0 is definedbythetorsion
ρ
∈ \{ }
τ(C (X;V ))ofC (X;V ).
ρ ρ
∗ ∗
Remark2.4.
(1) Wedefineτ (X) = 0 fora non-acyclicrepresentationρ.
ρ
(2) Thedefinitionof τ (X) dependson severalchoices. However itis
ρ
well known thatitisa piecewiselinearinvariantinthecaseof
SL(2;C)-representations.
3. Johnson’stheory
LetT(p,q) S3 bea(p,q)-torusknotwithcoprimeintegers p,q. Nowwe
⊂
write M to aclosedorientable3-manifoldobtainedby a 1-Dehn surgery
n n
alongT(p,q). HerethefundamentalgroupofS3 T(p,q)has the
\
presentationas follows;
π (S3 T(p,q)) = x,y xp = yq .
1
\ h | i
Furthermoreπ (M )admitsthepresentationas follows;
1 n
π (M ) = x,y xp = yq,mln = 1
1 n
h | i
wherem = x rys (r,s Z, ps qr = 1)is ameridianofT(p,q)and
−
similarlyl = x pmpq =∈y qmpq−isalongitude.
− −
Itis seen [1,5]that theset oftheconjugacyclassesoftheirreducible
representationsofπ (M )inSL(2;C)is finite. Anyconjugacyclasscan be
1 n
representedbyρ : π (M ) SL(2;C)forsometriple(a,b,k)suchthat
(a,b,k) 1 n
→
(1) 0 < a < p,0 < b < q,a b mod2,
(2) 0 < k < N = pqn+1,k≡ namod2,
(3) tr(ρ (x)) =| 2cos aπ|, ≡
(a,b,k) p
(4) tr(ρ (y)) = 2cos bπ,
(a,b,k) q
(5) tr(ρ (m)) = 2cos kπ.
(a,b,k) N
FurthermoreJohnsoncomputedτ (M ) asfollows.
ρ(a,b,k) n
Theorem 3.1 (Johnson).
(1) Arepresentationρ is acylicif andonlyifa b 1.
(a,b,k)
≡ ≡
4 TERUAKIKITANO
(2) Foranyacyclicrepresentationρ with a b 1, then onehas
(a,b,k)
≡ ≡
1
τ (M ) = .
ρ(a,b,k) n 2 1 cos aπ 1 cos bπ 1+cos pqkπ
− p − q N
(cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17)
4. Maintheorem
Inthissection wegiveaformulaofthetorsionpolynomialσ (t)for
(p,q,n)
M = Σ(p,q,N)obtainedbya 1-Dehn surgery alongT(p,q). Nowwe
n n
definetorsionpolynomialsas follows.
Definition4.1. A onevariablepolynomialσ (t)iscalledthetorsion
(p,q,n)
polynomialof M if thezeroset coincideswith theset ofall nontrivial
n
values 1 τ (M ) , 0 and itsatisfiesthefollowingnormalization
τρ(Mn) | ρ n
conditionnas o
(N 1)p(q 1)
( 1) − 8 − p iseven,qis odd,
−
σ(p,q,n)(0) =  ((−−11))((NN−−11))8((pp8−−11))q(q−q1)ips,eqveanr,eqoidsdo,dnd,iseven,
N(p 1)(q 1)
where N = pqn+1.  (−1) −8 − p,qareodd,nisodd
| |
Remark4.2.
(1) For M = S3,thetorsionpolynomialσ (t)isdefined by
0 (p,q,0)
σ (t) = 1.
(p,q,0)
(2) In thecasethat p = 2p is even and p is odd,thenthis
′ ′
normalizationconditioncoincideswith theonein [5].
Fromhere assumen , 0. Recall Johnson’sformula
1 aπ bπ pqkπ
= 2 1 cos 1 cos 1+cos
τ (M ) − p ! − q ! N !
ρ(a,b,k) n
where0 < a < p,0 < b < q,a b 1 mod2,k n mod2. Here by
≡ ≡ ≡
putting
aπ bπ
C = 1 cos 1 cos ,
(p,q,a,b)
− p ! − q !
onehas
1 1 pqkπ
= 4C 1+cos .
(p,q,a,b)
τ (M ) · 2 N !
ρ(a,b,k) n
Mainresultisthefollowing.
Theorem 4.3. Thetorsionpolynomialof M is given by
n
σ (t) = Y (t)
(p,q,n) (n,a,b)
Y
(a,b)
A POLYNOMIALDEFINEDBYREIDEMEISTERTORSION 5
where
TN+1(s) TN 1(s) (p orq is even,n > 0),
2(s2−1)2−
 TN+1(−s) TN 1(s) (p or q iseven,n < 0),
Here Y(n,a,b)(t) =  T−−TNNT+(N12(+s(ss122))2(((−s−ss)T221−−(−−N)T2p−11N1)),22(−−sq1)(,s)n((appr,,eqqoaadrrdee)o.odddd,,nniisseevveenn,,nn><00)),.
T (x) isthel-thTchebychevpolynomialofthefirstkind.
l
•
√t
s = .
• 2 C
(p,q,a,b)
C p= 1 cos aπ 1 cos bπ .
• (p,q,a,b) − p − q
apairof in(cid:16)tegers(a,b(cid:17))(cid:16)is satisfyin(cid:17)gthefollowingconditions;
•
– 0 < a < p,0 < b < q,
– a b 1 mod2.
≡ ≡
Remark4.4. Recall thatthel-thTchebychevpolynomialT (x)is defined
l
byT (cosθ) = cos(lθ).
l
Proof. Weconsiderthefollowing;
TN+1(x) TN 1(x) (n > 0)
X (x) = 2(x−2 1)−
n  TN+1(x)−TN 1(x) (n < 0).
 − 2(x−2 1)−
−
Xn′(x) = TN(x).
Firstweassume p = 2p iseven. Forthecase that p isodd,then itis
′ ′
provedin [5]. Thenwesupposethat p iseven. Here N = 2p qn+1 is
′ ′
| |
alwaysodd.
Case1:p = 2p , p is evenandn > 0
′ ′
1
Wemodifyonefactor (1+cos 2p′qkπ)of asfollows. See[5]for the
N τ (M )
ρ n
proof.
Lemma4.5. Theset cos 2p′qkπ 0 < k < N,k n mod 2 isequaltotheset
{ N | ≡ }
cos 2p′kπ 0 < k < N 1 .
−
{ N | 2 }
Nowwecan modify
1 2p kπ 1 2p kπ
1+cos ′ = 2cos2 ′
2 N ! 2 · 2N
p kπ
= cos2 ′ .
N
6 TERUAKIKITANO
Weput
p kπ
z = cos ′ (1 k N 1).
k
N ≤ ≤ −
By thedefinition,it isseen
p (N k)π
z = cos ′ −
N k
− N
p kπ
= cos(p π ′ )
′
− N
= z
k
because p is even.
′
Thereforeit isenoughto consideronlyz (1 k N 1).
Nowwesubstitute x = z to T (x). Thekn on≤eha≤s 2−
k N+1
p kπ
T (z ) = cos (N +1) ′
N+1 k
N !
p kπ
= cos ′
N
= z
k
and
p kπ
T (z ) = cos (N 1) ′
N 1 k
− − N !
p kπ
= cos ′
N
= z .
k
Henceitholds
T (z ) T (z ) = 0.
N+1 k N 1 k
− −
By propertiesofTchebychevpolynomials,itisseen that
T (1) T (1) = 0,
N+1 N 1
• T ( 1−) T− ( 1) = 0.
N+1 N 1
• − − − −
Weremark thatthedegreeof X (x) = TN+1(x) TN 1(x) is N 1 and
n 2(x−2 1)− −
z , ,z arezeros. Because bothofT (−x) and T (x)are even
1 N 1 N+1 N 1
··· 2− −
functions,then z , , z are alsozeros of X (x). Hence X (x) isa
1 N 1 n n
− ··· − 2−
functionsof x2. Hereby replacing xby √t , thedegreeofY (t)is
(n,a,b)
2√C(p,q,a,b)
N 1, and therootsofY (t)are
2− (n,a,b)
πk N 1
4C z2 = 4C cos2 0 < k < − ,
(p,q,a,b) k (p,q,a,b) N 2 !
whichare allnontrivialvaluesof 1 .
τρ(a,b,k)(Mn)
A POLYNOMIALDEFINEDBYREIDEMEISTERTORSION 7
Herewecheck thenormalizationcondition. By thedefinitionofY (t)
(n,a,b)
andproperties ofT (x),T (x), onehas
N+1 N 1
−
T (0) T (0)
Y(n,a,b)(0) = N+1 − N−1
2(0 1)
−
= ( 1)N2+1 ( 1)N2−1
− − −
− 2
= ( 1)N2−1.
−
Henceitcan beseen
σ(p,q,n)(0) = ( 1)N2−1
−
Y
(a,b)
p(q 1)
= ( 1)N2−1 4−
−
Y(a,b)(cid:16) (cid:17)
= ( 1)(N−1)8p(q−1).
−
Thereforeweobtaintheformula.
Case2:p = 2p and n < 0
′
Inthiscase wemodify N = 2p qn+1 = 2p qn 1. By thesame
′ ′
| | | |−
arguments,itiseasy to seetheclaim ofthetheorem isproved.
Nextassumebothof p,qare oddintegers.
Case3:p,qare oddand n iseven
Ifn iseven,then N = pqn+1 isodd. Thenthesimilarargumentsin[5]
| |
workwell. Thenit can beproved.
Case4:p,qare oddand n isodd
Supposen ispositive. Firstnotethat N = pqn+1 iseven. Wecan modify
1 | |
onefactor (1+cos pqkπ)of as follows. It isclear because
N τ (M )
ρ n
(q,N) = 1.
Lemma 4.6. Theset cos pqkπ 0 < k < N,k n mod2 isequalto theset
{ N | ≡ }
cos pkπ 0 < k < N,k 1mod 2 .
{ N | ≡ }
Nowwecan modify
1 pkπ 1 pkπ
1+cos = 2cos2
2 N ! 2 · 2N
pkπ
= cos2 .
2N
Weput
pkπ
z = cos (1 k N 1, k 1 mod2).
′k 2N ≤ ≤ − ≡
8 TERUAKIKITANO
Herewesubsitute x = z (1 k N 1, k 1mod2)to T (x). Then one
′k ≤ ≤ 2− ≡ N
has
N(pkπ)
T (z ) = cos
N ′k 2N !
pkπ
= cos
2 !
= 0
because pk isodd.
Similarlyit can bealsoseen that
T ( z ) = 0.
N − ′k
Wementionthat thedegreeof X (x) = T (x) is N and z , , z are
n′ N ± ′1 ··· ± ′N 1
thezeros. Because X (x) isafunctionsof x2. Here byreplacing xby−
n′
√t , Hereit holdsthatitsdegreeofY (t)is N 1, andtherootsof
2√C(p,q,a,b) (n,a,b) 2−
Y (t)are
(n,a,b)
πk N 1
4C z 2 = 4C cos2 0 < k < − ,
(p,q,a,b) ′k (p,q,a,b) N 2 !
whichare allnontrivialvaluesof 1 .
τρ(a,b,k)(Mn)
Finallywecan check thenormalizationconditionas follows. By the
definitionofY (t), onehas
(n,a,b)
Y (0) = T (0)
(n,a,b) N
= ( 1)N2
−
and
σ(p,q,n)(0) = ( 1)N2
−
Y
(a,b)
(p 1)(q 1)
= ( 1)N2 − 4 −
−
(cid:16) (cid:17)
= ( 1)N(p−18)(q−1).
−
Thereforeweobtaintheformula.
Inthecase thatn isnegative,then itcan beprovedbysimilararguments.
Thereforethiscompletestheproof. (cid:3)
Remark4.7. Bydefiningas X (t) = 1,it impliesY (t) = 1. Then the
0 (0,a,b)
abovestatementis trueforn = 0.
By direct computation,oneobtainsthefollowingcorollary.
A POLYNOMIALDEFINEDBYREIDEMEISTERTORSION 9
Corollary4.8. Thedegreedeg(σ (t))is givenby
(p,q,n)
(N 1)p(q 1) (p even,qodd),
− −
8
deg(σ(p,q,n)(t)) =  ((NN−−11))8((pp8−−11))(qq−(1)p o(pd,dq,qodevde,nn)e,ven),
Wementionthe3-termrelationsN.(Fp−o1r8)(eq−a1c)h(fpac,tqoordodf,Yn odd()t.)ofσ (t),
(n,a,b) (p,q,n)
thereexiststhefollowingrelation.
Proposition4.9.
(1) Assumeoneof p and qis even. Foranyn, itholdsthat
Y (t) = D(t)Y (t) Y (t)
(n+1,a,b) (n,a,b) (n 1,a,b)
− −
where D(t) = 2T √t .
pq
2√Cp,q,a,b!
(2) Assumebothof p,q areodd. Foranyn,it holdsthat
Y (t) = D(t)Y (t) Y (t)
(n+2,a,b) (n,a,b) (n 2,a,b)
− −
where D(t) = 2T √t .
2pq
2√C2p,q,a,b!
Proof. Herewe need toconsider N = pqn+1 is afunctionofn Z for
| | ∈
fixed p,q. Thenwewrite N(n) for N inthisproof.
Theproofforthefirst caseisessentiallythesameoneforthe3-term
relations[5]. Wegivetheproofonlyforthesecond case.
Recall thefollowingpropertyofTchebychevpolynomials
2T (x)T (x) = T (x)+T (x)
m n m+n m n
−
foranym,n Z.
∈
Case1: nis even
Ifn > 0 onehas
T (x) T (x)
2T2pq(x)Xn(x) = 2T2pq(x) N(n)+12(x2− 1N)(n)−1
−
T (x)+T (x) (T (x)+T (x))
= (pqn+1)+1+2pq (pqn+1)+1−2pq − (pqn+1)−1+2pq (pqn+1)−1−2pq
2(x2 1)
−
T (x) T (x)+T (x) T (x)
= pq(n+2)+1+1 − pq(n+2)+1−1 pq(n−2)+1+1 − pq(n−2)+1−1
2(x2 1)
−
T (x) T (x)+T (x) T (x)
= N(n+2)+1 − N(n+2)−1 N(n−2)+1 − N(n−2)−1
2(x2 1)
−
= X (x)+X (x).
n+2 n 2
−
10 TERUAKIKITANO
Thereforeit can beseen that
X (x) = 2T (x)X (x) X (x)
n+2 2pq n n 2
− −
and
√t
Y (t) = 2T Y (t) Y (t).
(n+2,a,b) 2pq2 C  (n,a,b) − (n−2,a,b)
Ifn < 0, itcan bealso provedbypth(e2pa,qb,ao,bv)eargument.
Case2: nis odd
Ifn > 0, onehas
2T (x)X (x) = 2T (x)T (x)
2pq n′ 2pq N(n)
= T (x)+T (x)
pqn+1+2pq pqn+1 2pq
−
= T (x)+T (x)
pq(n+2)+1 pq(n 2)+1
−
= T (x)+T (x)
N(n+2) N(n 2)
−
= X (x)+X (x).
n′+2 n′ 2
−
Thereforeit can beseen that
X (x) = 2T (x)X (x) X (x)
n′+2 2pq n′ − n′ 2
−
and
√t
Y (t) = 2T Y (t) Y (t).
(n+2,a,b) 2pq2 C  (n,a,b) − (n−2,a,b)
Ifn < 0, itcan bealso proved. p (2p,q,a,b)
Thiscompletestheproofofthisproposition. (cid:3)
5. examples
Finallywegivesomeexamples.
Example5.1. Put p = 4,q = 3. Now N = 12n+1. In thiscase
(a,b) = (1,1),(3,1). Byapplyingthemain|theorem|,onehas
σ (t) = 34359738368t10 77309411328t9+66840428544t8
(4,3, 1)
− −
28655484928t7+6677331968t6 882900992t5+66371584t4
− −
2723840t3+55680t2 480t+1.
− −
σ (t) = 1.
(4,3,0)
σ (t) = 4398046511104t12 12094627905536t11+13434657701888t10
(4,3,1)
−
7859790151680t9+2670664351744t8 552909930496t7
− −
[phone]t6 5727322112t5+278757376t4
−
7741440t3+110208t2 672t+1.
− −