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

Algebraic Geometry

By Maria R. Gonzalez-Dorrego

This monograph reviews 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 varied from 2). This Kummer floor is a quartic floor with 16 nodes as its purely singularities. those nodes provide 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 aspect 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 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 category of Kummer surfaces with specific equations and specific descriptions in their singularities. moreover, the attractive connections to the idea of K3 surfaces and abelian types are studied.

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Consider a hexagon in a plane. Then a conic can be inscribed in it if and only if the three lines connecting the three pairs of opposing vertices pass through the same point. We apply Brianchon's theorem to the conies number 1, 3 and 4 in (A). We get that the lines P1P4, P2P5, ft A pass through the same point; P1P4, P2P3, P5P6 pass through the same point and P1P4, P2P6, P3P5 pass through the same point. In other words, all of the points P2P5 Pi P 3 P 6 , P2P3 0 P5Pe and P2Pe fl P3P5 He on the line P1P4.

Then any point in the orbit of (a, 6, c, d) under W gives an isomorphic (16,6) configuration. Note that W C iV, where N = N(F0) denotes the normalizer of F 0 in PGL4(k). 61. Let N := N(Fo) be as above. Then N preserves U (since the set U can be defined purely in terms of Fo). Two points (a, 6, c, ef), (a', &', c', d') G U give rise to projectively isomorphic (16,6) configurations if they belong to the same orbit of N. Indeed, let M G N be a transformation such that M(a, 6, c, d) = (a', 6', C1 ,d').

55. 1), no three are coUinear. Proof. - We give a proof by contradiction. Suppose that the Pi are collinear. Let us introduce the following notation for the sets of indices: M i : = {11,12,13} M2 := M 3 := {J1J2J3} {kuk2,fa}. First of all, we observe that no two of the sets Mi are disjoint. 1). Consider two cases. Case 1. # ( M i fl M2) = 2. ) Without loss of generality we may take i^ = j i and i2 = j 2 . Since the Pi are collinear and Pi,P2 G w^ D Wi2, we have P 3 G w^ D Wi2. 1) (note that since the w^s are in general position, (16,6) CONFIGURATIONS AND GEOMETRY OF KUMMER SURFACES IN F 3 .

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