By Eriko Hironaka

ISBN-10: 082182564X

ISBN-13: 9780821825648

This paintings stories abelian branched coverings of gentle complicated projective surfaces from the topological perspective. Geometric information regarding the coverings (such because the first Betti numbers of a tender version or intersections of embedded curves) is expounded to topological and combinatorial information regarding the bottom house and department locus. precise consciousness is given to examples during which the bottom area is the complicated projective airplane and the department locus is a configuration of strains.

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**Additional info for Abelian Coverings of the Complex Projective Plane Branched Along Configurations of Real Lines**

**Sample text**

Recall that the curves in p~l(C) for any curve C in C are in one to one correspondence with cosets of the stabilizer subgroup He associated to C. Thus our goal now is to find the stabilizer subgroups explicitly. 4 PROPOSITION . Consider G as the quotient of the free abelian TLjnTLmodule Ak of rank k with basis # i , . . , #* by the submodule generated by g\ -f 92 4\- 9k> For each curve C C C we have: (1) ifC = Li, for i = 1 , . . , t , then He is the submodule ofG generated by the relation-free elements for all peLiC) S; (2) if C = Equ, for u = 1 , .

Take any two intersecting curves C and D in £. If C and D are the proper transforms L\ and L2, then Ic = {gLi) and ID = (<7L2). These intersect in the identity in G, since there is at least one more generator in G. If one of C and D is an exceptional curve, say C = Ep, and D is the proper transfoinkL, then Ic = ( £ 9L>) PZL'CC and ID = (9L). 34 ERIKO HIRONAKA In order for these to intersect nontrivially there must be a nontrivial relation among the gi' where V ranges over lines in C passing through p.

The only point in p~x(p) H Vq is g, so a(g) = g. But this implies that a is in Ip which is contained in He, so or(Ci) = Ci and 92 G Ci. This means that C\ and C2 intersect in p~1(p,)> D u t P; w a s chosen arbitrarily in [/, so Ci and C% intersect at all points in the open set Vq C\p~l{C). Therefore, C\ and C2 must be the same curve. 3 LEMMA. If C' is any irreducible component of p""1(C)J then the self intersection C" equals wc- Proof. Consider C as a divisor on Y and let p*C be its pullback. 6, pp.

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