By Eriko Hironaka

This paintings reviews abelian branched coverings of tender complicated projective surfaces from the topological standpoint. Geometric information regarding the coverings (such because the first Betti numbers of a delicate version or intersections of embedded curves) is said to topological and combinatorial information regarding the bottom house and department locus. exact consciousness is given to examples during which the bottom area is the complicated projective aircraft and the department locus is a configuration of strains.

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**Read e-book online Abelian Coverings of the Complex Projective Plane Branched PDF**

This paintings experiences abelian branched coverings of gentle advanced projective surfaces from the topological perspective. Geometric information regarding the coverings (such because the first Betti numbers of a gentle version or intersections of embedded curves) is expounded to topological and combinatorial information regarding the bottom house and department locus.

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The matrix M defined as above is the point/curve incidence matrix for the curves in C and the points of intersection S in C ordered as above. THE COMPLEX PROJECTIVE PLANE 57 B. Shift matrix for £. To find the shift matrix for C we begin by finding one for C. IV. 2 Definition. Let a , j , i = 1, . . , s ; j = 1 , . . ,k be the entries of M. Define the shift matrix Sh(£) with entries 6,-j inductively on i as follows. (1) Row 1: bltj = 0 for all j = 1 , . . , Jb. (2) Rowi: bi-ij if a$J- = 0 or j = i2 or j = i?

9 in more detail. Let ti,... ,<* be the locally ordered points in 7\ with respect to the fiber P~x(l). 9 that the local monodromy around the fiber over 0 fixes the locally ordered points numbered 1 , . . ,£ + d — 1 counterclockwise by 180 degrees. 11. It is important to notice where the center of rotation of this disk is in relation to the points £ , . . , £ -f d — 1. Let R be the first (global) index so that the line corresponding to IR has positive slope and the line corresponding to IR+\ has negative slope.

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. 20 - 21). Each component of p*C counts with multiplicity \Ic\. 2, so we have |G|C2 = |/c|2 £ C'\ 28 ERIKO HIRONAKA The number of irreducible components in p~l{C) is the index oi He in G. Since the covering is Galois, all the components have the same self intersection. Therefore, \G\C* = for a given C" C p~l(C). C>\ Multiplying both sides of this equation by \Hc\ \IcV\G\ finishes the proof.