Group actions, the Mattila integral and continuous sum-product problems

Transcription

1 Group actions, the Mattila integral and continuous sum-product problems Bochen Liu University of Rochester Acknowledgment: This work is done when the author was a research assistant in the Chinese University of Hong Kong, supported by Professor Ka-Sing Lau. 1 / 19

2 Falconer distance problem and Mattila integral Given E R d, d 2, denote the set of distances generated by E as (E) = { x y : x, y E}. In 1985, Falconer made the following conjecture. Conjecture (Falconer distance conjecture) For any E R d, d 2 dim H (E) > d 2, its distance set (E) has positive Lebesgue measure. Falconer also gives examples to show that d 2 is necessary and d 2 is the best dimensional threshold. We will discuss about his counterexamples later. 2 / 19

3 Falconer distance problem and Mattila integral The best known results are due to Wolff (1999) for d = 2 and Erdogan (2005) for d 3. They proved that (E) > 0 if dim H (E) > d The main tool they used is the following theorem due to Mattila. Theorem (Mattila integral, 1987) Let µ be a Borel measure on E and define ν on (E) by f dν = f ( x y ) dµ(x) dµ(y), i.e., ν = Φ (µ µ) where Φ(x, y) = x y. Then ν L 2 (R) if and only if ( ) 2 M(µ) = µ(rω) 2 dω r d 1 dr <. S d 1 In particular, in this case (E) has positive Lebesgue measure. 3 / 19

4 Erdős distance problem and group actions Falconer distance problem is in fact a continuous analog of the famous Erdős distance problem, which says Conjecture (Erdős distance conjecture, 1946) Let P R d be a set of N points, then for any ɛ > 0, there exists a constant C ɛ > 0 such that #( (P)) C ɛ N 2 d ɛ. This conjecture has been solved by Guth and Katz (2010) on the plane, while it is still open in higher dimensions. 4 / 19

5 Erdős distance problem and group actions In Guth-Katz argument, there is a very important observation which is originally due to Elekes and Sharir. Notice that for any quadruplet x 1, x 2, y 1, y 2 R d, x 1 x 2 = y 1 y 2 if and only if there exists θ O(d) such that x 1 θy 1 = x 2 θy 2. So we can work on O(d) to estimate of repetition of the distances. Since Falconer distance problem is a continuous analog of Erdős distance problem, a natural question at this point is, does this idea of groups actions help on Falconer distance problem? 5 / 19

6 Generalized Mattila integral In 2015, Greenleaf, Iosevich, B.L. and Palsson observed that, the Mattila integral can be interpreted as ( ) 2 ( ) ˆµ(rω) 2 dω r d 1 dr = c d ˆµ(ξ) 2 ˆµ(θξ) 2 dθ dξ. 0 S d 1 O(d) We also generalized the classical Mattila integral to study the set of side lengths of k-simplex set of E, e.g., when k = 2, we may ask when T 2 (E) = { ( x 1 x 2, x 2 x 3, x 3 x 1 ) R 3 : x i E} has positive 3-dimensional Lebesgue measure. Both Mattila s proof and GILP s proof of the Mattila integral are complicated and only valid on distances. In this talk we shall prove a more general result whose proof is, however, very simple. 6 / 19

7 A new derivation of the Mattila integral For simplicity, let s just consider the distance problem. Since x y = x y if and only if there exists g E(d), such that gx = x, gy = y, we have two ways to define ν on the distance set, ν(t) = µ(x) µ(y) dσ t (x, y), x y =t and, since {(x, y) : x y = t} = {(gx t, gy t ) : g E(d)} is an orbit, ν(t) µ(gx t ) µ(gy t ) dg, E(d) where x t y t = t, arbitrary but fixed. Therefore ( ( ) ) ν(t) 2 µ(x) µ(y) µ(gx t ) µ(gy t ) dg dσ t (x, y) dt E(d) x y =t 7 / 19

8 A new derivation of the Mattila integral Since x y = t and the integral is independent on the choice of (x t, y t ), we can take x t = x, y t = y. Then by co-area formula (change of variables), it becomes 2 2 µ(x)µ(gx) dx dg = µ(x)µ(θx + z) dx dz dθ. E(d) By Plancherel in x, = O(d) R d O(d) R d µ(ξ) e 2πiz ξ µ(θξ) dξ R d By Plancherel in z, ( ) = µ(ξ) 2 µ(θξ) 2 dξ dθ O(d) R d ( ) 2 = µ(rω) 2 dω r d 1 dr. S d 1 2 dz dθ. 8 / 19

9 A new derivation of the Mattila integral Suppose E 1,..., E k+1 R d with Borel measures µ j on E j. Suppose Φ(x 1, x 2,..., x k+1 ) = Φ(y 1, y 2,..., y k+1 ) if and only if for some g G, a LCT group, (y 1, y 2,..., y k+1 ) = (gx 1, gx 2,..., gx k+1 ). Let µ ɛ (x) = µ φ ɛ (x) C 0, where φ C 0, φ = 1, φ ɛ ( ) = ɛ d φ( ɛ ). Theorem (B.L., 2017) k+1 G j=1 ( uniformly in ɛ implies that the set R d µ ɛ j (x) µ ɛ j (gx) dx ) dg < Φ (E 1,..., E k+1 ) := {Φ(x 1,..., x k+1 ) : x j E j } has positive Lebesgue measure. 9 / 19

10 Sum-product problems Given any A R of N points, one can define its sum set, product set by A + A = {a 1 + a 2 : a 1, a 2 A}; AA = {a 1 a 2 : a 1, a 2 A}. The Erdős-Szemerédi conjecture states that for any ɛ > 0, Two extremal cases are A = {1, 2, 3,..., N}; A = {1, 2, 4,..., 2 N }. max{#(a + A), #(AA)} ɛ N 2 ɛ. Roughly speaking, this conjecture reflects that a set of real numbers cannot be structured in both additive and multiplicative sense simultaneously. The best currently known result in R is due to Konyagin and Shkredov. They prove max{#(a + A), #(AA)} ɛ N ɛ. 10 / 19

11 Sum-product problems There are various formulations of sum-product estimates. For example, Balog conjectured that #(A(A + A)) ɛ N 2 ɛ and the best currently known result is given by Murphy, Roche-Newton and Shkredov. They prove #(A(A + A)) ɛ N ɛ. One can also consider sum-product estimates on A F p, a finite field, where p is a prime. It was first studied by Bourgain, Katz, Tao and later studied by different authors. 11 / 19

12 Continuous sum-product problem Discrete version Erdős distance conjecture Continuous version Falconer distance conjecture #( (P)) C ɛ N 2 d ɛ (E) > 0 if dim H (E) > d 2 Sum-product conjectures max{#(a + A), #(AA)} ɛ N 2 ɛ? #(A(A + A)) ɛ N 2 ɛ Counterexamples show that for any s [0, 1], there exists A, B [0, 1], dim H (A) = dim H (A + A) = s, dim H (B) = dim H (BB) = s. Theorem (B.L., 2017) Suppose E, F, H R 2, dim H (E) + dim H (F ) + dim H (H) > 4. Then E (F + H) := {x (y + z) : x E, y F, z H} has positive Lebesgue measure. It is sharp, with E = {0}, F = H = [0, 1] / 19

13 Continuous sum-product problem Taking E = A {0}, F = B [0, 1], H = C [0, 1], we obtain the following sum-product estimate on R. Corollary (B.L., 2017) Suppose A, B, C R and dim H (A) + dim H (B) + dim H (C) > 2, then A(B + C) > 0. This dimensional exponent is generally optimal. In particular, for any A R, A(A + A) > 0 whenever dim H (A) > 2 3. To see the sharpness, one can take A = {0}, B = C = [0, 1]. 13 / 19

14 Sketch of the proof For any x = (x 1, x 2 ) R 2, denote x = (x 2, x 1 ) and E = {x : x E}. Without loss of generality, we may work on E (F + H). Notice x y = x y if and only if there exists g SL 2 (R) such that x = gx, y = gy. Therefore the corresponding Mattila integral is ( ) ( ) µ ɛ 1(x) µ ɛ 1(gx) dx µ ɛ 2(x) µ ɛ 2(gx) dx dg, R d R d SL 2 (R) where µ 1 = µ E, µ 2 = µ F µ H. By Plancherel, it suffices to show ( ) ( ) µ E (ξ) µ E (gξ) dξ µ F (η) µ H (η) µ F (gη) µ H (gη) dη dg <. R d R d We may take µ E, µ F, µ H as Frostman measures and it follows from harmonic analysis techniques. 14 / 19

15 Continuous sum-product problem Although the theorem and corollary above are generally sharp, one can still expect a better dimensional exponent for E (E + E), E R 2 and A(A + A), A R. For instance, it is not hard to show Theorem Suppose E, F, H R 2 and dim H (E) > 1, dim H (F ) + dim H (H) > 2, then In particular, E (F H) > 0. E (E ± E) > 0 if dim H (E) > 1, where the dimensional threshold is optimal. 15 / 19

16 Sketch of the proof The proof is super simple if we only consider E (E E). By Marstrand projection theorem, for almost all e S 1, and π e (A) > 0 (1) {t R : dim H (E π 1 e (t)) dim H (E) 1} > 0. (2) Then one can choose two distinct points x, y E such that x y is parallel to e where π e (A) > 0. Hence E (E E) E (x y) = π e (E) > 0. For general E (F H), we need to apply a radial projection theorem due to Mattila and Orponen to show F H contains a lot of directions. Then there must exist one direction e in F H such that E (F H) π e (E) > / 19

17 Continuous sum-product problem Corollary A(A + A) + A(A + A) > 0 whenever A R, dim H (A) > 1 2. This dimensional exponent is sharp. Now what we know is, for A R, A(A + A) > 0 if dim H (A) > 2 3 ; A(A + A) + A(A + A) > 0 if dim H (A) > 1 2, which is sharp. Then it is reasonable to make the following conjecture as an analog of Balog s conjecture. Conjecture Suppose A R, dim H (A) > 1 2, then A(A + A) has positive Lebesgue measure. 17 / 19

18 Counterexamples Let {q i } be a positive sequence such that q i+1 q i i. Take A i R as the q 1 s i neighborhood of and denote [0, 1] 1 q i Z A = A i. It is known that dim H (A) = s while for each i = 1, 2,..., A(A + A), i=1 (A + A)(A + A), etc. are contained in the q 1 s i neighborhood of [0, 1] 1 q 2 i Z whose Lebesgue measure is q 2 1 s i 0 if s < 1 2. In fact, instead of a constant s, we can choose s i 1 2 such that q 2 1 s i i 0. Then dim H (A) = 1 2 while A(A + A) = = / 19

19 Thank You! 19 / 19