Some Results on Uniform Mixing on Abelian Cayley Graphs

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Let A be the adjacency matrix of Γ, i.e., 1, if (u, v) E, A = (auv)u,v V, where auv = ∈ ∈ ( 0, otherwise. A continuous random walk on Γ is determined by a family of matrices of the form M(t), indexed by the vertices of Γ and parameterized by a real positive time t. The(u, v)-entry of M(t) represents the probability of starting at vertex u and reaching vertex v at time t. Define a continuous random walk on Γ by setting M(t) = exp(t(A D)), − where D is a diagonal matrix. Then each column of M(t) corresponds to a probability density of a walk whose initial state is the vertex indexing the column. For quantum computations, Fahri and Gutmann [24] proposed an analogue continuous quantum walk. For a connected simple graph Γ with adjacency matrix A, they defined the transfer matrix of Γ as the following n n matrix: × + ∞ (ıtA)s H(t) = HΓ(t) = exp(ıtA) = = (Hg,h(t))g,h V , t R, s! ∈ ∈ Xs=0 where ı = √ 1, R is the field of real numbers, and n = V(Γ) is the number of vertices in Γ. − | | Definition 1. Let Γ be a graph. For u, v V(Γ), we say that Γ has perfect state transfer (PST) ∈ from u to v at the time t(> 0) if the (u, v)-entry of H(t), denoted by H(t)uv, has absolute value 1. Further, whenever H(t)uu = 1 we say that Γ is periodic at u with period t. If Γ is periodic with period t at every point,| then| Γ is named periodic. Since H(t) is a unitary matrix, if PST happens in the graph from u to v, then the entries in the u-th row and the entries in the v-th column are all zero except for the (u, v)-th entry. That is, the probability starting from u to v is absolutely 1 which is an ideal model of state transferring. From the beginning of this century, the research on perfect state transfer on graphs becomes very active, see for example, [1, 3, 4, 5, 7, 9, 12, 18, 19, 21, 22, 23, 31]. A graph is called integral if all its eigenvalues are integers. Angeles-Canul et al. [5] investigated PST in integral circulant graphs and the join of graphs. Godsil [26, 27, 28, 29] explained the close relationship between the existence of perfect state transfer on certain graphs and algebraic combinatorics such as the spectrum of the adjacency matrix, association schemes and so on. In [18, 19], Christandl et al. showed that perfect quantum state transfer occurs in some spin networks. Moreover, existence of PST has been obtained for distance-regular graphs [21, 23], complete bipartite graphs [45] (for discrete-time quantum walks), and Hadamard diagonalizable graphs [35]. Cayley graphs are good candidates for exhibiting PST due to their nice algebraic structure [9, 10, 17, 27, 28, 29, 50]. Among these results, Baˇsićet al, [6, 7, 8], Cheung and Godsil [17] presented a criterion on circulant graphs and cubelike graphs having PST. Remarkably, Coutinho et al. [22] showed that one can decide whether a graph admits PST in polynomial time with respect to the size of the graph. In a previous paper [49], we presented a characterization on connected simple Cayley graphs Γ = Cay(G, S ) having PST, where G is an abelian group, and we gave a unified interpretation to many known results. Contrary to PST on graphs in a sense, Moore and Russel introduced the notation of uniform mixing property of continuous-time quantum walks, which they named it as instantaneous uniform mixing [41]. 2 Definition 2. We say that a complex matrix A is a flat matrix if each of its entries has the same modulus. A graph Γ admits uniform mixing at time t if its transfer matrix H(t) is flat at time t. Equivalently, a graph Γ admits uniform mixing if and only if 1 H(t)uv = | | √n for each pair of vertices u and v in Γ, where n is the size of the vertex set V. From the definition, a graph admitting uniform mixing confirms the uniform probability dis- tribution of the walk at time t. Up to date, the known graphs having uniform mixing are rare, we list some of them as follows:

• the star K1,3, [33];

• complete graphs K2, K3, K4, [4]; • Hamming graphs H(d, 2), H(d, 3) and H(d, 4), [14, 33, 41]; • the Paley graph of order nine, [32]; • some strongly regular graphs from regular symmetric Hadamard matrices, [32]; • d d d some linear Cayley graphs over Z2, Z3 and Z4, [2, 33]; • the Cartesian product of graphs which admit uniform mixing at the same time, [33]; As far as we know, there is no general characterization on which graphs have uniform mixing. Even though there are some necessary and sufficient conditions for some specific graphs to have uniform mixing in [32, 33], no classification of graphs having uniform mixing is known in literature. Thus, a basic question for us is how to find explicit characterizations on graphs that have uniform mixing. Moreover, more concrete constructions or classification of graphs having uniform mixing are always desirable. In this paper, we give a characterization of abelian Cayley graphs having uniform mixing. We show that uniform mixing on abelian Cayley graphs are closely related to bent functions over finite abelian groups (see Theorem 1, which provides a necessary and sufficient condition for an abelian Cayley graph to admit uniform mixing). We also present some concrete constructions of graphs having uniform mixing (see Example 1). Particularly, for cubelike graphs (i.e, the n underlying group is F2, the n-dimensional vector space over the finite field F2), we show that every bent function leads to a cubelike graph having uniform mixing (see Theorem 11). Since there are plenty constructionsof bent functions, we can obtain many (some infinite family) graphs having uniform mixing. We also discuss the integrality of graphs which have uniform mixing. Using a result of Kronecker, we prove that, with certain constrains, the time at which an integral abelian Cayley graph has uniform mixing corresponds to a root of unity (see Theorem 2). For integral p-graphs, we show that it has uniform mixing if its eigenvalues induce a difference- balanced mapping (see Theorem 4). For integral abelian Cayley graphs, it is shown that the graph has uniform mixing at some time for some connection set S if and only if G is an elementary 3- d r d group or an elementary 2-group, or Z4, or Z2 Z4, where r 2, d 1, see Theorem 9. Thus the classification of integral abelian Cayley graphs⊗ having uniform≥ mixing≥ is established. In [33], Godsil et al. conjectured that if an abelian Cayley graph has uniform mixing, then the graph should be integral. Once this conjecture is confirmed, the classification of abelian Cayley graphs having uniform mixing is then completed by our result (Theorem 9). 3 2. Preliminaries In this section, we give some notation and definitions which are needed in our discussion. Throughout this paper, we use N, Z, Q, R, and C to stand for the set of non-negative integers, the ring of integers, the fields of rational numbers, real numbers and complex numbers, respectively. Zn = 0, 1, 2, , n 1 is the ring of integers modulo n. The elements of Zn are sometimes viewed as{ integers··· if there− } is no confusion. We use to stand for a multi-set. s s+1 {∗···∗} For any integer n and a prime number p, if p n and p n, s 0, then we write vp(n) = s. | 6| ≥ 2.1. Bent functions over finite abelian groups The notation of Boolean bent functions was introduced by Rothaus in 1976. Some properties and constructions of such functions were presented[47]. In [40], McFarland set up a connection between bent functions and some other combinatorial objects such as difference sets, strongly regular graphs and so on. Bent functionsalso find applications in coding theory, communications, and some other fields, see [38, 51] for example. The concept of bent function over finite abelian groups was introduced by Logachev et al. in [37]. Let G be an abelian group with order n, the exponent of G (the maximal order of elements of G) be m. Denote CG = f : f is a complex-valued function on G . { } G For f1, f2 C , define ∈ f , f = f (x) f (x), h 1 2i 1 2 Xx G ∈ G where f2(x) is the complex conjugation of f2(x). Then C is a Hilbert space with the above scalar product. Let G be the character group of G consisting of the homomorphisms

m b χ : G Tm = z C : z = 1 . → { ∈ } It is well-known that G G as groups and χg(h) = χh(g) for all g, h G. Write G = χa : a G . Then the characters of G satisfy the following orthogonal relation: ∈ { ∈ } b b n, if a = b, χa, χb = (1) h i ( 0, otherwise.

1 G In other words, the set χa : a G form an orthonormal basis of C . { √n ∈ } For any f CG, its Fourier transform is deﬁned by ∈

f (a) = f (x)χa(x) = f, χa . h i Xx G b ∈ The inverse transform is determined by 1 1 1 f (x) = f (a)χa(x) = f, χa χa(x). n h i Xa G Xa G √n √n ∈ b ∈ Definition 3. [37] Let G be an abelian group of order n. A function f CG is called bent on G if for every a G, we have ∈ ∈ f (a) = √n. | | For more properties on bent functions onb abelian and non-abelian groups, we refer the reader to [25, 48]. 4 2.2. Cayley graphs over abelian groups Let G be a finite abelian group of order n. The operation of G is addition, the identity element of G is 0. Let S be a subset of G with S = d 1. The Cayley graph Γ = Cay(G, S ) is defined by | | ≥

V(Γ) = G, the set of vertices, E(Γ) = (u, v) : u, v G, u v S , the set of edges. { ∈ − ∈ } We assume that 0 < S and S = S := s : s S (which means that Γ is a simple graph) and G = S (G is generated by S which− means{− that∈ Γ}is connected). The adjacency matrix of Γ is h i deﬁned by A = A(Γ) = (ag,h)g,h G, where ∈ 1, if (g, h) E(Γ), ag,h = ∈ ( 0, otherwise.

For any symmetric real matrix A of n by n, assume that its eigenvalues are λi, 1 i n, ≤ ≤ not necessarily distinct. We can form an orthogonal matrix P = (p , , pn), where pi is an 1 ··· eigenvector of λi (1 i n). So that we have the following spectral decomposition: ≤ ≤

A = λ E + + λnEn, (2) 1 1 ··· where Ei = pi p (1 i n) satisfy i∗ ≤ ≤

Ei, if i = j, EiE j = (3) ( 0, otherwise.

Here we use the symbol superscript ∗ to denote the conjugate transpose of a matrix. Therefore, we have the decomposition of the transfer matrix

H(t) = exp(ıλ t)E + + exp(ıλnt)En. (4) 1 1 ··· Particularly, when A is the adjacency matrix of an abelian Cayley graph Γ = Cay(G, S ), we 1 can take P = (χg(h))g,h G, where n = G and χg Gˆ. Then P is a unitary matrix and it is easy √n ∈ | | ∈ to check that P∗AP = diag(λg; g G), ∈ where λg = χg(s) =: χg(S ), g G (5) ∈ Xs S ∈ are the eigenvalues of Γ. Suppose that Γ= Cay(G, S ) is a connected integral graph. Denote

M = gcd(d λg :0 , g G), d = S , (6) − ∈ | | Mh = gcd(λx+h λx : x G), 0 , h G, (7) − ∈ ∈ DG = gcd(Mh :0 , h G). (8) ∈ Since Γ is connected, M is a positive integer by the well-known Perron-Frobenius Theorem. The following result has its own independent interesting in graph theory. 5 Lemma 1. Let notation be deﬁned as above. Then we have (1) M G ; | | (2) Mh G (d λh), d = S , 0 , h G. | | − | | ∈ (3) If G is an abelian p-group, then DG is a power of p.

Proof. (1) Since M = gcd(d λg :0 , g G), we can write − ∈

d λg = Mtg, g G and tg Z. (9) − ∈ ∈ Then we have, for every z S , by (1), ∈

λgχg(z) = χg(s)χg(z) = χg(s)χg(z) = G . (10) | | Xg G Xg G Xs S Xs S Xg G ∈ ∈ ∈ ∈ ∈ Meanwhile, λgχg(z) = (d Mtg)χg(z) = M tgχg(z). (11) − − Xg G Xg G Xg G ∈ ∈ ∈ Thus, G /M = tgχg(z). (12) | | − Xg G ∈ The left hand side of (12) is a rational number, while the right hand side of it is an algebraic integer. Thus both of them are integers. (2) Write λx+h λx = Mhθh(x), x G, 0 , h G. Then θh(x) Z and for every z S , we have − ∈ ∈ ∈ ∈ (λx+h λx)χx(z) = Mh χx(z)θh(x). − Xx G Xx G ∈ ∈ That is χh(s) χx(s)χx(z) χx(s)χx(z) = Mh χx(z)θh(x). − Xs S Xx G Xs S Xx G Xx G ∈ ∈ ∈ ∈ ∈ By the orthogonality of characters, we have

G (χh( z) 1) = Mh χx(z)θh(x). | | − − Xx G ∈ In the above equation, let z run through S and add them together. Then we obtain that

G (λh d) = Mh θh(x)λx. | | − Xx G ∈ (3) It is a direct consequence of (1) and (2). This completes the proof.

3. A characterization of abelian Cayley graphs having uniform mixing

Let G be an abelian group with order n. Let S be a subset of G with S = d 1, 0 < S = S and G = S . Suppose that Γ = Cay(G, S ) is the Cayley graph with the| connection| ≥ set S . Take− h i

6 1 P = (χg(h))g,h G and Ex = px p∗x as before, where px is the x-th column of P. Then the √n ∈ adjacency matrix A of Γ has the following spectral decomposition:

A = λgEg. Xg G ∈ Meanwhile, the transfer matrix H(t) has the following decomposition:

H(t) = exp(ıλgt)Eg. Xg G ∈ Thus we have, for every pair u, v G, ∈ 1 H(t)u,v = exp(ıλgt)(Eg)u,v = exp(ıλgt)χg(u v). (13) n − Xg G Xg G ∈ ∈ Therefore, 1 H(t)u,v = exp(ıλgt)χg(u v) = √n. | | ⇔ − √n Xg G ∈

Hence, Γ has uniform mixing at time t if and only if for every a G, it holds that ∈

exp(ıλ t)χ (a) = √n. (14) g g Xg G ∈

Thus we have proved the following result. Theorem 1. Let Γ be a simple abelian Cayley graph. For each ﬁxed real number t, deﬁne a complex-valued function on G by Gt(g) = exp(ıλgt).

Then Γ has uniform mixing at time t if and only if Gt is a bent function on G. We note that the result in Theorem 1 can also be deduced from Proposition 2.3 of [2]. Equivalently, we have Corollary 1. Let notation be deﬁned as above. Then Γ has uniform mixing at time t if and only if for every h(, 0) G, it holds that the correlation function ∈

Rt(h) := exp(ı(λg+h λg)t) = 0. (15) − Xg G ∈ Corollary 2. Deﬁne an n by n matrix

W = (exp(ıλx+yt))x,y G. (16) ∈ Then Γ has uniform mixing at time t if and only if W is a symmetric complex Hadamard matrix, T namely, W = W andWW∗ = nIn. Corollary 1 and Corollary 2 are equivalent, we just give the proof of the second one. 7 Proof. Denote 1 zg := exp(ıλxt)χx(g), g G. (17) ∀ ∈ √n Xx G ∈ Then by (14), we have that Γ has uniform mixing at time t if and only if zg = 1 for all g G. From (17), we get | | ∈ 1 exp(ıλxt) = zgχg(x), x G. (18) ∀ ∈ √n Xg G ∈ Thus, if zg = 1 holds for all g G, then | | ∈ 1 exp(ıλ t)exp(ıλ t) = z χ (x + g) z¯ χ (y + g) x+g y+g n u u v v Xg G Xg G Xu G Xv G ∈ ∈ ∈ ∈ 2 = zu χu(y x) | | − Xu G ∈ = χu(y x) − Xu G ∈ 0, if y , x, = ( n, if y = x. Thus W is a symmetric complex Hadamard matrix. Conversely, if W is a complex Hadamard matrix, then by the above computation, we have

2 0, if a , 0, zu χu(a) = | | ( n, if a = 0. Xu G ∈ The desired result follows from the inverse Fourier transformation.

( ) For a matrix A with all nonzero entries, the Schur inverse of A, denoted by A − , is given by

( ) 1 (A − )uv = . Auv An n n complex matrix A is called a Type-II matrix if × ( ) T A A − = nIn. (19) ( ) Let W be deﬁned as (16). Since λg is a real number for all g G, we know that W∗ = W − . Thus we have ∈ Corollary 3. Γ has uniform mixing at time t if andonly if W is a symmetric type-II matrix, where W is deﬁned by (16).

4. Classification of integral abelian Cayley graphs having uniform mixing In this section, we consider the integrality of Γ= Cay(G, S ), where G is a finite abelian group. We divide this section into three subsections. The first subsection, Subsect. 4.1, studies the linear dependence of some transcendental numbers. The second subsection, Subsect. 4.2, investigates integral Cayley p-graphs having uniform mixing. In the third subsection, Subsect. 4.3, we will give a complete classification of integral abelian Cayley graphs having uniform mixing. 8 4.1. Linear dependence Before going to present our next results, we need a lemma due to Lindemann first.

Lemma 2. [43, Theorem 9.1] Given any m distinct algebraic numbers a , , am, the values 1 ··· exp(a ), , exp(am) are linearly independent over the ﬁeld of algebraic numbers. 1 ··· From Lemma 2, we have the following result. Proposition 1. Let Γ = Cay(G, S ) be an abelian Cayley graph. Then Γ cannot have uniform mixing at any algebraic number t.

Proof. Suppose that Γ has uniform mixing at time t, where t is an algebraic number. Then by Corollary 1, we have that

Rt(h) = exp(ı(λg+h λg)t) = 0 (20) − Xg G ∈ holds true for any nonzero h. Since ı(λg+h λg)t, g G are algebraic numbers, and not all of them are equal, we get a contradiction with− Lemma 2.∈ We need further another lemma which is named Gelfond-Schneider Theorem. Lemma 3. [13] If x and y are two non-zero complex numbers with x irrational, then at least one of the numbers x, exp(y), or exp(xy) is transcendental. A simple consequence of Lemma 3 is the following:

Corollary 4. If there are elements g, h , 0 G such that λg+h λg is irrational, then exp(ı(λg+h 2rπ ∈ − − λg) ) is transcendental for any positive integers r, N 2. N ≥ Proof. Putting 2πr x = λg+h λg, y = ı − N and applying Lemma 3, since x and exp(y) are algebraic numbers, we get the desired result. Next, we consider integral abelian Cayley graphs that have uniform mixing at some time t. Suppose that Γ = Cay(G, S ) is an integral abelian Cayley graph having uniform mixing at time t. Denote λg+h λg 1 Ah(X) := X − Z[X, X− ], h(, 0) G. ∈ ∀ ∈ Xg G ∈ Let dh dh = max λg+h λg : g G , and Bh(X) = X Ah(X). { − ∈ } Denote a(X) := gcd(Bh(X):0 , h G). (21) ∈ Based on the above preparations, we prove the following result:

Theorem 2. Let notation be defined as above. Suppose that Γ= Cay(G, S ) is an integral abelian Cayley graph. Then the numberof the time t in the interval (0, 2π) such that Γ has uniform mixing at t is upper boundedby 2 min dh. Moreover, assume that the a(X) in (21) has all its roots on the 0,h G unit circle. If Γ= Cay(G, S ) has∈ uniform mixing at a time t, then γ = eıt is a root of unity. 9 In order to prove this result, we need the following lemma due to Kronecker. Lemma 4. [34] If an algebraic number whose conjugates all have absolute value 1, then the algebraic number is a root of unity. Proof. Suppose that Γ = Cay(G, S ) is an integral abelian Cayley graph having uniform mixing at time t. Denote Nh,u = g G : λg+h λg = u , 0 , h G, u Z. |{ ∈ − }| ∈ ∈ Since Cay(G, S ) is an integral abelian graph, we have λg = z S χg(z) = z S χ g(z) = λ g for all g G. Thus ∈ ∈ − − ∈ P P λg+h λg = u λ (g+h)+h λ (g+h) = u. − ⇔ − − − − Hence Nh,u = Nh, u. − Therefore, λg+h λg u u Ah(X) := X − = Nh,u(X + X− ) Xg G Xu N ∈ ∈ d and Bh(X) = X h Ah(X) is a self-reciprocal polynomial in Z[X]. ıt Denote γ = e . Then by Corollary 1, γ isa rootof Bh0 (X) which implies that γ is an algebraic integer, where dh = max dh : 0 , h G . Note that Bh (X) is a monic polynomial in Z[X] of 0 { ∈ } 0 degree 2dh0 . Then γ is a root of a(X). Since Bh(X) is a self-reciprocal polynomial for every 0 , h G, so does a(X). The first statement of Theorem 2 is proved. ∈ Let m(X) be the minimal polynomial of γ over Q. Then m(X) a(X). By assumption, all the roots of m(X) are on the unit circle, thus all the conjugates of γ |have unit norm, and then by Lemma 2, γ is a root of unity. Remark 1. We note that there are extensive researches on those self-reciprocal polynomials with all their roots on the unit circle, see [46] and the references therein. In view of Proposition 1 and Corollary 4, we consider the possibility of integral abelian 2πr Cayley graphs having uniform mixing at time t = N for some integers r and N with gcd(r, N) = 1. In this case, exp(ıt)isa N-th root of unity. In fact, our numerical experiments indicate that if an integral abelian Cayley graph has uniform mixing at a time t, then exp(ıt) is a root of unity. 2π We recall some basic facts about cyclotomic fields. Let ωn = exp(ı n ) and K = Q(ωn) be the corresponding cyclotomic field. The Galois group of the extension K/Q is

Gal(K/Q) = σℓ : ℓ Z∗ Z∗, { ∈ n} n where Zn∗ = ℓ Zn : gcd(ℓ, n) = 1 is the unit group of the ring Zn = Z/(n) and σℓ is deﬁned by ℓ{ ∈ } σℓ(ωn) = ωn. Utilizing the cyclotomic ﬁeld Q(ωn), we have the following criterion about the integrality of abelian Cayley graphs. Lemma 5. Suppose that G is an abelian group of order n and Γ= Cay(G, S ) is a Cayley graph. Then the following statements are equivalent: (1) Γ is an integral graph; (2) for every positive integer ℓ, 1 ℓ n, and gcd(ℓ, n) = 1, it holds that λℓg = λg; (3) for every positive integer ℓ with≤ gcd(≤ ℓ, n) = 1, it holds that ℓS := ℓs : s S = S. 10 { ∈ } Proof. (1) (2) ⇔ Since λg is algebraic, it is rational if and only if it is integral. Thus, λg is an integer if and only if for every σℓ Gal(K/Q), it holds that σℓ(λg) = λg. Now, we have by (5), ∈

σ (λ ) = σ χ (s) = χ (g)ℓ = χ (ℓg) = λ . ℓ g ℓ  g  s s ℓg Xs S  Xs S Xs S  ∈  ∈ ∈   Note that in the above equation, we make use of the property that χa(b) = χb(a) for every a, b in an abelian group. Thus λg is integral if and only if for every positive integer ℓ with gcd(ℓ, n) = 1, we have λℓg = λg. (1) (3) This fact is known to the community and has been proved in [49],[30]. ⇔ As an application of Theorem 1 and Lemma 5, we can prove the following result. Theorem 3. Let p, q be two diﬀerent odd prime integers andG be the abelian group of order pq. If Γ= Cay(G, S ) is an integral graph with connection set S, then Γ cannot admit uniform mixing at any time t.

Proof. Without loss of generality, we assume that p > q. It is obvious that G Zpq Zp Zq. The group × Z∗ = ℓ :1 ℓ pq 1, gcd(ℓ, pq) = 1 pq { ≤ ≤ − } acts on G as the following way: ℓ(x, y) = (ℓx,ℓy). The orbit containing (x, y) is denoted by [(x, y)]. The orbits of the above action are:

[(0, 0)], [(1, 0)], [(0, 1)], [(1, 1)].

By Lemma 5, S should be a union of some orbits, we divide the discussion into the following cases: Case 1: S = [(1, 0)] or S = [(0, 1)]. In this cases, it is easy to see that Γ is not connected. Case 2: S = [(1, 1)]. In this case, since the dual group of G is Zp Zq, we have that the eigenvalues of Γ are: × ap x c caqy λ(x,y) = ζp ζq 1 aXp 1 1 aXq 1 ≤ p≤ − ≤ q≤ − (p 1)(q 1), if x = 0, y = 0, (p− 1), − if x = 0, y , 0, =  , (x, y) G.  −(q − 1), if x , 0, y = 0,  ∈  1−, − if x , 0, y , 0.   Take h = (1, 0). Then  p(q 1), if x = 0, y = 0, − −  p(q 1), if x = 1, y = 0,  0, − if x , 0−, 1, y = 0, λ λ =  (x+1,y) (x,y)  p, if x = 0, y−, 0 −   = ,  p, if x 1, y 0  − , − ,  0, if x 0, 1, y 0.  11 −  That is, the multi-set

(q(p 2)) (1) (1) (q 1) (q 1) λ x+ ,y λ x,y :(x, y) G = 0 − , (p(q 1)) , ( p(q 1)) , p − , ( p) − , {∗ ( 1 ) − ( ) ∈ ∗} { − − − − } where the exponents in the right hand side of the above equation stand for the frequency of the numbers in the multiset. Therefore, we have

Rh(t) = q(p 2) + exp( ıtp(q 1)) + exp(ıtp(q 1)) + (q 1)(exp( ıtp) + exp(ıtp)). − − − − − −

It is obvious that Rh(t) q(p 2) 2 2(q 1) = (p 4)q. Thus Rh(t) , 0 for all p 5, q 3. Case 3: S = [(1| , 0)]|≥ [(0,−1)]. − − − − ≥ ≥ In this case, the spectra∪ of Γ are

ap x aqy λ(x,y) = ζp + ζq 1 aXp 1 1 aXq 1 ≤ p≤ − ≤ q≤ − p + q 2, if x = 0, y = 0, p 2,− if x = 0, y , 0, =  (x, y) G.  q − 2, if x , 0, y = 0,  ∈  2−, if x , 0, y , 0.  −  Take h = (1, 0). Then we have 

(q(p 2)) (q) (q) λ x+ ,y λ x,y :(x, y) G = 0 − , p , ( p) . {∗ ( 1 ) − ( ) ∈ ∗} { − } Thus Rh(t) = q(p 2) + q(exp(ıtp) + exp( ıtp)) (p 4)q. | | | − − |≥ − Thus Rh(t) , 0 for all p 5, q 3. Case 4: S = [(1, 0)] ≥ [(1, 1)]≥ or [(0, 1)] [(1, 1)]. We just consider the∪ subcase of S = [(1∪ , 0)] [(1, 1)], the other subcase can be proved similarly. ∪ For every (x, y) G, the corresponding eigenvalue is ∈ ap x ap x aqy λ(x,y) = ζp + ζp ζq 1 aXp 1 1 aXp 1 1 aXq 1 ≤ p≤ − ≤ p≤ − ≤ q≤ − (p 1)q, if x = 0, y = 0, − =  0, if y , 0,  q, if x , 0, y = 0.  −  Take h = (1, 0). Then we have 

(pq 2) (1) (1) λ x+ ,y λ x,y :(x, y) G = 0 − , (pq) , ( pq) . {{ ( 1 ) − ( ) ∈ }} { − } Hence Rh(t) = pq 2 + (exp(ıtpq) + exp( ıtpq)) pq 4 > 0. | | | − − |≥ − Case 5: S = [(1, 0)] [(0, 1)] [(1, 1)] = G (0, 0) . ∪ ∪ \ { } In this case, Γ is the complete graph Kpq. It is known that Γ does not admit uniform mixing. In summary, Rh(t) , 0 for any real number t when h = (1, 0), thus Γ does not exhibit uniform mixing at any time t. This completes the proof. 12 4.2. Integral Cayley p-graphs If G is an p-group and Γ = Cay(G, S ) is an integral Cayley graph, we call Γ is an integral Cayley p-graph. A k-term sum x1 + x2 + + xk is called a vanishing sum of order k if its value is 0. For vanishing sums of N···-th root of unity, Lam and Leung have the following result.

e1 er Lemma 6. [36] Suppose that p1, , pr are distinct prime numbers and N = p1 pr , then there is a vanishing sum of order n,··· composed from N-th roots ofunityif andonlyif n··· is a natural linear combination of the numbers pi, i.e.

n = p1N + + prN := p1a1 + + prar : a1, , ar N . ··· { ··· ··· ∈ } A direct consequence of Lemma 6 and Corollary 1 is the following: Corollary 5. Let Γ= Cay(G, S ) be an integral abelian Cayley p-graph. If Γ has uniform mixing 2πr at time t = e for some prime number p′, then p = p′. p′ ′ For any abelian group G and an abelian group H, a mapping f from G to H is called a difference balanced function if for every a(, 0) G and every b H, the number of solutions to the following equation ∈ ∈ f (x + a) f (x) = b − is independent on the choice of a and b. Difference balanced functions find their applications in finite geometry, combinatorial theory, and coding theory etc, see for example, [44]. In words of difference balanced function, we have the following result. Theorem 4. Let G be an abelian group of an odd prime power order and Γ = Cay(G, S ) be an e 1 integral Cayley p-graph. Let e′ be a positive integer such that p ′− (λg+h λg) for every g, h G. e′ 1 e′ 1 | − 2∈πr Let H = p − Zp = p − j : j = 0, 1, , p 1 . Then Γ has uniform mixing at time t = if { ··· − } pe′ and only if the mapping λ : G H,isadifference balanced function. → Proof. “ ” By Corollary 1, Γ has uniform mixing at time t if and only if Rt(h) = 0 for all ⇒ 0 , h G. Denote N = pe′ and define a multi-set ∈

Ωh = λg+h λg (mod N) : g G . { ∗ − ∈ ∗ } Then z Rt(h) = ωN . zXΩ ∈ h e 1 (pe 1 ) We now show that if Rt(h) = 0, then Ωh = (p ′− j) − , j Zp , and λ is a diﬀerence balanced function from G to H. { ∈ } By the assumption, we can write

(t0) e′ 1 (t1) e′ 1 (tp 1) Ωh = 0 , (p − ) , , ((p 1)p − ) − . { ··· − } (t ) Here we use j j to indicate that the multiplicity of j in Ωh is t j( 0). Then ≥ p 1 − e 1 p ′ − j Rt(h) = t jωN = 0. (22) Xj=0 13 e 1 p ′− Since ωN = ωp is a pth root of unity, we can rewrite (22) as

p 1 − j Rt(h) = t jωp = 0. (23) Xj=0

It is obvious that p 1 − e t j = G := p . (24) | | Xj=0 2π The minimal polynomial of ωp = exp(ı p ) is the cyclotomic polynomial φp(x), i.e.,

p 1 p 2 φp(x) = x − + x − + + x + 1. ··· Hence combing (23) and (24), we have

e 1 t0 = = tp 1 = p − . (25) ··· −

“ ” Obvious. ⇐ Remark 2. For an integral Cayley p-graph Cay(G, S ), we denote integers M and Mh as in (6), (7), respectively. By DG we denote gcd(Mh : 0 , h G). Then Lemma 1 indicates that both M ∈ e 1 and DG are powers of p. Hence the number e′ in Theorem 4 is then determined by DG = p ′− . If the underlying group of an integral Cayley p-graph is an elementary group, then we have the following explicit characterization.

r Theorem 5. Let p be an odd prime number and G = Zp the elementary commutative p-group. If Γ= Cay(G, S ) is an integral graph, then Γ has uniform mixing for some subset S if and only if p = 3. Proof. If r = 1, by the assumption that Γ is an integral graph, we know that Γ is the complete graph Kp, the required result is well known. r Below we prove that if p > 3 and r 2, then any integral graph Γ= Cay(Zp, S ) cannot have ≥ r r uniform mixing for any subset S . For every x = (x1, , xr) Zp, the group Z∗p acts on Zp as the following way: ··· ∈ ℓ x = (ℓx , ,ℓxr). 1 ··· ¯ Since Γ is integral, S should be a union of some orbits under the action of Z∗p, let S be the set consisting of the representatives of the orbits. Denote s¯ = S¯ . Then it is easy to see that pr 1 r | | r s¯ p −1 . For every x = (x1, , xr) Zp, the corresponding eigenvalue is ≤ ≤ − ··· ∈ p 1 − x ℓs λ(x1, ,xr) = ζp· = δ(x s), ··· · Xℓ=1 Xs S¯ Xs S¯ ∈ ∈ where the function δ(x) is deﬁned by

p 1, x = 0, δ(x) = − ( 1, x , 0. −14 For every nonzero h Zr , we have ∈ p

λ x+h λx = [δ(x s + h s) δ(x s)]. ( ) − · · − · s S¯X,h s,0 ∈ · It is easy to see that when h s , 0, · p, if x s = 0, − · δ(x s + h s) δ(x s) =  p, if x s + h s = 0, · · − ·  · ·  0, otherwise.  Next we count the probability of zero in the diﬀerence δ(x s + h s) δ(x s) when x runs over r · · − · Zp, i.e., P(s) := Prob[δ(x s + h s) δ(x s) = 0, x Zr ]. · · − · ∈ p r Since S = Zp, there exist s1, , sr S which are linear independent (viewed as vectors of h i r r ··· ∈ the space Z ). Choosing h Z such that h si = 0, 1 i r 1, then h sr , 0, and for every p ∈ p · ≤ ≤ − · r ( j) ( j) , s j S , write s j = i=1 ki si. If kr 0, we can choose another element in the same orbit such ∈ ( j) ( j) ( j) that k = 1. Then Ph s j = k h sr, k = 0 or 1. Thus r · r · r ( j) 0, if k = 0, or x s j , 0, h sr, r · − · δ(x s j + h s j) δ(x s j) =  p, if x s j = h sr, · · − ·  · − ·  p, if x s j = 0.  − ·   r 1 It is obvious that the linear system x s j = 0, for some r j s¯, has at most p solutions, so · ≤ ≤ − does the linear system x s j = h sr, for some r j s¯. The more of the members of those · − · ≤ ≤ j, the less of the number of x (it depends on the p-rank of the matrix consisting the related s j’s r as the row vectors). Therefore, the probability of zero in the multi-set λx+h λx : x Zp r r 1 { ∗ − ∈ ∗ } p 2p − 2 1 is at least −pr = 1 p > 2 if p > 4. Thus Rh(t) > 0 for all t. This completes the proof of Theorem 5. − | | For cyclic p-groups, we have the following result.

Theorem 6. Let p be a prime number and r be a positive integer. Let Γ = Cay(Zpr , S ) be an integral Cayley graph. Then Γ admits uniform mixing at some time t if and only if p = 2, 3 and r = 1, that is Γ= K2 or K3,or p = 2 and r = 2.

r Proof. The group Z∗pr = ℓ : 1 ℓ p 1, gcd(ℓ, p) = 1 acts on Zpr by the natural way. The orbits of this action are: { ≤ ≤ − } r 1 [0], [1], [p], , [p − ], ··· i i r i where [p ] = p j :1 j p − 1, gcd( j, p) = 1 . By Lemma 5, S should be a union of some orbits. { ≤ ≤ − } When r = 1, then Γ is the complete graph Kp since Γ is integral, S = [1]. It is known that for a prime p, Kp has uniform mixing at some time t if and only if p = 2, 3. When r = 2 and p = 2, π we can take S = [1], then Γ is in fact the cycle C4 and has uniform mixing at time t = 4 . This proves the suﬃciency. For the necessity, we give an iterative proof of the desired result. Before going to do that, we recall a result on Ramanujan functions. 15 For any pair of integers k, n, the Ramanujan function c(k, n) is deﬁned by

c(k, n) = exp (2πkx/n) . (26) 1 x n,Xgcd(x,n)=1 ≤ ≤ For the computation of c(k, n), one has the following formula, see [39, Corollary 2.4]. φ(n)µ(m) c(k, n) = , where m = n , (27) φ(m) gcd(k,n) φ is the Euler-phi function and µ is the Mobius function. For every 0 , x Zpr , we write ∈

vp(x) r v (x) x = p x′, where 0 vp(x) r 1, 1 x p p 1, and gcd(x , p) = 1. (28) ≤ ≤ − ≤ ′ ≤ − − ′ Case 1: S = [1]. r 1 In this case, the spectra of Γ are λ = p − (p 1) and 0 − x j r λx = ω r , 0 < x < p 1. p − 1 j pr X1,gcd( j,p)=1 ≤ ≤ − For every 1 j pr 1, gcd( j, p) = 1, we write ≤ ≤ − r v (x) v (x) r v (x) j = j p − p + j , 0 j p p 1, 1 j p − p 1, gcd( j , p) = 1. 1 2 ≤ 1 ≤ − ≤ 2 ≤ − 2 Thus vp(x) j2 x′ vp(x) r vp(x) λx = p ω r vp (x) = p c(1, p − ). p − 1 j pr vp(xX) 1,gcd( j ,p)=1 ≤ 2≤ − − 2 By (27), we have r 1 p − , if vp(x) = r 1, λx = − − (29) ( 0, otherwise. r 1 Take h = p − . It is a routine to check that

r r (1) r (1) (pr 2) λx+h λx :0 x p 1 = (p ) , ( p ) , 0 − . { ∗ − ≤ ≤ − ∗ } { − } r r r Thus when r > 1, Rh(t) = (p 2) + exp(ıtp ) + exp( ıtp ) , 0 for all real number t and odd prime p. − − Case 2: [1] * S . In this case, the graph Γ is not connected. The general cases for r > 1 can be proved uniformly as follows. Since Γ is integral and connected, the connection set S has the following form:

d S = [1] d Ie [p ] , ∪ ∪ ∈ where Ie is a propersubsetof j :1 j r 1 . Thus, forevery x G, we have the corresponding eigenvalue { ≤ ≤ − } ∈

ℓx ℓ′ x λx = ζ r + ζ r d . p p − Z Z ℓX∗pr Xd Ie ℓ′ X∗r d ∈ ∈ ∈ p − 16 v (x) r v (x) If x , 0, then we can write x = p p x′, where 1 x′ p − p 1 and gcd(x′, p) = 1. Then r 1 vp(x) r 1 vp(x) ≤ ≤ r −1 vp(x) x + p − = p (x′ + p − − ). If vp(x) < r 1, then gcd(x′ + p − − , p) = 1 and there exists r 1 − an ℓ” Z∗pr such that x + p − = ℓ”x. When vp(x) = r 1, then 1 x′ < p 1. The relation r∈1 − r ≤ r 1 − x + p − = ℓ”x still holds for some ℓ” Z∗pr . Thus when x , 0, p p − , there always exists r 1 ∈ − ℓ” Z∗ r such that x + p − = ℓ”x. It follows that ∈ p

r 1 r 1 ℓ(x+p − ) ℓ′(x+p − ) λx+pr 1 = ζ r + ζ r d − p p − Z Z ℓX∗pr Xd Ie ℓ′ X∗r d ∈ ∈ ∈ p −

ℓℓ”x ℓ′ x = ζ r + ζ r d p p − Z Z ℓX∗pr Xd Ie ℓ′ X∗r d ∈ ∈ ∈ p − = λx.

We then have

r r (1) r (1) (pr 2) λx+h λx :0 x p 1 = (p ) , ( p ) , 0 − . { ∗ − ≤ ≤ − ∗ } { − }

Thus when r > 1, Rh(t) , 0. In summary, when r > 1, Γ cannot have uniform mixing at a time t except for the case of r = 2 and p = 2. For general abelian p-groups, we have the following result.

Theorem 7. Suppose that p is an odd prime number and r , , rn are positive integers (not 1 ··· necessarily distinct). Let G = Zpr1 Zprn be an abelian group. If max r1, , rn > 1 and for some subset S in G, Γ= Cay(G,×···×S ) is integral, then it cannot admit uniform{ ··· mixing.}

Proof. Suppose that max r1, , rn = r > 1. For every (x1, , xn) G, the corresponding eigenvalue is { ··· } ··· ∈ r r r rn p − 1 s1 x1+ +p − sn xn λ(x1, ,xn) = ζpr ··· . ··· (s ,X,s ) S 1 ··· n ∈ Thus we deﬁne a new scalar product on G by

r r1 r rn x y = p − x y + + p − xnyn, x = (x , , xn), y = (y , , yn) G. · 1 1 ··· ∀ 1 ··· 1 ··· ∈ n The group Z∗Pr acts on G = Zpr naturally. Since Γ is an integral graph, S should be a union of some orbits. Let S¯ be the set consisting of the representatives of the orbits and denotes ¯ = S¯ . | | Then the eigenvalue λ(x , ,x ) can be rewritten as 1 ··· n ℓx s r λ(x1, ,xn) = ζpr · = c(x s, p ), ··· · s=(s X, ,s ) S¯ ℓXZ∗ r s=(s X, ,s ) S¯ 1 ··· n ∈ ∈ P 1 ··· n ∈ where c(x s, pr) is the Ramanujun function, see (26). By (27), we have ·

0, if vp(x s) r 2, r r 1 · ≤ − c(x s, p ) =  p − , if vp(x s) = r 1, ·  −r 1 · − r  p − (p 1), if x s 0 (mod p ).  − · ≡  17 It is known that the ring Zpr is a local ring, its unique maximum ideal is (p). One has the following ideals chain: r 1 r 2 0 (p − ) (p − ) (p) Zpr . ⊂ ⊂ ⊂···⊂ ⊂ The quotient ring Zpr /(p) is a field isomorphic to Zp. Denote by τi : Zpri Zpr /(p) the natural n → homomorphism. Extend τi to G by τ(G) = τ1(Zpr1 ) τn(Zprn ) = Zp. Since G = S , there ¯ ×···× n r 1 h i n are s1, , sn S such that τ(s1), , τ(sn) form a basis of Zp. Take h = p − v, where v Zp ··· ∈ ··· r 1 ∈ such that v τ(si) = 0, 1 i n 1, and v τ(sn) , 0. Then (x+h) s j = x s j + p v τ(s j) = x s j · ≤ ≤ − · · · − · · n ( j) ( j) , for all 1 j n 1. For j n, write τ(s j) = i=1 ki τ(si). If kn 0, we can assume that ( j) ≤ ≤ − ≥ kn = 1 by replacing the element with another elementP in the same orbit. Then vτ(s j) = v τ(sn), r 1 · j n. Hence (x + h) s j = x s j + p − v τ(sn). ≥ · · r · r r Case (1). If x s j 0 (mod p ), then obviously c((x + h) s j, p ) , c((x + h) s j, p ). · ≡ r 1 · · Case (2). If vp(x s j) = r 1, write x s j = p − α(x, j), where α(x, j) 1, 2, , p 1 . In · r − r · ∈ { ··· − } this case, c((x + h) s j, p ) , c(x s j, p ) if and only if α(x, j) = v τ(sn). · · r + +r n − · In Case (1), Since τ : x τ(x) is an p 1 ··· n− -to-one mapping, we know that the number 7→ r n 1 r + +r n r + +r 1 of x G such that x s 0 (mod p ) holds some j, is at most p − p 1 ··· n− = p 1 ··· n− . The more∈ of the member ·j, the≡ less of the total number of x. The exact value depends on the p-rank of the matrix consisting of the row vectors τ(s j) for the relevant j. In Case (2), α(x, j) = v τ(sn) means that x s j is a fixed number for those j. Thus in this − · · r r case, we know that the number of x G such that c((x + h) s j, p ) , c(x s j, p ) holds some j, r + +r 1 ∈ · · is at most p 1 ··· n− as well. r + +r r + +r 1 Therefore, the number of x G such that λx+h = λx is at least p 1 ··· n 2p 1 ··· n− . In ∈ − 2 other words, the probability of zero in the multi-set λx+h λx : x G is at least 1 p . If p > 3, then an analogue argument as in the proof of{ Theorem ∗ − 5 leads∈to the∗ } desired conclusion.− If p = 3, then we need some refined technique. In fact, the above estimation is based on the r property that x s j 0 (mod p ) implies τ(x) τ(s j) 0 (mod p). However, the inverse is not true in general.· ≡ When r > 1, we can use the· p-adic≡ representation of the numbers. Let x = (x , , xn), s = (s , , sn) G. We assume, without loss of generality, that r = r. Write 1 ··· 1 ··· ∈ 1 (i) (i) (i) ri 1 (i) (i) (i) ri 1 xi = x0 + x1 p + + xr 1 p − , si = s0 + s1 p + + sr 1 p − , ··· i− ··· i− (i) (i) r where x = τi(xi), s = τi(si). Then x s 0 (mod p ) induces that 0 0 · ≡ τ(x) τ(s) 0 (mod p), · n n n n ≡ (30) ( (s(1) x(1) + s(1) x(1)) + + (s( ) x( ) + s( ) x( )) + possible terms 0 (mod p). 0 1 1 0 ··· 0 1 1 0 ≡ r + +r 2 For the fixed sn, the linear system (30) has r + +rn variables, it has at most p 1 n solutions. 1 ··· ··· − Following the foregoing discussion, we know that the number of x G such that λx+h = r + +r r + +r 2 ∈ λx is at least p 1 ··· n 2p 1 ··· n− . This means that the probability of zero in the multi-set − 2 1 , λx+h λx : x G is at least 1 p2 > 2 . Thus Rh(t) 0 for all t. This completes the {proof. ∗ − ∈ ∗ } − Remark 3. (1) In the above proof of Theorem 7, we have actually proved that there exists an element 0 , h G such that there are more than one half of elements x’s in G satisfying the property that the∈ Ramanujun function takes the same value on each orbit when we change x to x + h. This fact will be employed to prove Theorem 9 in the Appendix 6. (2) Theorem 7 is equivalent to saying that, if G is an abelian p-group with exp(G) 5, then any integral Cayley graph with the underlying group G cannot admit uniform mixing. ≥ 18 4.3. General integral abelian Cayley graphs For any abelian group G, there are the following possibilities: n n d 1. G = Z2, Z3, Z4. d 2. G = Z2 Z4, d 1. r × d ≥ 3. G = Z2 Z4, r 2, d 1. 4. exp(G) × 5, where≥ exp(≥G) is the exponent of G, namely, the greatest order of the elements in G. ≥ n n d For the first case, i.e., G = Z2, or Z3, or Z4. It is known that there are some suitable subset S such that Cay(G, S ) has uniform mixing. Now we prove that for the case of Item 2, Cay(G, S ) admits uniform mixing for some S G. ⊂ d Theorem 8. Suppose that G = Z2 Z4, where d 1. Then Γ = Cay(G, S ) exhibits uniform π × ≥ mixing at time t = 4 for some subset S in G.

Proof. By the Cartesian product construction, it is sufficient to prove that G = Z2 Z4 admits π × uniform mixing at time t = 4 . To this end, we take S = (10), (11), (13), (02) G. Then for every (x, y) G, the corresponding eigenvalue is { } ⊂ ∈ x y x y 3y λ x,y = ( 1) + ( 1) + ( 1) (ı + ı ). ( ) − − − Thus λ = 4, λ = λ = λ = 0, λ = 2, λ = λ = λ = 2. (00) (01) (02) (03) (12) (10) (11) (13) − Thus, by (13), we have π 1 π xa1 ya2 Huv = exp ı λ x,y ( 1) ı (where u v = (a , a )) 4 8 4 ( ) − − 1 2 (x,y)XZ Z ∈ 2× 4 1 = (ıa2 + 1)(( 1)a1+1ı + ( 1)a2 ) + (ıa2 1)(1 ( 1)a1+a2 ı) 8 h − − − − − i 1 ı = ± ± . 4 π √ Thus Huv 4 = 1/(2 2) for every u, v G as required. | | ∈ r d Using Cartesian product, we know that G = Z2 Z4 admits uniform mixing for some subset S , where r 2, d 1. × Finally,≥ for the≥ Item 4, by using the techniques involved in Theorem 3, 5, 6, and Theorem 8, one can get the following main result in this section whose sufficiency has been proved as before, the necessity is proved in the Appendix 6. Theorem 9. LetG be an abeliangroup and S be a subsetofG. If Γ= Cay(G, S ) is integral and has uniform mixing at some time t, then exp(G) 4, in other words, either G is an elementary d r d ≤ p-group, where p = 2 or 3,orG = Z4, Z2 Z4 for some integers r 1, d 1. Conversely, if G is an abelian group with exp(G) 4, then there× exists a subset S in G≥ such≥ that Cay(G, S ) admits uniform mixing. ≤ Remark 4. It is actually conjectured in [33] that if an abelian Cayley graph has uniform mixing, then the graph should be integral. If this conjecture is confirmed, then by Theorem 9, we can say that if G is an abelian group, then there is a suitable connection set S G such that Cay(G, S ) d d⊂ d r d admits uniform mixing if and only if the underlying group G is Z2, Z3, Z4 or Z2 Z4, where r 2, d 0. × ≥ ≥ 19 5. Cubelike graphs

n Let G be the additive group of F2, here F2 = Z2 which is the finite field with two elements. n < For any subset S of F2,0 S , Γ= Cay(G, S ) is an integral graph and the order of any non-zero element in G is two. We note that Chan [2] found some examples of such graphs that admit uniform mixing. It is proved that for k 2, there exist graphs in the Bose-Mesner algebra of the Hamming graph H(2k+2 8, 2) that have≥ uniform mixing at time π/2k. Best et al. [11] gave a characterization of certain− quotients of a Hamming graph that has uniform mixing. Mullin [42], Godsil et al. [33] also utilized graph quotients to find graphs that have uniform mixing. In this section, we use Boolean functions to get cubelike graphs having uniform mixing. We show that when n = 2m is even, Γ = Cay(G, S ) has uniform mixing at time t = π/2m if f is a bent function, where f is the characteristic function of S , that is, every bent function leads to a cubelike graph having uniform mixing. Since there are many constructions of bent functions (many infinite families), one can obtain some infinite families of cubelike graphs admitting uni- n form mixing. When n = 2m + 1 is odd, there is no bent function on F2. In this case, we can use some special Boolean functions to get some cubelike graphs having uniform mixing also. An existence result of such graphs that admit uniform mixing is also presented. Firstly, we recall some basic facts about finite fields. The character group of G is

n n Gˆ = Fˆ = χz : z F , 2 { ∈ 2} n where for g = (g1, , gn), z = (z1, , zn) F , ··· ··· ∈ 2 n z g χz(g) = ( 1) · , z g = z jg j F2. − · ∈ Xj=1 n n If we view F2 as the additive group of the ﬁnite ﬁeld Fq with q = 2 , then

Gˆ = (\Fq, +) = χz : z Fq , { ∈ } Tr(zg) where for g, z Fq, χz(g) = ( 1) ,andTr : Fq F2 is the trace mapping. Let f be the∈ characteristic− function of S , namely,→ 1, if x S , f (x) = ∈ ( 0, otherwise. In this case, S is also called the supporting set (or simply the support) of f . For every g(, 0) G, the corresponding eigenvalue of Γ is ∈

f (x) 1 ( 1) Tr(gx) 1 f (x)+Tr(gx) 1 λg = χg(s) = − − ( 1) = ( 1) = F(g), (31) 2 − −2 − −2 Xs S Xx G Xx G ∈ ∈ ∈ b f (x) where F(x) = ( 1) and F is the Fourier transform of F. Note that λ0 = S = d. The sum f (x)+Tr(−gx) | | x Fm ( 1) is usually denoted by W f (g) and termed as the Walsh-Hadamard Transfor- ∈ 2 − b mationP of f at g. Recall the number Mh defined in (7) and DG defined in (8). By Lemma 1, we have ν DG = gcd(Mh :0 , h G) = 2 . (32) ∈ for some nonnegative integer ν. 20 Theorem 10. Let G be an abelian 2-group and Γ = Cay(G, S ) be an integral simple Cayley graph. Let ν be defined as in (32). For every 0 , h G, define the following sets: ∈

N (h) = g G : v (λg+h λg) > ν + 1 , 2 { ∈ 2 − } N (h) = g G : v (λg+h λg) = ν + 1 , 1 { ∈ 2 − } N (h) = g G : v (λg+h λg) = ν , 0 { ∈ 2 − } π and denote ni(h) = Ni(h) , i = 0, 1, 2. Then Γ has uniform mixing at time t = ν+1 if and only if | | 2 for every 0 , h F2n ,n2(h) = n1(h). ∈ Proof. It is obvious that, for every g G, ∈ λg+h λg g N2(h) − Z, ∈ ⇔ 2ν+2 ∈ λg+h λg g N (h) − is an odd number, ∈ 1 ⇔ 2ν+1 λg+h λg g N (h) − is an odd number. ∈ 0 ⇔ 2ν

λ + λ Moreover, when g N (h), the numbers g h− g appear in pairs with diﬀerent signs. Thus ∈ 0 2ν π Rh + 2ν 1 π = exp ı (λg+h λg) 2ν+1 − Xg G ∈ λg+h λg λg+h λg π λg+h λg = exp ı2π − + exp ıπ − + exp ı − 2ν+2 ! 2ν+1 ! 2 2ν ! g XN (h) g XN (h) g XN (h) ∈ 2 ∈ 1 ∈ 0 = n (h) n (h). 2 − 1 The desired result now follows with Corollary 1.

5.1. n is even Assume that n = 2k and v (d) k 1. If f (x) is a bent function, namely, 2 ≤ − F(g) = 2k for every g G. | | ∈ If F(x) = ( 1) f (x) is bent, we alsob say f is bent for convenience. It is well-known that bent − 2k function exists on F2 for every k. See for example, [38, Chapter 4, pp.69-88]. 2k In this subsection, we will show that the Cayley graph Cay(F2 ; S ) has uniform mixing if f is bent, where f is the characteristic function of S . When f is bent, we can write F(g) = 2k( 1) f(g). − e The function f is called the dual of f . Itb is well-known that f is bent if and only if f is bent (see for example, [38], line 12, page 72). Thus we have e e k 1 f(g) λg = 2 − ( 1) , g , 0. (33) − − 21 e Then k 1 f(g) v2(d) k v2(d) 1 f(g) d λg = d + 2 − ( 1) = 2 d′ + 2 − − ( 1) , − − e − e v (d) n 1 k 1 where d = 2 2 d′, d′ is odd. Since f is a bent function, we know that d = 2 − 2 − (see [38, n 1 k 1 ± Theorem 4.3.1]). Without loss of generality, we assume that d = 2 − 2 − , thus v2(d) = k 1 k 2π , − − and M = 2 , i.e., ℓ = k. Taking t = 2M , then we have for every h 0, the correlation function

Rt(h) = exp(ı(λg+h λg)t) − Xg G ∈ = exp(ı(λh d)t) + exp(ı(d λh)t) + exp(ı(λg+h λg)t) − − − g,0X,h,g G π ∈ = exp(ı (( 1) f(g+h) ( 1) f(g))). 2 − − − Xg G e e ∈ Since F(0) = ( 1) f (x) = 2n 2 S = 2k, − − | | Xx G b ∈ we know that f (0) = 0. Noticing that ( 1) f = 1 2 f for any boolean function f , we get − − f(g+h) f (g) e Rt(h) = exp( ıπ( f (g + h) f (g))) = ( 1) − . (34) − − − Xg G Xg G e e ∈ e e ∈ Thus Rt(h) = 0 for all h , 0 if f is bent (see, for example [38], page 75). It is known that f is bent if and only if f is bent. Therefore, we have the following result: e e 2k Theorem 11. Let G = (F2 , +) be an elementary 2-group and S be a subset ofG. Let f be the Γ= = π characteristic function of S . Then Cay(G, S ) has uniform mixing at time t 2k if f isabent function on G. Below we present an example. 4 Example 1. Let G = (F2, +) and

f (x1, x2, x3, x4) = x1 x2 + x3 x4. Then the support of f is S = (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 0, 1), (0, 0, 1, 1), (1, 0, 1, 1), (0, 1, 1, 1) , { } π and Γ= Cay(G, S ) has uniform mixing at time t = 4 . f (X) 4 Proof. Let F(X) = ( 1) , X = (x1, x2, x3, x4). For every Y = (y1, y2, y3, y4) F , we have − ∈ 2 F(Y) = ( 1)x1x2+x3 x4+x1y1+x2y2+x3y3+x4y4 − x1,x2X,x3,x4 F2 b ∈ = ( 1)x1x2+x1y1+x2y2+x3y3 ( 1)x4(x3+y4) − − x ,xX,x F xXF 1 2 3∈ 2 4∈ 2 = 2( 1)y3y4 ( 1)x1y1 ( 1)(x1+y2)x2 − − − xXF xXF 1∈ 2 2∈ 2 = 4( 1)y1y2+y3y4 . − 22 4 Thus f is a self-dual bent function. For every Y = (y1, y2, y3, y4) F2, the corresponding eigenvalue of Γ is (see (5)) ∈

y1+y2 y1+y2+y3 y1+y2+y4 y3+y4 y1+y3+y4 y2+y3+y4 λY = ( 1) + ( 1) + ( 1) + ( 1) + ( 1) + ( 1) . − − − − − − y y +y y Since both the functions F(y1, y2, y3, y4) = ( 1) 1 2 3 4 and λ(y1, y2, y3, y4) are invariant under the action of the Klein group K = (1), (12)(34)− , (14)(23), (13)(24) ,one can check that { } y1y2+y3y4 λY = λ(y , y , y , y ) = 2( 1) , Y = (y , y , y , y ) , (0, 0, 0, 0). 1 2 3 4 − − 1 2 3 4 One can also use (33) to get the above equation. Thus by (13), the (u, v)-th entry of H(t) at time π/4 is π 1 π H = exp(ı λg)χg(a)(a = u v) 4 uv 16 4 − Xg G ∈

1 π π g g +g g g a = exp(ı6 ) + exp(ı ( 2( 1) 1 2 3 4 ))( 1) · 16  · 4 · 4 − − −   Xg,0      1 g g +g g g a = ı ı ( 1) 1 2 3 4 ( 1) · ( by exp(ıx) = cos x + ı sin x) 16 − − − −   Xg,0    1  = ı F(a) − 16 1 b = ı( 1) f(a). −4 − e Therefore, we get 1 H(t)uv = | | 4 as required. Remark 5. Godsil et al. [32] have a full characterization of strongly regular graphs that admit 2k 2k uniform mixing. It is obviousthat if f is benton F2 , then Cay(F2 , S ) is a strongly regular graph, where S = supp( f ). It must be noted, however, that not all strong regular graphs exhibit uniform mixing. Our result (Theorem 11) shows that every bent function leads to a cubelike graph having uniform mixing.

5.2. n is odd n When n = 2m + 1 is odd, it is well-known that there is no bent functions on F2. In this case, 2m < 2m we choose a bent function f on F2 whose supporting set is (0 )S 1 F2 . We assume that, 2m 1 m 1 n ⊆ without loss of generality, S 1 = 2 − 2 − . Deﬁne a subset S F by | | − ⊆ 2 S = (1, z) : z S (0, x) : x S , (35) { ∈ 1} ∪ { ∈ 1} 2m n where S 1 = F2 S 1 is the complement of S 1. Let Γ = Cay(F2; S ) be the Cayley graph with connection set S .\ Then we have the following result.

2m Theorem 12. Assume that n = 2m + 1 and f is a bent function on F2 whose supporting set is Γ= n = π S 1. Let S be deﬁned as in (35). Then Cay(F2; S ) has uniform mixing at time t 2m+1 if and only if m = 1. 23 2m n 2m Proof. It is obvious that S = d = 2 . For every g = (g1, g2) F2, where g1 F2, g2 F2 , the corresponding eigenvalue| is| as follows: ∈ ∈ ∈ Case 1: g1 = 0, g2 , 0.

g2 s1 g2 s2 g2 s2 λg = χg(s) = ( 1) · + ( 1) · ( 1) · = 0. − − − − Xs S sX1 S 1 sXF2m sX2 S 1 ∈ ∈ 2∈ 2 ∈

Case 2: g1 = 1, g2 , 0.

g2 s1 g2 s2 m f(g2) λg = χg(s) = ( 1) · ( 1) · = F(g2) = 2 ( 1) . − − − − − − e Xs S sX1 S 1 sXS ∈ ∈ 2∈ 1 b

Case 3: g1 = 1, g2 = 0.

g2 s1 g2 s2 2m m λg = χg(s) = ( 1) · ( 1) · = S S = 2 + 2 S = 2 . − − − | 1|−| 1| − | 1| − Xs S sX1 S 1 sXS ∈ ∈ 2∈ 1

Thus for every h = (h1, h2) , 0, we have

2m 2 , if g1 = h1, g2 = h2,  0, if g1 = h1, g2 , h2, λg+h = (36)  m f(g2+h2) , ,  2 ( 1) , if g1 h1, g2 h2,  − m − e 2 , if g1 , h1, g2 = h2.  −   = π = n = π From (36), one can see that if t 2ℓ , ℓ m 1, then Rt(h) 2 for all h G. If t 2m , then 2m π ≤ − ∈ Rt(h) = 2 for all h , 0. For t = m+1 , we have ± 2

exp ı(λg+h λg)t − Xg G ∈ = δm exp ıλ h ,h t + exp ıλ h ,g )t − ( 1 2) − ( 1 2) gX2,h2 f(g2+h2) ı ( 1) exp ıλ(1+h1,g2))t ı exp( ıλ(1+h1,h2)t) − − e − − − gX2,h2

f(g2+h2) = (δm 1) exp ıλ h ,h t + exp( ıλ h ,g t) ı ( 1) exp( ıλ +h ,g t), − − ( 1 2) − ( 1 2) − − − (1 1 2) gXF2m gXF2m e 2∈ 2 2∈ 2 where δm = 1 if m = 1, and 1 if m > 1. Now, if m > 1, then −

exp( ıλ ,g t) = exp( ıλ , t) + exp(ıλ ,g t) − (0 2) − (0 0) (0 2) gXF2m gX2,0 2∈ 2 = exp( ı22mt) + exp(0) − gX2,0 = 1 + (22m 1) − = 22m,

24 and when h2 , 0,

f(g2+h2) ( 1) exp( ıλ ,g t) − − (1 2) gXF2m e 2∈ 2 f(h2) f(g2+h2) = ( 1) exp( ıλ(1,0)t) + ( 1) exp( ıλ(1,g2)t) − e − − e − gX2,0 = ( 1) f(h2) exp( ı( 2m)t) + ( 1) f(g2+h2) exp( ı( 2m( 1) f(g2)t) − e − − − e − − − e gX2,0 f(h ) f(g +h ) f (g ) = ı( 1) 2 + ı ( 1) 2 2 − 2 − e − e e gX2,0 f(g +h ) f (g ) = ı ( 1) 2 2 − 2 − gXF2m e e 2∈ 2 = 0 (since f is bent).

2m Thus when m > 1 and h = (0, h2), h2 , 0, we have Rt(h) = 2 , and Γ cannot have uniform = π mixing at time t 2m+1 . In fact, we can use Theorem 10 to give a simple proof of the above statement as follows: Since it is obvious that ν = m and if we take h = (0, h2), h2 , 0, then from (36), one can check that n2(h) , n1(h). Thus Γ cannot have uniform mixing at that time. When m = 1, we take f (x , x ) = x x , then f is bent, its supporting set is S = (11) . A 1 2 1 2 1 { } slight modiﬁcation of the precede procedure, one can show that Rπ/4(h) = 0 for all h , 0, thus Γ π has uniform mixing at time t = 4 . Below, we provide a direct proof of this fact. Indeed, from the above computation, we have that the eigenvalues of Γ are

λ = 4, λ = λ = λ = 0, λ = λ = λ = 2, λ = 2. (000) (001) (010) (011) (100) (101) (110) − (111) The diﬀerence table of the eigenvalues are presented in the following Table 2.

Table 2. λg+h λg − h g 000 001 010 011 100 101 110 111 \ 001 -4 4 0 0 0 0 4 -4 010 -4 0 4 0 0 4 0 -4 011 -4 0 0 4 4 0 0 -4 100 -6 -2 -2 2 6 2 2 -2 101 -6 -2 2 -2 2 6 2 -2 110 -6 2 -2 -2 2 2 6 -2 111 -2 -2 -2 -2 2 2 2 2 For each of the first three rows in Table 2, the corresponding correlation is π π Rt(h) = 4 + 2 exp(ı4 ) + 2 exp( ı4 ) = 0. × 4 − × 4 For each of the middle three rows, the corresponding correlation is 6π 6π π π Rt(h) = exp( ı ) + exp(ı ) + 3 exp(ı2 ) + 3 exp( ı2 ) = 0. − 4 4 × 4 − × 4 25 For the last row, π π Rt(h) = 4 exp(ı2 ) + 4 exp( ı2 ) = 0. × 4 − × 4 π By Corollary 1, Γ has uniform mixing at t = 4 . Moreover, from Table 2, one can see that Γ cannot have uniform mixing at time less than π/4. We note that the case for m = 1 also illustrates Theorem 10. Indeed, from Table 2, it is easy to see that gcd(λg+h λg : g G, 0 , h G) = 2 and then the corresponding number ν in − ∈ ∈ Theorem 10 is ν = 1. For the first three rows of Table 2, we have n2(h) = n1(h) = 4, and for the rest four rows, we have n2(h) = n1(h) = 0. Thus the conditions for Theorem 10 are satisfied and π then Γ has uniform mixing at time t = 4 . As a consequence of Theorem 8, 12 and Example 1, we can construct more cublelike graphs having uniform mixing by using Cartesian product.

Conclusion Remarks

In this paper, we mainly present the following results: 1. A necessary and sufficient condition for an abelian Cayley graph having uniform mixing is provided (see Theorem 1, Corollary 1, Corollary 2). 2. For an integral abelian Cayley graph Γ having uniform mixing at time t, we show that exp(ıt) is a root of unity under certain constrictions (see Theorem 2). We hope that some of the constrains can be removed. 3. For an abelian integral Cayley p-graph,we show that it has uniform mixing at a time of the form 2πr/pe′ if and only if its eigenvalues induces a certain difference-balanced mapping (see Theorem 4). 4. For an abelian group G, we show that if Cay(G, S ) is integral and has uniform mixing at d r d some time t, then G is an elementary p-group, where p = 2 or 3, or G = Z4, Z2 Z4 for some integers r 1, d 1. Conversely, if G is an elementary p-group, where p×= 2 d ≥r d≥ or 3, or if G = Z4, Z2 Z4 for some integers r 1, d 1, then Cay(G, S ) has uniform mixing at some time t for× some connection set S ≥. See Theorem≥ 9. This gives a complete classification of integral abelian Cayley graphs having uniform mixing. 2m m 5. For cubelike graphs, we show that Γ = Cay(F2 ; S ) has uniform mixing at time t = π/2 if f is a bent function, where f is the characteristic function of S (see Theorem 11). 6. Some concrete constructions of graphs having uniform mixing are presented (see Theorem 5, Example 1, Theorem 11, and Theorem 12.) For further research, we list the following open questions: Open question 1: Find more properties of the characteristic function of S such that the n n Cayley graph Cay(Z3; S ) (or Cay(Z4; S )) has uniform mixing at some time t. n n Open question 2: Find some lower boundson the time t at which Cay(Z3; S ) (or Cay(Z4; S )) has uniform mixing. Open question 3: If Γ= Cay(G, S ) is an abelian Cayley graph having uniform mixing, prove or disprove that Γ is integral. Open question 4: If Γ = Cay(G, S ) is an abelian Cayley graph having uniform mixing at time t, prove or disprove that exp(ıt) is a root of unity.

26 Acknowledgements

The authors would like to express their grateful thankfulness to the editor and the referee for their valuable comments and suggestions. They also thank A. Chan, P. Sin and C. Godsil for helpful discussing on related topics.

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6. Appendix: Proof of Theorem 9

Proof. We ﬁrst prove the following result by generating the idea of Theorem 3.

Theorem 13. Suppose that G = G G is a direct product of two abelian groups with exp(Gi) = 1 × 2 ei > 1, i = 1, 2 and gcd(e1, e2) = 1. If Γ= Cay(G, S ) is an integral graph, then it cannot exhibit uniform mixing at any time t.

Proof. Firstly, observe that the dual group of G is the product of the dual group of G1 and G2, (i) respectively, i.e., G = G1 G2. For every χx G1 (resp. χy G2), and g Gi, i = 1, 2, (1) (2) × ∈ ∈ ∈ χx(g ) Q(ζe ), χy(g ) Q(ζe ). Denote e = e1e2. Then the characters of G take values ∈ 1 b c ∈ c 2 c b in the cyclotomic ﬁeld Q(ζe). Γ is integral if and only its eigenvalues are invariant under the Galois group of Q(ζe). Since Gal(Q(ζe)/Q) Ze∗, and by the Chinese Reminder Theorem,

28 Z∗ Z∗ Z∗ , we know that for every (a, b) G1 G2 and ℓ Z∗, e e1 × e2 ∈ × ∈ e

ℓ(a, b) = (ℓ1a,ℓ2b), (37) where ℓi ℓ (mod ei), and ℓi Z∗ , i = 1, 2. Conversely, for every pair of ℓi Z∗ , i = 1, 2, ≡ ∈ ei ∈ ei there is a ℓ Z∗ satisfying (37). Therefore, if we denote the orbits of action of Z∗ (resp. Z∗ ) on ∈ e e1 e2 G (resp. G ) by [0], [g(1)], , [g(1)] and [0], [g(2)], , [g(2)], respectively, then the orbits of G 1 2 1 ··· m 1 ··· n under the action of Ze∗ are as follows. [(g(1), 0)], [(0, g(2))], [(g(1), g(2))], where 1 i m, 1 j n. (38) i j i j ≤ ≤ ≤ ≤ Thus the connection set S should be a union of those orbits listed in (38). Say, without loss of generality, = (1) (2) (1) (2) S i I1 [(gi , 0)] j J1 [(0, g j )] i I2, j J2 [(gi , g j )] , (39) ∪ ∈ ∪ ∪ ∈ ∪ ∪ ∈ ∈ where I , I are subset of 1, 2, , m and J , J are subsets of 1, 2, , n . Note that 1 2 { ··· } 1 2 { ··· } [(g(1), 0)] = (ℓ g(1), 0):1 ℓ ord(g(1)), gcd(ℓ , ord(g(1))) = 1 , i { 1 i ≤ 1 ≤ i 1 i } [(0, g(2))] = (0,ℓ g(2)):1 ℓ ord(g(2)), gcd(ℓ , ord(g(2))) = 1 , j { 2 j ≤ 2 ≤ j 2 j } [(g(1), g(2))] = (ℓ g(1),ℓ g(2)) : gcd(ℓ , ord(g(1))) = 1, gcd(ℓ , ord(g(2))) = 1 . i j { 1 i 2 j 1 i 2 j } We assume that e is odd. For every (x, y) G, the corresponding eigenvalue is 1 ∈ (1) (2) λ(x,y) = χx(ℓ1gi ) + χy(ℓ2g j ) Z Z Xi I1 ℓ1 X∗ (1) Xj J1 ℓ2 X∗ (2) ord(g ) ord(g ) ∈ ∈ i ∈ ∈ j (1) (2) + χx(ℓ1gi ) χy(ℓ2g j ) Z Z i IX2, j J2 ℓ1 X∗ (1) ℓ2Xe∗ ord(g ) 2 ∈ ∈ ∈ i ∈ (2) = χ (1) (ℓ1x) + χy(ℓ2g ) gi j Z Z Xi I1 ℓ1 X∗ (1) Xj J1 ℓ2 X∗ (2) ord(g ) ord(g ) ∈ ∈ i ∈ ∈ j (2) + χ (1) (ℓ1x) χy(ℓ2g ). gi j Z Z i IX2, j J2 ℓ1 X∗ (1) ℓ2 X∗ (2) g g ∈ ∈ ∈ ord( i ) ∈ ord( j )

Note that in the above computation, we make use of the property χa(b) = χb(a) for every a, b G . Thus for every h G , we have ∈ 1 ∈ 1

λ(x+h,y) λ(x,y) = (χ (1) (ℓ1(x + h)) χ (1) (ℓ1x)) gi gi − Z − Xi I1 ℓ1 X∗ (1) ord(g ) ∈ ∈ i (2) + χy(ℓ2g ) (χ (1) (ℓ1(x + h)) χ (1) (ℓ1x)). j gi gi Z Z − Xj J2 ℓ2 X∗ (2) Xi I2 ℓ1 X∗ (1) ord(g ) ord(g ) ∈ ∈ j ∈ ∈ i

Let G = Syl (G), where Syl (G) is the Sylow p-subgroup of G. We can assume p prime, p G p p | | G1 is a SylowQ p-subgroup of G, otherwise, one can replace G1 with a Sylow subgroup and follow r the discussion above. Suppose that G1 = p . Repeating the procedure of the proof of Theorem | | 29 r r 1 7, we know that there exists an element 0 , h G1 such that there are at least p 2p − many x’s satisfying ∈ −

Z χ (1) (ℓ1(x + h)) = Z χ (1) (ℓ1x), i I1 ℓ1 ∗ (1) g i I1 ℓ1 ∗ (1) g ∈ ∈ ord(g ) i ∈ ∈ ord(g ) i  P P i P P i Z (χ (1) (ℓ1(x + h)) = Z (χ (1) (ℓ1 x).  i I2 ℓ1 ∗ (1) g i I2 ℓ1 ∗ (1) g  ∈ ∈ ord(g ) i ∈ ∈ ord(g ) i  i i  P P P P  See also Remark 3 (1). Note that the summation Z χ (1) (ℓ1x) is exactly the Ramanu- ℓ1 ∗ (1) g g i ∈ ord( i ) (1) P jun function on the orbit containing gi . Consequently, the probability of zero in the multi-set 2 λ(x+h,y) λ(x,y) :(x, y) G is 1 p . As a result, if G either has a Sylow p-subgroup with { ∗ − ∈ ∗ } − r p > 3, or the Sylow p-subgroup is neither an elementary p-group, where p = 2, 3, nor Z4 for some r, then G does not exhibit uniform mixing for any connection S . Therefore, there are two cases left which are needed to deal with. a b Case (1): G = Z2 Z3, where a, b > 1. a × b Case (2): G = Z4 Z3, where a, b > 1. We just give the× proof of Case 2, since the proof of Case 1 is similar. We assume that a b a b a 1. Then G Z12 Z3− . The case of a b 1 can be handled analogously. ≥Since≥ Γ is an integral graph,× we can assume that≥ ≥ the connection set S is as follows.

S = (ℓu,ℓv) : u I, v J,ℓ Z∗ , { ∈ ∈ ∈ 12} a b a a b a where I (resp. J) is a subset of Z12 (resp. Z3− ). For every x Z12, y Z3− , the corresponding eigenvalue is ∈ ∈

ℓu x ℓv y ℓ(u x+4v y) λ x,y = ζ · ζ · = ζ · · = c(u x + 4v y, 12). ( ) 12 3 12 · · Xu I Xv J ℓXZ Xu I Xv J ℓXZ Xu I Xv J ∈ ∈ ∈ 12∗ ∈ ∈ ∈ 12∗ ∈ ∈ a Take κ = Z12 and regard it as a ring. Then H := Z12 is a finite generated κ-module. It is obvious that rankκ(H) = a. Suppose that β , , βa I is a generating set of H. Then we can { 1 ··· } ⊆ find an nonzero element γ H such that βi γ = 0, 1 i a 1. For every β j, a j I , a ( j) ( j) ∈ · ( j) ≤ ≤ − ≤ ≤ | | write β j = k βi, k κ. Then β j γ = k βa γ. Moreover, if β j γ . 1 (mod 3), then we i=1 i i ∈ · a · · can replacePβ j in the generating set with β j. In other words, we can assume that β j γ 0, 1 (mod 3) for all j a. By (27), we have that− for every integer k, · ≡ ≥ 4, if k 0 (mod12), 0, if k ≡ 1 (mod2),  ≡ c(k, 12) =  4, if k 6 (mod12), (40)  − ≡  2, if k 2, 10 (mod 12),  ≡  2, if k 4, 8 (mod12).  − ≡  Thus for every β j with β j γ . 0 (mod 12),wehave c((x+4γ) β j +4v y, 12) , c(x β j +4v y, 12) · · · b a · · only if x β j (mod 12) 0, 2, 6, 10 . Let x run through H and y through Z − . By the Chinese · ∈ { } 3 Reminder Theorem, the total number of x such that x β j (mod 12) 0, 2, 6, 10 for some j 1 a · ∈ { 2 a } b a 1 with β j γ . 0 (mod 12) is at most 12 . Consequently, there are at least 12 3 − > G · 3 · 3 · · 2 | | many x’s such that λ(x+4γ,y) = λ(x,y). Thus the correlation function satisfies Rγ(t) > 0 for all t. By Corollary 1, Γ does not exhibit uniform mixing at any time t. | | Combining Theorem 5, 6, 7, and 13, we get the proof of Theorem 9. 30