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On Fundamental Groups of Galois Closures of Generic Projections

So we obtain the following commutative diagramK d,n → Kn,d∗↓ ↓ ∼ =K(F d , n) ↩→ F dnWe already know that the map from K d,n to K(F d , n) is surjective. To show thatit is also injective it is enough to prove that the homomorphism from K d,n to Kd,n∗is injective.To achieve this we define a series **of** groups lying in between K d,n and Kd,n ∗ :We define K ≤kd,n to be the group generated by K d,n and the elements t a with a ≤ ksubject to the relations (†1) and (†2). Of course, only those relations that involvet a ’s and t b ’s with a, b ≤ k are imposed. Hence we obtain the following groupsand homomorphismsK n,d = K ≤0n,d → K≤1 n,d→ ... → K≤dn,d = K∗ n,d .By Lemma 5.8 each **of** these homomorphisms is injective and so the compositehomomorphism from K n,d to Kn,d ∗ is injective.□Lemma 5.7 The mapµ : K ∗ n,d → F dndefined in the pro**of** **of** Proposition 5.6 is an isomorphism **of** groups.It is clear that µ defines a surjective homomorphism.We define a map ˆµ viaPROOF.ˆµ : F dn→ K ∗ n,df an↦→ t af ai↦→ t a · f ainIf we can show that ˆµ defines a homomorphism **of** groups it will be the inverse **of**µ and it follows that µ is an isomorphism.The group F d n is generated by the elements f a i with a = 1, ..., d and i =1, ..., n subject to the commutator relations [f a i , f b j ] = 1 for all i ≠ j .First we establish two further sets **of** relations that hold true inside K ∗ d,n :in f a · t b · (f in a ) −1 =−1 t a · t b · t a i ≠ n (†3)in jn f a · f b · (f in a ) −1 =−1 jn t a · f b · t a i ≠ j, and i, j ≠ n (†4)The relation (†3) can be seen by applying (†2) to the right hand side **of** (†1) withj = n. The relation (†4) is only a reformulation **of** (†1).48

First, assume that i = n. ThenNow assume that i ≠ n. Thenˆµ([f a i , f b j ])ˆµ([f a n , f b j ])jn −1 = t a · t b f b · t a · (t b f jn b ) −1jn −1 = t a · t b f b · t a · (f jn b ) −1} {{ }apply (†3)= t a t b · t −1 b t −1 −1a t b · t b= 1t b−1in jn = t a f a · t b f b · (t a f in a ) −1 · (t b f jn b ) −1= t a f in a t b · ((f in a ) −1 f} {{ }in a ) · f jn b (f in a ) −1 t −1 a (f jn b ) −1 · (t a t} {{ }−1 −1a ) · t bapply (†3)apply (†4)= t a · t −1 a t b t a · f in a f jn b (f in a ) −1 · f in a (f jn b ) −1 (f in a ) −1 · t −1 −1 a t b= 1Hence ˆµ defines a homomorphism and so we are done.□Lemma 5.8 Keeping the notations introduced in the pro**of** **of** Proposition 5.6there is an isomorphismK ≤kn,d∼ =K≤k−1n,d⋊ ¡where the infinite cyclic ¡ group is generated by t k . In particular, the map fromto K ≤kn,dconsidered in the pro**of** **of** Proposition 5.6 is injective.K ≤k−1n,dPROOF.We want to define a map from K ≤k−1n,dto itself viaϑ : K ≤k−1n,d→ K ≤k−1n,df aij↦→ f knm · f aij · (f k nm ) −1 m ≠ i, j, nt a ↦→ f knm · t a · (f k nm ) −1 m ≠ nFirst we have to show that ϑ does not depend on the choice **of** m in the definition **of**ϑ: For the definition **of** ϑ(f a ij ) this means we have to check that for m, m ′ ≠ i, j, nf knm · f aij · (f k nm ) −1 = f knm ′ · f aij · (f knm ′ ) −1holds true. If i, j ≠ n then both expressions are equal to f ij a by relation (∗4). Ifnmi = n then we conjugate this expression with f ′k and after applying (∗2) wemmare done since f ′ k and f ij a commute using (∗4).49

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On Fundamental GroupsofGalois Closu

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ContentsIntroductioniii1 A short re

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IntroductionSchon winkt der Wein im

- Page 9 and 10: is defined by a line bundle L on X.
- Page 11 and 12: depends only on G and n and not on
- Page 13 and 14: 1 A short reminder on fundamental g
- Page 15 and 16: Miyaoka [Mi] gave a construction of
- Page 17 and 18: 2 Generic projections and their Gal
- Page 19 and 20: 2.2 Galois closures of generic proj
- Page 21 and 22: Proposition 2.12 Let L be a suffici
- Page 23 and 24: 3 Semidirect products by symmetric
- Page 25 and 26: Remark 3.2 The assumption n ≥ 3 i
- Page 27 and 28: Corollary 3.5 Let n ≥ 2.1. If G i
- Page 29 and 30: The assertions about the torsion an
- Page 31 and 32: For j ≠ i, j ≥ 2 the group X τ
- Page 33 and 34: since the product over all componen
- Page 35 and 36: 4 A first quotient of π 1 (X gal )
- Page 37 and 38: 4.2 The quotient for the étale fun
- Page 39 and 40: of K. The normalisation of X inside
- Page 41 and 42: Hence the induced homomorphism from
- Page 43 and 44: zero. Hence the proof also works al
- Page 45 and 46: an isomorphism between these two gr
- Page 47 and 48: In particular, for H = 1 we obtain
- Page 49 and 50: For an element g of the inertia gro
- Page 51 and 52: eThis automorphism has order e i an
- Page 53 and 54: Hence the homomorphisms from π top
- Page 55 and 56: 5 A generalised symmetric group5.1
- Page 57 and 58: Theorem 5.3 For n ≥ 5 there exist
- Page 59: Proposition 5.6 (Rowen, Teicher, Vi
- Page 63 and 64: We note that for affine subgroups n
- Page 65 and 66: Every element of H 2 (G) maps to an
- Page 67 and 68: PROOF. We let F d /N ∼ = G and F
- Page 69 and 70: 5.4 ExamplesWe now compute ˜K(−,
- Page 71 and 72: Theorem 5.22 For a K(G, 1)-complex
- Page 73 and 74: Example 5.24 The homomorphism ψ ma
- Page 75 and 76: 6 ConclusionJetzt nehmt den Wein! J
- Page 77 and 78: This can be also formulated as foll
- Page 79 and 80: Severi claimed that a curve with on
- Page 81 and 82: where ϕ denotes the splitting of
- Page 83 and 84: Corollary 6.3 Under the assumptions
- Page 85 and 86: Proposition 6.5 Under the isomorphi
- Page 87 and 88: Remark 6.8 Proposition 6.5 shows us
- Page 89 and 90: We remark that the group on the lef
- Page 91: E(π top1 (Z), n). By Corollary 3.3
- Page 94 and 95: 7.3 Surfaces in 3Let X m be a smoot
- Page 96 and 97: If we denote by Π g the fundamenta
- Page 98 and 99: NotationsVarieties and morphismsf :
- Page 100 and 101: [GR1][GR2][GH][SGA1]H. Grauert, R.