Zariski-Van Kampen Method

Transkript

1 Zariski-Van Kampen Method Purpose: Obtain a presentation for the fundamental group of the complement of a plane projective curve in P 2. We will put together several ingredients, among which, the Van Kampen Theorem is key.

2 Let π : X M be a locally trivial fibration with section s : M X. Consider p M and x 0 F p.

3 Let π : X M be a locally trivial fibration with section s : M X. Consider p M and x 0 F p. Theorem π 1 (X, x 0 ) = π 1 (F p, x 0 ) π 1 (M, p), where the action of π 1 (M, p) on π 1 (F p, x 0 ) is given by the monodromy of π.

4 Let π : X M be a locally trivial fibration with section s : M X. Consider p M and x 0 F p. Theorem π 1 (X, x 0 ) = π 1 (F p, x 0 ) π 1 (M, p), where the action of π 1 (M, p) on π 1 (F p, x 0 ) is given by the monodromy of π. Proposition Meridians around the same irreducible components of B are conjugate in π 1 (M \ B). Moreover, the conjugacy class of a meridian coincides with the set of homotopy classes of meridians around the same irreducible component.

5 Let π : X M be a locally trivial fibration with section s : M X. Consider p M and x 0 F p. Theorem π 1 (X, x 0 ) = π 1 (F p, x 0 ) π 1 (M, p), where the action of π 1 (M, p) on π 1 (F p, x 0 ) is given by the monodromy of π. Proposition Meridians around the same irreducible components of B are conjugate in π 1 (M \ B). Moreover, the conjugacy class of a meridian coincides with the set of homotopy classes of meridians around the same irreducible component. Proposition The inclusion M \ B M induces a surjective morphism, whose kernel is given by the smallest normal subgroup of π 1 (M \ B) containing meridians of all the irreducible components of B.

6 Zariski-Van Kampen Theorem Let C P 2 be a projective plane curve. Consider P = [0 : 1 : 0] P 2 \ C.

7 Zariski-Van Kampen Theorem Let C P 2 be a projective plane curve. Consider P = [0 : 1 : 0] P 2 \ C.

8 Zariski-Van Kampen Theorem Let C P 2 be a projective plane curve. Consider P = [0 : 1 : 0] P 2 \ C. Project π : P 2 \ {P} P 1 from P

9 Zariski-Van Kampen Theorem Let C P 2 be a projective plane curve. Consider P = [0 : 1 : 0] P 2 \ C. Project π : P 2 \ {P} P 1 from P z n z n 1 z 2 z 1

10 Zariski-Van Kampen Theorem Let C P 2 be a projective plane curve. Consider P = [0 : 1 : 0] P 2 \ C. Project π : P 2 \ {P} P 1 from P L n L n 1 L 2 L 1 z n z n 1 z 2 z 1

11 Zariski-Van Kampen Theorem L n L n 1 L 2 L 1 z n z n 1 z 2 z 1 Remark (1) Let X = P 2 \ (C L), then π X : X P 1 \ Z n is a locally trivial fibration.

12 Zariski-Van Kampen Theorem L n L n 1 L 2 L 1 L 0 z n z n 1 z 2 z 1 z 0 Remark (1) Let X = P 2 \ (C L), then π X : X P 1 \ Z n is a locally trivial fibration. Moreover, its fiber is P 1 \ Z d, where d := deg C.

13 Zariski-Van Kampen Theorem xx xxxx xxxxxx xxxxxxx xxxxxxxxx xxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxx xxxxxxxxx xxxxxxxxx xxxxxxx xxxxxx xxxxx xxxx xxx xx x Remark (2) By (2.1), π 1 (X, x 0 ) = π 1 (F z0, x 0 ) π 1 (P 1 \ Z n, z 0 ). Action is given by the monodromy of π 1 (P 1 \ Z n, z 0 ) on π 1 (F z0, x 0 ).

14 Zariski-Van Kampen Theorem x 0 xx xxxx xxxxxx xxxxxxx xxxxxxxxx xxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxx g xxxxxxxx xxxxxxxxx 1 xxxxxxxx xxxxxxxx g xxxxxxxxx xxxxxxxx 2 xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx. xxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxxxxxxx xxxxxxxx g xxxxx xxxxxxxxxxx xxxxxxxx d xxx xxxxxxxxx xxxxxxxxx xxxxxxx xxxxxx xxxxx xxxx xxx xx x γ n γ n 1 γ 2 γ 1 z 0 Remark (3) Note that π 1 (F z0, x 0 ) = g 1,..., g d : g d g d 1 g 1 = 1 and π 1 (P 1 \ Z n, z 0 ) = γ 1,..., γ n : γ n γ 1 = 1.

15 Zariski-Van Kampen Theorem x 0 xx xxxx xxxxxx xxxxxxx xxxxxxxxx xxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxx g xxxxxxxx xxxxxxxxx 1 xxxxxxxx xxxxxxxx g xxxxxxxxx xxxxxxxx 2 xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx. xxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxxxxxxx xxxxxxxx g xxxxx xxxxxxxxxxx xxxxxxxx d xxx xxxxxxxxx xxxxxxxxx xxxxxxx xxxxxx xxxxx xxxx xxx xx x γ n γ n 1 γ 2 γ 1 z 0 Theorem π 1 (X, x 0 ) admits the following presentation: g 1,..., g d, γ 1,..., γ n : g d g d 1 g 1 = γ n γ 1 = 1, g γ j i = γ 1 j g i γ j

16 Zariski-Van Kampen Theorem x 0 xx xxxx xxxxxx xxxxxxx xxxxxxxxx xxxxxxxxxxx xxxxxxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxx g xxxxxxxx xxxxxxxxx 1 xxxxxxxx xxxxxxxx g xxxxxxxxx xxxxxxxx 2 xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx. xxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxxxxxxx xxxxxxxx g xxxxx xxxxxxxxxxx xxxxxxxx d xxx xxxxxxxxx xxxxxxxxx xxxxxxx xxxxxx xxxxx xxxx xxx xx x γ n γ n 1 γ 2 γ 1 z 0 Theorem π 1 (P 2 \ C) admits the following presentation: g 1,..., g d : g d g d 1 g 1 = 1, g γ j i = g i

17 Remark Let C = C 1... C r the decomposition of C in its irreducible components, then H 1 (P 2 \ C) = Z r 1 Z/(d 1,..., d r), where d i := deg C.

18 Remark Let C = C 1... C r the decomposition of C in its irreducible components, then where d i := deg C. H 1 (P 2 \ C) = Z r 1 Z/(d 1,..., d r), It two curves are in a connected family of equisingular curves, then they are isotopic

19 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ.

20 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. For computation purposes it is more convenient to use a non-generic projection.

21 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. For computation purposes it is more convenient to use a non-generic projection. Use for instance C := {F = 0}, where F(X, Y, Z) = X d + Y d Z d. P = [0 : 1 : 0] / C and F Y = dy d 1.

22 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. For computation purposes it is more convenient to use a non-generic projection. Use for instance C := {F = 0}, where F(X, Y, Z) = X d + Y d Z d. P = [0 : 1 : 0] / C and F Y = dy d 1. Let us compute the local monodromy of x = y d. Consider γ(t) = e 2πt 1 a loop around x = 0. The fiber at γ(t) is given by: ξ i d

23 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. The monodromy around x = 0 looks as follows:

24 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. Corresponds to the braid σ 1 σ 2 σ d 1

25 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. Note that the global part of the monodromy has no contribution:

26 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. Applying the Zariski-Van Kampen Theorem to these generators:

27 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. Applying the Zariski-Van Kampen Theorem to these generators: One obtains: g i = g (σ 1σ 2 σ d 1 ) i = ( g d i = 1 g 1 d g i 1 g d i 1

28 Zariski-Van Kampen Theorem C smooth of degree d π 1 (P 2 \ C) = Z/dZ. Applying the Zariski-Van Kampen Theorem to these generators: One obtains: ( g i = g (σ 1σ 2 σ d 1 ) g i = d i = 1 g 1 d g i 1 g d i 1 hence g 2 = g 1 d g 1 g d = g 1, and by induction g 1 =... = g d = g. Finally, g 1 g d = 1 becomes g d = 1 π 1 (P 2 \ C) = g : g d = 1 = Z/dZ.

29 (Zariski-Harris, Cheniot) C nodal π 1 (C) is abelian.

30 (Zariski-Harris, Cheniot) C nodal π 1 (C) is abelian. C 1 and C 2 intersect transversally π 1 (C) = π 1 (C 1 ) π 1 (C 2 )

31 (Zariski-Harris, Cheniot) C nodal π 1 (C) is abelian. C 1 and C 2 intersect transversally π 1 (C) = π 1 (C 1 ) π 1 (C 2 ) Remark (Harris) The space of irreducible nodal curves with given number of nodes is connected

32 (Zariski-Harris, Cheniot) C nodal π 1 (C) is abelian. C 1 and C 2 intersect transversally π 1 (C) = π 1 (C 1 ) π 1 (C 2 ) Remark (Harris) The space of irreducible nodal curves with given number of nodes is connected (Zariski) Let C be a general nodal rational curve of degree d. Consider C its dual. Note that C is a rational curve of degree 2(d 1), 2(d 2)(d 3) nodes, and 3(d 2) cusps. The fundamental group of C coincides with the fundamental group of the unordered configuration space of d points in S 2, that is, B d (S 2 ) = g 1,..., g d 1 : g i g j = g j g i, g i g i+1 g i = g i+1 g i g i+1, g 1 g d 2 g 2 d 1 g d 2 g 1 = 1.

33 Non-Generic Projections P C that is, existence of asymptotes.

34 Non-Generic Projections P C that is, existence of asymptotes. Very special fibers.

35 Local Braid Monodromy Can be obtained from the Puiseux Series (local parametrization) of the curve around a singular point.

36 Local Braid Monodromy Can be obtained from the Puiseux Series (local parametrization) of the curve around a singular point. Computational methods are generically effective.

37 Global Braid Monodromy Most difficult part of monodromy computations.

38 Global Braid Monodromy Most difficult part of monodromy computations. Real arrangements, real curves.

39 Global Braid Monodromy Most difficult part of monodromy computations. Real arrangements, real curves. Computational methods are effective essentially over Z[ 1].

40 Consider the following quartic:

41 Consider the following quartic and project from [0 : 1 : 0]

42 Compute the braid monodromy:

43 Compute the braid monodromy: σ 8 1,

44 Compute the braid monodromy: σ 8 1, σ 2,

45 Compute the braid monodromy: σ 8 1, σ 2, σ 1 3 σ 1 1 σ 2σ 1 σ 3.

46 Compute the braid monodromy: σ 8 1, σ 2, σ 1 3 σ 1 1 σ 2σ 1 σ 3.

47 σ 8 1 :

48 σ 8 1 : g σ8 1 1 = (g 2 g 1 ) 4 g 1 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 1 ] = 1

49 σ 8 1 : g σ8 1 1 = (g 2 g 1 ) 4 g 1 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 1 ] = 1 g σ8 1 2 = (g 2 g 1 ) 4 g 2 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 2 ] = 1

50 σ 8 1 : g σ8 1 1 = (g 2 g 1 ) 4 g 1 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 1 ] = 1 g σ8 1 2 = (g 2 g 1 ) 4 g 2 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 2 ] = 1 g σ8 1 3 = g 3

51 σ 8 1 : g σ8 1 1 = (g 2 g 1 ) 4 g 1 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 1 ] = 1 g σ8 1 2 = (g 2 g 1 ) 4 g 2 (g 2 g 1 ) 4 [(g 2 g 1 ) 4, g 2 ] = 1 g σ8 1 3 = g 3 g σ8 1 4 = g 4

52 σ 2 :

53 σ 2 : g σ 2 1 = g 1

54 σ 2 : g σ 2 1 = g 1 g σ 2 2 = g 3 g 2 = g 3

55 σ 2 : g σ 2 1 = g 1 g σ 2 2 = g 3 g 2 = g 3 g σ 2 3 = g 3 g 2 g 1 3 g 2 = g 3

56 σ 2 : g σ 2 1 = g 1 g σ 2 2 = g 3 g 2 = g 3 g σ 2 3 = g 3 g 2 g 1 3 g 2 = g 3 g σ 2 4 = g 4

57 σ 1 3 σ 1 1 σ 2σ 1 σ 3 :

58 σ 1 3 σ 1 1 σ 2σ 1 σ 3 : g σ 1 σ 1 σ σ 1 σ 3 1 = g 1 2 g 4 g 2 g 4 = g 2 g 1 g 1 2

59 σ 1 3 σ 1 1 σ 2σ 1 σ 3 : g σ 1 σ 1 σ σ 1 σ 3 1 = g 1 2 g 4 g 2 g 4 = g 2 g 1 g 1 2 g σ 1 σ 1 σ σ 1 σ 3 2 = g 2

60 σ 1 3 σ 1 1 σ 2σ 1 σ 3 : g σ 1 σ 1 σ σ 1 σ 3 1 = g 1 2 g 4 g 2 g 4 = g 2 g 1 g 1 2 g σ 1 σ 1 σ σ 1 σ 3 2 = g 2 g σ 1 σ 1 σ σ 1 σ 3 3 = g 3

61 σ 1 3 σ 1 1 σ 2σ 1 σ 3 : g σ 1 σ 1 σ σ 1 σ 3 1 = g 1 2 g 4 g 2 g 4 = g 2 g 1 g 1 2 g σ 1 σ 1 σ σ 1 σ 3 2 = g 2 g σ 1 σ 1 σ σ 1 σ 3 3 = g 3 g σ 1 σ 1 σ σ 1 σ 3 4 = g 4 g 2 g 1 g 1 2 g 1 4 g 4 = g 2 g 1 g 1 2