icerm.brown.edu · Created Date: 4/13/2018 7:06:56 PM

Sets avoiding norm 1 in Rn

Christine Bachoc

Universite de Bordeaux IMB

Computation and Optimization of Energy Packing and CoveringICERM Brown University April 9-13 2018

Christine Bachoc (Universite de Bordeaux IMB) Sets avoiding norm 1 in Rn 1 20

The chromatic number of the plane

The Hadwiger-Nelson problem 1950 What is the least number of colors neededto color R2 such that two points at Euclidean distance 1 receive different colors

In 1950 Nelson introduced this number χ(R2) and together with Isbell proved that

4 le χ(R2) le 7

On April 8 2018 Aubrey de Grey posted on arXiv a paper proving χ(R2) ge 5 Heconstructs a unit distance graph in the plane with chromatic number 5 and with1585 vertices (independently verified with SAT solvers)


The unit distance graph on Rn

The unit distance graph has vertices Rn and edges x y where 983042x minus y983042 = 1 Its chromatic number is denoted χ(Rn) Its independent sets are sets avoiding distance 1

A sub Rn avoids distance 1 if 983042x minus y983042 ∕= 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

m1(Rn) = supδ(A) A measurable avoids distance 1 For the measurable chromatic number χm(Rn) the color classes are assumed to

be measurable and we have

χm(Rn) ge 1m1(Rn)

Obviously χm(Rn) ge χ(Rn) Falconer 1981 χm(Rn) ge n + 3 for all n ge 2


Sets avoiding distance 1

A set avoiding distance 1 in the plane open disks of diameter 1 whose centersare hexagonal lattice points with pairwise minimal distance 2

δ(A) = ∆24 = π

8radic

3asymp 0226

Best known lower bound for m1(R2) Croft 1967Hexagonal arrangement of tortoises (disks cut out by hexagons) of densityδ asymp 0229


The combinatorial upper bounds for m1(Rn)

Let G = (V E) a finite subgraph embedded in Rn (ie the edges are the pairs ofvertices at distance 1 apart)Let α(G) be its independence number

α(G) = 2

We havem1(Rn) le α(G)

|V |Proof let A be a subset of Rn 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 dimensionsImproved by Szekely Wormald 1989Frankl Wilson 1981 Raigorodskii 2000

m1(Rn) 983253 (1239)minusn


The eigenvalue upper bound for m1(Rn)

Oliveira Vallentin 2010 if ω is the normalized surface measure of Snminus1

m1(Rn) le minusmin 983141ω(u)1 minusmin 983141ω(u)

This is a continuous analog of Hoffman bound for finite d-regular graphs

α(G)

|V | le minusλmin(AG)

d minus λmin(AG)

The Fourier transform of the surface measure on Snminus1 expresses in terms of theBessel function Jn2minus1

983141ω(u) = Ωn(983042u983042) = Γ(n2)(2983042u983042)n2minus1Jn2minus1(983042u983042)



Asymptotically the eigenvalue bound is not as good as the combinatorial bound

minusmin 983141ω(u)1 minusmin 983141ω(u) asymp (

983155e2)minusn asymp (1165)minusn

It can be strengthened through extra constraints leading to the best knownbounds for 2 le n le 24

Oliveira Vallentin 2018 A general framework using the cone of boolean quadraticconstraints (Fernandorsquos talk last monday)

Combined with combinatorial constraints it also leads to [B Passuello Thiery2013]

m1(Rn) 983253 (1268)minusn


Non euclidean norms

What about other normsIn particular what about norms defined by a convex symmetric polytope P

Examples 983042 983042infincorresponds to the hypercube 983042 9830421corresponds to thecrosspolytope

In general 983042 983042P is defined by

983042x983042P = minλ | x isin λP

and we have similar notions of mP(Rn) χP(Rn) 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 toanalyze if the packing problem is easy or even trivial


Polytopes that tile space

If the polytope P tiles space by translations then the density 12n is attained by

A = cupxisinL(x + P2)

Conjecture

(B Sinai Robins) If P is a convex polytope that tiles Rn by translations then

mP(Rn) = 12n

The 2n translates of A provide an admissible measurable coloring of Rn so theconjecture also implies that χPm(Rn) = 2n


Convex symmetric polytopes that tile space by translations

Lattices give rise to such polytopes their Dirichlet-Voronoı cells In dimension 2 two combinatorial types the rectangle and the hexagon

In dimension 3 there are 5 combinatorial types

Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3

Voronoı conjecture 1908 a translative convex polytope is the affine image of theVoronoı cell of a lattice Proved by Delone for n le 4


Methods

The combinatorial boundmP(Rn) le α(G)

|V | Example the hypercube

G is the complete graph rArr mP(Rn) = 12n

The eigenvalue-Fourier bound Let micro be a measure supported on partP

mP(Rn) le minusmin 983141micro(u)983141micro(0)minusmin 983141micro(u)

Recall983141micro(u) =

983133

partPe2iπ(xmiddotu)dmicro(x) 983141micro(0) = micro(partP)


Joint work with Thomas Bellitto Philippe Moustrou Arnaud Pecher(2017)

TheoremIf P tiles the plane then

mP(R2) = 14

TheoremIf P is the Dirichlet-Voronoı cell of the root lattice An n ge 2 then

mP(Rn) = 12n

If P is the Dirichlet-Voronoı cell of the root lattice Dn n ge 4 then

mP(Rn) le 1((34)2n + n minus 1)

An = Zn+1 cap n983131

i=0

xi = 0 Dn = (x1 xn) isin Zn n983131

i=1

xi = 0 mod 2


The Fourier-eigenvalue bound (P can be any symmetric convex body)

For all micro supported on partP mP(Rn) le minusmin 983141micro(u)983141micro(0)minusmin 983141micro(u)

Let A be 1-avoiding and L-periodic The auto-correlation function associated to A

fA(x) =1

vol(L)

983133

RnL1A(x + y) 1A(y)dy

Let m = min 983141micro(u) and

ν = microminus mδ0n We have 983141ν = 983141microminus m ge 0

We compute in two different ways983133

fA(x)dν(x) = minusmfA(0n) = minusmδ(A)

=983131

uisinL

983141fA(u)983141ν(u) ge 983141fA(0n)983141ν(0n) = δ(A)2(983141micro(0)minus m)


The Fourier-eigenvalue bound for polytopesOn going work with Philippe Moustrou and Sinai Robins


Without loss of generality micro can be chosen invariant under the orthogonal groupof P (by convexity argument)

For P = Snminus1 it leaves only one possibility up to scaling the surface measure ω For other P eg polytopes lots of possibilities (is it a good or a bad news) How can we optimize over micro We will see that point measures boil down to polynomial optimization problems

when the points have rational coordinates (the polytope having vertices in Zn) 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 tosolving a polynomial optimization problem


A toy example

The square

micro = 14

983123δ(plusmn1plusmn1) +

12

983123(δ(plusmn10) + δ(0plusmn1))

We have

983141micro(u) = 14

983131e2πi(plusmnu1plusmnu2) +

12(983131

e2πi(plusmnu1) +983131

e2πi(plusmnu2))

= cos(2πu1) cos(2πu2) + cos(2πu1) + cos(2πu2)

= (cos(2πu1) + 1)(cos(2πu2) + 1)minus 1

Leading to

983141micro(0) = 3 min 983141micro(u) = minus1 bound =1

3 + 1=

14


The hypercube

The centers of the k-dimensional faces are up to permutation of the coordinates(0 0plusmn1 plusmn1) with k zeroes

The measure micro supported on these points weighted by 12k

Xj = cos(2πuj) 983141micro(u) =n983132

j=1

(Xj + 1)minus 1

and has total volume 2n minus 1 and minimum minus1 leading to the sharp bound 2minusn In general if P is invariant under plusmn1n the Fourier transform of a measure

supported on points with rational coordinates can be expressed as a polynomial inthe variable Xj = cos(2πujk) if k is a common denominator of the coordinates


The crosspolytope CPn = (x1 xn) isin Rn | |x1|+ middot middot middot+ |xn| le 1

We consider measures supported on

partCPn cap1kZn = x isin 1

kZn | |x1|+ middot middot middot+ |xn| = 1

The orbits of kx = d under the action of plusmn1nSn are represented by thepartitions of k in at most n parts

d = (d1 dn) d1 ge d2 ge middot middot middot ge dn ge 0 d1 + middot middot middot+ dn = k

Let us denote this set Pkn The Fourier transform of a measure invariant underplusmn1Sn and with support contained in partCPn cap 1

k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)

Using the Chebyshev polynomials Tℓ we obtain a polynomial

983141micro(u) =983131

disinPkn

λd Td1(X1) Tdn (Xn) Xj = cos(2πuj

k)


The crosspolytope

It remains to minimize over the variables X and over the weights λ a polynomialoptimization problem that can be treated through sums of squares techniques

min983153 983131

disinPkn

λd Td1(X1) Tdn (Xn) minus1 le Xj le 1983131

λd = 2n minus 1983154

Numerical results for n = 3

k Nb of pts min 983141micro Bound1 6 minus7 052 18 minus14 016664 42 minus13253 015926 122 minus13201 015868 258 minus13195 01586

18 1298 minus13156 01582


The optimal measures on CP3

The distribution of weights in the numerically optimal measure for k = 4 6 8 18


More numerical results

The crosspolytopes

Dimension Division Minimum Bound3 18 minus13156 015824 4 minus15213 0092085 8 minus19742 005988

The Voronoı cells of Dn 983042x983042P = maxi ∕=j|xi |+|xj |

2


The optimal support appears to be the vertices and the middle of two verticesbelonging to a common facet

The Voronoı cells of Dn 983042x983042P = max

983153983042x983042infin

2 983042x9830421n

983154

Dimension Division Minimum Bound3 4 minus14143 016805 2 minus22202 006684


The chromatic number of the plane

The Hadwiger-Nelson problem 1950 What is the least number of colors neededto color R2 such that two points at Euclidean distance 1 receive different colors

In 1950 Nelson introduced this number χ(R2) and together with Isbell proved that

4 le χ(R2) le 7

On April 8 2018 Aubrey de Grey posted on arXiv a paper proving χ(R2) ge 5 Heconstructs a unit distance graph in the plane with chromatic number 5 and with1585 vertices (independently verified with SAT solvers)








χm(Rn) ge 1m1(Rn)





δ(A) = ∆24 = π

8radic

3asymp 0226





α(G) = 2




m1(Rn) 983253 (1239)minusn






α(G)


d minus λmin(AG)











m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154









χm(Rn) ge 1m1(Rn)





δ(A) = ∆24 = π

8radic

3asymp 0226





α(G) = 2




m1(Rn) 983253 (1239)minusn






α(G)


d minus λmin(AG)











m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154





δ(A) = ∆24 = π

8radic

3asymp 0226





α(G) = 2




m1(Rn) 983253 (1239)minusn






α(G)


d minus λmin(AG)











m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154





α(G) = 2




m1(Rn) 983253 (1239)minusn






α(G)


d minus λmin(AG)











m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154







α(G)


d minus λmin(AG)











m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154










m1(Rn) 983253 (1268)minusn


Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154



Non euclidean norms




983042x983042P = minλ | x isin λP







Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154






Conjecture


mP(Rn) = 12n






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154






Z3 A2 perp Z A3

983091

9831072 0 10 2 11 1 3

983092

983108 A3



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154



Methods







983133





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154





mP(R2) = 14


mP(Rn) = 12n



An = Zn+1 cap n983131

i=0


i=1

xi = 0 mod 2





fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154






fA(x) =1

vol(L)

983133






=983131

uisinL










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154










A toy example

The square

micro = 14


12


We have

983141micro(u) = 14


12(983131


e2πi(plusmnu2))



Leading to


3 + 1=

14


The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154



The hypercube




j=1

(Xj + 1)minus 1











k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154










k Zn

983141micro(u) =983131

disinPkn

λd cos(2πd1u1

k) cos(2π

dnun

k)


983141micro(u) =983131

disinPkn


k)


The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154



The crosspolytope


min983153 983131

disinPkn


λd = 2n minus 1983154



18 1298 minus13156 01582






The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154







The crosspolytopes



2




983153983042x983042infin

2 983042x9830421n

983154



icerm.brown.edu · Created Date: 4/13/2018 7:06:56 PM

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