Gauss periods and related Combinatorics

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2 Fundamentals of characters of abelian groups A character ψ of G is a homomorphism from G to C Ĝ: the set of all characters of G e: the exponent of G Remark The image of ψ is an eth root of unity since ψ(x) e = ψ(x e ) = ψ(1 G ) = 1 Note that ψ(1 G ) = 1 by ψ(1 G ) 2 = ψ(1 G )

3 Fundamentals of characters of abelian groups Remark Define ψ 0 (g) := 1 for g G Then ψ 0 is a character, called the trivial character Theorem Define ψ 1 (g) := ψ(g) 1 for a character ψ Then ψ 1 is a character, called the inverse of ψ Define ψ 1 ψ 2 (g) := ψ 1 (g)ψ 2 (g) for characters ψ 1, ψ 2 Then ψ 1 ψ 2 is a character Ṭhe set Ĝ forms a group isomorphic to G

4 Example Example: Z 3 Possible cases: ψ((0, 1, 2)) = (1, 1, 1), (1, 1, ω), (1, 1, ω 2 ), (1, ω, 1), (1, ω 2, 1), (1, ω, ω), (1, ω, ω 2 ), (1, ω 2, ω 2 ), (1, ω 2, ω) By noting that ψ(1)ψ(2) = ψ(1 + 2) = ψ(0) = 1, Only ψ((0, 1, 2)) = (1, 1, 1), (1, ω, ω 2 ), (1, ω 2, ω) are possible These three are all characters of Z 3

5 Orthogonal relations Theorem (1) For ψ, ψ Ĝ, ψ(g)ψ (g) = δ ψ1,ψ 2 G, g G where δ ψ,ψ = { 1 if ψ = ψ, 0 if ψ ψ (2) For g, h G, ψ(g)ψ(h) = δ g,h G ψ Ĝ where δ g,h = { 1 if g = h, 0 if g h

6 Orthogonal relations Proof of (1): Put ϕ = ψψ 1 If ϕ = ψ 0, ϕ(g) = g G 1 = G g G If ϕ ψ 0, ϕ(g ) ϕ(g) = ϕ(g )ϕ(g) = ϕ(g g) = ϕ(g), g G g G which implies that g G ϕ(g) = 0 g G g G

7 Definition Definition: Cayley graph G: a finite abelian group D: an inverse-closed subset of G (0 D and D = D) E := {(x, y) x, y G, x y D} (G, E) is called a Cayley graph, denoted by Cay(G, D) D is called the connection set of (G, E)

8 Paley graph G = Z 3 Z 3, D = {(0, 1), (0, 2), (2, 1), (1, 2)}

9 Eigenvalues of Cayley graphs Γ: a Cayley graph on an abelian group G with connection set D Ĝ: the character group of G M: the character table of G (Each of rows and columns are labeled by the elements of Ĝ and the elements of G, respectively The (ψ, g)-entry is defined by ψ(g)) A: the adjacency matrix of Γ (Each row and column are labeled similar to the columns of M) Theorem: Eigenvalues and character sums M AM T G = diag x D ψ(x) ψ Ĝ ie, the eigenvalues of A are given by ψ(d), ψ Ĝ,

10 Proof (MM T ) ψ,ψ = ψ(h)ψ (h) = ψψ 1 (h) = h G h G This implies that M/ G is an orthogonal matrix By (M A) ψ,g = ψ(h) = ψ(e + g), we have (M AM T ) ψ,ψ = = = h G;h g D e D { G if ψ = ψ, 0 if ψ ψ, ψ(e + g)ψ (g) = ψ(e) ψ(g)ψ (g) g G e D ψ(e) ψψ 1 (g) e D g G e D { G e D ψ(e) if ψ = ψ, 0 if ψ ψ g G

11 Cyclotomic classes F q : the finite field of order q F q : the multiplicative group of F q C : F q st C = C Each coset (called a cyclotomic coset) of F q /C is expressed as C (N,q) i = γ i γ N, 0 i N 1, where N q 1 is a positive integer and γ is a fixed primitive element of F q

12 Characters of finite fields There are two kinds of characters for finite fields, which are additive characters and multiplicative characters Lemma For a fixed primitive element γ F q, χ j : F q C, 0 j q 2, defined by χ j (γ k ) := ζ jk q 1 are all multiplicative characters of F q, where ζ q 1 = e 2πi Define the trace Tr q m /q from F q m to F q by Tr q m /q(x) = x + x q + x q2 + + x qm 1, which is a homomorphism from (F q m, +) to (F q, +) q 1

13 Characters of finite fields Lemma The function ψ j : F q C, j F q, defined by are all additive characters of F q ψ j (x) = ζ Tr q/p( jx) p It holds that ψ j (x + y) = ψ j (x)ψ j (y) since Tr is a homomorphism from F q to F p ψ 1 is called canonical Note that ψ a (x) = ψ 1 (ax) and ψ(x) = ψ( x)

14 Gauss sums Definition For the canonical additive character ψ of F q and a nontrivial multiplicative character χ of F q, the sum G(χ) := ψ(x)χ(x) is called a Gauss sum x F q Lemma For any nontrivial multiplicative character χ of F q, G(χ)G(χ) = q The lemma above implies that G(χ) = q

15 G(χ)G(χ) = q Proof: G(χ)G(χ) = ψ(x)ψ( y)χ(x)χ 1 (y) = x,y F q ψ(x y)χ(xy 1 ) (1) x,y F q Write z = xy 1 Then, (1) = χ(z)ψ(y(z 1)) = y,z F q χ(z) ψ(y(z 1)) χ(z) = q y F q z F q z F q

16 Eigenvalues and Gauss periods Definition: Gauss period For the canonical additive character ψ of F q, η i := ψ(x), 0 i N 1, x C (N,q) i are called Gauss periods of order N of F q Gauss periods are all eigenvalues of Cay(F q, C (N,q) ) i

17 Gauss periods and Gauss sums Lemma: Gauss periods and Gauss sums (i) ψ(c (N,q) ) = 1 N 1 G(χ h )χ h (γ i ), i N (ii) G(χ) = N 1 h=0 where χ is a multiplicative character of order N of F q i=0 ψ(c (N,q) )χ(γ i ), i

18 Eigenvalues of cyclotomic schemes ψ(c (N,q) ) = 1 i N = 1 N = = = 1 N ψ(γ i x N ) x F q x F q 1 (q 1)N 1 (q 1)N χ C 0 1 q 1 where C 0 is the subgroup of F q ψ(y) χ(γ i x N )χ(y) y F q x F q χ F q χ F q G(χ 1 )χ(γ i x N ) G(χ 1 )χ(γ i ) χ(x N ) χ F q G(χ 1 )χ(γ i ), x F q consisting of all χ trivial on C(N,q) 0

19 Evaluated Gauss sums (Small order) Gauss sums of order N 24 have been partially evaluated (Berndt et al, 1997) (Pureness) When Gauss sums take the form ζ N q was determined (Aoki, 2004, 2012) In particular, if 1 p (mod N), then G(χ N ) takes a rational value (Baumert et al, 1982) (Index 2 or 4 case) In the case where [Z : p ] = 2, Gauss N sums have been completely evaluated (Yang et al, 2010) In the case where [Z : p ] = 4, Gauss sums have been N partially evaluated (Feng et al, 2005)

20 Remarks The computation of Gauss periods is equivalent to that of weight distributions of certain cyclic codes, called irreducible cyclic codes Two-weight irreducible cyclic code is related to the problem of finding cyclic difference sets in design theory Gauss periods are eigenvalues of cyclotomic schemes A 2-class fusion (strongly regular graph) of a cyclotomic scheme is related to the problem of finding two-intersection sets in finite projective geometry

21 Cyclic codes An (m, r)-code C is an r-dimensional subspace of F m q If (c m 1, c 0,, c m 2 ) C for any (c 0, c 1,, c m 1 ) C, then C is called cyclic One may identify each vector (c 0, c 1,, c m 1 ) F m p with h(x) F p [x] st h(x) m 1 i=0 c i x i (mod x m 1) by using the isomorphism between the principal ideal ring A := F p [x]/(x m 1)F p [x] and F m p Note that every ideal of A is given by Ag(x) for some g(x) x m 1 If deg(g(x)) = m r, then C = Ag(x) forms a cyclic (m, r)-code

22 Irreducible cyclic codes Definition: Irreducible cyclic code f(x): an irreducible divisor of x m 1 F p [x], where gcd (m, p) = 1 The cyclic code of length m over F p generated by (x m 1)/ f(x) is called an irreducible cyclic code (This code has no proper cyclic subcodes) f: the order of p modulo m q := p f = 1 + km γ: a primitive root of F q f(x) := f 1 i=0 (x γkpi ) F p [x] irreducible over F p g(x) := l S(x γ kl ), where S = {l 0 l m 1, l a power of p (mod m)}

23 Irreducible cyclic codes Lemma C: the cyclic code generated by g(x) The q codewords in C are given by h α (x) := (Tr(α), Tr(αγ k ), Tr(αγ 2k ),, Tr(αγ (m 1)k )), α F q Proof: For any l S, h α (γ kl ) = h α (x) := m 1 j=0 m 1 j=0 Tr(αγ jk )x j, α F q f 1 Tr q/p (αγ jk )γ kl j = Hence, g(x) h α (x), ie, h α (x) C i=0 m 1 α pi j=0 γ jk(l pi) = 0

24 Proof Since C = q, it remains to show that h α (x) are all distinct Assume h α (x) = h β (x) Then, for ω := α β F q Tr(ω) = Tr(ωγ k ) = Tr(ωγ 2k ) = = Tr(ωγ (m 1)k ) = 0 For any choice of a j F p, 0 = f 1 j=0 f 1 a j Tr(ωγ jk ) = Tr(ω a j γ jk ) Since {1, γ k,, γ ( f 1)k } is a basis of F q over F p, the above is impossible j=0

25 Weight-distribution Theorem (McEliece) Let N := gcd (k, (q 1)/(p 1)) Then, w(h α (x)) = m(p 1) p 1 p ) pk ψ(αc(n,pf 0 Proof: Let χ be a mult character of order k of F q w(h α (x)) =m 1 p m 1 i=0 x F p ψ(xαγ ki ) m(p 1) = 1 m 1 p p ψ(xαγ ki ) i=0 x F p m(p 1) = 1 k 1 p pk j=0 x F p ) G(χ j )χ j (xα)

26 Weight-distribution Since for any y F p χ j (x) = χ j (yx) = χ j (y) χ j (x), x F p x F p x F p x F p χ j (x) = 0 iff χ j is nontrivial on F p Let χ be a mult character of order N of F q Then, m(p 1) w(h α (x)) = p 1 N 1 G(χ j )χ j (α) p pk m(p 1) = p 1 p j=0 ) pk ψ(αc(n,pf 0 Problem Characterize all two or three weight irreducible cyclic codes )

27 Two-valued Gauss periods Two-valued Gauss period was studied by Schmidt-White (2002) In this case, Cay(F p f, C (N,pf ) ) has exactly two nontrivial 0 eigenvalues, ie, it is strongly regular Theorem (Schmidt-White, 2002) Assume that N pf 1 p 1 Then, Gauss periods take exactly two values iff k N st (i) k N 1 (ii) kp sθ ±1 (mod N), (iii) k(n k) = (N 1)p s(m 2θ), where f = ms, m = ord N (p), and p θ G p m(χ) = x F p m χ(x)ψ(x) for all nontrivial multiplicative characters χ of exponent N of F p f

28 Schmidt-White conjecture Conjecture (Schmidt-White, 2002) Assume that N pf 1 If Gauss periods take exactly two values, p 1 either one of the following holds (subfield case) C (N,pf ) = F 0 pe where e f, (semi-primitive case) 1 p Z N, (exceptional case) it has either of the following parameters: No N p f No N p f

29 Difference set Definition: abelian difference set G: an abelian group of order v A k-subset D G is called a (v, k, λ) difference set (DS) if {xy 1 x, y D; x y} covers every nonzero element of G exactly λ times Note that D is a difference set in G iff ie, DD ( 1) = (k λ) [1 G ] + λ G, χ(d)χ 1 (D) = k λ for any nontrivial character χ of G

30 Designs appeared in the index set Assume that ψ(c (N,q) a )(= ψ(γ a C (N,q) )), a = 0, 1,, N 1, take 0 exactly two values, say α 1 and α 2 Set I i = { a ψ(γ a C (N,q) ) = α i, 0 a N 1 }, i = 1, 2 0 I i forms a difference set in Z N, called a subdifference set Why?

31 Designs appeared in index sets Then, G q (χ) = α 1 χ(γ a ) + α 2 χ(γ a ) = (α 1 α 2 ) χ(γ a ) a I 1 a I 2 a I 1 By G q (χ)g q (χ 1 ) = q, χ(γ a ) χ 1 (γ a ) a I 1 a I 1 is constant, ie, I 1 is a difference set in Z N

32 Subdifference sets In subfield case, the corresponding subdifference sets are the Singer difference sets In the sporadic case, the corresponding subdifference sets are listed below: No v k λ Name Quadratic residue DS Quadratic residue DS Twin-prime DS Biquadratic residue DS Hall s sextic DS Quadratic residue DS Quadratic residue DS Hall s sporadic DS Quadratic residue DS Twin-prime DS Quadratic residue DS

33 Definition Definition: Translation scheme Γ i := (G, E i ), 1 i d: Cayley graphs on an abelian group G R i : connection sets of (G, E i ) R 0 := {0} (G, {R i } d ) is called a translation scheme (TS) i=0 if (G, {Γ i } d ) is an association scheme (AS) i=0 In other words d R i=0 i = G, R i R j = {z (x, z) E i, (y, z) E j } is const according to l st (x, y) E l {z x z R i, y z R j } is const according to l st x y R l (R i + x y) R j is const according to l st x y R l (R i + w) R j is constant according to l st w R l

34 Dual of translation schemes R 0 = {0}, R 1, R 2,, R d : an (inverse-closed) partition of G This partition induces a partition S 0 = {ψ 0 }, S 1, S 2,, S e, of Ĝ: ψ, ϕ Ĝ \ {ψ 0 } are in the same S j iff ψ(r i ) = ϕ(r i ) for 1 i d Theorem (Bridges-Mena, 1982) It holds that d e In particular, (G, {R i } d ) forms a TS iff d = e i=0 R 0 R 1 R 2 R 3 ψ 0 S 0 1 R 1 R 2 R 3 ψ S 1 1 a 1 a 2 a 3 ψ S 2 1 b 1 b 2 b 3 ψ S 3 1 c 1 c 2 c 3 If (G, {R i } d ) forms a TS, then so does (Ĝ, {S i } d ), which is called i=0 i=0 the dual of (G, {R i } d ) i=0 G P 1 is the first eigenmatrix of (Ĝ, {S i } d ) i=0 for the first eigenmatrix P of (G, {R i } d i=0 )

35 Cyclotomic scheme F q : the finite field of order q F q : the multiplicative group of F q C : F q st C = C Lemma: Cyclotomic scheme The partition F q /C of F q gives a TS on (F q, +), called a cyclotomic scheme Each coset (called a cyclotomic coset) of F q /C is expressed as C (N,q) i = γ i γ N, 0 i N 1, where N q 1 is a positive integer and γ is a fixed primitive element of F q For w C (N,q), l p l i, j = (C(N,q) i Hence, p l is depending on l not w i, j + w) C (N,q) = (C (N,q) + 1) C (N,q) j i l j l

36 Fusion schemes Given a d-class AS (X, {R i } d ), we can take union of classes to i=0 form graphs with larger edge sets (this process is called a fusion) Big problem! Given an N-class cyclotomic scheme on F q, determine its fusion schemes X j, j = 1, 2,, d: a partition of Z N The Bridges-Mena theorem (more generally, the Bannai-Muzychuk criterion) implies that i X j C (N,q) s forms a TS i iff a partition Y h, h = 1, 2,, d, of Z N st each ψ(γ a i X j C (N,q) ) is const according to a Y i h

37 Gauss sum and trace zero We consider 2-class fusion schemes (strongly regular graphs) of a cyclotomic scheme of order N = qm 1 q 1 Proposition Let χ be a mult character of order N of F q m Let S 0 := {log γ x (mod N) Tr q m /q(x) = 0, x 0} Then, G(χ) = q χ(ω i ) i S 0 L :=a system of representatives of F q m/f q G(χ) = a F x L q χ(xa)ζ Tr q m /p(xa) p = =(q 1) χ(γ i ) i S 0 χ(x) x L a F q χ(γ i ) = q χ(γ i ) i L\S 0 i S 0 ζ Tr q/p(atr q m /q (x)) p

38 Geometric understanding X: a subset of Z N When is Γ = Cay( i X C (N,qm ) ) strongly regular? i (Γ is strongly regular iff ψ(γ a i X C (N,qm ) ), a = 0, 1,, N 1, i take exactly two values) ψ(γ a i X C (N,qm ) ) = 1 i N = q N i X χ G(χ 1 )χ(γ a+i ) X χ χ 0 N i X =q X (S 0 a) X, χ(γ a+i j ) X (1 + q S 0 ) N j S 0 where χ ranges through all mult characters of exponent N of F q m

39 Geometric understanding Proposition (Delsarte, 1972) Cay( i X C (N,qm ) ) is strongly regular iff X (S i 0 a), a Z N, take exactly two values What is S 0 a? Consider the (m 1)-dimensional projective space PG(m 1, q) on the point set Z N F q m/f q All hyperplanes ((m 1)-dimensional subspaces of F q m) of PG(m 1, q) are given by S 0 a, a Z N

40 Geometric understanding Problem Find a subset X of PG(m 1, q), which has two intersection numbers with the hyperplanes of PG(m 1, q) (X is called a two-intersection set in PG(m 1, q)) See Calderbank-Kantor (1986) for more on the geometric aspect of strongly regular graphs on F q

41 Problems on three-valued Gauss periods Problems Necessary condition Sufficient condition Examples Related combinatorial structures When do the index sets in Z N yield an interesting design? When does the partition of F q derived from the index sets form a three-class (translation) association scheme?

42 Designs appeared in the index sets Assume that ψ(γ a C (N,q) ), a = 0, 1,, N 1, take exactly three 0 values α 1, α 2, α 3 Set I i = { a ψ(γ a C (N,q) ) = α i, 0 a N 1 }, i = 1, 2, 3 0 By the orthogonality of characters, G q (χ) = α 1 χ(γ a ) + α 2 χ(γ a ) + α 3 χ(γ a ) a I 1 a I 2 a I 3 = (α 1 α 2 ) χ(γ a ) + (α 3 α 2 ) χ(γ a ) a I 1 a I 3

43 Designs appeared in the index sets By G q (χ)g q (χ 1 ) = q, we have ((α 1 α 2 )I 1 +(α 3 α 2 )I 3 )((α 1 α 2 )I 1 +(α 3 α 2 )I 3 ) ( 1) = k[0]+λz N Then, A = (α 1 α 2 )I 1 + (α 3 α 2 )I 3 Z[Z N ] forms a cyclic weighted design In particular, if α 1 α 2 = α 2 α 3, then A/(α 1 α 2 ) gives a cyclic balanced ternary design Furthermore, if λ = 0, then A/(α 1 α 2 ) gives a circulant weighing matrix

44 Designs appeared in the index sets Definition: balanced weighted design A square matrix M with entries from Z st MM T = (k λ)i + λj is called a weighted design If its entries are from {0, 1, 1}, it is called a balanced ternary design Furthermore, if λ = 0, it is called a weighing matrix

45 Remarks I 1 + I 2 + I 3 = N, α 1 I 1 + α 2 I 2 + α 3 I 3 = N 1 ψ(c i=0 i) = 1, Since (α 1 I 1 +α 2 I 2 +α 3 I 3 )(α 1 I ( 1) 1 +α 2 I ( 1) 2 by comparing the coefficient of [0] of both sides, Thus, +α 3 I ( 1) ) = q [0] q 1 3 N Z N, α 2 1 I 1 + α 2 2 I 2 + α 2 3 I 3 = q q 1 N I 1 = α 2α 3 (q 1) + k(q k + α 2 + α 3 ), k(α 1 α 2 )(α 3 α 1 ) I 2 = α 1α 3 (q 1) + k(q k + α 1 + α 3 ), k(α 1 α 2 )(α 2 α 3 ) I 3 = α 1α 2 (q 1) + k(q k + α 1 + α 2 ) k(α 2 α 3 )(α 3 α 1 )

46 Related association schemes Theorem Assume that gcd (p 1, N) = 1, and ψ(γ a C (N,q) ), 0 a = 0, 1,, N 1, take exactly three values α 1, α 2, α 3 Take the partition of F q : R 0 = {0}, R 1 =, R 2 =, R 3 = C (N,q) i i I 1 C (N,q) i i I 2 C (N,q) i i I 3 where I i = { a ψ(γ a C (N,q) } ) = α i, i = 1, 2, 3 If I1 = 1 or I 3 = 1, 0 then (F q, {R i } N 1 ) forms a self-dual 3-class TS i=0,

47 Example Let (p, f, N) = (7, 3, 19) Then, the Gauss periods take three values { 3 12, 4 6, 11}, and we have I 1 = {1, 2, 3, 4, 5, 6, 7, 9, 11, 14, 16, 17}, I 2 = {8, 10, 12, 13, 15, 18}, I 3 = {0} This satisfies the AP property, ie, I 1 I 3 Z[Z 19 ] gives a cyclic BT design For R 0 = {0}, R 1 = C (N,q) i i I 1, R 2 = (F q, {R i } N 1 ) forms a 3-class TS i=0 C (N,q) i i I 2, R 3 = C (N,q) i i I 3,

48 Sketch of proof We can assume that I 1 = 1, and let I 1 = {l} Since each I j is invariant under the multiplication by p, we have l = 0 Then, ψ(γ a C (N,q) ) = i i I 1 Next, we must compute ψ(γ a i I 2 C (N,q) ) i α 1 if a I 1 ; α 2 if a I 2 ; α 3 if a I 3

49 Sketch of proof Since the characteristic function f 2 of I 2 is given by f 2 (x) = (ψ(γ x C (N,q) ) α 1 )(ψ(γ x C (N,q) ) α 3 ) 0 0 (α 2 α 1 )(α 2 α 3 ) and ψ(γ a i I 2 C (N,q) ) = i x Z N f 2 (x)ψ(γ a+x C (N,q) ), we need to 0 compute ψ(γ x+a C (N,q) )ψ(γ x C (N,q) ) x Z N Let τ(x) = (ψ(γ x C (N,q) ) α 2 )/t, where t = gcd (α 3 α 2, α 2 α 1 ) 0 Then, the above is equivalent to compute τ(x + a)τ(x) 2 x Z N

50 Sketch of proof Let u = (α 1 α 2 )/t and v = (α 3 α 2 )/t By I 1 = 1, we have N { k if a = 0; τ(x + a)τ(x) = ux 1 + vx i = λ if a 0, x Z N i=l where x i = u, 0, or v (x i = u for exactly one i) 0th u 0 0 v v v ath x 1 x 2 x l 1 x l x l+1 x N Then, for a 0 we obtain x Z N τ(x + a)τ(x) 2 = u 2 x 1 + v = N vx i = u 2 x 1 + v(λ + ux 1 ) i=l { vλ if x1 = 0, ie, a I 2 ; vλ + v(vu + u 2 ) if x 1 = v, ie, a I 3

51 Examples of three-valued Gauss periods No parameters AP AS CW 1 p = 2, q = p 6 f, N = p3 f 1 p f 1 2 p: odd, q = p 6 f, N = p3 f 1 p f 1 3 q = p 3 f, N = p3 f 1 3(p f 1), pf 1 (mod 3) 4 q = p fe, k := pfe 1 N, k pf 1, Cay(F p f, C ( (pf 1)N p fe 1,pf ) 0 p f 1 k ) is an SRG 5 q = p lcm(e, f), e/ gcd (e, f) = 3, C (N,q) 0 6 N = p 1, [Z N pf 1 p 1 = F p e F p f, p ] = 2, f = d p for any d 7 N = p 1 p 2, [Z : p ] = 2, f = p N

52 Remark Remark We searched the existence of three-valued GPs by computer for q 2 25 with p < 330 and N < 1001 Most of examples with the AS property are covered by our examples, except for (p, f, N) = (7, 7, 29), (2, 11, 89) All examples with the CW property are covered by the 1st and 2nd classes

53 Open examples p f N Gauss periods AP , 4 6, , 16 18, , 16 18, , 16 18, , 18 20, , 30 32, , 46 48, , 45 47, , 48 50, , 58 60, , 64 66, , 84 86, , 46 48, 99 3 p f N Gauss periods AP , , , , , 88 90, , 60 62, , , , , , , , , , 94 96, , 72 74, , , , 74 76, , , Table: Computer search for p < 330, p f < 2 25, 6 < N < 1001, N pf 1 p 1 except for known examples

54 Necessary and sufficient condition for a partial case Theorem Assume that there are four positive integers u, v, r, s satisfying (i) t( ur + vs) 1 (mod N); (ii) (N 1)q + t 2 ( ur + vs) 2 = Nt 2 (u 2 r + v 2 s), where t is the largest power of p dividing G q (χ) for all nontrivial multiplicative characters of exponent N If u = v = r = 1 and s + 1 < N, or u = v = s = 1 and r + 1 < N, then ψ(γ a C (N,q) ), 0 a N 1, take exactly three values; in this 0 case, the AP property is satisfied This allows us an efficient computer search for N < 5000: Fix N and h := s + r Determine p and f by (ii) Determine t by Stickelberger s Thm Check (i) New examples: (p, f, N) = (13, 13, 53), (2, 36, 247)

55 An analogy of Schmidt-White s conjecture Conjecture If Gauss periods take three values satisfying the AP property and I 1 = 1 or I 3 = 1, either one of the following holds (i) q = p 3 f, N = p3 f 1 3(p f 1), and pf 1 (mod 3); or (ii) it has either of the following parameters: No N p f No N p f

56 Examples of three-valued Gauss periods Theorem If q = p 6 f, N = p3 f 1, then GPs take exactly three values In this p f 1 case, the index sets form a circulant weighing matrix Proof: We need to compute G p 6 f (χ) for a nontrivial character χ of exponent N of F q By the Hasse-Davenport thm, we have G p 6 f (χ) = G p 3 f (χ ) 2 for some character χ of exponent N of F p 3 f Since G p 3 f (χ ) = p f i S χ (γ i ), where S = {i Z N : Tr 3 f/ f (γ i ) = 0}, we need to compute S 2 in Z[Z N ]

57 Examples of three-valued Gauss periods The coefficient of [a] in S 2 is equal to the size of {i Z N : Tr 3 f/ f (γ i ) = 0, Tr 3 f/ f (γ i+a ) = 0} = Q (S a), where Q = {i Z N : Tr 3 f/ f (γ i ) = 0} Note that {x : Tr 3 f/ f (x 1 ) = 0} = {x : x q3 1 q 1 Tr 3 f/ f (x 1 ) = 0} = {x : Tr 3 f/ f (x 1+q ) = 0} Since Q is a conic in PG(2, p f ) and S a is a line of PG(2, p f ), we have Q (S a) = 0, 1, or 2, according to S a is passant, tangent, or secant

58 Thanks Thank you very much for your attention!

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