By Idun Reiten, Sverre O. Smalø, Øyvind Solberg
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Extra resources for Algebras and Modules One
Geometry of Numbers l 45 ( Fig. 9. lO. Example of a fundamental domain for all points P from IP. "" XN) are the coordinates of P. The integral is defined by [ f(P)dP= [00 ... [00 f(Xl, ... ,XN)dxl ... dxN, Loo 1-00 lIP and more generally by [ f(P)dP= [00 ... [00 CM(Xl, ... ,XN)f(Xl, ... ,XN)dxl ... dxN 1M Loo 1-00 The transformations 46 3. 5). , f(P)dP = 1. (~f(P +9)) 1. 5) would give a contradiction to the assumption. If in this example one restricts attention to f measurable set, the volume of M is given by = CM, where M is a bounded L cM(P)dP = vol(M) .
13. Convex sets then K contains at least one lattice point Q from the point lattice P + (5 based at P. = P + g, distinct from P, 9 E (5, Proof. The idea of the proof is to apply Blichfeldt's theorem for the set M of all P + a, as P + a runs through all points from K. Since ! vol(M) 1 = 2N vol(K) > 1 2N . 2N . d((5) = d((5) by Proposition 4 at least two points of the form R + g', R + g" lie in M for + (PR + g'), P + (PR + g") lie in M, by definition the points P + 2(PR + g') and P + 2(PR + g") are in K.
XNeNi IXll + ... + Ix N I ~ r . p For N = lone of course has that vol(Kr ) = vol(KriN = 1) = 2r: recursively one obtains the volume of the N -dimensional octahedron for N > 1 from vol(Kri N) = j ... j j j dXl ... dXN Ixd+ ... +lxN I~r = j ... j dXN' Ixd+ .. ·+lxN-d~r-lxNI IXNI~r vol(Kr-lxNliN = dXl ... dxN-l IXNI~r j = vol(KliN -1)· -1). e. because vol(Kli N = 1) = 2, If, noting that vol(Kr) (2r)N = N! one applies Minkowski's theorem, the essential condition reads (2r )N > 2N . d(~) N! e. r ~ VN!