Lattices and Minkowski’s Theorem

Chapter 2

Geometry of Numbers

Number Theory

Preface

A lattice point is a point in Rd with integer coordinates.

Later we will talk about general lattice point.

Lattice Point

Let C ⊆ Rd be symmetric around the origin, convex, bounded and suppose that volume(C)>2d. Then C contains at least one lattice point different from 0.

Minkowski’s Theorem

Definitions* A C set is convex whenever x,y∊C  implies

segment xy∊C .*  An object C  is centrally around the origin if

whenever (0,0) ∊ C and if x∊C then -x∊C.

Examples (d=2)Vol=2*2=4<22=4Vol=4*4=16>22=4

Proof

}.:2

1{

2

1C' :define sLet' CxxC

Claim

intersect.ctor integer vean by C'

of translateand C' i.e, .v)(C'C'that

}0{\ctor vinteger ve nonzero a exists There d

such

C’

C’+v

Proof –Claim(1)

}],[:'{

:Q Define

number.integer large a be MLet

false. is claim theSuppose ion.contradictBy

ddMMvvCQ

C’

C’+v

2M

2M

C

Proof –Claim(2)

d

d

M

DCvol

CvolMCvolQ

)12

121()'(

)'()12()'(2D)(2Mvol(K)

:Hence C'. ofdiameter thedenotes D where

D]MD,-[-MK cube enlarged in the contained all are

They well.asdisjoint are s translate theseof every two thusand

,assumption the toaccording C' fromdisjoint is latesuch transEach

d

d

Volume(cube) Possibilites of v in [-

M,M]d

K2M+2D

Upperbound

Proof –Claim(3)

ion.contradictA

M. oft independenamount certain aby

1 exceedingnumber fixed a is 1)(2)vol(C'hand,other On the

M. largely sufficientfor

1 toclose arbitrarly is side hand-right on the expression The

1)12

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d-

Cvol

M

DCvol Md

Proof-Minkowski’s Theorem

theorem.sMinkowski' proves which C, that vmeans

'.2

1)(

2

1

2

1 have weso and too,C'in lies x)-x(v

segment theofmidpoint theconvex, is C' Since C'.x-obtain v we

symmetric, is C' since and ,C'v- xhave Then we

v).(C'C'point x a choose uslet claim, the toAccording

This

Cvxvx

C’

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x

ExampleLet K be a circle of diameter 26 meters centered at the origin. Trees of diameter 0.16 grow at each lattice point within K except for the origin, which is where Shrek is standing. Prove Shrek can’t see outside this mini forest.

Proof

theorem.sMinkowski'

scontradict which 4,16.40.16*26(C)But volume

origin. but the points lattice no contains

convex) itself is setsconvex ofon intersecti (The SKCset convex symetric theMeaning,

sight). thebolcked have would treea (otherwise

origin for theexcept K in point lattice no contains line middle theas

l with 0.16 width of S strip that themeans This origin. ethrough th

passing l line some along outside see couldShrek than Suppose

K

D=26m

D=0.16m

S

l

PropositionApproximating an irrational number by a fraction

nN

1

n

m- and Nn

such that n m, numbers natural ofpair a exists Then there

number. natural a N andnumber real a be (0,1)Let

Note: This proposition implies that there are infinitely many pairs m,n such that:

2

1||

nm

m

Proof

.1

n

m- Meaning,

.m)y n,(x 1

m-n and Nn gives C of definition The 0.n

assumemay wesymmetry,By m).(n,point latticeinteger nonzero some containsit

theorem,sMinkowski' toaccording thereforeand 4N

21)(2N area ofset convex symetric a is This

}1

,2

1

2

1-N- :Ry){(x,C

:Define

2

nN

N

NyxNx

General Lattices

}.Z)i,...,i,(i :i...iz{i),...,,(z

ts.coefficieninteger with z theof

nscombinatiolinear all ofset theas },...,,z { basis with lattice thedefine We

.Rin t vectorsindependenlinearly of tuple-d a be ,...,,zLet

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]}1,0[),...(:...{

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columns. as ,...z vectors with theMmatrix dd a form usLet

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llyGeometrica

z

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TheoremMinkowski’s theorem for general lattices

0. fromdifferent ofpoint a contains CThen .det2)with vol(C

setconvex symmetric a be RClet ,Rin lattice a be Let d

dd

and

Proof

theorem.in the aspoint desired theis f(v)

,ZC' vector vnonzero a provides theorem'

.2vol(C)/det)ol(C'set with vconvex symetric a is This ).(fC'

.precision)arbitrary with

cubes smallly sufficient ofunion disjoint aby edapproximat becan set convex (A

)vol(X).det((f(X)) volhave weX,set convex any

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R:f mappinglinear a define We. of basis a be }z,...{zLet

d

d1-

ddd2211d21

dd1

andsMinkowski

C

For

Rd

f

Discrete subgroup of Rd

0.number real positive fixed somefor ,least at is

of pointsdistinct any two of distance thesuch that and

,y- xalso then ,y,whenever x

such that Rset a as R of subgroup discrete a define We dd

TheoremLattice basis theorem

).z,...,z,z(such that Rz,...,z,z

t vectoresindependenlinearly d exsits thereis, that basis; a has Then

.R isspan linear whoseR of subgroup discrete a be R

d21d

d21

ddd

Let

Proof(1)

theorem. theofstatement thegives this1,diFor .z,...,z,z

of nscombinatiolinear integer as written becan Fin lying of points

all then ,z,...,z,zby spanned subspace ldimensiona 1)-(i thedenotes F

If d.constructebeen already haveproperty following with thez,...,z,z

t vectorsindependenlinearly that suppose 1,di1 i, someFor induction.

1-i21

1-i

1-i211-i

1-i21

By

Proof(2)

property. required thehave

z,...,z,z that show toremainsIt .smallest with theone is zthen

,z...zz form in the written are P of points theif that Note

.zit call and Fnearest one choose ,Fin not but Pin lying of points theall Among

]}.1,0[,...,:z...zz{P

:by and z,...,z,zby determined pedparallelpi ldimensiona-i thebe PLet

.F subspace in the lyingnot vector a exists there,R generates d.ian consider So

i21ii

i1-i1-i2211

i1-i1-i

i1i1-i1-i2211

1-i21

1-id

Proof(3)

ts.coefficieninteger fact in are theall So .0

other theallget that also we,hypothesis inductive by the and ,F v'Hence

. than F nearer to be would v'otherwisefor 0 havemust

weTherefore, P. ipedparallelep in the lies v'10 Since .in lies also '

....v'

i.1,2,...,j ,

:Define .,..., numbers real some

for ... vcan write We.Fin lyingpoint a be let v

j

1-i

1-ii

j

i2211

j

1

i2211i

j

i

i

jj

i

i

z

v

zzz

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v

v’

Question…

How efficiently can one actually compute a nonzero lattice point in a symmetric convex body?

hard.-NP be known to is problem theinput, theofpart a as considered is d If

time.polynomialin solved becan problemsuch constant, a as considered is d If

An application in Number Theory

Theorem

.,

,ap :squares twoof sum a as written becan 4) (mod 1p primeEach 22

Zba

b

LemmaIf p is a prime with p ≡ 1(mod 4) then -1 is a quadric residue modulo p.

  For a given positive integer n, two integers a and b are called congruent modulo n, written a ≡ b (mod n) if a-b is divisible by n.For example, 37≡57(mod 10) since 37-57=-20 is a multiple of 10.

Definitions-Number Theory

.nonresidue quadratic a is a Otherwise,

p). (mod with x*Fan x exists

thereif p modulo residue quadratic a called is *Faelement An

{0}.\FF*let and number, prime a is p where

p, modulo classes residue of field for the stand GF(p)FLet

2 a

Example: 42≡6(mod 10) so 6 is a quadratic residue (mod 10).

Proof(Theorem)

.a Therefore

).(mod0)1(2)(a calculate We

).,z of definition(by jpiqb and ia that meanswhich

Z,ji, somefor izb)(a, time,same At the .2a0 have We

{0}.\b)(a,point a contains C so and ,4det 4pp2 is C of area The

2p}.yx:Ry){(x, Cdisk the

for lattices generalfor theoremsMinkowski' use Wep.det have We

p).(0,z and q)(1, z where),,(z lattice heConsider t

).(mod1qsuch that qnumber a choosecan welemma, By the

22

22222222222

21

2122

222

2121

2

pb

pqipjiqjpqiijiqib

z

jzpb

z

p

2pC

0≣q2≣-1(mod p)