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On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

obtain the following two

obtain the following two short exact sequences1 → π top1 (X gal ) → π top1 (X gal , S n ) → S n → 1|| ↑ ↑1 → π top1 (X gal ) → π top1 (X gal , S (1)n−1 ) → S(1) n−1 → 1The arrows upwards are injective. We now fix a universal cover ˜X gal of X gal andalso do not mention base points unless it is important for our considerations.The quotient S n \X gal is isomorphic to2 and the inertia groups generate thekernel of the homomorphism from π top1 (X gal , S n ) onto π top1 (S n \X gal ). Since thislatter group is trivial it follows that the inertia groups generate π top1 (X gal , S n ).The quotient S (1)n−1\X gal is isomorphic to X. So the kernel of the surjectivehomomorphism onto the fundamental groups of X is the the subgroup normallygenerated by the inertia groups contained in π top1 (X gal , S (1)n−1).By Proposition 2.12 the ramification divisor R gal of f gal : X gal →2 is theunion of the curves R τ where τ runs through the transpositions of S n . We denoteby ˜p : ˜X gal → X gal the universal cover of X gal . Then we let ˜R τ be a connectedcomponent of ˜p −1 (R τ ). We have seen in the previous section that there is a uniqueinertia automorphism of the universal S n -cover ˜X gal × S n that sends ˜R τ × {1}to ˜R τ × {τ}. Since the inertia group of R τ is ¡ 2 this automorphism is the onlynon-trivial inertia automorphism of ˜R τ .We let τ 1 and τ 2 be two transpositions of S n and choose two components˜R τ1 and ˜R τ2 of ˜p −1 (R τ1 ) and ˜p −1 (R τ2 ), respectively. For the non-trivial inertiaelements r 1 and r 2 of their inertia groups we set (cf. Definition 2.11)⎧⎪⎨1 if τ 1 = τ 2c(r 1 , r 2 ) := r 1 r 2 r1 −1⎪⎩r 2 −1 if τ 1 and τ 2 are disjointr 1 r 2 r 1 r −1 2 r −1 −1 1 r 2 if τ 1 and τ 2 are cuspidal.Then we define C proj to be the subgroup normally generated by all the c(r 1 , r 2 )’sinside π top1 (X gal , S n ) where the τ i ’s run through all transpositions of S n and ther i ’s run through all inertia groups of all components of the ˜p −1 (R τi )’s.Lemma 4.5 The subgroup C proj is contained in π top1 (X gal ) and in the followingkernels:ker( π top1 (X gal , S n ) ↠ π top1 ( 2 ) )ker( π top1 (X gal , S (1)n−1 ) ↠ πtop 1 (X) )The proof is completely analogous to the proof of Lemma 4.1 and therefore leftto the reader.□40

Hence the homomorphisms from π top1 (X gal , S n ) onto π top1 ( 2 ) and the mapfrom π top1 (X gal , S (1)n−1) onto π top1 (X) factor over the quotient by C proj . Moreover,we get the following two short exact sequences1 → π top1 (X gal ) → π top1 (X gal , S n ) → S n → 1↓ ↓ ||1 → π top1 (X gal )/C proj → π top1 (X gal , S n )/C proj → S n → 1 (∗)where the arrows downwards are surjective.Proposition 4.6 We can split the short exact sequence (∗) using inertia groups.With respect to this splitting there are the following isomorphisms(π top1 (X gal )/C proj ) / (π top1 (X gal )/C proj ) Sn∼ = πtop1 ( 2 ) = {1}(π top1 (X gal )/C proj ) / (π top1 (X gal )/C proj ) ∼(1) S = πtop1 (X)n−1where the notations are the ones introduced in Section 3.1If Question 2.14 has an affirmative answer for the universal cover ˜X gal of X galthen the group C proj is trivial.PROOF. The proof is analogous to the one of Proposition 4.2:For every transposition (1 k) we choose a component of ˜p −1 (R (1 k) ) and denoteby r k the non-trivial element of its inertia group. We denote by ¯r k the image of r kinside π top1 (X gal , S n )/C proj . As in the proof of Proposition 4.2 we conclude thatthese ¯r k ’s fulfill the Coxeter relations of the symmetric group and so they provideus with a splitting s : S n → π top1 (X gal , S n )/C proj .As in the proof of Proposition 4.2 there are the following equalities for thekernel N of the homomorphism from π top1 (X gal , S n ) onto π top1 ( 2 ):andN = (π top1 (X gal )/C proj ) Sn · s(S n )N ∩ (π top1 (X gal )/C proj ) Sn = (π top1 (X gal )/C proj ) SnApplying the second isomorphism theorem of group theory we obtain the firststatement. Again, we leave the second identity to the reader.Now suppose that the components of ˜p −1 (R gal ) fulfill the connectivity propertiesof Question 2.14 with respect to the universal cover ˜p : ˜X gal → X gal .For two disjoint transpositions τ 1 and τ 2 we choose components ˜R 1 and ˜R 2 of˜p −1 (R 1 ) and ˜p −1 (R 2 ), respectively. We let r 1 and r 2 be the non-trivial elements oftheir inertia groups. We know that these components intersect in a point z. Thereis an inclusion ¡ of 2 ¡ × 2 into π top1 (X gal , S n , ˜p(z)). This group is generated byr 1 and r 2 . Hence these two elements commute and c(r 1 , r 2 ) is equal to 1.41

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