16, 6 Configurations and Geometry of Kummer Surfaces in P3 - download pdf or read online

By Maria R. Gonzalez-Dorrego

ISBN-10: 0821825747

ISBN-13: 9780821825747

This monograph experiences the geometry of a Kummer floor in ${\mathbb P}^3_k$ and of its minimum desingularization, that's a K3 floor (here $k$ is an algebraically closed box of attribute various from 2). This Kummer floor is a quartic floor with 16 nodes as its merely singularities. those nodes supply upward thrust to a configuration of 16 issues and 16 planes in ${\mathbb P}^3$ such that every airplane comprises precisely six issues and every element belongs to precisely six planes (this is named a '(16,6) configuration').A Kummer floor is uniquely made up our minds via its set of nodes. Gonzalez-Dorrego classifies (16,6) configurations and reviews their manifold symmetries and the underlying questions about finite subgroups of $PGL_4(k)$. She makes use of this data to offer a whole class of Kummer surfaces with particular equations and particular descriptions in their singularities. additionally, the gorgeous connections to the idea of K3 surfaces and abelian forms are studied.

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Additional resources for 16, 6 Configurations and Geometry of Kummer Surfaces in P3

Sample text

To prove (a), it remains to exhibit an element a oi I which has order at least 4 modulo W (by this we mean that an £ W for n < 4). 2) Vo 0 z 0 0 0 0 0 i 0^ 0 1 0, One can easily check directly that a £ I and that the order of a modulo W is 4 (we have

To show that ip is an isomorphism, it is sufficient to show that the groups on the left and the righthandside have the same order, that is, #N = 2 8 • 3 2 • 5. For that it is sufficient to show two things: (a) The stabilizer / in N of an isotropic line in Pj. 64 will imply that jr = 5 4 x ¥2 [4, p. [4], line 44]). (b) N acts transitively on the set of isotropic lines. Since there are 15 isotropic lines, (a) and (b) together will show that 2 4 • 3 • 15 < # ^ - . Since we already know that # - ~ - | # 5 6 , we will obtain that # ^ r = # S 6 and ip is an isomorphism.

To prove that H and / belong only to those special planes which are prescribed by the (16,6) configuration, it is enough to check that H (fc 2 , 3 . Since all such arguments are very similar to each other, we go over them briefly. 2, 1' and I do not have a line in common (otherwise we would have 12' fl PR ^ 0), hence 2 fl 1' fl I = 2'. If H e 2 then fr = 2 n i ' n l = 2' which is a contradiction since 2' £ 1 but H e 1. (16,6) CONFIGURATIONS AND GEOMETRY OF KUMMER SURFACES IN P 3 . 25 Similarly, 1, 3 and I do not have a line in common, hence 1 H 3 fl I = 2.

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16, 6 Configurations and Geometry of Kummer Surfaces in P3 by Maria R. Gonzalez-Dorrego

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