DISCRETE MINIMAL ENERGY PROBLEMS

Transcription

1 DISCRETE MINIMAL ENERGY PROBLEMS Lecture III E. B. Saff Center for Constructive Approximation Vanderbilt University University of Crete, Heraklion May, 2017

2 Random configurations ΩN = {X1, X2,..., XN }: N independent samples chosen according to a probability measure µ supported on A random points What about the s-energy? 3000 points near optimal for s = 2

3 Random configurations: Expected s-energy E [E s (Ω N )] = = i j = i j 1 x i x j s dµ(x 1) dµ(x N ) i j 1 x i x j s dµ(x 1) dµ(x N ) 1 x i x j s dµ(x i)dµ(x j ). where E [E s (Ω N )] = N(N 1)I s (µ), I s (µ) := 1 x y s dµ(x)dµ(y).

4 As N grows large, what is the expected separation and expected covering, say on S d?

5 As N grows large, what is the expected separation and expected covering, say on S d? Best possible order that can be achieved by N points is 1 N 1/d.

6 Random configurations on S d Theorem (Brauchart, Reznikov, S, Sloan, Wang, Womersley) 2016 For N random, independently and uniformly distributed points (w.r.t. surface area), E(δ(S d, Ω N )) 2 1/d 1 Γ(1 + 1/d)C(d), N, N2/d where C(d) := Area(S d )/Vol(B d ), B d is d-dim unit ball. Proof uses estimates of Cai, Fan, and Jiang (2013).

7 Random configurations on S d Theorem (Brauchart, Reznikov, S, Sloan, Wang, Womersley) 2016 For N random, independently and uniformly distributed points (w.r.t. surface area), E(δ(S d, Ω N )) 2 1/d 1 Γ(1 + 1/d)C(d), N, N2/d where C(d) := Area(S d )/Vol(B d ), B d is d-dim unit ball. Proof uses estimates of Cai, Fan, and Jiang (2013). Theorem (Reznikov, S) 2015 For N random, independently and uniformly distributed points, ( ) 1/d log N E(ρ(S d, Ω N )) C(d), N, N

8 Minimal s-energy points on S d For 0 < s < d, K s (x, y) = 1 x y s is strictly positive definite on S d. The Riesz s-potential of normalized surface area σ d satisfies d+1 U σ Γ( d 2 )Γ(d s) s (x) = I s [σ d ] = Γ( d s+1 2 )Γ(d s/2), x Sd. σ d is the unique Riesz s-equilibrium measure. E s (S d, N) lim N N 2 = W s (S d ) = I s [σ d ]<, and any sequence of optimal N-point s-energy configurations is asymptotically uniformly distributed on S d. Optimal order separation 1/N 1/d holds for minimal s-energy configurations when d 2 < s < d (Dragnev, S) 2006

9 What About s d? Not difficult to show For s d, W s (S d ) =. Moreover, for s > d, there exist C 1, C 2 > 0 such that C 1 N 1+s/d E s (S d, N) C 2 N 1+s/d, for all N 2.

10 What About s d? Not difficult to show For s d, W s (S d ) =. Moreover, for s > d, there exist C 1, C 2 > 0 such that Questions: Does C 1 N 1+s/d E s (S d, N) C 2 N 1+s/d, for all N 2. E s (S d, N) lim N N 1+s/d exist? What about the limiting distribution of N-point minimal s-energy configurations?

11 More generally what happens on d-dimensional manifolds when s d? We begin with d = 1.

12 d = 1: Min. Riesz s-energy Points on a Curve N = s s s = 0.2 s = 2.0 For 0 s < 1, limit distribution is the measure minimizing continuous (integral) s-energy.

13 Minimal Energy Points on Curves (d = 1) Theorem (M-F, M, R, S) If Γ is a rectifiable Jordan arc or curve in R p, then for s = 1, E 1 (Γ, N) lim N N 2 log N = 2 L, L := length(γ), and for s > 1 E s (Γ, N) lim N N 1+s = 2ζ(s) L s. where ζ(s) denotes classical Riemann zeta function. Moreover, for s 1, the limit distribution of optimal s-energy configurations is uniform with respect to arclength measure on Γ.

14 Why the Zeta Function? x i d i,j x j Let ω N = {x k } N 1 be points on a Jordan arc listed in consecutive order. d i,j = length of subarc (x i, x j ) Then E s (ω N ) = 2 i<j 1 x i x j s 2 i<j 1 d s i,j N 1 N k = 2 k=1 i=1 1 d s i,i+k

15 Why the Zeta Function? So N 1 E s (ω N ) 2 Convexity of t s gives k=1 Ê k, where Ê k := N k i=1 1 d s i,i+k ( ) s N k N k Ê k = (d i,i+k ) s 1 (N k) d i,i+k N k i=1 = i=1 (N k) 1+s ( N k ) s i=1 d i,i+k = N1+s L s k s (N k)1+s (kl) s [ 1 k ] 1+s N

16 Why the Zeta Function? Thus, E s (ω N ) 2 N1+s L s N 1 k=1 [ 1 k s 1 k ] 1+s. N

17 Why the Zeta Function? Thus, E s (ω N ) 2 N1+s L s N 1 k=1 [ 1 k s 1 k ] 1+s. N Letting 0 < r < 1, we get E s (ω N ) 2 N1+s L s rn k=1 [ 1 k s 1 rn ] 1+s 2 N1+s rn N L s k=1 1 k s (1 r)1+s,

18 Why the Zeta Function? Thus, E s (ω N ) 2 N1+s L s N 1 k=1 [ 1 k s 1 k ] 1+s. N Letting 0 < r < 1, we get E s (ω N ) 2 N1+s L s and so for s > 1, lim inf N E s (ω N ) N 1+s Now let r 0, rn k=1 lim inf N [ 1 k s 1 rn ] 1+s 2 N1+s rn N L s k=1 lim inf N rn 2 (1 r)1+s Ls k=1 E s (A, N) N 1+s 1 k s (1 r)1+s, 1 k s = 2 L s (1 r)1+s ζ(s). 2ζ(s) L s.

19 Completing the proof for s > 1 To prove lim sup N E s (A, N) N 1+s 2ζ(s) L s. use "equally-spaced points" on Γ with appropriate modifications at corners and cusps to obtain upper bounds on energy. Remains to show: If {ωn } is a sequence of N-point minimal s-energy configurations on Γ, then ν(ωn) λ, N, where λ is normalized arc length on Γ.

20 What about higher dimensional sets?

21 An Example: Torus (Bagel), N = 1000

22 From the Heraklion Archaeological Museum

23 Let A R p. Hausdorff Measure and Dimension Definition: Let α > 0. H α (A) := lim ɛ 0 + inf where K := diam(k ). K j α : A j=1k j, K j ɛ, j=1 d = dim H (A) = inf{α : H α (A) = 0} = sup{α : H α (A) = }. We shall normalize so that embedding of [0, 1] d in R p satisfies H d ( [0, 1] d ) = 1.

24 Rectifiable Sets in R p A function Φ : T R p, T R d, is said to be a Lipschitz mapping on T if for some L > 0, Definition Φ(x) Φ(y) L x y, x, y T. A R p is a d-rectifiable set if A is the image of a bounded set in R d under a Lipschitz mapping.

25 E s (A, N) lim N N 2 = W s (A) = I s (µ A )<, for 0 < s < d = dim H (A).

26 E s (A, N) lim N N 2 = W s (A) = I s (µ A )<, for 0 < s < d = dim H (A). Poppy-Seed Bagel Theorem (HS (2005) and BHS (2008)) Suppose s d and A R p is a d-rectifiable set (i.e. Lipschitz image of a compact set in R d ). When s = d we further assume A is a subset of a d-dimensional C 1 manifold. Then E d (A, N) lim N N 2 log N = H d(b d ) H d (A), E s (A, N) C s,d lim = N N 1+s/d [H d (A)], s > d. s/d

27 E s (A, N) lim N N 2 = W s (A) = I s (µ A )<, for 0 < s < d = dim H (A). Poppy-Seed Bagel Theorem (HS (2005) and BHS (2008)) Suppose s d and A R p is a d-rectifiable set (i.e. Lipschitz image of a compact set in R d ). When s = d we further assume A is a subset of a d-dimensional C 1 manifold. Then E d (A, N) lim N N 2 log N = H d(b d ) H d (A), E s (A, N) C s,d lim = N N 1+s/d [H d (A)], s > d. s/d Furthermore, if H d (A) > 0, then optimal s-energy configurations for s d are asymptotically (as N ) uniformly distributed with respect to d-dimensional Hausdorff measure.

28 E s (A, N) lim N N 2 = W s (A) = I s (µ A )<, for 0 < s < d = dim H (A). Poppy-Seed Bagel Theorem (HS (2005) and BHS (2008)) Suppose s d and A R p is a d-rectifiable set (i.e. Lipschitz image of a compact set in R d ). When s = d we further assume A is a subset of a d-dimensional C 1 manifold. Then E d (A, N) lim N N 2 log N = H d(b d ) H d (A), E s (A, N) C s,d lim = N N 1+s/d [H d (A)], s > d. s/d Furthermore, if H d (A) > 0, then optimal s-energy configurations for s d are asymptotically (as N ) uniformly distributed with respect to d-dimensional Hausdorff measure. C s,1 = 2ζ(s), s > 1; unknown for d > 1.

29 Poppy-Seed Bagel Theorem, continued If {ωn } N=2 is a sequence of minimal s-energy configurations on the d-rectifiable set A with s > d, then the sequence has "optimal order separation"; i.e., δ(ωn) 1, N. N1/d Furthermore, if A is also d-regular, then the sequence {ωn } provides "optimal order covering"; i.e., ρ(ω N, A) := max y A min x ω N y x 1, N. N1/d By "d-regular" we mean there is a positive measure µ supported on A and c 1, c 2 > 0 such that c 1 r d µ(b(x, r)) c 2 r d, x A, 0 < r < diam(a).

30 Proof that lim N E s ([0, 1] d, N)/N 1+s/d exists for s > d Let Q d := [0, 1] d, the unit cube in R d. Start with N minimal s-energy points for Q d. Let 0 < γ < 1 and m a positive integer. Shrink Q d by a factor γ/m, creating m d disjoint subcubes separated by (1 γ)/m. E s (Q d, m d N) m d E s ( (γ/m)q d, N ) + cube-cube interactions where K := k Z d, k 0 γ s m s m d E s ( Q d, N ) + K (1 γ) s m s+d N 2, k s < for s > d.

31 Proof that lim N E s ([0, 1] d, N)/N 1+s/d exists for s > d Hence E s (Q d, m d N) (m d N) E s(q d, N) K (1 1+s/d γ s γ) s +. N 1+s/d N 1+s/d Now let (m 1) d N < n m d N. Then E s (Q d, n) E s(q d, m d N) n 1+s/d (m d N) 1+s/d ( γ s E s(q d, N) + N 1+s/d ( m ) d+s m 1 K (1 γ) s N 1+s/d ) ( ) d+s m. m 1

32 Proof that lim N E s ([0, 1] d, N)/N 1+s/d exists for s > d So lim sup n E s (Q d, n) n 1+s/d ( γ s E s(q d, N) + N 1+s/d ) K (1 γ) s. N 1+s/d Choose N appropriately, then γ close to 1 to get lim sup n E s (Q d, n) n 1+s/d lim inf k E s (Q d, k) k 1+s/d. So limit exists.

33 About the Proof To extend the result to d-rectifiable manifolds we need the following basic result (see Federer, 1969) of geometric measure theory: Lemma Suppose A R p is a d-dimensional rectifiable set and ɛ > 0. Then there exists a countable collection {K i i = 1, 2,...} of compact subsets of R d and bi-lipschitz mappings ψ i : K i R p, i = 1, 2,..., with constant (1 + ɛ) such that ψ 1 (K 1 ), ψ 2 (K 2 ), ψ 3 (K 3 ),... are pairwise disjoint subsets of A covering H d -almost all of A.

34 The Constant C s,d What is the constant C s,d for s > d? d = 1: From optimality of the roots of unity on S 1, C s,1 = 2ζ(s) for s > 1, where ζ(s) is the classical Riemann zeta function.

35 The Constant C s,d What is the constant C s,d for s > d? d = 1: From optimality of the roots of unity on S 1, C s,1 = 2ζ(s) for s > 1, where ζ(s) is the classical Riemann zeta function. d = 2: From [KS] it is known that ( ) s/2 C s,2 3/2 ζl (s), s > 2, where ζ L (s) is the zeta function for the unit hexagonal lattice.

36 The Constant C s,2 2 ( ) s/2 Conjecture: C s,2 = 3/2 ζl (s), s > 2, where ζ L := v s, 0 v L L = {k 1 v 1 + k 2 v 2 : k 1, k 2 Z} This is equivalent to the conjecture that the equi-triangular lattice describes the ground state for particles in the plane interacting through a Riesz potential r s. Similar conjectures for d = 8 (E 8 lattice) and d = 24 (Leech lattice).

37 Connection to Best-packing Problem in R d The d-dimensional packing density d is the maximum fraction of d space that can be covered by a collection of non-overlapping balls of the same radius. Connection between C s,d and d : ( ) 1/d (C s,d ) 1/s Vol(Bd ) (1/2) as s,. d

38 Connection to Best-packing Problem in R d The d-dimensional packing density d is the maximum fraction of d space that can be covered by a collection of non-overlapping balls of the same radius. Connection between C s,d and d : 1 = 1, 2 = π/ 12 (Fejes-Toth), 3 = π/ 18 (Hales) ( ) 1/d (C s,d ) 1/s Vol(Bd ) (1/2) as s,. d

39 8 Solved The sphere packing problem in dimension 8 Maryna S. Viazovska March 15, v1 [math.nt] 14 Mar 2016 In this paper we prove that no packing of unit balls in Euclidean space R 8 has density greater than that of the E8-lattice packing. Keywords: Sphere packing, Modular forms, Fourier analysis AMS subject classification: 52C17, 11F03, 11F30 1 Introduction The sphere packing constant measures which portion of d-dimensional Euclidean space can be covered by non-overlapping unit balls. More precisely, let R d be the Euclidean vector space equipped with distance k kand Lebesgue measure Vol( ). For x 2 R d and r 2 R>0 we denote by Bd(x, r) the ball in R d with center x and radius r. Let X R d be a discrete set of points such that kx yk 2 for any distinct x, y 2 X. Then the union P = [ Bd(x, 1) x2x is a sphere packing. IfX is a lattice in R d then we say that P is a lattice sphere packing. The finite density of a packing P is defined as P(r) := Vol(P \Bd(0,r)), r > 0.

40 24 Solved THE SPHERE PACKING PROBLEM IN DIMENSION v1 [math.nt] 21 Mar 2016 HENRY COHN, ABHINAV KUMAR, STEPHEN D. MILLER, DANYLO RADCHENKO, AND MARYNA VIAZOVSKA Abstract. Building on Viazovska s recent solution of the sphere packing problem in eight dimensions, we prove that the Leech lattice is the densest packing of congruent spheres in twenty-four dimensions, and that it is the unique optimal periodic packing. In particular, we find an optimal auxiliary function for the linear programming bounds, which is an analogue of Viazovska s function for the eight-dimensional case. 1. Introduction The sphere packing problem asks how to arrange congruent balls as densely as possible without overlap. The density is the fraction of space covered by the balls, and the problem is to find the maximal possible density. This problem plays an important role in geometry, number theory, and information theory. See [4] for background and references on sphere packing and its applications. Although many interesting constructions are known, provably optimality is very rare. Aside from the trivial case of one dimension, the optimal density was previously known only in two [8], three [6, 7], and eight [9] dimensions, with the latter result being a recent breakthrough due to Viazovska. Building on her work, we prove the following theorem: Theorem 1.1. The Leech lattice achieves the optimal sphere packing density in R 24, and it is the only periodic packing in R 24 with that density. It is unknown in general whether optimal packings have any special structure, but

41 Adding Weight Can Be A Good Thing! Notation SLP weight: w : A A [0, ) is Symmetric and Lower semi-continuous on A A and Positive on the diag(a A) := {(a, a): a A}. CPD weight: SLP weight Continuous at every point in diag(a A). For s > 0 and ω N = {x 1,..., x N } A, E w s (ω N ):= i j w(x i, x j ) x i x j s. E w s (A, N) := min {E w s (ω N ) : ω N A, #ω N = N}.

42 Weighted Riesz Energy Theorem (BHS, 2008) Let A R p be d-rectifiable. If s > d and w is a CPD-weight on A A, then Es w (A, N) C s,d lim = N N 1+s/d [ H s,w d (A) ], s/d where C s,d is as before and H s,w d is weighted Hausdorff measure on Borel sets B, H s,w 1 d (B) := w(x, x) dh d(x). d/s Moreover, if H d (A) > 0, then any sequence of (w, s)-energy minimizing points has limit distribution B H s,w d ( ) A.

43 Weighted Riesz Energy Corollary To distribute points on A according to a positive and continuous density ρ(x) on a d-rectifiable set A, choose w(x, y) := (ρ(x)ρ(y)) s/2d and compute minimal weighted s-energy points for any s > d.

44 Example: A = S 2 ; N = 750; s = 3; nonuniform weight { 1 π/4 φ 3π/4 ρ(φ) = 10 cos(4φ) + 11 otherwise Π