Tight Wavelet Frame Sets in Finite Vector Spaces
aa r X i v : . [ m a t h . F A ] M a r Tight Wavelet Frame Sets in Finite Vector Spaces
Alex Iosevich, Chun-Kit Lai, Azita MayeliMarch 21, 2017
Abstract
Let q ≥ F dq , d ≥
1, be the vector space over the cyclic space F q .The purpose of this paper is two-fold. First, we obtain sufficient conditions on E ⊂ F dq suchthat the inverse Fourier transform of 1 E generates a tight wavelet frame in L ( F dq ). We callthese sets (tight) wavelet frame sets. The conditions are given in terms of multiplicative andtranslational tilings, which is analogous with Theorem 1.1 ([20]) by Wang in the setting offinite fields. In the second part of the paper, we exhibit a constructive method for obtainingtight wavelet frame sets in F dq , d ≥ q an odd prime and q ≡ Key words and phrases.
Prime fields; finite vector space; wavelet frames; wavelet frame sets;tight frames; translational tiling; multiplicative tiling; rotational tiling; spectral set; spectrum;spectral pair
A countable subset { x k } k ∈ I of a Hilbert space H is said to be a frame if there exists two positiveconstants A ≤ B such that for any x ∈ H A k x k H ≤ X k ∈ I |h x, x k i H | ≤ B k x k H . The positive constants A and B are called lower and upper frame bounds, respectively. Theframe is called a tight frame when we can take A = B and it is called a Parseval frame if A = B = 1. One of most significant features of the frames that makes them practical and usefulis their redundancy which has an important role, for example, in robustness. The frames alsoallow a localized representation of elements in the Hilbert space and they have been used fora number of years by engineers and applied mathematicians for purposes of signal processingand data compression. The notion of frames was first introduced by Duffin and Schaeffer [6].Amongst the frames, tight frames play a fundamental role in the applications of frames due totheir numerical stability. In this paper we aim to construct tight frames on the finite vectorspaces over the finite fields which arise from dilation and translation of a function whose Fouriertransform is characteristic function of a non-empty set.In the classical setting, a function ψ ∈ L ( R d ) is said to generate a orthonormal wavelet basis (resp. wavelet frame ) if there is a set of d × d matrices D ⊂ GL ( d, R ) and a subset T ⊂ R d such1that the family {| det( D ) | / ψ ( Dx − t ) : D ∈ D , t ∈ T } , (1.1)forms an orthonormal basis (resp. frame) for L ( R d ). Then we say ψ is an orthonormal wavelet(resp. frame wavelet) and every element | det( D ) | / ψ ( Dx − t ) is a dilation and translation copyof ψ with respect to the invertible matrix D and translation t , respectively. The structuresof D and T and associated to which there exists orthonormal and frame wavelets for L ( R d )have been studied by many authors, for example [9, 5, 19, 20]. See also [12] for an alternativeperspective on wavelets in vector spaces over finite fields and connections with combinatorialproblems.Let q be an odd prime and F q be the prime field with q elements. Then F dq is the vector space ofdimension d over the finite field F q . In analogue to R d , the purpose of this paper is to study tightwavelet frames on F dq and its subspace for d ≥
1. Let Aut( F dq ) be the set of all automorphismson F dq . Following the same spirit of the frame wavelets on R d , we say that a function ψ : F dq → C is a wavelet or a frame wavelet for L ( F dq ) if there exists a set of automorphisms A ⊂ A ut ( F dq )and a subset Λ ⊂ F dq such that the family { ψ ( ax − λ ) : a ∈ A , λ ∈ Λ } . (1.2)is an orthonormal basis (resp. frame) for L ( F dq ). Note that in the continuous case, for thematrix D , the factor ∆ D := | det( D ) | / makes the dilation map f → | det( D ) | / f ( Dx )an isometry. In the discrete case F dq the Haar measure on F dq is discrete and invariant under thedilation, therefore the dilation factor ∆ a is 1.A common way to construct a wavelet frame on R d is to choose a function whose Fouriertransform is the indicator of a measurable set, and then consider the system (1.1) of dilationsand translations of the function. This leads to the traditional definition of frame wavelet sets asfollows: A set Ω ⊂ R d is called a frame wavelet set with respect to D and T if for the function ψ with ˆ ψ = 1 Ω , the system (1.1) is a frame in L ( R d ). If the system is an orthonormal basis for L ( R d ), then the function ψ is called minimally supported frequency wavelet (MSF wavelet) andΩ is called a wavelet set . The wavelet sets and minimally supported frequency wavelets wereintroduced in [19] and studied exclusively, for example, in [10, 11] and by many other authors.The existence of wavelet sets in R d for any expansive matrix was given in [5]. The wavelettheory and wavelet sets are studied in a constructive way in locally compact abelian groups withcompact open subgroups, such as p -adic groups Q p , in [2, 3]A well-known example of a wavelet set is the Shannon set given byΩ = [ − π, − π ] ∪ [ π, π ) . The orthonormal wavelet for Ω is then given by ψ ( x ) = 2 sinc(2 x − − sinc( x ) with b ψ = 1 Ω ,where 1 Ω is the indicator of set Ω . For more examples and constructions of wavelet sets in R d we invite the reader to see [1, 3, 4, 15, 16].In [20], Wang tied the existence of wavelet sets with the notion of spectral sets. We say that ameasurable set of positive measure Ω is a spectral set if there exists a countable set Γ such thatthe collection of exponentials { e πiγ · x : γ ∈ Γ } forms an orthonormal basis for L (Ω). In thiscase we say that (Ω , Γ) forms a spectral pair . Spectral sets were first introduced by Fuglede [8]and he proposed an infamous conjecture asserting that spectral sets are exactly translationaltiles on R d . However, this conjecture was proven to be false in its full generality by Tao [18].Nowadays, the exact relationship between spectral sets and translational tiles are mysterious.We refer to [7, 13] for some recent progress.Wang proved the following result in the classical setting which characterizes the wavelet sets bymultiplicative and translational tiling. Let D T denote the transpose of matrix D , and ˇ g denotethe inverse Fourier transform of g . Theorem 1.1 (Theorem 1.1, [20]) . Let
D ⊂ GL ( d, R ) and Γ ⊂ R d . Let Ω ⊂ R d with positiveand finite Lebesgue measure. If { D t (Ω) : D ∈ D} is a tiling of R d and (Ω , Γ) is a spectral pair,then ψ = ˇ1 Ω is a wavelet with respect to the dilation set D and the translation set Γ . Conversely,if ψ = ˇ1 Ω is a wavelet with respect to D and Γ and ∈ Γ , then { D T (Ω) : D ∈ D} is a tiling of R d and (Ω , Γ) is a spectral pair. Inspired by the result of the theorem, it is natural for us to ask for what degree one can extendthe notion and concept of wavelet sets and multiplicative tiling in F dq . In what follows, weshall study this. For the multiplicative tiling purpose, we have to remove the origin and let Y := F dq \ { } . Definition 1.2 (Multiplicative and translational tiling) . Let E be a subset of F dq . We say E is a multiplicative tiling set for F dq if there is a set of automorphisms A in Aut( F dq ) such that E tiles Y multiplicatively by A , i.e., Y = [ α ∈A α ( E ) (disjoint union) . As a result, a multiplicative tiling set does not include the origin ~
0. This is a natural requirementsince α ( ~
0) = ~ α .We say that a subset F ⊆ F dq is a translational tiling set for F dq if there exists Λ ⊆ F dq such that F dq = [ λ ∈ Λ ( F + λ ) (disjoint union) . We say a set E has a spectrum L if the characters { χ l } l ∈ L is an orthonormal basis for L ( E ).In this case, we say E is a spectral and ( E, L ) is a spectral pair.Our first result shows that there exists no wavelet sets in the traditional sense for L ( F dq ).More precisely, there is no Parseval frame (thus no orthonormal basis) of type (1.2) for L ( F dq )generated by any function of type ψ := ˇ1 E (Theorem 2.3). However, later in this paper weprove the existence of subsets E in F dq for which ψ := ˇ1 E generates a tight wavelet frame for asubspace of L ( F dq ). We shall call these sets tight wavelet frame sets . We will then present anexplicit construction of a class of wavelet frame sets when q ≡ F dq and we study necessary and sufficient conditions for a set E such that E is atight wavelet frame set for a subspace of L ( F dq ). These results are collected in Theorems 2.9and 2.10. In this section we also provide a counter example where the disjointness of the setsdoes not necessarily hold for tight wavelet frame sets. In Section 3, we will show the existenceof a multiplicative tiling set in F dq for q prime and q ≡ F dq ,when d = 2, q prime and q ≡ E is non trivial, i.e., 1 < ♯E < q d . Here we first review the Fourier transform on F dq . If f is a function on F dq , then the Fouriercoefficients, ˆ f ( ξ ), of f are given by b f ( ξ ) = q − d X m ∈ F dq a m χ m ( ξ ) ξ ∈ F dq where χ m ( ξ ) = e πi m · ξq is the character and m · ξ is the usual inner product and a m = f ( m ).The function ˆ f is called the Fourier transform of f . With the above notations, f ( x ) = X m ∈ F dq c m χ m ( x ) , where c m = ˆ f ( m ). By the above definition, the Fourier transform F : L ( F dq ) → L ( F dq ) given by f → ˆ f is a unitary map. For any g ∈ L ( F dq ), we shall denote by ˇ g the inverse Fourier transformof g .Let Aut( F dq ) be the set of all automorphisms on F dq . Let ψ be a function defined on F dq . For a ∈ A ut ( F dq ) and t ∈ F dq define the associated dilation and translation operators δ a and τ t by δ a ψ ( x ) = ψ ( ax ) , and τ t ψ ( x ) = ψ ( x − t ) , respectively. These operators are unitary . Lemma 2.1.
Let ψ : F dq → C . Then for given automorphism a ∈ Aut ( F dq ) and t ∈ F dq we have \ δ a τ t ψ ( m ) = χ a − t ( m ) ˆ ψ ( a ∗ m ) , (2.1) where a ∗ = ( a t ) − = ( a − ) t is the inverse transpose of a . If ˆ ψ = 1 E , then \ δ a τ t ψ ( m ) = χ a − t ( m )1 a t ( E ) ( m ) . (2.2) Proof.
Let t ∈ F dq and a ∈ Aut( F dq ). By applying the Fourier transform and using a change ofvariable in the definition of the Fourier transform, for all m ∈ F dp we have d τ t ψ ( m ) = q − d X n ∈ F dq ψ ( n − t ) χ m ( n ) (2.3)= q − d X n ∈ F dq ψ ( n ) χ m ( n + t )= χ m ( t ) q − d X n ∈ F dq ψ ( n ) χ m ( n ) = χ m ( t ) ˆ ψ ( m )= χ t ( m ) ˆ ψ ( m ) , and d δ a ψ ( m ) = q − d X n ∈ Z dp ψ ( an ) χ m ( n ) (2.4)= q − d X n ∈ Z dp ψ ( n ) χ m ( a − n )= q − d X n ∈ Z dp ψ ( n ) χ a ∗ m ( n ) (2.5)= ˆ ψ ( a ∗ m ) . (2.6)Now, a combination of (2.4) and (2.3) yields the assertion of the lemma: \ δ a τ t ψ ( m ) = d τ t ψ ( a ∗ m ) = χ a ∗ m ( t ) ˆ ψ ( a ∗ m ) = χ a − t ( m ) ˆ ψ ( a ∗ m ) . (2.7)By the equality 1 E ( a ∗ m ) = 1 a t ( E ) ( m ), the second part of the lemma also holds true.The proof of the following result is straightforward using the Fourier transform. Lemma 2.2.
Let
A ⊆
Aut ( F q ) and T ⊆ F dq , and the family { δ a τ t ψ : a ∈ A , t ∈ T } is an orthonormal basis for L ( F dq ) if and only if the family { χ a − t ( m ) ˆ ψ ( a ∗ m ) : t ∈ T, a ∈ A} is an orthonormal basis for L ( F dq ) . Here, m is the generic variable. The next result proves the existence of no Parseval wavelet frame for L ( F dq ) generated by ψ := ˇ1 E . Theorem 2.3.
There exists no non-empty subset E ( F dq such that ψ := ˇ1 E , the inverse Fouriertransform of the indicator function E , is the generator of a Parseval wavelet frame for L ( F dq ) .Proof. We shall prove this theorem by a contradiction argument. Assume that for a subset E there is an automorphism set A ⊂
Aut( F dq ) and a subset Λ of F dq such that the set { δ a τ λ ˇ1 E : a ∈ A , λ ∈ Λ } is a Parseval frame for L ( F dq ). By Lemma 2.2, this is equivalent to say that the family { χ a ∗ m ( λ )1 a t ( E ) ( m ) : λ ∈ Λ , a ∈ A} = { χ a − λ ( m )1 a t ( E ) ( m ) : λ ∈ Λ , a ∈ A} (2.8)is a Parseval frame for L ( c F dq ) = L ( F dq ). (Here, a t is the transpose of a .) Let ˆ g = 1 { ~ } ∈ L ( c F dq )be the indicator function for the set { ~ } . Then1 = k ˆ g k = X a ∈A ,λ ∈ Λ |h g, δ a τ λ ψ i| (2.9)= X a ∈A ,λ ∈ Λ |h ˆ g, \ δ a τ λ ψ i| = X a ∈A ,λ ∈ Λ | X m ∈ F dq ˆ g ( m )1 a t ( E ) ( m ) χ a ∗ m ( λ ) | = X a ∈A ,λ ∈ Λ | a t ( E ) (0) | = ♯ (Λ) X a ∈A | a t ( E ) (0) | . Here, we shall consider two cases: If 0 ∈ E , then 0 ∈ a t ( E ) for all a ∈ A . Thus the abovecalculation implies that 1 = ♯ ( A ) ♯ (Λ) . This means that the wavelet system must have onlyone element. Let A = { a } and Λ = { λ } . Then all the vectors in L ( F dq ) must be a constantmultiple of δ a τ λ ψ . We show that this is not the case since E = F dq . Let f = 0 in L ( F dq ) suchthat supp ( ˆ f ) ∩ a t ( E ) = ∅ . Such function exists since E is not the whole F dq . Then there is nocosntant c such that f = cδ a τ λ ψ . This shows that the Parseval wavelet frame can not have onlyone element when E = F dq , thus ♯ ( A ) ♯ (Λ) > E . By the equations in (2.9) we obtain 1 = 0 which isimpossible. This completes the proof our assertion.As a result of Theorem 2.3, we have the following corollary. Corollary 2.4.
There is no orthonormal basis of form { δ a τ λ ψ } a ∈A ,λ ∈ Λ for L ( F dq ) where ˆ ψ = 1 E and E ( F dq .Remark. Notice when E = F dq , by a similar calculation (2.9) for ˆ g = 1 F dq , we can conclude thatthe translation set Λ must contains 0 and ♯ ( A ) = 1. Then by the divisibility we must haveΛ = F dq . This implies that the set { τ λ ψ : λ ∈ F dq } , with ˆ ψ = 1 F dq , forms an orthonormal basis for L ( F dq ). But this is already well-known by the Fourier transform. Therefore, it is reasonable toassume in Theorem 2.3 that E = F dq .As we observed above, Corollary 2.4 implies the existence of no orthonormal basis of form (2.8)for L ( F dq ). However, in the following theorem, we prove that if we choose E appropriately in F dq , then the family (2.8) for E ∗ = E \ { } is a tight wavelet frame for L (( F dq ) ∗ ).For the rest, we use the notation Y := ( F dq ) ∗ . Definition 2.5.
Given E and L subsets of F dq , we say ( E, L ) is a (tight) frame spectral pair ifthe set of “exponentials” { χ l : l ∈ L } is a (tight) frame for L ( E ). Theorem 2.6.
Let E and L be subsets of F dq and ( E, L ) is a spectral pair. Assume that E ∗ is amultiplicative tiling set with respect to a set of automorphisms A ⊂
Aut ( F dq ) . Then the followinghold true:(1) ( E ∗ , L ) is tight frame spectral pair with the frame bound ♯ ( E ) .(2) ∀ a ∈ A , ( a ( E ∗ ) , ( a − ) t ( L )) is tight frame spectral pair with the frame bound ♯ ( E ) .(3) The family { a ( E ∗ ) χ ( a − ) t ( l ) : l ∈ L, a ∈ A} is a tight frame for L ( Y ) with the frame bound ♯ ( E ) , where Y = ( F dq ) ∗ .(4) If E , then { ( ♯E ) − / a ( E ) χ ( a − ) t ( l ) : l ∈ L, a ∈ A} is an orthonormal basis for L ( Y ) .Proof. Assume that E has a spectrum L . Then { ♯ ( E ) − / χ l : l ∈ L } is an orthonormal basisfor L ( E ). To proof (1), note that the map f → f E ∗ is a projection of L ( E ) onto L ( E ∗ ),thus the image of the orthonormal basis { ♯ ( E ) − / χ l : l ∈ L } by this map is a Parseval framefor L ( E ∗ ). This proves the statement (1). To prove (2), we instead prove the following: Let F ⊂ F dq and { χ l : l ∈ L } be a frame for L ( F ) with the frame bounds 0 < A ≤ B < ∞ . Then { χ ( a − ) t ( l ) : l ∈ L } is a frame for L ( a ( F )) with the unified frame bounds, A and B . To provethis, define the map T a : L ( F ) → L ( a ( F )) by f → f ◦ a − . The image of { χ l : l ∈ L } under T a is { a ( F ) χ ( a − ) t ( l ) : l ∈ L } and the map is unitary. Therefore { a ( F ) χ ( a − ) t ( l ) : l ∈ L } forms aframe for L ( a ( F )) with the same frame bounds.To prove (3), note that by the assumption on the multiplicative tiling property of E ∗ we have L ( Y ) = ⊕ a ∈A L ( a ( E ∗ )) . To complete (3), we shall prove the following, instead.Let X be measurable set and A be an index set such that X = ∪ a ∈ A X a (disjoint). Assume thatfor all a ∈ A , L ( X a ) has a tight frame { f n,a : n ∈ I a } with frame bound C . We claim that thefamily { f n,a : a ∈ A, n ∈ I a } is a tight frame for L ( X ) with the frame bound C . To provethat, let g ∈ L ( X ). Sine { X a : a ∈ A } is a partition for X , then g = ⊕ a ∈ A g a , g a := g X a , andwe have k g k = X a ∈ A k g a k L ( X a ) (2.10)= X a ∈ A C − X n ∈ I |h g a , f n,a i L ( X a ) | ! = C − X a ∈ A,n ∈ I a |h g, X a f n,a i L ( X ) | . This completes the proof of the assertion, thus (3). For (4), note that E = E ∗ when 0 E .Then the proof can be obtained directly from the fact that any Parseval frame with normalizedframe elements is an orthonormal basis.We state the following result from [14] which characterizes tiling set and spectral sets in F q andproves the Fuglede conjecture for F q . Theorem 2.7 (Fuglede conjecture for F q ) . A subset ∅ 6 = E of F q tiles F q with its translationsif and only if there is a set L ⊆ F q such that ( E, L ) is a spectral pair. As the corollary of Theorems 2.7 and Theorem 2.6 (3) we have the following result.
Corollary 2.8.
Assume that E is a translation tiling for F q and E ∗ is a multiplicative tilingwith respect to the automorphisms A ⊂
Aut ( F q ) . Then there is a set L in F q such that the family { ( ♯E ) − / χ a − ( l ) a ( E ∗ ) : l ∈ L, a ∈ A} is a Parseval frame for L ( Y ) . The system is an orthonormal basis if E . Given a subset F of F dq we say a function f ∈ L ( F dq ) belongs to P W F (The Paley-Wiener space)if its Fourier transform ˆ f has support in F . Then P W F := { f ∈ L ( F dq ) : ˆ f ( m ) = 0 ∀ m F } . (2.11)Note that if F = ( F dq ) ∗ , then P W Y contains the all functions f ∈ L ( F dq ) for which P m ∈ F dq f ( m ) =0.Our next result is analogous with Theorem 1.1. in [20] in F dq . For A ⊂
Aut( F dq ), we denote by A t the set of transpose of all matrices in A . Theorem 2.9.
Assume that E ⊆ F dq has a spectrum L and ∅ 6 = E ∗ is multiplicative tiling withrespect to A t for some A ⊆ A ut ( F dq ) . Take ψ := ( ♯E ) − / ˇ1 E ∗ and Y := ( F dq ) ∗ . Then the family W := { δ a τ l ψ : a ∈ A , l ∈ L } is an Parseval frame for P W Y . The system W is an orthonormalbasis if E = E ∗ . Conversely, if ψ = ( ♯E ) − / ˇ1 E is a Parseval wavelet for P W Y with respect todilation set A and translation set L , then Y = ∪ a ∈A a t ( E ∗ ) . The sets a t ( E ∗ ) are disjoint, then ( E ∗ , L ) is a tight frame spectral pair.Proof. Assume that (
E, L ) is a spectral pair and E ∗ is a multipilicative tiling with respect to A t for some A ⊆ A ut ( F dq ). By an application of the Fourier transform, to prove that W is aParseval frame for P W Y is equivalent to say that c W := { ( ♯E ) − / χ a − ( l )1 at ( E ∗ ) : l ∈ L, a ∈ A} isa Parseval frame for L ( Y ). By Theorem 2.6 (3) we know that c W is a Parseval frame for L ( Y ).Therefore by inverse of the Fourier transform, which is unitary, we conclude that the waveletsystem W is a Parseval frame for P W Y and this completes the proof of ” ⇐ ”.Now assume that W is a Parseval frame for L ( Y ). To prove the union of the sets a t ( E ∗ ) , a ∈ A ,covers F , we use a contradiction argument. Assume that there is a non empty set M ( Y suchthat M ∩ a t ( E ∗ ) = ∅ for all a ∈ A . Take f := 1 M . Then k f k = ( ♯E ) − / X a ∈A ,l ∈ L |h f, a t ( E ∗ ) χ a − ( l ) i| (2.12)Since M ∩ a t ( E ∗ ) = ∅ for all a ∈ A , then the right side in the preceding equation must bezero, while the left side is ♯ ( M ). This is a contradiction to our assumption that M is non-empty, therefore Y = ∪ a ∈A a t ( E ∗ ). To show that the pair ( E ∗ , L ) is a tight frame spectral pair,note that by the hypothesis on the disjointness of the sets a t ( E ∗ ), for any a ∈ A the system { χ a − ( l ) : l ∈ L } is a tight frame for L ( a t ( E ∗ )) with the frame bound ( ♯E ) − / . The dilationoperator T a : L ( a t ( E ∗ )) → L ( E ∗ ) given by f → f ◦ a t , f ◦ a t ( x ) = f ( a t ( x )), x ∈ E ∗ , is unitary,therefore { χ l : l ∈ L } , image of { χ a − ( l ) : l ∈ L } under T a , is a tight frame for L ( E ∗ ) with theunified frame bound and we are done.The disjointness of the sets a t ( E ∗ ) , a ∈ A , in Theorem 2.9 can be obtained under some additionalassumptions on W . Theorem 2.10.
If the system W is an orthogonal basis for P W Y and ∈ L , then the sets a t ( E ∗ ) , a ∈ A , are mutual disjoint and ( E ∗ , L ) is a spectral pair:Proof. Now assume that W is an orthogonal basis and 0 ∈ L . Take l = 0. Then the functions inthe family { a t ( E ∗ ) : a ∈ A} are orthogonal and ♯ ( a t ( E ∗ ) ∩ a ( E ∗ )) = 0 for any distinct a and a . From the other side, the exponentials { χ a − ( l ) : l ∈ L } are orthogonal basis for L ( a t ( E ∗ )).By a similar argument as above, we conclude that the family { χ l : l ∈ L } is also an orthogonalbasis for L ( E ∗ ). This completes the proof of the theorem.Question: Must 0 ∈ E c when the system is an orthogonal system?We conclude this section with an example of a tight wavelet frame associated to a set E andautomorphisms A , where the sets a t ( E ), a ∈ A , are not necessarily disjoint. Let W be a tightwavelet frame for P W Y with frame bound A , where ˆ ψ = 1 E . Take W = W ∪ W . Then W isa tight frame for P W Y with the frame bound A/ a t ( E ∗ ) are not disjoint. F dq Let q be an odd prime and d ≥
1. In this section we shall prove the existence of non-trivial (non-trivial here simply means that the set E is neither the whole space nor one point) multiplicativetiling sets in the finite vector space F dq when q ≡ d = 1, non-trivial multiplicative tiling set on F q , q odd, exists. Notice that F q can beidentified as { − ( q − , ..., − , , , ..., q − } and q − α and α to be α ( x ) = x and α ( x ) = − x . Take E = { , ..., ( q − / } . Then we immediatelysee that F q \ { } = α ( E ) ˙ ∪ α ( E ) . The union is disjoint and this yields naturally a non-trivial multiplicative tiling. If d >
1, it isnot immediately clear that why non-trivial multiplicative tiling sets exist. We first prove thisfor d = 2.Let us first consider the problem of multiplicative tiles on R and gain some motivations. Indeed,if we define R θ be the rotation matrix of angle θ and E p be the sector without the origin withaperture 2 π/p , then R \ { } is naturally partitioned into R \ { } = p − [ k =0 R πk/p ( E p ) . We can make the set compact by considering annulus of sectors with inner and outer radii equalto 1 and 2, respectively, and taking also dilation matrices into account. However, sectors cannever be a translational tile and hence we cannot produce wavelet sets on R using sectors.Nonetheless, we will see our construction of wavelet sets are produced by rotation on the finitefield and it can also be a translational tile on F q .For r ∈ F q , we consider the circle S r on F q with radius r as follows: S r := (cid:26)(cid:18) xy (cid:19) ∈ F q : x + y = r. (cid:27) We recall that a is a quadratic residue (mod q ) if x ≡ a (mod q ) has a solution in F p , otherwise,it is called a quadratic non-residue . On F q , there exist exactly ( q − / q − / Lemma 3.1.
Let q ≡ (mod 4). Then S r = (cid:26) , if r = 0 ; q + 1 , if r = 0 Proof.
By Theorem 1.2 in [17], every quadratic residue [non-residue] can be written as a sum oftwo quadratic residues [non-residues] in exactly d q − d q ways,where d q = q + 14 , when q ≡ mod r is a quadratic residue mod q . Then each sum of r as quadratic residues a + a induces four points in S r . Indeed, there are a = x has two solutions x , x and a = x hastwo solutions y , y . Thus there are four distinct points ( x , y ), ( x , y ), ( x , y ) and ( x , y ).1Furthermore, as r = x also has two solutions z , z . It induces 4 more solutions ( z , z , , z ) and (0 , z ) on the axes. Hence, by the theorem, when q ≡ S r = 4( d q −
1) + 4 = 4 d q = q + 1 . (3.1)Suppose that r is a quadratic non-residue of q . Then r can be written as exactly d q ways assum of quadratic residues. As each sum induces 4 pairs, S r = 4 d q which is the same answeras in (3.1).Finally, it follows directly that S = q − q − X r =1 S r = q − ( q − d q ) = 1 , as required.In the following lemma we prove that there exists an orthogonal matrix such that the multipli-cation of its exponents with a vector in a circle S r , r = 0, generates the circle. Lemma 3.2.
Suppose that q ≡ (mod 4). Then there exists an orthogonal matrix R = (cid:18) a − bb a (cid:19) such that a + b = 1 (mod q ), R q +1 = I , and for any e ∈ S r , R e , R e , ..., R S r e generates S r for all r = 0 .Proof. Since q ≡ − F q . In particular, we define i to bethe imaginary solution of i = − (cid:18) xy (cid:19) in F q as x + yi . Inthis sequel, F q is isomorphic to the finite field of q elements, denoted as F q . Note that themultiplicative group F × q is a cyclic group and the circle S = { x + yi : x + y = 1 } is a subgroup of F × q . As S = q + 1, there exists a + bi ∈ S such that S = { , a + bi, ( a + bi ) , ..., ( a + bi ) q } . In other words, a + bi generates the group S .Then it follows that for any e ∈ S r we can write e = c + di , and we have S r = { c + di, ( a + bi )( c + di ) , ..., ( a + bi ) q ( c + di ) } . Define the matrix R by (cid:18) a − bb a (cid:19) . Observe that R e = ( ac − bd, bc + ad ) T in F q is equal to( ac − bd ) + ( ad + bc ) i = ( a + bi )( c + di ) in F q . Then we have S r = { e , R e , ..., R q e } . This completes the proof of the lemma.2Our next result proves the existence of non-trivial multiplicative tiling sets F dq for d > Theorem 3.3.
There exists multiplicative tiling set E in F dq for q ≡ (mod 4).Proof. We first consider d = 2. Take the automorphisms to be { I, R, R , ..., R q } where R isdefined in Lemma 3.2. Define the set E by taking one point from each S r , for r = 0. ThenLemma 3.2 shows that F q \ { } = E ∪ R ( E ) ∪ ... ∪ R q ( E ) and the unions are disjoint since each R j ( E ) intersects S r exactly once.If d >
2, we may take e E := E × F d − q and e R := (cid:18) R OO I (cid:19) , where I is ( d − × ( d −
2) identitymatrix. Then e E is a multiplicative tiling set associated to the automorphisms { e R j : 1 ≤ j ≤ q + 1 } .We will call the multiplicative tiling in Proposition 3.3 rotational tiling . Therefore, any rotationaltiling is a set of q − As we will see in this section, the construction of tight wavelet frame sets on F q requires us tofind a set E such that 0 ∈ E , E ∗ is a multiplicative tiling and E tiles F q by translations. Herewe consider rotational tilings. We recall the following characterization of translational tiles on F p . Theorem 4.1. [14] Let E be a set that tiles F q by translation. Then ♯E = 1 , q or q and E isa graph if ♯E = q , i.e. E = { x e + f ( x ) e : x ∈ F q } for some basis e , e in F q and function f : F q → F q . Note that, by our construction in Proposition 3.3, a rotational tiling set has q − F q that are simultaneously rotationaltiling sets. We will provide a systematic way to construct such sets as we prove Theorem 4.4.First we need some key lemmas. Lemma 4.2.
There exists k ∈ F q , < k ≤ q − such that k is a quadratic non-residue.Proof. Assume that such k dose not exists. This means that 1 + k are quadratic residues forall 0 < k ≤ q − . We note that all 1 + k are in distinct residue classes (mod q ), otherwise1 + k ≡ k ′ would imply k = k ′ or k = − k ′ ≡ q − k ′ (mod q ). The latter is not possible since0 < k, k ′ ≤ q − . Since ♯QR = q − and 1 + k are distinct for different k , then we must have QR = { k : 0 < k ≤ q − } . However, we also know that 1 ∈ QR . Then for some k we have1 + k ≡ k = 0 or k = q , which contradicts the assumption.3As an example for k , if q = 7 ,
19, or 23, then k equals to 1 ,
1, or 2, respectively. The existenceof k in Lemma 4.2 allows us to represent the quadratic non-residue numbers as follows. Lemma 4.3.
For the k defined in Lemma 4.2,QNR = { (1 + k ) x : ( q + 1) / ≤ x ≤ q − } , and QR = { x : 0 ≤ x ≤ ( q − / } andProof. There is nothing to prove about the statement about QR. For the statement about
QN R ,due to the multiplicative property of Legendre symbol we have (cid:18) (1 + k ) x q (cid:19) = (cid:18) k q (cid:19) (cid:18) x q (cid:19) = ( − − a/q ) ≡ a q − = 1 if a isa quadratic residue and ( a/q ) = − a is a quadratic non-residue). Hence, all (1 + k ) x arequadratic non-residue. Moreover, they are all distinct since if (1 + k ) x = (1 + k ) y and x = y ,then x = q − y , which means x, y can’t be in { ( q + 1) / , ..., q − } at the same time. Hence,the set { (1 + k ) x : ( q + 1) / ≤ x ≤ p − } contains all ( q − / Theorem 4.4.
Assume that q is an odd prime congruent to (mod 4). Then there exists asubset E of F dq such that E is a translational tiling, has a spectrum and E ∗ is a multiplicativetiling in F dq .Proof. We first prove the case d = 2. By Theorem 4.1 and Theorem 3.3, it is sufficient toconstruct a set E in F q for q ≡ E = { ( x, f ( x )) : x ∈ F q } for some function f : F q → F q and ♯ ( E ∩ S r ) = 1 for all r ∈ F q . For this, we construct E withthe following properties: • ~ , ∈ E • ( x, ∈ E where 0 < x ≤ q − • For k in Lemma 4.2, let ( x, kx ) ∈ E where q + 12 ≤ x < q .Set E is clearly a graph of function f : F q → F q with f (0) = 0, f ( x ) = 0 when 0 < x ≤ q − , and f ( x ) = kx when q +12 ≤ x < q . So, it tiles F q by translations with respect to some coordinatesystem and the tiling partner A = { (0 , t ) : t ∈ F q } , and by Theorem 4.1 it has a spectrum.We show that E is a rotational tiling by showing that ♯ ( E ∩ S r ) = 1, r = 0. Indeed, if r is aquadratic residue, then there exists unique x satisfying 0 < x ≤ q − such that x + 0 = r . If r is a quadratic non-residue, then Lemma 4.3 implies the existence of the unique x satisfying4 q +12 < x ≤ q such that x + ( kx ) = (1 + k ) x = r . Hence, ♯ ( E ∩ S r ) = 1. This completes theproof for d = 2.When d >
2, we take E in F q be the set we just constructed above and define e E = E × F d − q .Then e E is a multiplicative tiling set on F dq by Theorem 3.3. Moreover, e E is also a translationaltile with respect to a coordinate system. For example if we choose e = (0 , , , · · · , { r e : 0 ≤ r ≤ q − } . To prove that e E has a spectrum, let L be a spectrumfor E . A simple calculation shows that L × F d − q is a spectrum for e E , and this completes theproof of the theorem. Remark.
Notice in Theorem 4.4 the tiling and spectral sets for d > e E = E × F d − q .However, other natural candidate is e E := E × E × · · · × E × F kq where 0 ≤ k ≤ d −
2. Clearly,the assertions of the theorem holds for e E .As a corollary of Theorem 4.4 and Theorem 2.9 we have the following result. Corollary 4.5.
Let q be an odd prime congruent to (mod 4) and d ≥ . Let Y := ( F dq ) ∗ . Thenthere exists tight wavelet frame sets for P W Y . Our result settles the existence of wavelet sets when q is an odd prime and q ≡ q ≡ S , contains more than one point. Proposition 4.6. If q ≡ (mod 4), there cannot be wavelet sets obtained by rotational tilingsand translations in F q .Proof. Take Y := F q \ { } . If E is a non-trivial set in F q which tiles Y by a set of rotations andtranslations, then E = q and E is a graph. Also, ♯ ( E \ { } ) = q − q − E \ { } . Taking union of all possiblerotations, the rotational tiling covers only points of non-zero radius. Therefore the points in thecircle with zero radius S are not covered. Note that for q ≡ ♯S = 2 q −
1. Thereforethere cannot be rotational tiling sets thus wavelet sets obtained by rotation and translationwhen q ≡ We end this paper with two open questions.
Question : Does there exist tight frame wavelet sets when q ≡ q ≡ mod
4) we exploited the fact that circle of 0 radius has only oneelement. We would have to come to grips with such circles to extend our results to the case q ≡ mod Question : To what extent is it possible to generalize the results of this paper to the case F dq , for q = p α , α > q is prime, congruent to 1 mod 4, we limited theimpact of arithmetic intricacies on the problem. The situation becomes quite fascinating when α > Acknowledgement
The second author was supported by Minigrant (Grant NO: ST659) ofORSP of San Francisco State University. The authors would like to thank the Graduate Centerof the City University of New York where this project was initiated.
References [1] L. Baggett, H. Medina, and K. Merrill,
Generalized multiresolution analyses and a con-struction procedure for all wavelet sets in R n , J. Fourier Anal. Appl. 5 (1999), 563–573[2][2] J. J. Benedetto, R. L. Benedetto, A theory for local fields and related groups , The Journalof Geometric Analysis, Vol. 14, No. 3, 2004. [2][3] J. J. Benedetto, R. L. Benedetto,
The Construction of Wavelet Sets , Wavelets and Multi-scale Analysis, Applied and Numerical Harmonic Analysis, pp 17-56, 2011 [2][4] J. Benedetto and M. Leon,
The construction of multiple dyadic minimally supported fre-quency wavelets on R n , Contemp. Math. 247 (1999), 43–74 [2][5] X. Dai, D. R. Larson, D. M. Speegle, Wavelet sets in R n , The Journal of Fourier Analysisand Applications 3 (1997), no. 4, 451–456. [2][6] R. J. Duffin, A. C. Schaeffer, A class of nonharmonic Fourier series , Trans. Amer. Math.Soc. 72 (1952), 341–366. [1][7] D. Dutkay and C.-K. Lai,
Some reductions of the spectral set conjecture to integers , MathProc Cambridge Phil Soc., 156 (2014), 123-135. [3][8] B. Fuglede,
Commuting self-adjoint partial differential operators and a group theoretic prob-lem , J. Funct. Anal. 16(1974), 101-121 [3][9] D. Han, D. Larson,
Frames, Bases and Group Representations , Memoirs of the AmericanMathematical Society, 2000. [2][10] E. Hernandez, X.-H Wang, G. Weiss,
Smoothing minimally supported frequency wavelets , I,J. Fourier Analysis and Applications, 2 (1996), 329–340. [2][11] E. Hernandez, X.-H Wang, G. Weiss,
Smoothing minimally supported frequency wavelets ,II, J. Fourier Analysis and Applications, 3 (1997), 23–4. [2][12] A. Iosevich, A. Liu, A. Mayeli and J. Pakianathan
Wavelet decomposition and band-width of functions defined on vector spaces over finite fields , (submitted for publication),(https://arxiv.org/pdf/1601.03473.pdf) (2016). [2][13] A. Iosevich, A. Mayeli,
Exponential bases, Paley-Wiener spaces, and applications ; Journalof Functional Analysis, Volume 268, Issue 2, 15 January 2015, Pages 363–375. [3]6[14] A. Iosevich, A. Mayeli, J. Pakianathan,
The Fuglede Conjecture holds in Z p × Z p , to appearin Analysis and PDE. (https://arxiv.org/pdf/1505.00883.pdf) [8, 12][15] K. D. Merrill, Simple wavelet sets for scalar dilations in L ( R ), in Wavelets and Frames:A Celebration of the Mathematical Work of Lawrence Bagget (P. Jorgensen, K. Merrill,and J. Packer, eds.), Birkhaeuser, Boston, 2008, 177–192 [2][16] K. D. Merrill, Simple wavelet sets for matrix dilations in R , Numer. Funct. Anal. Optim.33(7-9) (2012), 1112-1125. [2][17] C. Monico, M. Elia, Notes on an Additive Characterization of Quadratic Residues Modulop , Journal of Combinatorics, Information and System Sciences, 31 (2006), 209–215. [10][18] T. Tao,
Fuglede’s conjecture is false in 5 or higher dimensions , Math. Res. Letter, 11(2004),251-258. [3][19] X. Fang, X. Wang,
Construction of Minimally-Supported-Frequencies Wavelets , the Journalof Fourier Analysis and Application 2 (1996), no. 4, 315-327. [2][20] Y. Wang,