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Fuglede’s spectral set conjecture on cyclic groups
Romanos Diogenes Malikiosis
TU Berlin
Frame Theory and Exponential Bases4-8 June 2018
ICERM, Providence
Joint work with M. Kolountzakis (U. of Crete)& work in progress
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 2: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/2.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 3: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/3.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 4: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/4.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 5: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/5.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 6: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/6.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 7: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/7.jpg)
Fourier Analysis on domains Ω ⊆ Rn
Question
On which measurable domains Ω ⊆ Rn with µ(Ω) > 0 can we doFourier analysis, that is, there is an orthonormal basis of
exponential functions
1µ(Ω)e
2πiλ·x : λ ∈ Λ
in L2(Ω), where
Λ ⊆ Rn discrete?
Definition
If Ω satisfies the above condition it is called spectral, and Λ is thespectrum of Ω.
The n-dimensional cube C = [0, 1]n.
Parallelepipeds AC , where A ∈ GL(n,R).
Hexagons on R2.
Not n-dimensional balls! (n ≥ 2) (Iosevich, Katz, Pedersen,’99)
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 8: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/8.jpg)
Fuglede’s conjecture
Definition
A set Ω ⊆ Rn of positive measure is called tile of Rn if there isT ⊆ Rn such that Ω⊕ T = Rn.
Conjecture (Fuglede, 1974)
A set Ω ⊆ Rn of positive measure is spectral if and only if it tilesRn.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 9: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/9.jpg)
Fuglede’s conjecture
Definition
A set Ω ⊆ Rn of positive measure is called tile of Rn if there isT ⊆ Rn such that Ω⊕ T = Rn.
Conjecture (Fuglede, 1974)
A set Ω ⊆ Rn of positive measure is spectral if and only if it tilesRn.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 10: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/10.jpg)
Basic properties
Let eλ(x) = e2πiλ·x . Wlog, µ(Ω) = 1. Inner product and norm onL2(Ω):
〈f , g〉Ω =
∫Ωf g , ‖f ‖2
Ω =
∫Ω|f |2.
It holds 〈eλ, eµ〉Ω = 1Ω(µ− λ).
Lemma
Λ is a spectrum of Ω if and only if
1Ω(λ− µ) = 0, ∀λ 6= µ, λ, µ ∈ Λ
and∀f ∈ L2(Ω) : ‖f ‖2
Ω =∑λ∈Λ
|〈f , eλ〉|2.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 11: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/11.jpg)
Basic properties
Let eλ(x) = e2πiλ·x . Wlog, µ(Ω) = 1. Inner product and norm onL2(Ω):
〈f , g〉Ω =
∫Ωf g , ‖f ‖2
Ω =
∫Ω|f |2.
It holds 〈eλ, eµ〉Ω = 1Ω(µ− λ).
Lemma
Λ is a spectrum of Ω if and only if
1Ω(λ− µ) = 0, ∀λ 6= µ, λ, µ ∈ Λ
and∀f ∈ L2(Ω) : ‖f ‖2
Ω =∑λ∈Λ
|〈f , eλ〉|2.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 12: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/12.jpg)
Special cases
Theorem (Fuglede, ’74)
Let Ω ⊆ Rn be an open bounded set of measure 1 and Λ ⊆ Rn bea lattice with density 1. Then Ω⊕ Λ = Rn if and only if Λ? is aspectrum of Ω.
Theorem (Kolountzakis, ’00)
Let Ω ⊆ Rn, n ≥ 2, be a convex asymmetric body. Then Ω is notspectral.
Theorem (Iosevich, Katz, Tao, ’01)
Let Ω ⊆ Rn, n ≥ 2, be a convex symmetric body. If ∂Ω is smooth,then Ω is not spectral. The same holds for n = 2 when ∂Ω ispiecewise smooth possessing at least one point of nonzerocurvature.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 13: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/13.jpg)
Special cases
Theorem (Fuglede, ’74)
Let Ω ⊆ Rn be an open bounded set of measure 1 and Λ ⊆ Rn bea lattice with density 1. Then Ω⊕ Λ = Rn if and only if Λ? is aspectrum of Ω.
Theorem (Kolountzakis, ’00)
Let Ω ⊆ Rn, n ≥ 2, be a convex asymmetric body. Then Ω is notspectral.
Theorem (Iosevich, Katz, Tao, ’01)
Let Ω ⊆ Rn, n ≥ 2, be a convex symmetric body. If ∂Ω is smooth,then Ω is not spectral. The same holds for n = 2 when ∂Ω ispiecewise smooth possessing at least one point of nonzerocurvature.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 14: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/14.jpg)
Special cases
Theorem (Fuglede, ’74)
Let Ω ⊆ Rn be an open bounded set of measure 1 and Λ ⊆ Rn bea lattice with density 1. Then Ω⊕ Λ = Rn if and only if Λ? is aspectrum of Ω.
Theorem (Kolountzakis, ’00)
Let Ω ⊆ Rn, n ≥ 2, be a convex asymmetric body. Then Ω is notspectral.
Theorem (Iosevich, Katz, Tao, ’01)
Let Ω ⊆ Rn, n ≥ 2, be a convex symmetric body. If ∂Ω is smooth,then Ω is not spectral. The same holds for n = 2 when ∂Ω ispiecewise smooth possessing at least one point of nonzerocurvature.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 15: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/15.jpg)
Special cases
Theorem (Fuglede, ’74)
Let Ω ⊆ Rn be an open bounded set of measure 1 and Λ ⊆ Rn bea lattice with density 1. Then Ω⊕ Λ = Rn if and only if Λ? is aspectrum of Ω.
Theorem (Kolountzakis, ’00)
Let Ω ⊆ Rn, n ≥ 2, be a convex asymmetric body. Then Ω is notspectral.
Theorem (Iosevich, Katz, Tao, ’01)
Let Ω ⊆ Rn, n ≥ 2, be a convex symmetric body. If ∂Ω is smooth,then Ω is not spectral. The same holds for n = 2 when ∂Ω ispiecewise smooth possessing at least one point of nonzerocurvature.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 16: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/16.jpg)
Convex polytopes
According to the theorems of Venkov (’54) and McMullen (’80),the above do not tile Rn.
Theorem (Greenfeld, Lev, ’17)
Let K ⊆ Rn be a convex symmetric polytope, which is spectral.Then its facets are also symmetric. Also, if n = 3, any spectralconvex polytope tiles the space.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 17: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/17.jpg)
Convex polytopes
According to the theorems of Venkov (’54) and McMullen (’80),the above do not tile Rn.
Theorem (Greenfeld, Lev, ’17)
Let K ⊆ Rn be a convex symmetric polytope, which is spectral.Then its facets are also symmetric. Also, if n = 3, any spectralconvex polytope tiles the space.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 18: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/18.jpg)
Tao’s counterexample
“A cataclysmic event in the history of this problem took place in2004 when Terry Tao disproved the Fuglede Conjecture byexhibiting a spectral set in R12 which does not tile.”The Fuglede Conjecture holds in Zp × Zp, Iosevich, Mayeli,Pakianathan, 2017.
Theorem (Tao, ’04)
There are spectral subsets of R5 of positive measure that do nottile R5.
Theorem (Farkas-Kolountzakis-Matolcsi-Mora-Revesz-Tao, ’04-’06)
Fuglede’s conjecture fails for n ≥ 3 (both directions).
The conjecture is still open for n ≤ 2. Tao’s counterexample is aunion of unit cubes. It comes from a spectral subset of Z5
3 of size 6.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 19: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/19.jpg)
Tao’s counterexample
“A cataclysmic event in the history of this problem took place in2004 when Terry Tao disproved the Fuglede Conjecture byexhibiting a spectral set in R12 which does not tile.”The Fuglede Conjecture holds in Zp × Zp, Iosevich, Mayeli,Pakianathan, 2017.
Theorem (Tao, ’04)
There are spectral subsets of R5 of positive measure that do nottile R5.
Theorem (Farkas-Kolountzakis-Matolcsi-Mora-Revesz-Tao, ’04-’06)
Fuglede’s conjecture fails for n ≥ 3 (both directions).
The conjecture is still open for n ≤ 2. Tao’s counterexample is aunion of unit cubes. It comes from a spectral subset of Z5
3 of size 6.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 20: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/20.jpg)
Tao’s counterexample
“A cataclysmic event in the history of this problem took place in2004 when Terry Tao disproved the Fuglede Conjecture byexhibiting a spectral set in R12 which does not tile.”The Fuglede Conjecture holds in Zp × Zp, Iosevich, Mayeli,Pakianathan, 2017.
Theorem (Tao, ’04)
There are spectral subsets of R5 of positive measure that do nottile R5.
Theorem (Farkas-Kolountzakis-Matolcsi-Mora-Revesz-Tao, ’04-’06)
Fuglede’s conjecture fails for n ≥ 3 (both directions).
The conjecture is still open for n ≤ 2. Tao’s counterexample is aunion of unit cubes. It comes from a spectral subset of Z5
3 of size 6.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 21: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/21.jpg)
Tao’s counterexample
“A cataclysmic event in the history of this problem took place in2004 when Terry Tao disproved the Fuglede Conjecture byexhibiting a spectral set in R12 which does not tile.”The Fuglede Conjecture holds in Zp × Zp, Iosevich, Mayeli,Pakianathan, 2017.
Theorem (Tao, ’04)
There are spectral subsets of R5 of positive measure that do nottile R5.
Theorem (Farkas-Kolountzakis-Matolcsi-Mora-Revesz-Tao, ’04-’06)
Fuglede’s conjecture fails for n ≥ 3 (both directions).
The conjecture is still open for n ≤ 2. Tao’s counterexample is aunion of unit cubes. It comes from a spectral subset of Z5
3 of size 6.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 22: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/22.jpg)
Tao’s counterexample
“A cataclysmic event in the history of this problem took place in2004 when Terry Tao disproved the Fuglede Conjecture byexhibiting a spectral set in R12 which does not tile.”The Fuglede Conjecture holds in Zp × Zp, Iosevich, Mayeli,Pakianathan, 2017.
Theorem (Tao, ’04)
There are spectral subsets of R5 of positive measure that do nottile R5.
Theorem (Farkas-Kolountzakis-Matolcsi-Mora-Revesz-Tao, ’04-’06)
Fuglede’s conjecture fails for n ≥ 3 (both directions).
The conjecture is still open for n ≤ 2. Tao’s counterexample is aunion of unit cubes. It comes from a spectral subset of Z5
3 of size 6.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 23: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/23.jpg)
Passage to finite groups
Definition
Let G be an Abelian group. We write (S-T(G )) if every boundedspectral subset of G is also a tile, and (T-S(G )) if every boundedtile of G is spectral.
Theorem (Dutkay, Lai, ’14)
The following hold:
(T-S(Zn))∀n ∈ N⇔ (T-S(Z))⇔ (T-S(R))
and(S-T(R))⇒ (S-T(Z))⇒ (S-T(Zn))∀n ∈ N.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 24: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/24.jpg)
Passage to finite groups
Definition
Let G be an Abelian group. We write (S-T(G )) if every boundedspectral subset of G is also a tile, and (T-S(G )) if every boundedtile of G is spectral.
Theorem (Dutkay, Lai, ’14)
The following hold:
(T-S(Zn))∀n ∈ N⇔ (T-S(Z))⇔ (T-S(R))
and(S-T(R))⇒ (S-T(Z))⇒ (S-T(Zn))∀n ∈ N.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 25: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/25.jpg)
Passage to finite groups
The last hold in both directions if every bounded spectral subset ofR has a rational spectrum. Some results in the positive direction:
If Ω = A + [0, 1] ⊆ R is spectral, with |A| = N andA ⊆ [0,M − 1] with M < 5N/2, then Ω has a rationalspectrum ( Laba, ’02).
If Fuglede’s conjecture holds in R, then every boundedspectral set has a rational spectrum (Dutkay, Lai, ’14).
If Ω = A + [0, 1] ⊆ R is spectral, and A− A contains certainpatterns (flags), then Ω has rational spectrum (Bose, Madan,’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 26: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/26.jpg)
Passage to finite groups
The last hold in both directions if every bounded spectral subset ofR has a rational spectrum. Some results in the positive direction:
If Ω = A + [0, 1] ⊆ R is spectral, with |A| = N andA ⊆ [0,M − 1] with M < 5N/2, then Ω has a rationalspectrum ( Laba, ’02).
If Fuglede’s conjecture holds in R, then every boundedspectral set has a rational spectrum (Dutkay, Lai, ’14).
If Ω = A + [0, 1] ⊆ R is spectral, and A− A contains certainpatterns (flags), then Ω has rational spectrum (Bose, Madan,’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
![Page 27: Romanos Diogenes Malikiosis · 2020. 4. 24. · Romanos Diogenes Malikiosis TU Berlin Frame Theory and Exponential Bases 4-8 June 2018 ICERM, Providence Joint work with M. Kolountzakis](https://reader036.vdocuments.pub/reader036/viewer/2022062606/5fe907f330d91526ec2c6bef/html5/thumbnails/27.jpg)
Passage to finite groups
The last hold in both directions if every bounded spectral subset ofR has a rational spectrum. Some results in the positive direction:
If Ω = A + [0, 1] ⊆ R is spectral, with |A| = N andA ⊆ [0,M − 1] with M < 5N/2, then Ω has a rationalspectrum ( Laba, ’02).
If Fuglede’s conjecture holds in R, then every boundedspectral set has a rational spectrum (Dutkay, Lai, ’14).
If Ω = A + [0, 1] ⊆ R is spectral, and A− A contains certainpatterns (flags), then Ω has rational spectrum (Bose, Madan,’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Passage to finite groups
The last hold in both directions if every bounded spectral subset ofR has a rational spectrum. Some results in the positive direction:
If Ω = A + [0, 1] ⊆ R is spectral, with |A| = N andA ⊆ [0,M − 1] with M < 5N/2, then Ω has a rationalspectrum ( Laba, ’02).
If Fuglede’s conjecture holds in R, then every boundedspectral set has a rational spectrum (Dutkay, Lai, ’14).
If Ω = A + [0, 1] ⊆ R is spectral, and A− A contains certainpatterns (flags), then Ω has rational spectrum (Bose, Madan,’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Passage to finite groups
The last hold in both directions if every bounded spectral subset ofR has a rational spectrum. Some results in the positive direction:
If Ω = A + [0, 1] ⊆ R is spectral, with |A| = N andA ⊆ [0,M − 1] with M < 5N/2, then Ω has a rationalspectrum ( Laba, ’02).
If Fuglede’s conjecture holds in R, then every boundedspectral set has a rational spectrum (Dutkay, Lai, ’14).
If Ω = A + [0, 1] ⊆ R is spectral, and A− A contains certainpatterns (flags), then Ω has rational spectrum (Bose, Madan,’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Non-cyclic groups
The properties (S-T(G )) and (T-S(G )) are hereditary, that is,they hold for every subgroup of G .It suffices then to examine groups of the form Zd
N . For d ≥ 2 weget the following results:
There is a spectral subset of Z38 that does not tile
(Kolountzakis, Matolcsi, ’06).
There is a tile of Z324 that is not spectral (Farkas, Matolcsi,
Mora, ’06).
Fuglede’s conjecture holds in Z2p, p prime (Iosevich, Mayeli,
Pakianathan, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Non-cyclic groups
The properties (S-T(G )) and (T-S(G )) are hereditary, that is,they hold for every subgroup of G .It suffices then to examine groups of the form Zd
N . For d ≥ 2 weget the following results:
There is a spectral subset of Z38 that does not tile
(Kolountzakis, Matolcsi, ’06).
There is a tile of Z324 that is not spectral (Farkas, Matolcsi,
Mora, ’06).
Fuglede’s conjecture holds in Z2p, p prime (Iosevich, Mayeli,
Pakianathan, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Non-cyclic groups
The properties (S-T(G )) and (T-S(G )) are hereditary, that is,they hold for every subgroup of G .It suffices then to examine groups of the form Zd
N . For d ≥ 2 weget the following results:
There is a spectral subset of Z38 that does not tile
(Kolountzakis, Matolcsi, ’06).
There is a tile of Z324 that is not spectral (Farkas, Matolcsi,
Mora, ’06).
Fuglede’s conjecture holds in Z2p, p prime (Iosevich, Mayeli,
Pakianathan, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Non-cyclic groups
The properties (S-T(G )) and (T-S(G )) are hereditary, that is,they hold for every subgroup of G .It suffices then to examine groups of the form Zd
N . For d ≥ 2 weget the following results:
There is a spectral subset of Z38 that does not tile
(Kolountzakis, Matolcsi, ’06).
There is a tile of Z324 that is not spectral (Farkas, Matolcsi,
Mora, ’06).
Fuglede’s conjecture holds in Z2p, p prime (Iosevich, Mayeli,
Pakianathan, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Non-cyclic groups
The properties (S-T(G )) and (T-S(G )) are hereditary, that is,they hold for every subgroup of G .It suffices then to examine groups of the form Zd
N . For d ≥ 2 weget the following results:
There is a spectral subset of Z38 that does not tile
(Kolountzakis, Matolcsi, ’06).
There is a tile of Z324 that is not spectral (Farkas, Matolcsi,
Mora, ’06).
Fuglede’s conjecture holds in Z2p, p prime (Iosevich, Mayeli,
Pakianathan, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Laba’s work on tiles and spectral subsets A ⊆ Z with |A| = pn orpnqm, along with the results of Coven-Meyerowitz on tiling subsetsof Z, has the following consequences for for cyclic groups G = ZN :
If N is a prime power, then both (S-T(ZN)) and (T-S(ZN))hold (also by Fan, Fan, Shi, ’16, and Kolountzakis, M, ’17).
If N = pnqm, then (T-S(ZN)) (also Kolountzakis, M, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Laba’s work on tiles and spectral subsets A ⊆ Z with |A| = pn orpnqm, along with the results of Coven-Meyerowitz on tiling subsetsof Z, has the following consequences for for cyclic groups G = ZN :
If N is a prime power, then both (S-T(ZN)) and (T-S(ZN))hold (also by Fan, Fan, Shi, ’16, and Kolountzakis, M, ’17).
If N = pnqm, then (T-S(ZN)) (also Kolountzakis, M, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Laba’s work on tiles and spectral subsets A ⊆ Z with |A| = pn orpnqm, along with the results of Coven-Meyerowitz on tiling subsetsof Z, has the following consequences for for cyclic groups G = ZN :
If N is a prime power, then both (S-T(ZN)) and (T-S(ZN))hold (also by Fan, Fan, Shi, ’16, and Kolountzakis, M, ’17).
If N = pnqm, then (T-S(ZN)) (also Kolountzakis, M, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Laba’s work on tiles and spectral subsets A ⊆ Z with |A| = pn orpnqm, along with the results of Coven-Meyerowitz on tiling subsetsof Z, has the following consequences for for cyclic groups G = ZN :
If N is a prime power, then both (S-T(ZN)) and (T-S(ZN))hold (also by Fan, Fan, Shi, ’16, and Kolountzakis, M, ’17).
If N = pnqm, then (T-S(ZN)) (also Kolountzakis, M, ’17).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Moreover,
If N = pnq, then (S-T(ZN)) (Kolountzakis, M, ’17).
If N is square-free, then (T-S(ZN)) ( Laba and Meyerowitz,answering a question in Tao’s blog, also, Shi ’18).
If N = pnqm, with m or n ≤ 6, then (S-T(ZN)) (M, work inprogress).
If N = pnqm, with m, n ≥ 7 and pn−6 < q3 with p < q, then(S-T(ZN)) (M, work in progress).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Moreover,
If N = pnq, then (S-T(ZN)) (Kolountzakis, M, ’17).
If N is square-free, then (T-S(ZN)) ( Laba and Meyerowitz,answering a question in Tao’s blog, also, Shi ’18).
If N = pnqm, with m or n ≤ 6, then (S-T(ZN)) (M, work inprogress).
If N = pnqm, with m, n ≥ 7 and pn−6 < q3 with p < q, then(S-T(ZN)) (M, work in progress).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Moreover,
If N = pnq, then (S-T(ZN)) (Kolountzakis, M, ’17).
If N is square-free, then (T-S(ZN)) ( Laba and Meyerowitz,answering a question in Tao’s blog, also, Shi ’18).
If N = pnqm, with m or n ≤ 6, then (S-T(ZN)) (M, work inprogress).
If N = pnqm, with m, n ≥ 7 and pn−6 < q3 with p < q, then(S-T(ZN)) (M, work in progress).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Cyclic groups
Moreover,
If N = pnq, then (S-T(ZN)) (Kolountzakis, M, ’17).
If N is square-free, then (T-S(ZN)) ( Laba and Meyerowitz,answering a question in Tao’s blog, also, Shi ’18).
If N = pnqm, with m or n ≤ 6, then (S-T(ZN)) (M, work inprogress).
If N = pnqm, with m, n ≥ 7 and pn−6 < q3 with p < q, then(S-T(ZN)) (M, work in progress).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The mask polynomial
Definition (Coven-Meyerowitz, ’98)
Let A ⊆ ZN . The mask polynomial A is given by∑a∈A
X a ∈ Z[X ]/(XN − 1).
It holds1A(d) = A(ζdN),∀d ∈ ZN .
Λ is a spectrum of A if and only if |A| = |Λ| and
A(ζord(`−`′)) = 0, ∀`, `′ ∈ Λ, ` 6= `′.
Moreover, A⊕ T = ZN if and only if
A(X )T (X ) ≡ 1 + X + X 2 + · · ·+ XN−1 mod (XN − 1).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The mask polynomial
Definition (Coven-Meyerowitz, ’98)
Let A ⊆ ZN . The mask polynomial A is given by∑a∈A
X a ∈ Z[X ]/(XN − 1).
It holds1A(d) = A(ζdN),∀d ∈ ZN .
Λ is a spectrum of A if and only if |A| = |Λ| and
A(ζord(`−`′)) = 0, ∀`, `′ ∈ Λ, ` 6= `′.
Moreover, A⊕ T = ZN if and only if
A(X )T (X ) ≡ 1 + X + X 2 + · · ·+ XN−1 mod (XN − 1).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The mask polynomial
Definition (Coven-Meyerowitz, ’98)
Let A ⊆ ZN . The mask polynomial A is given by∑a∈A
X a ∈ Z[X ]/(XN − 1).
It holds1A(d) = A(ζdN),∀d ∈ ZN .
Λ is a spectrum of A if and only if |A| = |Λ| and
A(ζord(`−`′)) = 0, ∀`, `′ ∈ Λ, ` 6= `′.
Moreover, A⊕ T = ZN if and only if
A(X )T (X ) ≡ 1 + X + X 2 + · · ·+ XN−1 mod (XN − 1).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The properties (T1) and (T2)
Definition
Let A(X ) ∈ Z[X ]/(XN − 1), and let
SA = d | N : d prime power,A(ζd) = 0.
We define the following properties:
(T1) A(1) =∏
s∈SA Φs(1)
(T2) Let s1, s2, . . . , sk ∈ SA be powers of different primes. ThenΦs(X ) | A(X ), where s = s1 · · · sk .
Remark
When N is a prime power, (T2) holds vacuously. If N = pnqm,then (T2) is simply
A(ζpk ) = A(ζq`) = 0⇒ A(ζpkq`) = 0
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The properties (T1) and (T2)
Definition
Let A(X ) ∈ Z[X ]/(XN − 1), and let
SA = d | N : d prime power,A(ζd) = 0.
We define the following properties:
(T1) A(1) =∏
s∈SA Φs(1)
(T2) Let s1, s2, . . . , sk ∈ SA be powers of different primes. ThenΦs(X ) | A(X ), where s = s1 · · · sk .
Remark
When N is a prime power, (T2) holds vacuously. If N = pnqm,then (T2) is simply
A(ζpk ) = A(ζq`) = 0⇒ A(ζpkq`) = 0
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Example
Let A ⊆ ZN , N = p4q4r3, such that
A(ζp) = A(ζp3) = A(ζq2) = A(ζr3) = 0,
and A(X ) has no other root of order a power of p, q, or r . Then,(T1) is equivalent to |A| = p2qr , and (T2) is equivalent to
A(ζpq2) = A(ζp3q2) = A(ζpr3) = A(ζp3r3) = A(ζq2r3) =
= A(ζpq2r3) = A(ζp3q2r3) = 0.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Tiling, spectrality, and (T1), (T2)
The following are consequences of the works of Coven-Meyerowitz(’98) and Laba (’02); also Kolountzakis-Matolcsi (’07).
Theorem
If A ⊆ ZN satisfies (T1) and (T2), then it tiles ZN . If A tiles ZN ,then it satisfies (T1); if in addition N = pnqm, then A satisfies(T2) as well.
Theorem
If A ⊆ ZN satisfies (T1) and (T2), then it is spectral. If N = pn
and A is spectral, then it satisfies (T1).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Tiling, spectrality, and (T1), (T2)
The following are consequences of the works of Coven-Meyerowitz(’98) and Laba (’02); also Kolountzakis-Matolcsi (’07).
Theorem
If A ⊆ ZN satisfies (T1) and (T2), then it tiles ZN . If A tiles ZN ,then it satisfies (T1); if in addition N = pnqm, then A satisfies(T2) as well.
Theorem
If A ⊆ ZN satisfies (T1) and (T2), then it is spectral. If N = pn
and A is spectral, then it satisfies (T1).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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(T-S(ZN)), N square-free
Let A⊕ T = ZN , with |A| = m. Then, also A⊕mT = ZN , due to
(A− A) ∩ (T − T ) = 0.
The mask polynomial of mT is T (Xm) mod (XN − 1), so ifp1, . . . , pk | m, we have
A(ζpj ) = 0, 1 ≤ j ≤ k ,
since pj - |T |,
and
T (ζmp1···pk ) = T (1) 6= 0,
thusA(ζp1···pk ) = 0,
and A satisfies (T2).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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(T-S(ZN)), N square-free
Let A⊕ T = ZN , with |A| = m. Then, also A⊕mT = ZN , due to
(A− A) ∩ (T − T ) = 0.
The mask polynomial of mT is T (Xm) mod (XN − 1), so ifp1, . . . , pk | m, we have
A(ζpj ) = 0, 1 ≤ j ≤ k ,
since pj - |T |, and
T (ζmp1···pk ) = T (1) 6= 0,
thusA(ζp1···pk ) = 0,
and A satisfies (T2).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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(T-S(ZN)), N square-free
Let A⊕ T = ZN , with |A| = m. Then, also A⊕mT = ZN , due to
(A− A) ∩ (T − T ) = 0.
The mask polynomial of mT is T (Xm) mod (XN − 1), so ifp1, . . . , pk | m, we have
A(ζpj ) = 0, 1 ≤ j ≤ k ,
since pj - |T |, and
T (ζmp1···pk ) = T (1) 6= 0,
thusA(ζp1···pk ) = 0,
and A satisfies (T2).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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(T-S(ZN)), N square-free
Let A⊕ T = ZN , with |A| = m. Then, also A⊕mT = ZN , due to
(A− A) ∩ (T − T ) = 0.
The mask polynomial of mT is T (Xm) mod (XN − 1), so ifp1, . . . , pk | m, we have
A(ζpj ) = 0, 1 ≤ j ≤ k ,
since pj - |T |, and
T (ζmp1···pk ) = T (1) 6= 0,
thusA(ζp1···pk ) = 0,
and A satisfies (T2).
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Primitive subsets of ZN
Definition
A subset A ⊆ G is called primitive if it is not contained in a propercoset of G .
Lemma
Let G = ZN with N = pnqm, and A ⊆ ZN primitive. Then(A− A) ∩ Z?N 6= ∅.
Proof.
Let a ∈ A. Since a− A * pZN or qZN , there are a′, a′′ ∈ A suchthat a− a′ /∈ pZN , a− a′′ /∈ qZN . If either a− a′ /∈ qZN ora− a′′ /∈ pZN , then we’re done, so wlog q | a− a′ and p | a− a′′,which yields a′′ − a′ ∈ Z?N .
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Primitive subsets of ZN
Definition
A subset A ⊆ G is called primitive if it is not contained in a propercoset of G .
Lemma
Let G = ZN with N = pnqm, and A ⊆ ZN primitive. Then(A− A) ∩ Z?N 6= ∅.
Proof.
Let a ∈ A. Since a− A * pZN or qZN , there are a′, a′′ ∈ A suchthat a− a′ /∈ pZN , a− a′′ /∈ qZN . If either a− a′ /∈ qZN ora− a′′ /∈ pZN , then we’re done, so wlog q | a− a′ and p | a− a′′,which yields a′′ − a′ ∈ Z?N .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Primitive subsets of ZN
Definition
A subset A ⊆ G is called primitive if it is not contained in a propercoset of G .
Lemma
Let G = ZN with N = pnqm, and A ⊆ ZN primitive. Then(A− A) ∩ Z?N 6= ∅.
Proof.
Let a ∈ A. Since a− A * pZN or qZN , there are a′, a′′ ∈ A suchthat a− a′ /∈ pZN , a− a′′ /∈ qZN . If either a− a′ /∈ qZN ora− a′′ /∈ pZN , then we’re done, so wlog q | a− a′ and p | a− a′′,which yields a′′ − a′ ∈ Z?N .
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Primitive spectral pairs, N = pnqm
Corollary
Let (A,B) be a spectral pair in ZN , such that both A and B areprimitive. Then,
A(ζN) = B(ζN) = 0.
Remark
If A is not primitive, then A ⊆ pZN (say), which implies(B − B) ∩ N
p ZN = 0. Then, (A,B) is a spectral pair in ZN/p,
where p · A = A. Moreover, if A satisfies (T1) and (T2) in ZN/p,then A satisfies the same properties in ZN .
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Primitive spectral pairs, N = pnqm
Corollary
Let (A,B) be a spectral pair in ZN , such that both A and B areprimitive. Then,
A(ζN) = B(ζN) = 0.
Remark
If A is not primitive, then A ⊆ pZN (say), which implies(B − B) ∩ N
p ZN = 0. Then, (A,B) is a spectral pair in ZN/p,
where p · A = A. Moreover, if A satisfies (T1) and (T2) in ZN/p,then A satisfies the same properties in ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Vanishing sums of roots of unity
Lemma
Let rad(N) = pq and A(X ) ∈ Z[X ] with nonnegative coefficients,such that A(ζdN) = 0, for some d | N. Then,
A(X d) ≡ P(X d)Φp(XN/p) + Q(X d)Φq(XN/q) mod (XN − 1),
where P(X ),Q(X ) ∈ Z[X ] can be taken with nonnegativecoefficients.
The polynomial A(X d) is the mask polynomial of the multisetd · A.
Φp(XN/p) is the mask polynomial of the subgroup Np ZN . Its
cosets are called p-cycles.
The above Lemma shows that if A(ζN) = 0, then A is thedisjoint union of p- and q-cycles.
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Vanishing sums of roots of unity
Lemma
Let rad(N) = pq and A(X ) ∈ Z[X ] with nonnegative coefficients,such that A(ζdN) = 0, for some d | N. Then,
A(X d) ≡ P(X d)Φp(XN/p) + Q(X d)Φq(XN/q) mod (XN − 1),
where P(X ),Q(X ) ∈ Z[X ] can be taken with nonnegativecoefficients.
The polynomial A(X d) is the mask polynomial of the multisetd · A.
Φp(XN/p) is the mask polynomial of the subgroup Np ZN . Its
cosets are called p-cycles.
The above Lemma shows that if A(ζN) = 0, then A is thedisjoint union of p- and q-cycles.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Vanishing sums of roots of unity
Lemma
Let rad(N) = pq and A(X ) ∈ Z[X ] with nonnegative coefficients,such that A(ζdN) = 0, for some d | N. Then,
A(X d) ≡ P(X d)Φp(XN/p) + Q(X d)Φq(XN/q) mod (XN − 1),
where P(X ),Q(X ) ∈ Z[X ] can be taken with nonnegativecoefficients.
The polynomial A(X d) is the mask polynomial of the multisetd · A.
Φp(XN/p) is the mask polynomial of the subgroup Np ZN . Its
cosets are called p-cycles.
The above Lemma shows that if A(ζN) = 0, then A is thedisjoint union of p- and q-cycles.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Vanishing sums of roots of unity
Lemma
Let rad(N) = pq and A(X ) ∈ Z[X ] with nonnegative coefficients,such that A(ζdN) = 0, for some d | N. Then,
A(X d) ≡ P(X d)Φp(XN/p) + Q(X d)Φq(XN/q) mod (XN − 1),
where P(X ),Q(X ) ∈ Z[X ] can be taken with nonnegativecoefficients.
The polynomial A(X d) is the mask polynomial of the multisetd · A.
Φp(XN/p) is the mask polynomial of the subgroup Np ZN . Its
cosets are called p-cycles.
The above Lemma shows that if A(ζN) = 0, then A is thedisjoint union of p- and q-cycles.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Remark
If A is the disjoint union of p-cycles only, then
A ∩
0, 1, . . . ,N
p− 1
and
1
pB0 mod p
is a spectral pair in ZN/p.
We reduce to the case where both A and B are nontrivial unions ofp- and q-cycles. This implies
A(ζp) = A(ζq) = B(ζp) = B(ζq) = 0.
As a consequence,
|Aj mod p| = |Bj mod p| =1
p|A|, |Ai mod q| = |Bi mod q| =
1
q|A|,
for all i , j .
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Remark
If A is the disjoint union of p-cycles only, then
A ∩
0, 1, . . . ,N
p− 1
and
1
pB0 mod p
is a spectral pair in ZN/p.
We reduce to the case where both A and B are nontrivial unions ofp- and q-cycles. This implies
A(ζp) = A(ζq) = B(ζp) = B(ζq) = 0.
As a consequence,
|Aj mod p| = |Bj mod p| =1
p|A|, |Ai mod q| = |Bi mod q| =
1
q|A|,
for all i , j .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Remark
If A is the disjoint union of p-cycles only, then
A ∩
0, 1, . . . ,N
p− 1
and
1
pB0 mod p
is a spectral pair in ZN/p.
We reduce to the case where both A and B are nontrivial unions ofp- and q-cycles. This implies
A(ζp) = A(ζq) = B(ζp) = B(ζq) = 0.
As a consequence,
|Aj mod p| = |Bj mod p| =1
p|A|, |Ai mod q| = |Bi mod q| =
1
q|A|,
for all i , j .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a primitive spectral pair in ZN , N = pnqm, such thatneither A nor B is a union of p-(or q-)cycles exclusively. Then,both A(X ) and B(X ) vanish at
ζN , ζpn , ζqm ,
ζp, ζq, ζpq,
ζpnq, ζpqm .
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Proposition
Let (A,B) be a primitive spectral pair in ZN , N = pnqm, such thatneither A nor B is a union of p-(or q-)cycles exclusively. Then,both A(X ) and B(X ) vanish at
ζN , ζpn , ζqm ,
ζp, ζq, ζpq,
ζpnq, ζpqm .
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Proposition
Let (A,B) be a primitive spectral pair in ZN , N = pnqm, such thatneither A nor B is a union of p-(or q-)cycles exclusively. Then,both A(X ) and B(X ) vanish at
ζN , ζpn , ζqm ,
ζp, ζq, ζpq,
ζpnq, ζpqm .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a primitive spectral pair in ZN , N = pnqm, such thatneither A nor B is a union of p-(or q-)cycles exclusively. Then,both A(X ) and B(X ) vanish at
ζN , ζpn , ζqm ,
ζp, ζq, ζpq,
ζpnq, ζpqm .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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A special case
Proposition
If N = pmqn and A ⊆ ZN is spectral satisfying
A(ζp) = A(ζp2) = · · · = A(ζpm) = 0
then A tiles ZN .
Sketch of proof.
By hypothesis, |Aj mod pm | = 1pm |A|.
Each Aj mod pm(X ) hasprecisely the same roots of the form ζqk with A(X ). So, A satisfies(T1). Next, if A(ζqk ) = 0, then for each 1 ≤ i ≤ m we have
A(ζpiqk ) =
pm−1∑j=0
Aj mod pm(ζpiqk ) =
pm−1∑j=0
ζ jpiqk
σ(ζ−jqkAj mod pm(ζqk )) = 0
for some σ ∈ Gal(Q(ζN)/Q), so it A satisfies (T2) as well.
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A special case
Proposition
If N = pmqn and A ⊆ ZN is spectral satisfying
A(ζp) = A(ζp2) = · · · = A(ζpm) = 0
then A tiles ZN .
Sketch of proof.
By hypothesis, |Aj mod pm | = 1pm |A|. Each Aj mod pm(X ) has
precisely the same roots of the form ζqk with A(X ). So, A satisfies(T1).
Next, if A(ζqk ) = 0, then for each 1 ≤ i ≤ m we have
A(ζpiqk ) =
pm−1∑j=0
Aj mod pm(ζpiqk ) =
pm−1∑j=0
ζ jpiqk
σ(ζ−jqkAj mod pm(ζqk )) = 0
for some σ ∈ Gal(Q(ζN)/Q), so it A satisfies (T2) as well.
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A special case
Proposition
If N = pmqn and A ⊆ ZN is spectral satisfying
A(ζp) = A(ζp2) = · · · = A(ζpm) = 0
then A tiles ZN .
Sketch of proof.
By hypothesis, |Aj mod pm | = 1pm |A|. Each Aj mod pm(X ) has
precisely the same roots of the form ζqk with A(X ). So, A satisfies(T1). Next, if A(ζqk ) = 0, then for each 1 ≤ i ≤ m we have
A(ζpiqk ) =
pm−1∑j=0
Aj mod pm(ζpiqk ) =
pm−1∑j=0
ζ jpiqk
σ(ζ−jqkAj mod pm(ζqk )) = 0
for some σ ∈ Gal(Q(ζN)/Q), so it A satisfies (T2) as well.
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A special case
Proposition
If N = pmqn and A ⊆ ZN is spectral satisfying
A(ζp) = A(ζp2) = · · · = A(ζpm) = 0
then A tiles ZN .
Sketch of proof.
By hypothesis, |Aj mod pm | = 1pm |A|. Each Aj mod pm(X ) has
precisely the same roots of the form ζqk with A(X ). So, A satisfies(T1). Next, if A(ζqk ) = 0, then for each 1 ≤ i ≤ m we have
A(ζpiqk ) =
pm−1∑j=0
Aj mod pm(ζpiqk ) =
pm−1∑j=0
ζ jpiqk
σ(ζ−jqkAj mod pm(ζqk )) = 0
for some σ ∈ Gal(Q(ζN)/Q), so it A satisfies (T2) as well.
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Theorem
Let A ⊆ ZN be spectral, with N = pnqm, m ≤ 2. Then A tiles ZN .
Proof.
Wlog, A and a spectrum B are both primitive and nontrivial unionsof p- and q-cycles, so using the above reductions we may assumeA(ζq) = A(ζq2) = 0, which by the previous Proposition yields thatA tiles ZN .
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Theorem
Let A ⊆ ZN be spectral, with N = pnqm, m ≤ 2. Then A tiles ZN .
Proof.
Wlog, A and a spectrum B are both primitive and nontrivial unionsof p- and q-cycles, so using the above reductions we may assumeA(ζq) = A(ζq2) = 0, which by the previous Proposition yields thatA tiles ZN .
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The absorption-equidistribution property
Definition
We say that a subset A ⊆ ZN satisfies theabsorption-equidistribution property, if for every d | N and p primesuch that pd | N, either every subset Aj mod d is equidistributedmodpd , that is
|Aj+kd mod pd | =1
p|Aj mod d |,∀k ∈ 0, 1, . . . , p − 1,
or every Aj mod d is absorbed modpd , i. e. there isk ∈ 0, 1, . . . , p − 1 such that
Aj mod d = Aj+kd mod pd .
Question
Is the absorption-equidistribution property equivalent to (T1) &(T2)?
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The absorption-equidistribution property
Definition
We say that a subset A ⊆ ZN satisfies theabsorption-equidistribution property, if for every d | N and p primesuch that pd | N, either every subset Aj mod d is equidistributedmodpd , that is
|Aj+kd mod pd | =1
p|Aj mod d |,∀k ∈ 0, 1, . . . , p − 1,
or every Aj mod d is absorbed modpd , i. e. there isk ∈ 0, 1, . . . , p − 1 such that
Aj mod d = Aj+kd mod pd .
Question
Is the absorption-equidistribution property equivalent to (T1) &(T2)?
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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The absorption-equidistribution property
Definition
We say that a subset A ⊆ ZN satisfies theabsorption-equidistribution property, if for every d | N and p primesuch that pd | N, either every subset Aj mod d is equidistributedmodpd , that is
|Aj+kd mod pd | =1
p|Aj mod d |,∀k ∈ 0, 1, . . . , p − 1,
or every Aj mod d is absorbed modpd , i. e. there isk ∈ 0, 1, . . . , p − 1 such that
Aj mod d = Aj+kd mod pd .
Question
Is the absorption-equidistribution property equivalent to (T1) &(T2)?
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ;
in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1,
orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ,
such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1).
Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let (A,B) be a spectral pair in ZN , N = pnqm, such that Aj mod pk
is absorbed modpk+1 for every j , some k < n. Then, there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN ; in addition, every Aj mod pk is equidistributed modpk+1, orequivalently,
A(ζpk+1) = 0.
Corollary
Let (A,B) be a spectral pair in ZN , N = pnqm. Then there areS ,T ⊆ ZN such that (A,B) = (A⊕ S ,B ⊕ T ) is a spectral pair inZN , such that for every k < n (resp. ` < m), there is j such thatAj mod pk (resp. Aj mod q`) is not absorbed modpk+1 (resp.
modq`+1). Such subsets will be called absorption-free.
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Remark
With the above Corollary, we may further reduce to spectral(A,B), where both A, B are absorption-free.
This is used to prove:
Theorem (M)
Let A ⊆ ZN be spectral, N = pnqm, satisfying (T1). Then, it alsosatisfies (T2), hence A tiles ZN .
Therefore it suffices to confirm (T1) for a spectral A ⊆ ZN .Actually, (T1) can be replaced by a weaker condition:
Definition
Let A ⊆ ZN . We say that A satisfies (wT1) if there is a primep | N, such that pk || |A|, where A(X ) has exactly k roots of theform ζpν .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Remark
With the above Corollary, we may further reduce to spectral(A,B), where both A, B are absorption-free.
This is used to prove:
Theorem (M)
Let A ⊆ ZN be spectral, N = pnqm, satisfying (T1). Then, it alsosatisfies (T2), hence A tiles ZN .
Therefore it suffices to confirm (T1) for a spectral A ⊆ ZN .Actually, (T1) can be replaced by a weaker condition:
Definition
Let A ⊆ ZN . We say that A satisfies (wT1) if there is a primep | N, such that pk || |A|, where A(X ) has exactly k roots of theform ζpν .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Remark
With the above Corollary, we may further reduce to spectral(A,B), where both A, B are absorption-free.
This is used to prove:
Theorem (M)
Let A ⊆ ZN be spectral, N = pnqm, satisfying (T1). Then, it alsosatisfies (T2), hence A tiles ZN .
Therefore it suffices to confirm (T1) for a spectral A ⊆ ZN .Actually, (T1) can be replaced by a weaker condition:
Definition
Let A ⊆ ZN . We say that A satisfies (wT1) if there is a primep | N, such that pk || |A|, where A(X ) has exactly k roots of theform ζpν .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let A ⊆ ZN be spectral, N = pnqm. If pn | |A|, then A satisfies(wT1).
Theorem
Let A ⊆ ZN be spectral, N = pnq3. Then A tiles ZN .
Sketch of proof.
Wlog, A is primitive, nontrivial union of p- and q-cycles,absorption-free set. We know
A(ζq) = A(ζq3) = A(ζp) = A(ζpq) = A(ζpq3) = 0.
If A(ζq2) = 0 then A tiles ZN , so we assume A(ζq2) 6= 0. The factthat A is absorption-free forces A(ζpq2) = 0, which implies thateach Aj mod q is either absorbed or equidistributed modq2, bothphenomena appearing. So, there is j such that Aj mod q3 = 1
q3 |A|,so A satisfies (wT1), and it tiles ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let A ⊆ ZN be spectral, N = pnqm. If pn | |A|, then A satisfies(wT1).
Theorem
Let A ⊆ ZN be spectral, N = pnq3. Then A tiles ZN .
Sketch of proof.
Wlog, A is primitive, nontrivial union of p- and q-cycles,absorption-free set. We know
A(ζq) = A(ζq3) = A(ζp) = A(ζpq) = A(ζpq3) = 0.
If A(ζq2) = 0 then A tiles ZN , so we assume A(ζq2) 6= 0. The factthat A is absorption-free forces A(ζpq2) = 0, which implies thateach Aj mod q is either absorbed or equidistributed modq2, bothphenomena appearing. So, there is j such that Aj mod q3 = 1
q3 |A|,so A satisfies (wT1), and it tiles ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let A ⊆ ZN be spectral, N = pnqm. If pn | |A|, then A satisfies(wT1).
Theorem
Let A ⊆ ZN be spectral, N = pnq3. Then A tiles ZN .
Sketch of proof.
Wlog, A is primitive, nontrivial union of p- and q-cycles,absorption-free set. We know
A(ζq) = A(ζq3) = A(ζp) = A(ζpq) = A(ζpq3) = 0.
If A(ζq2) = 0 then A tiles ZN , so we assume A(ζq2) 6= 0. The factthat A is absorption-free forces A(ζpq2) = 0, which implies thateach Aj mod q is either absorbed or equidistributed modq2, bothphenomena appearing.
So, there is j such that Aj mod q3 = 1q3 |A|,
so A satisfies (wT1), and it tiles ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let A ⊆ ZN be spectral, N = pnqm. If pn | |A|, then A satisfies(wT1).
Theorem
Let A ⊆ ZN be spectral, N = pnq3. Then A tiles ZN .
Sketch of proof.
Wlog, A is primitive, nontrivial union of p- and q-cycles,absorption-free set. We know
A(ζq) = A(ζq3) = A(ζp) = A(ζpq) = A(ζpq3) = 0.
If A(ζq2) = 0 then A tiles ZN , so we assume A(ζq2) 6= 0. The factthat A is absorption-free forces A(ζpq2) = 0, which implies thateach Aj mod q is either absorbed or equidistributed modq2, bothphenomena appearing. So, there is j such that Aj mod q3 = 1
q3 |A|,so A satisfies (wT1), and it tiles ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Proposition
Let A ⊆ ZN be spectral, N = pnqm. If pn | |A|, then A satisfies(wT1).
Theorem
Let A ⊆ ZN be spectral, N = pnq3. Then A tiles ZN .
Sketch of proof.
Wlog, A is primitive, nontrivial union of p- and q-cycles,absorption-free set. We know
A(ζq) = A(ζq3) = A(ζp) = A(ζpq) = A(ζpq3) = 0.
If A(ζq2) = 0 then A tiles ZN , so we assume A(ζq2) 6= 0. The factthat A is absorption-free forces A(ζpq2) = 0, which implies thateach Aj mod q is either absorbed or equidistributed modq2, bothphenomena appearing. So, there is j such that Aj mod q3 = 1
q3 |A|,so A satisfies (wT1), and it tiles ZN .
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups
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Thank you
R. D. Malikiosis Fuglede’s spectral set conjecture on cyclic groups