 Sharedwww / gross / lattice-aug16-01.texOpen in CoCalc
Author: William A. Stein
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\documentclass{article}
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\usepackage{fullpage}
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\title{One More Lattice Computation}
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\author{William Stein}
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\begin{document}
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\maketitle
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\tableofcontents
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\section{The Question}
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\begin{verbatim}
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Subject: one more
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From: gross@math.harvard.edu
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To: was@abel.math.harvard.edu
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Date: Thu, 16 Aug 2001 10:42:16 -0400 (EDT)
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Dear William,
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I want to request one more computation, like the one you did for me last
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week. It's in a simpler case, but I need a little more data.
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y^5 - 5*y^3 + 4*y -1
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which has discriminant 38569, and 5 real roots. A root y generates the
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ring A = Z[y] of integers in a totally real 5th degree numberfield,
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with class number 1 and strict class number 2. I calculated totally
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positive generators of the inverse different, up to squares of units,
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and got:
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d = (y^4 -4*y^2 + 1)/(5*y^4 - 15*y^2 + 4)
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d' = (y+2)*d
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if I did it right. Consider the lattices L = A with form Tr(dxy), and
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L' = A with form Tr(d'xy). Both are integral, unimodular, positive
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definite of rank 5, so must be isomorphic to Z^5. Hence both have
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10 short vectors of lenght 1, or five short vectors up to sign.
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I need the short vectors m of L and m' of L' up to sign, together
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with the 5 by 5 matrix of their inner products with respect to the
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first form: <m,m'> = Tr(dmm'). Can MAGMA handle this? I promise not
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to bother you again (for a while...).
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Thanks, Dick
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\end{verbatim}
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\section{The Input}
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\begin{verbatim}
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R<y> := PolynomialRing(RationalField());
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f := y^5 - 5*y^3 + 4*y -1;
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print "f has discriminant", Discriminant(f);
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RR := PolynomialRing(RealField());
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roots := [r : r in Roots(RR!f)];
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print "roots = ", roots;
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K<a> := NumberField(f); // a is image of y in K.
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A := MaximalOrder(K);
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"h_A = ", ClassNumber(A);
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function EmbeddingSigns(z)
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h := RR!Eltseq(K!z);
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return [Sign(Evaluate(h, roots[i])) : i in [1..#roots]];
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end function;
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d := (a^4 -4*a^2 + 1)/(5*a^4 - 15*a^2 + 4);
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dprime := (a+2)*d;
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print "\nEmbeddingSigns(d) =", EmbeddingSigns(d);
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print "EmbeddingSigns(dprime) =", EmbeddingSigns(dprime);
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print "\nNorm(d) =",Norm(d);
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print "Norm(dprime) =",Norm(dprime);
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print "Norm(1/f'(y)) =", Norm(1/Evaluate(Derivative(f),a));
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B_d := MatrixAlgebra(Integers(),5)!
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[Trace(d*A.r*A.c) : r in [1..5], c in [1..5]];
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B_dprime := MatrixAlgebra(Integers(),5)!
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[Trace(dprime*A.r*A.c) : r in [1..5], c in [1..5]];
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print "\nB_d =", B_d;
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print "B_dprime =", B_dprime;
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print "\nDeterminant(B_d) =", Determinant(B_d);
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print "Determinant(B_dprime) =", Determinant(B_dprime);
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L_d := LatticeWithGram(B_d);
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L_dprime := LatticeWithGram(B_dprime);
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short_d := [v : v in ShortVectors(L_d,1)];
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short_dprime := [v : v in ShortVectors(L_dprime,1)];
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print "\nThe length 1 vectors for L_d up to sign, are:", short_d;
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print "\nThe length 1 vectors for L_dprime up to sign, are:", short_dprime;
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print "\nThe inner-product matrix of the short vectors with respect";
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print "to the first form <m,m'> = Tr(dmm') is:";
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C := MatrixAlgebra(Rationals(),5)!
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[InnerProduct(short_d[r], L_d!Eltseq(short_dprime[c])) :
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r in [1..5], c in [1..5]];
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print "Short product matrix:", C;
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print "which has determinant", Determinant(C);
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\end{verbatim}
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\section{The Output}
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\begin{verbatim}
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f has discriminant 38569
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roots = [ 0.27583419331709204817213899775777366753,
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0.79073430354452036503335025361110446084,
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-1.1509841732749715814109750623618883790,
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2.0384952910367372139914775431571571379,
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-1.9540796146233780457859917321641468873 ]
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h_A = 1
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EmbeddingSigns(d) = [ 1, 1, 1, 1, 1 ]
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EmbeddingSigns(dprime) = [ 1, 1, 1, 1, 1 ]
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Norm(d) = 1/38569
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Norm(dprime) = 1/38569
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Norm(1/f'(y)) = 1/38569
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B_d =
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[ 1 0 1 0 2]
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[ 0 1 0 2 1]
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[ 1 0 2 1 6]
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[ 0 2 1 6 6]
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[ 2 1 6 6 22]
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B_dprime =
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[ 2 1 2 2 5]
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[ 1 2 2 5 8]
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[ 2 2 5 8 18]
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[ 2 5 8 18 34]
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[ 5 8 18 34 72]
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Determinant(B_d) = 1
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Determinant(B_dprime) = 1
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The length 1 vectors for L_d up to sign, are: [
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( 1 -2 -1 1 0),
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( 1 0 -1 0 0),
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(1 0 0 0 0),
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( 2 -1 -4 0 1),
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(0 1 0 0 0)
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]
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The length 1 vectors for L_dprime up to sign, are: [
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( 1 -3 1 2 -1),
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( 1 -2 -1 1 0),
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( 1 -2 1 2 -1),
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( 1 1 -3 -1 1),
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( 0 2 -1 -2 1)
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]
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Short product matrix:
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[ 2 1 0 -1 0]
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[ 1 0 0 0 0]
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[ 2 1 0 -1 1]
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[-1 0 0 1 0]
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[-2 -1 1 1 -1]
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which has determinant 1
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\end{verbatim}
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\end{document}
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