Tiling a Flat Torus
Stephen Hetterich, Corey Manack
April 7, 2013 Important: wallpaper spans the entire plane!
What is a wallpaper?
A wallpaper, A, is a subset of R2, (C), whose translation subgroup LA is a rank 2 discrete subgroup of R2. What is a wallpaper?
A wallpaper, A, is a subset of R2, (C), whose translation subgroup LA is a rank 2 discrete subgroup of R2. Important: wallpaper spans the entire plane! Wallpaper:
Not a wallpaper:
What is a wallpaper? Examples: Not a wallpaper:
What is a wallpaper? Examples:
Wallpaper: What is a wallpaper? Examples:
Wallpaper:
Not a wallpaper: Let FA be the set of all isometries of A.
FA is called a wallpaper group.
LA ⊂ FA Two wallpapers are equivalent if their corresponding wallpaper groups are isomorphic. 17 different wallpaper groups in total.
What is a wallpaper? Examples:
Now, assume A is a wallpaper. FA is called a wallpaper group.
LA ⊂ FA Two wallpapers are equivalent if their corresponding wallpaper groups are isomorphic. 17 different wallpaper groups in total.
What is a wallpaper? Examples:
Now, assume A is a wallpaper.
Let FA be the set of all isometries of A. LA ⊂ FA Two wallpapers are equivalent if their corresponding wallpaper groups are isomorphic. 17 different wallpaper groups in total.
What is a wallpaper? Examples:
Now, assume A is a wallpaper.
Let FA be the set of all isometries of A.
FA is called a wallpaper group. 17 different wallpaper groups in total.
What is a wallpaper? Examples:
Now, assume A is a wallpaper.
Let FA be the set of all isometries of A.
FA is called a wallpaper group.
LA ⊂ FA Two wallpapers are equivalent if their corresponding wallpaper groups are isomorphic. What is a wallpaper? Examples:
Now, assume A is a wallpaper.
Let FA be the set of all isometries of A.
FA is called a wallpaper group.
LA ⊂ FA Two wallpapers are equivalent if their corresponding wallpaper groups are isomorphic. 17 different wallpaper groups in total. Wallpaper groups Wallpaper groups Wallpaper groups In other words, L is the set of all integer linear combinations of two linearly independent vectors ~ ~ l1, l2.
Given a wallpaper A, LA can be identified with some lattice L (for example, identify a translational ~ ~ symmetry in the direction of l1 with l1 itself).
We will use LA and L interchangeably.
What is a lattice?
A lattice L is a (rank 2) discrete additive subgroup of R2 (or C). Given a wallpaper A, LA can be identified with some lattice L (for example, identify a translational ~ ~ symmetry in the direction of l1 with l1 itself).
We will use LA and L interchangeably.
What is a lattice?
A lattice L is a (rank 2) discrete additive subgroup of R2 (or C). In other words, L is the set of all integer linear combinations of two linearly independent vectors ~ ~ l1, l2. We will use LA and L interchangeably.
What is a lattice?
A lattice L is a (rank 2) discrete additive subgroup of R2 (or C). In other words, L is the set of all integer linear combinations of two linearly independent vectors ~ ~ l1, l2.
Given a wallpaper A, LA can be identified with some lattice L (for example, identify a translational ~ ~ symmetry in the direction of l1 with l1 itself). What is a lattice?
A lattice L is a (rank 2) discrete additive subgroup of R2 (or C). In other words, L is the set of all integer linear combinations of two linearly independent vectors ~ ~ l1, l2.
Given a wallpaper A, LA can be identified with some lattice L (for example, identify a translational ~ ~ symmetry in the direction of l1 with l1 itself).
We will use LA and L interchangeably. Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus? Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) So we can look at pretty pictures.
What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 What is a flat torus?
Definition: T = R2/Λ, meaning T is the quotient of R2 by some lattice Λ =< v~1, v~2 >,. Several descriptions of T Description 1 - The lattice Λ itself. So we can make calculations. Description 2 - Schematically: A parallelogram with opposite edges identified (donut shape, cut and laid flat.) Description 3: The fundamental domain F: the convex hull of v~1, v~2, v~1 + v~2 So we can look at pretty pictures. A wallpaper tiles T if translations of F ∩ A by Λ recover A. (A ∩ F) + Λ = A Think of Λ as a set of translational symmetries of the plane. BIG Thm. Up to symmetry, A tiles T iff Λ is a rank 2 subgroup of LA.
Nutshell.
Given a wallpaper A, and a torus T = R2/Λ, we wish to know if the wallpaper “fits nicely” on the surface of the flat torus. (A ∩ F) + Λ = A Think of Λ as a set of translational symmetries of the plane. BIG Thm. Up to symmetry, A tiles T iff Λ is a rank 2 subgroup of LA.
Nutshell.
Given a wallpaper A, and a torus T = R2/Λ, we wish to know if the wallpaper “fits nicely” on the surface of the flat torus. A wallpaper tiles T if translations of F ∩ A by Λ recover A. BIG Thm. Up to symmetry, A tiles T iff Λ is a rank 2 subgroup of LA.
Nutshell.
Given a wallpaper A, and a torus T = R2/Λ, we wish to know if the wallpaper “fits nicely” on the surface of the flat torus. A wallpaper tiles T if translations of F ∩ A by Λ recover A. (A ∩ F) + Λ = A Think of Λ as a set of translational symmetries of the plane. Nutshell.
Given a wallpaper A, and a torus T = R2/Λ, we wish to know if the wallpaper “fits nicely” on the surface of the flat torus. A wallpaper tiles T if translations of F ∩ A by Λ recover A. (A ∩ F) + Λ = A Think of Λ as a set of translational symmetries of the plane. BIG Thm. Up to symmetry, A tiles T iff Λ is a rank 2 subgroup of LA. Look at p2 vs p6m p2 tiles square T , but p6m does not tile T .
Nutshell Look at p2 vs p6m p2 tiles square T , but p6m does not tile T .
Nutshell Nutshell
Look at p2 vs p6m p2 tiles square T , but p6m does not tile T . Answer: BIG Thm. Whenever Λ ⊂ LA. Counting: Each wallpaper A has a lattice
LA =< l1, l2 >
The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. If A tiles T , how many copies of P are contained in T ?
Answer: Area condition det(v1, v2)/ det(l1, l2).
Questions
Existence: For which pairs A, T , does A tile T ? Counting: Each wallpaper A has a lattice
LA =< l1, l2 >
The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. If A tiles T , how many copies of P are contained in T ?
Answer: Area condition det(v1, v2)/ det(l1, l2).
Questions
Existence: For which pairs A, T , does A tile T ?
Answer: BIG Thm. Whenever Λ ⊂ LA. The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. If A tiles T , how many copies of P are contained in T ?
Answer: Area condition det(v1, v2)/ det(l1, l2).
Questions
Existence: For which pairs A, T , does A tile T ?
Answer: BIG Thm. Whenever Λ ⊂ LA. Counting: Each wallpaper A has a lattice
LA =< l1, l2 > If A tiles T , how many copies of P are contained in T ?
Answer: Area condition det(v1, v2)/ det(l1, l2).
Questions
Existence: For which pairs A, T , does A tile T ?
Answer: BIG Thm. Whenever Λ ⊂ LA. Counting: Each wallpaper A has a lattice
LA =< l1, l2 >
The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. Answer: Area condition det(v1, v2)/ det(l1, l2).
Questions
Existence: For which pairs A, T , does A tile T ?
Answer: BIG Thm. Whenever Λ ⊂ LA. Counting: Each wallpaper A has a lattice
LA =< l1, l2 >
The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. If A tiles T , how many copies of P are contained in T ? Questions
Existence: For which pairs A, T , does A tile T ?
Answer: BIG Thm. Whenever Λ ⊂ LA. Counting: Each wallpaper A has a lattice
LA =< l1, l2 >
The parallelogram P spanned by l1, l2 captures the smallest piece of wallpaper that can be translated (via LA) to cover the entire plane. If A tiles T , how many copies of P are contained in T ?
Answer: Area condition det(v1, v2)/ det(l1, l2). The lattice L coming from a wallpaper group. The lattice Λ coming from a torus. We are interested in how Λ is situated inside of L 5 different types of planar lattices.
Remember...
We seek a relationship between two different lattices: The lattice Λ coming from a torus. We are interested in how Λ is situated inside of L 5 different types of planar lattices.
Remember...
We seek a relationship between two different lattices: The lattice L coming from a wallpaper group. We are interested in how Λ is situated inside of L 5 different types of planar lattices.
Remember...
We seek a relationship between two different lattices: The lattice L coming from a wallpaper group. The lattice Λ coming from a torus. 5 different types of planar lattices.
Remember...
We seek a relationship between two different lattices: The lattice L coming from a wallpaper group. The lattice Λ coming from a torus. We are interested in how Λ is situated inside of L Remember...
We seek a relationship between two different lattices: The lattice L coming from a wallpaper group. The lattice Λ coming from a torus. We are interested in how Λ is situated inside of L 5 different types of planar lattices. Remember...
We seek a relationship between two different lattices: The lattice L coming from a wallpaper group. The lattice Λ coming from a torus. We are interested in how Λ is situated inside of L 5 different types of planar lattices. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. √ ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct (Equilateral) triangular lattice: √ 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: 1 3 iπ/3 ω = 2 + i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. Translation subgroup LH = L is just the centers of hexagons of H.
One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus
Construct Hexagonal Wallpaper. Regard Euclidean plane as C. First construct√ (Equilateral) triangular lattice: ω = 1 + i 3 = eiπ/3. 2 √2 1 3 −iπ/3 ω = 2 − i 2 = e . Triangular lattice L∆ = Z[ω] is generated by ~ ~ l1 = 1, l2 = ω Consider the sublattice of L∆ defined as L = {x + yω | x + 2y ≡ 0 mod 3} Remove vertices in L from L∆, adjoin edges between a vertex and its nearest neighbors. This is the the hexagonal wallpaper, H Translation subgroup LH = L is just the centers of hexagons of H. One extreme: Hexagonal Wallpaper, (Equilateral) Triangular torus Can be described by a parallelogram with acute π angle 3 , and equal sidelengths. So, cutting along the shortest diagonal forms two equilateral triangles.
Equilateral Triangular torus (ETT)
Equilateral Triangular Torus (ETT): So, cutting along the shortest diagonal forms two equilateral triangles.
Equilateral Triangular torus (ETT)
Equilateral Triangular Torus (ETT): Can be described by a parallelogram with acute π angle 3 , and equal sidelengths. Equilateral Triangular torus (ETT)
Equilateral Triangular Torus (ETT): Can be described by a parallelogram with acute π angle 3 , and equal sidelengths. So, cutting along the shortest diagonal forms two equilateral triangles. ETT
An equilateral triangular torus tiled by n=13 hexagons: ETT
An equilateral triangular torus tiled by n=13 hexagons: iff n is a product of powers of 3, perfect squares (l2, l ∈ N), and primes congruent to 1 (mod 3).
Tiling an ETT by Hexagons
Theorem: n unit hexagons tiles an ETT T Tiling an ETT by Hexagons
Theorem: n unit hexagons tiles an ETT T iff n is a product of powers of 3, perfect squares (l2, l ∈ N), and primes congruent to 1 (mod 3). By BIG thm, we know that Λ ⊂ L. Since (Area of T )=(Area of n hexagons) must be true c2 + cd + d 2 = 3n. where c + dω is one side of our ETT T Z[ω] is a UFD, 3 factors as a product of irreducibles (1 + ω), (1 + ω) in Z[ω] so a solution to c2 + cd + d 2 = 3n implies that there must be an integers a, b that solve a2 + ab + b2 = n
Tiling an ETT by Hexagons: Idea of Proof
“→” Assume that the hexagonal wallpaper tiles some ETT T , and the number of hexagons contained in T is n. Since (Area of T )=(Area of n hexagons) must be true c2 + cd + d 2 = 3n. where c + dω is one side of our ETT T Z[ω] is a UFD, 3 factors as a product of irreducibles (1 + ω), (1 + ω) in Z[ω] so a solution to c2 + cd + d 2 = 3n implies that there must be an integers a, b that solve a2 + ab + b2 = n
Tiling an ETT by Hexagons: Idea of Proof
“→” Assume that the hexagonal wallpaper tiles some ETT T , and the number of hexagons contained in T is n. By BIG thm, we know that Λ ⊂ L. Z[ω] is a UFD, 3 factors as a product of irreducibles (1 + ω), (1 + ω) in Z[ω] so a solution to c2 + cd + d 2 = 3n implies that there must be an integers a, b that solve a2 + ab + b2 = n
Tiling an ETT by Hexagons: Idea of Proof
“→” Assume that the hexagonal wallpaper tiles some ETT T , and the number of hexagons contained in T is n. By BIG thm, we know that Λ ⊂ L. Since (Area of T )=(Area of n hexagons) must be true c2 + cd + d 2 = 3n. where c + dω is one side of our ETT T implies that there must be an integers a, b that solve a2 + ab + b2 = n
Tiling an ETT by Hexagons: Idea of Proof
“→” Assume that the hexagonal wallpaper tiles some ETT T , and the number of hexagons contained in T is n. By BIG thm, we know that Λ ⊂ L. Since (Area of T )=(Area of n hexagons) must be true c2 + cd + d 2 = 3n. where c + dω is one side of our ETT T Z[ω] is a UFD, 3 factors as a product of irreducibles (1 + ω), (1 + ω) in Z[ω] so a solution to c2 + cd + d 2 = 3n Tiling an ETT by Hexagons: Idea of Proof
“→” Assume that the hexagonal wallpaper tiles some ETT T , and the number of hexagons contained in T is n. By BIG thm, we know that Λ ⊂ L. Since (Area of T )=(Area of n hexagons) must be true c2 + cd + d 2 = 3n. where c + dω is one side of our ETT T Z[ω] is a UFD, 3 factors as a product of irreducibles (1 + ω), (1 + ω) in Z[ω] so a solution to c2 + cd + d 2 = 3n implies that there must be an integers a, b that solve a2 + ab + b2 = n by factoring n in Z[ω] (UFD), there exist integer solutions a, b to
a2 + ab + b2 = n
iff n is a product of powers of 3 3k = (1 + ω)k (1 + ω)k perfect squares. If n = l2, set a = l, b = 0 primes ≡ 1(mod 3) (quadratic reciprocity).
Tiling an ETT by Hexagons: Idea of Proof
Observe, a2 + ab + b2 factors as (a + bω)(a + bω), So, powers of 3 3k = (1 + ω)k (1 + ω)k perfect squares. If n = l2, set a = l, b = 0 primes ≡ 1(mod 3) (quadratic reciprocity).
Tiling an ETT by Hexagons: Idea of Proof
Observe, a2 + ab + b2 factors as (a + bω)(a + bω), So, by factoring n in Z[ω] (UFD), there exist integer solutions a, b to
a2 + ab + b2 = n
iff n is a product of perfect squares. If n = l2, set a = l, b = 0 primes ≡ 1(mod 3) (quadratic reciprocity).
Tiling an ETT by Hexagons: Idea of Proof
Observe, a2 + ab + b2 factors as (a + bω)(a + bω), So, by factoring n in Z[ω] (UFD), there exist integer solutions a, b to
a2 + ab + b2 = n
iff n is a product of powers of 3 3k = (1 + ω)k (1 + ω)k primes ≡ 1(mod 3) (quadratic reciprocity).
Tiling an ETT by Hexagons: Idea of Proof
Observe, a2 + ab + b2 factors as (a + bω)(a + bω), So, by factoring n in Z[ω] (UFD), there exist integer solutions a, b to
a2 + ab + b2 = n
iff n is a product of powers of 3 3k = (1 + ω)k (1 + ω)k perfect squares. If n = l2, set a = l, b = 0 Tiling an ETT by Hexagons: Idea of Proof
Observe, a2 + ab + b2 factors as (a + bω)(a + bω), So, by factoring n in Z[ω] (UFD), there exist integer solutions a, b to
a2 + ab + b2 = n
iff n is a product of powers of 3 3k = (1 + ω)k (1 + ω)k perfect squares. If n = l2, set a = l, b = 0 primes ≡ 1(mod 3) (quadratic reciprocity). Then there exists a, b such that (a + bω)(a + bω) = n. Muliply both sides by 3 = (1 + ω)(1 + ω):
(a + bω)(1 + ω)(a + bω)(1 + ω) = 3n
A strightforward check shows that the parallelogram
with sides (√a + bω)(1 + ω), (a + bω)(1 + ω)ω is an ETT, 3 3 with area 2 n which is exactly the area of n unit hexagons, and whose lattice Λ satisfies Λ ⊂ L. QED.
Tiling an ETT by Hexagons: Sketch of Proof
“←” Suppose n is a product of primes ≡ 1(mod3), perfect squares and powers of 3. Muliply both sides by 3 = (1 + ω)(1 + ω):
(a + bω)(1 + ω)(a + bω)(1 + ω) = 3n
A strightforward check shows that the parallelogram
with sides (√a + bω)(1 + ω), (a + bω)(1 + ω)ω is an ETT, 3 3 with area 2 n which is exactly the area of n unit hexagons, and whose lattice Λ satisfies Λ ⊂ L. QED.
Tiling an ETT by Hexagons: Sketch of Proof
“←” Suppose n is a product of primes ≡ 1(mod3), perfect squares and powers of 3. Then there exists a, b such that (a + bω)(a + bω) = n. A strightforward check shows that the parallelogram
with sides (√a + bω)(1 + ω), (a + bω)(1 + ω)ω is an ETT, 3 3 with area 2 n which is exactly the area of n unit hexagons, and whose lattice Λ satisfies Λ ⊂ L. QED.
Tiling an ETT by Hexagons: Sketch of Proof
“←” Suppose n is a product of primes ≡ 1(mod3), perfect squares and powers of 3. Then there exists a, b such that (a + bω)(a + bω) = n. Muliply both sides by 3 = (1 + ω)(1 + ω):
(a + bω)(1 + ω)(a + bω)(1 + ω) = 3n √ 3 3 with area 2 n which is exactly the area of n unit hexagons, and whose lattice Λ satisfies Λ ⊂ L. QED.
Tiling an ETT by Hexagons: Sketch of Proof
“←” Suppose n is a product of primes ≡ 1(mod3), perfect squares and powers of 3. Then there exists a, b such that (a + bω)(a + bω) = n. Muliply both sides by 3 = (1 + ω)(1 + ω):
(a + bω)(1 + ω)(a + bω)(1 + ω) = 3n
A strightforward check shows that the parallelogram with sides (a + bω)(1 + ω), (a + bω)(1 + ω)ω is an ETT, Tiling an ETT by Hexagons: Sketch of Proof
“←” Suppose n is a product of primes ≡ 1(mod3), perfect squares and powers of 3. Then there exists a, b such that (a + bω)(a + bω) = n. Muliply both sides by 3 = (1 + ω)(1 + ω):
(a + bω)(1 + ω)(a + bω)(1 + ω) = 3n
A strightforward check shows that the parallelogram
with sides (√a + bω)(1 + ω), (a + bω)(1 + ω)ω is an ETT, 3 3 with area 2 n which is exactly the area of n unit hexagons, and whose lattice Λ satisfies Λ ⊂ L. QED. Square wallpaper, L = Z2 Square Torus T , with corresponding lattice Λ =< (a, b), (−b, a) >, a, b ∈ Z. Theorem: n unit squares tile a square torus iff n is a product of powers of 2, perfect squares l2, or primes congruent to 1(mod4).
Square Tiling of a Square Torus
Similar Question Square Torus T , with corresponding lattice Λ =< (a, b), (−b, a) >, a, b ∈ Z. Theorem: n unit squares tile a square torus iff n is a product of powers of 2, perfect squares l2, or primes congruent to 1(mod4).
Square Tiling of a Square Torus
Similar Question Square wallpaper, L = Z2 Theorem: n unit squares tile a square torus iff n is a product of powers of 2, perfect squares l2, or primes congruent to 1(mod4).
Square Tiling of a Square Torus
Similar Question Square wallpaper, L = Z2 Square Torus T , with corresponding lattice Λ =< (a, b), (−b, a) >, a, b ∈ Z. Square Tiling of a Square Torus
Similar Question Square wallpaper, L = Z2 Square Torus T , with corresponding lattice Λ =< (a, b), (−b, a) >, a, b ∈ Z. Theorem: n unit squares tile a square torus iff n is a product of powers of 2, perfect squares l2, or primes congruent to 1(mod4). Given some torus T with corresponding lattice Λ, can we find Λ within L? Answer: Up to similarity we can find an approximation of Λ within L to any degree of accuracy . Similarity preserves R, the ratio of the sidelengths of T , and θ, the angle between the vectors that generate Λ.
Other extreme: Arb. Lattice, Arb. Torus
Fix any lattice L ⊂ R2. Answer: Up to similarity we can find an approximation of Λ within L to any degree of accuracy . Similarity preserves R, the ratio of the sidelengths of T , and θ, the angle between the vectors that generate Λ.
Other extreme: Arb. Lattice, Arb. Torus
Fix any lattice L ⊂ R2. Given some torus T with corresponding lattice Λ, can we find Λ within L? Similarity preserves R, the ratio of the sidelengths of T , and θ, the angle between the vectors that generate Λ.
Other extreme: Arb. Lattice, Arb. Torus
Fix any lattice L ⊂ R2. Given some torus T with corresponding lattice Λ, can we find Λ within L? Answer: Up to similarity we can find an approximation of Λ within L to any degree of accuracy . Other extreme: Arb. Lattice, Arb. Torus
Fix any lattice L ⊂ R2. Given some torus T with corresponding lattice Λ, can we find Λ within L? Answer: Up to similarity we can find an approximation of Λ within L to any degree of accuracy . Similarity preserves R, the ratio of the sidelengths of T , and θ, the angle between the vectors that generate Λ. Asymptotic Result
Theorem: Given a lattice L, an angle θ ∈ [0, π/2], and a ratio R ≥ 1, R ∈ R, then ∀ > 0, ∃ points v~1 = (x1, y1), v~2 = (x2, y2) ∈ L such that the angle between v~1 and v~2, φ, satisfies |φ − θ| < and the ratio of lengths satisfies | ||(x2,y2)|| − R| < ||(x1,y1)|| Pick any two rays subtended by θ. Generate epsilon cones around each ray. Travel along these rays in such a way that ratio R is preserved, scaling a fundamental domain of L until at least one lattice point is contained within the scaled fundamental domain and cones around each ray.
Asymptotic Results: Sketch of Proof
Idea: Story of three t’s: tθ, tL, tR. Generate epsilon cones around each ray. Travel along these rays in such a way that ratio R is preserved, scaling a fundamental domain of L until at least one lattice point is contained within the scaled fundamental domain and cones around each ray.
Asymptotic Results: Sketch of Proof
Idea: Story of three t’s: tθ, tL, tR. Pick any two rays subtended by θ. Travel along these rays in such a way that ratio R is preserved, scaling a fundamental domain of L until at least one lattice point is contained within the scaled fundamental domain and cones around each ray.
Asymptotic Results: Sketch of Proof
Idea: Story of three t’s: tθ, tL, tR. Pick any two rays subtended by θ. Generate epsilon cones around each ray. Asymptotic Results: Sketch of Proof
Idea: Story of three t’s: tθ, tL, tR. Pick any two rays subtended by θ. Generate epsilon cones around each ray. Travel along these rays in such a way that ratio R is preserved, scaling a fundamental domain of L until at least one lattice point is contained within the scaled fundamental domain and cones around each ray. Asymptotic Results: Sketch of Proof
Idea: Story of three t’s: tθ, tL, tR. Pick any two rays subtended by θ. Generate epsilon cones around each ray. Travel along these rays in such a way that ratio R is preserved, scaling a fundamental domain of L until at least one lattice point is contained within the scaled fundamental domain and cones around each ray. Let r~1, r~2 be unit vectors subtended by θ.
Parametrize rays ~ρ1 = tr~1 and ~ρ2 = tr~2, t ≥ 0.
Asymptotic Results: Sketch of Proof: tθ ~ ~ L is defined by vectors l1, l2, and we may assume ~ ~ ||l1|| ≤ ||l2||. Parametrize rays ~ρ1 = tr~1 and ~ρ2 = tr~2, t ≥ 0.
Asymptotic Results: Sketch of Proof: tθ ~ ~ L is defined by vectors l1, l2, and we may assume ~ ~ ||l1|| ≤ ||l2||.
Let r~1, r~2 be unit vectors subtended by θ. Asymptotic Results: Sketch of Proof: tθ ~ ~ L is defined by vectors l1, l2, and we may assume ~ ~ ||l1|| ≤ ||l2||.
Let r~1, r~2 be unit vectors subtended by θ.
Parametrize rays ~ρ1 = tr~1 and ~ρ2 = tr~2, t ≥ 0. Asymptotic Results: Sketch of Proof: tθ ~ ~ L is defined by vectors l1, l2, and we may assume ~ ~ ||l1|| ≤ ||l2||.
Let r~1, r~2 be unit vectors subtended by θ.
Parametrize rays ~ρ1 = tr~1 and ~ρ2 = tr~2, t ≥ 0. √Similarly, scale another fundamental domain (of L) by Rt, then translate to the point Rtr~2 on the ray ~ρ2.
Asymptotic Results: Sketch of Proof: tθ √ Scale the fundamental domain (of L) by t, then translate to the point tr~1 on the ray ~ρ1. Asymptotic Results: Sketch of Proof: tθ √ Scale the fundamental domain (of L) by t, then translate to the point tr~1 on the ray ~ρ1. √Similarly, scale another fundamental domain (of L) by Rt, then translate to the point Rtr~2 on the ray ~ρ2. Asymptotic Results: Sketch of Proof: tθ √ Scale the fundamental domain (of L) by t, then translate to the point tr~1 on the ray ~ρ1. √Similarly, scale another fundamental domain (of L) by Rt, then translate to the point Rtr~2 on the ray ~ρ2. Asymptotic Results: Sketch of Proof: tθ
Consider pairs of points tr~1, Rtr~2, t ∈ R and the regions,P1(t) P2(t) defined as the convex hull of points as described in the figure below: Asymptotic Results: Sketch of Proof: tθ
Consider pairs of points tr~1, Rtr~2, t ∈ R and the regions,P1(t) P2(t) defined as the convex hull of points as described in the figure below: ~ ~ l1+l2 √ || 2 || sin < tθ
Asymptotic Results: Sketch of Proof: tθ
√ ~ ~ l1+l2 Consider all tθ satisfying tθ|| 2 || < tθ sin , or Asymptotic Results: Sketch of Proof: tθ
√ ~ ~ l1+l2 Consider all tθ satisfying tθ|| 2 || < tθ sin , or ~ ~ l1+l2 √ || 2 || sin < tθ Asymptotic Results: Sketch of Proof: tθ
√ ~ ~ l1+l2 Consider all tθ satisfying tθ|| 2 || < tθ sin , or ~ ~ l1+l2 √ || 2 || sin < tθ Similarly, any point p ∈ P2(tθ) has an angle γ < with ~ρ2. stays inside respective -cones.
Asymptotic Results: Sketch of Proof: tθ
Through some simple geometry, any point p ∈ P1(tθ) makes an angle γ < with ~ρ1. Asymptotic Results: Sketch of Proof: tθ
Through some simple geometry, any point p ∈ P1(tθ) makes an angle γ < with ~ρ1. Similarly, any point p ∈ P2(tθ) has an angle γ < with ~ρ2. stays inside respective -cones. Asymptotic Results: Sketch of Proof: tθ
Through some simple geometry, any point p ∈ P1(tθ) makes an angle γ < with ~ρ1. Similarly, any point p ∈ P2(tθ) has an angle γ < with ~ρ2. stays inside respective -cones. ~ Observing all lines parallel to l1 and translated by ~ multiples of l2.
For tL sufficiently large, ∃ (a, b) ∈ L such that (a, b) ∈ P1(t), for all t > tL
As the area of P2(t) > area of P1(t), a similar argument holds for P2(t).
Asymptotic Results: Sketch of Proof: tL
Examining the same rays ~ρ1, ~ρ2, consider t > 0, tL ∈ R such that the area of P1(t) is greater than the area of the fundamental domain of L. For tL sufficiently large, ∃ (a, b) ∈ L such that (a, b) ∈ P1(t), for all t > tL
As the area of P2(t) > area of P1(t), a similar argument holds for P2(t).
Asymptotic Results: Sketch of Proof: tL
Examining the same rays ~ρ1, ~ρ2, consider t > 0, tL ∈ R such that the area of P1(t) is greater than the area of the fundamental domain of L. ~ Observing all lines parallel to l1 and translated by ~ multiples of l2. As the area of P2(t) > area of P1(t), a similar argument holds for P2(t).
Asymptotic Results: Sketch of Proof: tL
Examining the same rays ~ρ1, ~ρ2, consider t > 0, tL ∈ R such that the area of P1(t) is greater than the area of the fundamental domain of L. ~ Observing all lines parallel to l1 and translated by ~ multiples of l2.
For tL sufficiently large, ∃ (a, b) ∈ L such that (a, b) ∈ P1(t), for all t > tL Asymptotic Results: Sketch of Proof: tL
Examining the same rays ~ρ1, ~ρ2, consider t > 0, tL ∈ R such that the area of P1(t) is greater than the area of the fundamental domain of L. ~ Observing all lines parallel to l1 and translated by ~ multiples of l2.
For tL sufficiently large, ∃ (a, b) ∈ L such that (a, b) ∈ P1(t), for all t > tL
As the area of P2(t) > area of P1(t), a similar argument holds for P2(t). Asymptotic Results: Sketch of Proof: tL
Examining the same rays ~ρ1, ~ρ2, consider t > 0, tL ∈ R such that the area of P1(t) is greater than the area of the fundamental domain of L. ~ Observing all lines parallel to l1 and translated by ~ multiples of l2.
For tL sufficiently large, ∃ (a, b) ∈ L such that (a, b) ∈ P1(t), for all t > tL
As the area of P2(t) > area of P1(t), a similar argument holds for P2(t). Asymptotic Results: Sketch of Proof: tL √ circle centered at t on ~ρ1 with radius √t sin and circle centered at Rt on ~ρ2 with radius Rt sin . The largest and shortest possible√ sidelengths of the shorter√ side (lying on ~ρ1) are t + t sin and t − t sin , respectively The largest and shortest possible√ sidelengths of the longer√ side (lying on ~ρ2) are Rt + Rt sin and Rt − Rt sin , respectively
Asymptotic Results: Sketch of Proof: tR
Lastly, to satisfy the ratio requirement, | ||(x2,y2)|| − R| < , consider t > 0 ||(x1,y1)|| The largest and shortest possible√ sidelengths of the shorter√ side (lying on ~ρ1) are t + t sin and t − t sin , respectively The largest and shortest possible√ sidelengths of the longer√ side (lying on ~ρ2) are Rt + Rt sin and Rt − Rt sin , respectively
Asymptotic Results: Sketch of Proof: tR
Lastly, to satisfy the ratio requirement, | ||(x2,y2)|| − R| < , consider t > 0 ||(x1,y1)|| √ circle centered at t on ~ρ1 with radius √t sin and circle centered at Rt on ~ρ2 with radius Rt sin . The largest and shortest possible√ sidelengths of the longer√ side (lying on ~ρ2) are Rt + Rt sin and Rt − Rt sin , respectively
Asymptotic Results: Sketch of Proof: tR
Lastly, to satisfy the ratio requirement, | ||(x2,y2)|| − R| < , consider t > 0 ||(x1,y1)|| √ circle centered at t on ~ρ1 with radius √t sin and circle centered at Rt on ~ρ2 with radius Rt sin . The largest and shortest possible√ sidelengths of the shorter√ side (lying on ~ρ1) are t + t sin and t − t sin , respectively Asymptotic Results: Sketch of Proof: tR
Lastly, to satisfy the ratio requirement, | ||(x2,y2)|| − R| < , consider t > 0 ||(x1,y1)|| √ circle centered at t on ~ρ1 with radius √t sin and circle centered at Rt on ~ρ2 with radius Rt sin . The largest and shortest possible√ sidelengths of the shorter√ side (lying on ~ρ1) are t + t sin and t − t sin , respectively The largest and shortest possible√ sidelengths of the longer√ side (lying on ~ρ2) are Rt + Rt sin and Rt − Rt sin , respectively Asymptotic Results: Sketch of Proof: tR √ √ Rt−sin √Rt and Rt+sin √Rt t+sin t t−sin t
By calc 1, tR can√ be chosen sufficiently large√ so that R − < Rt−sin √Rt ≤ actual ratio ≤ Rt+sin √Rt < R + t+sin t t−sin t for all t > tR.
Asymptotic Results: Sketch of Proof: tR
Thus the possible ratio of sidelengths of any pair of points, one chosen from P1(t),one chosen from P2(t), is between By calc 1, tR can√ be chosen sufficiently large√ so that R − < Rt−sin √Rt ≤ actual ratio ≤ Rt+sin √Rt < R + t+sin t t−sin t for all t > tR.
Asymptotic Results: Sketch of Proof: tR
Thus the possible ratio of sidelengths of any pair of
points, one chosen√ from P1(t),one√ chosen from P2(t), is between Rt−sin √Rt and Rt+sin √Rt t+sin t t−sin t Asymptotic Results: Sketch of Proof: tR
Thus the possible ratio of sidelengths of any pair of
points, one chosen√ from P1(t),one√ chosen from P2(t), is between Rt−sin √Rt and Rt+sin √Rt t+sin t t−sin t
By calc 1, tR can√ be chosen sufficiently large√ so that R − < Rt−sin √Rt ≤ actual ratio ≤ Rt+sin √Rt < R + t+sin t t−sin t for all t > tR. Asymptotic Results: Sketch of Proof
∗ Choosing any t > max{tL, tθ, tR}, we can points in the lattice, whose lattice paralellogram approximates the ratio and angle measurements up to any degree of accuracy .