LECTURES ON ZARISKI VAN-KAMPEN THEOREM 1. Introduction ...

LECTURES ON ZARISKI VAN-KAMPEN THEOREM 25
Hence, by the corollary, the fundamental group π 1 (X) is isomorphic to G defined
by the presentation below:
〈
m = l q−1 l q−2 ···l 1 l 0 ,
〉
G := m, l j (j ∈ Z)
l j+q = ml j m −1 ,
.
∣ l j = l j+p (the monodromy relation)
Theorem 6.0.4. Suppose that p and q are prime to each other. Then the group G
has a simpler presentation G ′ := 〈 α, β | α p = β q 〉.
Proof. For any integer j, let (A j ,B j ) be a pair of integers satisfying A j q + B j p = j.
From the relations l j+q = ml j m −1 and l j = l j+p , we have
l j+q−1 ...l j = l Ajq+q−1 ...l Ajq = m Aj (l q−1 ...l 0 )m −Aj = m.
We define an element n ∈ G by
n = l p−1 l p−2 ···l 1 l 0 .
Then we have
m p = l qp−1 ...l 1 l 0 = n q .
Hence we can define a homomorphism ϕ : G ′ → G by
α ↦→ m, β ↦→ n.
We have
m Aj l 0 m −Aj = l Ajq = l Ajq+B jp = l j .
Let us calculate m A1 n B1 . Since m A1+kp n B1−kq = m A1 n B1 for any integer k, we
can assume B 1 > 0 and A 1 < 0. Then we have
m A1 n B1 =(l |A1|q ...l 1 ) −1 (l B1p−1 ...l 0 )=l 0 ,
because B 1 p − 1=|A 1 |q. Hence any l j can be written as a word of m and n. I n
particular, ϕ is surjective. It can be easily checked that
m ↦→ α, and l j ↦→ α Aj (α A1 β B1 )α −Aj
define an inverse homomorphism ϕ −1 : G → G ′ . Note that, from α p = β q , the righthand
side does not depend on the choice of (A j ,B j ). Thus ϕ is an isomorphism.
Corollary 6.0.5. The local fundamental group of the ordinary cusp x 2 − y 3 =0is
〈 α, β | α 2 = β 3 〉.
7. Fundamental groups of complements to projective plane curves
7.1. Zariski-van Kampen theorem for projective plane curves. Let C ⊂ P 2
be a complex projective plane curve defined by a homogeneous equation
Φ(X, Y, Z) =0
of degree d. Suppose that C is reduced; that is, Φ does not have any multiple factor.
We consider the fundamental group π 1 (P 2 \ C). (Since P 2 \ C is path-connected,
π 1 (P 2 \ C) does not depend on the choice of the base point.) We choose a point
a ∈ P 2 \ C. By a linear coordinate transformation, we can assume that
a =[0:0:1].
Since a/∈ C, the coefficient of Z d in Φ is not zero. Let L ⊂ P 2 be the line defined
by Z = 0. For a point p ∈ L, let pa ⊂ P 2 be the line connecting p and a. We put
X := { (p, q) ∈ L × P 2 | q ∈ pa },