Sets avoiding norm 1 in R n Christine Bachoc Universit´ e de Bordeaux, IMB Computation and Optimization of Energy, Packing, and Covering ICERM, Brown University, April 9-13, 2018 Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 1 / 20
The chromatic number of the plane ◮ The Hadwiger-Nelson problem 1950: What is the least number of colors needed to color R 2 such that two points at Euclidean distance 1 receive different colors? ◮ In 1950 Nelson introduced this number χ ( R 2 ) and together with Isbell proved that: 4 ≤ χ ( R 2 ) ≤ 7 ◮ On April 8, 2018, Aubrey de Grey posted on arXiv a paper proving χ ( R 2 ) ≥ 5. He constructs a unit distance graph in the plane with chromatic number 5 and with 1585 vertices (independently verified with SAT solvers). Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 2 / 20
The unit distance graph on R n ◮ The unit distance graph has vertices R n and edges { x , y } where x − y = 1 ◮ Its chromatic number is denoted χ ( R n ) . ◮ Its independent sets are sets avoiding distance 1. A ⊂ R n avoids distance 1 if x − y ∕ = 1 for all x , y in A . Example: the color classes of an admissible coloring. ◮ If A is measurable, it has a (upper) density δ ( A ) . Let m 1 ( R n ) := sup { δ ( A ) , A measurable, avoids distance 1 } ◮ For the measurable chromatic number χ m ( R n ) , the color classes are assumed to be measurable and we have: 1 χ m ( R n ) ≥ m 1 ( R n ) . ◮ Obviously χ m ( R n ) ≥ χ ( R n ) . Falconer 1981: χ m ( R n ) ≥ n + 3 for all n ≥ 2. Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 3 / 20
Sets avoiding distance 1 ◮ A set avoiding distance 1 in the plane: open disks of diameter 1, whose centers are hexagonal lattice points with pairwise minimal distance 2. δ ( A ) = ∆ 2 π 3 ≈ 0 . 226 4 = √ 8 ◮ Best known lower bound for m 1 ( R 2 ) : Croft 1967 Hexagonal arrangement of tortoises (disks cut out by hexagons) of density δ ≈ 0 . 229 Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 4 / 20
The combinatorial upper bounds for m 1 ( R n ) ◮ Let G = ( V , E ) a finite subgraph embedded in R n (ie the edges are the pairs of vertices at distance 1 apart). Let α ( G ) be its independence number. α ( G ) = 2 ◮ We have: m 1 ( R n ) ≤ α ( G ) | V | Proof: let A be a subset of R n avoiding 1. A translated copy of G has on average | V | δ ( A ) vertices in A , but also at most α ( G ) vertices in A . ◮ Larman Rogers 1972: good graphs for small dimensions. Improved by Szekely Wormald 1989. Frankl Wilson 1981, Raigorodskii 2000: m 1 ( R n ) ( 1 . 239 ) − n Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 5 / 20
The eigenvalue upper bound for m 1 ( R n ) ◮ Oliveira, Vallentin 2010: if ω is the normalized surface measure of S n − 1 , ω ( u ) − min m 1 ( R n ) ≤ 1 − min ω ( u ) ◮ This is a continuous analog of Hoffman bound for finite d -regular graphs: α ( G ) − λ min ( A G ) ≤ | V | d − λ min ( A G ) ◮ The Fourier transform of the surface measure on S n − 1 expresses in terms of the Bessel function J n / 2 − 1 : ω ( u ) = Ω n ( u ) = Γ ( n / 2 )( 2 / u ) n / 2 − 1 J n / 2 − 1 ( u ) Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 6 / 20
The eigenvalue upper bound for m 1 ( R n ) ◮ Asymptotically the eigenvalue bound is not as good as the combinatorial bound: ω ( u ) − min e / 2 ) − n ≈ ( 1 . 165 ) − n ω ( u ) ≈ ( 1 − min ◮ It can be strengthened through extra constraints, leading to the best known bounds for 2 ≤ n ≤ 24. Oliveira Vallentin 2018: A general framework using the cone of boolean quadratic constraints (Fernando’s talk last monday). ◮ Combined with combinatorial constraints it also leads to: [B., Passuello, Thiery 2013] m 1 ( R n ) ( 1 . 268 ) − n Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 7 / 20
Non euclidean norms ◮ What about other norms? In particular what about norms defined by a convex symmetric polytope P ? ◮ Examples: ∞ corresponds to the hypercube; 1 corresponds to the crosspolytope. ◮ In general, P is defined by x P = min { λ | x ∈ λ P } and we have similar notions of m P ( R n ) , χ P ( R n ) . ◮ The 1-avoiding problem appears to be related to (and more difficult than) the packing problem, in particular if there is a tight packing. Maybe it is easier to analyze if the packing problem is easy or even trivial. Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 8 / 20
Polytopes that tile space ◮ If the polytope P tiles space by translations then the density 1 / 2 n is attained by A = ∪ x ∈ L ( x + P / 2 ) Conjecture (B., Sinai Robins) If P is a convex polytope that tiles R n by translations then m P ( R n ) = 1 / 2 n ◮ The 2 n translates of A provide an admissible measurable coloring of R n so the conjecture also implies that χ P , m ( R n ) = 2 n . Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 9 / 20
Convex symmetric polytopes that tile space by translations ◮ Lattices give rise to such polytopes: their Dirichlet-Vorono¨ ı cells. ◮ In dimension 2, two combinatorial types: the rectangle and the hexagon. ◮ In dimension 3, there are 5 combinatorial types: 2 0 1 Z 3 A # A 2 ⊥ Z A 3 0 2 1 3 1 1 3 ◮ Vorono¨ ı conjecture 1908: a translative convex polytope is the affine image of the Vorono¨ ı cell of a lattice. Proved by Delone for n ≤ 4. Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 10 / 20
Methods ◮ The combinatorial bound: m P ( R n ) ≤ α ( G ) | V | ◮ Example: the hypercube G is the complete graph ⇒ m P ( R n ) = 1 / 2 n ◮ The eigenvalue-Fourier bound: Let µ be a measure supported on ∂ P , µ ( u ) − min m P ( R n ) ≤ µ ( u ) . µ ( 0 ) − min Recall: e 2 i π ( x · u ) d µ ( x ) µ ( u ) = µ ( 0 ) = µ ( ∂ P ) ∂ P Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 11 / 20
Joint work with Thomas Bellitto, Philippe Moustrou, Arnaud Pˆ echer (2017) Theorem If P tiles the plane, then m P ( R 2 ) = 1 / 4 Theorem If P is the Dirichlet-Vorono¨ ı cell of the root lattice A n , n ≥ 2 then m P ( R n ) = 1 / 2 n If P is the Dirichlet-Vorono¨ ı cell of the root lattice D n , n ≥ 4 , then m P ( R n ) ≤ 1 / (( 3 / 4 ) 2 n + n − 1 ) n n A n := Z n + 1 ∩ { D n := { ( x 1 , . . . , x n ) ∈ Z n : x i = 0 } x i = 0 mod 2 } i = 0 i = 1 Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 12 / 20
The Fourier-eigenvalue bound ( P can be any symmetric convex body) µ ( u ) − min m P ( R n ) ≤ For all µ supported on ∂ P , µ ( 0 ) − min µ ( u ) ◮ Let A be 1-avoiding and L -periodic. The auto-correlation function associated to A : 1 f A ( x ) = 1 A ( x + y ) 1 A ( y ) dy . vol( L ) R n / L ◮ Let m := min µ ( u ) and ν := µ − m δ 0 n . We have µ − m ≥ 0 . ν = ◮ We compute in two different ways f A ( x ) d ν ( x ) = − mf A ( 0 n ) = − m δ ( A ) ν ( u ) ≥ f A ( 0 n ) ν ( 0 n ) = δ ( A ) 2 ( f A ( u ) µ ( 0 ) − m ) . = u ∈ L # Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 13 / 20
The Fourier-eigenvalue bound for polytopes On going work with Philippe Moustrou and Sinai Robins µ ( u ) − min m P ( R n ) ≤ For all µ supported on ∂ P , µ ( 0 ) − min µ ( u ) ◮ Without loss of generality, µ can be chosen invariant under the orthogonal group of P (by convexity argument). ◮ For P = S n − 1 , it leaves only one possibility up to scaling: the surface measure ω . ◮ For other P , e.g. polytopes, lots of possibilities! (is it a good or a bad news?) ◮ How can we optimize over µ ? ◮ We will see that point measures boil down to polynomial optimization problems when the points have rational coordinates (the polytope having vertices in Z n ). ◮ Moreover, in this case the weights can be viewed as additional polynomial variables, so optimizing over the weights for a fixed finite support amounts again to solving a polynomial optimization problem. Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 14 / 20
A toy example ◮ The square δ ( ± 1 , ± 1 ) + 1 ( δ ( ± 1 , 0 ) + δ ( 0 , ± 1 ) ) µ = 1 4 2 ◮ We have µ ( u ) = 1 e 2 π i ( ± u 1 ± u 2 ) + 1 e 2 π i ( ± u 1 ) + e 2 π i ( ± u 2 ) ) 2 ( 4 = cos( 2 π u 1 ) cos( 2 π u 2 ) + cos( 2 π u 1 ) + cos( 2 π u 2 ) = (cos( 2 π u 1 ) + 1 )(cos( 2 π u 2 ) + 1 ) − 1 ◮ Leading to 3 + 1 = 1 1 µ ( 0 ) = 3 , µ ( u ) = − 1 , bound = min 4 Sets avoiding norm 1 in R n Christine Bachoc (Universit´ e de Bordeaux, IMB) 15 / 20
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