Lattice Packings of Spheres - Yale Universitynk354/papers/latticepackings.pdf · Lattices Lattice Packings in Dimensions Two and Three Hermite, Mordell, and Voronoi Poisson Summation

LatticePackings ofSpheres

Kaplan

Background

Packings inTwo andThreeDimensions

Lattices

LatticePackings inDimensionsTwo andThree

Hermite,Mordell, andVoronoi

PoissonSummationand n = 24

Lattice Packings of Spheres

Nathan Kaplan

Harvard University

Cambridge, MA. December 2, 2010

Kaplan Lattice Packings of Spheres


Kaplan

Background


Lattices




Sphere Packings

How can we most efficiently pack spheres of fixed radius inn-dimensional space?

A packing is completely described by its set of centers. Theradius of our spheres is one half of the minimum distancebetween two of these points.

Let CN denote an n-dimensional cube of radius N.

Density = limN→∞

Vol(Spheres ∩ CN)

Nn.

For a nonperiodic packing this limit is not guaranteed to exist,but we can get an upper limit.



Kaplan

Background


Lattices




Sphere Packings





Vol(Spheres ∩ CN)

Nn.




Kaplan

Background


Lattices




Sphere Packings





Vol(Spheres ∩ CN)

Nn.




Kaplan

Background


Lattices




Examples

For n = 1, packing spheres in R, we can pack all of space withspheres of radius 1/2 by taking our set of centers to be theintegers.

For n = 2, a first guess is to put a circle of radius 1/2 on everylattice point of a grid.

We can see that we are tiling space with 1× 1 squares, each ofwhich contains exactly one circle of radius 1/2. This givesdensity π

4 .



Kaplan

Background


Lattices




Examples

For n = 1, packing spheres in R, we can pack all of space withspheres of radius 1/2 by taking our set of centers to be theintegers.For n = 2, a first guess is to put a circle of radius 1/2 on everylattice point of a grid.

We can see that we are tiling space with 1× 1 squares, each ofwhich contains exactly one circle of radius 1/2. This givesdensity π

4 .



Kaplan

Background


Lattices




Hexagonal Lattice

We do better by tiling space with regular hexagons of sidelength 1 and putting a circle of radius 1/2 on each vertex, andin the center.

We are tiling space with rhombuses, of side length 1 and angles60◦ and 120◦, each of which contains exactly one circle. Thisgives density π/4√

32= π√

12.



Kaplan

Background


Lattices




Hexagonal Lattice

We do better by tiling space with regular hexagons of sidelength 1 and putting a circle of radius 1/2 on each vertex, andin the center.

We are tiling space with rhombuses, of side length 1 and angles60◦ and 120◦, each of which contains exactly one circle. Thisgives density π/4√

32= π√

12.



Kaplan

Background


Lattices




Thue’s Theorem

Theorem (Thue, 1890/1910, Fejes-Toth, 1940)

The hexagonal lattice gives the densest sphere packing in twodimensions.

Casselman’s Interactive proof (source for next two pictures):http://math.sunysb.edu/∼tony/whatsnew/column/pennies-1200/cass1.html

Definition

The Voronoi cell of a point x0 in our collection of points P is

Vx0(P) = {x ∈ Rn | ∀y ∈ P, ||x − x0|| ≤ ||x − y ||}.

Voronoi cells partition space, and the density of a packing is atmost the maximum density inside of a Voronoi cell of thatpacking.



Kaplan

Background


Lattices




Thue’s Theorem




Definition


Vx0(P) = {x ∈ Rn | ∀y ∈ P, ||x − x0|| ≤ ||x − y ||}.




Kaplan

Background


Lattices




Thue’s Theorem




Definition


Vx0(P) = {x ∈ Rn | ∀y ∈ P, ||x − x0|| ≤ ||x − y ||}.




Kaplan

Background


Lattices




Sketch of Thue’s Theorem

The Voronoi cell of each disc in the hexagonal packing is aregular hexagon around the circle.

We prove a stronger statement, that this regular hexagon givesthe densest possible Voronoi cell in two dimensions. We canclearly assume that there is no gap between circles largeenough to add another circle (our packing is saturated).

We consider when three circles come together. There is aunique triple point equidistant from all of them. We draw thepair of tangent lines to each circle, giving a kind of dunce cap.The entire cap will be contained in each Voronoi cell.



Kaplan

Background


Lattices





The Voronoi cell of each disc in the hexagonal packing is aregular hexagon around the circle.

We prove a stronger statement, that this regular hexagon givesthe densest possible Voronoi cell in two dimensions. We canclearly assume that there is no gap between circles largeenough to add another circle (our packing is saturated).

We consider when three circles come together. There is aunique triple point equidistant from all of them. We draw thepair of tangent lines to each circle, giving a kind of dunce cap.The entire cap will be contained in each Voronoi cell.



Kaplan

Background


Lattices





We consider a circle circumscribed each of ours with radius 2√3

times as large. We consider when two discs do not intersect,but their circumscribed discs do. These discs intersect at twopoints. We draw the radii of each disc to these points, forminga rhombus. Half of each rhombus will be contained in eachVoronoi cell.

We determine the maximum density of a Voronoi cell bypartitioning it into three regions and giving the maximumdensity of each:The points outside the circumscribed disc, the points in therhombus associated to a neighboring cell, and the points in thedisc but not in any rhombus.



Kaplan

Background


Lattices





We consider a circle circumscribed each of ours with radius 2√3

times as large. We consider when two discs do not intersect,but their circumscribed discs do. These discs intersect at twopoints. We draw the radii of each disc to these points, forminga rhombus. Half of each rhombus will be contained in eachVoronoi cell.We determine the maximum density of a Voronoi cell bypartitioning it into three regions and giving the maximumdensity of each:The points outside the circumscribed disc, the points in therhombus associated to a neighboring cell, and the points in thedisc but not in any rhombus.



Kaplan

Background


Lattices




The Kepler Conjecture, n = 3

For n = 3, the face centered cubic packing gives density π√18

.

This packing is built up by stacking layers of the hexagonalpacking where centers of spheres lie above the deepest holes.

The Voronoi cells are all rhombic dodecahedra, but these areno longer the densest possible cells (for example regulardodecahedra are denser). However, one cannot tile R3 withthese denser cells.



Kaplan

Background


Lattices





For n = 3, the face centered cubic packing gives density π√18

.

This packing is built up by stacking layers of the hexagonalpacking where centers of spheres lie above the deepest holes.

The Voronoi cells are all rhombic dodecahedra, but these areno longer the densest possible cells (for example regulardodecahedra are denser). However, one cannot tile R3 withthese denser cells.



Kaplan

Background


Lattices





Theorem (Hales, 1998)

The face centered cubic lattice gives the densest spherepacking in R3.

Rogers: ‘Many mathematicians believe and all physicists knowthat the density cannot exceed π√

18’.

This extremely difficult theorem requires a more difficultpartition of space than just into Voronoi cells, and also theconsideration of many locally dense configurations.



Kaplan

Background


Lattices








18’.




Kaplan

Background


Lattices








18’.




Kaplan

Background


Lattices




Definitions

Definition

A lattice Λ is a free abelian group of rank n with a positivedefinite symmetric pairing.

A lattice is a discrete subgroup Λ ⊂ Rn such that Rn/Λ iscompact. Rn/Λ is a fundamental domain of the lattice, and wecan tile space with its translates.It is a free abelian group of rank n, and is given by

n∑j=1

cjvj , such that cj ∈ Z, {v1, . . . , vn} generate Rn

.



Kaplan

Background


Lattices




Definitions

Definition

A lattice Λ is a free abelian group of rank n with a positivedefinite symmetric pairing.A lattice is a discrete subgroup Λ ⊂ Rn such that Rn/Λ iscompact. Rn/Λ is a fundamental domain of the lattice, and wecan tile space with its translates.

It is a free abelian group of rank n, and is given byn∑

j=1


.



Kaplan

Background


Lattices




Definitions

Definition

A lattice Λ is a free abelian group of rank n with a positivedefinite symmetric pairing.A lattice is a discrete subgroup Λ ⊂ Rn such that Rn/Λ iscompact. Rn/Λ is a fundamental domain of the lattice, and wecan tile space with its translates.It is a free abelian group of rank n, and is given by

n∑j=1


.



Kaplan

Background


Lattices




Definitions

Definition

The generator matrix M has columns v1, . . . , vn.The Gram matrix A = MMT . Its (i , j) entry is given by 〈vi , vj〉.

If we choose another set of generators given by BM, whereB ∈ GLn(Z), our new Gram matrix is BTAB and is positivedefinite if and only if A is.

Suppose v =∑n

i=1 mivi , where m = (m1, . . . ,mn) ∈ Zn. Then

〈vi , vj〉 =n∑

i=1

n∑j=1

Aijmimj ,

which is the value of a quadratic form given by the Grammatrix A at m.



Kaplan

Background


Lattices




Definitions

Definition

The generator matrix M has columns v1, . . . , vn.The Gram matrix A = MMT . Its (i , j) entry is given by 〈vi , vj〉.

If we choose another set of generators given by BM, whereB ∈ GLn(Z), our new Gram matrix is BTAB and is positivedefinite if and only if A is.

Suppose v =∑n

i=1 mivi , where m = (m1, . . . ,mn) ∈ Zn. Then

〈vi , vj〉 =n∑

i=1

n∑j=1

Aijmimj ,

which is the value of a quadratic form given by the Grammatrix A at m.



Kaplan

Background


Lattices




Density of a Lattice

We define the determinant of a lattice,

det(Λ) = det(A) = det(M)2 = Vol(Rn/Λ)2.

Let N(Λ) denote the minimal value of 〈v , v〉 among all v ∈ Λ.The density of a point set is the number of points per unitvolume. For a lattice, it is 1

Vol(Rn/Λ) .The density of a lattice is

∆(Λ) =Vol(

Sphere of Radius N(Λ)1/2

2

)det(Λ)1/2

.

If we want to find dense lattice packings, we want to maximize

N(Λ)n/2

det(Λ)1/2.



Kaplan

Background


Lattices








Vol(Rn/Λ) .

The density of a lattice is

∆(Λ) =Vol(


2

)det(Λ)1/2

.


N(Λ)n/2

det(Λ)1/2.



Kaplan

Background


Lattices









∆(Λ) =Vol(


2

)det(Λ)1/2

.


N(Λ)n/2

det(Λ)1/2.



Kaplan

Background


Lattices









∆(Λ) =Vol(


2

)det(Λ)1/2

.


N(Λ)n/2

det(Λ)1/2.



Kaplan

Background


Lattices




Topology of Lattices

We say that two lattices are isometric if we get one fromanother from an orthogonal linear transformation of Rn.Two lattices are similar if there is some α > 0 such thatΛ′ = αΛ.

Lattices are isometric if and only if

M ′ = UMB, B ∈ GLn(Z), U ∈ On(R),

and homothetic if and only if

M ′ = αUMB, α > 0, B ∈ GLn(Z), U ∈ On(R).

We can choose α such that | det(M)| = 1, and we can chooseU,M with determinant 1.



Kaplan

Background


Lattices





We say that two lattices are isometric if we get one fromanother from an orthogonal linear transformation of Rn.Two lattices are similar if there is some α > 0 such thatΛ′ = αΛ.Lattices are isometric if and only if

M ′ = UMB, B ∈ GLn(Z), U ∈ On(R),

and homothetic if and only if

M ′ = αUMB, α > 0, B ∈ GLn(Z), U ∈ On(R).

We can choose α such that | det(M)| = 1, and we can chooseU,M with determinant 1.



Kaplan

Background


Lattices





Let Λn denote homothecy classes of lattices in Rn. We haveidentified it with

SOn(R)\SLn(R)/GLn(Z).

We see that this space has dimension(n2 − 1)−

(n2

)= (n − 1)(n + 2)/2.

It also has a natural metric, so we can ask if the homothecyclass of Λ has some property locally.



Kaplan

Background


Lattices





Let Λn denote homothecy classes of lattices in Rn. We haveidentified it with

SOn(R)\SLn(R)/GLn(Z).

We see that this space has dimension(n2 − 1)−

(n2

)= (n − 1)(n + 2)/2.

It also has a natural metric, so we can ask if the homothecyclass of Λ has some property locally.



Kaplan

Background


Lattices




Examples

The Hexagonal lattice has generator matrix

(1 −1

2

0√

32

), and

Gram matrix

(1 −1

2−12 1

), which after scaling vector lengths

by√

2 gives

(2 −1−1 2

).

In this case, N(Λ) = 2, and det(Λ) = 3, giving ∆(Λ) = π4·√

3.

The face centered cubic lattice has generator matrix 1 1 01 0 10 1 1

, and Gram matrix

2 1 11 2 11 1 2

.



Kaplan

Background


Lattices




Examples


(1 −1

2

0√

32

), and

Gram matrix

(1 −1

2−12 1


by√

2 gives

(2 −1−1 2

).


3.


, and Gram matrix

2 1 11 2 11 1 2

.



Kaplan

Background


Lattices




Examples


(1 −1

2

0√

32

), and

Gram matrix

(1 −1

2−12 1


by√

2 gives

(2 −1−1 2

).


3.


, and Gram matrix

2 1 11 2 11 1 2

.



Kaplan

Background


Lattices




Families of Lattices

These two examples are special cases of a more generalconstruction. We identify n-dimensional Euclidean space with ahyperplane in Rn+1,

{(x0, . . . , xn) ∈ Rn+1 such that∑

xi = 0}.

Now let

An = {(x0, . . . , xn) ∈ Zn+1 such that∑

xi = 0}.

The fcc lattice gives A3, which in this special case is isometricwith D3.

Dn = {(x1, . . . , xn) ∈ Zn such that∑

xi ≡ 0 (mod 2)}.



Kaplan

Background


Lattices




Families of Lattices

These two examples are special cases of a more generalconstruction. We identify n-dimensional Euclidean space with ahyperplane in Rn+1,

{(x0, . . . , xn) ∈ Rn+1 such that∑

xi = 0}.

Now let

An = {(x0, . . . , xn) ∈ Zn+1 such that∑

xi = 0}.

The fcc lattice gives A3, which in this special case is isometricwith D3.

Dn = {(x1, . . . , xn) ∈ Zn such that∑

xi ≡ 0 (mod 2)}.



Kaplan

Background


Lattices




Minkowski Reduced Forms

We say that two quadratic forms are equivalent if we get onefrom the other by multiplication by a matrix in SLn(Z).

Definition

Let f be a positive definite n dimensional quadratic form. Wesay that it is Minkowski reduced if it is expressed in terms of anintegral basis e1, . . . , en such that f (ei ) ≤ f (v) for all v suchthat e1, . . . , ei−1, v can be extended to an integer basis of Λ.

Proposition

Every quadratic form is equivalent to a Minkowski reducedform.

Choose vi inductively to minimize f (vi ). Then v1, . . . , vn forma basis. Let f1(x) = f (V · x) and claim that f1 is Minkowskireduced.



Kaplan

Background


Lattices






Definition


Proposition





Kaplan

Background


Lattices






Definition


Proposition





Kaplan

Background


Lattices






Definition


Proposition





Kaplan

Background


Lattices




Lattice Packings n = 2

We want to maximize N(Λ)n/2

det(Λ)1/2 . We work with quadratic forms

instead of lattices themselves.

Letf (x1, x2) = f11x2

1 + 2f12x1x2 + f22x22 ,

and suppose that it is Minkowski reduced.We have f11 ≤ f22 and can suppose f12 ≥ 0. Nowf (−1, 1) ≥ f (0, 1) implies 2f12 ≤ f11.

Let D = f11f22 − f 212 and consider

4D − 3f11f22 = f11f22 − 4f 212 ≥ f 2

11 − 4f 212 ≥ 0.

Equality holds if and only if f (x1, x,2) = f11(x21 + x1x2 + x2

2 ).This is the Gram matrix of the Hexagonal lattice.



Kaplan

Background


Lattices







instead of lattices themselves.Let

f (x1, x2) = f11x21 + 2f12x1x2 + f22x2

2 ,



4D − 3f11f22 = f11f22 − 4f 212 ≥ f 2

11 − 4f 212 ≥ 0.





Kaplan

Background


Lattices








f (x1, x2) = f11x21 + 2f12x1x2 + f22x2

2 ,



4D − 3f11f22 = f11f22 − 4f 212 ≥ f 2

11 − 4f 212 ≥ 0.





Kaplan

Background


Lattices








f (x1, x2) = f11x21 + 2f12x1x2 + f22x2

2 ,



4D − 3f11f22 = f11f22 − 4f 212 ≥ f 2

11 − 4f 212 ≥ 0.





Kaplan

Background


Lattices





Gauss proved that the face centered cubic lattice gives thedense lattice packing in R3.

Proposition

Let f (x) =∑

1≤i ,j≤3 fijxixj .

1 ∃ a nonzero u ∈ Z2 such that f (u) ≤ (2D)1/3.

2 f (x) is Minkowski reduced if and only if f11f22f33 ≤ 2D,

3 Equality holds if and only if the form is equivalent to theone given by fcc.

We note that the first statement follows from the others.Consider two cases based on the sign of f12f23f31. If theproduct is positive, we can suppose each is positive. LetΘij = fii − 2fij . If f is reduced, this is nonnegative. Next,consider 2D − f11f22f33. Using the Θij , this is nonnegative.



Kaplan

Background


Lattices





Gauss proved that the face centered cubic lattice gives thedense lattice packing in R3.

Proposition

Let f (x) =∑

1≤i ,j≤3 fijxixj .

1 ∃ a nonzero u ∈ Z2 such that f (u) ≤ (2D)1/3.

2 f (x) is Minkowski reduced if and only if f11f22f33 ≤ 2D,

3 Equality holds if and only if the form is equivalent to theone given by fcc.

We note that the first statement follows from the others.Consider two cases based on the sign of f12f23f31. If theproduct is positive, we can suppose each is positive. LetΘij = fii − 2fij . If f is reduced, this is nonnegative. Next,consider 2D − f11f22f33. Using the Θij , this is nonnegative.



Kaplan

Background


Lattices




Hermite’s Theorem

Theorem (Hermite)

Every lattice Λ has a basis e1, . . . , en such that

N(e1) · · ·N(en) ≤(

4

3

)n(n−1)/2

det(Λ).

Corollary

N(Λ)n/2

det(Λ)1/2≤(

4

3

)n(n−1)/4

.

Pick e1 minimal and project onto the intersection of Λ with theorthogonal complement of 〈e1〉. Induct on n.Hadamard’s theorem: det(Λ) ≤ N(e1) · · ·N(en).



Kaplan

Background


Lattices




Hermite’s Theorem

Theorem (Hermite)


N(e1) · · ·N(en) ≤(

4

3

)n(n−1)/2

det(Λ).

Corollary

N(Λ)n/2

det(Λ)1/2≤(

4

3

)n(n−1)/4

.

Pick e1 minimal and project onto the intersection of Λ with theorthogonal complement of 〈e1〉. Induct on n.

Hadamard’s theorem: det(Λ) ≤ N(e1) · · ·N(en).



Kaplan

Background


Lattices




Hermite’s Theorem

Theorem (Hermite)


N(e1) · · ·N(en) ≤(

4

3

)n(n−1)/2

det(Λ).

Corollary

N(Λ)n/2

det(Λ)1/2≤(

4

3

)n(n−1)/4

.

Pick e1 minimal and project onto the intersection of Λ with theorthogonal complement of 〈e1〉. Induct on n.Hadamard’s theorem: det(Λ) ≤ N(e1) · · ·N(en).



Kaplan

Background


Lattices




The Hermite Invariant

Definition

The Hermite invariant of Λ is

γ(Λ) =N(Λ)

det(Λ)1/n.

The Hermite constant for dimension n is γn = supΛ γ(Λ).

A lattice for which γ(Λ) = γn is called critical.

We have already seen that γ2 = 2√3

, and γ3 = (2D)1/3

D1/3 = 21/3.



Kaplan

Background


Lattices





Definition


γ(Λ) =N(Λ)

det(Λ)1/n.




, and γ3 = (2D)1/3

D1/3 = 21/3.



Kaplan

Background


Lattices





Definition


γ(Λ) =N(Λ)

det(Λ)1/n.




, and γ3 = (2D)1/3

D1/3 = 21/3.



Kaplan

Background


Lattices





Definition


γ(Λ) =N(Λ)

det(Λ)1/n.




, and γ3 = (2D)1/3

D1/3 = 21/3.



Kaplan

Background


Lattices




Mordell’s Theorem

In some cases, we can determine γn from γn−1.

Theorem (Mordell)

For any 2 ≤ m < n,

γn ≤ γ(n−1)/(m−1)m .

Definition

The dual lattice of Λ is

Λ∗ = {y ∈ Rn such that 〈x , y〉 ∈ Z, ∀ x ∈ Λ}.

Pick x minimal in Λ∗ and let L = Λ ∩ 〈x〉⊥, and then expressγ(M) in terms of γ(Λ), γ(Λ∗),N(M),N(Λ). We then do thesame with the roles of Λ and Λ∗ switched.



Kaplan

Background


Lattices




Mordell’s Theorem


Theorem (Mordell)


γn ≤ γ(n−1)/(m−1)m .

Definition






Kaplan

Background


Lattices




Mordell’s Theorem


Theorem (Mordell)


γn ≤ γ(n−1)/(m−1)m .

Definition






Kaplan

Background


Lattices




γ4 and γ8

We recall that γ3 = 21/3, so Mordell’s Theorem gives

γ4 ≤ γ3/23 =

√2.

Equality holds for the lattice D4, so γ4 =√

2.

We note that the distance from (1/2, . . . , 1/2) to a latticepoint of D8 is

√2, exactly the radius of one sphere. We can

add in a translated copy of our lattice, giving E8. Luckily, thisis a lattice as well!Let v be a minimal vector of E8. Let

E7 = {x ∈ E8 such that 〈x , v〉 = 0}.

We get γ(E7) = 26/7.If we show that in fact γ7 = 26/7 then Mordell implies γ8 ≤ 2,but equality holds for E8!



Kaplan

Background


Lattices




γ4 and γ8


γ4 ≤ γ3/23 =

√2.


2.




E7 = {x ∈ E8 such that 〈x , v〉 = 0}.




Kaplan

Background


Lattices




γ4 and γ8


γ4 ≤ γ3/23 =

√2.


2.



add in a translated copy of our lattice, giving E8. Luckily, thisis a lattice as well!

Let v be a minimal vector of E8. Let

E7 = {x ∈ E8 such that 〈x , v〉 = 0}.




Kaplan

Background


Lattices




γ4 and γ8


γ4 ≤ γ3/23 =

√2.


2.




E7 = {x ∈ E8 such that 〈x , v〉 = 0}.

We get γ(E7) = 26/7.

If we show that in fact γ7 = 26/7 then Mordell implies γ8 ≤ 2,but equality holds for E8!



Kaplan

Background


Lattices




γ4 and γ8


γ4 ≤ γ3/23 =

√2.


2.




E7 = {x ∈ E8 such that 〈x , v〉 = 0}.




Kaplan

Background


Lattices




Minkowski’s Convex Body Theorem

Theorem (Minkowski)

A bounded convex region in Rn symmetric about a lattice pointwith volume greater than 2n contains at least 3 lattice points.

Corollary

We have γn ≤ 4ω−2/nn , where ωn is the volume of the unit ball

in Rn.A careful consideration of ωn gives γn ≤ 1 + n/4 for all n.



Kaplan

Background


Lattices





Theorem (Minkowski)


Corollary


in Rn.

A careful consideration of ωn gives γn ≤ 1 + n/4 for all n.



Kaplan

Background


Lattices





Theorem (Minkowski)


Corollary


in Rn.A careful consideration of ωn gives γn ≤ 1 + n/4 for all n.



Kaplan

Background


Lattices




Local Density of Lattices

Korkine and Zolotareff used a different reduction of quadraticforms to compute γn for n ≤ 5. Blichfeldt extended theirmethods to determine γn for n = 6, 7, 8.

Since we have a metric on unimodular lattices, we can ask whenγ(Λ) has a local maximum. Such a lattice is called extreme.

Definition

Let min(Λ) = {x ∈ Λ such that 〈x , x〉 = N(Λ)}.Λ is perfect if {xxT =: πx , such that x ∈ min(Λ)} spans Rn×n

sym .Λ is eutactic if ∃{λx}, such that λx ≥ 0 for all x ∈ min(Λ) andIn =

∑x∈min(Λ) λx · πx .



Kaplan

Background


Lattices







Definition






Kaplan

Background


Lattices







Definition






Kaplan

Background


Lattices




Voronoi’s Theorem

Theorem (Voronoi)

Λ is extreme if and only if it is perfect and eutactic.

Proposition (K-Z)

A perfect lattice is proportional to an integral lattice.

Corollary

γnn is rational

Theorem (Voronoi)

There are only finitely many non-isometric unimodular latticesin a given dimension and there is an algorithm to computethem.

For n = 7 there are 33, but for n = 8 there are over 10, 000.



Kaplan

Background


Lattices




Voronoi’s Theorem

Theorem (Voronoi)


Proposition (K-Z)


Corollary

γnn is rational

Theorem (Voronoi)


For n = 7 there are 33, but for n = 8 there are over 10, 000.



Kaplan

Background


Lattices




Voronoi’s Theorem

Theorem (Voronoi)


Proposition (K-Z)


Corollary

γnn is rational

Theorem (Voronoi)


For n = 7 there are 33, but for n = 8 there are over 10, 000.Kaplan Lattice Packings of Spheres


Kaplan

Background


Lattices




Example of Voronoi’s Theorem

There are 6 minimal vectors in the hexagonal lattice. Let

x1 =

(10

), x1 =

(12√3

2

), and x3 =

(−1

2√3

2

).

We see that

πx1 =

(1 00 0

),

2

3(πx2 +πx3 ) =

(0 00 1

),

2√3

(πx2−πx3 ) =

(0 11 0

)so the hexagonal lattice is perfect and eutactic, hence extreme.



Kaplan

Background


Lattices




Example of Voronoi’s Theorem

There are 6 minimal vectors in the hexagonal lattice. Let

x1 =

(10

), x1 =

(12√3

2

), and x3 =

(−1

2√3

2

).

We see that

πx1 =

(1 00 0

),

2

3(πx2 +πx3 ) =

(0 00 1

),

2√3

(πx2−πx3 ) =

(0 11 0

)so the hexagonal lattice is perfect and eutactic, hence extreme.



Kaplan

Background


Lattices




Poisson Summation for Lattices

We know γn only for n ≤ 8 and n = 24. The last result isrecent work of Cohn and Kumar, building on techniques ofCohn and Elkies.

For f ∈ L1, f : Rn → R, f (t) =∫Rn f (x)e2πi〈t,x〉dx .

Proposition (Poisson Summation for Lattices)

Let f be a Schwartz function∑x∈Λ

f (x) =1

Vol(Rn/Λ)

∑t∈Λ∗

f (t).



Kaplan

Background


Lattices




Poisson Summation for Lattices

We know γn only for n ≤ 8 and n = 24. The last result isrecent work of Cohn and Kumar, building on techniques ofCohn and Elkies.For f ∈ L1, f : Rn → R, f (t) =

∫Rn f (x)e2πi〈t,x〉dx .

Proposition (Poisson Summation for Lattices)

Let f be a Schwartz function∑x∈Λ

f (x) =1

Vol(Rn/Λ)

∑t∈Λ∗

f (t).



Kaplan

Background


Lattices




Upper Bound for Packing Density

Theorem (Cohn, Elkies)

Let f : Rn → R be a Schwartz function, f (0) 6= 0. If f (x) ≤ 0for |x | ≥ 1 and f (t) ≥ 0 for all t, then the packing density is atmost

πn/2

2n · Γ(n/2 + 1)· f (0)

f (0).

It is enough to prove this for periodic packings, (unions oftranslates of lattices), since we can use these to approximategeneral packings. For lattices we sketch an argument.The density of centers per unit volume is 1

Vol(Rn/Λ) . We have

f (0) ≤∑

f (t) and also∑

f (x) ≤ f (0). Then 1Vol(Rn/Λ) ≤

f (0)

f (0).



Kaplan

Background


Lattices





Theorem (Cohn, Elkies)

Let f : Rn → R be a Schwartz function, f (0) 6= 0. If f (x) ≤ 0for |x | ≥ 1 and f (t) ≥ 0 for all t, then the packing density is atmost

πn/2

2n · Γ(n/2 + 1)· f (0)

f (0).

It is enough to prove this for periodic packings, (unions oftranslates of lattices), since we can use these to approximategeneral packings. For lattices we sketch an argument.The density of centers per unit volume is 1

Vol(Rn/Λ) . We have

f (0) ≤∑

f (t) and also∑

f (x) ≤ f (0). Then 1Vol(Rn/Λ) ≤

f (0)

f (0).



Kaplan

Background


Lattices





It is conjectured that for n = 2, 8, 24 there exists an f giving asharp bound.

Such a sharp f must vanish at every point of Λ, and f mustvanish at every point of Λ∗. If f were sharp, rotations of fwould be as well, so these functions vanish identically onspheres of radius equal to the radius of any lattice point. Thelengths of lattice points imply that neither f nor f havecompact support. It seems likely that f would have to be aradial function.

In practice, linear combinations of orthogonal polynomials areused to find good f , and double zeros are imposed in certainplaces to minimize its last sign change. Extensive linearprogramming led to an upper bound powerful enough to showthat no lattice in R24 is denser than the Leech lattice.



Kaplan

Background


Lattices





It is conjectured that for n = 2, 8, 24 there exists an f giving asharp bound.

Such a sharp f must vanish at every point of Λ, and f mustvanish at every point of Λ∗. If f were sharp, rotations of fwould be as well, so these functions vanish identically onspheres of radius equal to the radius of any lattice point. Thelengths of lattice points imply that neither f nor f havecompact support. It seems likely that f would have to be aradial function.

In practice, linear combinations of orthogonal polynomials areused to find good f , and double zeros are imposed in certainplaces to minimize its last sign change. Extensive linearprogramming led to an upper bound powerful enough to showthat no lattice in R24 is denser than the Leech lattice.


Lattice Packings of Spheres - Yale Universitynk354/papers/latticepackings.pdf · Lattices Lattice Packings in Dimensions Two and Three Hermite, Mordell, and Voronoi Poisson Summation

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