Steinhaus tiling sets Mihalis Kolountzakis University of Crete Pecs 2017 Joint work with M. Papadimitrakis 1/18
The classical Steinhaus question ◮ Steinhaus (1950s): Are there A , B ⊆ R 2 such that | τ A ∩ B | = 1 , for every rigid motion τ ? Are there two subsets of the plane which, no matter how moved, always intersect at exactly one point? 2/18
The classical Steinhaus question ◮ Steinhaus (1950s): Are there A , B ⊆ R 2 such that | τ A ∩ B | = 1 , for every rigid motion τ ? Are there two subsets of the plane which, no matter how moved, always intersect at exactly one point? Sierpi´ nski, 1958: Yes. 2/18
The classical Steinhaus question ◮ Steinhaus (1950s): Are there A , B ⊆ R 2 such that | τ A ∩ B | = 1 , for every rigid motion τ ? Are there two subsets of the plane which, no matter how moved, always intersect at exactly one point? Sierpi´ nski, 1958: Yes. ◮ Equivalent: � for all rotations ρ , x ∈ R 2 . 1 ρ A ( x − b ) = 1 , b ∈ B 2/18
The classical Steinhaus question ◮ Steinhaus (1950s): Are there A , B ⊆ R 2 such that | τ A ∩ B | = 1 , for every rigid motion τ ? Are there two subsets of the plane which, no matter how moved, always intersect at exactly one point? Sierpi´ nski, 1958: Yes. ◮ Equivalent: � for all rotations ρ , x ∈ R 2 . 1 ρ A ( x − b ) = 1 , b ∈ B ◮ In tiling language: ρ A ⊕ B = R 2 , for all rotations ρ . Every rotation of A tiles (partitions) the plane when translated at the locations B . 2/18
Fixing B = Z 2 : the lattice Steinhaus question ◮ Can we have ρ A ⊕ Z 2 = R 2 for all rotations ρ ? ◮ Equivalent: A is a fundamental domain of all ρ Z 2 . Or, A tiles the plane by translations at any ρ Z 2 . 3/18
Fixing B = Z 2 : the lattice Steinhaus question ◮ Can we have ρ A ⊕ Z 2 = R 2 for all rotations ρ ? ◮ Equivalent: A is a fundamental domain of all ρ Z 2 . Or, A tiles the plane by translations at any ρ Z 2 . ◮ Jackson and Mauldin, 2002: Yes. 3/18
Fixing B = Z 2 : the lattice Steinhaus question ◮ Can we have ρ A ⊕ Z 2 = R 2 for all rotations ρ ? ◮ Equivalent: A is a fundamental domain of all ρ Z 2 . Or, A tiles the plane by translations at any ρ Z 2 . ◮ Jackson and Mauldin, 2002: Yes. ◮ Can A be Lebesgue measurable? We interpret tiling almost everywhere. 3/18
Fixing B = Z 2 : the lattice Steinhaus question ◮ Can we have ρ A ⊕ Z 2 = R 2 for all rotations ρ ? ◮ Equivalent: A is a fundamental domain of all ρ Z 2 . Or, A tiles the plane by translations at any ρ Z 2 . ◮ Jackson and Mauldin, 2002: Yes. ◮ Can A be Lebesgue measurable? We interpret tiling almost everywhere. Results by Sierpi´ nski (1958), Croft (1982), Beck (1989), K. & Wolff (1999) : If such a measurable A exists then it must be large at infinity: � 27 + ǫ dx = ∞ . 46 | x | A 3/18
Fixing B = Z 2 : the lattice Steinhaus question ◮ Can we have ρ A ⊕ Z 2 = R 2 for all rotations ρ ? ◮ Equivalent: A is a fundamental domain of all ρ Z 2 . Or, A tiles the plane by translations at any ρ Z 2 . ◮ Jackson and Mauldin, 2002: Yes. ◮ Can A be Lebesgue measurable? We interpret tiling almost everywhere. Results by Sierpi´ nski (1958), Croft (1982), Beck (1989), K. & Wolff (1999) : If such a measurable A exists then it must be large at infinity: � 27 + ǫ dx = ∞ . 46 | x | A ◮ In higher dimension: K. & Wolff (1999), K. & Papadimitrakis (2002) : ⇒ No measurable Steinhaus sets exist for Z d , d ≥ 3. = 3/18
The lattice Steinhaus question in Fourier space ◮ For f to tile with Z 2 its periodization � f ( x − n ) n ∈ Z 2 must be constant. 4/18
The lattice Steinhaus question in Fourier space ◮ For f to tile with Z 2 its periodization � f ( x − n ) n ∈ Z 2 must be constant. f ( n ) = 0 for all n ∈ Z 2 \ { 0 } . ◮ Equivalently � 4/18
The lattice Steinhaus question in Fourier space ◮ For f to tile with Z 2 its periodization � f ( x − n ) n ∈ Z 2 must be constant. f ( n ) = 0 for all n ∈ Z 2 \ { 0 } . ◮ Equivalently � ◮ Applying to f = 1 ρ A for all rotations ρ we get that � 1 A must vanish on all circles through lattice points. 4/18
The lattice Steinhaus question in Fourier space, cont’d ◮ Successive circles in the zero set at distance R from the originat distance R from the origin are about √ 1 / R apart . 5/18
The lattice Steinhaus question in Fourier space, cont’d ◮ Successive circles in the zero set at distance R from the originat distance R from the origin are about √ 1 / R apart . ⇒ � ◮ Many zeros = 1 A must decay 5/18
The lattice Steinhaus question in Fourier space, cont’d ◮ Successive circles in the zero set at distance R from the originat distance R from the origin are about √ 1 / R apart . ⇒ � ◮ Many zeros = 1 A must decay ◮ Decay of � 1 A = ⇒ lack of concentration for 1 A , regularity 5/18
The lattice Steinhaus question in Fourier space, cont’d ◮ Successive circles in the zero set at distance R from the originat distance R from the origin are about √ 1 / R apart . ⇒ � ◮ Many zeros = 1 A must decay ◮ Decay of � 1 A = ⇒ lack of concentration for 1 A , regularity � 27 + ǫ dx = ∞ . 46 ◮ In dimension d = 2 this gives A | x | 5/18
The lattice Steinhaus question in Fourier space, cont’d ◮ Successive circles in the zero set at distance R from the originat distance R from the origin are about √ 1 / R apart . ⇒ � ◮ Many zeros = 1 A must decay ◮ Decay of � 1 A = ⇒ lack of concentration for 1 A , regularity � 27 + ǫ dx = ∞ . 46 ◮ In dimension d = 2 this gives A | x | ◮ In dimension d ≥ 3: better control of circle gap. We get 1 A is continuous (contradiction) 5/18
The lattice Steinhaus question for finitely many lattices ◮ Given lattices Λ 1 , . . . , Λ n ⊆ R d all of volume 1 can we find measurable A which tiles with all Λ j ? 6/18
The lattice Steinhaus question for finitely many lattices ◮ Given lattices Λ 1 , . . . , Λ n ⊆ R d all of volume 1 can we find measurable A which tiles with all Λ j ? ◮ Generically yes! If the sum Λ ∗ 1 + · · · + Λ ∗ n is di- rect then Kronecker-type den- sity theorems allow us to rear- range a fundamental domain of one lattice to accomodate the others. 6/18
Restated for the algebraically inclined ◮ If G is an abelian group and H 1 , . . . , H n subgroups of same index 7/18
Restated for the algebraically inclined ◮ If G is an abelian group and H 1 , . . . , H n subgroups of same index can we find a common set of coset representatives for the H j ? 7/18
Restated for the algebraically inclined ◮ If G is an abelian group and H 1 , . . . , H n subgroups of same index can we find a common set of coset representatives for the H j ? ◮ Always possible for two subgroups H 1 , H 2 (even in non-abelian case). 7/18
Restated for the algebraically inclined ◮ If G is an abelian group and H 1 , . . . , H n subgroups of same index can we find a common set of coset representatives for the H j ? ◮ Always possible for two subgroups H 1 , H 2 (even in non-abelian case). ◮ Fails in general: take G = Z 2 × Z 2 and the 3 copies of Z 2 therein. 7/18
Restated for the algebraically inclined ◮ If G is an abelian group and H 1 , . . . , H n subgroups of same index can we find a common set of coset representatives for the H j ? ◮ Always possible for two subgroups H 1 , H 2 (even in non-abelian case). ◮ Fails in general: take G = Z 2 × Z 2 and the 3 copies of Z 2 therein. ◮ No good condition is known! 7/18
An application in Gabor analysis ◮ Question: If K , L are two lattices in R d with vol K · vol L = 1 , can we find g ∈ L 2 ( R d ), such that the ( K , L ) time-frequency translates g ( x − k ) e 2 π i ℓ · x , ( k ∈ K , ℓ ∈ L ) form an orthogonal basis? 8/18
An application in Gabor analysis ◮ Question: If K , L are two lattices in R d with vol K · vol L = 1 , can we find g ∈ L 2 ( R d ), such that the ( K , L ) time-frequency translates g ( x − k ) e 2 π i ℓ · x , ( k ∈ K , ℓ ∈ L ) form an orthogonal basis? ◮ Han and Wang (2000): Since vol ( L ∗ ) = vol ( K ) let g = 1 E where E is a common tile for K , L ∗ . 8/18
An application in Gabor analysis ◮ Question: If K , L are two lattices in R d with vol K · vol L = 1 , can we find g ∈ L 2 ( R d ), such that the ( K , L ) time-frequency translates g ( x − k ) e 2 π i ℓ · x , ( k ∈ K , ℓ ∈ L ) form an orthogonal basis? ◮ Han and Wang (2000): Since vol ( L ∗ ) = vol ( K ) let g = 1 E where E is a common tile for K , L ∗ . ◮ Then L forms an orthogonal basis for L 2 ( E ). 8/18
An application in Gabor analysis ◮ Question: If K , L are two lattices in R d with vol K · vol L = 1 , can we find g ∈ L 2 ( R d ), such that the ( K , L ) time-frequency translates g ( x − k ) e 2 π i ℓ · x , ( k ∈ K , ℓ ∈ L ) form an orthogonal basis? ◮ Han and Wang (2000): Since vol ( L ∗ ) = vol ( K ) let g = 1 E where E is a common tile for K , L ∗ . ◮ Then L forms an orthogonal basis for L 2 ( E ). ◮ The space is partitioned in copies of E and on each copy L is an orthogonal basis. 8/18
B = Z × { 0 } or B a finite set ◮ B = Z × { 0 } : y The strip tiles with B x 9/18
B = Z × { 0 } or B a finite set ◮ B = Z × { 0 } : y The strip tiles with B x ◮ B is a finite set: y x The shaded set tiles with B 9/18
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