By Maria R. Gonzalez-Dorrego

This monograph reports the geometry of a Kummer floor in ${\mathbb P}^3_k$ and of its minimum desingularization, that is 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 simply singularities. those nodes provide upward push to a configuration of 16 issues and 16 planes in ${\mathbb P}^3$ such that every aircraft includes precisely six issues and every aspect belongs to precisely six planes (this is named a '(16,6) configuration').A Kummer floor is uniquely made up our minds by means 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 data to provide an entire class of Kummer surfaces with particular equations and specific descriptions in their singularities. additionally, 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**

**Sample text**

38 MARIA R. GONZALEZ-DORREGO Each of the ej is a diagonaHzable matrix with eigenvalues 1,-1,-1. 3) together with det ei = 1 imply that both e$ and e 4 commute with e2. 2), e 4 must commute with both e 2 and e3, hence e 4 must be a diagonal matrix with eigenvalues 1,-1,-1. But then e 4 belongs to the subgroup of SL^k) generated by e 2 and e3, which is a contradiction. The Sublemma is proved. • From now on we assume that e2j = 1, 1 < j < 4. Then each of the e7- is a diagonaHzable matrix with eigenvalues 1,1,-1,-1.

5) are in the span of {vjt }i

16,6) CONFIGURATIONS AND GEOMETRY OF KUMMER SURFACES IN P 3 . 48. The map 0 is injective. Proof. 1) together with their position in the incidence diagram completely determine our (16,6) configuration of type (*). Indeed, the following ten points of the configuration are uniquely determined, since each of them is an intersection point of three of our planes (and since by the previous lemma no three of our planes have a line in common). o o o o o o • • Each of the remaining 10 special planes contains three or more of the points marked by o above.