Transcript Lattices and Minkowski`s Theorem

Chapter 2
A lattice point is a point in Rd with integer coordinates.
Later we will talk about general 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.
* 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.
Vol=4*4=16>22=4
Vol=2*2=4<22=4
Let' s define : C' 
1
1
C  { x : x  C}.
2
2
T hereexistsa nonzerointeger vector v d \ {0} such
thatC'(C' v)  . i.e, C' and translateof
C' by an integer vectorintersect.
C’+v
C’
C’+v
By contradiction.Suppose theclaim is false.
Let M be a large integernumber.
DefineQ :
Q  {C 'v : v  [ M , M ]d   d }
C
2M
C’
2M
Each such translateis disjoint fromC' accordingto theassumption,
and thusevery two of thesetranslates are disjoint as well. T hey
are all containedin theenlargedcube K  [-M - D, M  D]d
where D denotesthediameterof C'. Hence:
vol(K) (2M  2D)d  Q vol(C ' )  (2M  1) d vol(C ' )
Volume(cube)
Upper
bound
Possibilites of
v in [-M,M]d
2D 1 d

 vol(C ' )  (1 
)
2M  1
2M+2D
K
vol(C ' )  (1 
2 D  1 d M  
) 1
2M  1
T heexpressionon theright - handside is arbitrarlyclose to1
for sufficiently large M.
On theotherhand,vol(C')  2-d vol(C )  1 is a fixed number exceeding1
by a certainamountindependent of M.
A contradiction.
Accordingto theclaim,let us choosea point x C'(C' v).
T hen wehave x - v  C' , and since C' is symmetric,
we obtain v- x  C'. Since C' is convex,themidpointof thesegment
1
1
1
x(v - x) lies in C' too,and so we have x  (v  x)  v  C '.
2
2
2
This means that v C, which provesMinkowski's theorem.
C’+v
x
C’
Let 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.
Suppose t hanShrek could see out side alongsome line l passing
t hrough t he origin.T hismeans t hat t hest ripS of widt h 0.16 wit h l
as t hemiddle line cont ainsno lat t icepointin K exceptfor t heorigin
(ot herwisea t ree would havebolcked t hesight ).
Meaning,t hesymet ricconvexset C  K  S (T heint ersecton
i of convexset s is it self convex)
cont ainsno lat t icepoint sbut t heorigin.
But volume(C)  26 * 0.16  4.16  4, which cont radicts
Minkowski's t heorem.
K
D=26m
D=0.16m
l
S
Let   (0,1)be a real number and N a naturalnumber.
T hen thereexistsa pair of naturalnumbersm, n such that
n  N and  -
m
1

n nN
Note: This proposition implies that there are infinitely many pairs
m,n such that:
| 
m
1
| 2
m n
Define:
C  {(x, y) R 2 : - N -
1
1
1
 x  N  , x  y  }
2
2
N
2
 4 and thereforeaccordingto Minkowski's theorem,
N
it containssome nonzerointegerlatt icepoint(n,m).By symmetry,we may assume
1
n  0. T hedefinitionof C gives n  N and n - m 
(x  n, y  m).
N
m
1
Meaning, - 
.
n nN
T hisis a symetricconvexset of area (2N  1)
Let z1 , z 2 ,..., z d be a d - tuple of linearlyindependent vectorsin R d .
We define thelatt icewith basis { z1 , z 2 ,..., z d } as theset of all linear combinat ions
of thez i with integercoefficients.
   (z1 , z 2 ,..., z d )  {i1z1  i 2 z 2  ...  i d z d : (i1 , i 2 ,...,i d )  Zd }.
Let us forma d  d mat rixM wit h t hevect orsz1 ,...z d as columns.
We define t hedet erminant of t helat t ice   (z1 ,...z d ) as det   det M .
Geom etrically, det  is t he volumeof t heparallelpiped
{1 z1   2 z 2  ...   d z d : (1 ,... d )  [0,1]}
Let  be a latticein R d , and let C  R d be a symmetricconvexset
with vol(C)  2d det . T henC containsa pointof  differentfrom0.
Let {z1 ,...z d } be a basis of . We define a linear mappingf : R d   R d
by f(x1 , x 2 ,.., x d )  x1z1  x 2 z 2  ...  x d z d . T henf is a bijectionand   f(Zd ).
For any convexset X, we have vol(f(X)) det( )vol(X).
(A convexset can be approximated by a disjoint union of sufficiently small cubes
with arbitraryprecision).
C'  f -1 (C ). T hisis a symetricconvexset with vol(C')  vol(C)/det  2d.
Minkowski' s theoremprovidesa nonzerovector v C'Zd , and
f(v)is thedesired pointas in t hetheorem.
f
We define a discretesubgroup of R d as a set   R d such that
whenever x, y  , thenalso x - y  ,
and such thatthedistanceof any twodistinct pointsof 
is at least  , for some fixed posit ivereal number   0.
Let   R d be a discretesubgroup of R d whose linear span is R d .
T hen has a basis; thatis, thereexsitsd linearlyindependent vect ores
z1 , z 2 ,...,z d  R d such t hat   (z1 , z 2 ,...,z d ).
By induction.For some i, 1  i  d  1, suppose thatlinearlyindependent vectors
z1 , z 2 ,...,z i-1   with thefollowingpropertyhavealready been constructed. If
Fi-1 denotesthe(i - 1) dimensional subspace spannedby z1 , z 2 ,...,z i-1 , thenall
pointsof  lyingin Fi-1 can be writtenas integerlinear combinations of
z1 , z 2 ,...,z i-1. For i  d  1, thisgives thestatementof the theorem.
So consider an i  d.  generatesR d , thereexistsa vector   not lyingin thesubspace Fi-1.
Let P be thei - dimensional parallelpiped determinedby z1 , z 2 ,...,z i-1 and by  :
P  {1z1   2 z 2  ...   i-1z i-1   i : 1 ,..., i  [0,1]}.
Amongall thepointsof  lyingin P but not in Fi-1 , chooseone nearest Fi-1 and call it z i .
Note thatif thepointsof   P are writtenin theform1z1   2 z 2  ...   i-1z i-1   i ,
then z i is one with thesmallest i . It remainsto show thatz1 , z 2 ,...,z i
have therequired property.
So let v  be a pointlyingin Fi . We can writ ev  1 z1   2 z 2  ...   i zi for
some real numbers 1 ,..., i . Define:
 j   j   j , j  1,2,...,i.
v'  1 z1   2 z 2  ...   i zi .
v' also lies in . Since 0   j  1 v' lies in t heparallelepiped P .T herefore,we
must have  i  0 for ot herwisev' would be nearer t oFi-1 t han zi .
Hence v'   Fi-1 , and by t heinduct ivehypot hesis, we also get t hatall t heot her
 j  0. So all t he j are in fact int egercoefficient s.
v
v’
How efficiently can one actually compute a nonzero lattice
point in a symmetric convex body?
If d is consideredas a constant,such problemcan be solvedin polynomialtime.
If d is consideredas a part of theinput,theproblemis known tobe NP - hard.
Theorem
Each primep  1 (mod4) can be writtenas a sum of two squares : p  a 2  b 2 ,
a, b  Z .
If 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.
Let F  GF(p) standfor thefield of residue classes modulo p,
where p is a primenumber,and let F*  F \ {0}.
An elementa  F * is called a quadratic residue modulo p if there
existsan x F * with x2  a (modp).
Otherwise,a is a quadratic nonresidue.
Example: 42≡6(mod 10) so 6 is a quadratic residue (mod 10).
By t helemma,we can choosea number q such t hatq 2  1(mod p ).
Consider the lattice   (z1 , z 2 ), where z1  (1,q) and z 2  (0,p).
We havedet   p. We use Minkowski's theoremfor generallatticesfor
thedisk C  {(x, y) R 2 : x 2  y 2  2p}.
T hearea of C is 2p  4p  4det , and so C containsa point(a, b)   \ {0}.
We have0  a 2  b 2  2 p. At thesame time,(a, b)  iz1  jz2 for somei, j  Z,
which means thata  i and b  iq  jp (by definitionof z1 , z 2 ).
We calculatea 2  b 2  i 2  (iq  j ) 2  i 2  i 2 q 2  2iqjp  j 2 p 2  i 2 (1  q 2 )  0(mod p ).
T hereforea 2  b 2  p.
≣0
q2≣-1(mod p)
2p
C