Simultaneous Diophantine Approximation with Excluded Primes László Babai Daniel Štefankovič
Dirichlet (1842) Simultaneous Diophantine Approximation Given reals integers α,,...,, 1 α 2 α n Q r1,..., r n q Q and such that and α q r Q i i 1/n q for all i α Q p 1/2 trivial i i
Simultaneous Diophantine Approximation with an excluded prime Given reals? α, α,..., α 1 2 n integers r1,..., r n prime and q p such that gcd( pq, ) = 1 and q α i r ε for all i i
Simultaneous diophantine -approximation excluding p ε Not always possible Example p = 3 If α = 1/3 1 then ε qα r = q/3 r 1/3 1 1 1
Simultaneous diophantine -approximation excluding p If α + 2α = 1/ p 1 2 then ε ε obstacle with 2 variables ε qα r 1 1 qα r 2 2 3 ε q( α + 2 α ) ( r + 2 r ) 1/ p 1 2 1 2
Simultaneous diophantine ε -approximation excluding p general obstacle If bα + bα +... + bα = 1/ p+ t 1 1 2 2 then n n ε b 1/ i p
Simultaneous diophantine ε -approximation excluding p Theorem: If there is no ε -approximation excluding p then there exists an obstacle with 3/2 b n / ε i Kronecker s theorem ( ): Arbitrarily good approximation excluding possible IFF no obstacle. p
Simultaneous diophantine ε -approximation excluding p obstacle with 3/2 b n / ε i necessary to prevent ε -approximation excluding p ε sufficient to prevent 3/2-approximation pn excluding p
Motivating example Shrinking by stretching
Motivating example set A ( Z/ mz ) arc length of A max a(mod m) a A stretching by x gcd( xm, ) = 1 Ax = { ax a A} a axmod m
Example of the motivating example A = 11-th roots of unity mod 11177
Example of the motivating example A = 11-th roots of unity mod 11177 168
Shrinking modulo a prime If m a prime then every small set can be shrunk
Shrinking modulo a prime m a prime d = A there exists x such that m 11/d arc-length of Ax proof: a a 1,..., d x : = q m m Q: = m 1 Dirichlet q;0< q Q qα p 1 i i 1/ n Q
Shrinking modulo any number m a prime every small set can be shrunk?
Shrinking modulo any number m a prime every small set can be shrunk m = 2 k A k 1 = {1,1 + 2 } If gcd( xm, ) = 1 then the arc-length of Ax 2 k 2
Where does the proof break? m = 2 k proof: a a 1,..., d x : = q m m Q: = m 1 Dirichlet q;0< q Q qα p 1 i i 1/ n Q
Where does the proof break? m = 2 k need: approximation excluding 2 proof: a a 1,..., d x : = q m m Q: = m 1 Dirichlet q;0< q Q qα p 1 i i 1/ n Q
Shrinking cyclotomic classes m a prime every small set can be 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 k Warring problem mod p intersection conditions modulo p k
Shrinking cyclotomic classes cyclotomic class can be shrunk
Shrinking cyclotomic classes cyclotomic class can be shrunk Show that there is no small obstacle!
Theorem: If there is no ε -approximation excluding p then there exists an obstacle with 3/2 b n / ε i
Lattice v1,..., v R n n linearly independent v 1 v 2
Lattice v1,..., v R n n vz+... + v Z 1 n
Lattice v1,..., v R n n vz+... + v Z 1 n Dual lattice * L = { u ( v L) v u Z} T
Banasczyk s technique (1992) gaussian weight of a set ρ ( A) = x A e π x 2 mass displacement function of lattice φ ρ ρ ( x) = ( L+ x)/ ( L) L
Banasczyk s technique (1992) mass displacement function of lattice φ ρ ρ ( x) = ( L+ x)/ ( L) L properties: 0 φ ( x) 1 L dist( xl, ) n φ ( x) 1/ 4 L
Banasczyk s technique (1992) discrete measure σ ρ ρ ( A) = ( L A)/ ( L) L relationship between the discrete measure and the mass displacement function of the dual σ L( x) = φ ( x) * L 1 σ 2 T L( x) = exp( π y ) exp(2 πiy x) ρ( L) y L
Banasczyk s technique (1992) discrete measure defined by the lattice σ ρ ρ ( A) = ( L A)/ ( L) L σ L( x) = φ ( x) * L ρ 1 ( L) x s * 1 * ρ ( L) x > s 1 σ 2 T L( x) = exp( π y ) exp(2 πiy x) ρ( L) y L
Banasczyk s technique (1992) α, α, α 1 2 3 1 0 0 0 1 0 0 0 1 0 0 0 α α α ν 1 2 3 n / ε there is no short vector with coefficient of the last column 0(mod p) w L
Banasczyk s technique (1992) there is no short vector with coefficient of the last column 0(mod p) w σ ( u ) 1/2 u: = en+ 1 L φ ( u ) 1/2 * L L p ν ε n dist( ul, * ) n obstacle QED
Lovász (1982) Simultaneous Diophantine Approximation Given rationals α,,...,, 1 α 2 α n Q can find in polynomial time integers p1,..., p 0< q Q n 2 n qα p i i for all i Q 1/ n 2 Factoring polynomials with rational coefficients.
Simultaneous diophantine -approximation excluding p - algorithmic ε Given rationals α α,,..., n 1 2 α,prime p 2C n can find in polynomial time pε + -approximation excluding p 1 where ε is smallest such that there exists ε-approximation excluding p Cn = 4 n2 n /2
Exluding prime and bounding denominator If there is no ε -approximation excluding p with q Q then there exists an approximate obstacle with 3/2 b n / ε i b + b +... + b = 1/ p+ t+ 1 1 2 2 n n κ n/ Q α α α κ
Exluding prime and bounding denominator the obstacle necessary to prevent ε -approximation excluding p with q Q sufficient to prevent 3/2 ε /(2 n p) -approximation excluding p with q Q/(2 p n)
Exluding several primes If there is no ε -approximation excluding p1,..., pk then there exists obstacle with 1/2 b n (max( n, k)) / ε i n bα = 1/ p + t i i j i= 1 j A [ k]
Show that there is no small obstacle! m=7 k m * primitive 3-rd root of unity know 2 k 1+ ω + ω 0(mod7 ) obstacle c cω t t k 1 0 + 1 = 7, gcd(,7) = 1
Show that there is no small obstacle! 2 k 1+ ω + ω 0(mod7 ) k 1 c0 cω + 1 = t7, gcd( t,7) = 1 2 Res(1 + x+ x, c0 + c1x) ε = 0 divisible by 2( c + c ) 4 ( 1)/2 7 k 2 2 0 1 7 k 1 There is g with all 3-rd roots 1/2 1/2 [ (4 7) m,(4 7) m ]
Dual lattice 1 0 0 0 0 1 0 0 0 0 1 0 ε / α α α 1 2 3 1 ν ν ν ν n
Algebraic integers? possible that a small integer combination with small coefficients is doubly exponentially close to 1/p