2 years ago

On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

Question 2.14 Is it true

Question 2.14 Is it true that for distinct transpositions τ 1 and τ 2 every irreduciblecomponent of p −1 (R τ1 ) intersects every irreducible component of p −1 (R τ2 ) ?In this thesis we want to compute the fundamental groups π top1 (X gal ) andπ top1 (Xgal aff ). The main result (Theorem 6.2) is that there is always a surjectivehomomorphismπ toptop) ↠ ˜K(π1 (X aff ), n)1 (Xgal affwhere ˜K(−, n) is the group-theoretic construction defined in Section 5.3.Now, if Question 2.14 has an affirmative answer for all topological covers ofthen the kernel of this surjective homomorphism is trivial.X affgal10

3 Semidirect products by symmetric groups3.1 Definition of K( − ,n) and E( − ,n)Ein voller Becher Weins zur rechten ZeitIst mehr wert, als alle Reiche dieser Erde!Dunkel ist das Leben, ist der Tod.Let G be an arbitrary group and n ≥ 2 be a natural number. We denote by θ thepermutation representation of the symmetric group S n on G n given byθ : S n → Aut(G n )σ ↦→ ( θ(σ) : (g 1 , ..., g n ) ↦→ (g σ −1 (1), ..., g σ −1 (n)) )Then we form the split extension of groups with respect to θ1 → G n → G n ⋊ θ S n → S n → 1and denote by s : S n → G n ⋊ θ S n the associated splitting.We define the subgroup E(G, n) of G n ⋊ θ S n to be the group generated byall conjugates of s(S n ) and define K(G, n) to be the intersection G n ∩ E(G, n).Hence we get a split extension1 → K(G, n) → E(G, n) → S n → 1.We give another characterisation of these groups in Proposition 3.3.More generally, let S be a subgroup of S n . Then we defineE(G, n) S := 〈⃗gs(σ)⃗g −1 |⃗g ∈ G n , σ ∈ S〉 ≤ E(G, n)K(G, n) S := E(G, n) S ∩ K(G, n) K(G, n) .These subgroups remain the same when passing to a G n -conjugate splitting. Wewill therefore suppress s in future. Clearly, K(G, n) S is always a normal subgroupof G n and K(G, n).In the notation introduced in Definition 2.10 we have the following equalitiesand isomorphisms:E(G, n) Sn = E(G, n)E(G, n) (i) S n−1∼ = E(G, n − 1) for n ≥ 3and similarly for K(−, n).Later on we have to deal with subgroups of K(G, n) that are generated byK(G, n)-conjugates of a subgroup S of S n rather than G n -conjugates. Fortunately,we have the following11

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