2 years ago

On Fundamental Groups of Galois Closures of Generic Projections

On Fundamental Groups of Galois Closures of Generic Projections

So we obtain the

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 proof 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 ofµ 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 proof 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 proof 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 ofϑ: 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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