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单词 ProofOfMinkowskisTheorem
释义

proof of Minkowski’s theorem


Theorem 1.

Let L be an arbitrary latticeMathworldPlanetmath in Rn and let Δ be the area of a fundamental parallelepipedMathworldPlanetmath. Any convex region K symmetrical about the origin with μ(K)>2nΔ contains a point of the lattice L other than the origin.

Proof. Let D be any fundamental parallelepiped. Then obviously

n=x(D+x)

(where means disjoint unionMathworldPlanetmathPlanetmath) and thus

12𝔎=x(12𝔎(D+x)).

Now, note that

12𝔎(D+x)=((12𝔎-x)D)-x

(draw a picture!) and thus, since measure is preserved by translationPlanetmathPlanetmath,

μ(12𝔎(D+x))=μ((12𝔎-x)D)

so that if all the 12𝔎-x are disjoint, we have

2-nμ(𝔎)=μ(12𝔎)=μ(x(12𝔎(D+x)))=xμ((12𝔎-x)D)μ(D)=Δ

which is a contradictionMathworldPlanetmathPlanetmath. Thus there must exist xy and c1,c2𝔎 such that

12c1-x=12c2-y.

Thus x-y=12(c2-c1)𝔎 since 𝔎 is convex and centrally symmetricMathworldPlanetmathPlanetmathPlanetmath, and certainly x-y, so we have found a nonzero element of 𝔎Λ.

Corollary 1.

Let L be an arbitrary lattice in Rn and let Δ be the area of a fundamental parallelepiped. Any compact convex region K symmetrical about the origin with μ(K)2nΔ contains a point of the lattice L other than the origin.

Note that this corollary requires that 𝔎 be compact in additionPlanetmathPlanetmath to being convex and centrally symmetric, but slightly relaxes the volume condition on 𝔎.

Proof. Apply the previous case to Cn=(1+1n)C, i.e. dilate C. This gives a sequencePlanetmathPlanetmath of points x1,x2,,xn, with xiΛCi-{0}. But Λ is discrete, so there must be a subsequence constant at a nonzero element

xΛ(i=1Ci-{0})=ΛC¯-{0}.

Since C is compact and thus closed, xC.

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更新时间:2025/5/5 5:09:12