Lattice packings: an upper bound on the number of perfect lattices - PowerPoint PPT Presentation

Lattice packings: an upper bound on the number of perfect lattices Wessel van Woerden, CWI, Amsterdam.

1 | 23 Sphere Packing Problem

1 | 23 Sphere Packing Problem

1 | 23 Sphere Packing Problem • Only solved in dimensions 2 , 3 , 8 and 24 ...

2 | 23 Lattice Packing Problem b 1 + b 2 b 2 0 b 1

2 | 23 Lattice Packing Problem b 1 + b 2 b 2 0 b 1 • Solved in dimensions ≤ 8 and 24 .

3 | 23 Solution space • Cone of positive definite quadratic forms:

4 | 23 Ryshkov Polyhedron • Spheres of radius at least 1 :

4 | 23 Ryshkov Polyhedron • Spheres of radius at least 1 : • Concave minimization problem = ⇒ optima at vertices.

4 | 23 Ryshkov Polyhedron • Spheres of radius at least 1 : • Concave minimization problem = ⇒ optima at vertices. • Finite number of non-similar vertices.

4 | 23 Ryshkov Polyhedron • Spheres of radius at least 1 : • Concave minimization problem = ⇒ optima at vertices. • Finite number of non-similar vertices. ← how many?

5 | 23 Voronoi’s Algorithm • How to solve the lattice packing problem in a fixed dimension: • Enumerate all non-similar vertices.

5 | 23 Voronoi’s Algorithm • How to solve the lattice packing problem in a fixed dimension: • Enumerate all non-similar vertices. • Pick the best one.

6 | 23 Space of quadratic forms • Let S d ⊂ R d × d be the space of real symmetric matrices.

6 | 23 Space of quadratic forms • Let S d ⊂ R d × d be the space of real symmetric matrices. • Dimension n := 1 2 d ( d + 1 ) .

6 | 23 Space of quadratic forms • Let S d ⊂ R d × d be the space of real symmetric matrices. • Dimension n := 1 2 d ( d + 1 ) . • We define the following inner product on S d : � A , B � := Tr ( A t B ) = � A ij B ij i , j

6 | 23 Space of quadratic forms • Let S d ⊂ R d × d be the space of real symmetric matrices. • Dimension n := 1 2 d ( d + 1 ) . • We define the following inner product on S d : � A , B � := Tr ( A t B ) = � A ij B ij i , j • Q ∈ S d defines a quadratic form by Q [ x ] := x t Qx = � Q , xx t � ∀ x ∈ R d

6 | 23 Space of quadratic forms • Let S d ⊂ R d × d be the space of real symmetric matrices. • Dimension n := 1 2 d ( d + 1 ) . • We define the following inner product on S d : � A , B � := Tr ( A t B ) = � A ij B ij i , j • Q ∈ S d defines a quadratic form by Q [ x ] := x t Qx = � Q , xx t � ∀ x ∈ R d • For a positive definite quadratic form (PQF) Q ∈ S d > 0 : λ ( Q ) := x ∈ Z d −{ 0 } Q [ x ] min Min ( Q ) := { x ∈ Z d : Q [ x ] = λ ( Q ) }

7 | 23 Hermite Constant • Lattice L = B Z d = ⇒ PQF Q = B t B ∈ S d > 0 .

7 | 23 Hermite Constant • Lattice L = B Z d = ⇒ PQF Q = B t B ∈ S d > 0 .

7 | 23 Hermite Constant • Lattice L = B Z d = ⇒ PQF Q = B t B ∈ S d > 0 . ∼ λ ( Q ) d / 2 ∼ det( Q ) 1 / 2

7 | 23 Hermite Constant • Lattice L = B Z d = ⇒ PQF Q = B t B ∈ S d > 0 . ∼ λ ( Q ) d / 2 ∼ det( Q ) 1 / 2 • Hermite invariant: λ ( Q ) (det Q ) 1 / d ∼ density ( L ) 2 / d γ ( Q ) =

7 | 23 Hermite Constant • Lattice L = B Z d = ⇒ PQF Q = B t B ∈ S d > 0 . ∼ λ ( Q ) d / 2 ∼ det( Q ) 1 / 2 • Hermite invariant: λ ( Q ) (det Q ) 1 / d ∼ density ( L ) 2 / d γ ( Q ) = • Lattice packing problem ⇔ determine Hermite’s constant: H d := sup γ ( Q ) Q ∈S d > 0

8 | 23 Ryshkov Polyhedra • For λ > 0 we define the Ryshkov Polyhedron P λ = { Q ∈ S d > 0 : λ ( Q ) ≥ λ }

8 | 23 Ryshkov Polyhedra • For λ > 0 we define the Ryshkov Polyhedron { Q ∈ S d : � Q , xx t � ≥ λ } ⊂ S d � P λ = > 0 x ∈ Z d \{ 0 }

8 | 23 Ryshkov Polyhedra • For λ > 0 we define the Ryshkov Polyhedron { Q ∈ S d : � Q , xx t � ≥ λ } ⊂ S d � P λ = > 0 x ∈ Z d \{ 0 } • Facets correspond to x ∈ Z d \ { 0 } .

8 | 23 Ryshkov Polyhedra • For λ > 0 we define the Ryshkov Polyhedron P λ = { Q ∈ S d > 0 : λ ( Q ) ≥ λ } • We have λ H d = Q ∈P λ det( Q ) 1 / d inf

8 | 23 Ryshkov Polyhedra • For λ > 0 we define the Ryshkov Polyhedron P λ = { Q ∈ S d > 0 : λ ( Q ) ≥ λ } • We have λ H d = Q ∈P λ det( Q ) 1 / d inf • Minkowski: det( Q ) 1 / d is (strictly) concave on S d > 0 = ⇒ Local optima at vertices of P λ .

9 | 23 Perfect forms • Q is perfect ⇔ Q is a vertex of P λ ( Q ) .

9 | 23 Perfect forms • Q is perfect ⇔ Q is a vertex of P λ ( Q ) . • Note that | Min Q | ≥ 2 n = d ( d + 1 )

10 | 23 Similarity • B and BU generate the same lattice for U ∈ GL d ( Z ) .

10 | 23 Similarity • B and BU generate the same lattice for U ∈ GL d ( Z ) . • Arithmetically equivalence: ∃ U s.t. Q ′ = U t QU .

10 | 23 Similarity • B and BU generate the same lattice for U ∈ GL d ( Z ) . • Arithmetically equivalence: ∃ U s.t. Q ′ = U t QU . • Note that λ 1 ( U t QU ) = λ 1 ( Q ) and det( U t QU ) = det( Q ) .

10 | 23 Similarity • B and BU generate the same lattice for U ∈ GL d ( Z ) . • Arithmetically equivalence: ∃ U s.t. Q ′ = U t QU . • Note that λ 1 ( U t QU ) = λ 1 ( Q ) and det( U t QU ) = det( Q ) . • Similarity: Arithmetical equivalence up to scaling.

Perfect Forms: how many?

11 | 23 Number of perfect forms • The exact set of perfect forms is known up to dimension 8 . • For d ≥ 6 Voronoi’s Algorithm was used. d # non-similar Perfect forms 2 1 (Lagrange, 1773) 3 1 (Gauss, 1840) 4 2 (Korkine & Zolotarev, 1877) 5 3 (Korkine & Zolotarev, 1877) 6 7 (Barnes, 1957) 7 33 (Jaquet, 1993) 8 10916 (DSV, 2005) 9 ≥ 500.000 (DSV, 2005) ≥ 23.000.000 (vW, 2018)

12 | 23 An improved upper bound • p d := number of non-similar d -dimensional perfect forms.

12 | 23 An improved upper bound • p d := number of non-similar d -dimensional perfect forms. • Known bounds for p d . p d < e O ( d 4 log( d )) ( C. Soul´ e , 1998 ) e Ω( d ) < p d < e O ( d 3 log( d )) ( R. Bacher , 2017 )

12 | 23 An improved upper bound • p d := number of non-similar d -dimensional perfect forms. • Known bounds for p d . p d < e O ( d 4 log( d )) ( C. Soul´ e , 1998 ) e Ω( d ) < p d < e O ( d 3 log( d )) ( R. Bacher , 2017 ) Theorem (This talk) p d < e O ( d 2 log( d ))

13 | 23 Outer Normal Cones polyhedron inside cone = ⇒ subdivision of cone 1 4 2 3 1 2 3 4

13 | 23 Inner Normal Cones polyhedron inside cone = ⇒ subdivision of cone 4 3 2 1 1 4 2 3

14 | 23 Subdivision for d = 2 Figure: Subdivision by normal cones of Ryshkov Polyhedron.

15 | 23 Voronoi Domain Definition For a PQF Q ∈ S d > 0 its Voronoi Domain V ( Q ) is V ( Q ) := cone ( { xx t : x ∈ Min Q } ) ⊂ S d ≥ 0 .

15 | 23 Voronoi Domain Definition For a PQF Q ∈ S d > 0 its Voronoi Domain V ( Q ) is V ( Q ) := cone ( { xx t : x ∈ Min Q } ) ⊂ S d ≥ 0 . • Q is perfect ⇔ V ( Q ) is full dimensional.

16 | 23 Proof strategy ( 1 , 1 ) ( 3 , 2 ) ( 2 , 3 ) ( 2 , 1 ) ( 1 , 2 ) ( 3 , 1 ) ( 1 , 3 ) ( 1 , 0 ) ( 0 , 1 ) ( − 3 , 1 ) ( − 1 , 3 ) ( − 2 , 1 ) ( − 1 , 2 ) ( − 3 , 1 ) ( − 2 , 3 ) ( − 1 , 1 )

16 | 23 Proof strategy ( 1 , 1 ) ( 3 , 2 ) ( 2 , 3 ) ( 2 , 1 ) ( 1 , 2 ) Can’t fit many large ( 3 , 1 ) ( 1 , 3 ) Voronoi domains. ( 1 , 0 ) ( 0 , 1 ) ( − 3 , 1 ) ( − 1 , 3 ) ( − 2 , 1 ) ( − 1 , 2 ) ( − 3 , 1 ) ( − 2 , 3 ) ( − 1 , 1 )

16 | 23 Proof strategy ( 1 , 1 ) ( 3 , 2 ) ( 2 , 3 ) ( 2 , 1 ) ( 1 , 2 ) Can’t fit many large ( 3 , 1 ) ( 1 , 3 ) Voronoi domains. Each similariy-class has at least one ( 1 , 0 ) ( 0 , 1 ) large representative. ( − 3 , 1 ) ( − 1 , 3 ) ( − 2 , 1 ) ( − 1 , 2 ) ( − 3 , 1 ) ( − 2 , 3 ) ( − 1 , 1 )

16 | 23 Proof strategy ( 1 , 1 ) ( 3 , 2 ) ( 2 , 3 ) ( 2 , 1 ) ( 1 , 2 ) Can’t fit many large ( 3 , 1 ) ( 1 , 3 ) Voronoi domains. Each similariy-class has at least one ( 1 , 0 ) ( 0 , 1 ) large representative. = ⇒ Can’t have many similarity-classes. ( − 3 , 1 ) ( − 1 , 3 ) ( − 2 , 1 ) ( − 1 , 2 ) ( − 3 , 1 ) ( − 2 , 3 ) ( − 1 , 1 )

17 | 23 Volumetric argument • Find a complete set of representatives P d such that: � � S d Vol ( V ( Q )) ≥ ℓ d Vol ≤ u d ≥ 0 ∀ Q ∈ P d

17 | 23 Volumetric argument • Find a complete set of representatives P d such that: � � S d Vol ( V ( Q )) ≥ ℓ d Vol ≤ u d ≥ 0 ∀ Q ∈ P d • Then p d = | P d | ≤ u d ℓ d .

17 | 23 Volumetric argument • Find a complete set of representatives P d such that: � � S d Vol ( V ( Q )) ≥ ℓ d Vol ≤ u d ≥ 0 = o ( 1 ) ∀ Q ∈ P d • Then p d = | P d | ≤ u d ℓ d . • To quantify the volume we restrict to the half space T d := { Q ∈ S d : Tr ( Q ) = � Q , I d � ≤ 1 } .

18 | 23 Volume simplex a 2 a 1 a 3 n -dimensional simplex: n ! · | det( � a i , a j � ) i , j | 1 / 2 Volume = 1 0

19 | 23 Volume Voronoi domain • Tr ( xx t ) = x t x .

19 | 23 Volume Voronoi domain • Tr ( xx t ) = x t x . • Can look at subcone: w.l.o.g. Min Q = {± x 1 , . . . , ± x n } . x 2 x t 2 x t 2 x 2 x 1 x t x 3 x t 1 3 x t x t 1 x 1 3 x 3 V ( Q ) ∩ T d 0