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 by the genus of f. In this paper, we show that for any integer n greater than or equal to 27, every n-regular (even form f is (even n-universal, that is, f represents all (even, respectively positive definite integral quadratic forms of rank n. As an application, we show that the minimal rank of n-regular forms has an exponential lower bound for n as it increases. 1. Introduction A positive definite integral quadratic form f is called regular if f represents all integers that are represented by the genus of f. Regular quadratic forms were first studied systematically by Dickson in [6] where the term regular was coined. In the last chapter of his doctoral thesis [20], Watson showed by arithmetic arguments that there are only finitely many equivalence classes of positive definite primitive regular ternary quadratic forms. He did so by providing explicit bounds on the prime power divisors of the discriminant of those regular ternary quadratic forms. The classification of such quadratic forms was done by Kaplansky and his collaborators [13]. They proved that there are at most 913 positive definite primitive regular ternary quadratic forms, 22 of which are still candidates. On the contrary, Earnest proved in [8] that there are infinitely many equivalence classes of positive definite primitive regular quaternary quadratic forms. In fact, every positive definite quadratic form with more than 4 variables is almost regular, that is, such a form f represents almost all positive integers that are represented by the genus of f (cf. [24]. Unless stated otherwise, by an integral form we shall always mean a positive definite quadratic form having an integer matrix. The rank of a quadratic form f is defined by the number of variable of f. The study of higher dimensional analogs of regular quadratic forms is first initiated by Earnest in [7]. An integral form f of rank m is called n-regular if f represents all quadratic forms of rank n that are represented by the genus of f. Hence every n-regular form f satisfies, so called, a local-global principle over Z in the following sense: for any quadratic form g of rank n, g is represented by f over Z if and only if g is represented by f over the p-adic integer ring Z p, for every prime p. If the class number h(f of a quadratic form f is one, then f is n-regular for any n such that 1 n rank(f. Kitaoka proved that a quadratic The author s work was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD (KRF-2005-070-C00004. 1

2 BYEONG-KWEON OH form f of rank n is (n 1-regular if and only if the class number h(f of f is one (cf. [14], Corollary 6.4.1. In fact, there are many examples of n-regular forms of rank n + 2 or n + 3 whose class number is bigger than 1 for 1 n 5. However if n is large, the situation is quite different. Note that the class number of every quadratic form with rank greater than 10 is bigger than one (cf. [23]. Furthermore, it seems to be quite difficult to find an n-regular form with minimal rank, for each n 11. On the other hand, every quadratic form satisfies a local-global principle over Z under some restrictions. In 1978, Hsia, Kitaoka and Kneser proved in [10] that if the minimum positive integer that is represented by g is sufficiently large and rank(f 2rank(g+3, then g is represented by f over Z if and only if g is represented by f over Z p for every prime p. Recently, Ellenberg and Venkatesh improved this theorem by showing that the rank condition in this theorem could be replaced by rank(f rank(g + 7, with one additional assumption that the discriminant of g is squarefree (cf. [9]. But their method is quite different from that of the former theorem. In both theorems stated above, the condition that the represented form g has sufficiently large minimum is crucial. Without this condition, it seems to be quite difficult to use some kinds of approximation theorems or analytic methods developed by several authors. In this paper, we consider the localglobal principle over Z without this minimum condition. Earnest showed in [7] that there exist only finitely many equivalence classes of 2-regular primitive integral quaternary forms. Actually, his method is an extension of Watson s analytic argument (see [21], which seems to be inadequate for other higher dimensional situations. Recently, Chan and the author proved in [4] that there exist only finitely many equivalence classes of primitive integral n-regular forms of rank n + 3, for any integer n 2. In that paper, we turned the stage back to an arithmetic setting and bring back Watson s transformations into the arsenal. To show that there is an n-regular form for every positive integer n, we need a notion of universal forms: An integral form f is called (even n-universal if it represents all (even, respectively integral forms of rank n. Here, an integral form is called even if every diagonal entry of its corresponding symmetric matrix is even, and is called odd otherwise. Every n-universal form is, in fact, an n-regular form that represents all integral forms of rank n over Z p, for every prime p. For the classification of n-universal forms with minimal rank, see the recent survey article [15]. Let {f 1, f 2,..., f h } be a set of representatives of all equivalence classes of odd unimodular forms of rank n + 3. Since every integral form g of rank n is represented by the genus of any unimodular form of rank n + 3, g is represented by f i for some i = 1, 2,..., h. Hence the unimodular form f 1 (y 11,..., y 1,n+3 + f 2 (y 21,..., y 2,n+3 + + f h (y h1,..., y h,n+3 of rank (n + 3h represents all integral forms of rank n. This is a typical example of an n-regular form. However, every known example of an n- regular form is a universal form or an even universal form, for every n greater than 10. We denote by R(n the minimal rank of n-regular quadratic forms. Since there is a quadratic from of rank n having class number one for every n less

POSITIVE DEFINITE n-regular QUADRATIC FORMS 3 than or equal to 10, and such a form is n-regular, we have R(n = n for every n = 1, 2,..., 10. As mentioned above, R(n n + 2 for every n bigger than 10. As far as the author knows, this was the best known lower bound for the minimal rank of n-regular forms. In this paper, it is proved that every (even n-regular form is, in fact, (even, respectively n-universal, for every integer n greater than or equal to 27. As an application of this result, we also show that R(n has an exponential lower bound for n as it increases (cf. Theorem 5.4. The subsequent discussion will be conducted in the more adapted geometric language of quadratic spaces and lattices, and any unexplained notations and terminologies can be found in [5], [14] or [18]. The term lattice will always refer to an integral Z-lattice on an m-dimensional positive definite quadratic space over Q. The scale and the norm ideal of a lattice L are denoted by s(l and n(l, respectively. It will be assumed that every Z-lattice L is primitive, i.e., s(l = Z. For any Z-lattice L, we denote by h(l the class number of L, and µ i (L the i-th successive minimum for every i = 1, 2,..., rank(l. For two Z-lattices l and L, if every Z-lattice in the genus of l is represented by L, we write gen(l L. Let {e 1, e 2,..., e n } be a basis of a Z-lattice L. For any real number a, al is denoted by the Z-lattice with a basis {ae 1, ae 2,..., ae n }. If M = (m ij is a matrix, then am is denoted by the matrix (am ij. If B(e i, e j = a i δ ij, then we will write L = a 1, a 2,..., a n for simplicity. For every odd prime p, p denotes a nonsquare unit in Z p. 2. γ-transformations and n-regular lattices Let L be any Z-lattice on the quadratic space V. For any positive integer m, we define Λ m (L = {x L : Q(x + z Q(z (mod m for all z L }. Note that x Λ m (L if and only if Q(x 2B(x, z 0 (mod m for every z L. In particular if m = 2, then we prefer to use the following notation L(e = {x L : Q(x 0 (mod 2}, instead of Λ 2 (L. Let λ m (L be the primitive lattice obtained from Λ m (L by scaling V by a suitable rational number. Note that the scaling factor depends only on the lattice structure of L p, for every prime p dividing m. These λ m -transformations were used by Watson in his study of regular ternary lattices [20] and class numbers [23], and more recently by various authors in [3] and [4] concerning lattices satisfying different kinds of regularity conditions. These references contain most of the basic properties of the λ m -transformations needed in this paper. For any Z p -lattice L, the Λ m -transformation Λ m (L is defined similarly. In particular, one may easily verify that for every Z-lattice L, (λ m (L p λ m (L p

4 BYEONG-KWEON OH for every prime p, where L p = L Z p is the Z p -lattice. For a Z-lattice L and a prime p, let L p = L p,0 L p,1 L p,tp be a Jordan decomposition such that s(l p,i = p i Z p or L p,i = 0 for every i = 1, 2,..., t p, and L p,tp 0. Throughout this paper, when we consider a Jordan decomposition, we always assume this condition. We define r p,i = r p,i (L p = rank(l p,i and r p (L p = (r p,0, r p,1,..., r p,tp. Note that these definitions are independent of a Jordan decomposition of L p. If p is odd, then we may easily check that Therefore we have ( r p ((λ p (L p = Λ p (L p pl p,0 L p,1 L p,tp. { (r p,1, r p,0 + r p,2, r p,3,..., r p,tp if r p,1 0, (r p,0 + r p,2, r p,3,..., r p,tp otherwise. Now assume that p = 2. If L is an odd Z-lattice, then Λ 2 (L 2 L 2,0 (e L 2,1 L 2,t2, and if L is even then Λ 4 (L 2 2L 2,0 ( 1 2 2 L 2,1 (e L 2,3 L 2,t2. For any odd unimodular Z 2 -lattice l of rank n 3, there is an even unimodular Z 2 -lattice U, and an odd unimodular Z 2 -lattice l 1 of rank less than 3 such that l U l 1. Hence l(e U l 1 (e. Note that if n is odd, then l 1 (e is a proper 4Z 2 - modular lattice of rank 1, otherwise l 1 (e is a 2Z 2 -modular lattice of rank 2. Therefore if L is odd, { (r 2,0 1, r 2,1, r 2,2 + 1, r 2,3,..., r 2,t2 if r 2,0 is odd, r 2 ((Λ 2 (L 2 = (r 2,0 2, r 2,1 + 2, r 2,2,..., r 2,t2 otherwise. If L is even, r 2 ((Λ 4 (L 2 equals to (0, r 2,1 1, r 2,0 + r 2,2, r 2,3 + 1,..., r 2,t2 if r 2,1 is odd, (0, r 2,1 2, r 2,0 + r 2,2 + 2, r 2,3,..., r 2,t2 if r 2,1 is even, n(l 2,1 = 2Z 2, (0, r 2,1, r 2,0 + r 2,2, r 2,3,..., r 2,t2 otherwise. Note that for any even Z-lattice L, L(e = λ 2 (L = L. modified λ p -transformation, which we call γ p -transformation. So we use a Definition 2.1. Let L be a Z-lattice. For every prime p, Γ p -transformation is defined as follows: If p is odd Γ p (L = Λ p (L, and { Λ 2 (L if L is odd, Γ 2 (L = Λ 4 (L otherwise. γ p (L is defined by the primitive lattice obtained from Γ p (L by scaling V by a suitable rational number.

POSITIVE DEFINITE n-regular QUADRATIC FORMS 5 Definition 2.2. Let p be a prime and l, L be Z p -lattices such that l L. Assume that a Jordan decomposition of L is L = L 0 L 1 L t. l is called a lower type of L if there is an integer u = u(l, L (0 u t such that a Jordan decomposition of l is l = L 0 L 1 L u 1 L u, where L u is a non-zero p u Z p -modular lattice such that n(l u = n(l u and L u L u L t. If l is a lower type of L, then we write l L. Lemma 2.3. Let p be a prime and L be a Z p -lattice with rank greater than one. Assume that a Jordan decomposition of L is L = L 0 L 1 L t. For some positive integer s, let k be any integer satisfying rank(l 0 L 1 L s 1 k < rank(l 0 L 1 L s. Then, there does not exist a Z p -lattice l with rank k such that l L if and only if p = 2, n(l s s(l s and k rank(l 0 L s 1 is odd. Proof. The proof is quite straightforward. Lemma 2.4. For any integer n greater than 5, let l and L be any Z 2 -lattices such that rank(l = n and l L. Assume that the rank of the unimodular component in a Jordan decomposition of l is less than [ n 2 2 ]. Then there is an integer k satisfying the following properties: (i γ2 i(l γi 2 (L for any i = 0, 1,..., max(0, k 2. [ ] n 2 (ii r 2,0 (γ2 k 1 (l <. 2 [ n 2 (iii r 2,0 (γ2 k(l 2 ]. (iv If γ2 k 1 (L = L 0 L 1 L t is a Jordan decomposition, then L 0 is even unimodular or rank(l 0 2. Furthermore, γk 1 2 (l γ2 k 1 (L or γ2 k 1 (l L 0 l 1 4ɛ, where l 1 is a 2Z 2-modular lattice with n(l 1 = n(l 1 and ɛ Z 2. (v For any i = 1, 2,..., k, n(γ 2 (γ i 1 2 (L=n(Γ 2 (γ i 1 2 (l and s(γ 2 (γ i 1 2 (L=s(Γ 2 (γ i 1 2 (l. Proof. We use an induction on u = u(l, L. Let L = L 0 L 1 L u L t, l = L 0 L 1 L u 1 l u be a Jordan decomposition for each Z 2 -lattice L and l. First, assume that u = 1. If L 0 is even or rank(l 0 2, then k = 1 and everything is trivial.

6 BYEONG-KWEON OH Assume that L 0 is an odd unimodular lattice with rank greater than 2. Then, γ 2 (L 0 = L 0 B, or γ 2 (L 0 = L 0 4ɛ according the parity of rank(l 0. Here L 0 is even unimodular, ɛ Z 2 B is a binary 2Z 2 -modular lattice. Hence γ 2 (l = L 0 l 1 4ɛ or γ 2 (l = L 0 (l 1 B. Therefore, k = 2 and the lemma follows from this. Now assume that u = 2. Recall that l = L 0 L 1 l 2 is a Jordan decomposition. Suppose that L 0 is even unimodular. If L 1 = 0 or rank(l 1 2 and n(l 1 = s(l 1, then clearly k = 1. If n(l 1 s(l 1 or rank(l 1 is even, then we may easily check that γ 2 (l γ 2 (L and u(γ 2 (l, γ 2 (L = 1. Therefore, the lemma follows from the induction hypothesis. Assume that n(l 1 = s(l 1 and rank(l 1 is odd greater than 1. Then Jordan decompositions for both γ 2 (l and γ 2 (L are of the form γ 2 (L = L 0 L 1 L t 1, γ 2 (l = L 0 l 1 4ɛ, where rank(l 0 = rank(l 1 1, rank(l 1 = rank(l 0 + rank(l 2, and rank(l 1 = rank(l 0 + rank(l 2. Note that L 0 is even unimodular and n(l 1 = n(l 1. Suppose that rank(l 0 < [ n 2 2 ]. Then, one may easily check that r 2,0 (γ2 2 n 2 (l [ 2 ]. Therefore, k = 2 in this case and the lemma follows from this. Suppose that L 0 is odd unimodular. If rank(l 0 2, then either k = 1 or γ 2 (l γ 2 (L and u(γ 2 (l, γ 2 (L < 2. If rank(l 0 3, then γ 2 (l γ 2 (L and k 1. So, this case can be reduced to the above case. Finally assume that u = u(l, L 3. If L 0 is even or rank(l 0 2, then it is easy to show that γ 2 (l γ 2 (L and u(γ 2 (l, γ 2 (L < u(l, L. Therefore the lemma follows from the induction hypothesis. If L 0 is an odd unimodular Z 2 -lattice with rank greater than 2, then we may consider γ 2 (l and γ 2 (L instead of the original lattices. The condition (v follows directly from the condition (iv. This completes the proof. Remark 2.5. In Lemma 2.4, assume that n 12 and γ2 k 1 (l γ2 k 1 (L. Let γ k 1 2 (L = L 0 L 1 L t and γ k 1 2 (l = L 0 L 1 l s be a Jordan decomposition for each Z 2 -lattice. If s is greater than 1, then one may easily show that rank(l 1 2 or rank(l 1 [ n 2 2 ] 2. Furthermore, if rank(l 1 2 then [ ] n 2 rank(l 0 + rank(l 2 2. 2 Note that if s = 1 or γ2 k 1 (l is not a lower type of γ2 k 1 (L, then rank(l 1 [ n+1 2 ]. Lemma 2.6. Let p be an odd prime and l, L be any Z p -lattices such that l L. Assume that the rank of the unimodular component in a Jordan decomposition of l is less than [ n+1 2 ], where rank(l = n. Then there is an integer k satisfying the following properties: and

POSITIVE DEFINITE n-regular QUADRATIC FORMS 7 (i γp(l i γp(l i for any i = 0, 1,..., k 1. [ ] n + 1 (ii r p,0 (γp k 1 (l <. 2 [ ] n + 1 (iii r p,0 (γp k (l. 2 (iv s(γ p (γp i 1 (L = s(γ p (γp i 1 (l for any i = 1, 2,..., k. Proof. If one use the rank equation ( given above, this lemma can be proved in a similar manner to Lemma 2.4. In some cases, the γ p -transformation preserves the n-regularity. More precisely, assume that a Z-lattice L is n-regular. If the unimodular component in a Jordan decomposition of L p is anisotropic for some prime p, then γ p (L is also n-regular. If L is an odd n-regular Z-lattice, then γ 2 (L is always n-regular without any assumption. But, this is not true in general for an odd prime p. For example, the quinary Z-lattice 1, 1, 1, 1, 7 is 1-regular, whereas γ 7 (L = 1, 7, 7, 7, 7 is not 1-regular. Lemma 2.7. Let p be a prime, and l, L be any primitive Z-lattices such that l L and n(l p = n(l p. If r p,0 (l p = r p,0 (L p, then Γ p (l Γ p (L. Proof. For any prime p, define L(p = {x L : B(x, L pz}. Note that L(p = Γ p (L for any odd prime p, whereas L(2 is not equal to Γ 2 (L in general. Since r p,0 (l p = r p,0 (L p, one may easily show that L = l + L(p. Assume that p is odd and x Γ p (l. For any z L, there are z 1 l and z 2 L(p such that z = z 1 + z 2. Hence B(x, z = B(x, z 1 + B(x, z 2 0 (mod p, which implies that x Γ p (L. Now assume that p = 2. If L is odd, then everything is trivial. If L is even, then the proof is quite similar to that of an odd prime case given above. A Z-lattice L is called almost n-regular if L represents almost all Z-lattices of rank n that are represented by the genus of L. Note that every almost n-regular Z-lattice is (n 1-regular. Lemma 2.8. For n 2, let L be an almost n-regular Z-lattice of rank greater than n + 1. Then there is a prime r such that gcd(r, 2dL = 1, and a sublattice l of L with rank n or n 1 satisfying the following properties: (i l p is a lower type of L p for any prime p r. (ii Every Z-lattice in the genus of l is represented by L. (iii dl r rz r. Proof. Let L be an almost n-regular Z-lattice of rank greater than n + 1. We define S = {N : rank(n = n, N gen(l and N L}. Note that the set S is finite and is empty if L is n-regular.

8 BYEONG-KWEON OH Choose an odd prime q such that for any N S, L q and N q are unimodular and = 1. Define ( dn q T = {p ord p (2dL 1} {q}. Let l (2 be a Z 2 -sublattice of L 2 with rank n or n 1 such that l (2 L 2. Here, we choose a sublattice l (2 of rank n 1 only in the case when n and L 2 satisfy the condition in Lemma 2.3. Now for each p T {2, q}, let l (p be any Z p -sublattice such that l (p L p and rank(l (p = rank(l (2. Finally, let l (q be a sublattice of L q such that l (q 1, 1,..., q and rank(l (q = rank(l (2. Then by Lemma 1.6 of [10], there is a Z-sublattice l of L such that l p l (p for any p T and for p T, d(l p Z p with precisely one exception p = r, where d(l r rz r. Clearly l p is a lower type of L p for any p r, and N gen(l for any N S. The definition of S implies that every Z-lattice in the genus of l is represented by L. The lemma follows from this. 3. Local structures of n-regular lattices for large n Throughout this and next section, we always assume that n 27. Under this assumption, we prove the following theorem. Theorem 3.1. Let L be an n-regular Z-lattice. Then for every prime p, ( r p,0 (L p 12, where r p,0 (L p is the rank of the unimodular component in a Jordan decomposition of the Z p -lattice L p. Lemma 3.2. Let p be an odd prime. Assume that there is an n-regular Z-lattice L such that r p,0 (L p < 12. Then there is an even Z-lattice L satisfying the following properties: (i r p,0 ( L p < 12. (ii r p,1 ( L p 12 or r p,1 ( L p = 0 and r p,0 ( L p + r p,2 ( L p 12. (iii For any even Z-lattice N with rank less than or equal to 9 such that N p L p, N is represented by L. Proof. Note that the rank of L is greater than n + 1 by Corollary 6.4.1 of [14]. Let l be a sublattice of L and r be a prime satisfying all conditions in Lemma 2.8. Define T 1 := {q : r q,0 (L q < 12} = {q 0 = p, q 1, q 2,..., q t }. For any prime q T 1 {p}, let k(q be an integer satisfying all conditions in Lemma 2.4 or 2.6. Then [ ] n 3 r q,0 (γq k(q (l q 12, 2 and γq k(q (l is represented by γq k(q (L by Lemma 2.7. Furthermore, since the γ q -transformation is a surjective map from gen(k to gen(γ q (K, for any Z-lattice K and any prime q (cf. [22], gen(γ k(q q (l γ k(q (L. q

Define POSITIVE DEFINITE n-regular QUADRATIC FORMS 9 l 1 = ( t i=1 γ k(q i q i (l and L 1 = ( t i=1 γ k(q i q i (L. Then gen(l 1 L 1 and r q,0 ((l 1 q 12 for any q p. Note that (L 1 p is isometric to L p up to unit scaling factor and (l 1 p (L 1 p. Now, by a similar reasoning to Lemma 2.6, there is an integer s such that (α (γp s 1 (l 1 p (γp s 1 (L 1 p and r p,0 ((γp s 1 (L 1 p < 12, (β r p,0 ((γp(l s 1 p r p,0 ((γp(l s 1 p 12, (γ gen(γp s 1 (l 1 γp s 1 (L 1. For K = l or L, we define { γp K s 1 (K 1 if γp s 1 (K 1 is even, = γ 2 (γp s 1 (K 1 otherwise. Note that the second condition in the statement of the lemma follows from conditions (α and (β. Let N be any even Z-lattice with rank less than or equal to 9 such that N p L p. Since r q,0 (γp s 1 ((l 1 q = r q,0 ((l 1 q 12, N q l q for any q p. Furthermore, from the fact that N p L p and (α and (β, one may easily show that N p l p. Therefore N gen( l, which implies that N L by (γ. This completes the proof. Lemma 3.3. Assume that there is an n-regular Z-lattice L such that r 2,0 (L 2 < 12. Then there is a Z-lattice L, not necessarily even, satisfying the following properties: (i r 2,0 ( L 2 < 12. (ii L 2 satisfies one of the following rank conditions: r 2,0 ( L 2 (type r 2,1 ( L 2 (type r 2,2 ( L 2 r 2,0 ( L 2 (type r 2,1 ( L 2 (type r 2,2 ( L 2 1(odd 12 1(odd 0 11 2(odd 10 a(even 0, 1(odd 12 a a(even 2(odd 10 a (even 12(even (even 13(odd Here, we say the 2 i Z 2 -modular component L 2,i in a Jordan decomposition of L 2 is of odd type if s( L 2,i = n( L 2,i, and of even type otherwise. (iii Let L 2 = L 2,0 L 2,1 L 2,t be a Jordan decomposition and i be an integer such that rank( L 2,0 L 2,1 L 2,i 1 < 12 rank( L 2,0 L 2,1 L 2,i. Let N be any Z-lattice with rank less than or equal to 10 such that N 2 L 2,0 L 2,1 L 2,i 1 M,

10 BYEONG-KWEON OH for any 2 i Z 2 -modular lattice M with same type to L 2,i, and rank(m = t rank( L 2,0 L 2,1 L 2,i 1, where t = 13 if L 2,i is of even type and rank( L 2,0 L 2,1 L 2,i 1 is odd, otherwise t = 12. Then N is represented by L. Proof. First, by a similar reasoning to the above lemma, we may assume that there are Z-lattices l 1 and L 1 such that [ n ] gen(l 1 L 1 and r p,0 ((l 1 p 13 for every odd prime p, 2 where rank(l 1 = rank(l, rank(l 1 = n or n 1, and (l 1 2 (L 1 2. Now, by a similar reasoning to Lemma 2.4, there is an integer s such that (α r 2,0 ((γ2 s 1 (L 1 2 < 12 and r 2,0 ((γp(l s 1 2 r 2,0 ((γp(l s 1 2 12. (β (γ2 s 1 (l 1 2 and (γ2 s 1 (L 1 2 satisfy the condition (iv of Lemma 2.4. (γ gen(γ2 s 1 (l 1 γ2 s 1 (L 1. We define L = γ2 s 1 (L 1 and l = γ2 s 1 (l 1. Then, by using conditions (α and (β, we may easily show that the rank condition (ii in this lemma is satisfied. Let N be any Z-lattice satisfying the condition (iii. Since r p,0 ( l p 13 for every odd prime p, N p l p. For p = 2, choose any 2 i Z 2 - modular sublattice M of the 2 i Z 2 -modular component of l 2 satisfying all conditions given in the lemma such that L 2,0 L 2,1 L 2,i 1 M is represented by l 2. Note that this is always possible by Lemma 2.4 (iv. Hence N 2 is represented by l 2 and N is represented by gen( l. Therefore by (γ, N is represented by L. This completes the proof. 4. Proof of Theorem 3.1 In this section, we prove Theorem 3.1. In fact, we do this by showing that there does not exist a Z-lattice L satisfying all conditions in Lemma 3.2 or 3.3. The following lemma is very useful. Lemma 4.1. Let L be a Z-lattice. If l is a sublattice of L and if x L Ql L, then d(l + Zx dl Q(x. Furthermore if the equality holds, then B(l, x = 0. Proof. See [1], p. 330. Assuming that Theorem 3.1 is false one has at least one prime p for which k = r p,0 (L p < 12. For such a prime p, Lemma 3.2 for an odd prime p or Lemma 3.3 for p = 2 yields the existence of a lattice L with the properties (i- (iii given in these Lemmata. Lemma 3.2 and Lemma 3.3 imply that lattices of rank less than or equal to 10 (or rank less than or equal to 9 for an odd prime p are represented by L globally if they are represented by L p for an odd p or they satisfy a suitable local condition at p if p = 2. This allows to first find a supply of indecomposable root lattices represented by L and then, using the uniqueness of the splitting of a root system into indecomposable components, some orthogonal sums of root lattices are represented by L.

POSITIVE DEFINITE n-regular QUADRATIC FORMS 11 In the easier cases this gives a unimodular sublattice of L p of dimension lager than k and hence a contradiction. In the more involved cases one has to add some more auxiliary lattices represented by L and construct in a similar way recursively a sequence E = (N 1, N 2,..., N t of lattices represented by L, i.e., having N 1, N 2,..., N i 1 allows to construct another lattice N i represented by L and not yet among N 1, N 2,..., N i 1. Finally one finds (starting from N t and using Lemma 4.1 a sublattice Ñ of L of small discriminant and large rank, which contracts the assumption on the small unimodular component of L. In many cases, the sequence E consists of root lattices. An even Z-lattice generated by vectors of norm 2 is called a root lattice. It is well known that every (even root lattice is an orthogonal direct sum of A n (n 1, D n (n 4 and E n (6 n 8. For the definitions of theses lattices, see [5]. Note that D 4 A n, A n D n+1 and A 8, D 8 E 8, for any positive integer n. We also define an indefinite root lattice E 9. First, consider the following graph: 9 8 1 2 3 4 5 6 7 Graph of E 9. The Z-lattice E 9 = Zx 1 + Zx 2 + + Zx 9 is defined by Q(x i = 2 for every i = 1, 2,..., 9, and { 1 if i is directly connected to j in the graph E 9, B(x i, x j = 0 otherwise. Note that d(e 9 = 6. We also denote by E 9 the corresponding matrix (B(x i, x j, if no confusion arises. For any even Z-lattice L, R L denotes the sublattice of L generated by vectors of norm 2. In general, R L is not necessarily a primitive sublattice of L. However, every root lattice that is represented by L is also represented by R L. E n (k, s = (e ij denotes the n n elementary matrix such that { 1 if i = k and j = s, e ij = 0 otherwise. Let a, b, k be any integers such that a + (k 1b > 0 and a > max(0, b. We denote by M k (a, b = Zx 1 + Zx 2 + + Zx k the Z-lattice of rank k satisfying Q(x i = a, B(x i, x j = b for any i, j such that 1 i j n.

12 BYEONG-KWEON OH Note that det(m k (a, b = (a + (k 1b(a b k 1. For any odd prime p such that p kb and ord p (a b = 1, one may easily show that M k (a, b p a, p, p,..., p, pɛ, for some ɛ Z p. Finally, for any Z- (or Z p - lattice L, we define L k = k-times {}}{ L L. Lemma 4.2. For an odd prime p, let L be a Z-lattice satisfying all conditions in Lemma 3.2. Then we have 1 r p,0 ( L p 4. Proof. First assume that p 3. Let r p,0 ( L p = k for some k = 5, 6,..., 11. Then by Lemma 3.2, L represents all even root lattices of rank min(k 1, 9 such that the prime factors of their discriminants are only 2 and 3. By direct computations, we have the following Table 3.1. Note that L represents every k root lattices of rank k 1 (E possible sublattices of L 5 A 2 A 2, D 4 D 4 A 2, D 6, E 6 6 A 5, D 5, A 2 2 A 1, A 3 A 2 D 6 A 1, E 6 A 1, E 7, A 5 D 5 7 E 6, D 6, D 4 A 2 E 8, E 7 A 2, E 6 D 6 8 E 7, D 7, A 2 2 A3 1 E 8 A 1, E 7 D 7 9 E 8, A 5 A 3 E 8 A 3 k 10 E 8, D 9 E 8 D 9 Table 3.1. root lattice in the left hand side of Table 3.1, for each k. As a representative case, we consider the case when k = 6. Since L represents both A 5 and D 5, it should represent either D 6 or E 6 or A 5 D 5. Assume that L represents D 6. Since A 2 A 2 A 1 is represented by L but is not represented by D 6, L represents either D 6 A 1 or E 7. Finally assume that L represents E 6. In this case, since A 3 A 2 is not represented by E 6, L should represent E 6 A 1 or E 7. For each k, by a similar reasoning to the above, L should represent at least one root lattice in the right hand side. However, every Z-lattice in the right hand side, whose rank is greater than k, are unimodular over Z q for any prime q 2, 3. This is a contradiction to the assumption that r p,0 ( L p = k. Now assume that p = 3 and r 3,0 ( L 3 = 5. Since A 4, D 4 L, we may assume that D 5 L. Note that if A 4 D 4 L, then r 3,0 ( L 3 8. Hence, A 4 A 1 is also represented by L. Therefore the only possibility is that E 6 is represented by L. This implies that r 3,1 ( L 3 12 by Lemma 3.2 (ii. Therefore A 2 A 2 A 2 A 2 L, which is a contradiction. Assume that r 3,0 ( L 3 = 6. Let L 3,0 be the unimodular component in a Jordan decomposition of L 3. Suppose that d( L 3,0 is a nonsquare unit in

POSITIVE DEFINITE n-regular QUADRATIC FORMS 13 Z 3. Then, D 5 A 1 and A 4 A 1 A 1 are represented by L. This implies that E 6 A 1 is also represented by L. By a similar reasoning to the above argument, A 5 2 L. This implies that r 3,0 ( L 3 7, which is a contradiction. For each remaining case, the sequence E in Table 3.2 gives a contradiction. rank( L 3,0 ( det( L 3,0 3 E 6 1 D 6, A 6 7 1 E 7, A 4 A 3 7 1 D 7, A 4 A 3 1 8 1 E 7, D 7 8 1 E 8, A 4 A 3 A 1 9 1 E 8 A 1, D 5 A 4 9 1 E 8, D 9 rank( L 3,0 10 ±1 E 8, D 9 Table 3.2. Every computation is quite similar to the above case. Proof of Theorem 3.1. Now we show that the inequality r p,0 ( L p 4 is also impossible for every odd prime p, and r 2,0 ( L 2 12. Note that L is always even when we consider an odd prime case. Case (1 p 29. First assume that r p,0 ( L p ( = 1 and 1 L p. Note that for every positive even integer a such that a p = 1, a is represented by L. If 2 is a square in Z p, then 2, 4 are represented by L. Choose vectors x, y L such that Q(x = 2 and Q(y = 4. Then the binary Z p -lattice Z p x + Z p y is unimodular, which is a contradiction. For the remaining cases, every computation is quite similar to this case. For each case, we may take ( 2, 4 if 2 (Z p 2, E = ( 4, 12 if 3 (Z p 2 and 2 (Z p 2, ( 4, 6 otherwise. Assume that r p,0 ( L p = 1 and p L p. If 2 (Z p 2, then we take E = ( 2, 12 when 3 (Z p 2, and E = ( 2, 6 otherwise. Now, we assume that 2 (Z p 2, that is, p ±1 (mod 8. First assume that p 1 (mod 8. If 3 (Z p 2, then 6, 12 L. Choose vectors x, y L such that Q(x = 6, Q(y = 12. Then the discriminant of the binary Z-lattice Zx + Zy is divisible by p. Furthermore d(zx + Zy 0, 4, 7 (mod 8. Hence 72 d(zx + Zy 4p, which is a contradiction. Therefore we may assume that 3 (Z p 2, that is, p 1 (mod 24. Let a be the smallest quadratic nonresidue positive integer

14 BYEONG-KWEON OH modulo p. Note that by [11], a < p 3 + 2. Clearly, a is odd and 2a, 4a L. Furthermore, by a similar reasoning to the above, we may conclude that 8a 2 4p. Combining this and the above inequality, we have p 238. Furthermore since p 1 (mod 24, p is one of the following primes 73, 97, 193. One may easily show by direct computations that L contains a binary sublattice whose discriminant is not divisible by p, by taking two suitable nonresidues for each remaining prime p. This is a contradiction. Now assume that p 7 (mod 8. By a similar reasoning to the above, we may assume that 3 (Z p 2, that is, p 23 (mod 24. Let a be the smallest quadratic nonresidue positive integer modulo p. Note that a < p 2 5 + 12p 1 5 + 33 by [12]. Since 2a and 6a are represented by L, 3a 2 p by a similar reasoning to the above argument. Therefore we have p 14 5. For each prime satisfying this inequality, one may show by direct computations that there is a nonresidue a such that 3a 2 < p, except the following primes 47, 71, 311, 479. For each exceptional prime p, one may take { 6, 12 } if p = 41, { 14, 22 } if p = 71, E = { 22, 34 } if p = 311, { 26, 34 } if p = 479. Assume that r p,0 ( L p = 2. Since 2, 4 are represented by L, at least one of the following binary Z-lattices 2 0 2 1 2 0 2 1,,, 0 2 1 2 0 4 1 4 is represented by L. Each binary lattice given above does not represent, respectively, 6, 4, 10, 6. Furthermore, the discriminant of every ternary Z- sublattice of L is divisible by 2p 58. Therefore, the third binary lattice is the only possible candidate that is represented by L. Assuming this, one 4 0 may easily show that the binary lattice is also represented by 0 6 L. This implies that the third successive minimum µ 3 ( L is less than or equal to 6. Hence, L contains a ternary sublattice whose discriminant is less than or equal to 8 6 = 48 by Lemma 4.1. This is a contradiction. If r p,0 ( L p = 3, then both A 1 A 1 and A 2 are represented by L. Hence A 2 A 1 L or A 3 L. For the former case, at least one of the following ternary lattices A 1 A 1 A 1, A 3, A 2 4 is represented by L, and for the latter case, at least one of the following ternary lattices A 1 A 1 A 1, A 1 A 2, A 2 4

POSITIVE DEFINITE n-regular QUADRATIC FORMS 15 is represented by L. Therefore µ 4 ( L 4 for both cases, which implies that L contains a quaternary sublattice whose discriminant is less than or equal to 24. This is a contradiction. Finally assume that r p,0 ( L p = 4. Since A 3, A 2 A 1 and A 1 A 1 A 1 are represented by L, A 3 A 1 is represented by L. Furthermore, since there does not exist a root lattice of rank 5 whose discriminant is divisible by p 29, the root sublattice of R L L is A 3 A 1. Hence A 4, A 2 A 1 A 1 and A 2 A 2 are not represented by L. Therefore we may assume that From this follows ( 2 p 2, 4 ( 3 p = ( 5 p = 1. 4 1 1 4 L, which implies that µ 5 (L 4. This is a contradiction to the fact that the discriminant of any quinary sublattice of L is divisible by 2p. Case (2 5 p 23. For each prime p, we give Tables 3.3 3.9 on the sequence E. rank( L 23,0 ( det( L 23,0 23 E 1 1 10, 14, S 6 23 = γ 23(A 4 A 1 230[21 1 10 ], M 6(20, 3 1 1 A 1, 4 2 1 4 1 2 1, 1 4 1 4 2 1 A 2, A 2 1 3 1 A 2, A 2 1 3 1 A 3, A 1 A 2 4 1 A 4, A 3 1 4 1 A 1 A 3, A 2 2 Table 3.3. p = 23. In those tables, Sp n is a Z-lattice of rank n such that det(sp n = p n 1 and r p,0 ((Sp n p = 1. For the definition of these lattices, see [5]. In every case containing Sp n, the pair of the first two integers, which are represented by L, shows that r p,1 ( L p 0. Hence by Lemma 3.2, Sp n is represented by L. Furthermore, since the Z-lattice M k (a, b on the same line is also represented by L but not by Sp n, there is a vector x L φ(sp n such that Q(x = a. Here φ is any representation from S n p to L. Therefore L has a sublattice K of rank n + 1 such that φ(s n p K and d(k p n 1 a. Since n is even in every case, d(k is divisible by 2p n. This gives a contradiction.

16 BYEONG-KWEON OH rank( L 19,0 ( det( L 19,0 19 E 1 1 A 1, 10, 12 1 1 4, 6 2 1 A 2, A 1 4 2 1 2 1 A 2 1, 1 4 3 1 A 3 1, A 2 4 3 1 A 3, A 1 A 2 4 1 A 1 A 3, A 2 1 A 2 4 1 A 4, A 2 2 Table 3.4. p = 19. rank( L 17,0 ( det( L 17,0 17 E 1 1 6, 10, S 4 17 = γ 17(A 2 A 1 102[11 1 6 ], M 4(14, 3 1 1 A 1, 4 2 1 2 1 A 2, 1 4 2 1 A 2 1, A 1 4 2 1 3 1 A 1 A 2, A 1, A 1 4 2 4 3 1 A 3, A 3 1 4 1 A 4, A 3 1 4 1 A 1 A 3, A 2 2 Table 3.5. p = 17. rank( L 13,0 ( det( L 13,0 13 E 1 1 A 1, 6 1 1 4, 10, S13 4 = γ 13(A 3 52[1 1 4 ], M 5(12, 1 2 1 1 2 1 2 1 2 1, A 1 4 1 4,, 1 10 4 1 8 1 4 10 2 1 A 2, A 2 1 3 1 A 1 A 2, A 3 1 2 1 4 2 3 1 A 3,, 4 1 4 2 4 4 1 A 4, A 1 A 3 4 1 D 4, A 1 A 2 Table 3.6. p = 13. To show how each table works for each remaining case, we provide a proof of the case when p = 5, rank( L 5,0 = 2 and = 1, as a ( det( L5,0 5

POSITIVE DEFINITE n-regular QUADRATIC FORMS 17 det( L 11,0 11 ( rank( L 11,0 E 1 1 8 3 A 1, 6, 3 8 1 1 4, 12, S11 6 = γ 11(D 5 44[1 1 4 ], M 6(12, 1 2 1 2 1, A 1 4 1 4 2 1 A 2, A 2 1 2 1 3 1 A 1 A 2, 4 1 4 2 1 3 1 A 3, A 1 1 4 2 1 2 1 4 1 4 1 A 1 A 3, A 2 1, 1 4 1 4 1 4 4 1 A 4, A 2 A 2 Table 3.7. p = 11. rank( L 7,0 ( det( L 7,0 7 E 1 1 6, 10, S7 6 = γ 7(A 6, M 7 (10, 3 4 2 2 1 1 A 1, 4, 2 8 1 2 1 8 4 2 2 1 A 2, 2 4 2 1 6 1 2 1 A 2 1, A 1 4,, 12 1 8 1 6 4 2 3 1 A 1 A 2, A 1 2 4 3 1 A 3, A 3 1 4 1 A 4, A 2 6, A 2 4 1 A 1 A 3, A 2 2 Table 3.8. p = 7. 4 1 1 4 representative one. Since A 2 and A 1 4 are represented by L in this case, K(3 = 2 1 0 1 2 1 L. 0 1 4 Note that 4 2 2 4 L and 4 2 K(3. 2 4 This implies that µ 4 ( L 4, and hence 2 1 0 1 K(4 = 1 2 1 1 0 1 4 2 L. 1 1 2 4

18 BYEONG-KWEON OH rank( L 5,0 ( det( L 5,0 5 E 1 1 A 1, 12, S 8 5 := γ 5(E 7 10[1 1 2 ], M 8(8, 3 1 1 4, 6, S5 4 := γ 5(A 4, M 6 (6, 1 4 2 2 1 6 1 2 1 A 2, A 1 4,, 2 4 1 8 1 6 2 1 4 1 2 1 A 2 1,, 1 6 1 4 3 1 A 2, A 2 1 3 1 A 3, A 1 A 2 4 1 A 1 A 3, A 2 1 A 2, A 2 6, 6 4 1 D 4, A 2 2 Note that d(k(4 = 5 2. Since 2 1 6 1 1 8 1 6 L, Table 3.9. p = 5. 2 1 1 8 6 1 K(4, 1 6 L contains a quinary sublattice whose determinant is less than or equal to 5 2 8 by Lemma 4.1. This is a contradiction to the fact that the discriminant of every quinary sublattice of L is divisible by 2 5 3. Case (3 p = 3. Note that r 3,1 ( L 3 12 or r 3,1 ( L 3 = 0 and r 3,0 ( L 3 + r 3,2 ( L 3 12. Hence if r 3,1 ( L 3 0, then we may assume that r 3,1 ( L 3 12. For each case, the sequence E that gives a contradiction is given by Table 3.10. The Z-lattices T 9 and T 8 (3 14 appearing in Table 3.10 are defined as follows: The Z-lattice T 8 (3 14 = Zx 1 + Zx 2 + + Zx 8 is defined by { Q(x 1 = 4, Q(x 2 = 16, B(x 1, x 2 = 1, Q(x i = 10 and B(x i, x j = 1, B(x 2, x i = 2B(x 1, x i = 4 for 3 i j 8. Note that d(t 8 (3 14 = 3 14 and T 8 (3 14 is a sublattice of E 8 such that (T 8 (3 14 3 1, 3 2, 3 2, 3 2, 3 2, 3 2, 3 2, 3 2. T 9 = Zx 1 + Zx 2 + + Zx 9 is the Z-lattice whose corresponding matrix is 2E 9 (4, 4 + 3E 9. Note that d(t 9 = 2 5 3 9. Since (T 9 3 has one dimensional unimodular component, T 9 L in the first case of Table 3.10. Since the other cases can be easily checked, we only provide the proof of the first case in Table 3.10. Suppose that there is a Z-lattice L such that A 2, T 9 L and the unimodular component in a Jordan decomposition of L 3 is isometric to 3. The assumptions on L imply that it has these properties. Then the root sublattice is isometric to A R L 2. Let L = Zx 1 + Zx 2 + Zx 3 + + Zx N

POSITIVE DEFINITE n-regular QUADRATIC FORMS 19 rank( L 3,0 ( det( L 3,0 3 r 3,1 ( L 3 = 0? E 1 1 No A 2, T 9 8 1 1 1 Yes A 1, 1 8 1 1 No γ 3 (E 6, M 7 (4, 1 1 1 Yes T 8 (3 14, M 8 (10, 1 2 1 No A 2 γ 3 (E 6, M 8 (4, 1 2 0 1 2 1 Yes 0 4 2 4 1, 1 6 1 2 6 2 1 No A 2 2, M 4(4, 1 2 1 2 1 Yes A 2 1, 1 4 3 1 No A 3 2, M 7(4, 1 2 1 3 1 Yes A 3 1, A 1 1 4 3 1 No A 2 2, A 3 2 1 3 1 Yes A 3, 4 1 4 4 1 No A 5, A 3 2 4 1 Yes A 4, A 3 1 4 1 No A 4 2, D 4 2 1 4 1 Yes D 4, A 2 1 1 4 Table 3.10. p = 3. and K = Zx 1 + Zx 2 A 2. One may easily show that for every z L such that Q(z = 6, z K or B(K, z = 0. Assume that M = Zy 1 + Zy 2 + + Zy 9 is a sublattice of L such that (B(y i, y j = 2E 9 (4, 4 + 3E 9. Since vertices (1, 2, 3, (3, 8, 9 and (5, 6, 7 are connected in the graph E 9, y i K = {w L : B(w, K = 0} for every i 4. Note that Q(y 4 = 8 and y 4 K. The corresponding matrix of K + Zy 4 is of the form 2 1 a 1 2 b for some a, b such that 3 a, b 3, a b 8 and d(k + Zy 4 is divisible by 9. Hence K + Zy 4 A 2 6. Therefore there are vectors u 1 K and u 2 K such that Q(u 1 = 2, Q(u 2 = 6 and y 4 = u 1 + u 2. Therefore the sublattice Zy 1 + Zy 2 + Zy 3 + Zu 2 + Zy 5 + + Zy 9 of L is isometric to 3E 9, which is indefinite. This is a contradiction.

20 BYEONG-KWEON OH In the remaining case for p = 2, we will use notations M a (b and N a (b. These notations represent certain Z-lattices of rank a with discriminant b. In particular, if we need a Z-lattice satisfying only some local properties, we prefer to use the notation N a (b. In this case, we only check that there exists a genus containing N a (b. Case (4 p = 2. For any even unimodular Z 2 -lattice l of rank n, n 0 (mod 2 and dl n + 1 (mod 4. If dl 1, 3 (mod 8 we say l is of type 1, and of type 2 otherwise. Let L be a Z-lattice satisfying all conditions in Lemma 3.3 and let L 2 L 2,0 L 2,1 L 2,2 L 2,t be a Jordan decomposition of L 2. Note that r 2,0 ( L 2 2, or L 2,0 is an even Z 2 -lattice with rank less than or equal to 10 by Lemma 3.3. Subcase (4-1 8 r 2,0 ( L 2 10. First, assume that r 2,0 ( L 2 = 10. Since E 8 and A 6 A 2 are represented by L, E 8 A 2 L. Therefore A 10 is represented by L, which is a contradiction. Assume that r 2,0 ( L 2 = 8 and L 2,0 is of type 1. Then E 8 L, and hence there is a Z-lattice M such that L E 8 M and s(m 2Z. Since A 2 2 1 2 1 2 1 4 1 4 L and A 2 2 1 2 1 2 E 1 4 1 4 8, M should represent 2. Therefore A 2 A 6 L, which is a contradiction. Now assume that r 2,0 ( L 2 = 8 and the unimodular component is always of type 2 for any Jordan decomposition of L 2. Then either L 2,1 = 0 or L 2,1 is of even type. But this is impossible for E 7 and A 2 A 6 are represented by L. Subcase (4-2 2 r 2,0 ( L 2 6 and L 2,0 is even. Note that, by Lemma 3.3 (ii, every possible structure of L 2 up to the 4Z 2 -modular component can be given by Table 3.11. In this table, for example, L 2,2 (o is the 4Z 2 - modular component of odd type and L 2,2 (e is the 4Z 2 -modular component of even type. In particular, L 2,0 (e, i is the even unimodular component of type i for i = 1, 2. Note that the 4Z 2 -modular component might be nonzero in Number 10, 11 and 14 cases in Table 3.11. For those cases, the rank of the 2Z 2 -modular component is greater than or equal to 12 by Lemma 3.3 (ii. Subcase (4-2-1 r 2,0 ( L 2 = 6. In this case, we have Table 3.12. In this table, 2 1 1 1 2 1 M 2 (7 = and M 1 4 4 (16 5 = 1 4 0 0 1 0 4 0. 1 0 0 4

POSITIVE DEFINITE n-regular QUADRATIC FORMS 21 Name L2,0 L 2,1 L 2,2 Name L2,0 L 2,1 L 2,2 1 L2,0 (e, 1 L 2,2 (o 2 L2,0 (e, 1 L 2,2 (e 3 L2,0 (e, 1 2 L 2,2 (o 4 L2,0 (e, 1 2 L 2,2 (e 5 L2,0 (e, 1 6 L 2,2 (e 6 L2,0 (e, 1 2, 2 L 2,2 (o 7 L2,0 (e, 1 2, 2 L 2,2 (e 8 L2,0 (e, 1 2, 6 L 2,2 (e 9 L2,0 (e, 1 6, 6 L 2,2 (e 10 L2,0 (e, 1 L 2,1 (o 11 L2,0 (e, 1 L 2,1 (e 12 L2,0 (e, 2 L 2,2 (o 13 L2,0 (e, 2 L 2,2 (e 14 L2,0 (e, 2 L 2,1 (e Table 3.11. Possible structures of L 2,0 L 2,1 L 2,2. Name E Name E 1 E 6, A 2 1 A 4 2 E 6 (8 4 8 2 [1 1 1 3 3 ], A 2 M 2 (7 2 3 E 7, A 1 E 6 4 E 6, A 6 5 E 7, A 2 2 M 2(7 6 E 7 A 1, A 2 1 E 6 7 E 6, A 6 8 E 7 A 1, A 1 A 2 2 M 2(7 9 E 7, D 7 10 E 7, A 2 2 D 4 11 E 6, A 2 2 D 4 12 D 7, A 7 13 A 7 8[4 1 2 ], A2 2 M 4(16 5 14 A 7, D 4 D 4 Table 3.12. r 2,0 ( L 2 = 6. Since other cases can be done in a similar manner, we only provide a proof of Number 2 case. Note that d(e 6 (8 4 8 2 [1 1 1 3 3 ] = 16 (for the precise definition of this lattice, see [5] and [ (E 6 (8 4 8 2 1 1 ] 1 2 1 2 1 2 1 8 4. 3 3 1 2 1 2 1 2 4 8 2 Therefore it is represented by L for this case. Let φ be a representation from E 6 (8 4 8 2 [1 1 1 3 3 ] to L. Since [ A 2 M 2 (7 2 L and A 2 M 2 (7 2 E 6 (8 4 8 2 1 1 ] 1, 3 3 there is a vector x L φ(e 6 (8 4 8 2 [1 1 3 3 ] such that Q(x = 4. This implies that L has a sublattice of rank 9 whose discriminant is less than or equal to 16 4 by Lemma 4.1. This is a contradiction to the fact that the discriminant of any sublattice of L with rank 9 is divisible by 16 8. Subcase (4-2-2 r 2,0 ( L 2 = 4. In this case we have Table 3.13. 1

22 BYEONG-KWEON OH Name E Name E 1 A 2 3, M 4(16 5 2 2 N 8 (2 8, M 4 (16 13 M 4 (16 5 3 A 5, A 2 A 3 4 A 1 A 4, A 1 A 2 2 5 N 7 (2 5, M 4 (16 5 2 6 D 6, A 1 A 5 7 D 5, A 2 A 3 8 N 7 (2 5 A 1, A 1 M 4 (16 5 2 9 D 6, M 2 (7 M 2 (15 4, 4 10 D 6, A 3 D 4 11 D 2 4, A9 1 12 D 5, A 2 M 4 (16 5 13 N 8 (28, M 4 (16 5 M 4 (16 13 14 D 5, A 2 1 D 4 Table 3.13. r 2,0 ( L 2 = 4. 4 1 In this table M 2 (15 =, 1 4 2 1 1 0 2 1 0 1 M 4 (16 5 = 1 2 0 1 1 0 6 1 and M 4(16 13 = 1 6 1 3 0 1 6 3. 0 1 1 6 1 3 3 6 N 8 (2 8 and N 8 (28 are Z-lattices of discriminant 2 8 such that 2 1 2 1 8 4 8 4 (N 8 (2 8 2, ( 1 2 ( 1 2 ( 4 8 ( 4 8 2 1 0 1 8 4 0 4 (N 8 (28 2. 1 2 1 0 4 8 4 0 Note that each Z-lattice given above exists as a sublattice of E 8. The Z- lattice N 7 (2 5 is a sublattice of E 7 with discriminant 2 5 such that (N 7 (2 5 2 1 2 1 8 4 2 6. 1 2 1 2 4 8 We only provide a proof of Number 1 case. In this case, note that 2 1 2 1 L 2 1 2 1 2 L 2,2 (o M, where r 2,2 ( L 2 = rank( L 2,2 (o 8 and s(m 8Z 2. Therefore R L = A 3 A 3. Note that R L is a primitive sublattice of L. Let L = Zx 1 + Zx 2 + + Zx 6 + + Zx N and K = Zx 1 + Zx 2 + + Zx 6 A 3 A 3. Assume that there is a vector x L such that Q(x = 4. If x K then the rank of K = Zx 1 +Zx 2 + + Zx 6 + Zx is 7, and the discriminant dk is less than or equal to 4 3. Since the discriminant of any sublattice of L with rank 7 is divisible by 4 3, we have dk = 4 3. This also implies that B(x i, x = 0 for 1 i 6. Therefore, every vector in L of norm 4 is either contained in K or is orthogonal to K. Assume that there is a binary sublattice Zw 1 + Zw 2 L such that 2 1 (B(w i, w j =. 1 4

POSITIVE DEFINITE n-regular QUADRATIC FORMS 23 Since = K, we have w R L 1 K. Therefore w 2 K by the above observation. Now since 2 1 1 1 2 1 1 1 M 4 (16 5 M 4 (16 5 = 1 4 0 0 1 0 4 0 1 4 0 0 1 0 4 0 L, 1 0 0 4 1 0 0 4 M 4 (16 5 M 4 (16 5 K. This is a contradiction. Subcase (4-2-3 r 2,0 ( L 2 = 2 and L 2,0 is even. In this case, we have Table 3.14. Name E Name E 1 A 3, M 5 (2 9 2 N 8 (2 12, M 9 (6, 2 3 A 1 A 3, M 4 (16 5 4 A 1 A 2, M 4 (16 5 5 N 7 (2 9, M 3 (2 5 5 M 4 (16 13 6 A 2 1 A 3, A 2 1 M 3(12 7 A 3, A 2 1 A 2 8 N 8 (2 10, A 1 M 4 (16 13 M 3 (2 5 5 9 N 6 (2 6, M 4 (16 13 6, 6 10 D 4, A 5 1 6[11111 1 2 ] 11 D 4, A 5 1 12 N 8 (212, M 9 (6, 2 13 N 8 (2 6, A 9 1 14 M 5 (2 6, M 4 (16 7 In Number 1 case, Table 3.14. r 2,0 ( L 2 = 2 and L 2,0 is even. 4 0 0 0 2 M 5 (2 9 0 4 0 0 2 = 0 0 4 0 2 0 0 0 4 2, 2 2 2 2 6 which is represented by L. The proof of this case is as follows: Let L = Zx 1 + Zx 2 + Zx 3 + Zx 4 + + Zx N and K = Zx 1 + Zx 2 + Zx 3 A 3. We can show, by a similar method to the Subcase (4-2-2, that for any vector z L of norm 4, z is contained in K K. Let Zw 1 + Zw 2 L be a binary Z-lattice such that 4 2 (B(w i, w j =. 2 6 Assume that w 1 K, i.e., w 1 K. Then w 2 K. Hence K = K + Zw 2 is a quaternary sublattice of L. Since dk is divisible by 16, K A 3 4. This implies that there are vectors z 1 K and z 2 K with Q(z 1 = 2 and Q(z 2 = 4 such that w 2 = z 1 + z 2. Hence 4 2 K + Zw 1 + Zz 2 A 3 2 4 L, which is a contradiction. Consequently w 1 K. From this and the fact that M 5 (2 9 L, A 3 represents a quaternary Z-lattice 4, 4, 4, 4. This is a contradiction.

24 BYEONG-KWEON OH In Number 2 and 12 cases, N 8 (2 12 and N 8 (212 are sublattices of E 8 such that 2 1 8 4 8 4 8 4 (N 8 (2 12 2, ( 1 2 ( 4 8 ( 4 8 ( 4 8 0 1 8 4 8 4 0 4 (N 8 (212 2. 1 0 4 8 4 8 4 0 Also note that (M 9 (6, 2 2 6 8 4 4 8 8 4 4 8 8 4 4 8 8 4. 4 8 In Number 5 and 6 cases, N 7 (2 9 is a sublattice of E 7 such that (N 7 (2 9 2 1 8 4 8 4 2 6 1 2 4 8 4 8 and M 3 (2 5 5 = 6 2 2 2 6 2, M 3 (12 = 2 0 1 0 2 1. 2 2 6 1 1 4 In Number 8 case, N 8 (2 10 is a sublattice of E 8 such that (N 8 (2 10 2 1 8 4 8 4 2 2, 6. 1 2 4 8 4 8 In Number 9 case, N 6 (2 6 is a sublattice of I 6 such that (N 6 (2 6 2 1 8 4 2 6, 6. 1 2 4 8 In Number 13 case, N 8 (2 6 is a sublattice of E 8 such that (N 8 (2 6 0 1 0 2 0 2 0 2 2. 1 0 2 0 2 0 2 0 In Number 14 case, 2 0 0 0 1 M 5 (2 6 0 2 0 0 1 4 2 1 2 = 0 0 4 0 2 0 0 0 4 2, M 4(16 7 = 2 4 0 0 1 0 4 0. 2 0 0 4 1 1 2 2 4 Note that (M 5 (2 6 2 0 1 28, 28, 28 and (M 1 0 4 (16 7 2 0 1 20, 20. 1 0 Hence they are represented by L in this case. One may easily show that every vector x L of norm 4 is contained in M 5 (2 6 M 5 (2 6. This gives a contradiction. Subcase (4-3 r 2,0 ( L 2 = 2 and L 2,0 is odd. In this case, r 2,1 ( L 2 = rank( L 2,1 10. We may easily show that for each case, the sequence E can

POSITIVE DEFINITE n-regular QUADRATIC FORMS 25 be given as follows: ( 1, 1, A 1 A 1 A 1 6[111 1 2 ] if L2,0 L 2,1 1, 1 L 2,1 (o, ( 1, 1, 2, 2, 2 if L2,0 E = L 2,1 1, 1 L 2,1 (e, ( 1 N7 (2 6, 1 M 8 (3, 1 if L2,0 L 2,1 1, 3 L 2,1 (e, (A 6 1 [111111], A7 1 if L2,0 L 2,1 3, 3 L 2,1 (e. Here N 7 (2 6 is a sublattice of I 7 such that (N 7 (2 6 4 2 2 3 2 4 4 2 2 4 4 2. 2 4 Subcase (4-4 r 2,0 ( L 2 = 1. In this case, we have the following Table 3.15 on the possible local structures up to 4Z 2 -modular component and the sequence E. L 2,0 L 2,1 L 2,2 E L2,0 L 2,1 L 2,2 E 1 L 2,1 (o L 2,2 1, M 5 (3 2 4 1 L 2,1 (e L 2,2 1, S 9 (2 3 L 2,1 (e L 2,2 N 7 (2 6, M 8 (3, 1 1 L 2,2 (o 1, S 9 (2 In the first case, 3 L 2,2 (o N 7 (2 12, 3 1 L 2,2 (e 1, S 9 (2 3 L 2,2 (e M 3 (2 4, 4 4 5 L 2,2 (e N 5 (2 8, 4 6 7 L 2,2 (e N 7 (2 12, 4 8 Table 3.15. r 2,0 ( L 2 = 1. 3 2 2 2 2 M 5 (3 2 4 2 4 2 0 0 = 2 2 4 0 0 2 0 0 4 2. 2 0 0 2 4 If 3 is replaced by 2 in the above matrix, then it is indefinite. S 9 (2 is the Z- lattice whose corresponding matrix is E 9 (3, 3 + 2E 9. Note that d(s 9 (2 = 2 8 3 11 and (S 9 (2 2 1 4 2 2 4 4 2 2 4 4 2 2 4 4 2. 2 4 S 9 (2 is the Z-lattice whose corresponding matrix is E 9(3, 3 + 4E 9 Note that d(s 9 (2 = 2 16 3 7 and (S 9(2 8 4 8 4 8 4 0 4 2 1. 4 8 4 8 4 8 4 0 N 7 (2 12 is a sublattice of I 7 such that (N 7 (2 12 0 4 2 7 4 0 0 4 4 0 0 4. 4 0

26 BYEONG-KWEON OH N 5 (2 8 is a sublattice of I 5 such that Finally which is a sublattice of I 3. (N 5 (2 8 2 5 8 4 4 8 M 3 (2 4 = 3 1 1 1 3 1, 1 1 3 5. n-regular Z-lattices 0 4. 4 0 Recall that a Z-lattice L is called (even n-universal if L represents all (even, respectively Z-lattices of rank n. Lemma 5.1. Let n be any integer greater than or equal to 27. For every n-regular Z-lattice L and every prime p, r p,0 (L p n + 4. Proof. Let L be an n-regular Z-lattice. In Section 4 we proved that r p,0 (L p is greater than or equal to 12, for every prime p. This implies that L is even 9-universal, that is, L represents all even Z-lattices of rank 9. Let m be an integer such that 8m < n 8(m + 1. Note that D 8k [1] is an indecomposable even unimodular Z-lattice of rank 8k, for every positive integer k. Since E 8 = D 8 [1] is represented by L, there is a Z-lattice L such that L E 8 L. Furthermore since D 9 is also represented by L, we may assume that D 9 L, which follows from the fact that every indecomposable splitting of root systems is unique. From this, D 16 [1] is locally represented by L and hence is represented by L globally by regularity of L. Therefore we have E 8 D 16 [1] L by a similar reasoning to the above. Now, by an induction argument, we have E 8 D 16 [1] D 8m [1] L. Therefore r p,0 (L 4m(m + 1 n + 4, for every prime p. Theorem 5.2. For any n 27, every n-regular Z-lattice is, in fact, an (even, if L is even n-universal Z-lattice. Proof. Note that every (even unimodular Z p -lattice of rank n+3 represents all (even, respectively Z p -lattices of rank n, for every prime p. Hence the theorem follows directly from Theorem 5.1. Corollary 5.3. For n 28, let L be an almost n-regular Z-lattice, that is, L represents almost all Z-lattices of rank n that are represented by the genus of L. Then L represents almost all (even, if L is even Z-lattices of rank n. Proof. Note that every almost n-regular Z-lattice is (n 1-regular. Hence r p,0 (L n + 3 by Theorem 5.1, which implies that the Z p -lattice L p represents all (even, if L 2 is even Z p -lattices of rank n, for every prime p. The corollary follows from this.