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

By Maria R. Gonzalez-Dorrego

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 diversified from 2). This Kummer floor is a quartic floor with 16 nodes as its in simple terms singularities. those nodes supply upward thrust to a configuration of 16 issues and 16 planes in ${\mathbb P}^3$ such that every aircraft comprises precisely six issues and every element belongs to precisely six planes (this is termed a '(16,6) configuration').A Kummer floor is uniquely decided by way of its set of nodes. Gonzalez-Dorrego classifies (16,6) configurations and experiences their manifold symmetries and the underlying questions on finite subgroups of $PGL_4(k)$. She makes use of this knowledge to provide a whole category of Kummer surfaces with specific equations and particular descriptions in their singularities. moreover, the gorgeous connections to the idea of K3 surfaces and abelian kinds are studied.

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

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43. I V . A n o n - d e g e n e r a t e (16,6) configuration in P 3 is of t h e form (a, b, c, d) of ( 1 . 4 . 1 ) . 44. We say that a set of planes is in general linear p o s i t i on if no four of them pass through the same point. 45. 6, up to an automorphism of P 3 . in P 3 is of the form de- Proof. 1). As usual, let us denote a plane in P 3 by a 4-tuple of elements of k. We may view this 4-tuple as a point in P 3 . Let V denote the open subset of (P 3 ) 6 consisting of all the ordered 6-tuples of planes in general linear position.

Lift the ej's in an arbitrary way to elements ej G SL±(k), 1 < j < 4. Then e^ = ± 1 or ± i (where z is an element of A; such that i2 = —1, which we fix once and for all; here and below we identify an element a G k with a times the identity matrix in SL^k)). Multiplying by z, if necessary, we may assume that e2- = 1 or i. The first step is to rule out the case e2- = i. 53. e) ^ i, 1 < j < 4. Proof. Suppose the contrary. Say, e\ — i. Then e\ is a diagonalizable matrix whose eigenvalues are ± \ A (this notation should cause no confusion; again, we choose and fix once and for all an element \Ji of k whose square is i).

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