Simultaneous Diophantine Approximation with Excluded Primes László Babai Daniel Štefankovič
Dec 22, 2015
Dirichlet (1842) Simultaneous Diophantine Approximation
1 2, ,..., ,n Q
1,..., nr r
Given reals
integers
1/ 2i iQ p trivial
for all i
and
q Q
q
1/ ni iq r Q
such that and
Simultaneous Diophantine Approximationwith an excluded prime
1 2, ,..., n
1,..., nr r
Given reals
integers
i iq r for all i
and q
gcd( , ) 1p q
prime p
?such that and
Simultaneous diophantine -approximationexcluding
1 1/ 3
1 1 1| | | / 3 | 1/ 3q r q r
p
Not always possible
Example 3p
If
then
Simultaneous diophantine -approximationexcluding
1 1| |q r
p
obstacle with 2 variables
1 22 1/ p
1 2 1 23 | ( 2 ) ( 2 ) | 1/q r r p
2 2| |q r
If
then
Simultaneous diophantine -approximationexcluding
p
general obstacle
1 1 2 2 ... 1/n nb b b p t
| | 1/ib p
If
then
Simultaneous diophantine -approximationexcluding
p
Theorem:
If there is no -approximationexcluding p then there exists an obstacle with
3/ 2| | /ib n Kronecker’s theorem ():
Arbitrarily good approximation excluding possible IFF no obstacle.
p
Simultaneous diophantine -approximationexcluding
p
obstacle with 3/ 2| | /ib n
3/ 2pn
pnecessary to prevent -approximation
excluding
psufficient to prevent -approximation
excluding
Motivating example
set
stretching by
/A m Z Z
gcd( , ) 1x m
moda ax m{ | }Ax ax a A
x
arc length of A
max | (mod ) |a A
a m
m a prime
1 ,..., da a
m m
proof:
; 0q q Q 1/
1i i n
q pQ
Dirichlet: 1Q m
:x q
| |d A
1 1/ dm there exists such thatx
arc-length of Ax
Shrinking modulo a prime
Shrinking modulo any number
m a prime every small set canbe shrunk
gcd( , ) 1x m
2km 1{1,1 2 }kA
If
then the arc-length of Ax22k
1 ,..., da a
m m
proof:
; 0q q Q 1/
1i i n
q pQ
Dirichlet: 1Q m
:x q
Where does the proof break?
2km
1 ,..., da a
m m
proof:
; 0q q Q 1/
1i i n
q pQ
Dirichlet: 1Q m
:x q
Where does the proof break?
2km
approximation excluding 2
need:
Shrinking cyclotomic classes
m a prime every small set canbe shrunk
set of interest – cyclotomic class(i.e. the set of r-th roots of unity mod m)
•locally testable codes•diameter of Cayley graphs•Warring problem mod p•intersection conditions modulo p k
k
Banasczyk’s technique (1992)
2|| ||( ) x
x A
A e
gaussian weight of a set
( ) ( ) / ( )L x L x L mass displacement function of lattice
Banasczyk’s technique (1992)
( ) ( ) / ( )L x L x L mass displacement function of lattice
0 ( ) 1L x
dist( , ) ( ) 1/ 4Lx L n x
properties:
Banasczyk’s technique (1992)
( ) ( ) / ( )L A L A L discrete measure
*( ) ( )LL
x x
21( ) exp( || || ) exp(2 )
( )T
L
y L
x y iy xL
relationship between the discrete measure and the mass displacement function of the dual
Banasczyk’s technique (1992)
( ) ( ) / ( )L A L A L discrete measure defined by the lattice
*( ) ( )LL
x x
21( ) exp( || || ) exp(2 )
( )T
L
y L
x y iy xL
|| ||
1*
( ) x sL
|| ||
1*
( ) x sL
Banasczyk’s technique (1992)
1
2
3
1 0 0
0 1 0/
0 0 1
0 0 0
n
1 2 3, ,
there is no short vectorwith coefficient of thelast column
w L
0(mod )p
Banasczyk’s technique (1992)there is no short vectorwith coefficient of thelast column
w L
0(mod )p
( ) 1/ 2L u 1: nu ep n
* ( ) 1/ 2Lu *dist( , )u L n
obstacleQED
Lovász (1982) Simultaneous Diophantine Approximation
1 2, ,..., ,n Q
1,..., 0<np p q Q
Given rationals
integers
2
1/
2ni i n
q pQ
for all i
can find in polynomial time
Factoring polynomials with rational coefficients.
Simultaneous diophantine -approximationexcluding
p - algorithmic
1 2, ,..., n Given rationals
can find in polynomial time
,prime p
12 nC p -approximation excluding p
where is smallest such that thereexists -approximation excluding p
/ 24 2nnC n
Exluding prime and bounding denominator
If there is no -approximationexcluding pthen there exists an
approximate obstacle with 3/ 2| | /ib n
with q Q
1 1 2 2 ... 1/n nb b b p t | | /n Q
Exluding prime and bounding denominator
necessary to prevent -approximationexcluding p with q Q
the obstacle
sufficient to prevent3/ 2/(2 )n p -approximation
excluding p with /(2 )q Q pn
Exluding several primes
If there is no -approximationexcluding 1,..., kp pthen there exists
obstacle with 1/ 2| | (max( , )) /ib n n k
1 [ ]
1/n
i i ji j A k
b p t
Show that there is no small obstacle!
m=7k
m*
primitive 3-rd root of unity
10 1 7 , gcd( ,7) 1kc c t t
obstacle
know21 0 (mod 7 )k
Show that there is no small obstacle!
10 1 7 , gcd( ,7) 1kc c t t
21 0 (mod 7 )k
20 1Res(1 , )x x c c x
0divisible by 17k
2 20 12( )c c
( 1) / 2
4
7 k
There is g with all 3-rd roots
1/ 2 1/ 2[ (4 7) ,(4 7) ]m m