Regular positive ternary quadratic forms. Byeong-Kweon Oh. Department of Applied Mathematics. Sejong University. 22 Jan. 2008

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1 Regular positive ternary quadratic forms Byeong-Kweon Oh Department of Applied Mathematics Sejong University Jan. 008

2 ABSTRACT A positive definite quadratic form f is said to be regular if it globally represents all integers that are represented by the genus of f. In 997, Jagy, Kaplansky and Schiemann provided a list of 93 candidates of primitive positive definite regular ternary quadratic forms, and all but of them are verified to be regular. In this talk we show that 8 forms among candidates are, in fact, regular. At the end of the talk, we show some finiteness result on ternary forms that represent every eligible integer in some arithmetic progression.

3 [] Quadratic Forms (. Definition of quadratic forms (i S m (Z = {Symmetric matrices of rank m over Z} (ii L = Zx + +Zx m with a non-degenerate symmetric bilinear form B(x i, x j = m ij Z and Q(x := B(x, x. (iii f L (x, x..., x m = i,j m ij x i x j L M L := (B(x i, x j f L

4 (. Representation of quadratic forms (i For M S m (Z and N S n (Z, N M iff T M m,n (Z such that T t MT = N. (ii For two Z-lattices l and L, l L iff σ : l L such that B(x, y = B(σ(x, σ(y for all x, y l. (iii For f(x,..., x m, g(y,..., y n, g f iff T M m,n (Z such that f(t y = g(y.

5 (.3 We define R(l, L = {σ : l L} and O(L = R(L, L If l and L is positive definite, then R(l, L is a finite set. (.4 If l L and L l, we write l L. (.5 A Z-lattice L is called even if Q(x Z for every x L. (.6 Throughout this talk, we always assume that every Z-lattice L is primitive and positive definite.

6 [] Well Known Theorems L : Z-lattice, Z p : p-adic integer ring (. Scalar extension of a Z-lattice L We define L p = L Z p for every prime p. If x L p such that Q(x = a, then we write a L p. Note that a L p if and only if there is an x n L such that Q(x n a (mod p n, for every non-negative integer n.

7 (. Genus and class number gen(l := {L L p L p for every prime p}, h(l := gen(l/ (.3 Local information If a L p for every prime p (we write a gen(l, then a L for some L gen(l. Hence if h(l =, then We define a L if and only if a gen(l. Q(L = {a Z a L} and Q(gen(L = {a Z a gen(l}. Every integer a Q(gen(L is called eligible and a Q(gen(L Q(L is called exceptional.

8 (.4 Definition of regular lattices A Z-lattice L is called regular if it represents every integer a that is represented by the genus of L. Note that L is regular Q(L = Q(gen(L If h(l =, L is regular. The converse is not true in general. For example, gen(h = x + xy + y + 8z = {h, x + 6y + 6z + 6yz}. Both are regular forms (Hsia.

9 [3] Brief History (3. (Kitaoka If L is a binary Z-lattice, then L is regular if and only if h(l =. (3. (Earnest There are infinitely many quaternary regular Z- lattices. (3.3 (Kim classified all diagonal quaternary regular Z-lattices: x + y + z + dt : for d =, 3, 5, 7 or d = r+ x (x =,, 3. 3x + 6y + 48z + dt : for d = 6 3 r+ x (x =, 3x + 4y + 8z + dt : for d = 8,, 6, 0.

10 (3.4 There is no known example of a ternary Z-lattice L such that Q(L is completely determined and except the following result: Q(gen(L Q(L, (3.5 (Ono-Soundararajan (Assuming that GRH is true For the Ramanujan form f L = x + y + 0z, Q(gen(L Q(L = 3, 7,, 3, 33, 43, 67, 79, 87, 33, 7, 9, 3, 53, 307, 39, 679(Jones and Pall, 79(Gupta.

11 (3.4 (Legendre, Gauss and Dirichlet x + y + z = n has a solution if and only if n 4 k (8s + 7. (3.5 (Lebesque, Dirichlet, Liouville x + y + az is regular for a =, 3, 5. (3.6 ( Jones-Pall There are exactly 0 diagonal regular ternary Z-lattices (up to equivalence. Among them, 0 lattices have class number bigger than, for example, h(x + 48y + 44z = 4.

12 (3.7 (Watson There are only finitely many regular ternary Z-lattices. (3.8 ( Hsia, Kaplansky, Jagy, Jagy-Kaplansky-Schiemann There are at most 93 regular ternary Z-lattices. They remained the following ternary Z-lattices as candidates:

13 L( = L(4 = L(7 = L(0 = L(3 = L(6 = L(9 = L( = ( ( ( 0 4 0, L( = 0 3, L(3 = 0, ( ( ( , L(5 = ( ( ( ( , L(8 =, L( =, L(4 =, L(7 =., L(0 = ( ( ( , L(6 = ( ( ( , L(9 =, L( =, L(5 =, L(8 =, L( = ( ( ( ,, ( ( ( ,,,,

14 Exactly 794 lattices have class number among 93. (3.9 (O The following 8 Z-lattices among candidates are, in fact, regular: L(6 = L(8 = L( = ( ( ( , L( =, L(9 =, L( = ( ( , L(7 = 8 0, ( ( , L(0 = ( ,

15 [4] Watson Transformation Let L be a Z-lattice. (4. Gamma transformation Let ɛ = if L is even and p =, ɛ = otherwise. Γ p (L = {x L Q(x + z Q(z mod ɛp, z L}. γ p (L : scaling. primitive Z-lattice obtained from Γ p (L by a suitable

16 (4. A Z p -lattice L p is called isotropic if x L {0} such that Q(x = 0, otherwise L is called anisotropic. L p = L p,0 L p,t : Jordan decomposition such that (4.3 Lemma s(l p,i = p i Z p or L p,i = 0. Assume that L p,0 is anisotropic and additionally, L, = 0 only when p = and L is even. Then Q(L ɛpz = Q(Γ p (L and Q(gen(L ɛpz = Q(gen(Γ p (L. Hence if L is regular, then so is γ p (L. regular (that is, Γ p (L is regular, then Conversely, if γ p (L is (Q(gen(L Q(L ɛpz =.

17 If L p,0 is isotropic, γ p -transformation does not preserve the regularity in general. For example f L = x + y + z + 7t is regular, but f γ7 (L = x + 7y + 7z + 7t is not regular. (4.4 Note that γ 3 (L(8 = L(0, γ 5 (L(9 = L(.

18 (4.5 Sketch of proof Let L = L(7 with f L = 7x + 8y + 0z + 4xy + 4xz. (i l = γ 3 (L with f l = 3x + 7y + 7z + xy + xz + 6yz is regular. Hence (ii Since 3l L l, (Q(gen(L Q(L 3Z =. Q(gen(L Q(gen(l = Q(l. Assume that a Q(gen(L and 3 a. such that Q(x = a. Define Then there is an x l S ± = {y l/3l Q(y ± (mod 3}. Assume that a (mod 3. Then x (mod 3 S +.

19 (iii Show that there is an isometry σ y R(3l, L such that σ y (3y 3L, for every y S +. Hence a Q(L for every a (mod 3. If a (mod 3, there are some vectors y S such that σ(3y 3L for every σ R(3l, L. Let S be the set of such vectors and S = S S. Assume that x (mod 3 S. (iv Find isometries τ, τ R(3l, l such that 3 τ i(3x l for i =, and τ τ O(Ql has an infinite order. Note that Q( 3 τ i(3x = a. We may assume that 3 τ i(3x S. (v Define x 0 = 3 τ (3x and x n = 9 (τ τ n (9x. Show that if x n S, then x n+ l.

20 (vi Show that x n x m for every n m. If x n S for every n, then there are infinitely many vectors x n l such that Q(x n = a. This is a contradiction to the fact that R(a, l <. Therefore there is an integer m such that x m (mod 3 S and hence a Q(L. This completes the proof. We may replace 3 and l in the above with other integers and Z-lattices, respectively, to prove the regularity for some other Z-lattices.

21 (4.6 Numerical data for L = L(7 R(3l, L = S = 0, {( S + 0 = 0, 0, 0 S, 0, ( 3 0, 0 0 0, If x (0, ±, ± (mod 3, τ = ( , 0 0 0,, 0 0, ( 3 0 0, 0,,, = {(0, ±, ±, (0, ±, }., τ = ( ,, 0, ( } 3 0, 0 0, R(3l, l,,.

22 and if x (0, ±, (mod 3, τ = ( , τ = ( 4 4 R(3l, l. (4.7 Remaining candidates γ 3 (L(8 = L(4 and γ 3 (L(4 = L(, γ 3 (L( = L(4. γ 3 (L(3 = L(5, γ 3 (L(5 = L(3. There are exactly 4 candidates of regular ternary Z-lattices:

23 L( = L(7 = L(0 = L(4 = Candidates of regular ternary Z-lattices ( ( ( , L(3 = 0, L(5 = 0 0 5, ( ( ( , L(8 =, L( =, L(6 = ( ( ( , L(9 =, L(3 = ( ( ,, L( = ( ( ( , L(4 = 3 0 3, L(5 =

24 [5] Representations of an Arithmetic Progression S d,a = {dn + a n = 0,,... } (d Z +, a Z + {0} (5. Definitions (i A Z-lattice L is called S d,a -universal if S d,a Q(L. (ii A Z-lattice L is called S d,a -regular if S d,a Q(L = S d,a Q(gen(L. If L is S d,a -universal, then L is S d,a -regular. Assume that rank(l 3. If L is S d,a -regular, then for any m dividing 8 p dl,odd p, there is an integer a 0 such that L is S md,a0 -universal.

25 (5. S d,a -universal Z-lattices Unary case (Fermat s four squares theorem There are no four distinct squares in arithmetic progression. Binary case (Alaca-Alaca-Williams There is no binary S d,a -universal Z-lattice for any d, a. Ternary case There is no S d,0 -universal ternary Z-lattice for every positive integer d.

26 (i (Kaplansky There are exactly 5 ternary S, -universal Z-lattices. There are at most 8 ternary even S 4, -universal Z-lattices including 4 as candidates (one was confirmed by Jagy. (ii (Lemma If L is an S d,a -universal ternary Z-lattice, then dl 6d 3 (3d + a. Hence there are only finitely many ternary S d,a -universal Z-lattices for any d, a.

27 (5.3 Ternary S d,a -regular Z-lattices The Ramanujan form x + y + 0z is S 3, -regular and S 0,5 - regular though it is not regular. L( is S 4,0, S 6,6 and S 6,0 regular. (Main Theorem For any d, a, there is a polynomial f(d, a satisfying the following: For any S d,a -regular ternary Z-lattice L, det(l f(d, a. Therefore there are only finitely many S d,a -regular Z-lattices.

28 (5.4 Sketch of proof L : S d,a -regular lattice. (i Find d, a with d 0 (mod such that L is S d,a -regular and for any p d, S d,a Q(L p. (ii Show that for any prime p (gcd(p, d = such that the unimodular component of L p is anisotropic, γ p (L is also S d,a - regular for some integer a. (iii By using (ii, show that L p is isotropic for every large prime p. (iv If p is large enough, show that det(l is not divisible by p.

29 (v For any p det(l, by taking γ q -transformation to L, if necessary repeatedly, for any prime q p, show that ord p (det(l is bounded. (vi In each step, show that every constant has a polynomial growth to d, a.

30 (Duke and Schulze-Pillot For any ternary Z-lattice L, there is a constant C satisfying the following property: If an integer a satisfies (i a is primitively represented by the genus of L; (ii a is represented by the spinor genus of L; (iii a > C; then a is represented by L. (Corollary For any integers S and d, a with gcd(d, a =, there is a constant C satisfying the following property: For any Z-lattice L with det(l > C, (Q (gen(l Q(L {m m a (mod d} S.

POSITIVE DEFINITE n-regular QUADRATIC FORMS

POSITIVE DEFINITE n-regular QUADRATIC FORMS BYEONG-KWEON OH Abstract. A positive definite integral quadratic form f is called n- regular if f represents every quadratic form of rank n that is represented