Fractal zeta functions and geometric combinatorial problems for self-similar subsets of Z n. Ben Lichtin joint work with Driss Essouabri (Saint-Etienne) 1 / 27
2 examples: Pascal s triangle mod p { Pas(p) := (m 1, m 2 ) N 2 0 : ( m1 m 2 ) } 0 (mod p). Pascal s pyramid (mod p). For any α = (α 1,..., α n ) N n 0, first set ( ) α1 + + α n Multnom n (α) := α 1,..., α n = (α 1 + + α n )!. α 1!... α n! Define: M n (p) := {α N n 0 : Multnom n (α) 0 (mod p)}. 2 / 27
2 examples: Pascal s triangle mod p { Pas(p) := (m 1, m 2 ) N 2 0 : ( m1 m 2 ) } 0 (mod p). Pascal s pyramid (mod p). For any α = (α 1,..., α n ) N n 0, first set ( ) α1 + + α n Multnom n (α) := α 1,..., α n = (α 1 + + α n )!. α 1!... α n! Define: M n (p) := {α N n 0 : Multnom n (α) 0 (mod p)}. 2 / 27
Pas(2) : 3 / 27
Upper Minkowski dimension of discrete sets. Definition: q = positive definite quadratic form on R n x = q 1/2 (x). F = discrete subset of R n. F has finite upper Minkowski dimension if : where: log (#F B(R)) e(f) := lim [0, ). R log R B(R) := {m R n ; m R}. 4 / 27
Define the zeta function of F ζ(f; s) := m F m s (F = F \ {0}). Properties: converges absolutely in {R(s) > e(f)}. e(f) = abscissa of convergence. Always assume e(f) (0, ). { m : m F } = {λ 1 < λ 2 < }; r k = #{m : m = λ k } σ = Re s > e(f) = ζ(f, s) = k r k λ s k. 5 / 27
Define the zeta function of F ζ(f; s) := m F m s (F = F \ {0}). Properties: converges absolutely in {R(s) > e(f)}. e(f) = abscissa of convergence. Always assume e(f) (0, ). { m : m F } = {λ 1 < λ 2 < }; r k = #{m : m = λ k } σ = Re s > e(f) = ζ(f, s) = k r k λ s k. 5 / 27
Define the zeta function of F ζ(f; s) := m F m s (F = F \ {0}). Properties: converges absolutely in {R(s) > e(f)}. e(f) = abscissa of convergence. Always assume e(f) (0, ). { m : m F } = {λ 1 < λ 2 < }; r k = #{m : m = λ k } σ = Re s > e(f) = ζ(f, s) = k r k λ s k. 5 / 27
Tauberian Applications Define F x := F { m < x} = x ball of F. #F x = λ k <x r k = average of coefficients of ζ(f, s). Problem: Describe #F x as x. Tauberian theory is useful if information outside halfplane of absolute convergence is available. Key Point: Self-similarity of F helps provide such information. 6 / 27
Tauberian Applications Define F x := F { m < x} = x ball of F. #F x = λ k <x r k = average of coefficients of ζ(f, s). Problem: Describe #F x as x. Tauberian theory is useful if information outside halfplane of absolute convergence is available. Key Point: Self-similarity of F helps provide such information. 6 / 27
Tauberian Applications Define F x := F { m < x} = x ball of F. #F x = λ k <x r k = average of coefficients of ζ(f, s). Problem: Describe #F x as x. Tauberian theory is useful if information outside halfplane of absolute convergence is available. Key Point: Self-similarity of F helps provide such information. 6 / 27
Compatible self-similar discrete sets. Similarity = affine map x c T x + b c > 1 and T is invariant by O(q) Finite set f = (f i ) r 1 (f i = c i T i + b i ) of similarities is compatible if T i T j = T j T i i, j Definition. F is compatible self-similar {T i } r 1 compatible s.t. defining f i = c i T i + b i : f i (F) f i (F) finite if i i F r i=1 f i(f) is finite ( symmetric difference). 7 / 27
Compatible self-similar discrete sets. Similarity = affine map x c T x + b c > 1 and T is invariant by O(q) Finite set f = (f i ) r 1 (f i = c i T i + b i ) of similarities is compatible if T i T j = T j T i i, j Definition. F is compatible self-similar {T i } r 1 compatible s.t. defining f i = c i T i + b i : f i (F) f i (F) finite if i i F r i=1 f i(f) is finite ( symmetric difference). 7 / 27
Compatibility implies: B = (e 1,..., e n ) = orthonormal basis of C n (= C R R n ) s.t. each T j is diagonal. i λ i = (λ i,1,..., λ i,n ) (S 1 ) n s.t. Scal(f) := {c i } r 1. Ti (e j ) = λ i,j e j j (Ti = adjoint). Theorem 1 [E-L, Adv. Math. 2013]. Compatible self-similar F R n satisfies: ζ(f; s) = m F 1 m s has a meromorphic continuation to C with moderate growth and poles in: P(F) = { } r s k : λ β i c s i = 1 (λ β := λ β k k ). β N n 0 k N 0 i=1 k 8 / 27
Compatibility implies: B = (e 1,..., e n ) = orthonormal basis of C n (= C R R n ) s.t. each T j is diagonal. i λ i = (λ i,1,..., λ i,n ) (S 1 ) n s.t. Scal(f) := {c i } r 1. Ti (e j ) = λ i,j e j j (Ti = adjoint). Theorem 1 [E-L, Adv. Math. 2013]. Compatible self-similar F R n satisfies: ζ(f; s) = m F 1 m s has a meromorphic continuation to C with moderate growth and poles in: P(F) = { } r s k : λ β i c s i = 1 (λ β := λ β k k ). β N n 0 k N 0 i=1 k 8 / 27
Publicity for forthcoming article. Compatibility hypothesis not needed for our results. Arguments are fairly different from compatibility case. 9 / 27
Fractal dimension. Theorem 2. [E-L, 2013] F R n = compatible self-similar set. Then e(f) = fractal dimension of F = the unique real solution of 2 types of self similar sets: r i=1 c s i = 1. Latticelike: Z{log c i } i = discrete subgroup of R; E.G. Pas(p), M n (p) are latticelike. Non latticelike: Z{log c i } i is dense in R. 10 / 27
Fractal dimension. Theorem 2. [E-L, 2013] F R n = compatible self-similar set. Then e(f) = fractal dimension of F = the unique real solution of 2 types of self similar sets: r i=1 c s i = 1. Latticelike: Z{log c i } i = discrete subgroup of R; E.G. Pas(p), M n (p) are latticelike. Non latticelike: Z{log c i } i is dense in R. 10 / 27
Properties of Pas(p): Similarities: f i = p Id 2 + b i b i { (u, v) {0, 1,..., p 1} 2}. Set θ p = log( p(p+1) 2 ) log p. Theorem 3 [E.- Jap. J. of Math. (2005)]. ζ (Pas(p); s) converges in {σ > θ p }. e(pas(p)) = θ p. ζ (Pas(p); s) has a meromorphic continuation to C with moderate growth and simple poles in: { θ p k + 2πih } log p : (k, h) N 0 Z 0. 11 / 27
Properties of Pas(p): Similarities: f i = p Id 2 + b i b i { (u, v) {0, 1,..., p 1} 2}. Set θ p = log( p(p+1) 2 ) log p. Theorem 3 [E.- Jap. J. of Math. (2005)]. ζ (Pas(p); s) converges in {σ > θ p }. e(pas(p)) = θ p. ζ (Pas(p); s) has a meromorphic continuation to C with moderate growth and simple poles in: { θ p k + 2πih } log p : (k, h) N 0 Z 0. 11 / 27
Pascal s pyramid (mod p). Define Theorem 4 [E-L, 2013] log θ n,p = ζ(m n (p), s) = ( (p+n 1 p 1 log p α M n(p) ) ). α s has a meromorphic continuation to C with moderate growth and simple poles in: { θ n,p k + 2πih } log p : (k, h) N 0 Z 0, and e(m n (p)) = θ n,p. 12 / 27
(I) σ > θ p = log( p(p+1) 2 ) log p = ζ(pas(p), s) = k r k (p) λ s k where: {λ k } k := { (m 1, m 2 ) : (m 1, m 2 ) Pas(p)} r k (p) := #{(m 1, m 2 ) Pas(p) : (m 1, m 2 ) = λ k } Average of coefficients: #(Pas(p) x ) := λ k <x r k (p) Theorem 5.[E-L, 2013]. η > 0 and a convergent Fourier series f p (u) 0 s.t. ( ) logx #(Pas(p) x ) = x θp f p + O(x θp η ) x. logp 13 / 27
(I) σ > θ p = log( p(p+1) 2 ) log p = ζ(pas(p), s) = k r k (p) λ s k where: {λ k } k := { (m 1, m 2 ) : (m 1, m 2 ) Pas(p)} r k (p) := #{(m 1, m 2 ) Pas(p) : (m 1, m 2 ) = λ k } Average of coefficients: #(Pas(p) x ) := λ k <x r k (p) Theorem 5.[E-L, 2013]. η > 0 and a convergent Fourier series f p (u) 0 s.t. ( ) logx #(Pas(p) x ) = x θp f p + O(x θp η ) x. logp 13 / 27
(II) ( σ > θ n,p = log ( p+n 1 p 1 ) ) log p = ζ(m n (p), s) = k R k (p) Λ s k where: {Λ k } k := { α : α M n (p)} R k (p) := #{α M n (p) : α = Λ k } Average of coefficients: #(M n (p) x ) := Λ k <x R k (p) A new proof of result of Barbolosi-Grabner (1996). Theorem 6.[E-L, 2013]. η > 0 and convergent Fourier series F n,p (u) 0 s.t. ( ) log x #(M n (p) x ) = x θn,p F n,p + O(x θn,p η ). log p 14 / 27
(II) ( σ > θ n,p = log ( p+n 1 p 1 ) ) log p = ζ(m n (p), s) = k R k (p) Λ s k where: {Λ k } k := { α : α M n (p)} R k (p) := #{α M n (p) : α = Λ k } Average of coefficients: #(M n (p) x ) := Λ k <x R k (p) A new proof of result of Barbolosi-Grabner (1996). Theorem 6.[E-L, 2013]. η > 0 and convergent Fourier series F n,p (u) 0 s.t. ( ) log x #(M n (p) x ) = x θn,p F n,p + O(x θn,p η ). log p 14 / 27
Averages for other weights Metric Invariants Examples of metric invariants (w.r.t. O(q)): (k = 2) : Distinct distances, angles (k 3) : k point configurations, k chains, volumes of n simplices (k = n + 1). k Configuration = {( x 1 x 2, x 1 x 3,..., (x k 1 x k ) : x i F i }. k Chain = {( x 1 x 2, x 2 x 3,..., x k 1 x k ) : x i F i }. Geometric combinatorial problems: For metric invariant on (R n ) k find a lower bound for : # distinct values of invariant determined by elements of F k x (x 1). 15 / 27
Averages for other weights Metric Invariants Examples of metric invariants (w.r.t. O(q)): (k = 2) : Distinct distances, angles (k 3) : k point configurations, k chains, volumes of n simplices (k = n + 1). k Configuration = {( x 1 x 2, x 1 x 3,..., (x k 1 x k ) : x i F i }. k Chain = {( x 1 x 2, x 2 x 3,..., x k 1 x k ) : x i F i }. Geometric combinatorial problems: For metric invariant on (R n ) k find a lower bound for : # distinct values of invariant determined by elements of F k x (x 1). 15 / 27
Averages for other weights Metric Invariants Examples of metric invariants (w.r.t. O(q)): (k = 2) : Distinct distances, angles (k 3) : k point configurations, k chains, volumes of n simplices (k = n + 1). k Configuration = {( x 1 x 2, x 1 x 3,..., (x k 1 x k ) : x i F i }. k Chain = {( x 1 x 2, x 2 x 3,..., x k 1 x k ) : x i F i }. Geometric combinatorial problems: For metric invariant on (R n ) k find a lower bound for : # distinct values of invariant determined by elements of F k x (x 1). 15 / 27
Erdös distinct distance problem (1946): q(x) := i x 2 i x := ( i x 2 i )1/2. If X is a finite subset of R n, then the set of its distinct distances (X ) := { x y ; x, y X } should satisfy: # (X ) ε #X 2 n ε as #X. n = 2: Guth-Katz (2014) (with slightly better exponent of Erdòs). 16 / 27
Erdös distinct distance problem (1946): q(x) := i x 2 i x := ( i x 2 i )1/2. If X is a finite subset of R n, then the set of its distinct distances (X ) := { x y ; x, y X } should satisfy: # (X ) ε #X 2 n ε as #X. n = 2: Guth-Katz (2014) (with slightly better exponent of Erdòs). 16 / 27
Distance set in any x ball of F : (F x ) := { m 1 m 2 : m i F x i = 1, 2}. Theorem 7 [E-L, PLMS 2015]. F = compatible self-similar subset of Z n (n 2); e(f) > n 2. n 2 1 = ε > 0, # (F x ) ε #(F x ) e(f) ε (x ). 17 / 27
Corollaries. For (compatible) self-similar F Z n : (1) n = 2 and e(f) > 0 implies Erdös distance conjecture but only for all sufficiently large x balls: ε > 0, # (F x ) ε #(F x ) 1 ε (x ). (2) θ p > 0 = # (Pas(p) x ) ε [ #Pas(p)x ] 1 ε. (3) a n s.t. p > a n = θ n,p > n 2. p > a n = # (M n (p) x ) ε [ #Mn (p) x ] 1 n 2 θn,p ε. 18 / 27
Corollaries. For (compatible) self-similar F Z n : (1) n = 2 and e(f) > 0 implies Erdös distance conjecture but only for all sufficiently large x balls: ε > 0, # (F x ) ε #(F x ) 1 ε (x ). (2) θ p > 0 = # (Pas(p) x ) ε [ #Pas(p)x ] 1 ε. (3) a n s.t. p > a n = θ n,p > n 2. p > a n = # (M n (p) x ) ε [ #Mn (p) x ] 1 n 2 θn,p ε. 18 / 27
Corollaries. For (compatible) self-similar F Z n : (1) n = 2 and e(f) > 0 implies Erdös distance conjecture but only for all sufficiently large x balls: ε > 0, # (F x ) ε #(F x ) 1 ε (x ). (2) θ p > 0 = # (Pas(p) x ) ε [ #Pas(p)x ] 1 ε. (3) a n s.t. p > a n = θ n,p > n 2. p > a n = # (M n (p) x ) ε [ #Mn (p) x ] 1 n 2 θn,p ε. 18 / 27
Sketch of proof of Theorem 7. Step 1: Introduce 2 variable zeta-function: ζ dis (F, s) = (m 1,m 2 ) (F ) 2 m 1 m 2 2 m 1 s 1 m2 s 2. Converges absolutely in {s C 2 ; σ i > D i} (D := e(f) + 2). Set: (F x ) := { m 1 m 2 ; (m 1, m 2 ) Fx 2 } := {δ 1,..., δ Nx } (N x = # (F x )) { m : m F x } := {λ 1,..., λ Mx } ν i,k1,k 2 := #{(m 1, m 2 ) : m j = λ kj j, m 1 m 2 = δ i }. σ 1, σ 2 > D = ζ dis (F, s) = (k 1,k 2 ) i δ2 i ν i,k 1,k 2 λ s 1 k1 λ s 2. k2 19 / 27
Sketch of proof of Theorem 7. Step 1: Introduce 2 variable zeta-function: ζ dis (F, s) = (m 1,m 2 ) (F ) 2 m 1 m 2 2 m 1 s 1 m2 s 2. Converges absolutely in {s C 2 ; σ i > D i} (D := e(f) + 2). Set: (F x ) := { m 1 m 2 ; (m 1, m 2 ) Fx 2 } := {δ 1,..., δ Nx } (N x = # (F x )) { m : m F x } := {λ 1,..., λ Mx } ν i,k1,k 2 := #{(m 1, m 2 ) : m j = λ kj j, m 1 m 2 = δ i }. σ 1, σ 2 > D = ζ dis (F, s) = (k 1,k 2 ) i δ2 i ν i,k 1,k 2 λ s 1 k1 λ s 2. k2 19 / 27
Sketch of proof of Theorem 7. Step 1: Introduce 2 variable zeta-function: ζ dis (F, s) = (m 1,m 2 ) (F ) 2 m 1 m 2 2 m 1 s 1 m2 s 2. Converges absolutely in {s C 2 ; σ i > D i} (D := e(f) + 2). Set: (F x ) := { m 1 m 2 ; (m 1, m 2 ) Fx 2 } := {δ 1,..., δ Nx } (N x = # (F x )) { m : m F x } := {λ 1,..., λ Mx } ν i,k1,k 2 := #{(m 1, m 2 ) : m j = λ kj j, m 1 m 2 = δ i }. σ 1, σ 2 > D = ζ dis (F, s) = (k 1,k 2 ) i δ2 i ν i,k 1,k 2 λ s 1 k1 λ s 2. k2 19 / 27
Step 2: Upper bound for coefficient averages A(x) := j λ kj <x i δ2 i ν i,k1,k 2. Claim 1: A(x) ε N x x e(f)+n+ɛ. µ i (x) := #{(m 1, m 2 ) F 2 x : m 1 m 2 = δ i }. A(x) = N x i=1 δ2 i µ i(x). Uniform bound for µ i (x): (via integral points on spheres). m 2 F x : #{m 1 Z n ; m 1 m 2 2 = t 2, m 1 x} ε t n 2+ε where implied constant is uniform in m 2. Setting t = δ i = µ i (x) ε m 2 F x t n 2+ε ε x e(f)+n 2+ε. A(x) ε N x x e(f)+n+ε. 20 / 27
Step 2: Upper bound for coefficient averages A(x) := j λ kj <x i δ2 i ν i,k1,k 2. Claim 1: A(x) ε N x x e(f)+n+ɛ. µ i (x) := #{(m 1, m 2 ) F 2 x : m 1 m 2 = δ i }. A(x) = N x i=1 δ2 i µ i(x). Uniform bound for µ i (x): (via integral points on spheres). m 2 F x : #{m 1 Z n ; m 1 m 2 2 = t 2, m 1 x} ε t n 2+ε where implied constant is uniform in m 2. Setting t = δ i = µ i (x) ε m 2 F x t n 2+ε ε x e(f)+n 2+ε. A(x) ε N x x e(f)+n+ε. 20 / 27
Step 3: Lower bound for A(x) - Harder part Claim 2: A(x) ε x 2e(F)+2 ε. So: Claims 1 + 2 = x 2e(F)+2 ε ε A(x) ε x e(f)+n+ε N x N x := # (F x ) ε x e(f) n+2 ε n 2 1 ε (#F x ) e(f) ε. 21 / 27
Step 3: Lower bound for A(x) - Harder part Claim 2: A(x) ε x 2e(F)+2 ε. So: Claims 1 + 2 = x 2e(F)+2 ε ε A(x) ε x e(f)+n+ε N x N x := # (F x ) ε x e(f) n+2 ε n 2 1 ε (#F x ) e(f) ε. 21 / 27
Idea behind Proof of Claim 2 Write ζ dis (F, s) = ζ(f, s 1 )ζ(f, s 2 2) + ζ(f, s 1 2)ζ(F, s 2 ) 2ζ 3 (s) = ζ 1 + ζ 2 2ζ 3 where m 1, m 2 n ζ 3 (s) := m 1 s 1 m2 s = ζ 2 3,i (s 1 )ζ 3,i (s 2 ). (m 1,m 2 ) F F i=1 ζ 1 is analytic in {σ 1 > D 2, σ 2 > D} ζ 2 is analytic in {σ 1 > D, σ 2 > D 2} ζ 3 is analytic in {σ i > D 1 i}. Each ζ i is meromorphic on C 2 with moderate growth. 22 / 27
Idea behind Proof of Claim 2 Write ζ dis (F, s) = ζ(f, s 1 )ζ(f, s 2 2) + ζ(f, s 1 2)ζ(F, s 2 ) 2ζ 3 (s) = ζ 1 + ζ 2 2ζ 3 where m 1, m 2 n ζ 3 (s) := m 1 s 1 m2 s = ζ 2 3,i (s 1 )ζ 3,i (s 2 ). (m 1,m 2 ) F F i=1 ζ 1 is analytic in {σ 1 > D 2, σ 2 > D} ζ 2 is analytic in {σ 1 > D, σ 2 > D 2} ζ 3 is analytic in {σ i > D 1 i}. Each ζ i is meromorphic on C 2 with moderate growth. 22 / 27
( Initial ) Poles of ζ 1, ζ 2, ζ 3 : ζ 1 has first pole (real part) at Q 1 = (D 2, D); ζ 2 has first pole (real part) at Q 2 = (D, D 2); Real part of first pole (u i, v i ) of ζ 3,i (s 1 )ζ 3,i (s 2 ) lies in or below Int [Q 1, Q 2 ]. 23 / 27
Step 4: Perron formula - if Latticelike ξ = (ξ, ξ) R 2 set (ξ) := (ξ) (ξ). Then ξ > D(= e(f) + 2) = H(x 1, x 2 ) := = {m i F : m i <x i i} (ξ) (ζ 1 + ζ 2 2ζ 3 ) m 1 m 2 2 i x s i i s i (s i + 1) ds 1ds 2 Next: Iterate Cauchy residue method. For ζ 1 resp. ζ 2 move to (Q 1 η) resp. (Q 2 η) (η 1) etc. = A, B s.t. A + B < 2D 2 and i ( 1 m i x i ) H(x 1, x 2 ) = x1 D 2 x2 D φ 1 (log x 1, log x 2 ) + x1 D x2 D 2 φ 2 (log x 1, log x 2 ) t + ±x u i 1 x v i 2 ψ i(log x 1, log x 2 ) + O(x1 A ε x2 B ε ) i=1 where φ 1, φ 2, ψ i are non zero almost periodic functions. 24 / 27
Step 4: Perron formula - if Latticelike ξ = (ξ, ξ) R 2 set (ξ) := (ξ) (ξ). Then ξ > D(= e(f) + 2) = H(x 1, x 2 ) := = {m i F : m i <x i i} (ξ) (ζ 1 + ζ 2 2ζ 3 ) m 1 m 2 2 i x s i i s i (s i + 1) ds 1ds 2 Next: Iterate Cauchy residue method. For ζ 1 resp. ζ 2 move to (Q 1 η) resp. (Q 2 η) (η 1) etc. = A, B s.t. A + B < 2D 2 and i ( 1 m i x i ) H(x 1, x 2 ) = x1 D 2 x2 D φ 1 (log x 1, log x 2 ) + x1 D x2 D 2 φ 2 (log x 1, log x 2 ) t + ±x u i 1 x v i 2 ψ i(log x 1, log x 2 ) + O(x1 A ε x2 B ε ) i=1 where φ 1, φ 2, ψ i are non zero almost periodic functions. 24 / 27
Step 5: Monomialisation-if Latticelike Polygon from H monomials: vertices at Q 1, Q 2. Choose, say, Q 1 = (D 2, D). κ > κ > 1 show: x k = (x k,1, x k,2 ) s.t. xk,1 κ < x k,2 < xk,1 κ, φ ( log x k,1, log x k,2 ) ) 1 and xk,1 D x D 2 k,2 + i x u i k,1 x v i D 2 k,2 = o(xk,1 x k,2 D ) = o(xq 1 k ) H(x k,1, x k,2 ) = x Q 1 k φ(log x k,1, log x k,2 ) + o(x Q 1 k ) xq 1 k. A(x) ε x 2e(F)+2 ε (1) Non latticelike case: Variant of monomialisation proves (1) but uses screens - needed due to possible pole clustering on σ = D, D 2. 25 / 27
Step 5: Monomialisation-if Latticelike Polygon from H monomials: vertices at Q 1, Q 2. Choose, say, Q 1 = (D 2, D). κ > κ > 1 show: x k = (x k,1, x k,2 ) s.t. xk,1 κ < x k,2 < xk,1 κ, φ ( log x k,1, log x k,2 ) ) 1 and xk,1 D x D 2 k,2 + i x u i k,1 x v i D 2 k,2 = o(xk,1 x k,2 D ) = o(xq 1 k ) H(x k,1, x k,2 ) = x Q 1 k φ(log x k,1, log x k,2 ) + o(x Q 1 k ) xq 1 k. A(x) ε x 2e(F)+2 ε (1) Non latticelike case: Variant of monomialisation proves (1) but uses screens - needed due to possible pole clustering on σ = D, D 2. 25 / 27
Step 5: Monomialisation-if Latticelike Polygon from H monomials: vertices at Q 1, Q 2. Choose, say, Q 1 = (D 2, D). κ > κ > 1 show: x k = (x k,1, x k,2 ) s.t. xk,1 κ < x k,2 < xk,1 κ, φ ( log x k,1, log x k,2 ) ) 1 and xk,1 D x D 2 k,2 + i x u i k,1 x v i D 2 k,2 = o(xk,1 x k,2 D ) = o(xq 1 k ) H(x k,1, x k,2 ) = x Q 1 k φ(log x k,1, log x k,2 ) + o(x Q 1 k ) xq 1 k. A(x) ε x 2e(F)+2 ε (1) Non latticelike case: Variant of monomialisation proves (1) but uses screens - needed due to possible pole clustering on σ = D, D 2. 25 / 27
Step 5: Monomialisation-if Latticelike Polygon from H monomials: vertices at Q 1, Q 2. Choose, say, Q 1 = (D 2, D). κ > κ > 1 show: x k = (x k,1, x k,2 ) s.t. xk,1 κ < x k,2 < xk,1 κ, φ ( log x k,1, log x k,2 ) ) 1 and xk,1 D x D 2 k,2 + i x u i k,1 x v i D 2 k,2 = o(xk,1 x k,2 D ) = o(xq 1 k ) H(x k,1, x k,2 ) = x Q 1 k φ(log x k,1, log x k,2 ) + o(x Q 1 k ) xq 1 k. A(x) ε x 2e(F)+2 ε (1) Non latticelike case: Variant of monomialisation proves (1) but uses screens - needed due to possible pole clustering on σ = D, D 2. 25 / 27
Forthcoming article proves Erdös distinct distance problem for x balls of any self-similar subset of Z 2. Need use Ingham s Tauberian theorem (1965) instead of Cauchy Residues and screens if non latticelike. 26 / 27
Angles For any m 1, m 2 F set: θ(m 1, m 2 ) := angle formed between m 1, m 2. Ang(x) := #{cos ( θ(m 1, m 2 ) ) ; m 1, m 2 F x }. Find bound on e(f) s.t. Ang(x) +. Theorem 8 [PLMS 2015]. e(f) > n 2 and n 4 = Uses Angle Zeta function: n 2 1 Ang(x) ε [#F(x)] e(f) ε (x 1). ζ ang (F; s 1, s 2 ) := m 1,m 2 F m1 m 1, m 2 2 m 2 m 1 s 1 m2 s. 2 27 / 27
Angles For any m 1, m 2 F set: θ(m 1, m 2 ) := angle formed between m 1, m 2. Ang(x) := #{cos ( θ(m 1, m 2 ) ) ; m 1, m 2 F x }. Find bound on e(f) s.t. Ang(x) +. Theorem 8 [PLMS 2015]. e(f) > n 2 and n 4 = Uses Angle Zeta function: n 2 1 Ang(x) ε [#F(x)] e(f) ε (x 1). ζ ang (F; s 1, s 2 ) := m 1,m 2 F m1 m 1, m 2 2 m 2 m 1 s 1 m2 s. 2 27 / 27