monodromia de curvas algébricas planas · lê dung˜ tráng em [lê1], deu uma resposta...

Monodromia de curvas algebricas planas

Silas Fantin

SERVICO DE POS-GRADUACAO DO ICMC-USP

Data de Deposito: 12/07/2007

Assinatura:

Monodromia de curvas algebricas planas

Silas Fantin

Orientador: Prof. Dr. Abramo Hefez

Co-orientador: Prof. Dr. Daniel Levcovitz

Tese apresentada ao Instituto de Ciencias Matematicas e deComputacao - ICMC-USP, como parte dos requisitos necessariosa obtencao do tıtulo de Doutor em Ciencias - Area: Matematica.

USP - Sao Carlos

Julho/2007

Agradecimentos

A Deus, por estar sempre presente em minha vida.

Ao professor Abramo Hefez, por ter me aceito como seu aluno de doutorado, e

pela sugestao do tema da tese, permitindo-me trabalhar com liberdade e tranqui-

lidade.

A Andrea Gomes Guimaraes, pela sua ajuda e pelo seu entusiasmo no decorrer

do doutorado, e ao Billy ( Carlos ) pela amizade, e pelos diagramas feitos na tese.

A Ana Paula e Laura, na secretaria de pos-graduacao do ICMC-USP, e a Ma-

riana e Lea, na secretaria da pos-graduacao da UFF, pela ajuda prestada no de-

correr deste perıodo.

Aos meus pais e irmaos, pelo apoio e incentivo constantes. A minha mae, que

mais desejava ver este trabalho concluido, presto minha homenagem postuma.

A minha esposa Regina Freitas, pela paciencia, apoio, estımulo e compreensao

neste perıodo de doutorado.

Ao departamento de estruturas matematicas da UERJ, e a CAPES, pelo su-

porte financeiro, que permitiu a realizacao deste trabalho.

Aos professores que aceitaram participar da banca examinadora.

Aos demais funcionarios, professores e alunos do ICMC-USP, neste momento

anonimos, o meu (nao menor) muito obrigado.

Resumo

Em 1968, J. Milnor introduziu a monodromia local de Picard-Lefschetz de uma

hipersuperfıcie complexa com singularidade isolada. Em seguida, E. Brieskorn

perguntou se esta monodromia e sempre finita. Em 1972, Le Dung Trang provou

que a resposta e positiva no caso de germes de curvas planas analıticas irredutıveis.

Na epoca, ja eram conhecidos exemplos de curvas planas com dois ramos e mon-

odromia finita. Em 1973, N. A’Campo produziu o primeiro exemplo de germe

de curva plana com dois ramos e monodromia infinita. Portanto, a questao mais

simples, e ainda em aberto, que se coloca neste contexto, e a determinacao da

finitude da monodromia para germes de curvas planas com dois ramos. O presente

trabalho, consiste em determinar, em varias situacoes, o polinomio mınimo da

monodromia de germes de curvas analıticas planas com dois ramos, cujos generos

sao menores ou iguais a dois, o que permite decidir a sua finitude.

Abstract

In 1968, J. Milnor introduced the Picard-Lefschetz monodromy of a complex

hypersurface with an isolated singularity. Subsequently, E. Brieskorn asked if this

monodromy is always finite. In 1972, Le Dung Trang proved that the answer is

positive in the case of irreducible analytic germs of plane curves. At this time,

examples of plane curves with two branches and finite monodromy were known. In

1973, N. A’Campo produced the first example of a germ of plane curve with two

branches and infinite monodromy. Therefore, the simplest and still open problem

in this context is to determine whether the monodromy of a plane curve with two

branches is finite or infinite. The present work consists in determining, in several

situations, the minimal polynomial of the monodromy for germs of plane analytic

curves with two branches, whose genera are less or equal than two, wich allows us

to decide its finiteness.

Indice

Introducao 1

Capıtulo 1: Preliminares

1.1 A monodromia local ........................................................................... 3

1.2 Polinomios de Alexander e finitude da monodromia .......................... 5

1.3 Apresentacao do grupo de um link algebrico ..................................... 8

Capıtulo 2: Ramos de genero 1

2.1 Ramos com pares de Puiseux distintos ............................................... 11

2.2 Ramos com pares de Puiseux iguais .................................................. 31

Capıtulo 3: Ramos de genero 2

3.1 Ramos com pares de Puiseux distintos .............................................. 44

3.2 Ramos com pares de Puiseux iguais .................................................. 85

3.3 Epılogo ............................................................................................... 99

Apendice

A Calculos para ramos de genero 1 ......................................................... 100

B Calculos para ramos de genero 2 ....................................................... 110

Referencias bibliograficas 136

Introducao

Seja f : ( ICn+1, 0 ) → ( IC, 0 ) um germe de funcao analıtica com f(0) = 0 tal que

C = f−1(0) tem uma singularidade isolada em 0. A geometria local de C proximo

de 0 e completamente descrita pelo link algebrico L := C∩S2n+1ε da singularidade,

onde S2n+1ε e a esfera de raio ε em IR2(n+1) (' ICn+1 ).

Estamos interessados no seguinte problema: Dado o germe f, como descrito

acima, determinar se a sua monodromia algebrica e de ordem finita ou infinita.

O problema torna-se mais facil quando n = 1, uma vez que, nesta situacao, a

geometria foi completamente descrita nos trabalhos de Brauner [Bra], Kahler [Ka]

e Reeve [Re] no inıcio do Seculo 20. Neste caso, chamamos C de curva plana e

o link algebrico L tem r componentes, correspondendo aos ramos de C na ori-

gem. Cada componente de L e um no torico iterado, e tanto as iteracoes em cada

componente do link, quanto os entrelacamentos entre as varias componentes, sao

completamente especificados pelo desenvolvimento de Puiseux de cada ramo.

No caso em que n = 1 e o germe f e analiticamente irredutıvel, ou seja, r = 1,

a sua monodromia e sempre finita ( cf. [Le1], [A’c1] ou [SW1] ). Se r = 2, existem

exemplos nos quais a monodromia e finita ou infinita (cf. [A’c1] e [Wo] ). Este

trabalho se propoe de estudar, para n = 1 e r = 2, em varias situacoes, o proble-

ma da finitude da monodromia, com enfoque no calculo do segundo polinomio de

Alexander associado ao link algebrico da singularidade.

Apresentamos a seguir uma descricao sucinta dos capıtulos que compoem este

trabalho.

No primeiro capıtulo, encontram-se as ferramentas gerais para o desenvolvi-

mento do trabalho. Fazemos tambem um apanhado de alguns resultados que se

encontram na literatura, para contextualizar o nosso problema.

No segundo capıtulo, determinamos o segundo polinomio de Alexander, no caso

de curvas com dois ramos de genero 1, isto e, cada ramo tem um unico par de Pui-

seux, para com isso decidir sobre a finitude da monodromia.

1

Finalmente, no terceiro capıtulo, abordamos a mesma questao no caso de ramos

de genero 2, isto e, cada ramo tem dois pares de Puiseux, onde determinamos o

segundo polinomio de Alexander, ou damos explicitamente a matriz de Alexander.

Nos apendices apresentamos os desenvolvimentos algebricos omitidos no decor-

rer do texto, para tornar a leitura mais amena e manter o foco no problema.

2

Capıtulo 1 - Preliminares

O objetivo deste capıtulo e introduzir as notacoes, as definicoes, e os fatos

basicos que serao utilizados nos capıtulos seguintes.

1.1 A monodromia local

Seja f : (IC2, 0) → (IC, 0) um germe de funcao analıtica com um ponto crıtico

isolado na origem. Denotaremos por Bε a bola fechada de IC2 centrada na origem

e de raio ε > 0. Consideremos Sε = ∂Bε, Cε = Bε ∩ f−1(0) e Lε = Cε ∩ Sε.

Se ε e suficientemente pequeno, sabe-se que Lε e uma subvariedade C∞ de Sε e

que o conhecimento da topologia do par (Sε, Lε) e equivalente ao conhecimento da

topologia do par (Bε, Cε) (cf. [Mi]). Portanto, para compreender a topologia do

par (Bε, Cε), bastara compreender a topologia de Lε e de Sε \Lε. A este proposito

tem-se o seguinte resultado:

Teorema da Fibracao de Milnor Seja f : ( IC2, 0) → ( IC, 0) um germe de

funcao analıtica na origem de IC2 com ponto crıtico isolado. Entao existe ε0 > 0

tal que, para todo 0 < ε ≤ ε0, a aplicacao

ϕε : Sε \ Lε −→ S1

z 7−→ f(z)

| f(z) |e a projecao de uma fibracao C∞ localmente trivial, com fibra Fθ = ϕ−1

ε (e2πiθ)

uma variedade de dimensao real 2 do tipo de homotopia de um bouquet de µ

cırculos colados em um ponto, onde µ e a codimensao sobre IC do ideal jacobiano

J (f) = < ∂f∂z1

, ∂f∂z2

> no anel das series de potencias convergentes IC{z1, z2}, sendo

chamado de numero de Milnor de f .

Este resultado implica que a homologia da fibra de Milnor esta concentrada

na dimensao 1 e que a mesma e um grupo abeliano livre com µ geradores; isto e,

3

H1 (Fθ) ∼= ZZµ. Como para ε suficientemente pequeno, nem o mergulho de Lε em

Sε, nem a fibracao ϕε dependem de ε, denotaremos Sε por S, Lε por L e ϕε por ϕ,

quando este for o caso.

Como ϕ : S \ L → S1 e uma fibracao, podemos levantar o arco e2πit para

0 ≤ t ≤ s ≤ 1 e construir assim uma famılia contınua de homeomorfismos hs :

Fθ → Fθ+2πs tal que h0 = IdFθ. O homeomorfismo h = h1 : Fθ → Fθ e chamado de

monodromia geometrica de C na origem. A monodromia geometrica h induz

no grupo de homologia da fibra Fθ, com coeficiente num grupo G (no nosso caso

ZZ ou IC), o automorfismo

T = (h)∗ : H1 (Fθ, G) −→ H1 (Fθ, G),

chamado de operador de monodromia algebrica local, ou de operador de mo-

nodromia de Picard-Lefschetz.

A investigacao do operador de monodromia algebrica local, quando G = IC, teve

inıcio com a prova por E. Brieskorn em [Br] do famoso Teorema de Monodromia

para hipersuperfıcies com singularidades isoladas, que enunciaremos a seguir no

caso de curvas.

Teorema de Monodromia Todos os autovalores de T sao raızes da unidade,

os blocos de Jordan da forma canonica de T tem no maximo ordem 2, e os blocos

de Jordan com autovalor 1 tem ordem 1.

No mesmo trabalho, Brieskorn fez a seguinte pergunta: A monodromia algebrica

local T , para uma hipersuperficie com singularidade isolada, tem sempre ordem

finita?

A pergunta de Brieskorn equivale a saber se existe um numero natural d tal

que T d = Id. Portanto, em vista do Teorema de Monodromia, a finitude da

monodromia algebrica local e equivalente ao fato do operador T ser diagonalizavel,

ou ainda, equivalente ao fato do polinomio mınimo de T nao ter raızes multiplas.

Logo, uma maneira de decidir tal questao, e determinar o polinomio mınimo de T .

Le Dung Trang em [Le1], deu uma resposta afirmativa a questao no caso de

curvas planas analiticamente irredutıveis, calculando o polinomio mınimo de T , e

mostrando que todas as suas raızes sao simples.

A’Campo em [A’c1], exibiu o primeiro exemplo de uma curva plana com dois

ramos, cuja monodromia e infinita. Portanto, a questao mais simples que se coloca

e a determinacao da finitude da monodromia para curvas planas com dois ramos.

4

Esta e a questao a qual e dedicado o presente trabalho. Mais precisamente, de-

terminaremos em algumas situacoes, o polinomio mınimo da monodromia para

germes de curvas analıticas planas com dois ramos. E possıvel tratar, com essas

mesmas tecnicas, o caso de curvas com um numero maior de ramos. Mas nao o

faremos, dado a complexidade dos calculos envolvidos.

1.2 Polinomios de Alexander e finitude da monodromia.

Sabe-se que, no caso de uma curva plana C, o polinomio caracterıstico da

monodromia e o primeiro polinomio de Alexander ∆1 (t) de C, e que o polinomio

mınimo λ (t), e o quociente do primeiro polinomio de Alexander ∆1 (t) pelo segundo

polinomio de Alexander ∆2 (t) de C. (cf. [SW2], pagina 129).

O primeiro polinomio de Alexander de uma curva e calculado na literatura,

de varios modos (cf. [Bu2], [FC], [SW2], [CDG1]), inclusive para um numero

arbitrario de ramos. Entretanto, nao ha registro do calculo do polinomio ∆2 (t)

para curvas redutıveis (no caso irredutıvel, Le Dung Trang mostra que ∆2 (t) = 1).

O nosso trabalho e centrado no calculo de ∆2 (t) quando C tem dois ramos,

com o objetivo de encontrar o polinomio minimal da monodromia e consequen-

temente decidir sobre a sua finitude. Os resultados conhecidos mais abrangentes

nesta direcao, se encontram nos artigos [SW2] e [Wo] que exploram apenas proprie-

dades de ∆1 (t), nao sendo por isso completos, mas fornecendo algumas condicoes

necessarias e outras suficientes, para a finitude da monodromia. O resultado mais

significativo dos trabalhos acima citados, e baseado na seguinte observacao:

Observacao Se T tem ordem finita e δ e uma raiz pm-esima da unidade para

algum primo p e algum inteiro positivo m, entao a multiplicidade de δ como raiz

de ∆1 (t) e no maximo r − 1, onde r e o numero de ramos da curva.

A estrategia que adotaremos neste trabalho e semelhante aquela utilizada no ar-

tigo [Le1]. O calculo do polinomio de Alexander ∆2 (t) sera feito aqui pelo metodo

desenvolvido por R. Fox em [F2] e [F5], por meio de uma matriz cujas entradas

sao polinomios em ZZ [ t ], chamada de matriz de Alexander, e que descreveremos a

seguir.

Seja π1 (X) = {x1, . . . , xn; r1, . . . , rm} uma apresentacao com n geradores e

m relacoes com n > m, do grupo fundamental de X = S \ L. Seja Fn o grupo

livre gerado por x1, . . . , xn. Considere Dj = ∂∂xj

: ZZ [ Fn ] → ZZ [ Fn ] o operador de

5

derivacao no anel de grupo ZZ [ Fn ], possuindo as seguintes propriedades:

Se ξ1, ξ2 ∈ ZZ [ Fn ] e g, g1, . . . , gk ∈ Fn, entao

(1) Dj (ξ1 + ξ2) = Dj (ξ1) + Dj (ξ2).

(2) Dj (g1 . . . gk) = Dj g1 + g1 Dj g2 + g1 g2 Dj g3 + · · · + g1 . . . gk−1 D gk.

(3) Dj g−1 = − g−1 Dj g,

(4) Dj gn = gn−1g −1

Dj g,

(5) Dj g−n = − g−n gn−1g −1

Dj g.

A matriz de Alexander de π1 (X) e a matriz Φ

(∂ri

∂xj

), onde a aplicacao

Φ : ZZ [ π1 (X) ] −→ ZZ [ π1 (S1) ] ' ZZ [ t, t−1 ]

e induzida pela fibracao ϕ de Milnor.

O primeiro polinomio de Alexander ∆1 (t) de π1 (X) e definido como sendo o

maximo divisor comum dos determinantes das submatrizes de ordem m da matriz

de Alexander, enquanto que o segundo polinomio de Alexander ∆2 (t) e o maximo

divisor comum dos determinantes das submatrizes de ordem (m− 1).

Sendo a classe de conjugacao da monodromia de C um invariante topologico

(cf. [Mi]), e sendo os pares de Puiseux de cada ramo de C, juntamente com o con-

tato entre os seus respectivos desenvolvimentos de Puiseux, invariantes topologicos

completos de C, o nosso problema consistira em determinar a matriz de Alexander

de um link algebrico de dois ramos, dadas as expansoes em series de Puiseux. Com

isso, poderemos responder se o operador de monodromia associada a este link, tem

ordem finita ou infinita.

A medida que o numero de pares de Puiseux associados aos ramos aumenta,

a quantidade de menores envolvidos nos calculos aumenta, tornando mais difıcil

descrever explicitamente o polinomio mınimo da monodromia.

O polinomio ∆1 (t) para um link algebrico com duas componentes (ou mais), foi

calculado por Burau na decada de 30 em [Bu2] usando outras tecnicas de topologia

algebrica. A seguir, damos explicitamente a expressao de ∆1 (t) como se encontra

em [SW2]. Seja C uma curva com dois ramos de equacao f = f1 f2 e considere o

desenvolvimento de Puiseux de cada ramo, que podemos supor finitos:

(? ) y1 =

s1∑i =1

a1 εixεi e y2 =

s2∑j =1

a2 δjxδj

6

onde para i = 1, . . . , s1 e j = 1, . . . , s2, temos que

εi =m1i

n11 . . . n1i

e δj =m2j

n21 . . . n2j

As informacoes topologicas de C podem ser extraıdas das representacoes para-

metricas acima. A determinacao dos pares de Puiseux de cada ramo e padrao, e

o seu numero e chamado de genero do ramo. O contato entre os dois ramos se

mede pelo grau i de coincidencia dos dois desenvolvimentos:

i = max { j ; a1,εk= a2,δk

em1,k

n1,k

=m2,k

n2,k

para todo k ≤ j }.

Sem perda de generalidade, assumiremos quem1,i+1

n1,i+1≥ m2,i+1

n2,i+1.

Polinomio Caracterıstico da Monodromia ([SW2], Th. 7.6) O polinomio ca-

racterıstico da monodromia para um link algebrico com dois ramos e

∆1 (t) = (t− 1)i∏

j =1

(tw1,j .e0,j − 1

tw1,j .e0,j+1 − 1

te0,j+1 − 1

te0,j − 1

){

tw2,i+1.e0,i+1 − 1

te0,i+1 − 1

2∏k = 1

sk∏j = i+k

tnk,j .ek,j − 1

tek,j − 1

}onde wi e definido recursivamente para cada 1 ≤ j ≤ sk como segue{

wk1 = mk1

wkj = mkj −mk,j−1 nkj + wk,j−1 nk,j−1 nkj, 2 ≤ j ≤ sk

onde

eq,j =

b1,j,s1 + b2,j,s2 q = 0w2,i+1 . b2,i+2,s2 . b1,i+1,j−1 + w1,j . b1,j+1,s1 q = 1w2,i+1 . b1,i+1,s1 . b2,i+2,j−1 + w2,j . b2,j+1,s2 q = 2

com

bk,l,m =m∏

j = l

nk,j e bk,l,m = 1 se l > m.

Para calcular o polinomio minimal λ (t) da monodromia, so nos resta calcular

∆2 (t), o que faremos usando a matriz de Alexander, para cuja determinacao se

faz necessario uma apresentacao do grupo π1 (X) que exibiremos a seguir.

7

1.3 Apresentacao do grupo de um link algebrico

A apresentacao que descreveremos abaixo e devida a O. Neto e P. C. Silva em

[NS]. Seja C como na secao anterior com parametrizacoes de seus ramos dadas por

(?). Seja S = {(1, 0), (d, ε); d ∈ {1, 2} e a1,ε 6= 0 ou a2,ε 6= 0} ∪ {(d′, ε); d, d′ ∈{1, 2} d′ 6= d e a1,γ = a2,γ para 0 ≤ γ < ε}. Associamos a C uma arvore Ξ,

cujos vertices sao classes de equivalencia [d, ε] em S, segundo a seguinte relacao de

equivalencia:

(1, ε) ∼ (2, ε) ⇔ a1,γ = a2,γ para 0 ≤ γ ≤ ε

A classe φ = [1, 0] e chamada de raiz de Ξ. Diremos que w = [d, δ] e um filho

de z = [d, ε], escrevendo w > z, se δ > ε e e minimal com esta propriedade.

Diremos que [d, ε] e uma haste se ad,ε = 0. Chamaremos de terminais os vertices

que nao possuem filhos. A arvore e formada ligando cada vertice aos seus filhos.

Seja z = [d, η] um vertice nao terminal de Ξ, onde η = mdk

nd1...ndk. Ao vertice z

sao associados dois numeros inteiros positivos νz e µz definidos como segue:

Como z nao e terminal, existe um filho w′ = [d′, ρ] de z que nao e haste.

Logo, ρ =md′k+1

nd′1...nd′k+1. Pomos νz = nd′k+1 e µz = ld′k+1 onde ld′1 = md′1 e

ld′j = md′j − md′j−1nd′j + md′j−1nd′j−1nd′j para j ≥ 2. Os inteiros νz e µz sendo

coprimos, existem inteiros rz e sz tais que rz µz = sz νz + 1.

Grupo do Link: ([NS], Th. 1.2) O grupo fundamental π1 (X), e apresentado

pelos geradores αz, βz com z percorrendo os vertices da arvore Ξ e pelas relacoes

βφ = 1, [αz, βz] = 1 para todo z, alem das relacoes abaixo, separadas em 4 tipos:

(R1) ανz µzw βw = αµz

z βνzz se w > z e w nao e uma haste

(R2) αµzw βνz

w = αµzz βνz

z se w > z e w e uma haste

(R3) (γz αszz βrz

z )νz = (αµzz βνz

z )rz se z e nao terminal sem filho haste

(R4) γz αszz βrz

z = αszz0

βrzz0

se z e nao terminal com filho haste z0

onde γz e o produto dos αz′ , com z′ um filho nao haste de z.

8

Exemplo 1: Seja f uma curva plana com dois ramos cujos desenvolvimentos de

Puiseux, sao respectivamentes:{y1 = x3/2 + x7/4

y2 = x3/2 + x5/2

Representaremos os vertices de Ξ que correspondem as hastes pelos cırculos brancos

e os restantes pelos cırculos negros, conforme a Figura 1.

Figura 1:

Temos que {φ, 1, 10} sao os vertices nao terminais da arvore, onde:εφ = 0, ε0 = ε1 = 3/2, ε11 = ε10 = 7/4, ε101 = 5/2,

(µφ, νφ) = (3, 2), (µ1, ν1) = (13, 2), (µ1,0, ν1,0) = (8, 1)

(rφ, sφ) = (1, 1), (r1, s1) = (1, 6), (r10, s10) = (1, 7)

Logo, o grupo do link π1 (X) e apresentado por αz, βz com z percorrendo o conjunto

V = {φ, 1, 11, 10, 101}, verificando as seguintes relacoes βφ = 1 e [αz, βz] = 1 para

todo z e as relacoes associadas a cada vertice nao terminal da arvore Ξ acimaφ : α3

φ β2φ = α6

1 β1 (α1 αφ βφ)2 = (α3

φ β2φ)

1

1 : α131 β2

1 = α2611 β11 = α13

10 β210 α11 α6

1 β1 = α610 β10

10 : α810 β10 = α8

101 β101 α101 α710 β10 = α8

10 β10

9

Capıtulo 2 - Ramos de genero 1

Nos proximos dois capıtulos, descreveremos explicitamente a matriz de Alexan-

der em funcao dos desenvolvimentos de series de Puiseux dos dois ramos da curva

C, para com ela calcular efetivamente o segundo polinomio de Alexander, e com

isto, encontrar o polinomio mınimo da monodromia.

Convem ressaltar, que o metodo desenvolvido pelo R. Fox, para o calculo dos

polinomios de Alexander, conforme descrito na Secao 1.2, necessita de uma apre-

sentacao do grupo do link com mais geradores do que relacoes. Como utilizamos

neste trabalho a apresentacao do grupo do link dada por O. Neto e P. C. Silva, e

em tal apresentacao nem sempre temos mais geradores do que relacoes, seremos

conduzidos a impor condicoes sobre os pares de Puiseux dados, para estarmos na

situacao desejada.

Alem disso, a apresentacao do grupo do link de O. Neto e P. C. Silva e descrita

por arvores cujos tipos variam dentro de um conjunto finito de possibilidades que

dependem dos dados topologicos da curva. Isso naturalmente nos conduz ao des-

dobramento da analise em varios casos, correspondendo aos tipos de arvores que

ocorrem.

Neste capıtulo, estudaremos especificamente a monodromia de um germe de

curva plana com dois ramos, ambos de genero 1.

O caso de dois ramos se diferencia do caso irredutıvel pela variedade de arvores

que se apresentam. No caso irredutıvel, qualquer que seja o genero da curva, ha

apenas um tipo de arvore, o que facilita o processo indutivo do calculo da matriz

de Alexander, como realizado em [Le1].

Figura 2: Arvore no caso irredutıvel de genero 3

No caso de germes de curvas planas com dois ramos, a quantidade de tipos de

arvores cresce muito a medida que o genero das curvas aumenta, o que torna o

10

processo indutivo mais complexo. Na situacao em foco, temos os seguintes tipos

de arvores:

Figura 3: Arvores para ramos de genero 1 com pares de Puiseux distintos

Figura 4: Arvores para ramos de genero 1 com pares de Puiseux iguais

2.1 Ramos com pares de Puiseux distintos

Caso 1: Nessa situacao, os ramos da nossa curva, podem ser representados pelas

seguintes expansoes de Puiseux: {y1 = x

m11n11

y2 = xm21n21

onde (mi1, ni1) e o unico par de Puiseux do ramo fi para i = 1, 2. Isto conduz

especificamente a seguinte arvore, dada na Figura 5.

11

Figura 5:

Com as notacoes da Secao 1.3,{µφ = m21, νφ = n21 e m21rφ = n21sφ + 1µ0 = m11, ν0 = n11 e m11r0 = n11s0 + 1

Portanto, o grupo do link π1 (X) e gerado por αz, βz com z percorrendo o

conjunto dos vertices V = {φ, 0, 01, 1} da arvore Ξ, satisfazendo as relacoes que

exibiremos a seguir1.

Relacoes de (φ):

(R1) 1 > φ e 1 nao e haste ⇒ ανφµφ

1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m21 n211 αm21

φ

(R2) 0 > φ e 0 e uma haste ⇒ αµφ

0 βνφ

0 = αµφ

φ βνφ

φ

⇒ αm210 βn21

0 = αm21φ

(R4) φ e nao terminal com haste 0 ⇒ α1 αsφ

φ βrφ

φ = αsφ

0 βrφ

0

⇒ α1 = αsφ

0 βrφ

0 α−sφ

φ

Relacoes de (0):

(R1) 01 > 0 e 01 nao e haste ⇒ αν0µ0

01 β01 = αµ0

0 βν00

⇒ β01 = α−m11 n1101 αm11

0 βn110

1Os desenvolvimentos algebricos omitidos encontram-se no apendice A, segundo as referenciasespecificadas.

12

(R3) 0 e nao terminal sem haste ⇒ (α01 αs00 βr0

0 )ν0 = (αµ0

0 βν00 )r0

⇒ (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

Relacoes de comutacao:

[αφ, βφ] = 1 ⇔ αφ βφ = βφ αφ ⇔ αφ = αφ

[α0, β0] = 1 ⇒ [α1, β1] = 1 veja (A.1)

[α1, β1] = 1

[α01, β01] = 1 ⇔ [ αm110 βn11

0 , α01 ] = 1 veja (A.2)

Portanto, π1 (X) pode ser apresentado com geradores α0, β0, αφ, α01, sujeitos as

relacoes:

(1) [α0, β0] = 1

(2) αm210 βn21

0 = αm21φ

(3) (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

(4) [ αm110 βn11

0 , α01 ] = 1

Como temos interesse em grupos π1 (X) que tem uma apresentacao na qual

existem mais geradores do que relacoes (veja Secao 1.2), fazemos a hipotese restri-

tiva m11 = λ n11 + 1, para algum λ ∈ ZZ, o que implica que r0 = 1. Adotaremos

como novos geradores:{x1 = α0,x2 = β0,

{x3 = α01 αs0

0 βr00 ,

x4 = αφ.

Observe que a relacao (4) e equivalente a relacao [αm110 βn11

0 , x3] = 1, e que

esta e trivializada devido a relacao (3). Isto conduz a nova apresentacao:

π1 (X) = { x1, x2, x3, x4; r1, r2, r3 }, onder1 = x1 x2 x−1

1 x−12

r2 = xm111 xn11

2 x−n113

r3 = xm211 xn21

2 x−m214

Para determinar a matriz de Alexander associada a esta apresentacao, devemos

determinar Φ (xi) ∈ ZZ [ t, t−1 ] para i = 1, 2, 3, 4. onde Φ e a aplicacao induzida

13

pela fibracao de Milnor. Como ainda nao sabemos a priori os valores de Φ (xi),

temporariamente escrevemos:

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7→ ta

x2 7→ tb

x3 7→ tc

x4 7→ td

Note que as relacoes ri, para i = 2, 3, fornecem relacoes entre a, b, c, d, a saber:

(∗){

n11 c = m11 a + n11 bm21 d = m21 a + n21 b

Para determinar a matriz de Alexander, faremos o seguinte calculo de Φ

(∂ri

∂xj

).

Como r1 = x1 x2 x−11 x−1

2 , temos que

∂r1

∂x1

= 1−x1 x2 x−11 ⇒ Φ

(∂r1

∂x1

)= 1−Φ (x2) = 1−tb.

∂r1

∂x2

= x1−x1.x2x−11 x−1

2 ⇒ Φ

(∂r1

∂x2

)= Φ (x1)−1 = ta−1

∂r1

∂x3

= 0 ⇒ Φ

(∂r1

∂x3

)= 0

∂r1

∂x4

= 0 ⇒ Φ

(∂r1

∂x4

)= 0

Como r2 = xm111 xn11

2 x−n113 , temos que

∂r2

∂x1

=xm11

1 − 1

x1 − 1⇒ Φ

(∂r2

∂x1

)=

tm11a − 1

ta − 1.

∂r2

∂x2

= xm111

xn112 − 1

x2 − 1⇒ Φ

(∂r2

∂x2

)= tm11a tn11b − 1

tb − 1

∂r2

∂x3

= − xn113 − 1

x3 − 1⇒ Φ

(∂r2

∂x3

)= − tn11c − 1

tc − 1

∂r2

∂x4

= 0 ⇒ Φ

(∂r2

∂x4

)= 0


2 x−m214 , temos que

∂r3

∂x1

=xm21

1 − 1

x1 − 1⇒ Φ

(∂r3

∂x1

)=

tm21a − 1

ta − 1

14

∂r3

∂x2

= xm211

xn212 − 1

x2 − 1⇒ Φ

(∂r3

∂x2

)= tm21a tn21b − 1

tb − 1

∂r3

∂x3

= 0 ⇒ Φ

(∂r3

∂x3

)= 0

∂r3

∂x4

= − xm214 − 1

x4 − 1⇒ Φ

(∂r3

∂x4

)= − tm21d − 1

td − 1

Portanto, a matriz de Alexander e dada por

Φ

(∂r

∂x

)=

−(tb − 1) (ta − 1) 0 0tm11a−1ta −1

tm11a tn11b−1tb −1

− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

.

Por definicao, o primeiro polinomio de Alexander e dado por

∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4) },

onde Ai e a matriz Φ

(∂r

∂x

)na qual foi retirada a i-esima coluna. Desse modo,

det (A1) =tn11c − 1

tc − 1

tm21d − 1

td − 1(ta−1)

det (A2) = − tn11c − 1

tc − 1

tm21d − 1

td − 1(tb−1)

det (A3) =tn11c − 1

tc − 1

tm21d − 1

td − 1(tc−1) veja (A.3)

det (A4) = − tn11c − 1

tc − 1

tm21d − 1

td − 1(td−1) veja (A.4)

Logo,

∆1 (t) = mdc { (ta − 1), (tb − 1), (tc − 1), (td − 1) } tn11c − 1

tc − 1

tm21d − 1

td − 1.

Por outro lado, da Secao 1.2, sabemos que:

∆1 (t) =tm21 (n11+n21) − 1

t(n11+n21) − 1

tn11(m11+m21) − 1

t(m11+m21) − 1(t− 1).

Igualando entre si essas duas expressoes de ∆1 (t), e colocando

(tr − 1) = mdc {(ta − 1), (tb − 1), (tc − 1), (td − 1)},

segue que

(tr − 1)tn11c − 1

tc − 1

tm21d − 1

td − 1= (t− 1)

tn11(m11+m21) − 1

t(m11+m21) − 1

tm21 (n11+n21) − 1

t(n11+n21) − 1.

15

Portanto,

(tr − 1) (tn11c − 1) (tm21d − 1) (t(m11+m21) − 1) (t(n11+n21) − 1)

= (t− 1) (tn11(m11+m21) − 1) (tm21 (n11+n21) − 1) (tc − 1) (td − 1). (∗∗)

Identificando os termos de menor grau em t, na igualdade acima, vemos que r = 1.

Simplificando (t − 1) e notando que a possibilidade d ≥ c gera um absurdo em

vista de (∗∗) e (∗), segue que d < c. Analisando os termos de menor grau em t

na igualdade resultante, temos que

d = n11 + n21.

Em virtude das relacoes (∗), concluimos que

c = m11 + m21, a = n11 e b = m21.

Por definicao, o segundo polinomio de Alexander e dado por

∆2 (t) = mdc

det (A12), det (A13), det (A14), det (A23), det (A24), det (A34),det (B12), det (B13), det (B14), det (B23), det (B24), det (B34),det (C12), det (C13), det (C14), det (C23), det (C24), det (C34)

,

onde A, B, C sao obtidas de Φ

(∂r

∂x

)retirando-se a terceira, segunda ou primeira

linhas respectivamente, e Aij, Bij e Cij denotam as matrizes formadas pelas colunas

i e j de A, B, e C respectivamente.

Temos que

A =

[−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0

].

det (A12) = − (tn11c−1) veja (A.3)

det (A13) =tn11c − 1

tc − 1(tb−1)

det (A14) = 0

det (A23) = − tn11c − 1

tc − 1(ta−1)

det (A24) = 0

det (A34) = 0

Por outro lado,

B =

[−(tb − 1) (ta − 1) 0 0

tm21a−1ta −1


0 − tm21d−1td −1

].

16

det (B12) = − (tm21d−1) veja (A.4)

det (B13) = 0

det (B14) =tm21d − 1

td − 1(tb−1)

det (B23) = 0

det (B24) = − tm21d − 1

td − 1(ta−1)

det (B34) = 0

Finalmente, temos que

C =

[tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

].

det (C12) =tm21d − 1

tb − 1

tm11a − 1

ta − 1− tn11c − 1

tb − 1

tm21a − 1

ta − 1veja (A.5)

det (C13) =tm21a − 1

ta − 1

tn11c − 1

tc − 1

det (C14) = − tm11a − 1

ta − 1

tm21d − 1

td − 1

det (C23) =tn21b − 1

tb − 1

tn11c − 1

tc − 1tm21a

det (C24) = − tn11a − 1

tb − 1

tm21d − 1

td − 1(tm11a−1)

det (C34) =tn11c − 1

tc − 1

tm21d − 1

td − 1

Concluimos entao que, neste caso,

∆2 (t) = mdc { tn11c−1tc −1

, tm21d−1td −1

, tm21d−1tb −1

tm11a−1ta −1

− tn11c−1tb −1

tm21a−1ta −1

},

onde a = n11, b = m21, c = m11 + m21 e d = n11 + n21

Exemplo 1 Sejam f e g dois germes de curvas planas redutıveis com dois ramos

de genero 1, onde f = (y4 − x25)(x4 − y5) e g = (y4 − x25)(x5 − y4).

17

Para a curva f , temos que {y1 = x

254 = x

m11n11

y2 = x45 = x

m21n21

Denotando por φi o i-esimo polinomio ciclotomico, temos que

∆1 (t) =tm21 (n11+n21) − 1

t(n11+n21) − 1

tn11(m11+m21) − 1

t(m11+m21) − 1(t− 1)

=t4(4+5) − 1

t(4+5) − 1

t4(25+4) − 1

t(25+4) − 1(t− 1)

=t36 − 1

t9 − 1

t116 − 1

t29 − 1(t− 1)

= φ36 φ18 φ12 φ6 φ116 φ58 φ29 φ24 φ2

2 φ1.

Logo, a monodromia de f e infinita devido a observacao feita na Secao 1.2. Isto

tambem pode ser verificado com o calculo do polinomio minimal da monodromia

que faremos a seguir. Como neste caso, a = 4, b = 4, c = 29, d = 9, temos que

∆2 (t) = mdc

{tn11c − 1

tc − 1,

tm21d − 1

td − 1,

tm21d − 1

tb − 1

tm11a − 1

ta − 1− tn11c − 1

tb − 1

tm21a − 1

ta − 1

}

= mdc

{t116 − 1

t29 − 1,

t36 − 1

t9 − 1,

t36 − 1

t4 − 1

t100 − 1

t4 − 1− t116 − 1

t4 − 1

t16 − 1

t4 − 1

}= mdc

{φ116 φ58 φ4 φ2, φ36 φ18 φ12 φ6 φ4 φ2,

t36 − 1

t4 − 1

t100 − 1

t4 − 1− t116 − 1

t4 − 1

t16 − 1

t4 − 1

}= mdc

{φ4 φ2,

t36 − 1

t4 − 1

t100 − 1

t4 − 1− t116 − 1

t4 − 1

t16 − 1

t4 − 1

}= 1

Consequentemente,

λ (t) =∆1 (t)

∆2 (t)=

φ116 φ58 φ36 φ29 φ18 φ12 φ6 φ24 φ2

2 φ1

1

o que nos permite concluir que a monodromia e infinita nesta situacao.

Para a curva g, temos que {y1 = x

254 = x

m11n11

y2 = x54 = x

m21n21

18

e portanto,

∆1 (t) =tm21 (n11+n21) − 1

t(n11+n21) − 1

tn11(m11+m21) − 1

t(m11+m21) − 1(t− 1)

=t5(4+4) − 1

t(4+4) − 1

t4(25+5) − 1

t(25+5) − 1(t− 1)

=t40 − 1

t8 − 1

t120 − 1

t30 − 1(t− 1)

= φ240 φ2

20 φ10 φ5 φ120 φ60 φ12 φ8 φ4 φ1.

Como, neste caso, a = 4, b = 5, c = 30, e d = 8, temos que

∆2 (t) = mdc

{tn11c − 1

tc − 1,

tm21d − 1

td − 1,

tm21d − 1

tb − 1

tm11a − 1

ta − 1− tn11c − 1

tb − 1

tm21a − 1

ta − 1

}

= mdc

{t120 − 1

t30 − 1,

t40 − 1

t8 − 1,

t40 − 1

t5 − 1

t100 − 1

t4 − 1− t120 − 1

t5 − 1

t20 − 1

t4 − 1

}= mdc

{φ40 φ20,

t40 − 1

t5 − 1

t100 − 1

t4 − 1− t120 − 1

t5 − 1

t20 − 1

t4 − 1

}= φ40 φ20

Consequentemente,

λ (t) =∆1 (t)

∆2 (t)=

φ240 φ2

20 φ10 φ5 φ120 φ60 φ12 φ8 φ4 φ1

φ40 φ20

,

o que nos permite concluir, nesta situacao, que a monodromia e finita.

Caso 2: Neste caso, podemos escolher f = f1f2 onde temos as seguinte repre-

sentacoes por series de Puiseux:{y1 = x

m11n11

y2 = xm21n21 + x

m22n22

onde (m11, n11) e o par de Puiseux do ramo f1 e (m22, n22) e o par de Puiseux do

ramo f2. Isto conduz a seguinte arvore, dada na Figura 6.

19

Figura 6:

Com as notacoes da Secao 1.3:µφ = m21, νφ = n21 = 1 e m21rφ = n21sφ + 1µ0 = m11, ν0 = n11, e m11r0 = n11s0 + 1µ1 = m22, ν1 = n22 e m22r1 = n22 s1 + 1

Portanto, o grupo do link π1 (X), e gerado por αz, βz com z percorrendo o

conjunto dos vertices V = {φ, 0, 01, 1, 11} da arvore Ξ, satisfazendo as relacoes que

exibiremos a seguir.

Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m21 n211 αm21

φ


0 βνφ

0 = αµφ

φ βνφ

φ

⇒ β10 = α−m21

0 αm21φ


φ βrφ

φ = αsφ

0 βrφ

0

⇒ α1 = αsφ

0 βrφ

0 α−sφ

φ

Relacoes de (0):

(R1) 01 > 0 e 01 nao e haste ⇒ αν0µ0

01 β01 = αµ0

0 βν00

⇒ β01 = α−m11 n1101 αm11

0 βn110


0 )ν0 = (αµ0

0 βν00 )r0

⇒ (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

20

Relacoes de (1):

(R1) 11 > 1 e 1 nao e haste ⇒ αν1µ1

11 β11 = αµ1

1 βν11

⇒ β11 = α−m22 n2211 αm22

1 βn221


1 )ν1 = (αµ1

1 βν11 )r1

⇒ (α11 αs11 βr1

1 )n22 = (αm221 βn22

1 )r1



[α0, β0] = 1 ⇒ [α1, β1] = 1 veja (A.6)

[α1, β1] = 1

[α01, β01] = 1 ⇔ [ αm110 βn11

0 , α01 ] = 1 veja (A.7)

[α11, β11] = 1 ⇔ [ αm221 βn22

1 , α11 ] = 1 veja (A.8)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α0, β0, α1, β1, α01, α11,

sujeitos as relacoes:

(1) β1 = α−m21 n211 αm21

φ

(2) α1 = αsφ

0 βrφ

0 α−sφ

φ

(3) [α0, β0] = 1

(4) αm210 βn21

0 = αm21φ

(5) (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

(6) [ αm110 βn11

0 , α01 ] = 1

(7) (α11 αs11 βr1

0 )n22 = (αm221 βn22

1 )r1

(8) [ αm221 βn22

1 , α11 ] = 1

21

Para podermos ter mais geradores do que relacoes na apresentacao de π1 (X),

somos conduzidos a fazer as seguintes restricoes sobre os pares de Puiseux dos dois

ramos: {m11 = λ1 n11 + 1 para algum λ1 ∈ ZZm22 = λ2 n22 + 1 para algum λ2 ∈ ZZ

o que implica que r0 = r1 = 1. Adotaremos como novos geradoresx1 = α0,x2 = β0,x3 = α01 αs0

0 βr00 ,

x4 = αφ,

x5 = α1,x6 = β1,x7 = α11 αs1

1 βr11 .

Observe que a relacao (6) e equivalente a relacao [ αm110 βn11

0 , x3 ] = 1 e que

esta e trivializada devido a (5). Observe tambem que a relacao (8) e equivalente a

relacao [ αm221 βn22

1 , x7 ] = 1, e que esta e trivializada devido a (7). Ficamos assim

com as seguintes relacoes:

(1) x6 = x−m21 n215 xm21

4

(2) x5 = xsφ

1 xrφ

2 x−sφ

4

(3) [ x1, x2 ] = 1

(4) xm211 xn21

2 = xm214

(5) xn113 = xm11

1 xn112

(6) xn227 = xm22

5 xn226

Podemos eliminar os geradores x5 e x6 e as relacoes (1) e (2), adaptando conveni-

entemente a relacao (6). Ficamos assim com os 5 geradores {x1, x2, x3, x4, x7} e

as 4 relacoes a seguir:

(1) [ x1, x2 ] = 1

(2) xm211 xn21

2 = xm214

(3) xn113 = xm11

1 xn112

(4) xn227 = (x

sφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21 xm214 ]n22

22

Logo, renomeando x7 por x5, podemos escrever:

π1 (X) = { x1, x2, x3, x4, x5; r1, r2, r3, r4 }, onder1 = x1 x2 x−1

1 x−12

r2 = xm111 xn11

2 x−n113

r3 = xm211 xn21

2 x−m214

r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21xm214 ]n22 x−n22

5


determinar Φ (xi) ∈ ZZ [ t, t−1 ] para i = 1, 2, 3, 4, 5. Como ainda nao sabemos a

priori os valores de Φ (xi), temporariamente escrevemos

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7−→ ta

x2 7−→ tb

x3 7−→ tc

x4 7−→ td

x5 7−→ te

Note que a relacoes ri, para i = 2, 3, 4, fornecem relacoes entre a,b,c,d,e; a saber,

(∗)

n11 c = m11 a + n11 bm21 d = m21 a + n21 bn22 e = (sφa + rφb− sφd)m22 + [(sφa + rφb− sφd)(−m21n21) + m21d]n22

Faremos a seguir o calculo de Φ

(∂ri

∂xj

).

Como r1 = x1 x2 x−11 x−1

2 , temos que

∂r1

∂x1

= 1−x1 x2 x−11 ⇒ Φ

(∂r1

∂x1

)= 1−Φ (x2) = 1−tb.

∂r1

∂x2

= x1−x1x2 x−11 x−1

2 ⇒ Φ

(∂r1

∂x2

)= Φ (x1)−1 = ta−1

∂r1

∂x3

= 0 ⇒ Φ

(∂r1

∂x3

)= 0

∂r1

∂x4

= 0 ⇒ Φ

(∂r1

∂x4

)= 0



∂r2

∂x1

=xm11

1 − 1

x1 − 1⇒ Φ

(∂r2

∂x1

)=

tm11a − 1

ta − 1.

23

∂r2

∂x2

= xm111

xn112 − 1

x2 − 1⇒ Φ

(∂r2

∂x2

)= tm11a tn11b − 1

tb − 1

∂r2

∂x3

= − xn113 − 1

x3 − 1⇒ Φ

(∂r2

∂x3

)= − tn11c − 1

tc − 1

∂r2

∂x4

= 0 ⇒ Φ

(∂r2

∂x4

)= 0



∂r3

∂x1

=xm21

1 − 1

x1 − 1⇒ Φ

(∂r3

∂x1

)=

tm21a − 1

ta − 1

∂r3

∂x2

= xm211

xn212 − 1

x2 − 1⇒ Φ

(∂r3

∂x2

)= tm21a tn21b − 1

tb − 1

∂r3

∂x3

= 0 ⇒ Φ

(∂r3

∂x3

)= 0

∂r3

∂x4

= − xm214 − 1

x4 − 1⇒ Φ

(∂r3

∂x4

)= − tm21d − 1

td − 1

Como r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21 xm214 ]n22 x−n22

5 , temos que

∂r4

∂x1

=x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (A.9)

− (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ] − 1

xsφ

1 − 1

x1 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

⇒ Φ

(∂r4

∂x1

)=

tsφa − 1

ta − 1

t(sφa+rφb−sφd)m22 − 1

t(sφa+rφb−sφd) − 1

− t(sφa+rφb−sφd)m22t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1

t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1

tsφa − 1

ta − 1t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


∂r4

∂x2

= xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (A.10)

− (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

xsφ

1

xrφ

2 − 1

x2 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

24

⇒ Φ

(∂r4

∂x2

)= tsφa trφb − 1

tb − 1





tsφa trφb − 1

tb − 1t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


∂r4

∂x3

= 0 ⇒ Φ

(∂r4

∂x3

)= 0

∂r4

∂x4

= − (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (A.11)

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

xsφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(m21n21−1) (xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

(xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm21

4 − 1

x4 − 1

⇒ Φ

(∂r4

∂x4

)= − t(sφa+rφb−sφd) tsφd − 1

td − 1



+ t(sφa+rφb−sφd)m22t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1


tsφd − 1

td − 1t(sφa+rφb−sφd)(−(m21n21−1)) t(sφa+rφb−sφd)(m21n21) − 1



t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21) tm21d − 1

td − 1

∂r4

∂x5

= − xn225 − 1

x5 − 1veja (A.12)

⇒ Φ

(∂r4

∂x5

)= − tn22e − 1

te − 1


Φ

(∂r

∂x

)=

−(tb − 1) (ta − 1) 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) − tn22e−1

te −1

.


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4), det (A5) }

25


(∂r

∂x

)na qual foi retirada a i-esima coluna. Deste modo,

det (A1) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(ta−1)

det (A2) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(tb−1)

det (A3) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(tc−1) veja (A.3)

det (A4) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(td−1) veja (A.4)

det (A5) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(te−1) veja (A.13)

Logo,

∆1 (t) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1mdc { (ta−1), (tb−1), (tc−1), (td−1), (te−1) }.

Por outro lado, como a curva e dada por{y1 = x

m11n11

y2 = xm21

1 + xm22n22 ,

segue da Secao 1.2, com i = 0, s1 = 1, s2 = 2 e n21 = 1, que

∆1 (t) = (t− 1)tw21e01 − 1

te01 − 1

tn11e11 − 1

te11 − 1

tn22e22 − 1

te22 − 1.

ondew21 = m21

w22 = m22 −m21n22 + m21n21n22 = m22

e01 = b1,1,1 + b2,1,2 = (∏1

j=1 n1,j) + (∏2

j=1 n2,j) = n11 + n21 n22

e11 = w2,1.b2,2,2.b1,1,0 + w1,1.b1,2,1 = m21.n22.1 + m11.1 = m11 + m21 n22

e22 = w2,1.b1,1,1.b2,2,1 + w22.b2,3,2 = m21.n11.1 + w22.1 = m22 + m21 n11

Igualando entre si essas duas expressoes de ∆1 (t), e colocando

(tr − 1) = mdc {(ta − 1), (tb − 1), tc − 1), (td − 1), (te − 1)},

temos que

(tr−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1= (t−1)

tn11.e11 − 1

te11 − 1

tm21e01 − 1

te01 − 1

tn22.e22 − 1

te22 − 1.

26

Portanto,

(tr − 1) (tn11c − 1) (tm21d − 1) (tn22e − 1) (te11 − 1) (te01 − 1) (te22 − 1)

= (t−1) (tn11e11−1) (tm21e01−1) (tn22e22−1) (tc−1) (td−1) (te−1). (∗∗)

Identificando os termos de menor grau em t na igualdade acima, vemos que

r = 1. Simplificando (t− 1) e notando que a possibilidade d ≥ c e d > e gera um

absurdo, em vista de (∗∗) e (∗), temos que d < c e d < e. Analisando os termos

de menor grau em t, na igualdade resultante, temos que,d = e01 = n11 + n21 n22

c = e11 = m11 + m21 n22

e = e22 = m22 + m21 n11

Em vista das relacoes (∗), concluimos que:

a = n11 e b = m21 n22


∆2 (t) = mdc

det (A123), det (A124), det (A125), det (A134), det (A135),det (A145), det (A234), det (A235), det (A245), det (A345),det (B123), det (B124), det(B125), det (B134), det (B135),det (B145), det (B234), det(B235), det (B245), det (B345),det (C123), det (C124), det (C125), det (C134), det (C135),det (C145), det (C234), det (C235), det (C245), det (C345),det (D123), det (D124), det (D125), det (D134), det (D135),det (D145), det (D234), det (D235), det (D245), det (D345)

onde A, B, C, D sao obtidas de Φ

(∂r

∂x

)retirando-se a quarta, segunda, terceira

e primeira linhas respectivamente, e Aijk, Bijk, Cijk e Dijk denotam as matrizes

formadas pelas colunas i, j, k de A, B, C e D respectivamente.

Temos que

A =

−(tb − 1) (ta − 1) 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

.

det (A123) = − tn11c − 1

tc − 1

tm21d − 1

td − 1(td−1) veja (A.4)

det (A124) =tn11c − 1

tc − 1

tm21d − 1

td − 1(tc−1) veja (A.3)

27

det (A125) = 0

det (A134) = − tn11c − 1

tc − 1

tm21d − 1

td − 1(tb−1)

det (A135) = 0

det (A145) = 0

det (A234) = − tn11c − 1

tc − 1

tm21d − 1

td − 1(ta−1)

det (A235) = 0

det (A245) = 0

det (A345) = 0

Por outro lado, temos que,

B =

−(tb − 1) (ta − 1) 0 0 0(tm21a−1)(ta −1)

tm21a (tn21b−1)(tb −1)

0 − (tm21d−1)(td −1)

0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −1

.

det (B123) = 0

det (B124) = − tm21d − 1

td − 1(tn22e−1) veja (A.4)

det (B125) =tn22e − 1

te − 1(tm21d−1) veja (A.4)

det (B134) = 0

det (B135) = 0

det (B145) = − tm21d − 1

td − 1

tn22e − 1

te − 1(tb−1)

det (B234) = 0

det (B235) = 0

det (B245) =tm21d − 1

td − 1

tn22e − 1

te − 1(ta−1)

det (B345) = 0

Tambem, temos que,

C =

−(tb − 1) (ta − 1) 0 0 0tm11a−1ta −1


− tn11c−1tc −1

0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −1

.

28

det (C123) = − tn11c − 1

tc − 1

[(ta − 1) Φ

(∂r4

∂x2

)+ (tb − 1) Φ

(∂r4

∂x1

) ]det (C124) = − Φ

(∂r4

∂x1

)(tn11c−1) veja (A.3)

det (C125) =tn22e − 1

te − 1(tn11c−1) veja (A.3)

det (C134) =tn11c − 1

tc − 1Φ

(∂r4

∂x4

)(tb−1)

det (C135) = − tn11c − 1

tc − 1

tn22e − 1

te − 1(tb−1)

det (C145) = 0

det (C234) = − tn11c − 1

tc − 1Φ

(∂r4

∂x4

)(ta−1)

det (C235) =tn11c − 1

tc − 1

tn22e − 1

te − 1(ta−1)

det (C245) = 0

det (C345) = 0

Finalmente, temos que,

D =

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −1

.

det (D123) = − tn11c − 1

tc − 1

[tm21a − 1

ta − 1Φ

(∂r4

∂x2

)+ tm21a tn21b − 1

tb − 1Φ

(∂r4

∂x1

) ]det (D124) =

tm21d − 1

ta − 1

[tm11a − 1

ta − 1Φ

(∂r4

∂x2

)− tm11a tn11b − 1

tb − 1Φ

(∂r4

∂x2

) ]+ Φ

(∂r4

∂x4

) [tm11a − 1

ta − 1tm21a tn21b − 1

tb − 1− tm21a − 1

ta − 1tm11a tn11b − 1

tb − 1

]det (D125) = − tn22e − 1

te − 1

[tm11a − 1

ta − 1tm21a tn21b − 1

tb − 1− tm21a − 1

ta − 1tm11a tn11b − 1

tb − 1

]det (D134) = − tn11c − 1

tc − 1

[tm21a − 1

ta − 1Φ

(∂r4

∂x4

)+

tm21d − 1

td − 1Φ

(∂r4

∂x1

) ]det (D135) = − tn11c − 1

tc − 1

tm21a − 1

ta − 1

tn22e − 1

te − 1

det (D145) =tm21d − 1

td − 1

tm11a − 1

ta − 1

tn22e − 1

te − 1

29

det (D234) = − tn11c − 1

tc − 1

[tm21a tn21b − 1

tb − 1Φ

(∂r4

∂x4

)+

tm21d − 1

td − 1Φ

(∂r4

∂x2

) ]det (D235) = − tn11c − 1

tc − 1

tn22e − 1

te − 1

tn21b − 1

tb − 1tm21a

det (D245) =tn11c − 1

tc − 1

tn22e − 1

te − 1

tn21b − 1

tb − 1tm21a

det (D345) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

Concluimos entao que,

∆2 (t) = mdc { 25 determinantes nao nulos acima }

Exemplo 2 Seja f = f1f2 na qual{y1 = x

163 = x

m11n11

y2 = x21 + x

165 = x

m21n21 + x

m22n21n22

segue da Secao 1.2, com i = 0, s1 = 1, s2 = 2 e n21 = 1 quew21 = m21 = 2w22 = m22 = 16e01 = n11 + n21 n22 = 3 + 5 = 8e11 = m11 + m21 n22 = 16 + 10 = 26e22 = m22 + m21 n11 = 16 + 6 = 22

Temos que

∆1 (t) =tw21e01 − 1

te01 − 1

tn11e11 − 1

te11 − 1

tn22e22 − 1

te22 − 1(t− 1)

=t16 − 1

t8 − 1

t78 − 1

t26 − 1

t110 − 1

t22 − 1(t− 1)

= φ16 φ78 φ39 φ6 φ3 φ110 φ55 φ10 φ5 φ1.

Portanto, a monodromia e finita.

30

2.2 Ramos com pares de Puiseux iguais

Caso 1: Neste caso, podemos escolher f = f1 f2 = (xm − yn) (xn − ym), com

as seguintes representacoes por series de Puiseux{y1 = x

m11n11 = x

mn

y2 = xm21n21 = x

nm

onde (mi1, ni1) e o par de Puiseux do ramo fi para i = 1, 2. Vemos que este caso, e

um caso particular, da apresentacao que fizemos para ramos com pares de Puiseux

distintos. Neste caso,

∆1 (t) =tm21 (n11+n21) − 1

t(n11+n21) − 1

tn11(m11+m21) − 1

t(m11+m21) − 1(t− 1)

=tn(n+m) − 1

t(n+m) − 1

tn(m+n) − 1

t(m+n) − 1(t− 1)

=

[tn(n+m) − 1

t(n+m) − 1

]2

(t− 1)

Por outro lado,

∆2 (t) = mdc

{tn11c − 1

tc − 1,

tm21d − 1

td − 1,

tm21d − 1

tb − 1

tm11a − 1

ta − 1− tn11c − 1

tb − 1

tm21a − 1

ta − 1

}onde {

a = n11 = n c = m11 + m21 = n + mb = m21 = n d = n11 + n21 = n + m

Substituindo os valores acima na expressao de ∆2(t), obtemos

∆2 (t) = mdc

{tn(n+m) − 1

t(n+m) − 1,

tn(n+m) − 1

t(n+m) − 1,

tn(n+m) − 1

tn − 1

tmn − 1

tn − 1− tn(n+m) − 1

tn − 1

tnn − 1

tn − 1

}Observe que:

tn(n+m) − 1

t(n+m) − 1=

[tn(n+m) − 1

t(n+m) − 1

t − 1

tn − 1

]tn − 1

t − 1

tn(n+m) − 1

t(n+m) − 1

tmn − 1

tn − 1=

[tn(n+m) − 1

t(n+m) − 1

t − 1

tn − 1

]tmn − 1

tn − 1(tm+n−1)

tn(n+m) − 1

tn − 1

tnn − 1

tn − 1=

[tn(n+m) − 1

t(n+m) − 1

t − 1

tn − 1

]tnn − 1

tn − 1(tm+n−1)

Logo,

31

∆2 (t) =tn(n+m) − 1

t(n+m) − 1

t − 1

tn − 1

Consequentemente,

λ (t) =∆1 (t)

∆2 (t)=

tn(n+m) − 1

t(n+m) − 1(tn − 1) = φ2

n(t) q(t),

e portanto, temos monodromia infinita neste caso.

Caso 2: Neste caso, podemos escolher f = f1f2 com as seguintes representacoes

por series de Puiseux:{y1 = x

m11n11 = x

mn

y2 = xm21n21 + x

m22n22 = x

z1 + x

mn

onde (m11, n11) e o par de Puiseux do ramo f1 e (m22, n22) e o par de Puiseux do

ramo f2. Vemos que este caso, e um caso particular da apresentacao que fizemos

para ramos com pares de Puiseux distintos. Temos que

∆1 (t) = (t− 1)tn11(m11+m21n22) − 1

t(m11+m21n22) − 1

tm21(n11+n21n22) − 1

t(n11+n21n22) − 1

tn22(m22+m21n11) − 1

t(m22+m21n11) − 1.

Logo,

∆1 (t) = (t− 1)tn(m+zn) − 1

t(m+zn) − 1

tz(n+n) − 1

t(n+n) − 1

tn(m+zn) − 1

t(m+zn) − 1,

= (t− 1)

[tn(m+zn) − 1

t(m+zn) − 1

]2t2nz − 1

t2n − 1,

= φ2p(t) . q(t) com p primo divisor de n

e portanto, temos monodromia infinita neste caso.

32

Caso 3: Neste caso, podemos escolher f = f1f2 = (xm1 − yn1) (2xm1 − yn1) cujas

representacoes por series de Puiseux sao dadas por{y1 = x

m11n11 = x

m1n1

y2 = xm21n21 = 2 x

m1n1

onde (mi1, ni1) e o par de Puiseux do ramo fi para i = 1, 2. Isto conduz a seguinte

arvore, dada na Figura 7.

Figura 7:

Com as notacoes da Secao 1.3,{µφ = m1, νφ = n1 e m1rφ = n1sφ + 1

Portanto, o grupo do link π1 (X), e gerado por αz, βz com z percorrendo o

conjunto dos vertices V = {φ, 1, 2} da arvore Ξ, satisfazendo as relacoes que exi-

biremos a seguir.

Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ


2 β2 = αµφ

φ βνφ

φ

⇒ β2 = α−m1 n12 αm1

φ

(R3) φ e nao terminal sem haste ⇒ (α1 α2 αsφ

φ βrφ

φ )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 α2 αsφ

φ )n1 = (αm1φ )rφ

33



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α2, β2] = 1 ⇔ [αm1φ , α2] = 1 veja (A.15)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1 e α2, sujeitos as

relacoes: (α1 α2 α

sφ


[αm1φ , α1] = 1

[αm1φ , α2] = 1

Como temos 3 geradores αφ, α1 e α2, sujeitos a 3 relacoes, e como nao conse-

guimos fazer uma mudanca de variavel que reduza o numero de relacoes, o metodo

mencionado na Secao 1.2 para o calculo dos polinomios de Alexander, nao se aplica

a este caso.

Caso 4: Neste caso, podemos escolher f = f1 f2 = (xm1 − yn1) (xm1 − yn1 + xkyj),

com n1k + m1j > m1n1, cujas representacoes por series de Puiseux sao da forma:{y1 = x

m1n1

y2 = xm1n1 + x

m2n1 n2

onde (m1, n1) e o par de Puiseux de ambos os ramos. Isto conduz a seguinte arvore,

dada na Figura 8.

Figura 8:

Com as notacoes da Secao 1.3,{µφ = m1, νφ = n1 e m1 rφ = sφ n1 + 1µ1 = m2 −m1 + m1n1, ν1 = n2 = 1 e µ1 r1 = s1 n2 + 1

34

Note que, neste caso, temos que r1 = 1 e s1 = µ1 − 1.

Portanto, o grupo do link π1 (X) e gerado por αz, βz com z percorrendo o



Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ

(R3) φ e nao terminal sem haste ⇒ (α1 αsφ

φ βrφ

φ )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 αsφ


Relacoes de (1):

(R1) 11 > 0 e 11 nao e haste ⇒ αν1µ1

11 β11 = αµ1

1 βν11

⇒ β11 = α−µ1

11 αµ1

1 (α−m1n11 αm1

φ )

(R2) 10 > 1 e 10 e uma haste ⇒ αµ1

1 βν11 = αµ1

10βν110

⇒ β110 = α−µ1

10 αµ1

1 (α−m1n11 αm1

φ )1

(R4) 1 e nao terminal com haste 10 ⇒ α11 αs11 βr1

1 = αs110 βr1

10

⇒ α10 = α1 α−111 veja (A.16)



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α11, β11] = 1 ⇔ [ αµ1−m1n1

1 αm1φ , α11 ] = 1 veja (A.17)

[α10, β10] = 1 ⇔ [ αµ1−m1n1

1 αm1φ , α11 ] = 1 veja (A.18)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1 e α11, sujeitos as

relacoes:

(1) (α1 αsφ


35

(2) [αm1φ , α1] = 1

(3) [ αµ1−m1n1

1 αm1φ , α11 ] = 1


somos conduzidos a fazer a seguinte restricao:

m1 = λ n1 + 1, para algum λ ∈ ZZ.

Isto implica que rφ = 1. Adotaremos como novos geradores

x1 = αφ, x2 = α1 αsφ

φ , x3 = α11.

Observe que a relacao (2) se torna a relacao [xm11 , x2 x

−sφ

1 ] = 1, via mudanca de

geradores, e sendo esta trivializada, devido a nova relacao xn12 = xm1

1 decorrente

de (1). Assim, temos que π1 (X) = { x1, x2, x3 ; r1, r2 }, onde{r1 = xn1

2 x−m11

r2 = (x2 x−s1 )µ1−m1n1 xm1

1 x3 x−m11 (x2 x−s

1 )−(µ1−m1n1) x−13


determinar Φ (xi) ∈ ZZ [ t, t−1 ] para i = 1, 2, 3, onde Φ e a aplicacao induzida pela

fibracao de Milnor. Temporariamente, escrevemos

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7→ ta

x2 7→ tb

x3 7→ tc

Para efeito de simplicacao das contas no calculo do Fox da matriz de Alexander,

faremos uma nova mudanca de geradores; a saber,

y1 = x2 x−sφ

1 , y2 = x1, y3 = x3.

Consequentemente, temos que

π1 (X) = { y1, y2, y3 ; r1, r2 }, onder1 = ( y1 y

sφ

2 )n1 y−m12

r2 = yµ1−m1n1

1 ym12 y3 y−m1

2 y−(µ1−m1n1)1 y−1

3

36

Logo, a aplicacao Φ nos novos geradores, e dada por

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]y1 7→ tb−sφa

y2 7→ ta

y3 7→ tc

Note que a relacao r1 fornece a seguinte relacao entre a e b.

(∗){

m1 a = n1 b


(∂ri

∂yj

).

Como r1 = ( y1 ysφ

2 )n1 y−m12 , temos que

∂r1

∂y1

=(y1 y

sφ

2 )n1 − 1

(y1 ysφ

2 ) − 1⇒ Φ

(∂r1

∂y1

)=

tn1b − 1

tb − 1

∂r1

∂y2

= y1y

sφ

2 − 1

y2 − 1

(y1 ysφ

2 )(n1−1) − 1

(y1 ysφ

2 ) − 1− (y1 y

sφ

2 )n1−1 y1 y−(m1−sφ)2

y(m1−sφ)2 − 1

y2 − 1

⇒ Φ

(∂r1

∂y2

)= − tm1a − 1

ta − 1

tb−sφa − 1

tb − 1veja (A.19)

∂r1

∂y3

= 0 ⇒ Φ

(∂r1

∂y3

)= 0

Como r2 = y µ1−m1n1

1 ym12 y3 y−m1

2 y−(µ1−m1n1)1 y−1

3 , temos que

∂r2

∂y1

=y

(µ1−m1n1)1 − 1

y1 − 1− y

(µ1−m1n1)1 ym

2 y3 y−m12 y

−(µ1−m1n1)1

y(µ1−m1n1)1 − 1

y1 − 1

⇒ Φ

(∂r2

∂y1

)= − tµ1(b−sφa)−m1a − 1

t(b−sφa) − 1(tc−1) veja (A.20)

∂r2

∂y2

= y(µ1−m1n1)1

ym12 − 1

y2 − 1− y

(µ1−m1n1)1 ym1

2 y3 y−m12

ym12 − 1

y2 − 1

⇒ Φ

(∂r2

∂y2

)= − tµ1(b−sφa)−m1a tm1a − 1

ta − 1(tc−1) veja (A.20)

∂r2

∂y3

= y(µ1−m1n1)1 ym1

2 − 1 ⇒ Φ

(∂r2

∂y3

)= tµ1(b−sφa)−1 veja (A.20)


Φ

(∂r

∂y

)=

[tm1a−1tb −1

− tm1a−1ta −1

tb−sφa−1tb −1

0

− tµ1(b−sφa)−m1a−1

t(b−sφa) −1

(tc − 1) −tµ1(b−sφa)−m1a tm1a−1ta −1

(tc − 1) tµ1(b−sφa) − 1

].

37


∆1 (t) := mdc { det (A1), det (A2), det (A3) }

onde Ai a matriz Φ

(∂r

∂y


det (A1) = − tm1a − 1

ta − 1

tµ1(b−sφa) − 1

tb − 1(tb−sφa−1)

det (A2) =tm1a − 1

ta − 1


tb − 1(ta−1)

det (A3) = − tm1a − 1

ta − 1


tb − 1(tc−1) veja (A.21)

Logo,

∆1 (t) =tm1a − 1

ta − 1


tb − 1mdc { (t(b−sφa) − 1), (ta − 1), (tc − 1) }.

Por outro lado, como {y1 = x

m1n1

y2 = xm1n1 + x

m2n1n2


∆1 (t) = (t− 1)tw11 e01 − 1

tw11 e02 − 1

tw22 e02 − 1

te01 − 1.

onde w11 = m1

w22 = m2 −m1n2 + m1n1n2 = µ1

e01 = b1,1,1 + b2,1,2 = n1 + n1 n2 = 2 n1

e02 = b1,2,1 + b2,2,2 = 1 + n2 = 2

Substituindo esses valores na expressao logo acima de ∆1 (t), obtemos

∆1 (t) = (t− 1)tm1(2 n1) − 1

t2n1 − 1

t2µ1 − 1

t2m1 − 1.

Igualando entre si as duas expressoes de ∆1 (t) obtidas, e colocando

(tr − 1) = mdc{(tb−sφa − 1), (ta − 1), (tc − 1)},

segue que

(tr − 1)tm1a − 1

ta − 1

t(b−sφa)µ1 − 1

tb − 1= (t− 1)

tm1(2 n1) − 1

t2n1 − 1

t2µ1 − 1

t2m1 − 1.

38

Portanto,

(tr − 1) (tm1a − 1) (tµ1(b−sφa) − 1) (t2m1 − 1) (t2n1 − 1)

= (t− 1) (tm1(2n1) − 1) (t2µ1 − 1) (tb − 1) (ta − 1).

Identificando os termos de menor grau em t, na igualdade acima, vemos que

r = 1. Simplificando (t− 1) e em vista de (∗), temos a < b. Analisando os termos

de menor grau em t, na igualdade resultante, temos que

a = 2 n1

Em vista de (∗), concluimos que

m1 a = n1 b ⇒ b = 2 m1

b− sφ a = 2 m1 − sφ 2 n1 = 2 m1 − 2 (m1 rφ − 1) = 2 pois rφ = 1.


∆2 (t) = mdc { det (B1), det (B2), det (B3), det (B4), det (B5), det (B6) },

onde Bi sao as matrizes (1 x 1) obtidas de Φ

(∂r

∂y

).

Temos que

det (B1) =t2m1n1 − 1

t2n1 − 1

t2 − 1

t2m1 − 1

t2n1 − 1

t2 − 1

det (B2) = − t2m1n1 − 1

t2n1 − 1

t2 − 1

t2m1 − 1

det (B3) = 0

det (B4) = − t2(µ1−m1n1) − 1

t2 − 1(tc−1)

det (B5) = − t2m1n1 − 1

t2n1 − 1

t2 − 1

t2m1 − 1

t2m1 − 1

t2 − 1t2(µ1−m1n1) (tc−1)

det (B6) = t2µ1−1

Portanto,

∆2 (t) = mdc { det (B2), det (B4), det (B6) }.

Como

mdc {t2µ1−2m1n1 − 1, t2µ1 − 1} = t2 mdc{µ1, m1 n1 } − 1

e notando que

∆2 (t) = mdc { mdc{ det (B2), det (B6) }, mdc{ det (B4), det (B6) } }

39

obtemos facilmente que

∆2 (t) = mdc

{t2m1n1 − 1

t2m1 − 1

t2 − 1

t2n1 − 1, t2µ1 − 1

}.

Note que µ1 = m2 −m1 + m1n1 e exatamente o ındice de intersecao I, entre os

ramos f1 e f2. Com isso, provamos o seguinte resultado:

Teorema 1 Seja f um germe analıtico com dois ramos de genero 1, cuja classe

topologica e determinada pelas parametrizacoes de Puiseux y1 = xm1/n1 e y2 =

xm1/n1 + xm2/n1. Suponha que m1 = λ n1 + 1 para algum λ ∈ ZZ. Temos que os

polinomios de Alexander de f , sao dados por:

∆1 (t) =t2m1n1 − 1

t2m1 − 1

t − 1

t2n1 − 1(t2(m2−m1+m1n1) − 1)

∆2 (t) = mdc

{t2m1n1 − 1

t2m1 − 1

t2 − 1

t2n1 − 1, t2(m2−m1+m1n1) − 1

}.

Corolario Nas condicoes do Teorema, o polinomio mınimo da monodromia

λ (t) = ∆1 (t)/∆2 (t), nao possui fatores multiplos. Nesses casos, a monodro-

mia de f e finita.

A seguir, apresentamos tabelas, para ilustrar o uso dos resultados obtidos. Nes-

tas tabelas, I representa o ındice de intersecao entre os dois ramos da curva C,

e percorre todos os possıveis valores iniciais. O numero µ, representa o numero

de Milnor, e consequentemente o grau de ∆1(t), o polinomio caracterıstico da

monodromia. A notacao n.a., representa o numero de autovalores da monodromia,

(sem levar em consideracao suas multiplicidades). Alem disso, as curvas listadas,

sao apenas representantes de suas classes de equisingularidade.

40

I Polinomio µ n.a. ∆1 (t) ∆2 (t)4 f4 = (x2 + y3)(x3 + y2) 11 6 φ2

10.φ22.φ1 φ10

6 f6 = (x2 + y3)(x2 + y3 + x2) 15 11 φ212.φ6.φ4.φ3.φ1 φ12

7 f7 = (x2 + y3)(x2 + y3 + xy2) 17 17 φ14.φ12.φ7.φ1 18 f8 = (x2 + y3)(x2 + y3 + x2y) 19 19 φ16.φ12.φ8.φ4.φ1 19 f9 = (x2 + y3)(x2 + y3 + x3) 21 21 φ18.φ12.φ9.φ6.φ3.φ1 110 f10 = (x2 + y3)(x2 + y3 + x2y2) 23 23 φ20.φ12.φ10.φ5.φ4.φ1 111 f11 = (x2 + y3)(x2 + y3 + x3y) 25 25 φ22.φ12.φ11.φ1 112 f12 = (x2 + y3)(x2 + y3 + x4) 27 23 φ24.φ

212.φ8.φ6.φ4.φ3.φ1 φ12

13 f13 = (x2 + y3)(x2 + y3 + x3y2) 29 29 φ26.φ13.φ12.φ1 114 f14 = (x2 + y3)(x2 + y3 + x4y) 31 31 φ28.φ14.φ12.φ7.φ1 115 f15 = (x2 + y3)(x2 + y3 + x5) 33 33 φ30.φ15.φ12.φ10.φ6.φ5.φ3.φ1 116 f16 = (x2 + y3)(x2 + y3 + x4y2) 35 35 φ32.φ16.φ12.φ8.φ4.φ1 117 f17 = (x2 + y3)(x2 + y3 + x5y) 37 37 φ34.φ17.φ12.φ1 118 f18 = (x2 + y3)(x2 + y3 + x6) 39 35 φ36.φ18.φ

212.φ9.φ6.φ4.φ3.φ1 φ12

19 f19 = (x2 − y3)(x2 − y3 + x5y2) 41 41 φ12.(t38 − 1)/φ2 1

20 f20 = (x2 − y3)(x2 − y3 + x6y) 43 43 φ12.(t40 − 1)/φ2 1

21 f21 = (x2 − y3)(x2 − y3 + x7) 45 45 φ12.(t42 − 1)/φ2 1

22 f22 = (x2 − y3)(x2 − y3 + x6y2) 47 47 φ12(t44 − 1)/φ2 1

23 f23 = (x2 − y3)(x2 − y3 + x7y) 49 49 φ12.(t46 − 1)/φ2 1

24 f24 = (x2 − y3)(x2 − y3 + x8) 51 47 φ12.(t48 − 1)/φ2 φ12

25 f25 = (x2 − y3)(x2 − y3 + x7y2) 53 53 φ12.(t50 − 1)/φ2 1

26 f26 = (x2 − y3)(x2 − y3 + x8y) 55 55 φ12.(t52 − 1)/φ2 1

27 f27 = (x2 − y3)(x2 − y3 + x9) 57 57 φ12.(t54 − 1)/φ2 1

28 f28 = (x2 − y3)(x2 − y3 + x8y2) 59 59 φ12.(t56 − 1)/φ2 1

29 f29 = (x2 − y3)(x2 − y3 + x9y) 61 61 φ12.(t58 − 1)/φ2 1

30 f30 = (x2 − y3)(x2 − y3 + x10) 63 59 φ12.(t60 − 1)/φ2 φ12

Tabela 1

Note que na tabela acima, a linha correspondente a I = 4, nos da monodro-

mia infinita, pois se trata do primeiro exemplo de curvas planas redutıveis, cuja

monodromia e infinita, dada por A’Campo em [A’c1]. Nas demais linhas, temos

monodromia finita. O calculo de ∆2 (t) para I = 6, foi obtido usando o software

Singular. As linhas onde I = 6 + 6k com k ∈ IN, nos fornecem o resultado que a

monodromia e finita, que nao decorre de nenhum criterio anteriormente conhecido.

Note tambem que I pode percorrer todos os inteiros maiores do que 30.

41

I Polinomio µ n.a ∆1 (t) ∆2 (t)9 (x3 + y16)(x16 + y3) 77 39 φ2

57 φ23 φ1 φ57

18 y1 = x16/3 e y2 = x2/1 + x16/3 95 51 φ266 φ2

33 φ12 φ26 φ4 φ2

3 φ1 127 y1 = x16/3 e y2 = x3/1 + x16/3 113 63 φ2

75 φ18 φ215 φ9 φ2

3 φ1 136 y1 = x16/3 e y2 = x4/1 + x16/3 131 71 φ2

84φ242φ24φ

221φ

312φ8φ

26φ4φ

23φ1 1

45 y1 = x16/3 e y2 = x5/1 + x16/3 149 87 φ293 φ30 φ15 φ10 φ5 φ2

3 φ1 148 (x3 + y16)(x3 + y16 + x3) 155 95 φ96φ48φ24φ12(t

96 − 1)/φ2 φ96φ48φ24φ12

54 (x3 + y16)(x3 + y16 + x3y2) 167 163 φ96φ48φ24φ12 (t108 − 1) / φ2 φ12

60 (x3 + y16)(x3 + y16 + x3y4) 179 167 φ96φ48φ24φ12 (t120 − 1) / φ2 φ24 φ12

66 (x3 + y16)(x3 + y16 + x3y6) 191 187 φ96φ48φ24φ12 (t132 − 1) / φ2 φ12

72 (x3 + y16)(x3 + y16 + x3y8) 203 175 φ96φ48φ24φ12 (t144 − 1) / φ2 φ48φ24φ12

78 (x3 + y16)(x3 + y16 + x3y10) 215 211 φ96φ48φ24φ12 (t156 − 1) / φ2 φ12

84 (x3 + y16)(x3 + y16 + x3y12) 227 215 φ96φ48φ24φ12 (t168 − 1) / φ2 φ24 φ12

90 (x3 + y16)(x3 + y16 + x3y14) 239 235 φ96φ48φ24φ12 (t168 − 1) / φ2 φ12

96 (x3 + y16)(x3 + y16 + x6) 251 191 φ96φ48φ24φ12 (t192 − 1) / φ2 φ96φ48φ24φ12

Tabela 2

Podemos tomar os seguintes polinomios, como representantes das classes para

I = 18, 27, 36 e 45, respectivamente:f18 = (y3 − x16) (y3 − 3y2x2 + 3yx4 − x16 − x6)f27 = (y3 − x16) (y3 − 3y2x3 + 3yx6 − x16 − x9)f36 = (y3 − x16) (y3 − 3y2x4 + 3yx8 − x16 − x12)f45 = (y3 − x16) (y3 − 3y2x5 + 3yx10 − x16 − x15)

Note que na Tabela 2, as linhas correspondentes a I = 9, 18, 27, 36 e 45 nos

dao monodromia infinita. Nas demais linhas, temos monodromia finita. O calculo

de ∆2 (t) para I = 48, foi obtido usando o software Singular. As linhas onde

I = 48 + 6k com k ∈ IN, nos fornecem o resultado que a monodromia e finita, o

que nao decorre de nenhum criterio anteriormente conhecido. Note tambem que

para os I intermediarios omitidos na Tabela 2, temos que ∆1 (t) nao possui fatores

multiplos, e consequentemente, ∆2 (t) = 1 para estes valores de I. Alem disso, I

pode percorrer todos os inteiros maiores do que 96.

42

Resumindo, obtemos os seguintes resultados para ramos de genero 1:

Ramos com pares de Puiseux distintos

Caso 1: A matriz de Alexander foi determinada, ∆2 (t) foi calculado em funcao

dos pares de Puiseux e a monodromia pode ser finita ou infinita.


dos 25 determinantes menores e a monodromia pode ser finita ou infinita.

Ramos com pares de Puiseux iguais


dos pares de Puiseux e a monodromia e sempre infinita.


dos 25 determinantes menores e a monodromia e sempre infinita.

Caso 3: A matriz de Alexander nao foi determinada.


dos pares de Puiseux e a monodromia e sempre finita.

43

Capıtulo 3 - Ramos de genero 2

Neste capıtulo, estudaremos a monodromia de um germe de curva plana com

dois ramos, ambos de genero 2. A nossa analise sera naturalmente desdobrada

em varios casos, segundo as arvores que ocorrem quando os pares de Puiseux sao

distintos ou iguais.

3.1 Ramos com pares de Puiseux distintos

Nesta situacao, temos as seguintes possibilidades para os tipos de arvores que

podem ocorrer, conforme dado na Figura 9.

Figura 9: Arvores para ramos de genero 2 com pares de Puiseux distintos

Caso 1: Neste caso, podemos escolher f = f1f2, onde temos as seguintes repre-

sentacoes por series de Puiseux:{y1 = x

m11n11 + x

m12n11n12

y2 = xm21n21 + x

m22n21n22

onde (m1i, n1i) sao os pares de Puiseux do ramo f1 e (m2i, n2i) sao os pares de

Puiseux do ramo f2, para i = 1, 2. Isto conduz a seguinte arvore:

44

Figura 10:

Com as notacoes da Secao 1.3,µφ = m21, νφ = n21 e m21rφ = n21sφ + 1µ0 = m11, ν0 = n11 e m11r0 = n11s0 + 1µ1 = m22 −m21n22 + m21n21n22, ν1 = n22 e µ1 r1 = s1 ν1 + 1µ01 = m12 −m11n12 + m11n11n12, ν01 = n12 e µ01 r01 = s01 n12 + 1

Portanto, o grupo do link π1 (X) e gerado por αz, βz com z percorrendo o con-

junto dos vertices V = {φ, 0, 01, 1, 2, 3} da arvore Ξ, satisfazendo as relacoes que

exibiremos a seguir2.

Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m21 n211 αm21

φ


0 βνφ

0 = αµφ

φ βνφ

φ

⇒ αm210 βn21

0 = αm21φ


φ βrφ

φ = αsφ

0 βrφ

0

⇒ α1 = αsφ

0 βrφ

0 α−sφ

φ

Relacoes de (0):

(R1) 01 > 0 e 01 nao e haste ⇒ αν0µ0

01 β01 = αµ0

0 βν00

⇒ β01 = α−m11 n1101 αm11

0 βn110

2Os desenvolvimentos algebricos omitidos encontram-se no apendice B, segundo as referenciasespecificadas.

45


0 )ν0 = (αµ0

0 βν00 )r0

⇒ (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

Relacoes de (1):


2 β2 = αµ1

1 βν11

⇒ β2 = α−µ1 n22

2 αµ1

1 βn221


1 )ν1 = (αµ1

1 βν11 )r1

⇒ (α2 αs11 βr1

1 )n22 = (αµ1

1 βn221 )r1

Relacoes de (01):

(R1) 3 > 01 e 3 nao e haste ⇒ αν01µ01

3 β3 = αµ01

01 βν0101

⇒ β3 = α−µ01 ν01

3 αµ01

01 βν0101


01 )ν01 = (αµ01

01 βν0101 )r01

⇒ ( α3 αs0101 βr01

01 )n12 = αµ01 r01

01 βn2 r0101


[ αφ, βφ ] = 1 ⇔ αφ βφ = βφ αφ ⇔ αφ = αφ

[ α0, β0 ] = 1 ⇒ [α1, β1] = 1 veja (A.1)

[ α1, β1 ] = 1

[α01, β01] = 1 ⇔ [ αm110 βn11

0 , α01 ] = 1 veja (A.2)

[α2, β2] = 1 ⇔ [ αµ1

1 βn221 , α2 ] = 1 veja (B.1)

[α3, β3] = 1 ⇔ [ αµ01

01 βn1201 , α3 ] = 1 veja (B.2)

Portanto, π1(X) pode ser apresentado com geradores αφ, α0, β0, α1, β1, α2, α3, α01, β01, α11,

e β11, sujeitos as relacoes:

(1) [α0, β0] = 1

(2) αm210 βn21

0 = αm21φ

46

(3) (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

(4) [ αm110 βn11

0 , α01 ] = 1

(5) (α2 αs11 βr1

1 )n22 = (αµ1

1 βn221 )r1

(6) [αµ1

1 βn221 , α2] = 1

(7) ( α3 αs0101 βr01

01 )n12 = (αµ01

01 βn1201 )r01

(8) [ αµ01

01 βn1201 , α3 ] = 1

Para podermos ter mais geradores do que relacoes na apresentacao de π1(X),

somos conduzidos a fazer as seguintes restricoes, sobre os pares de Puiseux dos

dois ramos: m11 = n11 λ1 + 1 para algum λ1 ∈ ZZ,µ1 = n22 λ2 + 1 para algum λ2 ∈ ZZ,µ01 = n12 λ3 + 1 para algum λ3 ∈ ZZ,

o que implica que r0 = r1 = r01 = 1. Adotaremos como novos geradoresx1 = α0,x2 = β0,x3 = α01 αs0

0 βr00 ,

x4 = αφ,x5 = α2 αs1

1 βr11 ,

x6 = α3 αs0101 βr01

01 .

Observe que temos as seguintes equivalencias entre as relacoes:

(4′) [ αm110 βn11

0 , x3 ] = 1 ⇔ (4) [ αm110 βn11

0 , α01 ] = 1

(6′) [ αµ1

1 βn221 , x5 ] = 1 ⇔ (6) [ αµ1

1 βn221 , α2 ] = 1

(8′) [ αµ01

01 βn1201 , x6 ] = 1 ⇔ (8) [ αµ01

01 βn1201 , α3 ] = 1

Vemos que as relacoes (4′), (6′) e (8′) sao trivializadas pelas relacoes (3′), (5′) e

(7′) respectivamente. Desse modo, temos que

π1 (X) = { x1, x2, x3, x4, x5, x6; r1, r2, r3, r4, r5 }, onder1 = x1 x2 x−1

1 x−12

r2 = xm111 xn11

2 x−n113

r3 = xm211 xn21

2 x−m214

r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 xm21n224 x−n22

5

r5 = (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11 n12

2 x−n126

47


determinar Φ (xi) ∈ ZZ [ t, t−1 ] para i = 1, . . . , 6, onde Φ e a aplicacao induzida

pela fibracao de Milnor. Temporariamente, escrevemos

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]

x1 7−→ ta

x2 7−→ tb

x3 7−→ tc

x4 7−→ td

x5 7−→ te

x6 7−→ tf

Note que as relacoes ri, para i = 2, 3, 4, 5, fornecem as seguintes relacoes entre

a, b, c, d, e, f :

(∗)

n11 c = m11 a + n11 bm21 d = m21 a + n21 bn22 e = (sφ a + rφ b− sφ d) (m22 −m21 n22) + m21 n22 dn12 f = (c− r0 b− s0 a) (m12 − m11 n12) + m11 n12 a + n11 n12 b


(∂ri

∂xj

).

Como r1 = x1 x2 x−11 x−1

2 , temos que

∂r1

∂x1

= 1−x1 x2 x−11 ⇒ Φ

(∂r1

∂x1

)= 1 − Φ (x2) = 1 − tb.

∂r1

∂x2

= x1−x1.x2x−11 x−1

2 ⇒ Φ

(∂r1

∂x2

)= Φ (x1) − 1 = ta − 1

∂r1

∂x3

= 0 ⇒ Φ

(∂r1

∂x3

)= 0

∂r1

∂x4

= 0 ⇒ Φ

(∂r1

∂x4

)= 0

∂r1

∂x5

= 0 ⇒ Φ

(∂r1

∂x5

)= 0

∂r1

∂x6

= 0 ⇒ Φ

(∂r1

∂x6

)= 0



∂r2

∂x1

=xm11

1 − 1

x1 − 1⇒ Φ

(∂r2

∂x1

)=

tm11a − 1

ta − 1.

48

∂r2

∂x2

= xm111

xn112 − 1

x2 − 1⇒ Φ

(∂r2

∂x2

)= tm11a tn11b − 1

tb − 1

∂r2

∂x3

= − xn113 − 1

x3 − 1⇒ Φ

(∂r2

∂x3

)= − tn11c − 1

tc − 1

∂r2

∂x4

= 0 ⇒ Φ

(∂r2

∂x4

)= 0

∂r2

∂x5

= 0 ⇒ Φ

(∂r1

∂x4

)= 0

∂r2

∂x6

= 0 ⇒ Φ

(∂r2

∂x6

)= 0



∂r3

∂x1

=xm21

1 − 1

x1 − 1⇒ Φ

(∂r3

∂x1

)=

tm21a − 1

ta − 1.

∂r3

∂x2

= xm211

xn212 − 1

x2 − 1⇒ Φ

(∂r3

∂x2

)= tm21a tn21b − 1

tb − 1

∂r3

∂x3

= 0 ⇒ Φ

(∂r3

∂x3

)= 0

∂r3

∂x4

= − xm214 − 1

x4 − 1⇒ Φ

(∂r3

∂x4

)= − tm21d − 1

td − 1

∂r3

∂x5

= 0 ⇒ Φ

(∂r3

∂x5

)= 0

∂r3

∂x6

= 0 ⇒ Φ

(∂r3

∂x6

)= 0

Como r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 xm21n224 x−n22

5 , temos que

∂r4

∂x1

=x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

⇒ Φ

(∂r4

∂x1

)=

tsφa − 1

ta − 1

t(sφa+rφb−sφd)(m22−m21n22) − 1

(tsφa+rφb−sφd) − 1

∂r4

∂x2

= xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

⇒ Φ

(∂r4

∂x2

)= tsφa trφb − 1

tb − 1



∂r4

∂x3

= 0 ⇒ Φ

(∂r4

∂x3

)= 0

49

∂r4

∂x4

= − (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22xm21n22

4 − 1

x4 − 1

⇒ Φ

(∂r4

∂x4

)= − tsφa+rφb−sφd tsφd − 1

td − 1



+ t(sφa+rφb−sφd)(m22−m21n22) tm21n22d − 1

td − 1

∂r4

∂x5

= − xn225 − 1

x5 − 1

⇒ Φ

(∂r4

∂x5

)= − tn22e − 1

te − 1

∂r4

∂x6

= 0 ⇒ Φ

(∂r4

∂x6

)= 0

Como r5 = (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11 n12


∂r5

∂x1

= − (x3 x−r02 x−s0

1 )xs0

1 − 1

x1 − 1

(x3 x−r02 x−s0

1 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1

+ (x3 x−r02 x−s0

1 )m12−m11n12xm11n12

1 − 1

x1 − 1

⇒ Φ

(∂r5

∂x1

)= − tc−r0b−s0a ts0a − 1

ta − 1

t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

+ t(c−r0b−s0a)(m12−m11n12) tm11n12a − 1

ta − 1

∂r5

∂x2

= − x3 x−r02

xr02 − 1

x2 − 1

(x3 x−r02 x−s0

1 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1

+ (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121

xn11n122 − 1

x2 − 1

⇒ Φ

(∂r5

∂x2

)= − tc−r0b tr0b − 1

tb − 1

t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

+ t(c−r0b−s0a)(m12−m11n12) tm11n12a tn11n12b − 1

tb − 1

∂r5

∂x3

=(x3 x−r0

2 x−s01 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1⇒ Φ

(∂r5

∂x3

)=

t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

∂r5

∂x4

= 0 ⇒ Φ

(∂r5

∂x4

)= 0

50

∂r5

∂x5

= 0 ⇒ Φ

(∂r5

∂x5

)= 0

∂r5

∂x6

= − xn126 − 1

x6 − 1⇒ Φ

(∂r5

∂x6

)= − tn12f − 1

tf − 1


Φ

(∂r

∂x

)=

−(tb − 1) (ta − 1) 0 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) Φ ( ∂r5

∂x3) 0 0 − tn12f−1

tf −1

.


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4), det (A5), det (A6) },


(∂r

∂x


det (A1) = (ta−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1

det (A2) = − (tb−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1

det (A3) = (tc−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1veja (A.3)

det (A4) = − (td−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1veja (A.4)

det (A5) = (te−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1veja (B.6)

det (A6) = − (tf−1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1veja (B.9)

Logo,

∆1 (t) = mdc { (ta − 1), (tb − 1), (tc − 1), (td − 1), (te − 1), (tf − 1) }

tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1.

Por outro lado, recordando que a curva foi dada por{y1 = x

m11n11 + x

m12n11n12

y2 = xm21n21 + x

m22n21n22 ,

51

segue da Secao 1.2, com i = 0, s1 = 2, e s2 = 2, que

∆1 (t) = (t− 1)tw21 e01 − 1

te01 − 1

tn11 e11 − 1

te11 − 1

tn22 e22 − 1

te22 − 1

tn12 e12 − 1

te12 − 1

onde

w11 = m11

w12 = m12 − m11 n12 + m11 n11 n12

w21 = m21

w22 = m22 − m21 n22 + m21 n21 n22

e01 = b1,1,2 + b2,1,2 = (∏2

j=1 n1,j) + (∏2

j=1 n2,j) = n11 n12 + n21 n22

e11 = w2,1 b2,2,2 b1,1,0 + w1,1 b1,2,2 = m21 n22 1 + m11 n12 = m11 n12 + m21 n22

e12 = w2,1 b2,2,2 b1,1,1 + w12 b1,3,2 = m21 n22 n11 + w12

e22 = w2,1 b1,1,2 b2,2,1 + w22 b2,3,2 = m21 n11 n12 + w22


(tr − 1) = mdc {(ta − 1), (tb − 1), (tc − 1), (td − 1), (te − 1), (tf − 1)},

segue que

(tr − 1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1

= (t− 1)tw21 e01 − 1

te01 − 1

tn11 e11 − 1

te11 − 1

tn22 e22 − 1

te22 − 1

tn12 e12 − 1

te12 − 1.

Portanto,

(tr−1) (tn11c−1) (tm21d−1) (tn22e−1) (tn12f−1) (te01−1) (te11−1) (te22−1) (te12−1)

= (t−1) (tn11 e11−1) (tm21 e01−1) (tn22 e22−1) (tn12 e12−1) (tc−1) (td−1) (te−1) (tf−1). (∗∗)

Identificando, na igualdade acima, os termos de menor grau em t, vemos que

r = 1. Simplificando (t− 1) e analisando na igualdade resultante, em vista de (∗),os termos de menor grau em t, temos que

d = e01 = n11n12 + n21n22

c = e11 = m11n12 + m21n22

e = e22 = m21n11n12 + w22

f = e12 = m21n11n22 + w12

Em vista das relacoes de (∗), concluimos que

a = n11 n12 e b = m21 n22

52


∆2 (t) = mdc

det (A1234), det (A1235), det (A1236), det (A1345), det (A1346),det (A1456), det (A2345), det (A2346), det (A2356), det (A2456),det (A3456), det (B1234), det (B1235), det (B1236), det (B1345),det (B1346), det (B1456), det (B2345), det (B2346), det (B2356),det (B2456), det (B3456), det (C1234), det (C1235), det (C1236),det (C1345), det (C1346), det (C1456), det (C2345), det (C2346),det (C2356), det (C2456), det (C3456), det (D1234), det (D1235),det (D1236), det (D1345), det (D1346), det (D1456), det (D2345),det (D2346), det (D2356), det (D2456), det (D3456), det (E1234),det (E1235), det (E1236), det (E1345), det (E1346), det (E1456),det (E2345), det (E2346), det (E2356), det (E2456), det (E3456)

onde as matrizes A, B, C, D, E sao obtidas de Φ

(∂r

∂x

)retirando-se a quinta,

quarta, terceira, segunda e primeira linhas respectivamente, e Aijkl, Bijkl, Cijkl, Dijkl

e Eijkl denotam as matrizes formadas pelas colunas i, j, k e l de A, B, C,D e E,

respectivamente.

Temos que

A =

−(tb − 1) (ta − 1) 0 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10

.

det (A1234) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(te−1) veja (B.5)

det (A1235) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(td−1) veja (A.4)

det (A1236) = 0

det (A1345) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(tb−1)

det (A1346) = 0

det (A1456) = 0

det (A2345) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1(ta−1)

det (A2346) = 0

det (A2356) = 0

det (A2456) = 0

53

det (A3456) = 0

Temos que

B =

−(tb − 1) (ta − 1) 0 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) Φ ( ∂r5

∂x3) 0 0 − tn12f−1

tf −1

.

det (B1234) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn12f − 1

tf − 1(tf−1) veja (B.8)

det (B1235) = 0

det (B1236) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn12f − 1

tf − 1(td−1)

det (B1345) = 0

det (B1346) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn12f − 1

tf − 1(tb−1)

det (B1456) = 0

det (B2345) = 0

det (B2346) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn12f − 1

tf − 1(ta−1)

det (B2356) = 0

det (B2456) = 0

det (B3456) = 0

Temos que

C =

−(tb − 1) (ta − 1) 0 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) Φ ( ∂r5

∂x3) 0 0 − tn12f−1

tf −1

.

det (C1234) = Φ

(∂r4

∂x4

)tn11c − 1

tc − 1

tn12f − 1

tf − 1(tf−1) veja (B.7)

det (C1235) = − tn11c − 1

tc − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1(tf−1) veja (B.7)

det (C1236) = − tn11c − 1

tc − 1

tn12f − 1

tf − 1(ta−1) (tsφa+rφb−1)



det (C1345) = 0

54

det (C1346) = − Φ

(∂r4

∂x4

)tn11c − 1

tc − 1

tn12f − 1

tf − 1(tb−1)

det (C1456) = 0

det (C2345) = 0

det (C2346) = Φ

(∂r4

∂x4

)tn11c − 1

tc − 1

tn12f − 1

tf − 1(ta−1)

det (C2356) = − tn11c − 1

tc − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1(ta−1)

det (C2456) = 0

det (C3456) = 0

Temos que

D =

−(tb − 1) (ta − 1) 0 0 0 0

tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) Φ ( ∂r5

∂x3) 0 0 − tn12f−1

tf −1

.

det (D1234) = Φ

(∂r5

∂x3

)tm21d − 1

td − 1

tn22e − 1

te − 1(te−1)

det (D1235) = − Φ

(∂r5

∂x3

)tm21d − 1

td − 1

tn22e − 1

te − 1(td−1)

det (D1236) = 0

det (D1345) = − Φ

(∂r5

∂x3

)tm21d − 1

td − 1

tn22e − 1

te − 1(tb−1)

det (D1346) = 0

det (D1456) =tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1(tb−1)

det (D2345) = Φ

(∂r5

∂x3

)tm21d − 1

td − 1

tn22e − 1

te − 1(ta−1)

det (D2346) = 0

det (D2356) = 0

det (D2456) = − tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1(ta−1)

det (D3456) = 0

55


E =

tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) Φ ( ∂r5

∂x3) 0 0 − tn12f−1

tf −1

det (E1234) = Φ

(∂r5

∂x3

)tm21d − 1

td − 1(tn22e−1)

+tn11c − 1

tc − 1

tm21d − 1

td − 1

[Φ

(∂r5

∂x2

)Φ

(∂r4

∂x1

)− Φ

(∂r5

∂x1

)Φ

(∂r4

∂x2

) ]− tn11c − 1

tc − 1Φ

(∂r4

∂x4

) [tm21a − 1

ta − 1Φ

(∂r5

∂x2

)− tm21a tn21b − 1

tb − 1Φ

(∂r5

∂x1

) ]det (E1235) =

tn11c − 1

tc − 1

tn22e − 1

te − 1

[tm21a − 1

ta − 1Φ

(∂r5

∂x2

)− tm21a tn21b − 1

tb − 1Φ

(∂r5

∂x1

) ]tn22e − 1

te − 1Φ

(∂r5

∂x3

) [tm21a tn21b − 1

tb − 1

tm11a − 1

ta − 1− tm11a tn11b − 1

tb − 1

tm21a − 1

ta − 1

]det (E1236) = − tn11c − 1

tc − 1

tn12f − 1

tf − 1

[tm21a − 1

ta − 1Φ

(∂r4

∂x2

)− tm21a tn21b − 1

tb − 1Φ

(∂r4

∂x1

) ]det (E1345) =

tm21d − 1

td − 1

tn22e − 1

te − 1

[tm11a − 1

ta − 1Φ

(∂r5

∂x3

)+

tn11c − 1

td − 1Φ

(∂r5

∂x1

) ]det (E1346) =

tn11c − 1

tc − 1

tn12f − 1

tf − 1

[tm21a − 1

ta − 1Φ

(∂r4

∂x4

)+

tm21d − 1

td − 1Φ

(∂r4

∂x1

) ]det (E1456) = − tm11a − 1

ta − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1

det (E2345) =tm21d − 1

td − 1

tn22e − 1

te − 1

[tm11a tn11b − 1

tb − 1Φ

(∂r5

∂x3

)+

tn11c − 1

tc − 1Φ

(∂r5

∂x2

) ]det (E2346) = − tn11c − 1

tc − 1

tn12f − 1

tf − 1

[tm21a tn21b − 1

tb − 1Φ

(∂r4

∂x4

)− tm21d − 1

td − 1Φ

(∂r4

∂x2

) ]det (E2356) =

tn21b − 1

tb − 1

tn11c − 1

tc − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1tm21a

det (E2456) = − tn11b − 1

tb − 1

tn11c − 1

tc − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1tm11a

det (E3456) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn12f − 1

tf − 1

Concluimos entao que

56

∆2 (t) = mdc { 31 menores nao nulos acima }.

Caso 2: Seja f = f1f2 com as seguintes representacoes por series de Puiseux:{y = x

m1n1 + x

m12n1n12

y = xm1n1 + x

m22n1n22

onde {(m1, n1), (mi2, ni2)} sao os pares de Puiseux do ramo fi para i = 1, 2. Isto

conduz a seguinte arvore:

Figura 11:

Com as notacoes da Secao 1.3,µφ = m1, νφ = n1 e m1rφ = n1sφ + 1µ1 = w22, ν1 = n22 e w22r1 = n22s1 + 1µ10 = w12, ν10 = n12 e w12r10 = n12 s10 + 1

Portanto, o grupo do link π1 (X), e gerado por αz, βz, com z percorrendo o

conjunto dos vertices V = {φ, 1, 10, 2, 3} da arvore Ξ, satisfazendo as relacoes que


Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ


φ βrφ

φ )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 αsφ


57

Relacoes de (1):


2 β2 = αµ1

1 βν11

⇒ β2 = α−w22 n222 αw22

1 (α−m1n11 αm1

φ )n22

(R2) 10 > 1 e 10 e uma haste ⇒ αµ1

1 βν11 = αµ1

10 βν110

⇒ αµ1

1 (α−m1n11 αm1

φ )n22 = αw2210 βn22

10


1 = αs110 βr1

10

⇒ α2 = αs110 βr1

10 β−r11 α−s1

1

Relacoes de (10):

(R1) 3 > 10 e 3 nao e haste ⇒ αν10µ10

3 β3 = αµ10

10 βν1010

⇒ β3 = α−w12 n123 αw12

10 βn1210


10 )ν10 = (αµ10

10 βν1010 )r10

⇒ (α3 αs1010 βr10

10 )n12 = (αw1210 βn12

10 )r10



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α10, β10] = 1 ⇒ [α2, β2] = 1 veja (B.10)

[α2, β2] = 1

[ α3, β3 ] = 1 ⇔ [ α w1210 βn12

10 , α3 ] = 1 veja (B.11)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1, α3, α10 e β10, sujeitos

as relacoes:

(1) (α1 αsφ


(2) [ αm1φ , α1 ] = 1

(3) αm22−m1n121 αm1n22

φ = αw2210 βn22

10

58

(4) (α3 αs1010 βr10

10 )n12 = (αw1210 βn12

10 )r10

(5) [ α w1210 βn12

10 , α3 ] = 1

(6) [α10, β10] = 1



ramos da curva: {m1 = n1 λ1 + 1 para algum λ1 ∈ ZZµ10 = n12 λ3 + 1 para algum λ3 ∈ ZZ.

o que implica que rφ = r10 = 1. Adotaremos como novos geradores:x1 = α10,x2 = β10,x3 = α1 α

sφ

φ ,

{x4 = αφ,x5 = α3 αs10

10 βr1010 .

Observe que a relacao (2), via mudanca, se torna a relacao [xm11 , x2 x

−sφ

1 ] = 1

que e equivalente a relacao [xm11 , x2] = 1. Sendo esta, trivializada devido a nova

relacao (1) xn12 = xm1

1 ; e que a relacao (5), via esta mudanca de geradores, se torna

a relacao [αw1210 βn12

10 , x5] = 1, a qual e trivializada pela relacao (4). Assim,

π1 (X) = { x1, x2, x3, x4, x5; r1, r2, r3, r4 }, onder1 = x1 x2 x−1

1 x−12

r2 = xw121 xn12

2 x−n125

r3 = xn13 x−m1

4

r4 = xw221 xn22

2 x−m1n224 (x3 x

−sφ

4 )−(m22−m1n12)



pela fibracao de Milnor. Temporariamente escrevemos:

59

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7→ ta

x2 7→ tb

x3 7→ tc

x4 7→ td

x5 7→ te

Note que as relacoes ri, para i = 2, 3, 4, fornecem relacoes entre a, b, c, d; a saber:

(∗)

n12 e = w12 a + n12 bm1 d = n1 c( c− sφd ) ( m22 − m1 n12 ) = w22 a + n22 b − m1 n22 d


(∂ri

∂xj

).

Como r1 = x1 x2 x−11 x−1

2 , temos que

∂r1

∂x1

= 1−x1 x2 x−11 ⇒ Φ

(∂r1

∂x1

)= 1 − Φ (x2) = − (tb − 1)

∂r1

∂x2

= x1−x1 x2 x−11 x−1

2 ⇒ Φ

(∂r1

∂x2

)= Φ (x1) − 1 = ta − 1

∂r1

∂x3

= 0 ⇒ Φ

(∂r1

∂x3

)= 0

∂r1

∂x4

= 0 ⇒ Φ

(∂r1

∂x4

)= 0

∂r1

∂x5

= 0 ⇒ Φ

(∂r1

∂x5

)= 0

Como r2 = xw121 xn12


∂r2

∂x1

=xw12

1 − 1

x1 − 1⇒ Φ

(∂r2

∂x1

)=

tw12a − 1

ta − 1

∂r2

∂x2

= xw121

xn122 − 1

x2 − 1⇒ Φ

(∂r2

∂x2

)= tw12a tn12b − 1

tb − 1

∂r2

∂x3

= 0 ⇒ Φ

(∂r2

∂x3

)= 0

∂r2

∂x4

= 0 ⇒ Φ

(∂r2

∂x4

)= 0

∂r2

∂x5

= − xn125 − 1

x5 − 1⇒ Φ

(∂r2

∂x5

)= − tn12e − 1

te − 1

60

Como r3 = xn13 x−m1

4 , temos que

∂r3

∂x1

= 0 ⇒ Φ

(∂r3

∂x1

)= 0

∂r3

∂x2

= 0 ⇒ Φ

(∂r3

∂x2

)= 0

∂r3

∂x3

=xn1

3 − 1

x3 − 1⇒ Φ

(∂r3

∂x3

)=

tn1c − 1

tc − 1

∂r3

∂x4

= − xm14 − 1

x4 − 1⇒ Φ

(∂r3

∂x4

)= − tm1d − 1

td − 1

∂r3

∂x5

= 0 ⇒ Φ

(∂r3

∂x5

)= 0


2 x−m1n224 (x3 x

−sφ

4 )−(m22−m1n12), temos que

∂r4

∂x1

=xw22

1 − 1

x1 − 1⇒ Φ

(∂r4

∂x1

)=

tw22a − 1

ta − 1

∂r4

∂x2

= xw221

xn222 − 1

x2 − 1⇒ Φ

(∂r4

∂x2

)= tw22a tn22b − 1

tb − 1

∂r4

∂x3

= − (x3 x−sφ

4 )(m22−m1n12) − 1

(x3 x−sφ

4 ) − 1veja (B.12)

⇒ Φ

(∂r4

∂x3

)= − t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1

∂r4

∂x4

= − xw221 xn22

2 x−m1n224

xm1n224 − 1

x4 − 1veja (B.13)

+ (x3 x−sφ

4 )m22−m1n12x

sφ

4 − 1

x4 − 1(x3 x

−sφ

4 )−(m22−m1n12−1) (x3 x−sφ

4 )m22−m1n12 − 1

(x3 x−sφ

4 ) − 1

⇒ Φ

(∂r4

∂x4

)= − tw22a+n22b−m1n22d tm1n22d − 1

td − 1+ tc−sφd tsφd − 1

td − 1

t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1

∂r4

∂x5

= 0 ⇒ Φ

(∂r4

∂x5

)= 0


Φ

(∂r

∂x

)=

−(tb − 1) (ta − 1) 0 0 0

tw12a−1ta −1

tw12a tn12b−1tb −1

0 0 − tn12e−1te −1

0 0 tn1c−1tc −1

− tm1d−1td −1

0tw22a−1ta −1


− t(c−sφd)(m22−m1n12)−1

t(c−sφd) −1

Φ ( ∂r4

∂x4) 0

.

61


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4), det (A5) },


(∂r

∂x


det (A1) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(ta−1) veja (B.16)

det (A2) = − tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(tb−1) veja (B.16)

det (A3) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(tc−1) veja (B.14)

det (A4) = − tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(td−1) veja (B.14)

det (A5) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(te−1) veja (B.17)

Logo,

∆1 (t) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1mdc { (ta−1), (tb−1), (tc−1), (td−1), (te−1) }.


m1n1 + x

m12n1n12

y2 = xm1n1 + x

m22n1n22

segue da Secao 1.2, com i = 1, s1 = 2, e s2 = 2, que

∆1 (t) =tm1 e01 − 1

te01 − 1

tw22e02 − 1

tm1 e02 − 1

tn12e12 − 1

te12 − 1(t− 1)

onde

w11 = m1

w12 = m12 −m1 n12 + m1 n1 n12

w22 = m22 −m1 n22 + m1 n1 n22

e01 = b1,1,2 + b2,1,2 = (∏2

j=1 n1,j) + (∏2

j=1 n2,j) = n11n12 + n21n22 = n1(n12 + n22)

e02 = b1,2,2 + b2,2,2 = n12 + n22

e12 = w2,2.b2,3,2.b1,2,1 + w12.b2,3,2 = w22 + w12


(tr − 1) = mdc {(ta − 1), (tb − 1), (tc − 1), (td − 1), (te − 1)},

62

segue que

tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1(tr−1) =

tm1 e01 − 1

te01 − 1

tw22e02 − 1

tm1 e02 − 1

tn12e12 − 1

te12 − 1(t−1)

Portanto,

(tm1d − 1) (tw22a+n22b − 1) (tn12e − 1)(te01 − 1) (tm1 e02 − 1) (te12 − 1) (tr − 1)

= (tm1e01 − 1) (tw22n12+n22w22 − 1) (tn12e12 − 1) (td − 1) (tc − 1) (te − 1) (t− 1)


r = 1. Simplificando (t−1) e notando que a possibilidade d ≥ c ou c ≥ e geram um

absurdo, em vista de (∗∗) e (∗), temos que d < c < e. Analisando, na igualdade

resultante, os termos de menor grau em t, temos qued = e01 = n1 (n12 + n22)c = m1 e02 = m1 (n12 + n22)a = n12

b = w22

e = w12 + w22


∆2 (t) = mdc

det (A123), det (A124), det (A125), det (A134), det (A135),det (A145), det (A234), det (A235), det (A245), det (A345),det (B123), det (B124), det(B125), det (B134), det (B135),det (B145), det (B234), det(B235), det (B245), det (B345),det (C123), det (C124), det (C125), det (C134), det (C135),det (C145), det (C234), det (C235), det (C245), det (C345),det (D123), det (D124), det (D125), det (D134), det (D135),det (D145), det (D234), det (D235), det (D245), det (D345)

onde A, B, C, D sao obtidas de Φ

(∂r

∂x

)retirando-se a quarta, terceira, segunda

e primeira linhas respectivamente, e onde Aijk, Bijk, Cijk e Dijk denotam as matri-

zes formadas pelas colunas i, j e k, de A, B, C e D, respectivamente.

Temos que

A =

−(tb − 1) (ta − 1) 0 0 0tw12a−1ta −1


0 0 − tn12e−1te −1

0 0 tn1c−1tc −1

− tm1d−1td −1

0

.

det (A123) = − tn1c − 1

tc − 1

tn12e − 1

te − 1(te−1)

63

det (A124) =tm1d − 1

td − 1

tn12e − 1

te − 1(te−1)

det (A125) = 0

det (A134) = 0

det (A135) = − tn1c − 1

tc − 1

tn12e − 1

te − 1(tb−1)

det (A145) =tm1d − 1

td − 1

tn12e − 1

te − 1(tb−1)

det (A234) = 0

det (A235) =tn1c − 1

tc − 1

tn12e − 1

te − 1(ta−1)

det (A245) = − tm1d − 1

td − 1

tn12e − 1

te − 1(ta−1)

det (A345) = 0

Temos que

B =

−(tb − 1) (ta − 1) 0 0 0tw12a−1ta −1


0 0 − tn12e−1te −1

tw22a−1ta −1


− t(c−sφd)(m22−m1n12)−1

t(c−sφd) −1

Φ ( ∂r4

∂x4) 0

.

det (B123) =t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1(tn12e−1)

det (B124) = − Φ

(∂r4

∂x4

)(tn12e−1)

det (B125) = − tn12e − 1

te − 1(tw22a+n22b−1)

det (B134) = 0

det (B135) =tn12e − 1

te − 1

t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1(tb−1)

det (B145) = − tn12e − 1

te − 1Φ

(∂r4

∂x4

)(tb−1)

det (B234) = 0

det (B235) = − tn12e − 1

te − 1

t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1(ta−1)

det (B245) =tn12e − 1

te − 1Φ

(∂r4

∂x4

)(ta−1)

det (B345) = 0

64

Temos que

C =

−(tb − 1) (ta − 1) 0 0 0

0 0 tn1c−1tc −1

− tm1d−1td −1

0tw22a−1ta −1


− t(c−sφd)(m22−m1n12)−1

t(c−sφd) −1

Φ ( ∂r4

∂x4) 0

.

det (C123) =tn1c − 1

tc − 1(tw22a+n22b−1)

det (C124) = − tm1d − 1

td − 1(tw22a+n22b−1)

det (C125) = 0

det (C134) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1(tb−1)

det (C135) = 0

det (C145) = 0

det (C234) = − tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1(ta−1)

det (C235) = 0

det (C245) = 0

det (C345) = 0


D =

tw12a−1ta −1


0 0 − tn12e−1te −1

0 0 tn1c−1tc −1

− tm1d−1td −1

0tw22a−1ta −1


− t(c−sφd)(m22−m1n12)−1

t(c−sφd) −1

Φ ( ∂r4

∂x4) 0

.

det (D123) =tn1c − 1

tc − 1

[tw22a tn22b − 1

tb − 1

tw12a − 1

ta − 1− tw12a tn12b − 1

tb − 1

tw22a − 1

ta − 1

]det (D124) = − tm1d − 1

td − 1

[tw22a tn22b − 1

tb − 1

tw12a − 1

ta − 1− tw12a tn12b − 1

tb − 1

tw22a − 1

ta − 1

]det (D125) = 0

det (D134) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tw12a − 1

ta − 1

det (D135) =tn12e − 1

te − 1

tn1c − 1

tc − 1

tw22a − 1

ta − 1

det (D145) = − tn12e − 1

te − 1

tm1d − 1

td − 1

tw22a − 1

ta − 1

65

det (D234) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1tw12a tn12b − 1

tb − 1

det (D235) =tn12e − 1

te − 1

tn1c − 1

tc − 1tw22a tn12b − 1

tb − 1

det (D245) = − tn12e − 1

te − 1

tm1d − 1

td − 1tw22a tn22b − 1

tb − 1

det (D345) =tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

tn12e − 1

te − 1

Concluimos entao que

∆2 (t) = mdc { 26 menores nao nulos acima }

Caso 3: Seja f = f1f2 com as seguintes representacoes por series de Puiseux:{y1 = x

m1n1 + x

m12n1n12

y2 = xm1n1 + x

m22n1 + x

m23n1n23

onde {(m1, n1), (m12, n12)} sao os pares de Puiseux do ramo f1 e {(m1, n1), (m23, n23)}sao os pares de Puiseux do ramo f2. Isto conduz a arvore, dada na Figura 12.

Figura 12:

Com as notacoes da Secao 1.3,µφ = m1, νφ = n1 e m1rφ = n1sφ + 1µ1 = w22 = m22 −m1n22 + m1n1n22, ν1 = n22 = 1 e w22r1 = n22s1 + 1µ10 = w12 = m12 −m1n12 + m1n1n12, ν10 = n12 e w12r10 = n12 s10 + 1µ11 = w23 = m23 ν11 = n23 e w23r11 = n23s11 + 1

66

Note que (r1, s1) = (1, µ1 − 1). Portanto, o grupo do link π1 (X), e gerado por

αz, βz, com z percorrendo o conjunto dos vertices V = {φ, 1, 10, 11, 2, 3} da arvore

Ξ, satisfazendo as relacoes que exibiremos a seguir.

Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ


φ βrφ

φ )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 αsφ


Relacoes de (1):

(R1) 11 > 1 e 11 nao e haste ⇒ αν1µ1

11 β11 = αµ1

1 βν11

⇒ β11 = α−µ1

11 αµ1

1 β1

(R2) 10 > 1 e 10 e uma haste ⇒ αµ1

1 βν11 = αµ1

10 βν110

⇒ β10 = α−µ1

10 αµ1

1 β1


1 = αs110 βr1

10

⇒ α11 = αs110 βr1

10 β−r11 α−s1

1

Relacoes de (10):

(R1) 3 > 10 e 3 nao e haste ⇒ αν10µ10

3 β3 = αµ10

10 βν1010

⇒ β3 = α−w12 n123 αw12

10 βn1210


10 )ν10 = (αµ10

10 βν1010 )r10

⇒ (α3 αs1010 βr10

10 )n12 = (αµ10

10 βn1210 )r10

Relacoes de (11):

(R1) 2 > 11 e 2 nao e haste ⇒ αν11 µ11

2 β2 = αµ11

11 βν1111

⇒ β2 = α−µ11 n23

2 αµ11

11 βn2311

67


11 )ν11 = (αµ11

11 βν1111 )r11

⇒ (α2 αs1111 βr11

11 )n23 = (αµ11

11 βn2311 )r11



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α10, β10] = 1 ⇔ [ αµ1

1 β1 , α10 ] = 1 veja (B.18)

[α11, β11] = 1 ⇔ [ αµ1

1 β1 , α10 ] = 1 veja (B.19)

[α2, β2] = 1 ⇔ [ αµ1

11 βn2311 , α2 ] = 1 veja (B.20)

[α3, β3] = 1 ⇔ [ α w1210 βn12

10 , α3 ] = 1 veja (B.21)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1, β1, α2, α3, α10, β10, α11,


(1) (α1 αsφ


(2) [ αm1φ , α1 ] = 1

(3) (α2 αs1111 βr11

11 )n23 = (αµ11

11 βn2311 )r11

(4) [ α µ11

11 βn2311 , α2 ] = 1

(5) (α3 αs1010 βr10

10 )n12 = (αµ10

10 βn1210 )r10

(6) [ α w1210 βn12

10 , α3 ] = 1

(7) [αµ1

1 β1, α10] = 1


somos conduzidos a fazer as seguintes restricoes:m1 = λ1 n1 + 1 para algum λ1 ∈ ZZµ11 = λ2 n23 + 1 para algum λ2 ∈ ZZµ10 = λ3 n12 + 1 para algum λ3 ∈ ZZ

o que implica que rφ = r10 = r11 = 1. Adotaremos como novos geradores,x1 = αφ,x2 = α1 α

sφ

φ ,

x3 = α2 αs1111 βr11

11 ,

{x4 = α3 αs10

10 βr1010 ,

x5 = α10.

68

Observe que a relacoes (2), (4) e (6) sao equivalentes as relacoes

[ xm11 , x2 ] = 1, [ αµ11

11 βn2311 , x3 ] = 1 e [ αw12

10 βn1210 , x4 ] = 1

respectivamente, e que as mesmas sao trivializadas pelas novas relacoes (1′), (3′) e

(5′), obtidas de (1), (3) e (5) respectivamente, apos a troca dos geradores. Temos

que π1 (X) e gerado por x1, x2, x3, x4 e x5, sujeitos as seguintes relacoes:

(1) xn12 = xm1

1

(2) xn233 = x

−(µ11−µ1n23)5 ( x2 x

−sφ

1 )µ11−m1 n1 n23 x1m1 n23 veja (B.22)

(3) xn124 = xµ10−µ1n12

5 ( x2 x−sφ

1 )n12(m22−m1) xm1n121 veja (B.23)

(4) [ ( x2 x−sφ

1 )m22−m1 xm11 , x5 ] = 1



pela fibracao de Milnor. Temporariamente escrevemos:

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7→ ta

x2 7→ tb

x3 7→ tc

x4 7→ td

x5 7→ te


faremos uma nova mudanca de geradores; a saber, y1 = x2 x−sφ

1 ,y2 = x1,y3 = x3,

{y4 = x4,y5 = x5.

Temos que

π1 (X) = { y1, y2, y3, y4 ; r1, r2, r3 }, onder1 = ( y1 y

sφ

2 )n1 y−m12

r2 = y−(µ11−µ1n23)5 yµ11−m1n1n23

1 ym1n232 y−n23

3

r3 = yµ10−µ1n12

5 yn12(m22−m1)1 ym1n12

2 y−n124

r4 = ym22−m11 ym1

2 y5 y−m12 y

−(m22−m1)1 y−1

5

69

Consequentemente, a aplicacao Φ nos novos geradores, e dada por


y2 7→ ta

y3 7→ tc

y4 7→ td

y5 7→ te

Note que as relacoes ri, para i = 1, 2, fornecem as seguintes relacoes entre a, b, c:

(∗)

m1 a = n1 bn23 c = −(µ11 − µ1 n23) e + (µ11 −m1 n1 n23) (b− sφa) + m1 n23 an12 d = (µ10 − µ1n12) e + n12 (m22 −m1) (b− sφa) + m1 n12 a


(∂ri

∂yj

).

Como r1 = ( y1 ysφ

2 )n1 y−m12 , temos que

∂r1

∂y1

=(y1 y

sφ

2 )n1 − 1

(y1 ysφ

2 ) − 1⇒ Φ

(∂r1

∂y1

)=

tn1b − 1

tb − 1

∂r1

∂y2

= y1y

sφ

2 − 1

y2 − 1

(y1 ysφ

2 )(n1−1) − 1

(y1 ysφ

2 ) − 1− (y1 y

sφ

2 )n1−1 y1 y−(m1−sφ)2

y(m1−sφ)2 − 1

y2 − 1

⇒ Φ

(∂r1

∂y2

)= − tm1a − 1

ta − 1

tb−sφa − 1

tb − 1veja (A.19)

∂r1

∂y3

= 0 ⇒ Φ

(∂r1

∂y3

)= 0

∂r1

∂y4

= 0 ⇒ Φ

(∂r1

∂y4

)= 0

∂r1

∂y5

= 0 ⇒ Φ

(∂r1

∂y5

)= 0

Como r2 = y−(µ11−µ1n23)5 yµ11−m1n1n23

1 ym1n232 y−n23

3 , temos que

∂r2

∂y1

= y−(µ11−µ1n23)5

yµ11−m1n1n23

1 − 1

y1 − 1

⇒ Φ

(∂r2

∂y1

)= t−(µ11−µ1n23)e t(µ11−m1n1n23)(b−sφa) − 1

t(b−sφa) − 1

∂r2

∂y2

= y−(µ11−µ1n23)5 yµ11−m1n1n23

1

ym1n232 − 1

y2 − 1

70

⇒ Φ

(∂r2

∂y2

)= t−(µ11−µ1n23)e +(µ11−m1n1n23)(b−sφa) tm1n23a − 1

ta − 1

∂r2

∂y3

= − y−(µ11−µ1n23)5 yµ11−m1n1n23

1 ym1n232 y−n23

3

yn233 − 1

y3 − 1⇒ Φ

(∂r2

∂y3

)= − tn23c − 1

tc − 1

∂r2

∂y4

= 0 ⇒ Φ

(∂r2

∂x4

)= 0

∂r2

∂y5

= −y−(µ11−µ1n23)5

y(µ11−µ1n23)5 − 1

y5 − 1⇒ Φ

(∂r2

∂y5

)= − t−(µ11−µ1n23)e t(µ11−µ1n23)e − 1

te − 1

Como r3 = yµ10−µ1n12

5 yn12(m22−m1)1 ym1n12

2 y−n124 , temos que

∂r3

∂y1

= yµ10−µ1n12

5

yn12(m22−m1)1 − 1

y1 − 1

⇒ Φ

(∂r3

∂y1

)= t(µ10−µ1n12)e tn12(m22−m1)(b−sφa) − 1

t(b−sφa) − 1

∂r3

∂y2

= yµ10−µ1n12)5 y

n12(m22−m1)1

ym1n122 − 1

y2 − 1

⇒ Φ

(∂r3

∂y2

)= t(µ10−µ1n12)e + n12(m22−m1)(b−sφa) tm1n12a − 1

ta − 1

∂r3

∂y3

= 0 ⇒ Φ

(∂r3

∂y3

)= 0

∂r3

∂y4

= − y(µ10−µ1n12)5 y

n12(m22−m1)1 ym1n12

2 y−n123

yn124 − 1

y4 − 1⇒ Φ

(∂r3

∂y4

)= − tn12d − 1

td − 1

∂r3

∂y5

=y

(µ10−µ1n12)5 − 1

y5 − 1⇒ Φ

(∂r3

∂y5

)=

t(µ10−µ1n12)e − 1

te − 1

Como r4 = ym22−m11 ym1

2 y5 y−m12 y

−(m22−m1)1 y−1

5 , temos que

∂r4

∂y1

=ym22−m1

1 − 1

y1 − 1− ym22−m1

1 ym12 y5 y−m1

2 y−(m22−m1)1

ym22−m11 − 1

y1 − 1

⇒ Φ

(∂r4

∂y1

)= − t(m22−m1)(b−sφa) − 1

tb−sφa − 1(te−1)

∂r4

∂y2

= ym22−m11

ym12 − 1

y2 − 1− ym22−m1

1 ym12 y5 y−m1

2

ym12 − 1

y2 − 1

⇒ Φ

(∂r4

∂y2

)= − t(m22−m1)(b−sφa) tm1a − 1

ta − 1(te−1)

∂r4

∂y3

= 0 ⇒ Φ

(∂r4

∂y3

)= 0

71

∂r4

∂y4

= 0 ⇒ Φ

(∂r4

∂y4

)= 0

∂r4

∂y5

= ym22−m11 ym1

2 − ym22−m11 ym1

2 y5 y−m12 y

−(m22−m1)1 y−1

5

⇒ Φ

(∂r4

∂y5

)= t(m22−m1)(b−sφa)+m1a−1


Φ

(∂r

∂y

)=

tn1b−1tb −1

− tm1a−1ta −1


0 0 0

Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) − tn23c−1

tc −10 −t−(µ11−µ1n23)e t(µ11−µ1n23)e−1

te −1

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) 0 − tn12d−1

td −1t(µ10−µ1n12)e−1te −1

Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2) 0 0 t(m22−m1)(b−sφa)+m1a − 1

.


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4), det (A5) }


(∂r

∂y


det (A1) = − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(tb−sφa−1)

det (A2) =tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(ta−1)

det (A3) = − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(tc−1) veja (B.24)

det (A4) =tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(td−1) veja (B.25)

det (A5) = − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(te−1) veja (B.26)

Logo,

∆1 (t) =tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1

mdc { (ta − 1), (tb−sφa − 1), (tc − 1), (td − 1), (te − 1) }.


m1n1 + x

m12n1n12

y2 = xm1n1 + x

m22n1 + x

m23n1n23

72


∆1 (t) = (t− 1)tw11e01 − 1

tw11e02 − 1

tw22e02 − 1

te01 − 1

tn12e12 − 1

te12 − 1

tn23e23 − 1

te23 − 1.

onde

w11 = m1

w12 = m12 −m1n12 + m1n1n12

w22 = m22 −m21n22 + m21n21n22 = m22 + m1 (n1 − 1)w23 = m23 −m22n23 + m22n22n23 = m23

e01 = b1,1,2 + b2,1,3 = n1 n12 + n1 n22 n23 = n1 (n12 + n23)e02 = b1,2,2 + b2,2,3 = n12 + n23

e12 = w2,2.b2,3,3.b1,2,1 + w1,2.b1,3,2 = w22 n23 + w12

e23 = w2,2.b1,2,2.b1,3,2 + w2,3.b2,4,3 = w22.n12 + w23


(tr − 1) = mdc{(ta − 1), (tb−sφa − 1), (tc − 1), (td − 1), (te − 1)}

segue que

(tr − 1)tm1a − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1

= (t− 1)tw11e01 − 1

tw11e02 − 1

tw22e02 − 1

te01 − 1

tn12e12 − 1

te12 − 1

tn23e23 − 1

te23 − 1.

Portanto,

(tr − 1) (tm1a − 1) (tm23c − 1) (tn12d − 1) (t(m22−m1)(b−sφa)+m1a − 1)

(tm1e02 − 1) (te01 − 1) (te12 − 1) (te23 − 1)

= (t− 1) (tm1e01 − 1) (tw22e02 − 1) (tn12e12 − 1) (tn23e23 − 1) (∗∗)

(ta − 1) (tb − 1) (tc − 1)(tc − 1)


r = 1. Simplificando (t− 1) e analisando na igualdade resultante, em vista de (∗),os termos de menor grau em t, temos que:

a = e01 = n1(n12 + n23)

Em vista das relacoes (∗), concluimos que:b = m1 (n12 + n23)b− sφa = n12 + n23

c = e23 = w22n12 + w23

d = e12 = w22n23 + w12

e = n12

73

Assim, determinamos a matriz de Alexander, com a qual podemos calcular ex-

plicitamente ∆2 (t), e portanto, o polinomio mınimo da monodromia neste caso.

Caso 4: Seja f = f1f2, com as seguintes representacoes por series de Puiseux:{y1 = x

m11n11 + x

m12n11n12

y2 = xm21n21 + x

m22n22 + x

m23n22n23

onde {(m1i, n1i)} sao os pares de Puiseux do ramo f1 para i = 1, 2 e {(m2j, n2j)}sao os pares de Puiseux do ramo f2 para j = 2, 3. Isto conduz a seguinte arvore:

Figura 13:

Com as notacoes da Secao 1.3,µφ = m21, νφ = n21 = 1 e m21rφ = n21sφ + 1µ0 = m11, ν0 = n11 e m11r0 = n11s0 + 1µ1 = m22, ν1 = n22 e m22r1 = n22s1 + 1µ11, ν11 = n23 e µ11r11 = n23s11 + 1µ01, ν01 = n12 e µ01r01 = n12s01 + 1

Portanto, o grupo do link π1 (X) e gerado por αz, βz, com z percorrendo o con-

junto dos vertices V = {φ, 0, 01, 1, 11, 2, 3} da arvore Ξ, satisfazendo as relacoes

que exibiremos a seguir.

Relacoes de (φ):


1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m21 n211 αm21

φ


0 βνφ

0 = αµφ

φ βνφ

φ

⇒ β10 = α−m21

0 αm21φ

74


φ βrφ

φ = αsφ

0 βrφ

0

⇒ α1 = αsφ

0 βrφ

0 α−sφ

φ

Relacoes de (0):

(R1) 01 > 0 e 01 nao e haste ⇒ αν0µ0

01 β01 = αµ0

0 βν00

⇒ β01 = α−m11 n1101 αm11

0 βn110


0 )ν0 = (αµ0

0 βν00 )r0

⇒ (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

Relacoes de (1):

(R1) 11 > 1 e 1 nao e haste ⇒ αν1 µ1

11 β11 = αµ1

1 βν11

⇒ β11 = α−µ1 n22

11 αµ1

1 βn221


1 )ν1 = (αµ1

1 βν11 )r1

⇒ (α11 αs11 βr1

1 )n22 = (αm221 βn22

1 )r1

Relacoes de (11):

(R1) 2 > 11 e 11 nao e haste ⇒ αν11 µ11

2 β2 = αµ11

11 βν1111

⇒ β2 = α−µ11 n23

2 αµ11

11 βn2311


11 )ν11 = (αµ11

11 βν1111 )r11

⇒ (α2 αs1111 βr11

11 )n23 = (αµ11

11 βn2311 )r11

Relacoes de (01):

(R1) 3 > 01 e 01 nao e haste ⇒ αν01 µ01

3 β3 = αµ01

01 βν0101

⇒ β3 = α−µ01 n12

3 αµ01

01 βn1201


01 )ν01 = (αµ01

01 βν0101 )r01

⇒ (α3 αs0101 βr01

01 )n12 = (αµ01

01 βn1201 )r01

75



[α0, β0] = 1 ⇒ [α1, β1] = 1 veja (A.1)

[α1, β1] = 1

[α01, β01] = 1 ⇔ [ αm110 βn11

0 , α01 ] = 1 veja (A.2)

[α11, β11] = 1 ⇔ [ αµ1

1 βn221 , α11 ] = 1 veja (B.27)

[α2, β2] = 1 ⇔ [ αµ11

11 βn2311 , α2 ] = 1 veja (B.28)

[α3, β3] = 1 ⇔ [ αµ01

01 βn1201 , α3 ] = 1 veja (B.29)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1, β1, α0, β0, α2, α3, α11, β11, α01,


(1) β1 = α−m21 n211 αm21

φ

(2) α1 = αsφ

0 βrφ

0 α−sφ

φ

(3) [α0, β0] = 1

(4) αm210 βn21

0 = αm21φ

(5) (α01 αs00 βr0

0 )n11 = (αm110 βn11

0 )r0

(6) [ αm110 βn11

0 , α01 ] = 1

(7) (α11 αs11 βr1

1 )n22 = (αm221 βn22

1 )r1

(8) [ αµ1

1 βn221 , α11 ] = 1

(9) (α2 αs1111 βr11

11 )n23 = (αµ11

11 βn2311 )r11

(10) [ αµ11

11 βn2311 , α2 ] = 1

(11) (α3 αs0101 βr01

01 )n12 = (αµ01

01 βn1201 )r01

(12) [ αµ01

01 βn1201 , α3 ] = 1

76



ramos da curva: m11 = λ1 n11 + 1 para algum λ1 ∈ ZZm22 = λ2 n22 + 1 para algum λ2 ∈ ZZµ11 = λ3 n23 + 1 para algum λ3 ∈ ZZµ01 = λ4 n12 + 1 para algum λ4 ∈ ZZ

o que implica que r0 = r1 = r11 = r01 = 1. Adotaremos como novos geradores:x1 = α0,x2 = β0,x3 = α01 αs0

0 βr00 ,

x4 = αφ,x5 = α1,

x6 = β1,x7 = α11 αs1

1 βr11 ,

x8 = α2 αs1111 βr11

11 ,x9 = α3 αs01

01 βr0101 .

Observe que as relacoes (6), (8), (10) e (12) sao equivalentes a{(6′) [ αm11

0 βn110 , x3 ] = 1,

(8′) [ αµ1

1 βn221 , x7 ] = 1,

{(10′) [ αµ11

11 βn2311 , x8 ] = 1

(12′) [ αµ01

01 βn1201 , x9 ] = 1

respectivamente, e que as mesmas sao trivializadas pelas relacoes (5′), (7′), (9′) e

(11′), obtidas de (5), (7), (9) e (11), respectivamente, apos a troca dos geradores.

Ficamos assim, com as seguintes relacoes:

(1) x6 = x−m21 n215 xm21

4

(2) x5 = xsφ

1 xrφ

2 x−sφ

4

(3) [ x1, x2 ] = 1

(4) xm211 xn21

2 = xm214

(5) xn113 = xm11

1 xn112

(6) xn227 = xm22

5 xn226

(7) xn238 = αµ11

11 βn2311

(8) xn229 = αµ01

01 βn1201

Substituindo as relacoes (1) e (2) na relacao (6) acima, podemos elimina-las, e

ficaremos com 7 geradores {x1, x2, x3, x4, x7, x8, x9} e as 6 relacoes a seguir:

77

(1) [ x1, x2 ] = 1

(2) xm211 xn21

2 = xm214

(3) xn113 = xm11

1 xn112

(4) xn227 = (x

sφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21 xm214 ]n22

(5) xn238 = [ x7 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 xm21n22n234 veja (B.30)

(6) xn129 = (x3 x−r0

2 x−s01 )m12−m11n12 xm11n12

1 xn11n122 veja (B.31)

Renomeando x7, x8, x9 por x5, x6, x7, respectivamente, temos que

π1 (X) = { x1, x2, x3, x4, x5, x6, x7; r1, r2, r3, r4, r5, r6 }, onder1 = x1 x2 x−1

1 x−12

r2 = xm111 xn11

2 x−n113

r3 = xm211 xn21

2 x−m214

r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21xm214 ]n22 x−n22

5

r5 = [ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 xm21n22n234 x−n23

6

r6 = (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11n12

2 x−n127

Observe que as quatro primeiras relacoes acima, sao as mesmas que aparecem

na apresentacao do grupo do link para curvas com dois ramos distintos de genero

1, como descrito no Caso 2, da Secao 2.1 (veja pagina 23). Portanto, a matriz de

Alexander deste link, tera um bloco consistindo da matriz de Alexander obtida na

Secao 2.1.


determinar Φ (xi) ∈ ZZ [ t, t−1 ] para i = 1, . . . , 7. Temporariamente, escrevemos

78

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]

x1 7→ ta

x2 7→ tb

x3 7→ tc

x4 7→ td

x5 7→ te

x6 7→ tf

x7 7→ tg

Note que as relacoes ri, para i = 2, 3, 4, 5, 6, fornecem as relacoes entre a,b,c,d,e,f,g:

(∗)

n11 c = m11 a + n11 bm21 d = m21 a + n21 bn22 e = (sφa + rφb− sφd)m22 + [(sφa + rφb− sφd)(−m21n21) + m21d]n22

n23 f = [ e + (sφ a + rφ b − sφ d) (m21 n21)−m21 d+ (sφ a + rφ b − sφ d) (−s1) ] (m23 −m22 n23)+ (sφ a + rφ b − sφ d) (m22 −m21 n21 n22) n23 + m21 n22 n23 d

n12 g = (c− r0 b− s0 a) (m12 −m11 n12) + m11 n12 a + n11 n12 b


(∂ri

∂xj

).

Pela observacao que fizemos acima, bastara derivar as relacoes r5 e r6.

Como

r5 = [ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 xm21n22n234 x−n23

6 ,

temos que

∂r5

∂x1

= [ x5x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

1 − 1

x1 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23x

sφ

1 − 1

x1 − 1

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

( xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (B.32)

79

⇒ Φ

(∂r5

∂x1

)= [ te

tsφa − 1

ta − 1

t(sφa+rφb−sφd)(m21n21) − 1

t(sφa+rφb−sφd) − 1− te+(sφa+rφb−sφd)(m21n21)−m21d

tsφa − 1

ta − 1t(sφa+rφb−sφd)(−s1) t(sφa+rφb−sφd)s1 − 1

tsφa+rφb−sφd − 1]

t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) − 1

t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] − 1

+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) tsφa − 1

ta − 1

t(sφa+rφb−sφd) (m22−m21n22)n23 − 1


∂r5

∂x2

= [ x5 xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

1

xrφ

2 − 1

x2 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 xsφ

1

xrφ

2 − 1

x2 − 1

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

( xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (B.33)

⇒ Φ

(∂r5

∂x2

)= [ te+sφa trφb − 1

tb − 1


tsφa+rφb−sφd − 1− te+(sφa+rφb−sφd)(m21n21)−m21d

tsφa trφb − 1

tb − 1t(sφa+rφb−sφd) (−s1) t(sφa+rφb−sφd)(s1) − 1

t(sφa+rφb−sφd) − 1]

t[ e+(sφa+rφb−sφd) (m21n21)−m21d+(sφa+rφb−sφd) (−s1) ](m23−m22n23) − 1

t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd) (−s1) ] − 1

+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d+(sφa+rφb−sφd) (−s1) ](m23−m22n23) tsφa trφb − 1

tb − 1

t(sφa+rφb−sφd)(m22−m21n22)n23 − 1


∂r5

∂x3

= 0 ⇒ Φ

(∂r5

∂x3

)= 0

∂r5

∂x4

= [ − x5 (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1veja (B.34)

− x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xm214 − 1

x4 − 1

80

+ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(s1−1) (xsφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

− [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

(xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23xm21n22n23

4 − 1

x4 − 1

⇒ Φ

(∂r5

∂x4

)= [ − te+sφa+rφb−sφd tsφd − 1

td − 1


tsφa+rφb−sφd − 1

− te+(sφa+rφb−sφd)(m21n21)−m21d tm21d − 1

td − 1+ te+(sφa+rφb−sφd)(m21n21)−m21d

tsφd − 1

td − 1t(sφa+rφb−sφd)(−(s1−1)) t(sφa+rφb−sφd)s1 − 1


t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23) − 1

t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] − 1

− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23)

tsφa+rφb−sφd tsφd − 1

td − 1



+ t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23)

t(sφa+rφb−sφd)(m22−m21n22)n23tm21n22n23d − 1

td − 1

∂r5

∂x5

=[x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

⇒ Φ

(∂r5

∂x5

)=



∂r5

∂x6

= − xn236 − 1

x6 − 1⇒ Φ

(∂r5

∂x6

)= − tn23f − 1

tf − 1

∂r5

∂x7

= 0 ⇒ Φ

(∂r5

∂x7

)= 0

Como r6 = (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11n12


∂r6

∂x1

= − (x3 x−r02 x−s0

1 )xs0

1 − 1

x1 − 1

(x3 x−r02 x−s0

1 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1

81

+ (x3 x−r02 x−s0

1 )m12−m11n12xm11n12

1 − 1

x1 − 1

⇒ Φ

(∂r6

∂x1

)= − tc−r0b−s0a ts0a − 1

ta − 1

t(c−r0b−s0a)(m12−m11n12) − 1

tc−r0b−s0a − 1

+ t(c−r0b−s0a)(m12−m11n12) tm11n12a − 1

ta − 1

∂r6

∂x2

= − x3 x−r02

xr02 − 1

x2 − 1

(x3 x−r02 x−s0

1 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1

+ (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121

xn11n122 − 1

x2 − 1

⇒ Φ

(∂r6

∂x2

)= − tc−r0b tr0b − 1

tb − 1

t(c−r0b−s0a)(m12−m11n12) − 1



tb − 1

∂r6

∂x3

=(x3 x−r0

2 x−s01 )m12−m11n12 − 1

(x3 x−r02 x−s0

1 ) − 1

⇒ Φ

(∂r6

∂x3

)=

t(c−r0b−s0a)(m12−m11n12) − 1


∂r6

∂x4

= 0 ⇒ Φ

(∂r6

∂x4

)= 0

∂r6

∂x5

= 0 ⇒ Φ

(∂r6

∂x5

)= 0

∂r6

∂x6

= 0 ⇒ Φ

(∂r6

∂x6

)= 0

∂r6

∂x7

= − xn127 − 1

x7 − 1⇒ Φ

(∂r6

∂x7

)= − tn12g − 1

tg − 1


Φ

(∂r

∂x

)=

−(tb − 1) (ta − 1) 0 0 0 0 0tm11a−1ta −1


− tn11c−1tc −1

0 0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0 0

Φ ( ∂r4

∂x1) Φ ( ∂r4

∂x2) 0 Φ ( ∂r4

∂x4) − tn22e−1

te −10 0

Φ ( ∂r5

∂x1) Φ ( ∂r5

∂x2) 0 Φ ( ∂r5

∂x4) Φ ( ∂r5

∂x5) − tn23f−1

tf −10

Φ ( ∂r6

∂x1) Φ ( ∂r6

∂x2) Φ ( ∂r6

∂x3) 0 0 0 − tn12g−1

tg −1

.


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4), det (A5), det (A6), det (A7) },

82


(∂r

∂x


det (A1) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(ta−1)

det (A2) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(tb−1)

det (A3) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(tc−1) veja (A.3)

det (A4) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(td−1) veja (A.4)

det (A5) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(te−1) veja (A.13)

det (A6) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(tf−1) veja (B.36)

det (A7) = − tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1(tg−1) veja (B.37)

Logo,

∆1 (t) =tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1

mdc { (ta−1), (tb−1), (tc−1), (td−1), (te−1), (tf−1), (tg−1) }.


m11n11 + x

m12n11n12

y2 = xm21n21 + x

m22n22 + x

m23n22n23

segue da Secao 1.2, com i = 0, s1 = 2 e s2 = 3, que

∆1 (t) = (t− 1)tw21e01 − 1

te01 − 1

tn11e11 − 1

te11 − 1

tn12e12 − 1

te12 − 1

tn22e22 − 1

te22 − 1

tn23e23 − 1

te23 − 1.

onde

w12 = m12 −m11n12 + m11n11n12

w21 = m21

w22 = m22 −m21n22 + m21n21n22 = m22

w23 = m23 −m22n23 + m22n22n23

e01 = b1,1,2 + b2,1,3 = (∏2

j=1 n1,j) + (∏3

j=1 n2,j) = n11 n12 + n21 n22 n23

e11 = w2,1.b2,2,3.b1,1,0 + w1,1.b1,2,2 = m21.n22.n23.1 + m11.n12 = m21 n22 n23 + m11 n12

e12 = w2,1.b2,2,3.b1,1,1 + w1,2.b1,3,2 = m21.n22.n23.n11 + w12.1 = m21 n22 n23 n11 + w12

e22 = w2,1.b1,1,2.b2,2,1 + w22.b2,3,3 = m21.n11.n121 + w22.n23 = m21 n11 n12 + m22 n23

e23 = w2,1.b1,1,2.b2,2,2 + w2,3.b2,4,3 = m21.n11.n12.n22 + w23.1 = m21 n11 n12 n22 + w23

83


(tr − 1) = mdc {(ta − 1), (tb − 1), (tc − 1), (td − 1), (te − 1), (tf − 1), (tg − 1)},

segue que

(tr − 1)tn11c − 1

tc − 1

tm21d − 1

td − 1

tn22e − 1

te − 1

tn23f − 1

tf − 1

tn12g − 1

tg − 1

= (t− 1)tn11.e11 − 1

te11 − 1

tm21e01 − 1

te01 − 1

tn22.e22 − 1

te22 − 1

tn23e23 − 1

te23 − 1

tn12e12 − 1

te12 − 1.

Portanto

(tr − 1) (tn11c − 1) (tm21d − 1) (tn22e − 1) (tn23f − 1) (tn12g − 1)

(te11 − 1) (te01 − 1) (te22 − 1) (te23 − 1) (te12 − 1)

= (t− 1) (tn11e11 − 1) (tm21e01 − 1) (tn22e22 − 1) (tn23e23 − 1) (tn12e12 − 1) (∗∗)

(tc − 1) (td − 1) (te − 1)(tf − 1) (tg − 1).


r = 1. Simplificando (t− 1) e analisando na igualdade resultante, em vista de (∗∗)e (∗), os termos de menor grau em t, concluimos que

d = e01 = n11 n12 + n21 n22 n23

c = e11 = m11 n12 + m21 n22 n23

e = e22 = m21 n11 n12 + m22 n23

f = e23 = m21 n11 n12 n22 + w23

g = e12 = m21 n22 n23 n11 + w12

Em vista das relacoes (∗), concluimos que

a = n11 n12 e b = m21 n22 n23.

Assim, determinamos a matriz de Alexander, com a qual podemos calcular ex-

plicitamente ∆2 (t), e portanto, o polinomio mınimo da monodromia neste caso.

84

3.2 Ramos com pares de Puiseux iguais

Nesta situacao, temos as seguintes possibilidades para os tipos de arvores que

podem ocorrer, conforme dado na Figura 14.

Figura 14: Arvores para ramos de genero 2 com pares de Puiseux iguais

Como os casos de 1 a 4, sao casos particulares dos casos estudados para ramos

com pares de Puiseux distintos, eles serao omitidos.

Caso 5: Neste caso, podemos escolher f = f1f2 com as seguintes representacoes

por series de Puiseux: {y1 = x

m1n1 + x

m2n1n2

y2 = xm1n1 + 2 x

m2n1n2

onde (mi, ni) sao os pares de Puiseux dos ramos fi para i = 1, 2. Isto conduz a

seguinte arvore:

85

Figura 15:

Com as notacoes da Secao 1.3,{µφ = m1, νφ = n1 e m1rφ = n1sφ + 1µ1 = m2 −m1n2 + m1n1n2, ν1 = n2 e µ1r1 = s1ν1 + 1

Portanto, o grupo do link π1(X), e gerado por αz, βz, com z percorrendo o



Relacoes de (φ):

(R1) 1 > φ e 1 nao e haste ⇒ ανφ µφ

1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ


φ βrφ

0 )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 αsφ


Relacoes de (1):


2 β2 = αµ1

1 βν11

⇒ β2 = α−µ1 n2

2 αµ1

1 βn21


3 β3 = αµ1

1 βν11

⇒ β3 = α−µ1 n2

3 αµ1

1 βn21

(R3) 1 e nao terminal sem haste ⇒ (α2 α3 αs11 βr1

φ )ν1 = (αµ1

1 βν11 )r1

⇒ (α2 α3 αs11 )n2 = (αµ1

1 βn21 )r1

86



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α2, β2] = 1 ⇔ [αµ1−m1n1n2

1 αm1n2φ , α2] = 1 veja (B.1)

[α3, β3] = 1 ⇔ [αµ1−m1n1n2

1 αm1n2φ , α3] = 1 veja (B.1)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1, α2 e α3, sujeitos as

relacoes:

(1) (α1 αsφ


(2) [αm1φ , α1] = 1

(3) (α2 α3 αsφ

φ )n1 = (αµ1−m1n1n2

1 αm1n2φ )r1

(4) [αµ1−m1n1n2

1 αm1n2φ , α2] = 1

(5) [αµ1−m1n1n2

1 αm1n2φ , α3] = 1

Como temos 4 geradores αφ, α1, α2, α3, sujeitos a 5 relacoes, e como nao

conseguimos fazer uma mudanca de geradores que reduza o numero de relacoes,

o metodo mencionado na Secao 1.2, para o calculo dos polinomios de Alexander,

nao se aplica neste caso.

87

Caso 6: Neste caso, podemos escolher f = f1f2, com as seguintes representacoes

por series de Puiseux: {y1 = x

m1n1 + x

m2n1n2

y2 = xm1n1 + x

m2n1n2 + x

m3n1n2n3

onde {(m1, n1), (m2, n2)} sao os pares de Puiseux de ambos os ramos. Isto conduz

a arvore, dada na Figura 16.

Figura 16:

Com as notacoes da Secao 1.3,µφ = m1 νφ = n1 e m1rφ = sφn1 + 1µ1 = m2 −m1n2 + m1n1n2, ν1 = n2 e µ1r1 = s1ν1 + 1µ2 = m3 −m2n3 + m2n2n3, ν2 = 1 e µ2r2 = s2ν2 + 1

Note que (r2, s2) = (1, µ2 − 1). Temos que, o grupo do link π1 (X), e gerado

por αz, βz, com z percorrendo o conjunto dos vertices V = {φ, 0, 1, 2, 20, 21} da

arvore Ξ, satisfazendo as relacoes que exibiremos a seguir.

Relacoes de (φ):

(R1) 1 > φ e 1 nao e haste ⇒ ανφ µφ

1 β1 = αµφ

φ βνφ

φ

⇒ β1 = α−m1 n11 αm1

φ


φ βrφ

0 )νφ = (αµφ

φ βνφ

φ )rφ

⇒ (α1 αsφ


Relacoes de (1):

88

(R1) 2 > 1 e 2 nao e haste ⇒ αν1 µ1

2 β2 = αµ1

1 βν11

⇒ β2 = α−µ1 n2

2 αm11 βn2

1


1 )ν1 = (αµ1

1 βν11 )r1

⇒ (α2 αs11 βr1

1 )n2 = (αµ1

1 βn21 )r1

Relacoes de (2):

(R1) 21 > 2 e 21 nao e haste ⇒ αν2 µ2

21 β21 = αµ2

2 βν22

⇒ β21 = α−µ2 121 αµ2

2 β12

(R2) 20 > 1 e 20 e uma haste ⇒ αµ2

2 βν22 = αµ2

20 βν220

⇒ β120 = α−µ2

20 αµ2

2 β2


2 = αs220 βr2

20

⇒ α20 = α2 α−121 veja (B.38)



[α1, β1] = 1 ⇔ [αm1φ , α1] = 1 veja (A.14)

[α2, β2] = 1 ⇔ [ αµ1

1 βν11 , α2 ] = 1 veja (B.1)

[α21, β21] = 1 ⇔ [ αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ , α21 ] = 1 veja (B.39)

[α20, β20] = 1 ⇔ [ αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ , α21 ] = 1 veja (B.40)

Portanto, π1 (X) pode ser apresentado com geradores αφ, α1, α2 e α21, sujeitos as

relacoes:

(1) (α1 αsφ


(2) [ αm1φ , α1 ] = 1

(3) (α2 αs11 βr1

1 )ν1 = (αµ1

1 βν11 )r1

(4) [ αµ1

1 βν11 , α2 ] = 1

89

(5) [ αµ2−µ1n2

2 αµ1−m1 n1 n2

1 αm1 n2φ , α21 ] = 1


somos conduzidos a fazer as seguintes restricoes:{m1 = λ1 n1 + 1 para algum λ1 ∈ ZZµ1 = λ2 n2 + 1 para algum λ2 ∈ ZZ

o que implica que rφ = r1 = 1. Adotaremos como novos geradores:{x1 = αφ,x2 = α1 α

sφ

φ ,

{x3 = α2 αs1

1 βr11 ,

x4 = α21.

Observe que a relacoes (2) e (4) sao equivalentes as relacoes

(2′) [ xm11 , x2 ] = 1 e (4′) [ αµ1

1 βν11 , x3 ] = 1

respectivamente, e que as mesmas sao trivializadas pelas relacoes (1′) e (3′), obtidas

de (1) e (3) respectivamente, apos a troca dos geradores. Temos que π1 (X) pode

ser apresentado com geradores x1, x2, x3, e x4, sujeitos as relacoes:

(1) xn12 = xm1

1

(2) xn23 = ( x2 x

−sφ

1 )µ1−m1 n1 n2 x1m1 n2

(3) [ (x3 x−m11 ( x2 x

−sφ

1 )m1 n1−s1 )µ2−µ1 n2 (x2 x−sφ

1 )µ1−m1n1n2 xm1n21 , x4] = 1


determinar Φ (xi) ∈ ZZ [ t, t−1 ], para i = 1, 2, 3, 4, onde Φ e a aplicacao induzida

pela fibracao de Milnor. Temporariamente, escrevemos

Φ : ZZ [ π1 (X) ] −→ ZZ [ t, t−1 ]x1 7→ ta

x2 7→ tb

x3 7→ tc

x4 7→ td


faremos uma nova mudanca de geradores, a saber,{y1 = x2 x

−sφ

1 ,y2 = x1,

{y3 = x3,y4 = x4.

90

Temos que

π1 (X) = { y1, y2, y3, y4 ; r1, r2, r3 }, onder1 = ( y1 y

sφ

2 )n1 y−m12

r2 = yµ1−m1n1n2

1 ym1n22 y−n2

3

r3 = ( y3 y−m12 ym1n1−s1

1 )µ2−µ1n2 yµ1−m1n1n2

1 ym1n22 y4 y−m1n2

2 y−(µ1−m1n1n2)1

( y3 y−m12 ym1n1−s1

1 )−(µ2−µ1n2) y−14

Consequentemente, a aplicacao Φ nos novos geradores, e dada por


y2 7→ ta

y3 7→ tc

y4 7→ td

Note que as relacoes ri, para i = 1, 2, fornecem as seguintes relacoes entre a, b, c,

(∗){

m1 a = n1 bn2 c = ( µ1 − m1 n1 n2 ) ( b − sφ a ) + m1 n2 a


(∂ri

∂yj

). Observe que a primeira linha desta matriz

se obtem pelos calculos feitos no caso 4 da Secao 2.2. Deste modo, basta determinar

as duas linhas restantes da matriz.

Como r2 = yµ1−m1n1n2

1 ym1n22 y−n2

3 , temos que

∂r2

∂y1

=yµ1−m1n1n2

1 − 1

y1 − 1⇒ Φ

(∂r2

∂y1

)=

t(µ1−m1n1n2)(b−sφa) − 1

t(b−sφa) − 1

∂r2

∂y2

= yµ1−m1n1n2

1

ym1n22 − 1

y2 − 1⇒ Φ

(∂r2

∂y2

)= t(µ1−m1n1n2)(b−sφa) tm1n2a − 1

ta − 1

∂r2

∂y3

= yµ1−m1n1n2

1 ym1n22 y−n2

3

yn23 − 1

y3 − 1⇒ Φ

(∂r2

∂y3

)= − tn2c − 1

tc − 1

∂r2

∂y4

= 0 ⇒ Φ

(∂r2

∂y4

)= 0

Como

r3 = ( y3 y−m12 ym1n1−s1

1 )µ2−µ1n2 yµ1−m1n1n2

1 ym1n22 y4 y−m1n2

2 y−(µ1−m1n1n2)1

91

( y3 y−m12 ym1n1−s1

1 )−(µ2−µ1n2) y−14 ,

fazendo calculos similares, temos que

Φ

(∂r3

∂y1

)= [ tc−m1a t(m1n1−s1)(b−sφa) − 1

tb−sφa − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

+ t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) t(m1n1−s1)(b−sφa) − 1

tb−sφa − 1] (1− td)

Φ

(∂r3

∂y2

)= [ tc−m1a t(m1n1−s1)(b−sφa) − 1

tb−sφa − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2)+(µ1−m1n1n2)(b−sφa) tm1n2a − 1

ta − 1] (td−1)

Φ

(∂r3

∂y3

)= − t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1(td−1)

Φ

(∂r3

∂y4

)= t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2)+n2c− 1


Φ

(∂r

∂y

)=

tn1b− 1tb − 1

− tn1b− 1ta − 1

tb−sφa− 1

tb − 10 0

t(µ1−m1n1n2)(b−sφa)−1

tb−sφa −1

t(µ1−m1n1n2)(b−sφa) tm1n2a−1ta −1

− tn2c−1tc −1

0

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) Φ ( ∂r3

∂y3) Φ ( ∂r3

∂y4)

.


∆1 (t) := mdc { det (A1), det (A2), det (A3), det (A4) },


(∂r

∂y


det (A1) =tn2c − 1

tc − 1

tn1b − 1

tb − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2)+n2c − 1

ta − 1(tb−sφa−1)

det (A2) = − tn2c − 1

tc − 1

tn1b − 1

tb − 1


ta − 1(ta−1)

det (A3) =tn2c − 1

tc − 1

tn1b − 1

tb − 1


ta − 1(tc−1)

veja (B.42)

det (A4) =tn2c − 1

tc − 1

tn1b − 1

tb − 1


ta − 1(td−1)

veja (B.43)

92

Logo,

∆1 (t) =tn2c − 1

tc − 1

tn1b − 1

tb − 1


ta − 1

mdc { (t(b−sφa) − 1), (ta − 1), (tc − 1), (td − 1) }.


m1n1 + x

m2n1n2

y2 = xm1n1 + x

m2n1n2 + x

m3n1n2n3


∆1 (t) =tw11 e01 − 1

tw11 e02 − 1

tw12 e02 − 1

tw12 e03 − 1

tw23 e03 − 1

te01 − 1(t− 1)

onde

w11 = m1

w12 = m2 −m1n2 + m1n1n2 = µ1

w23 = m3 −m2n3 + m2n2n3 = µ2

e01 = b1,1,2 + b2,1,3 = n1 n2 + n1 n2 n3 = 2 n1 n2

e02 = b1,2,2 + b2,2,3 = n2 + n2 n3 = 2 n2

e03 = b1,3,2 + b2,3,3 = 1 + n3 = 2

Logo: Substituindo, temos que

∆1 (t) =tn1(2m1n2) − 1

t2m1n2 − 1

tn2(2w12) − 1

t2w12 − 1

t2µ2 − 1

t2n1n2 − 1(t− 1).


(tr − 1) = mdc {(t(b−sφa) − 1), (ta − 1), (tc − 1), (td − 1)},

segue que

tn2c − 1

tc − 1

tn1b − 1

tb − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](I−µ1 n2)+n2c − 1

ta − 1=

tn2(2w12) − 1

t2w12 − 1

tn1(2m1n2) − 1

t2m1n2 − 1

t2I − 1

t2n1n2 − 1(t− 1).

Portanto,

(tn2c − 1) (tn1b − 1) (t[c−m1a+(m1 n1−s1)(b−sφa)](I−µ1 n2)+n2c − 1)(tr − 1)

(t2w12 − 1) (t2m1n2 − 1) (t2n1n2 − 1) =

(tn2(2w12) − 1) (tn1(2m1n2) − 1) (t2I − 1) (t− 1) (tc − 1) (tb − 1) (ta − 1).

93


r = 1. Simplificando (t − 1), e em vista de (*), temos que a < b. Analisando, na

igualdade resultante, os termos de menor grau em t, temos que,

a = 2 n1 n2

Em vista de (*), concluimos que:

(1) m1 a = n1 b ⇒ b =m1

n1

(2n1n2) = 2m1n2

(2) b− sφa = 2m1n2 − sφ (2n1n2) = 2m1n2 − 2n2(m1rφ − 1) = 2n2

(3) n2 c = (µ1 −m1n1n2) (b− sφa) + m1n2a

= (µ1 −m1n1n2) (2n2) + m1n2(2n1n2)

⇒ c = 2(µ1 −m1n1n2 + m1n1n2) = 2µ1 = 2w12

(4) [ c−m1a + (m1n1 − s1)(b− sφa) ] (µ2 − µ1n2) + n2c

= [ 2w12 −m1(2n1n2) + (m1n1 − s1)(2n2) ] (µ2 − w12n2) + n2(2w12)

= [ 2w12 − 2m1n1n2 + 2m1n1n2 − 2(n2s1)] (µ2 − w12n2) + 2w12n2

= [ 2w12 − 2(w12 − 1) ] (µ2 − w12n2) + 2w12n2 = 2µ2


∆2 (t) = mdc

det (A12), det (A13), det (A14), det (A23), det (A24), det (A34),det (B12), det (B13), det (B14), det (B23), det (B24), det (B34),det (C12), det (C13), det (C14), det (C23), det (C24), det (C34)

,

onde A, B e C sao obtidas de Φ

(∂r

∂y

)retirando-se a terceira, segunda e primeira

linhas respectivamente, e Aij, Bij e Cij denotam as matrizes formadas pelas colunas

i e j de A, B e C, respectivamente.

Temos que

A =

[tn1b− 1tb − 1

− tn1b− 1ta − 1

tb−sφa− 1

tb − 10 0

t(µ1−m1n1n2)(b−sφa)−1

tb−sφa −1


− tn2c−1tc −1

0

].

det (A12) =tn2c − 1

tc − 1

tn1b − 1

tb − 1

tc − 1

ta − 1veja (B.41)

det (A13) = − tn2c − 1

tc − 1

tn1b − 1

tb − 1

det (A14) = 0

94

det (A23) = − tn2c − 1

tc − 1

tn1b − 1

tb − 1

t2n2 − 1

ta − 1

det (A24) = 0

det (A34) = 0

Temos que

B =

[tn1b− 1tb − 1

− tn1b− 1ta − 1

tb−sφa− 1

tb − 10 0

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) Φ ( ∂r3

∂y3) Φ ( ∂r3

∂y4)

].

det (B12) = − tn1b − 1

tb − 1

td − 1

ta − 1

1

t2 − 1[ (t2µ1−1) (t2(µ2−µ1n2)−1)−(t2−1)(t2µ2−1)]

det (B13) =tn1b − 1

tb − 1

td − 1

t2 − 1(t2(µ2−µ1n2)−1)

det (B14) =tn1b − 1

tb − 1(t2µ2−1)

det (B23) =tn1b − 1

tb − 1

td − 1

ta − 1

tb−sφa − 1

t2 − 1(t2(µ2−µ1n2)−1)

det (B24) =tn1b − 1

tb − 1

t2n2 − 1

ta − 1(t2µ2−1)

det (B34) = 0


C =

[Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) − tn2c−1

tc −10

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) Φ ( ∂r3

∂y3) (t2µ2 − 1)

].

det (C12) = Φ

(∂r2

∂y1

)Φ

(∂r3

∂y2

)− Φ

(∂r2

∂y2

)Φ

(∂r3

∂y1

)det (C13) = Φ

(∂r2

∂y1

)Φ

(∂r3

∂y3

)− Φ

(∂r2

∂y3

)Φ

(∂r3

∂y1

)det (C14) =

t(µ1−m1n1n2)(2n2) − 1

t2n2 − 1(t2µ2−1)

det (C23) = Φ

(∂r2

∂y2

)Φ

(∂r3

∂y3

)− Φ

(∂r2

∂y3

)Φ

(∂r3

∂y2

)det (C24) =

tm1n2a − 1

ta − 1t(µ1−m1n1n2)(2n2) (t2µ2−1)

det (C34) = − tn2c − 1

tc − 1(t2µ2−1)

Definimos

∆ (t) := mdc

{tn2c − 1

tc − 1

tn1b − 1

tb − 1

t2 − 1

ta − 1, t2µ2 − 1

}.

95

Proposicao 2 Nas condicoes acima, temos que

∆2 (t) = mdc { ∆(t), det(C12), det(C13), det(C23) }

Prova: Para isto, basta mostrar as seguintes afirmacoes:

Afirmacao 1:

mdc ( det (A12), det (A23) ) =tn2c − 1

tc − 1

tn1b − 1

tb − 1

t2 − 1

ta − 1

De fato, como

det (A12) =tn2c − 1

tc − 1

tn1b − 1

tb − 1

tc − 1

ta − 1

det (A23) = − tn2c − 1

tc − 1

tn1b − 1

tb − 1

t2n2 − 1

ta − 1

segue que,

mdc ( det (A12), det (A23) ) =tn2c − 1

tc − 1

tn1b − 1

tb − 1

1

ta − 1mdc ( tc−1, t2n2−1 )

=tn2c − 1

tc − 1

tn1b − 1

tb − 1

1

ta − 1(tmdc(2µ1, 2n2) − 1)

=tn2c − 1

tc − 1

tn1b − 1

tb − 1

t2 − 1

ta − 1.

Afirmacao 2:

mdc ( det (B14), det (C34) ) = (t2µ2 − 1)

De fato, como

det (B14) =tn1b − 1

tb − 1(t2µ2−1) det (C34) =

tn2c − 1

tc − 1(t2µ2−1)

basta mostrarmos, que

mdc

(tn2c − 1

tc − 1,

tn1b − 1

tb − 1

)= 1

Suponha por absurdo que

mdc

(tn2c − 1

tc − 1,

tn1b − 1

tb − 1

)= mdc

∏j/n2c j 6 | c

φj,∏

i/n1b i6 | b

φi

=∏

i/n2c, i/n1b,i6 | b, i6 | c

φi

96

Deste modo, deve existir i satisfazendo a seguinte condicao:

i = c fator(n2) = b fator(n1)

pois i nao divide b e i nao divide c. Logo,

c fator(n2) = b fator(n1) ⇒c

bfator(n2) = fator(n1)

⇒[

2(m2 −m1n2 + m1n1n2)

2m1n2

]fator(n2) = fator(n1)

⇒[

m2

m1n2

+ (n1 − 1)

]2 ≤

[m2

m1n2

+ (n1 − 1)

]fator(n2) = fator(n1) ≤ n1

⇒ 2m2

m1n2

+ 2n1 − 2 ≤ n1

⇒ 2m2

m1n2

+ (n1 − 2) ≤ 0 o que e um absurdo pois n1 ≥ 2

Neste caso 6, temos a seguinte conjectura para ∆2 (t).

∆2 (t) = ∆ (t) = mdc

{t2µ1n2 − 1

t2µ1 − 1

t2m1n1n2 − 1

t2m1n2 − 1

t2 − 1

t2n1n2 − 1, t2µ2 − 1

}Exemplo 3 Seja f um germe de curva com as seguintes respresentacoes por series

de Puiseux {y1 = x

94 + x

198

y2 = x94 + x

198 + x

m38

onde {(9, 4), (19, 2)} sao os pares de Puiseux de ambos os ramos. Neste caso

∆1 (t) =t2m1n1n2 − 1

t2m1n2 − 1

t − 1

t2n1n2 − 1

t2µ1n2 − 1

t2µ1 − 1(t2µ2 − 1)

onde µ1 = 73, e µ2 = I(f1, f2) = I e o ındice de intersecao entre os ramos f1 e f2.

Consequentemente,

∆1 (t) =t144 − 1

t36 − 1

t − 1

t16 − 1

t292 − 1

t146 − 1(t2I − 1)

se I = 180, entao ∆1(t) = φ224 φ2

72 q1(t) e ∆2(t) = φ24 φ72

se I = 192, entao ∆1(t) = φ224 φ2

48 q2(t) e ∆2(t) = φ24 φ48

se I = 204, entao ∆1(t) = φ224 q3(t) e ∆2(t) = φ24

se I = 216, entao ∆1(t) = φ224 φ2

48 φ272 φ2

144 q4(t) e ∆2(t) = φ24 φ48 φ72 φ144

onde os qi(t), para i = 1, 2, 3, 4, sao polinomios sem fatores multiplos.

97

Resumindo, obtemos os seguintes resultados para ramos de genero 2:

Ramos com pares de Puiseux distintos

Caso 1: A matriz de Alexander foi determinada e ∆2 (t) foi calculado em funcao

de 31 determinantes menores.


de 26 determinantes menores.

Caso 3: A matriz de Alexander foi determinada explicitamente.

Caso 4: A matriz de Alexander foi determinada explicitamente.

Ramos com pares de Puiseux iguais

Casos 1 a 4: Sao analogos aos feitos para ramos com pares de Puiseux dis-

tintos.

Caso 5: A matriz de Alexander nao foi determinada.


dos pares de Puiseux.

98

Epılogo

O presente trabalho permite calcular explicitamente o polinomio mınimo da

monodromia em varias situacoes. Nao ha resultados analogos na literatura onde

este calculo e feito de modo tao explıcito para curvas com varios ramos. Com

isto, podemos decidir em muitos casos sobre a finitude da monodromia, obtendo

resultados que nao sao recuperados com criterios anteriormente obtidos. Por outro

lado, existe um algoritmo desenvolvido por E. Brieskorn e implementado por M.

Schulze no Singular, para o calculo da monodromia, mas cuja complexidade nao

permite executa-lo para curvas com dois ramos de genero maiores ou iguais a 2.

Em particular, nao permite fazer os calculos de nosso Exemplo 3.

Ao iniciarmos este trabalho tinhamos a expectativa de encontrar, em funcao

dos dados topologicos de um germe de curva plana com dois ramos quaisquer, a

expressao fechada para ∆2 (t), o que nao pode se concretizar por varios motivos.

O primeiro motivo, reside no fato, da apresentacao de Neto e Silva do grupo

fundamental do link da curva, ter um numero de geradores menor ou igual ao

numero de relacoes, impossibilitando o uso do metodo de Fox. Tal inconveniente

poderia ser sanado, se conseguissemos outra apresentacao mais apropriada, mas

isto envolveria outro tipo de opcao inicial, com o respectivo grau de incerteza de seu

sucesso. Uma outra dificuldade, consiste na diversidade de situacoes combinatorias

que aparecem a medida que aumenta a complexidade da topologia da curva.

Por esta razao, optamos por impor restricoes numericas sobre os dados to-

pologicos da curva, para recaırmos na boa situacao de ter mais geradores do que

relacoes na apresentacao do grupo, e assim nos limitamos, a analizar as possıveis

arvores que podem ocorrer para generos 1 e 2, para decidirmos sobre a finitude da

monodromia, nestas varias situacoes.

99

Apendice A

Neste apendice, apresentaremos os desenvolvimentos algebricos, que foram

omitidos no decorrer do texto no capıtulo 2 de ramos de genero 1.

(A.1)

[α0, β0] = 1 ⇒ α0 β0 = β0 α0

⇒ αsφ

0 βrφ

0 (αm210 βn21

0 ) = (αm210 βn21

0 ) αsφ

0 βrφ

0

⇒ αsφ

0 βrφ

0 αm21φ = αm21

φ αsφ

0 βrφ

0

⇒ ( αsφ

0 βrφ

0 α−sφ

φ ) αm21φ = αm21

φ (αsφ

0 βrφ

0 α−sφ

φ )

⇒ α1 αm21φ = αm21

φ α1

⇒ α1 (α−m21 n211 αm21

φ ) = (α−m21 n211 αm21

φ ) α1

⇒ α1 β1 = β1 α1

⇒ [α1, β1] = 1

(A.2)

[α01, β01] = 1 ⇔ α01 β01 = β01 α01

⇔ α01 (α−m11 n1101 αm11

0 βn110 ) = (α−m11 n11

01 αm110 βn11

0 ) α01

⇔ α01 αm110 βn11

0 = αm110 βn11

0 α01

⇔ [ αm110 βn11

0 , α01 ] = 1

(A.3)

det

[−(tb − 1) (ta − 1)

tm11a−1ta −1


]= − tm11a (tn11b−1) − (tm11a−1)

100

= − tm11a + n11b + tm11a − tm11a + 1

= − (tn11c−1)

(A.4)

det

[−(tb − 1) (ta − 1)

tm21a−1ta −1


]= − tm21a (tn21b−1) − (tm21a−1)

= − tm21a + n21b + tm21a − tm21a + 1

= − (tm21d−1)

(A.5)

det

[tm11a−1ta −1


tm21a−1ta −1


]

= tm21a tn21b − 1

tb − 1

tm11a − 1

ta − 1− tm11a tn11b − 1

tb − 1

tm21a − 1

ta − 1

=(tm21a+n21b − tm21a)

(tb − 1)

tm11a − 1

ta − 1− (tm11a+n11b − tm11a)

(tb − 1)

tm21a − 1

ta − 1

=(tm21d − 1)− (tm21a − 1)

(tb − 1)

tm11a − 1

ta − 1− (tn11c − 1)− (tm11a − 1)

(tb − 1)

tm21a − 1

ta − 1

=tm21d − 1

tb − 1

tm11a − 1

ta − 1− tm21a − 1

tb − 1

tm11a − 1

ta − 1

− tn11c − 1

tb − 1

tm21a − 1

ta − 1+

tm11a − 1

tb − 1

tm21a − 1

ta − 1

=tm21d − 1

tb − 1

tm11a − 1

ta − 1− tn11c − 1

tb − 1

tm21a − 1

ta − 1

(A.6)

[α0, β0] = 1 ⇒ α0 β0 = β0 α0

⇒ αsφ

0 βrφ

0 (αm210 βn21

0 ) = (αm210 βn21

0 ) αsφ

0 βrφ

0

⇒ αsφ

0 βrφ

0 αm21φ = αm21

φ αsφ

0 βrφ

0

⇒ ( αsφ

0 βrφ

0 α−sφ

φ ) αm21φ = αm21

φ (αsφ

0 βrφ

0 α−sφ

φ )

⇒ α1 αm21φ = αm21

φ α1

101

⇒ α1 (α−m21 n211 αm21

φ ) = (α−m21 n211 αm21

φ ) α1

⇒ α1 β1 = β1 α1

⇒ [α1, β1] = 1

(A.7)

[α01, β01] = 1 ⇔ α01 β01 = β01 α01

⇔ α01 (α−m11 n1101 αm11

0 βn110 ) = (α−m11 n11

01 αm110 βn11

0 ) α01

⇔ α01 αm110 βn11

0 = αm110 βn11

0 α01

⇔ [ αm110 βn11

0 , α01 ] = 1

(A.8)

[α11, β11] = 1 ⇔ α11 β11 = β11 α11

⇔ α11 (α−m22 n2211 αm22

1 βn221 ) = (α−m22 n22

11 αm221 βn22

1 ) α11

⇔ α11 αm221 βn22

1 = αm221 βn22

1 α11

⇔ α11 αm221 (α−m21 n21

1 αm21φ )n11 = αm22

1 (α−m21 n211 αm21

φ )n11 α11

⇔ α11 αm22−m21 n21 n111 αm21 n11

φ = αm22−m21 n21 n111 αm21 n11

φ α11

⇔ α11 ( αsφ

0 βrφ

0 α−sφ

φ )m22−m21 n21 n11 αm21 n11φ

= ( αsφ

0 βrφ

0 α−sφ

φ )m22−m21 n21 n11 αm21 n11φ α11

⇔ [ ( αsφ

0 βrφ

0 α−sφ

φ )m22−m21 n21 n11 αm21 n11φ , α11 ] = 1

(A.9)

Seja D1 = ∂∂x1

. Pelas propriedade (5) da derivacao, temos que:

D1 (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 = − (xsφ

1 xrφ

2 x−sφ

4 )−m21n21(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

xsφ

1 − 1

x1 − 1

Como r4 = (xsφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21 n21 xm214 ]n22 x−n22

5 temos que:

∂r4

∂x1

=(x

sφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

xsφ

1 − 1

x1 − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ] − 1

D1(xsφ

1 xrφ

2 x−sφ

4 )−m21n21

102

=x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

− (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ] − 1

xsφ

1 − 1

x1 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21 [(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

(A.10)

D2 (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 = − (xsφ

1 xrφ

2 x−sφ

4 )−m21n21(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1x

sφ

1

xrφ

2 − 1

x2 − 1

Logo

∂r4

∂x2

=(x

sφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1x

sφ

1

xrφ

2 − 1

x2 − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

D2(xsφ

1 xrφ

2 x−sφ

4 )−m21n21

= xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

− (xsφ

1 xrφ

2 x−sφ

4 )m22[ (x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

xsφ

1

xrφ

2 − 1

x2 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

(A.11)

D4 (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 = (xsφ

1 xrφ

2 x−sφ

4 )−(m21n21−1) (xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

xsφ

4 − 1

x4 − 1

∂r4

∂x4

= − (xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1(x

sφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

D4(xsφ

1 xrφ

2 x−sφ

4 )−m21n21

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

(xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm21

4 − 1

x4 − 1

103

Substituindo temos que

∂r4

∂x4

= − (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m22 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

xsφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(m21n21−1) (xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ (xsφ

1 xrφ

2 x−sφ

4 )m22[(x

sφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ]n22 − 1

[ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm214 ] − 1

(xsφ

1 xrφ

2 x−sφ

4 )−m21n21xm21

4 − 1

x4 − 1

(A.12)

∂r4

∂x5

= − (xsφ

1 xrφ

2 x−sφ

4 )m22 [ (xsφ

1 xrφ

2 x−sφ

4 )−m21n21 xm214 ]n22 x−n22

5

xn225 − 1

x5 − 1

= − xn225 x−n22

5

xn225 − 1

x5 − 1= − xn22

5 − 1

x5 − 1

(A.13)

det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4)

=

tn11c − 1

tc − 1

tm21d − 1

td − 1

[− (ta − 1) Φ

(∂r4

∂x1

)− (tb − 1) Φ

(∂r4

∂x2

) ]− tn11c − 1

tc − 1

tm21d − 1

td − 1(td−1) Φ

(∂r4

∂x4

)= − tn11c − 1

tc − 1

tm21d − 1

td − 1

[(ta − 1) Φ

(∂r4

∂x1

)+ (tb − 1) Φ

(∂r4

∂x2

)+ (td − 1) Φ

(∂r4

∂x4

)]Observe que:

(ta−1) Φ

(∂r4

∂x1

)+(tb−1) Φ

(∂r4

∂x2

)

= (tsφa−1)t(sφa+rφb−sφd)m22 − 1


104



(tsφa−1) t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


+ tsφa (trφb−1)t(sφa+rφb−sφd)m22 − 1




tsφa (trφb−1) t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


= (tsφa+rφb−1)t(sφa+rφb−sφd)m22 − 1




(tsφa+rφb−1) t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


Deste modo

(ta−1) Φ

(∂r4

∂x1

)+ (tb−1) Φ

(∂r4

∂x2

)+ (td−1) Φ

(∂r4

∂x4

)

= (tsφa+rφb−1)t(sφa+rφb−sφd)m22 − 1




(tsφa+rφb−1) t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1


− t(sφa+rφb−sφd) (tsφd−1)t(sφa+rφb−sφd)m22 − 1




(tsφd−1) t(sφa+rφb−sφd)(−(m21n21−1)) t(sφa+rφb−sφd)(m21n21) − 1



t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21) (tm21d−1)

= [ ( tsφa+rφb−1) + ( −tsφa+rφb + t(sφa+rφb−sφd) ) ]t(sφa+rφb−sφd)m22 − 1


+ [ −(tsφa+rφb−1 ) + ( tsφd−1) t(sφa+rφb−sφd) ] t(sφa+rφb−sφd)m22

105

t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1

t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1




= ( t(sφa+rφb−sφd)m22−1 )

+ [ − tsφa+rφb+1 + tsφa+rφb − t(sφa+rφb−sφd) ] t(sφa+rφb−sφd)m22

t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1

t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21) t(sφa+rφb−sφd)(m21n21) − 1






t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21)

( t(sφa+rφb−sφd)(m21n21)−1 )


t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1t(sφa+rφb−sφd)(−m21n21)(tm21d−1)


+ [ − ( t(sφa+rφb−sφd)(m21n21)−1 ) + (tm21d−1 ) ]

t(sφa+rφb−sφd)m22t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1



+ [ tm21d − t(sφa+rφb−sφd)(m21n21) ]

t(sφa+rφb−sφd)m22t[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − 1



+t(sφa+rφb−sφd)m22+[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − t(sφa+rφb−sφd)m22


t[ (sφa+rφb−sφd)(−m21n21)+m21d ]

− t(sφa+rφb−sφd)m22+[(sφa+rφb−sφd)(−m21n21)+m21d ]n22 − t(sφa+rφb−sφd)m22



+ (t[ (sφa+rφb−sφd)(−m21n21)+m21d ]−1)tn22e − t(sφa+rφb−sφd)m22

(t[(sφa+rφb−sφd)(−m21n21)+m21d] − 1)

= ( t(sφa+rφb−sφd)m22−1 )+(tn22e−t(sφa+rφb−sφd)m22)

106

= (tn22e−1)

Portanto:

det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4)

= − tn11c − 1

tc − 1

tm21d − 1

td − 1(tn22e−1)

(A.14)

[α1, β1] = 1 ⇔ α1 β1 = β1 α1

⇔ α1 (α−m1 n11 αm1

φ ) = (α−m1 n11 αm1

φ ) α1

⇔ α1 . αm1φ = αm1

φ α1

⇔ [αm1φ , α1] = 1

(A.15)

[α2, β2] = 1 ⇔ α2 β2 = β2 α2

⇔ α2 (α−m1n12 αm1

φ ) = (α−m1 n12 αm1

φ ) α2

⇔ α2 αm1φ = αm1

φ α2

⇔ [αm1φ , α2] = 1

(A.16)


1 = αs110 βr1

10

⇒ α11 αµ1−11 (α−m1 n1

1 αm1φ )1 = αµ1−1

10 β110

⇒ α11 αµ1−1−m1 n1

1 αm1φ = αµ1−1

10 (α−µ1

10 αµ1−m1n1

1 αm1φ )

⇒ α11 αµ1−1−m1 n1

1 = α−110 αµ1−m1n1

1

⇒ α10 = α1 α−111

(A.17)

[α11, β11] = 1 ⇔ α11 β11 = β11 α11

⇔ α11 ( α−µ1

11 αµ1

1 ( α−m1n11 αm1

φ ) ) = ( α−µ1

11 αµ1

1 ( α−m1n11 αm1

φ ) ) α11

107

⇔ α11 αµ1−m1n1

1 αm1φ = αµ1−m1n1

1 αm1φ α11

⇔ [ αµ1−m1n1

1 αm1φ , α11 ] = 1

(A.18)

[α10, β10] = 1 ⇔ α10 β10 = β10 α10

⇔ α10 ( α−µ1

10 αµ1

1 (α−m1n11 αm1

φ )1) = (α−µ1

10 αµ1

1 (α−m1n11 αm1

φ )1) α10

⇔ α10 αµ1−m1n1


1 αm1φ α10

⇔ ( α1 α−111 ) αµ1−m1n1


1 αm1φ ( α1 α−1

11 )

⇔ [ αµ1−m1n1

1 αm1φ , α11 ] = 1

(A.19)

∂r1

∂y2

= y1y

sφ

2 − 1

y2 − 1

(y1 ysφ

2 )(n1−1) − 1

(y1 ysφ

2 ) − 1− (y1 y

sφ

2 )n1−1 y1 y−(m1−sφ)2

y(m1−sφ)2 − 1

y2 − 1

temos que

Φ

(∂r1

∂y2

)= tb−sφa tsφa − 1

ta − 1

t(n1−1)b − 1

tb − 1− t(n1−1)b tb−sφa t−(m1−sφ)a t(m1−sφ)a − 1

ta − 1

= tb−sφa tsφa − 1

ta − 1

t(n1−1)b − 1

tb − 1

− tn1b−b+b−sφa−m1a+sφa t(m1−sφ)a − 1

ta − 1

tb − 1

tb − 1

=(tb − tb−sφa) (tn1b−b − 1) − (t(m1−sφ)a − 1) (tb − 1)

(ta − 1) (tb − 1)

=tn1b − tb − tb−sφa+n1b−b + tb−sφa − tm1a−sφa+b + tm1a−sφa + tb − 1

(ta − 1) (tb − 1)

=tn1b + tb−sφa − tm1a−sφa+b − 1

(ta − 1) (tb − 1)

=(tm1a − 1)− tb−sφa(tm1a − 1)

(ta − 1) (tb − 1)

= − tm1a − 1

ta − 1

tb−sφa − 1

tb − 1

108

(A.20)

(b−sφa) (µ1−m1n1) = µ1b − m1 (n1b) − µ1sφ a + m1 (n1sφ) a

= µ1(b−sφa) − m1 (m1a) + m1a (m1−1) = µ1(b−sφa) − m1a

(A.21)

det (A3) = − tm1a − 1

tb − 1tµ1(b−sφa)−m1a tm1a − 1

ta − 1(tc−1)

− tm1a − 1

ta − 1

tb−sφa − 1

tb − 1

tµ1(b−sφa)−m1a − 1

t(b−sφa) − 1(tc − 1)

= − tm1a − 1

tb − 1

tc − 1

ta − 1[ tµ1(b−sφa)−m1a (tm1a − 1) + (tµ1(b−sφa)−m1a − 1) ]

= − tm1a − 1

ta − 1

tc − 1

tb − 1( tµ1(b−sφa)−1 )

109

Apendice B

A seguir, apresentamos os desenvolvimentos algebricos que foram omitidos no

decorrer do texto, para os ramos de genero 2.

(B.1)

[α2, β2] = 1 ⇔ α2 β2 = β2 α2

⇔ α2 ( α−µ1n22

2 αµ1

1 βn221 ) = ( α−µ1n22

2 αµ1

1 βn221 ) α2

⇔ α2 αµ1

1 βn221 = αµ1

1 βn221 α2

⇔ [ αµ1

1 βn221 , α2 ] = 1

(B.2)

[α3, β3] = 1 ⇔ α3 β3 = β3 α3

⇔ α3 ( α−µ01n12

3 αµ01

01 βn1201 ) = ( α−µ01n12

3 αµ01

01 βn1201 ) α3

⇔ α3 αµ01

01 βn1201 = αµ01

01 βn1201 α3

⇔ [ αµ01

01 βn1201 , α3 ] = 1

(B.3)

αµ1

1 βn221 = αµ1

1 (α−m21n211 αm21

φ )n22

= αµ1

1 α−m21n21n221 αm21n22

φ

= (αsφ

0 βrφ

0 α−sφ

φ )m22−m21n22 αm21n22φ

= (xsφ

1 xrφ

2 x−sφ

4 )m22−m21n22 xm21n224

(B.4)

αµ01

01 βn1201 = αµ01

01 (α−m11 n1101 αm11

0 βn110 )n12

= αµ01

01 α−m11 n11 n1201 αm11 n12

0 βn11n120

110

= (x3 β−r00 α−s0

0 )m12−m11n12 αm11n120 βn11n12

0

= (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11 n12

2

(B.5)

det (A1234) = det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4)

= − tn11c − 1

tc − 1det

−(tb − 1) (ta − 1) 0tm21a−1ta −1


− tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) Φ( ∂r4

∂x4)

= − tn12f − 1

tf − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1

[−(tb − 1) Φ

(∂r4

∂x2

)− (ta − 1) Φ

(∂r4

∂x1

) ]+

tn12f − 1

tf − 1

tn11c − 1

tc − 1Φ

(∂r4

∂x4

)[ (tm21a − 1) + (tm21d − tm21a) ]

=tn12f − 1

tf − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1

[(ta − 1)Φ

(∂r4

∂x1

)+ (tb − 1)Φ

(∂r4

∂x2

)+ (td − 1)Φ

(∂r4

∂x4

)]mas

(ta−1) Φ

(∂r4

∂x1

)+ (tb−1) Φ

(∂r4

∂x2

)+ (td−1) Φ

(∂r4

∂x4

)

= ( tsφa − 1 )t(sφa+rφb−sφd)(m22−m21n22) − 1


+ ( tsφa+rφb − tsφa )t(sφa+rφb−sφd)(m22−m21n22) − 1


− tsφa+rφb−sφd (tsφd − 1)t(sφa+rφb−sφd)(m22−m21n22) − 1


+ t(sφa+rφb−sφd)(m22−m21n22) (tm21n22d − 1)

= ( tsφa+rφb − 1 )t(sφa+rφb−sφd)(m22−m21n22) − 1


+ ( − tsφa+rφb + tsφa+rφb−sφd )t(sφa+rφb−sφd)(m22−m21n22) − 1


+ t(sφa+rφb−sφd)(m22−m21n22) (tm21n22d − 1)

= ( tsφa+rφb−sφd − 1 )t(sφa+rφb−sφd)(m22−m21n22) − 1


111

+ t(sφa+rφb−sφd)(m22−m21n22)+m21n22d − t(sφa+rφb−sφd)(m22−m21n22)

= ( t(sφa+rφb−sφd)(m22−m21n22)+m21n22d − 1 )

= ( tn22e − 1 )

Logo

det (A1234) =tn11c − 1

tc − 1

tm21d − 1

td − 1(tn22e−1)

(B.6)

det (A5) = − tn12f − 1

tf − 1det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4)

= − tn12f − 1

tf − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1(tn22e−1)

(B.7)

det

−(tb − 1) (ta − 1) 0tm11a−1ta −1


− tn11c−1tc −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3)

=

tn11c − 1

tc − 1

[− (tb − 1) Φ

(∂r5

∂x2

)− (ta − 1) Φ

(∂r5

∂x1

) ]+ Φ

(∂r5

∂x3

) [−(tm11a+n11b − tm11a)− (tm11a − 1)

]= − tn11c − 1

tc − 1

[(tb − 1) Φ(

∂r5

∂x2

) + (ta − 1) Φ

(∂r5

∂x1

)+ (tc − 1) Φ

(∂r5

∂x3

) ]mas

(ta−1) Φ

(∂r5

∂x1

)+ (tb−1) Φ

(∂r5

∂x2

)+ (tc−1) Φ

(∂r5

∂x3

)=

= − ( tc−r0b − tc−r0b−s0a )t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

+ t(c−r0b−s0a)(m12−m11n12) (tm11n12a − 1)

− ( tc − tc−r0b )t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

112

+ t(c−r0b−s0a)(m12−m11n12) tm11n12a (tn11n12b − 1)

+ ( tc − 1 )t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

= (tc−r0b−s0a−1)t(c−r0b−s0a)(m12−m11n12) − 1

t(c−r0b−s0a) − 1

+ t(c−r0b−s0a)(m12−m11n12)+ m11n12a + n11n12b − t(c−r0b−s0a)(m12−m11n12)

= (tn12f−1)

Logo:

det

−(tb − 1) (ta − 1) 0tm11a−1ta −1


− tn11c−1tc −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3)

= − tn11c − 1

tc − 1(tn12f−1)

(B.8)

det (B1234) = det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3) 0

=tm21d − 1

td − 1det

−(tb − 1) (ta − 1) 0tm11a−1ta −1


− tn11c−1tc −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3)

= − tm21d − 1

td − 1

tn11c − 1

tc − 1(tn12f−1)

(B.9)

det (A6) = det

−(tb − 1) (ta − 1) 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) − tn22e−1

te −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3) 0 0

=tn22e − 1

te − 1det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x3) 0

= − tn22e − 1

te − 1

tm21d − 1

td − 1

tn11c − 1

tc − 1(tn12f−1)

113

(B.10)

[α2, β2] = 1 ⇔ α2 β2 = β2 α2

⇔ α2 ( α−w22 n22 αw22−m1n1n22

1 αm1n22φ ) = ( α−w22 n2

2 αw22−m1n1n221 αm1n22

φ ) α2

⇔ αm22−m1n221 αm1n22

φ α2 = α2 αm22−m1n221 αm1n22

φ

⇔ αm22−m1n221 αm1n22

φ αs110 βr1

10 β−r11 α−s1

1 =

αs110 βr1

10 β−r11 α−s1

1 αm22−m1n221 αm1n22

φ

⇔ αm22−m1n221 αm1n22

φ αs110 βr1

10 = αs110 βr1

10 αm22−m1n221 αm1n22

φ

⇔ αw2210 βn22

10 αs110 βr1

10 = αs110 βr1

10 αw2210 βn22

10

⇔ [α10, β10] = 1

(B.11)

[ α3, β3 ] = 1 ⇔ α3 β3 = β3 α3

⇔ α3 (α−w12 n123 αw12

10 βn1210 ) = (α−w12 n12

3 αw1210 βn12

10 ) α3

⇔ α3 αw1210 βn12

10 = αw1210 βn12

10 α3

⇔ [ α w1210 βn12

10 , α3 ] = 1

(B.12)


2 x−m1n224 (x3 x

−sφ

4 )−(m22−m1n12) Temos que

∂r4

∂x3

= xw221 xn22

2 x−m1n224 D3 (x3 x

−sφ

4 )−(m22−m1n12)

= − xw221 xn22

2 x−m1n224 (x3 x

−sφ

4 )−(m22−m1n12) (x3 x−sφ

4 )(m22−m1n12) − 1

(x3 x−sφ

4 ) − 1

= − (x3 x−sφ

4 )(m22−m1n12) − 1

(x3 x−sφ

4 ) − 1

(B.13)∂r4

∂x4= − xw22

1 xn222 x−m1n22

4x

m1n224 −1

x4 −1

+ xw221 xn22

2 x−m1n224 D4 (x3 x

−sφ

4 )−(m22−m1n12)

114

= − xw221 xn22

2 x−m1n224

xm1n224 − 1

x4 − 1

+ xw221 xn22

2 x−m1n224

xsφ

4 − 1

x4 − 1(x3 x

−sφ

4 )−(m22−m1n12−1) (x3 x−sφ

4 )m22−m1n12 − 1

(x3 x−sφ

4 ) − 1

= − xw221 xn22

2 x−m1n224

xm1n224 − 1

x4 − 1

+ (x3 x−sφ

4 )m22−m1n12x

sφ

4 − 1

x4 − 1(x3 x

−sφ

4 )−(m22−m1n12−1) (x3 x−sφ

4 )m22−m1n12 − 1

(x3 x−sφ

4 ) − 1

(B.14)

det

[−(tb − 1) (ta − 1)

tw22a−1ta −1


]= − [ tw22a+n22b − tw22a + tw22a − 1 ]

= − ( tw22a+n22b − 1 )

(B.15)

det

[−(tb − 1) (ta − 1)

tw12a−1ta −1


]= − [ tw12a+n12b − tw12a + tw12a − 1 ]

= − ( tw12a+n12b − 1 )

= − ( tn12e − 1 )

(B.16)

det

[tn1c−1tc −1

− tm1d−1td −1

Φ( ∂r4

∂x3) Φ( ∂r4

∂x4)

]=

tn1c − 1

tc − 1Φ(

∂r4

∂x4

) +tm1d − 1

td − 1Φ(

∂r4

∂x3

)

=tn1c − 1

tc − 1

1

td − 1[−tw22a+n22b+tw22a+n22b−m1n22d+(tc−tc−sφd)

t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1]

+tm1d − 1

td − 1

1

tc − 1[ − t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1(tc−1) ]

=tn1c − 1

tc − 1

1

td − 1[ − tw22a+n22b+tw22a+n22b−m1n22d−(tc−sφd−1)

t(c−sφd)(m22−m1n12) − 1

t(c−sφd) − 1]

=tn1c − 1

tc − 1

1

td − 1[ − tw22a+n22b + tw22a+n22b−m1n22d − t(c−sφd)(m22−m1n12)+1 ]

= − tn1c − 1

tc − 1

tw22a+n22b − 1

td − 1

115

= − tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1

(B.17)

det (A5) = det

−(tb − 1) (ta − 1) 0 0

tw12a−1ta −1


0 0

0 0 tn1c−1tc −1

− tm1d−1td −1

tw22a−1ta −1


− t(c−sφd)(m22−m1n12)−1

t(c−sφd) −1

Φ( ∂r4

∂x4)

= det

[−(tb − 1) (ta − 1)

tw22a−1ta −1


]det

[tn1c−1tc −1

− tm1d−1td −1

Φ( ∂r4

∂x3) Φ( ∂r4

∂x4)

]

=tm1d − 1

td − 1

tw22a+n22b − 1

tc − 1(tn12e−1)

(B.18)

[α10, β10] = 1 ⇔ α10 β10 = β10 α10

⇔ α10 ( α−µ1

10 αµ1

1 β1) = ( α−µ1

10 αµ1

1 β1) α10

⇔ [ αµ1

1 β1 , α10 ] = 1

(B.19)

[α11, β11] = 1 ⇔ α11 β11 = β11 α11

⇔ α11 ( α−µ1

11 αµ1

1 β1) = ( α−µ1

11 αµ1

1 β1) α11

⇔ α11 αµ1

1 β1 = αµ1

1 β1 α11

⇔ ( αs110 βr1

10 β−r11 α−s1

1 ) αµ1

1 β1 = αµ1

1 β1 ( αs110 βr1

10 β−r11 α−s1

1 )

⇔ αs110 βr1

10 αµ1

1 β1 = αµ1

1 β1 αs110 βr1

10

⇔ αs110 ( α−µ1

10 αµ1

1 β1 ) αµ1

1 β1 = αµ1

1 β1 αs110 ( α−µ1

10 αµ1

1 β1 )

⇔ αµ1−110 α−µ1

10 αµ1

1 β1 = αµ1

1 β1 αµ1−110 α−µ1

10

⇔ [ αµ1

1 β1 , α10 ] = 1

(B.20)

[α2, β2] = 1 ⇔ α2 β2 = β2 α2

⇔ α2 ( α−µ1n23

2 αµ1

11 βn2311 ) = ( α−µ1n23

2 αµ1

11 βn2311 ) α2

116

⇔ [ αµ1

11 βn2311 , α2 ] = 1

(B.21)

[α3, β3] = 1 ⇔ α3 β3 = β3 α3

⇔ α3 ( α−µ1n23

3 αw1210 βn12

10 ) = ( α−w12n123 αw12

10 βn1210 ) α3

⇔ [ αw1210 βn12

10 , α3 ] = 1

(B.22)

Observe que

αµ1 β1 = αµ1

1 α−m1n11 αm1

φ = αm22−m11 αm1

φ

como β11 = α−µ1

11 αµ1

1 β1 entao

⇒ βn2311 = (α−µ1

11 αµ1

1 β1)n23 = α−µ1n23

11 αµ1n23

1 βn231

⇒ αµ11

11 βn2311 = αµ11

11 α−µ1n23

11 αµ1n23

1 βn231

= (αs110 βr1

10 β−r11 α−s1

1 )µ11−µ1n23 αµ1n23

1 βn231

= αs1(µ11−µ1n23)10 β

r1(µ11−µ1n23)10 β

−r1(µ11−µ1n23)1 α

−s1(µ11−µ1n23)1 αµ1n23

1 βn231

= α(µ1−1)(µ11−µ1n23)10 (α−µ1

10 αµ1

1 β1 )(µ11−µ1n23) α−(µ1−1)(µ11−µ1n23)+µ1n23

1 β−(µ11−µ1n23)+n23

1

= α−(µ11−µ1n23)10 αµ11

1 βn231

= α−(µ11−µ1n23)10 αµ11

1 (α−m1n11 αm1

φ )n23

= α− (µ11−µ1n23)10 αµ11−m1n1n23

1 αm1n23φ

(B.23)

como β10 = αµ1

10 αµ1

1 β1 entao

βn1210 = ( α−µ1

10 αµ1

1 β1 )n12 = α−µ1n12

10 αµ1n12

1 βn121

⇒ αw1210 βn12

10 = αµ10

10 α−µ1n12

10 αµ1n12

1 βn121

= αµ10−µ1n12

10 αµ1n12

1 ( α−m1n11 αm1

φ )n12

= αµ10−µ1n12

10 αn12(µ1−m1n1)1 αm1n12

φ

Algumas contas que utilizaremos a seguir

117

(C.1)

det

[tn1b−1tb −1

− tm1a−1ta −1


Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2)

]

=tn1b − 1

tb − 1t−(µ11−µ1n23)e+(µ11−m1n1n23)(b−sφa) tm1n23a − 1

ta − 1

+tm1a − 1

ta − 1

1

tb − 1[ t−(µ11−µ1n23)e+(µ11−m1n1n23)(b−sφa) − t−(µ11−µ1n23)e ]

=tm1a − 1

ta − 1

1

tb − 1[ tn23c−t−(µ11−µ1n23)e ]

(C.2)

det

[tn1b−1tb −1

− tm1a−1ta −1


Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2)

]

=tn1b − 1

tb − 1

1

ta − 1[ t(µ10−µ1n12)e+n12(m22−m1)(b−sφa)+m1n12a − t(µ10−µ1n12)e+n12(m22−m1)(b−sφa) ]

+tm1a − 1

ta − 1

1

tb − 1[ t(µ10−µ1n12)e+n12(m22−m1)(b−sφa) − t(µ10−µ1n12)e ]

=tn1b − 1

tb − 1

1

ta − 1[ tn12d−t(µ10−µ1n12)e ]

(C.3)

det

[tn1b−1tb −1

− tm1a−1ta −1


Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2)

]

= − tn1b − 1

tb − 1

tm1a − 1

ta − 1t(m22−m1)(b−sφa) (te−1)

− tm1a − 1

ta − 1

te − 1

tb − 1(t(m22−m1)(b−sφa) − 1)

= − te − 1

ta − 1

tn1b − 1

tb − 1t(m22−m1)(b−sφa)+m1a − t(m22−m1)(b−sφa)+ t(m22−m1)(b−sφa)−1]

= − tn1b − 1

tb − 1

te − 1

ta − 1(t(m22−m1)(b−sφa)+m1a−1)

118

(B.24)

det (A3) = det

tn1b−1tb −1

− tm1a−1ta −1


0 0

Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) 0 −t−(µ11−µ1n23)e t(µ11−µ1n23)e−1

te −1

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) − tn21d−1

td −1t(µ10−µ1n12)e−1te −1

Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2) 0 t(m22−m1)(b−sφa)+m1a − 1

= − tn21d − 1

td − 1

tn1b−1tb −1

− tm1a−1ta −1


0

Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) −t−(µ11−µ1n23)e t(µ11−µ1n23)e−1

te −1

Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2) t(m22−m1)(b−sφa)+m1a − 1

= − tn21d − 1

td − 1{ (t(m22−m1)(b−sφa)+m1a−1)

(tm1a − 1

ta − 1

1

tb − 1[ tn23c − t−(µ11−µ1n23)e ]

)(C.1)

1− t−(µ11−µ1n23)e

te − 1

[− te − 1

tb − 1

tm1a − 1

ta − 1(t(m22−m1)(b−sφa)+m1a − 1)

]} (C.2)

= − tn21d − 1

td − 1(t(m22−m1)(b−sφa)+m1a−1)

tm1a − 1

ta − 1

1

tb − 1

[ tn23c − t−(µ11−µ1n23)e + t−(µ11−µ1n23)e − 1 ]

= − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(tc−1)

(B.25)

det (A4) =

tn1b−1tb −1

− tm1a−1ta −1


0 0

Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) − tn23c−1

tc −1−t(µ11−µ1n23)e t(µ11−µ1n23)e−1

te −1

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) 0 t(µ10−µ1n12)e−1

te −1

Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2) 0 t(m22−m1)(b−sφa)+m1a − 1

=

tn23c − 1

tc − 1{ (t(m22−m1)(b−sφa)+m1a−1)

[tn1b − 1

tb − 1

1

ta − 1(tn12d − t(µ10−µ1n12)e)

](C.2)

− t(µ10−µ1n12)e − 1

te − 1

[− te − 1

tb − 1

tm1a − 1

ta − 1(t(m22−m1)(b−sφa)+m1a − 1)

]} (C.3)

= − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(td−1)

(B.26)

det (A5) = det

tn1b−1tb −1

− tm1a−1ta −1


0 0

Φ ( ∂r2

∂y1) Φ ( ∂r2

∂y2) − tn23c−1

tc −10

Φ ( ∂r3

∂y1) Φ ( ∂r3

∂y2) 0 − tn12d−1

td −1

Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2) 0 0

119

=tn23c − 1

tc − 1

tn12d − 1

td − 1det

[tn1b−1tb −1

− tm1a−1ta −1


Φ ( ∂r4

∂y1) Φ ( ∂r4

∂y2)

]

= − tn1b − 1

tb − 1

tn23c − 1

tc − 1

tn12d − 1

td − 1

t(m22−m1)(b−sφa)+m1a − 1

ta − 1(te−1) (C.3)

(B.27)

[α11, β11] = 1 ⇔ α11 β11 = β11 α11

⇔ α11 (α−µ1 n22

11 αµ1

1 βn221 ) = (α−µ1 n22

11 αµ1

1 βn221 ) α11

⇔ α11 αµ1

1 βn221 = αµ1

1 βn221 α11

⇔ [ αµ1

1 βn221 , α11 ] = 1

(B.28)

[α2, β2] = 1 ⇔ α2 β2 = β2 α2

⇔ α2 (α−µ11 n23

2 αµ11

11 βn2311 ) = (α−µ11 n23

2 αµ11

11 βn2311 ) α2

⇔ α2 αµ11

11 βn2311 = αµ11

11 βn2311 α2

⇔ [ αµ11

11 βn2311 , α2 ] = 1

(B.29)

[α3, β3] = 1 ⇔ α3 β3 = β3 α3

⇔ α3 (α−µ01 n12

3 αµ01

01 βn1201 ) = (α−µ01 n12

3 αµ01

01 βn1201 ) α3

⇔ α3 αµ01

01 βn1201 = αµ01

11 βn1201 α3

⇔ [ αµ01

01 βn1201 , α3 ] = 1

(B.30)

β11 = α−µ1n22

11 αµ1

1 βn221 = α−µ1n22

11 αµ1

1 (α−m21n211 αm21

φ )n22

= α−µ1n22

11 αµ1

1 α−m21n21n221 αm21n22

φ = α−µ1n22

11 αµ1−m21n21n22

1 αm21n22φ

⇒ βn2311 = (α−µ1n22

11 αµ1−m21n21n22

1 αm21n22φ )n23

= α−µ1n22n23

11 α(m22−m21n22)n23

1 αm21n22n23φ

120

Como n21 = 1 temos que µ1 = m22 logo

αµ11

11 βn2311 = αµ11

11 α−m22n22n2311 α

(m22−m21n22)n23

1 αm21n22n23φ

= α(m23−m22n23)11 α

(m22−m21n22)n23

1 αm21n22n23φ

Por outro lado, como:

α1 = (αsφ

0 βrφ

0 α−sφ

φ ), β1 = ( α−m21n211 αm21

φ ) e r1 = 1

⇒ α11 = x7 β−r11 α−s1

1 = x7 ( αm21n211 α−m21

φ ) α−s11

= x7 (αsφ

0 βrφ

0 α−sφ

φ )m21n21 α−m21φ (α

sφ

0 βrφ

0 α−sφ

φ )−s1

Logo

αµ11

11 βn2311 = αm23−m22n23

11 α(m22−m21n22)n23

1 αm21n22n23φ

= [ x7 (αsφ

0 βrφ

0 α−sφ

φ )m21n21 α−m21φ (α

sφ

0 βrφ

0 α−sφ

φ )−s1 ]m23−m22n23

( αsφ

0 βrφ

0 α−sφ

φ )(m22−m21n22)n23 αm21n22n23φ

= [ x7 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 xm21n22n234

(B.31)

βn1201 = ( α−m11n11

01 αm110 βn11

0 )n12 = α−m11 n11 n1201 αm11 n12

0 βn11 n120

⇒ αµ01

01 βn1201 = αµ01

01 α−m11n11n1201 αm11n12

0 βn11n120

= α(m12−m11 n12)01 αm11 n12

0 βn11 n120

Como α01 = x3 β−r00 α−s0

0 , temos que

αµ01

01 βn1201 = αm12−m11n12

01 αm11n120 βn11n12

0

= (x3 β−r00 α−s0

0 )m12−m11n12 αm11n120 βn11n12

0

= (x3 x−r02 x−s0

1 )m12−m11n12 xm11n121 xn11n12

2

121

(B.32)

Como r5 = [ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 xm21n22n234 x−n23

6

Observe que

∂

∂x1

[ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]

= x5x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

1 − 1

x1 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

Logo:

∂r5

∂x1

= [ x5x

sφ

1 − 1

x1 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (B.32)

xsφ

1 − 1

x1 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23x

sφ

1 − 1

x1 − 1

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

( xsφ

1 xrφ

2 x−sφ

4 ) − 1

(B.33)

∂

∂x2

[ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]

= x5 xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

1

xrφ

2 − 1

x2 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

122

Logo:

∂r5

∂x2

= [ x5 xsφ

1

xrφ

2 − 1

x2 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1− x5 (x

sφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

1

xrφ

2 − 1

x2 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−s1(x

sφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 xsφ

1

xrφ

2 − 1

x2 − 1

( xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

( xsφ

1 xrφ

2 x−sφ

4 ) − 1

(B.34)

∂

∂x4

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 = − xsφ

1 xrφ

2 (x−14 + · · · + x

−sφ

4 )(x

sφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

= − (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

∂

∂x4

(xsφ

1 xrφ

2 x−sφ

4 )−s1 =x

sφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(s1−1) (xsφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

∂

∂x4

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 = (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

Observando que

∂

∂x4

[ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]

= x5∂

∂x4

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xm214 − 1

x4 − 1

+ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

∂

∂x4

(xsφ

1 xrφ

2 x−sφ

4 )−s1

= − x5 (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

− x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xm214 − 1

x4 − 1

123

+ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(s1−1) (xsφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

Logo

∂r5

∂x4

= [ − x5 (xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )m21n21 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

− x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xm214 − 1

x4 − 1

+ x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214

xsφ

4 − 1

x4 − 1(x

sφ

1 xrφ

2 x−sφ

4 )−(s1−1) (xsφ

1 xrφ

2 x−sφ

4 )s1 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1]

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23 − 1

[x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ] − 1

− [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

(xsφ

1 xrφ

2 x−sφ

4 )x

sφ

4 − 1

x4 − 1

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23 − 1

(xsφ

1 xrφ

2 x−sφ

4 ) − 1

+ [x5 (xsφ

1 xrφ

2 x−sφ

4 )m21n21 x−m214 (x

sφ

1 xrφ

2 x−sφ

4 )−s1 ]m23−m22n23

(xsφ

1 xrφ

2 x−sφ

4 )(m22−m21n22)n23xm21n22n23

4 − 1

x4 − 1

(B.35)

det

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) 0 Φ( ∂r5

∂x4)

=tn11c − 1

tc − 1

−(tb − 1) (ta − 1) 0tm21a−1ta −1


− tm21d−1td −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) Φ( ∂r5

∂x4)

=tn11c − 1

tc − 1

tm21d − 1

td − 1

[−(tb − 1) (ta − 1)Φ( ∂r5

∂x1) Φ( ∂r5

∂x2)

]

+tn11c − 1

tc − 1Φ(

∂r5

∂x4

)

[−(tb − 1) (ta − 1)

tm21a−1ta −1


]

= − tn11c − 1

tc − 1

tm21d − 1

td − 1

[(ta − 1) Φ

(∂r5

∂x1

)+ (tb − 1) Φ

(∂r5

∂x2

)+ (td − 1) Φ

(∂r5

∂x4

) ]124

= − tn11c − 1

tc − 1

tm21d − 1

td − 1

[(tn23f − 1) − (te − 1) Φ

(∂r5

∂x5

)]De fato, como

Φ

(∂r5

∂x1

)= [ te

tsφa − 1

ta − 1



tsφa − 1






ta − 1



Φ

(∂r5

∂x2

)= [ te+sφa trφb − 1

tb − 1

tsφa+rφb−sφd)(m21n21) − 1


tsφa trφb − 1





+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d+(sφa+rφb−sφd) (−s1) ](m23−m22n23) tsφa trφb − 1

tb − 1



segue que

(ta−1) Φ

(∂r5

∂x1

)+ (tb−1) Φ

(∂r5

∂x2

)

= (ta−1) { [ tetsφa − 1

ta − 1



tsφa − 1






ta − 1

125

t(sφa+rφb−sφd) (m22−m21n22)n23) − 1

t(sφa+rφb−sφd) − 1}

+ (tb−1) { [ te+sφa trφb − 1

tb − 1



tsφa trφb − 1





+ [t[ e+(sφa+rφb−sφd) (m21n21)−m21d+(sφa+rφb−sφd) (−s1) ](m23−m22n23) tsφa trφb − 1

tb − 1



= { [ (te+sφa−te)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (tsφa−1)t(sφa+rφb−sφd)s1 − 1


+ [ ( te+sφa+rφb−te+sφa)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd) (−s1) ] (tsφa+rφb−tsφa)t(sφa+rφb−sφd)(s1) − 1

t(sφa+rφb−sφd) − 1] }



+ { t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) (tsφa − 1)



+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d+(sφa+rφb−sφd) (−s1) ](m23−m22n23) (tsφa+rφb − tsφa)



= [ (te+sφa+rφb−te)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (tsφa+rφb−1)t(sφa+rφb−sφd)s1 − 1




+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) (tsφa+rφb − 1)


t(sφa+rφb−sφd) − 1.

126

Como,

Φ

(∂r5

∂x4

)= [ − te+sφa+rφb−sφd tsφd − 1

td − 1



− te+(sφa+rφ−sφd)(m21n21)−m21d tm21d − 1


tsφd − 1







td − 1





td − 1,

temos que

(td−1) Φ

(∂r5

∂x4

)=

= (td−1) { [ − te+sφa+rφb−sφd tsφd − 1

td − 1



− te+(sφa+rφb−sφd)(m21n21)−m21d tm21d − 1


tsφd − 1







td − 1





td − 1}

= [ ( − te+sφa+rφb + te+sφa+rφb−sφd)t(sφa+rφb−sφd)(m21n21) − 1


− te+(sφa+rφb−sφd)(m21n21) + te+(sφa+rφb−sφd)(m21n21)−m21d

127

+ te+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) (tsφd−1) t(sφa+rφb−sφd) t(sφa+rφb−sφd)s1 − 1





( tsφa+rφb − tsφa+rφb−sφd )t(sφa+rφb−sφd)(m22−m21n22)n23 − 1

t(sφa+rφb−sφd) − 1+ tn23f

− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23) + (sφa+rφb−sφd)(m22−m21n22)n23 .

Consequentemente,

(ta−1) Φ

(∂r5

∂x1

)+ (tb−1) Φ

(∂r5

∂x2

)+ (td−1) Φ

(∂r5

∂x4

)

= [ (te+sφa+rφb−te)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (tsφa+rφb−1)t(sφa+rφb−sφd)s1 − 1




+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) (tsφa+rφb−1)



+ [ (− te+sφa+rφb + te+sφa+rφb−sφd)t(sφa+rφb−sφd)(m21n21) − 1


− te+(sφa+rφ−sφd)(m21n21) + te+(sφa+rφ−sφd)(m21n21)−m21d

+ te+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) (tsφd−1) t(sφa+rφb−sφd) t(sφa+rφb−sφd)s1 − 1





( tsφa+rφb − tsφa+rφb−sφd )t(sφa+rφb−sφd)(m22−m21n22)n23 − 1

t(sφa+rφb−sφd) − 1+ tn23f

− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23)+ (sφa+rφb−sφd)(m22−m21n22)n23

= [ (te+sφa+rφb−sφd−te)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (tsφa+rφb−sφd−1)t(sφa+rφb−sφd)s1 − 1


128

− te+(sφa+rφb−sφd)(m21n21) + te+(sφa+rφb−sφd)(m21n21)−m21d ]



+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) (tsφa+rφb−sφd−1)


t(sφa+rφb−sφd) − 1+tn23f


= [ (te+sφa+rφb−sφd−te)t(sφa+rφb−sφd)(m21n21) − 1


− t[ e+(sφa+rφb−sφd)(m21n21)−m21d + t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ]

− te+(sφa+rφb−sφd)(m21n21) + te+(sφa+rφb−sφd)(m21n21)−m21d ]



+ t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23)+(sφa+rφb−sφd) (m22−m21n22)n23

− t[ e+(sφa+rφb−sφd) (m21n21)−m21d + (sφa+rφb−sφd)(−s1) ] (m23−m22n23) + tn23f


= [ te (t(sφa+rφb−sφd)(m21n21)−1)

+ t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] − te+(sφa+rφb−sφd)(m21n21) ]




= ( t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] − te )




= [ t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ]−1 ) − (te−1 ) ]




= (tn23f−1) − (te−1)t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] (m23−m22n23) − 1

t[ e+(sφa+rφb−sφd)(m21n21)−m21d+(sφa+rφb−sφd)(−s1) ] − 1.

129

Logo,

(ta−1) Φ

(∂r5

∂x1

)+ (tb−1) Φ

(∂r5

∂x2

)+ (td−1) Φ

(∂r5

∂x4

)= (tn23f−1) − (te−1) Φ

(∂r5

∂x5

).

(B.36)

det (A6) =

−(tb − 1) (ta − 1) 0 0 0 0tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) − tn22e−1

te −10

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) 0 Φ( ∂r5

∂x4) Φ( ∂r5

∂x5) 0

Φ( ∂r6

∂x1) Φ( ∂r6

∂x2) Φ( ∂r6

∂x3) 0 0 − tn12g−1

tg −1

= − tn12g − 1

tg − 1

−(tb − 1) (ta − 1) 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) − tn22e−1

te −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) 0 Φ( ∂r5

∂x4) Φ( ∂r5

∂x5)

= − tn12g − 1

tg − 1

{ tn22e − 1

te − 1

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) 0 Φ( ∂r5

∂x4)

+ Φ

(∂r5

∂x5

) −(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4)

}

= − tn12g − 1

tg − 1

{ − tn22e − 1

te − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1

[(tn12f − 1) − (te − 1) Φ

(∂r5

∂x5

) ]− Φ

(∂r5

∂x5

)tn11c − 1

tc − 1

tm21d − 1

td − 1(tn22e−1) } veja (A.13)

= − tn12g − 1

tg − 1

tn22e − 1

te − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1[− (tn23f − 1) + (te − 1) Φ

(∂r5

∂x5

)− (te − 1) Φ

(∂r5

∂x5

) ]130

Logo,

det (A6) =tn12g − 1

tg − 1

tn22e − 1

te − 1

tn11c − 1

tc − 1

tm21d − 1

td − 1(tn23f−1)

(B.37)

det (A7) =

−(tb − 1) (ta − 1) 0 0 0 0tm11a−1ta −1


− tn11c−1tc −1

0 0 0tm21a−1ta −1


0 − tm21d−1td −1

0 0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) tn22e−1

te −10

Φ( ∂r5

∂x1) Φ( ∂r5

∂x2) 0 Φ( ∂r5

∂x4) Φ( ∂r5

∂x5) − tn23f−1

tf −1

Φ( ∂r6

∂x1) Φ( ∂r6

∂x2) Φ( ∂r6

∂x3) 0 0 0

=tn23f − 1

tf − 1

−(tb − 1) (ta − 1) 0 0 0

tm11a−1ta −1


− tn11c−1tc −1

0 0tm21a−1ta −1


0 − tm21d−1td −1

0

Φ( ∂r4

∂x1) Φ( ∂r4

∂x2) 0 Φ( ∂r4

∂x4) tn22e−1

te −1

Φ( ∂r6

∂x1) Φ( ∂r6

∂x2) Φ( ∂r6

∂x3) 0 0

=tn23f − 1

tf − 1

tn22e − 1

te − 1

−(tb − 1) (ta − 1) 0 0

tm11a−1ta −1


− tn11c−1tc −1

0tm21a−1ta −1


0 − tm21d−1td −1

Φ( ∂r6

∂x1) Φ( ∂r6

∂x2) Φ( ∂r6

∂x3) 0

=

tn23f − 1

tf − 1

tn22e − 1

te − 1

tm21d − 1

td − 1

−(tb − 1) (ta − 1) 0tm11a−1ta −1


− tn11c−1tc −1

Φ( ∂r6

∂x1) Φ( ∂r6

∂x2) Φ( ∂r6

∂x3)

=

tn23f − 1

tf − 1

tn22e − 1

te − 1

tm21d − 1

td − 1

{ tn11c − 1

tc − 1

[−(tb − 1) (ta − 1)Φ( ∂r6

∂x1) Φ( ∂r6

∂x2)

]+ Φ

(∂r6

∂x3

) [−(tb − 1) (ta − 1)

tm11a−1ta −1


]}

= − tn23f − 1

tf − 1

tn22e − 1

te − 1

tm21d − 1

td − 1

tn11c − 1

tc − 1[(ta − 1) Φ

(∂r6

∂x1

)+ (tb − 1) Φ

(∂r6

∂x2

)+ (tc − 1) Φ

(∂r6

∂x3

) ]= − tn23f − 1

tf − 1

tn22e − 1

te − 1

tm21d − 1

td − 1

tn11c − 1

tc − 1(tn12g−1).

De fato, temos que

(ta−1) Φ

(∂r6

∂x1

)+ (tb−1) Φ

(∂r6

∂x2

)+ (tc−1) Φ

(∂r6

∂x3

)131

= (ta−1) [ − tc−r0b−s0a ts0a − 1

ta − 1

t(c−r0b−s0a)(m12−m11n12) − 1


+ t(c−r0b−s0a)(m12−m11n12) tm11n12a − 1

ta − 1]

+ (tb−1) [ − tc−r0b tr0b − 1

tb − 1

t(c−r0b−s0a)(m12−m11n12) − 1



tb − 1]

+ (tc−1) [t(c−r0b−s0a)(m12−m11n12) − 1

tc−r0b−s0a − 1]

= ( tc−r0b − tc−r0b−s0a )t(c−r0b−s0a)(m12−m11n12) − 1


+ t(c−r0b−s0a)(m12−m11n12)+m11n12a − t(c−r0b−s0a)(m12−m11n12)

+ ( tc − tc−r0b )t(c−r0b−s0a)(m12−m11n12) − 1


+ t(c−r0b−s0a)(m12−m11n12)+m11n12a+n11n12b − t(c−r0b−s0a)(m12−m11n12)+m11n12a

+ ( tc−1 )t(c−r0b−s0a)(m12−m11n12) − 1


= ( tc−r0b−s0a − tc )t(c−r0b−s0a)(m12−m11n12) − 1


− t(c−r0b−s0a)(m12−m11n12) + tn12g

+ ( tc−1 )t(c−r0b−s0a)(m12−m11n12) − 1


= ( tc−r0b−s0a −1 )t(c−r0b−s0a)(m12−m11n12) − 1


− t(c−r0b−s0a)(m12−m11n12) + tn12g

= (tn12g−1).

(B.38)


2 = αs220 βr2

20

⇒ α21 αµ2−12 β1

2 = αµ2−120 (α−µ2

20 αµ2

2 β12)

1

⇒ α21 = α−120 αµ2

2 β12 β−1

2 α−(µ2−1)2

⇒ α21 = α−120 α2

⇒ α20 = α2 α−121

132

(B.39)

[α21, β21] = 1 ⇔ α21 β21 = β21 α21

⇔ α21 ( α−µ2

21 αµ2

2 β2) = ( α−µ2

21 αµ2

2 β2) α21

⇔ α21 αµ2

2 β2 = αµ2

2 β2 α21

⇔ α21 αµ2

2 ( αµ1 n2

2 αµ1

1 βn21 ) = αµ2

2 ( αµ1 n2

2 αµ1

1 βn21 ) α21

⇔ α21 αµ2−µ1 n2

2 αµ1

1 βn21 = αµ2−µ1 n2

2 αµ1

1 βn21 α21

⇔ α21 αµ2−µ1 n2

2 αµ1

1 ( α−m1 n1 n21 αm1 n2

φ ) =

αµ2−µ1 n2

2 αµ1

1 ( α−m1 n1 n21 αm1 n2

φ ) α21

⇔ [ αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ , α21 ] = 1

(B.40)

[α20, β20] = 1 ⇔ α20 β20 = β20 α20

⇔ α20 ( α−µ2

20 αµ2

2 β2 ) = ( α−µ2

20 αµ2

2 β2 ) α20

⇔ α20 αµ2

2 β2 = αµ2

2 β2 α20

⇔ α20 αµ2

2 ( α−µ1 n2

2 αµ1

1 βn21 ) = αµ2

2 ( α−µ1 n2

2 αµ1

1 βn21 ) α20

⇔ α20 αµ2−µ1 n2

2 αµ1

1 βn21 = αµ2−µ1 n2

2 αµ1

1 βn21 α20

⇔ α20 αµ2−µ1 n2

2 αµ1

1 ( α−m1 n1 n21 αm1 n2

φ ) =

αµ2−µ1 n2

2 αµ1

1 ( α−m1 n1 n21 αm1 n2

φ ) α20

⇔ ( α2 α−121 ) αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ =

αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ )( α2 α−1

21 ) se r2 = 1

⇔ [ αµ2−µ1 n2

2 αµ1−m1 n1 n2

1 αm1 n2φ , α21 ] = 1

(B.41)

det

[tn1b−1tb −1

− tn1b−1ta −1

tb−sφa−1

tb −1t(µ1−m1n1n2)(b−sφa)−1

t(b−sφa) −1


]

=tn1b − 1

tb − 1t(µ1−m1n1n2)(b−sφa) tm1n2a − 1

ta − 1+

tn1b − 1

tb − 1

tb−sφa − 1

ta − 1

t(µ1−m1n1n2)(b−sφa) − 1

t(b−sφa) − 1

133

=tn1b − 1

tb − 1

1

ta − 1[ t(µ1−m1n1n2)(b−sφa)+m1n2a − t(µ1−m1n1n2)(b−sφa) + t(µ1−m1n1n2)(b−sφa)−1]

=tn1b − 1

tb − 1

tn2c − 1

ta − 1

(B.42)

det (A3) = det

tn1b− 1tb − 1

− tn1b− 1ta − 1

tb−sφa− 1

tb − 10

t(µ1−m1n1n2)(b−sφa)− 1

t(b−sφa) − 1

t(µ1−m1n1n2)(b−sφa) tm1n2a− 1ta − 1

0

Φ( ∂r3

∂y1) Φ( ∂r3

∂y2) Φ( ∂r3

∂y4)

= (t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2)+n2c − 1)

tn1b − 1

tb − 1

tn2c − 1

ta − 1.

Temos que

Φ

(∂r3

∂y2

)+

tb−sφa − 1

ta − 1Φ

(∂r3

∂y1

)

= [ tc−m1a tm1a − 1

ta − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2)+(µ1−m1n1n2)(b−sφa) tm1n2a − 1

ta − 1] (td − 1)

− tb−sφa − 1

ta − 1[ tc−m1a t(m1n1−s1)(b−sφa) − 1

tb−sφa − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) t(m1n1−s1)(b−sφa) − 1

tb−sφa − 1] (td − 1)

=td − 1

ta − 1[ (tc−tc−m1a)

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2)+(µ1−m1n1n2)(b−sφa)+m1n2a

+ t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2)+(µ1−m1n1n2)(b−sφa)

− (tc−m1a+(m1n1−s1)(b−sφa)−tc−m1a)t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2)+(m1n1−s1)(b−sφa)

+ t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) ]

=td − 1

ta − 1[ ( tc − tc−m1a+(m1n1−s1)(b−sφa) )

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1

− t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) + n2c + t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) ]

134

(B.43)

det (A4) = det

tn1b− 1tb − 1

− tn1b− 1ta − 1

tb−sφa− 1

tb − 10

t(µ1−m1n1n2)(b−sφa)− 1

t(b−sφa) − 1

t(µ1−m1n1n2)(b−sφa) tm1n2a− 1ta − 1

− tn2c−1tc −1

Φ( ∂r3

∂y1) Φ( ∂r3

∂y2) Φ( ∂r3

∂y3)

=tn2c − 1

tc − 1

[tn1b−1tb −1

− tn1b−1tb−1

tb−sφa−1)

(ta −1)

Φ( ∂r3

∂y1) Φ( ∂r3

∂y2)

]+ Φ

(∂r3

∂y3

)tn1b − 1

tb − 1

tn2c − 1

ta − 1

=tn1b − 1

tb − 1

tn2c − 1

tc − 1

[Φ

(∂r3

∂y2

)+ Φ

(∂r3

∂y1

)tb−sφa − 1

ta − 1

]− tn1b − 1

tb − 1

tn2c − 1

ta − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1(td−1)

=tn1b − 1

tb − 1

tn2c − 1

tc − 1

td − 1

ta − 1

[ ( tc − tc−m1a+(m1n1−s1)(b−sφa) )t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1


− tn1b − 1

tb − 1

tn2c − 1

tc − 1

td − 1

ta − 1

t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1(tc−1)

=tn1b − 1

tb − 1

tn2c − 1

tc − 1

td − 1

ta − 1

[ − ( tc−m1a+(m1n1−s1)(b−sφa)−1 )t[c−m1a+(m1 n1−s1)(b−sφa)](µ2−µ1 n2) − 1

tc−m1a+(m1 n1−s1)(b−sφa) − 1


= − tn2c − 1

tc − 1

tn1b − 1

tb − 1

t[ c−m1a+(m1 n1−s1)(b−sφa) ](µ2−µ1n2) + n2c − 1

ta − 1(td−1)

135

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