On the Two-Parabolic Subgroups of SL(2, C )

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Revista Colombiana de Matem´aticas Volumen 45(2011)1, p´aginas 37-50

On the Two-Parabolic Subgroups of SL(2, C )

Sobre los subgrupos dos-parab´olicos de SL(2,C)

Christian Pommerenke

¹

, Margarita Toro

²^,a

1

Technische Universit¨ at Berlin, Berlin, Germany

2

Universidad Nacional de Colombia, Medell´ın, Colombia

Abstract.We consider homomorphismsHt from the free groupF of rank 2 onto the subgroup of SL(2,C) that is generated by two parabolic matrices. Up to conjugation, Ht depends only on one complex parameter t. We study the possible relators, that is, the wordsw∈F withw6= 1 such thatHt(w) =I for somet∈C.

We find several families of relators. Of particular interest here are relators connected with 2-bridge knots, which we consider in a purely algebraic setting.

We describe an algorithm to determine whether a given word is a possible relator.

Key words and phrases. Representation, Parabolic, Wirtinger presentation, Two- generated groups, Homomorphism, Longitude.

2000 Mathematics Subject Classification.15A30, 57M05.

Resumen.Consideramos homomorfismosHtdel grupo libreFde rango 2 sobre el subgrupo de SL(2,C) que es generado por dos matrices parabólicas. Salvo conjugación, Ht depende sólo de un parámetro complejo t. Estudiamos los posibles relatores, esto es, las palabrasw∈F conw6= 1 tal queHt(w) =I para algúnt∈C.

Encontramos varias familias de relatores. De particular inter´es aqu´ı son los relatores asociados con nudos de 2 puentes, los cuales consideramos de forma puramente algebraica. Describimos un algoritmo para determinar cu´ando una palabra dada es un posible relator.

Palabras y frases clave. Representación, parabólico, presentación de Wirtinger, grupos dos-generados, homomorfismos, longitud.

aPartially supported by COLCIENCIAS, code 1118-521-28160.

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1. Introduction The subgroup of SL(2,C) generated by At =

1 0 t 1

and B = 1 1

0 1

has been studied by many mathematicians, for instance by R. Riley [16, 17], J.

Gilman [4] and P. Waterman [6]. It is of particular interest in knot theory [12, Chapter 4][2].

In terms of the corresponding Moebius transformations α and β it is, up to conjugation, the only subgroup of PSL(2,C) generated by two parabolic transformations with distinct fixed points. Indeed, we may assume thatα(0) = 0,β(∞) =∞, moreover thatβ(z) =z+ 1. Writing α(z) = (az+b)/(cz+d) with ad−bc= 1, we may also assume that trα=a+d= 2. Since b = 0 it follows thata=d= 1 and c=t∈C r{0} remains as a free parameter. It is convenient to allowt= 0.

LetF be the free group hx, yi. We consider the homomorphismsHt:F → SL(2,C) with Ht(x) =At and Ht(y) =B. For clarity we distinguish between the abstract group F and its image in the matrix group SL(2,C). Our main interest is to study the set of possible relators, that is the sets

R^±={r∈F, r6= 1|there iss∈CwithHs(r) =±I}.

If r∈R⁺ thenHs(F) has a presentationhx, y; r1, r2, . . .iwithr1=rand perhaps other relatorsr2, . . .

We shall exhibit various families of relators, some old, some new. An important family of relators comes from the presentations hx, y;xw = wyi of 2-bridge knots. Riley introduced the automorphism w∈F →we∈F induced by x→ x⁻¹ and y → y⁻¹. Our group has the special property that r ∈ R^± implies er ∈ R^±. At the end we give 8 examples to illustrate the results and show their scope.

We try to give a systematic account of the theory including some folklore results. We will use several results from Combinatorial Group Theory [13, 10]

and stress the connections to Knot Theory [2]. We have not been able to elu- cidate the role of palindromes, that is, words of F that read the same way forwards and backwards [8, 5].

We will not discuss the set orthogonal toR⁺, namely {s∈C|there existsr6= 1 such thatHs(r) =I},

the set ofswhereH_sis not injective. This set and its closure has been studied in [16, 4, 6] and, in a more general context, in [18, 14].

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C

2. Groups and Homomorphisms

Let SL(2,C) and PSL(2,C) be the groups with elements of the forms C=

a b c d

, γ(z) =az+b

cz+d, (a, b, c, d∈C, ad−bc= 1) respectively. Fort∈Cwe write

A_t= 1 0

t 1

, B =

1 1 0 1

, I=

1 0 0 1

. (1)

While the group PSL(2,C) of Moebius transformations is perhaps more important in analysis and geometry, the group SL(2,C) is more convenient for computations.

LetF be the abstract free grouphx, yi. There are [11, Theorem 8.04] unique homomorphisms

Ht:F →SL(2,C) with Ht(x) =At, Ht(y) =B. (2) ht:F →PSL(2,C) with ht(x) =αt, ht(y) =β, (3) whereα_t(z) =z/(tz+ 1) and β(z) =z+ 1 are both parabolic.

Every wordw6= 1 inF can be uniquely written as

w=xê⁰yê¹· · ·xê^m−1yê^m, eµ∈Z r{0} (µ= 1, . . . , m−1) (4) withe₀, e_m∈Zandm∈N. The exponent sums

σ_x(w) =e₀+e₂+· · ·+e_m−1, σ_y(w) =e₁+e₃+· · ·+e_m (5) are invariant under conjugations inF.As in [17, p.206] we writee1 = 1 and

we=x^−e⁰y^−e¹· · ·x^−e^m−1y^−e^m. (6) This defines an automorphism ofF. In formulas we write (·)e.

Now suppose that w ∈ F is not conjugate in F to x^k or y^k with k ∈ Z. Then the process of cyclic reduction shows that wis conjugate to a worduof the form

u=y^k¹x^j¹· · ·y^kⁿx^jⁿ, k_ν, j_ν∈Zr{0} (ν= 1, . . . , n) with n∈N. (7) Our standard form (7) allows us to arrange all words ofFup to conjugation in sequences. Let (jn) and (kn) be any given sequences withjn, kn ∈Z r{0}.

Then we defineu0= 1 and

un=y^k¹x^j¹· · ·y^kⁿx^jⁿ, for n∈N. (8)

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A conjugate of everyw∈F will appear in many such sequences. No u_n with n≥1 is conjugate tox^k ory^k.

Now we turn to the matrix elements. We often write H_t(u) =

a(t) b(t) c(t) d(t)

or

a b c d

; (9)

the elements ofH_t(u_n) will bea_n, . . . , d_n. The following proposition is folklore.

Proposition 1. Let (u_n)be given by (8). Then, for n≥0,

an+1=jn+1kn+1tan+an+jn+1tbn, (10)

b_n+1=k_n+1a_n+b_n, (11)

cn+1=jn+1kn+1tcn+cn+jn+1tdn, (12)

dn+1=kn+1cn+dn. (13)

Forn≥1, the an, . . . , dn are polynomials overZ of the forms

an=j1k1· · ·jnkntⁿ+· · ·, bn=j1k1· · ·j_n−1k_n−1kntⁿ⁻¹+· · · , (14) c_n=j₁j₂k₂· · ·j_nk_ntⁿ+· · · , d_n=j₁j₂k₂· · ·j_n−1k_n−1k_ntⁿ⁻¹+· · · . (15) The tracetrHt(un) =an+dn is non-constant forn≥1.

Proof. By (8), (1) and (2) we have

Ht(un+1) =Ht(uny^kⁿ⁺¹x^jⁿ⁺¹) =Ht(un)B^kⁿ⁺¹A^j_tⁿ⁺¹

=

an bn

c_n d_n

1 kn+1

0 1

1 0

j_n+1t 1

and (10)–(13) follow by (9); we havea0=d0= 1 andb0=c0= 0.

Now we prove the other assertions by induction. In each of the recursion formulas (10)–(13), all coefficients are inZ. Furthermore, the degree of the first term is always, by induction hypothesis, higher than the degree of the other terms. Hence, by (14), an+1 begins withjn+1kn+1t·j1k1· · ·jnkntⁿ and bn+1

begins with kn+1·j1k1· · ·jnkntⁿ, similarly forcn+1 anddn+1. The statement

about the trace follows from (14) and (15). X

Theorem 2. Letunsatisfy (8)with|jν|=|kν|= 1. Then the coefficientsan,m

of a_n(t)and so on satisfy

|an,m| ≤

n+m 2m

(0≤m≤n), |bn,m| ≤

n+m 2m+ 1

(0≤m≤n−1),

|cn,m| ≤

n+m−1 2m−1

(1≤m≤n), |dn,m| ≤

n+m−1 2m

(0≤m≤n−1).

If un = (yx)ⁿ then equality holds without taking absolute values.

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C

Compare [16, p. 233] for the last statement. The values for the case v_n = (yx)ⁿ were found by the method of generating functions. We have for instance

∞

X

n=0

a_n(t)zⁿ= 1−z (1−z)²−tz.

Proof. Letn≥0 and m≥0. Writinga_n,−1=· · ·=d_n,−1= 0, we obtain from (10)–(13) that

a_n+1,m=a_n,m+j_n+1k_n+1a_n,m−1+j_n+1b_n,m−1, bn+1,m=kn+1an,m+bn,m,

cn+1,m=cn,m+jn+1kn+1c_n,m−1+jn+1d_n,m−1, d_n+1,m=k_n+1c_n,m+d_n,m.

Now we verify the assertions by induction onn. The casen= 0 is clear because H_t(1) =I. We repeatedly use that ^α_β

+ _β−1^α

= ^α+1_β .

Since|jn+1|=|kn+1|= 1 the above recursion formulas show that

|an+1,m| ≤ |an,m|+|an,m−1|+|bn,m−1|

≤

n+m 2m

+

n+m−1 2m−2

+

n+m−1 2m−1

=

n+m 2m

+

n+m 2m−1

=

n+ 1 +m 2m

,

|bn+1,m| ≤ |an,m|+|bn,m| ≤

n+m 2m

+

n+m 2m+ 1

=

n+ 1 +m 2m+ 1

|cn+1,m| ≤ |cn,m|+|c_n,m−1|+|d_n,m−1|

≤

n+m−1 2m−1

+

n+m−2 2m−3

+

n+m−1 2m−2

=

n+m 2m−1

,

|dn+1,m| ≤ |cn,m|+|dn,m| ≤

n+m−1 2m−1

+

n+m−1 2m

=

n+m 2m

.

If u_n = (yx)ⁿ then j_n+1 = k_n+1 = 1 and all quantities are non-negative.

Hence we have equality in all the above inequalities. X Our homomorphism has an important property with respect to the auto- morphismu→eudefined in (6). The following proposition is well-known.

Proposition 3. Let u∈F,H_t(u) = a b

c d

andQ= i 0

0 −i

.Then

Ht(u) =e QHt(u)Q⁻¹=

a −b

−c d

. (16)

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Proof. It is easy to see that Q

a b c d

Q⁻¹=

a −b

−c d

. (17)

Hence it follows from (1) and (2) that QAtQ⁻¹=

1 0

−t 1

=A⁻¹_t =Ht(x⁻¹) =Ht(ex), QBQ⁻¹=

1 −1

0 1

=B⁻¹=H_t(y⁻¹) =H_t(y).e

This implies (16) because (uv)e=euev, (u⁻¹)e=eu⁻¹andH_tis a homomorphism.

X

3. Relators Our main interest is in the set of words

R^±={r∈F, r6= 1| there existss∈CwithHs(r) =±I}, (18) R={r∈F, r6= 1| there existss∈Cwithh_s(r) = id}. (19) LetN(r) denote the normal closure ofr, that is the smallest normal subgroup ofF withr∈N(r).

It follows from Proposition 3 thatHs(u) =IimpliesHs(eu) =I; see (6) for the definition of eu. Hence, for r∈ R⁺ or r ∈ R,the normal closure of {r,er}

also belongs toR⁺ or Rfor the sames. Thus we have

u=v1r1v₁⁻¹· · ·vmrmv_m⁻¹∈R⁺ or u∈R (20) wherevµ∈F and rµ ∈ {r, r⁻¹,r,eer⁻¹} forµ= 1, . . . , m andm∈N. It follows that the exponent sums (5) foru∈Nsatisfy

σx(u) =λσx(r), σy(u) =λσy(r) for some λ∈Z. (21) Ifr∈Rand thush_s(r) = id for somes∈Cthen, by the first isomorphism theorem, there is a homomorphism

h_r,s :hx, y;ri−−−→^onto h_s(F) (22) defined by h_r,s(w) := h_s(u) for any w ∈ uN(r). Now we show that h_r,s is in general not an isomorphism so that the representation of hx, y ; ri is not faithful. See Example 1.

Proposition 4. Ifhr,s is an isomorphism thenreis conjugate tor orr⁻¹.

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C

Note that, if er is conjugate to r, then σ_x(r) = σ_y(r) = 0. This result is related to [3, Theorem 3.1]. On the other hand, it is easy to see that even er=r⁻¹holds ifris a palindrome, that is, the wordrreads forwards the same as backwards. See [3, Propositions 3.4 and 3.2] for a fuller description.

Proof. Lethr,s be an isomorphism. Sincehs(er) = id it follows that er∈N(r).

Hence the normal closureN(r) ofe er, the smallest normal subset ofF containing er, satisfiesN(er)⊂N(r). Furthermoreer=v₁r^±1v₁⁻¹· · ·v_mr^±1v_m⁻¹and therefore

r= (er)e=ev1er^±1ev₁⁻¹· · ·vemer^±1ev⁻¹_m ∈N(er).

Hence N(r) ⊂ N(er) so that r and er have the same normal closure N(r). It follows [13, p. 261] [10, Proposition. 5.8, p. 106] that er is conjugate to r or

r⁻¹. X

Now we present a general method to obtain relators. See Examples 2 and 3.

Theorem 5. Let u∈F not be conjugate to x^k ory^k withk∈Z. Then

ueu∈R⁺, u²∈R⁻, (23)

uⁿ ∈R⁺∩R⁻, for n≥3, (24) thusueu∈R anduⁿ∈R forn≥2.

For uue we have to exclude the case that uis a palindrome because then ueu= 1, compare (18) and (19). Note that uue = 1 holds if and only if uis a palindrome.

Proof.

(a) We obtain from (9) and (16) that H_t(ueu) =

a b c d

a −b

−c d

=

1 +a(a−d) −b(a−d) c(a−d) 1−d(a−d)

.

Sincea−dis non-constant by Proposition 1, it follows thata(s)−d(s) = 0 for somes∈C. HenceHs(ueu) =I.

(b) First let n ≥ 2. There exists s such that trHs(u) = 2 cos(π/n). Then trHs(u) 6= ±2. Hence Hs(u) is conjugate to diag(e^iπ/n, e^−iπ/n) 6= I. It follows thatHs(uⁿ) =−I.

Now letn ≥ 3. There exists s such that trHs(u) = 2 cos(2π/n) so that, again, trH_s(u)6=±2. HenceH_s(u) is conjugate to diag(e^2πi/n, e^−2πi/n)6=I

so thatH_s(uⁿ) =I. X

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Proposition 6. Let u∈F have the form (7). If s6= 0andH_s(u) =I thens is an algebraic number of degree ≤(n−1)/2, and if H_s(u) =−I thens is an algebraic number of degree≤n/2. If|j_ν|=|k_ν|= 1forν = 1, . . . , nthen s is an algebraic integer.

This is common knowledge except for the sharp bounds (n−1)/2 andn/2 for the degrees, see Example 4. Note thatHs(F)⊂SL(2,Z[s]). We have shown that, up to conjugation,uhas the form (7) wheneveruis not conjugate tox^k ory^k.

Proof. We use the notation (9). IfH_s(u) =±Ithenc(s) = 0. Hence, by (15),s is an algebraic number which is an algebraic integer if|j_ν|=|k_ν|= 1. Now let p(t) be the minimal polynomial ofs. First letHs(u) =I ands6= 0. We write

a+d−2 =−(a−1)(d−1) +bc.

Sincea(s) =d(s) = 1 andb(s) =c(s) = 0, we conclude thatpdivides (a−1), (d−1), b and c. Hence p² divides a+d−2. Since s 6= 0, it follows that tp² dividesa+d−2, which is a polynomial of degreenby (14). Thusshas degree

≤(n−1)/2.

Now letHs(u) =−I. We write

a+d+ 2 = (a+ 1)(d+ 1)−bc.

Nowpdivides (a+ 1),(d+ 1), bandc. Hencep²divides the polynomiala+d+ 2

of degreen. Thusshas degree≤n/2. X

Now we describe an algorithm to determine whether u∈F belongs to R⁺ or R⁻. This is not the case if u is conjugate to x^k or y^k. Therefore we may assume thatuhas the form (7). We use the notation (9).

First we check whether it is possible thatb=c = 0 for somet∈C. To do this we calculate the polynomial

q0:= gcd(b, c)∈Z[t]. (25)

If degq0 = 0 then u /∈R⁺∪R⁻. If however degq0 >0 then we calculate the polynomials

q^±:= gcd(a∓1, q0). (26)

If degq⁺= 0 then u /∈R⁺, if degq⁻= 0 thenu /∈R⁻.

Now if degq^±>0 then there iss∈Csuch thata(s) =±1. It follows from (25) and (26) that b(s) = c(s) = 0 so that 1 = a(s)d(s)−b(s)c(s) = ±d(s).

Therefore we haveH_s(u) =±Iand thusu∈R^±. Additionally we may factorize q^± into irreducible polynomials overZ. If sis a zero of a factor then all other zeros t of this factor satisfy H_t(u) = ±I. The main computational difficulty of this algorithm is that very large integer coefficients may occur during the calculation of (25) and (26).

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4. The Wirtinger Relators and the Longitude

LetK be a knot inR³, see e.g. [2, 9]. The complement Ω =R³rV(K), where V(K) is a tubular neighborhood of K, is a multiply connected domain. The fundamental group Π1(Ω) is an important invariant of K though it does not completely determine the equivalence class ofK, although the prime knots are determined by their knot group [7]. A very well understood family of knots are the so called 2-bridge knots and links [20, 2, 15, 19].

The fundamental group of a 2-bridge knot admits a presentationhx, y; xw_n= wnyiwhere

w_n =y^kⁿx^kⁿ⁻¹· · ·y^k¹x^k¹y^k²· · ·y^kⁿ⁻¹x^kⁿ, k_ν∈ {1,−1}, nodd

w_n =y^kⁿx^kⁿ⁻¹· · ·y^k²x^k¹y^k¹· · ·y^kⁿ⁻¹x^kⁿ, k_ν∈ {1,−1}, neven (27) forn∈N, wherekν, ν= 1, . . . , n, satisfy some additional conditions [2, 1]. We inverted the usual order of exponents in order to have a recursive definition.

On the following we leave the context of knot theory and call any word of the form (27) aWirtinger word.

It follows from (27) that, with e defined in (6), wn+1= y^−kⁿ⁺¹wenx^−kⁿ⁺¹

e (28) Instead of (9) we now write

Wn =Ht(wn) =

a_n b_n cn dn

. (29)

It follows from (28), (1) and Proposition 3 that, withk=k_n+1, Wn+1=QB^−kW_n⁻¹A^−kQ⁻¹

=Q

1 −k

0 1

d_n −b_n

−cn an

1 0

−kt 1

Q⁻¹.

Using also (17) andk²= 1, we obtain Wn+1=

tan+ktbn+kcn+dn kan+bn

ktan+cn an

. (30)

Sinceb0=c0= 0 we deduce by induction the well-known formula [12, p. 141]

cn =tbn. (31)

Hence we obtain from (30) the recursion formulas

an+1=tan+ 2kn+1tbn+a_n−1, bn+1=kn+1an+bn.

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Now (27) is a special case of (7). Hence the estimates of Theorem 2 apply also with the new notation.

Now we drop the indexnand write W_t:=H_t(w) =

a(t) b(t) c(t) d(t)

. (32)

Theorem 7. If wsatisfies the Wirtinger condition (27)then

Hs(xw) =Hs(wye ⁻¹) (33) holds if and only if a(s) + 2b(s) = 0. Thus wywe⁻¹x∈R⁺.

Proof. By (16) and (31), the condition (33) is equivalent to a b

t(a+b) tb+d

=AtWt=fWtB⁻¹=

a −a−b

−tb tb+d

,

and this condition holds if and only if t satisfies a(t) + 2b(t) = 0. The non- constant polynomiala+ 2b has a roots. Hence

r=wywe⁻¹x=x⁻¹(xw)(wye ⁻¹)⁻¹x∈R⁺. X In knot theory the condition (33) is replaced by

Hs(xw) =Hs(wy), r=wy⁻¹w⁻¹x∈R⁺. (34) which holds if and only ifa(s) = 0, see e.g. [12, p. 141].

By Proposition 3 and conjugation, we see that (34) implies

Hs(xw) =e Hs(wy),e re=wye we⁻¹x⁻¹∈R⁺. (35) Condition (34) induces a homomorphism h_r,s from hx, y; ri into PSL(2,C);

see (22). Now (35) says that it automatically induces a homomorphism from hx, y; r,eri. In the case of a 2-bridge knot it is known [2, 1] that there exists a faithful discrete SL(2,C)-representation of a 2-bridge knot of type (p, q) with q6≡ ±1, so thatxw=wy impliesxwe=wy. But this is not true in general, seee Example 5.

The situation is different for (33) because re=wye ⁻¹w⁻¹x⁻¹ is conjugate to r⁻¹ so that rand erhave the same normal closure; compare Proposition 4.

Hence they induce the same group.

For 2-bridge knots the group G = hx, y; xw = wyi and its peripheral subgroup are important concepts to distinguish equivalence classes of knots.

This subgroup is generated by a meridian, sayy, and the longitude l=w⁻¹we

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C

(see [17, p. 206]). We omitted Riley’s factor y^2σ. It is easy to check that (r= 1,er= 1) is equivalent to (r= 1, ly=yl).

Now we study the longitude l =w⁻¹we in a more general context. We do not assume that the wordwcomes from knot theory and we do not assume the consequence (31) of the Wirtinger condition. For w∈F we obtain from (32) and (16) that

Ht(l) =W_t⁻¹Wft=

ad+bc −2bd

−2ac ad+bc

. (36)

We note thatl=w⁻¹weis a palindrome.

Theorem 8. Let wsatisfy (7)with |jν|=|kν|= 1 and leta(s) = 0. Then

Ls:=Hs(l) =

−1 −2b(s)d(s)

0 −1

. (37)

Ifa=c+dthenb(s)d(s) = 1. If the polynomialais irreducible and ifa6=c+d thenb(s)d(s)∈/ QandLs andB generate a free abelian group of rank 2.

Formulas similar to (37) follow from (36) if b(s) = 0, c(s) = 0 or d(s) = 0.

The 2-bridge knots of type (2n+ 1,1) have the Wirtinger word w= (yx)ⁿ. It follows from Theorem 2 thata=c+dso that b(s)d(s) = 1. See Examples 6, 7 and 8.

Proof. Sincead−bc= 1 we can writead+bc=−1 + 2ad. Hence (37) follows from (36). If a=c+d thenc(s) =−d(s) and thusb(s)d(s) =−b(s)c(s) = 1 becausea(s) = 0.

Now let q := b(s)d(s) and suppose that q ∈ Q. Since a(s) = 0, it follows from Proposition 1 thatsis an algebraic integer so thatq∈Z. It follows from (14) and (15) that

f(t) :=qc(t) +d(t) =qλtⁿ+· · · , λ=±1. (38) Sincea(s) = 0 impliesb(s)c(s) =−1 we have

b(s)f(s) =qb(s)c(s) +b(s)d(s) =−q+q= 0

so that f(s) = 0. Hence the irreducible polynomial a(t) divides f(t). Since a(t) = λtⁿ +· · · with the same λ, we conclude from (38) that q = 1 and thereforea=c+d. Ifa6=c+dwe therefore have−2b(s)d(s)∈/ Qso that L_s

andB are free abelian generators. X

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5. Examples

The words ofF in the following examples are generated by

z0=yx, z1=yx⁻¹, z2=y⁻¹x, z3=y⁻¹x⁻¹.

All polynomials will be written as the product of irreducible factors inZ[t]. The factorization used the program Kash3 developed by M. Pohst and his group, www.math.tu-berlin.de/~kant.

Example 1. The following two words

r1=z₀²z1z₃², σx(r1) =−1, σy(r1) = 1, r2=z₀¹⁰, σx(r2) = 10, σy(r2) = 10

are relators with the same minimal polynomial 1 + 3t+t². The normal closures satisfyr1 ∈/ N(r2) and r2 ∈/ N(r1) because the exponent sums do not satisfy (21). It follows that no homomorphism hr,s with s = −1/2±√

5/2 can be injective, see (22).

Example 2. Let u = z²₀z2 and r = uue = z₀²z2z₃²z1. The polynomials for u area(t) = 1 + 4t−t²−t³ andd(t) = 1−t−t². Now part (a) of the proof of Theorem 5 shows that Ht(r) = I if and only if a(s)−d(s) = s(5−s²) = 0.

Hencer∈R⁺.

Example 3. Letr=z²₀. Then Ht(r) =

1 + 3t+t² 2 +t 2t+t² 1 +t

6=

1 0 0 1

, for t∈C so thatr /∈R⁺. But (23) shows thatr∈R⁻.

Example 4. The following words belong toR⁺∩R⁻. Their minimal polynomials

u=z₀⁵: p⁺(t) = 5 + 5t+t², p⁻(t) = 1 + 3t+t², u=z₀⁶: p⁺(t) = 3 + 4t+t², p⁻(t) = 2 + 9t+ 6t²+t³ have the smallest degrees possible by Proposition 6.

Example 5. The Wirtinger wordw=z₀z₁z₁z₀does not come from a 2-bridge knot. Its relator is r =wy⁻¹w⁻¹xwith σ_x(r) = 1, σ_y(r) = −1. Furthermore er = wye we⁻¹x⁻¹ with σ_x(er) = −1, σ_y(r) = 1 so thate er is not conjugate to r.

Nowr⁻¹containsy⁻¹x⁻¹y⁻¹x⁻¹whereas no conjugate ofercontains this word.

Henceris not conjugate tor⁻¹either. Thus it follows from Proposition 4 that, withHs(r) =I, the homomorphism

hx, y; r,eri=hx, y; xw=wy, xwe=wyi →e SL(2,C) is not injective.

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C

Example 6. The Wirtinger word of the 2-bridge knot of type (9,1) isw=z₀⁴ and

a(t) = (1 +t)(1 + 9t+ 6t²+t³)

is reducible. It satisfiesa=c+dand thusb(s)d(s) = 1 by Theorem 8.

Example 7. Letw=z0z3z2z0. This is not a Wirtinger word because c6=tb.

It satisfies

a(t) = (1 +t)p(t), p(t) =−1 +t+ 2t²+t³, b(t)d(t)−1 = (1 +t)(−1−t−t²+ 2t³+ 2t⁴+t⁵), b(t)d(t) + 1 = (−1 +t+t²+t³)p(t).

Henceb(−1)d(−1) = 1 whereasb(s)d(s) =−1 ifp(s) = 0. Thus, in Theorem 8, the assumption thata(t) is irreducible can not be omitted.

Example 8. The 2-bridge knot of type (5,3) hasw=z1z2anda(t) = 1−t+t². This givess= (1 +i√

3)/2 andb(s)d(s) =±i√ 3.

Acknowledgment. We would like to thank the referees for the careful read- ing and helpful comments.

References

[1] W. Brumfield and H. M. Hilden, SL(2) Representations of Finitely Pre- sented Groups, Contemporary Math (Providence, United States), vol. 187, AMS, 1995.

[2] G. H. Burde and H. Zieschang, Knots, Walter de Gruyter, 1985.

[3] B. Fine, F. Levin, and G. Rosenberger,Faithful Complex Representations of one Relator Groups, N. Z. J. Math.26(1997), 45–52.

[4] J. Gilman,The Structure of Two-Parabolic Space: Parabolic Dust and It- eration, Geom. Dedicata 131(2008), 27–48.

[5] J. Gilman and L. Keen,Discreteness Criteria and the Hyperbolic Geometry of Palindromes, Conform. Geom. Dyn13(2009), 76–90.

[6] J. Gilman and P. Waterman, Classical T-Schottky Groups, J. Analyse Math.98(2006), 1–42.

[7] C. Gordon and J. Luecke, Knots are Determined by their Complements, Bull. Amer. Math. Soc.20(1989), 83–87.

[8] H. M. Hilden, D. M. Tejada, and M. M. Toro, Tunnel Number one Knots Have Palindrome Presentations, J. Knot Th. Ramif. 11 (2002), no. 5, 815–831.

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[9] A. Kawauchi, A Survey of Knot Theory, Birkh¨auser Verlag, 1996.

[10] R. C. Lyndon and P. E. Schupp,Combinatorial Group Theory, Springer, Berlin, Germany, 1977.

[11] I. D. Macdonald,The Theory of Groups, Clarendon Press, Oxford, 1968.

[12] C. Maclachlan and A. W. Reid,The Arithmetic of Hyperbolic 3-Manifolds, Springer, New York, United States, 2003.

[13] W. Magnus, A. Karrass, and D. Solitar,Combinatorial Group Theory, 2nd revised edition ed., Dover Publ., New York, United States, 1966.

[14] D. Mej´ıa and Ch. Pommerenke,Analytic Families of Homomorphisms into PSL(2,C), Comput. Meth. Funct. Th.10(2010), 81–96.

[15] T. Ohtsuki, R. Riley, and M. Sakuma, Epimorphisms between 2-Bridge Link Groups, Geom. Topol. Monogr.14(2008), 417–450.

[16] R. Riley,Parabolic Representations of Knot Groups I, Proc. London Math.

Soc.3(1972), no. 24, 217–242.

[17] , Nonabelian Representations of 2-Bridge Knot Groups, Quart. J.

Math. Oxford2(1984), no. 35, 191–208.

[18] ,Holomorphically Parametrized Families of Subgroups of SL(2,C), Mathematika32(1985), 248–264.

[19] ,Algebra for Heckoid Groups, Trans. Amer. Math. Soc.32(1994), no. 1, 389–409.

[20] H. Schubert,Knoten Mit Zwei Br¨ucken, Math. Z.65(1956), 133–170.

(Recibido en septiembre de 2010. Aceptado en febrero de 2011)

Institut f¨ur Mathematik MA 8-1 Technische Universit¨at Berlin D-10623, Berlin, Germany e-mail: [email protected]

Escuela de Matem´aticas Universidad Nacional de Colombia Sede Medell´ın Cra 59A #63-20, bloque 43 Medell´ın, Colombia e-mail:[email protected]