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 diverse from 2). This Kummer floor is a quartic floor with 16 nodes as its purely singularities. those nodes provide upward push to a configuration of 16 issues and 16 planes in ${\mathbb P}^3$ such that every airplane includes 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 reports their manifold symmetries and the underlying questions about finite subgroups of $PGL_4(k)$. She makes use of this knowledge to provide a whole category of Kummer surfaces with specific equations and specific descriptions in their singularities. additionally, the attractive connections to the speculation of K3 surfaces and abelian forms are studied.

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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). 2) ejti = or ±ietej. Let j» = 1 in the above commutation relations. 2) is impossible and the only matrices anticommuting with e1 have determinant 0). 3) where * denotes a possible non-zero entry. We may now assume that e2 = 1 for 2 < / < 4 (if not, multiply ej by t\ and by an appropriate element of Ker n).

23 Next, consider the case V D L\,V fl L2 ^ 0. Since L ^ V, L must either pass through V D L\ and be contained in the plane H generated by V and L2 or viceversa. Say, the former holds. But L must also pass through the point L$ C\ H, which is well defined and distinct from V D L\ by disjointness of the Lj's. This shows, again, that L is determined up to two possibilities. In both cases, we have proved that cri(L') fl (D^_ 1 cri(Li)) is a finite set, which means, by definition, that the four lines in question are independent.

60. 1) under ^ _ 1 (a; / )(cr) intersected with the original ten points contains four points which are not strongly coplanar but three of which are strongly coplanar. Proof. Let K denote the set {old 10 points} fl {new 10 points}. 1) then the four points in the first row satisfy the conclusion of the Lemma and there is nothing to prove. 1). 1) and at least two points in the image of the first row under <^-1(u/)(<7). If <^ -1 (u/)(