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 repre-sented by the genus of f . In this paper, we show that for any integern greater than or equal to 27, every n-regular (even) form f is (even)n-universal, that is, f represents all (even, respectively) positive definiteintegral quadratic forms of rank n. As an application, we show that theminimal rank of n-regular forms has an exponential lower bound for nas it increases.

1. Introduction

A positive definite integral quadratic form f is called regular if f representsall integers that are represented by the genus of f . Regular quadratic formswere first studied systematically by Dickson in [6] where the term “regular”was coined. In the last chapter of his doctoral thesis [20], Watson showed byarithmetic arguments that there are only finitely many equivalence classesof positive definite primitive regular ternary quadratic forms. He did so byproviding explicit bounds on the prime power divisors of the discriminant ofthose regular ternary quadratic forms. The classification of such quadraticforms was done by Kaplansky and his collaborators [13]. They proved thatthere are at most 913 positive definite primitive regular ternary quadraticforms, 22 of which are still candidates. On the contrary, Earnest provedin [8] that there are infinitely many equivalence classes of positive definiteprimitive regular quaternary quadratic forms. In fact, every positive definitequadratic form with more than 4 variables is almost regular, that is, sucha form f represents almost all positive integers that are represented by thegenus of f (cf. [24]).

Unless stated otherwise, by an “integral form” we shall always mean apositive definite quadratic form having an integer matrix. The rank of aquadratic form f is defined by the number of variable of f .

The study of higher dimensional analogs of regular quadratic forms is firstinitiated by Earnest in [7]. An integral form f of rank m is called n-regularif f represents all quadratic forms of rank n that are represented by thegenus of f . Hence every n-regular form f satisfies, so called, “a local-globalprinciple over Z” in the following sense: for any quadratic form g of rankn, g is represented by f over Z if and only if g is represented by f over thep-adic integer ring Zp, for every prime p.

If the class number h(f) of a quadratic form f is one, then f is n-regularfor 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 bythe 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) off is one (cf. [14], Corollary 6.4.1). In fact, there are many examples ofn-regular forms of rank n+ 2 or n+ 3 whose class number is bigger than 1for 1 ≤ n ≤ 5. However if n is large, the situation is quite different. Notethat the class number of every quadratic form with rank greater than 10 isbigger than one (cf. [23]). Furthermore, it seems to be quite difficult to findan n-regular form with minimal rank, for each n ≥ 11.

On the other hand, every quadratic form satisfies a local-global principleover Z under some restrictions. In 1978, Hsia, Kitaoka and Kneser provedin [10] that if the minimum positive integer that is represented by g issufficiently large and rank(f) ≥ 2rank(g)+3, then g is represented by f overZ if and only if g is represented by f over Zp for every prime p. Recently,Ellenberg and Venkatesh improved this theorem by showing that the rankcondition in this theorem could be replaced by rank(f) ≥ rank(g) + 7, withone 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 ghas sufficiently large minimum is crucial. Without this condition, it seemsto be quite difficult to use some kinds of approximation theorems or analyticmethods developed by several authors. In this paper, we consider the local-global principle over Z without this minimum condition.

Earnest showed in [7] that there exist only finitely many equivalenceclasses of 2-regular primitive integral quaternary forms. Actually, his methodis an extension of Watson’s analytic argument (see [21]), which seems to beinadequate for other higher dimensional situations. Recently, Chan and theauthor proved in [4] that there exist only finitely many equivalence classesof 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 bringback 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 correspond-ing symmetric matrix is even, and is called odd otherwise. Every n-universalform is, in fact, an n-regular form that represents all integral forms of rankn over Zp, for every prime p. For the classification of n-universal forms withminimal rank, see the recent survey article [15]. Let {f1, f2, . . . , fh} be aset of representatives of all equivalence classes of odd unimodular forms ofrank n + 3. Since every integral form g of rank n is represented by thegenus of any unimodular form of rank n+ 3, g is represented by fi for somei = 1, 2, . . . , h. Hence the unimodular form

f1(y11, . . . , y1,n+3) + f2(y21, . . . , y2,n+3) + · · ·+ fh(yh1, . . . , yh,n+3)

of rank (n + 3)h represents all integral forms of rank n. This is a typicalexample 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 ngreater than 10.

We denote by R(n) the minimal rank of n-regular quadratic forms. Sincethere 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 forevery n = 1, 2, . . . , 10. As mentioned above, R(n) ≥ n+ 2 for every n biggerthan 10. As far as the author knows, this was the best known lower boundfor 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 equalto 27. As an application of this result, we also show that R(n) has anexponential lower bound for n as it increases (cf. Theorem 5.4).

The subsequent discussion will be conducted in the more adapted geomet-ric language of quadratic spaces and lattices, and any unexplained notationsand terminologies can be found in [5], [14] or [18]. The term lattice will al-ways refer to an integral Z-lattice on an m-dimensional positive definitequadratic space over Q. The scale and the norm ideal of a lattice L are de-noted by s(L) and n(L), respectively. It will be assumed that every Z-latticeL is primitive, i.e., s(L) = Z.

For any Z-lattice L, we denote by h(L) the class number of L, and µi(L)the i-th successive minimum for every i = 1, 2, . . . , rank(L).

For two Z-lattices ` and L, if every Z-lattice in the genus of ` is representedby L, we write

gen(`)→ L.

Let {e1, e2, . . . , en} be a basis of a Z-lattice L. For any real number a, aLis denoted by the Z-lattice with a basis {ae1, ae2, . . . , aen}. If M = (mij)is a matrix, then aM is denoted by the matrix (amij). If B(ei, ej) = aiδij ,then we will write L = 〈a1, a2, . . . , an〉 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 integerm, we define

Λm(L) = {x ∈ L : Q(x+ z) ≡ Q(z) (mod m) for all z ∈ L }.Note that x ∈ Λm(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 Λ2(L). Let λm(L) be the primitive lattice obtained from Λm(L)by scaling V by a suitable rational number. Note that the scaling fac-tor depends only on the lattice structure of Lp, for every prime p dividingm. These λm-transformations were used by Watson in his study of regularternary lattices [20] and class numbers [23], and more recently by variousauthors in [3] and [4] concerning lattices satisfying different kinds of regu-larity conditions. These references contain most of the basic properties ofthe λm-transformations needed in this paper.

For any Zp-lattice L, the Λm-transformation Λm(L) is defined similarly.In particular, one may easily verify that for every Z-lattice L,

(λm(L))p ' λm(Lp)

4 BYEONG-KWEON OH

for every prime p, where Lp = L⊗ Zp is the Zp-lattice.For a Z-lattice L and a prime p, let

Lp = Lp,0 ⊥ Lp,1 ⊥ · · · ⊥ Lp,tp

be a Jordan decomposition such that s(Lp,i) = piZp or Lp,i = 0 for everyi = 1, 2, . . . , tp, and Lp,tp 6= 0. Throughout this paper, when we consider aJordan decomposition, we always assume this condition. We define

rp,i = rp,i(Lp) = rank(Lp,i) and rp(Lp) = (rp,0, rp,1, . . . , rp,tp).

Note that these definitions are independent of a Jordan decomposition ofLp . If p is odd, then we may easily check that

Λp(L)p ' pLp,0 ⊥ Lp,1 ⊥ · · · ⊥ Lp,tp .

Therefore we have

(∗) rp((λp(L))p) =

{(rp,1, rp,0 + rp,2, rp,3, . . . , rp,tp) if rp,1 6= 0,

(rp,0 + rp,2, rp,3, . . . , rp,tp) otherwise .

Now assume that p = 2. If L is an odd Z-lattice, then Λ2(L)2 ' L2,0(e) ⊥L2,1 ⊥ · · · ⊥ L2,t2 , and if L is even then

Λ4(L)2 ' 2L2,0 ⊥√

2

((1√2L2,1

)(e)

)⊥ L2,3 ⊥ · · · ⊥ L2,t2 .

For any odd unimodular Z2-lattice ` of rank n ≥ 3, there is an evenunimodular Z2-lattice U , and an odd unimodular Z2-lattice `1 of rank lessthan 3 such that

` ' U ⊥ `1.Hence `(e) ' U ⊥ `1(e). Note that if n is odd, then `1(e) is a proper 4Z2-modular lattice of rank 1, otherwise `1(e) is a 2Z2-modular lattice of rank2. Therefore if L is odd,

r2((Λ2(L))2) =

{(r2,0 − 1, r2,1, r2,2 + 1, r2,3, . . . , r2,t2) if r2,0 is odd,

(r2,0 − 2, r2,1 + 2, r2,2, . . . , r2,t2) otherwise.

If L is even, r2((Λ4(L))2) equals to(0, r2,1 − 1, r2,0 + r2,2, r2,3 + 1, . . . , r2,t2) if r2,1 is odd,

(0, r2,1 − 2, r2,0 + r2,2 + 2, r2,3, . . . , r2,t2) if r2,1 is even, n(L2,1) = 2Z2,

(0, r2,1, r2,0 + r2,2, r2,3, . . . , r2,t2) otherwise.

Note that for any even Z-lattice L, L(e) = λ2(L) = L. So we use amodified λp-transformation, which we call γp-transformation.

Definition 2.1. Let L be a Z-lattice. For every prime p, Γp-transformationis defined as follows: If p is odd Γp(L) = Λp(L), and

Γ2(L) =

{Λ2(L) if L is odd,

Λ4(L) otherwise.

γp(L) is defined by the primitive lattice obtained from Γp(L) by scaling V bya suitable rational number.

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 5

Definition 2.2. Let p be a prime and `, L be Zp-lattices such that `→ L.Assume that a Jordan decomposition of L is

L = L0 ⊥ L1 ⊥ · · · ⊥ Lt .

` is called a lower type of L if there is an integer u = u(`, L) (0 ≤ u ≤ t)such that a Jordan decomposition of ` is

` = L0 ⊥ L1 ⊥ · · · ⊥ Lu−1 ⊥ L′u ,

where L′u is a non-zero puZp-modular lattice such that

n(L′u) = n(Lu) and L′u→Lu ⊥ · · · ⊥ Lt .

If ` is a lower type of L, then we write ` ≺ L.

Lemma 2.3. Let p be a prime and L be a Zp-lattice with rank greater thanone. Assume that a Jordan decomposition of L is

L = L0 ⊥ L1 ⊥ · · · ⊥ Lt .

For some positive integer s, let k be any integer satisfying

rank(L0 ⊥ L1 ⊥ · · · ⊥ Ls−1) ≤ k < rank(L0 ⊥ L1 ⊥ · · · ⊥ Ls).

Then, there does not exist a Zp-lattice ` with rank k such that ` ≺ L if andonly if p = 2, n(Ls) 6= s(Ls) and k − rank(L0 ⊥ · · · ⊥ Ls−1) is odd.

Proof. The proof is quite straightforward. �

Lemma 2.4. For any integer n greater than 5, let ` and L be any Z2-latticessuch that rank(`) = n and ` ≺ L. Assume that the rank of the unimodularcomponent in a Jordan decomposition of ` is less than [n−2

2 ]. Then there isan integer k satisfying the following properties:

(i) γi2(`) ≺ γi2(L) for any i = 0, 1, . . . ,max(0, k − 2).

(ii) r2,0(γk−12 (`)) <

[n− 2

2

].

(iii) r2,0(γk2 (`)) ≥[n− 2

2

].

(iv) If γk−12 (L) = L′0 ⊥ L′1 ⊥ · · · ⊥ L′t is a Jordan decomposition, then

L′0 is even unimodular or rank(L′0) ≤ 2. Furthermore, γk−12 (`) ≺

γk−12 (L) or

γk−12 (`) ' L′0 ⊥ l′1 ⊥ 〈4ε〉,

where l′1 is a 2Z2-modular lattice with n(l′1) = n(L′1) and ε ∈ Z×2 .

(v) For any i = 1, 2, . . . , k,

n(Γ2(γi−12 (L)))=n(Γ2(γi−1

2 (`))) and s(Γ2(γi−12 (L)))=s(Γ2(γi−1

2 (`))).

Proof. We use an induction on u = u(`, L). Let

L = L0 ⊥ L1 ⊥ · · · ⊥ Lu ⊥ · · · ⊥ Lt , ` = L0 ⊥ L1 ⊥ · · · ⊥ Lu−1 ⊥ lu

be a Jordan decomposition for each Z2-lattice L and `. First, assume thatu = 1. If L0 is even or rank(L0) ≤ 2, then k = 1 and everything is trivial.

6 BYEONG-KWEON OH

Assume that L0 is an odd unimodular lattice with rank greater than 2.Then,

γ2(L0) = L′0 ⊥ B, or γ2(L0) = L′0 ⊥ 〈4ε〉according the parity of rank(L0). Here L′0 is even unimodular, ε ∈ Z×2 andB is a binary 2Z2-modular lattice. Hence

γ2(`) = L′0 ⊥ l1 ⊥ 〈4ε〉 or γ2(`) = L′0 ⊥ (l1 ⊥ B).

Therefore, k = 2 and the lemma follows from this.Now assume that u = 2. Recall that ` = L0 ⊥ L1 ⊥ l2 is a Jordan decom-

position. Suppose that L0 is even unimodular. If L1 = 0 or rank(L1) ≤ 2and n(L1) = s(L1), then clearly k = 1. If n(L1) 6= s(L1) or rank(L1) iseven, then we may easily check that γ2(`) ≺ γ2(L) and u(γ2(`), γ2(L)) = 1.Therefore, the lemma follows from the induction hypothesis. Assume thatn(L1) = s(L1) and rank(L1) is odd greater than 1. Then Jordan decompo-sitions for both γ2(`) and γ2(L) are of the form

γ2(L) = L′0 ⊥ L′1 ⊥ · · · ⊥ L′t−1, γ2(`) = L′0 ⊥ l′1 ⊥ 〈4ε〉,

where rank(L′0) = rank(L1) − 1, rank(L′1) = rank(L0) + rank(L2), andrank(l′1) = rank(L0) + rank(l2). Note that L′0 is even unimodular andn(L′1) = n(l′1). Suppose that rank(L′0) < [n−2

2 ]. Then, one may easily

check that r2,0(γ22(`)) ≥ [n−2

2 ]. Therefore, k = 2 in this case and the lemmafollows from this.

Suppose that L0 is odd unimodular. If rank(L0) ≤ 2, then either k = 1 orγ2(`) ≺ γ2(L) and u(γ2(`), γ2(L)) < 2. If rank(L0) ≥ 3, then γ2(`) ≺ γ2(L)and k 6= 1. So, this case can be reduced to the above case.

Finally assume that u = u(`, L) ≥ 3. If L0 is even or rank(L0) ≤ 2,then it is easy to show that γ2(`) ≺ γ2(L) and u(γ2(`), γ2(L)) < u(`, L).Therefore the lemma follows from the induction hypothesis. If L0 is an oddunimodular Z2-lattice with rank greater than 2, then we may consider γ2(`)and γ2(L) instead of the original lattices. The condition (v) follows directlyfrom the condition (iv). This completes the proof. �

Remark 2.5. In Lemma 2.4, assume that n ≥ 12 and γk−12 (`) ≺ γk−1

2 (L).Let

γk−12 (L) = L′0 ⊥ L′1 ⊥ · · · ⊥ L′t and γk−1

2 (`) = L′0 ⊥ L′1 ⊥ · · · ⊥ l′s

be a Jordan decomposition for each Z2-lattice. If s is greater than 1, then onemay easily show that rank(L′1) ≤ 2 or rank(L′1) ≥ [n−2

2 ] − 2. Furthermore,if rank(L′1) ≤ 2 then

rank(L′0) + rank(l′2) ≥[n− 2

2

]− 2.

Note that if s = 1 or γk−12 (`) is not a lower type of γk−1

2 (L), then rank(l′1) ≥[n+1

2 ].

Lemma 2.6. Let p be an odd prime and `, L be any Zp-lattices such that` ≺ L. Assume that the rank of the unimodular component in a Jordandecomposition of ` is less than [n+1

2 ], where rank(`) = n. Then there is aninteger k satisfying the following properties:

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 7

(i) γip(`) ≺ γip(L) for any i = 0, 1, . . . , k − 1.

(ii) rp,0(γk−1p (`)) <

[n+ 1

2

].

(iii) rp,0(γkp (`)) ≥[n+ 1

2

].

(iv) s(Γp(γi−1p (L))) = s(Γp(γ

i−1p (`))) for any i = 1, 2, . . . , k.

Proof. If one use the rank equation (∗) given above, this lemma can beproved in a similar manner to Lemma 2.4. �

In some cases, the γp-transformation preserves the n-regularity. More pre-cisely, assume that a Z-lattice L is n-regular. If the unimodular componentin a Jordan decomposition of Lp is anisotropic for some prime p, then γp(L)is also n-regular. If L is an odd n-regular Z-lattice, then γ2(L) is alwaysn-regular without any assumption. But, this is not true in general for anodd prime p. For example, the quinary Z-lattice 〈1, 1, 1, 1, 7〉 is 1-regular,whereas γ7(L) = 〈1, 7, 7, 7, 7〉 is not 1-regular.

Lemma 2.7. Let p be a prime, and `, L be any primitive Z-lattices suchthat ` ⊂ L and n(`p) = n(Lp). If rp,0(`p) = rp,0(Lp), then Γp(`) ⊂ Γp(L).

Proof. For any prime p, define

L(p) = {x ∈ L : B(x, L) ⊂ pZ}.

Note that L(p) = Γp(L) for any odd prime p, whereas L(2) is not equalto Γ2(L) in general. Since rp,0(`p) = rp,0(Lp), one may easily show thatL = `+L(p). Assume that p is odd and x ∈ Γp(`). For any z ∈ L, there arez1 ∈ ` and z2 ∈ L(p) such that z = z1 + z2. Hence

B(x, z) = B(x, z1) +B(x, z2) ≡ 0 (mod p),

which implies that x ∈ Γp(L). Now assume that p = 2. If L is odd, theneverything is trivial. If L is even, then the proof is quite similar to that ofan odd prime case given above. �

A Z-lattice L is called almost n-regular if L represents almost all Z-latticesof rank n that are represented by the genus of L. Note that every almostn-regular Z-lattice is (n− 1)-regular.

Lemma 2.8. For n ≥ 2, let L be an almost n-regular Z-lattice of rankgreater than n+ 1. Then there is a prime r such that gcd(r, 2dL) = 1, anda sublattice ` of L with rank n or n− 1 satisfying the following properties:

(i) `p is a lower type of Lp for any prime p 6= r.(ii) Every Z-lattice in the genus of ` is represented by L.

(iii) d`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 N9 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, Lq and Nq are unimod-

ular and(dNq

)= 1. Define

T = {p | ordp(2dL) ≥ 1} ∪ {q}.Let `(2) be a Z2-sublattice of L2 with rank n or n − 1 such that `(2) ≺ L2.Here, we choose a sublattice `(2) of rank n− 1 only in the case when n andL2 satisfy the condition in Lemma 2.3. Now for each p ∈ T − {2, q}, let`(p) be any Zp-sublattice such that `(p) ≺ Lp and rank(`(p)) = rank(`(2)).Finally, let `(q) be a sublattice of Lq such that `(q) ' 〈1, 1, . . . ,∆q〉 andrank(`(q)) = rank(`(2)). Then by Lemma 1.6 of [10], there is a Z-sublattice

` of L such that `p ' `(p) for any p ∈ T and for p 6∈ T , d(`p) ∈ Z×p with

precisely one exception p = r, where d(`r) ∈ rZ×r . Clearly `p is a lowertype of Lp for any p 6= r, and N 6∈ gen(`) for any N ∈ S. The definitionof S implies that every Z-lattice in the genus of ` is represented by L. Thelemma follows from this. �

3. Local structures of n-regular lattices for large n

Throughout this and next section, we always assume that n ≥ 27. Underthis assumption, we prove the following theorem.

Theorem 3.1. Let L be an n-regular Z-lattice. Then for every prime p,

(∗) rp,0(Lp) ≥ 12,

where rp,0(Lp) is the rank of the unimodular component in a Jordan decom-position of the Zp-lattice Lp.

Lemma 3.2. Let p be an odd prime. Assume that there is an n-regularZ-lattice L such that

rp,0(Lp) < 12.

Then there is an even Z-lattice L satisfying the following properties:

(i) rp,0(Lp) < 12.

(ii) rp,1(Lp) ≥ 12 or rp,1(Lp) = 0 and rp,0(Lp) + rp,2(Lp) ≥ 12.(iii) For any even Z-lattice N with rank less than or equal to 9 such that

Np→Lp, 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 ` be a sublattice of L and r be a prime satisfying all conditions inLemma 2.8. Define

T1 := {q : rq,0(Lq) < 12} = {q0 = p, q1, q2, . . . , qt}.For any prime q ∈ T1 − {p}, let k(q) be an integer satisfying all conditionsin Lemma 2.4 or 2.6. Then

rq,0(γk(q)q (`)q) ≥

[n− 3

2

]≥ 12,

and γk(q)q (`) is represented by γ

k(q)q (L) by Lemma 2.7. Furthermore, since

the γq-transformation is a surjective map from gen(K) to gen(γq(K)), forany Z-lattice K and any prime q (cf. [22]),

gen(γk(q)q (`))→γk(q)

q (L).

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 9

Define

`1 =

(t∏i=1

γk(qi)qi

)(`) and L1 =

(t∏i=1

γk(qi)qi

)(L).

Then gen(`1)→L1 and rq,0((`1)q) ≥ 12 for any q 6= p. Note that (L1)p isisometric to Lp up to unit scaling factor and (`1)p ≺ (L1)p.

Now, by a similar reasoning to Lemma 2.6, there is an integer s such that

(α) (γs−1p (`1))p ≺ (γs−1

p (L1))p and rp,0((γs−1p (L1))p) < 12,

(β) rp,0((γsp(L1))p) ≥ rp,0((γsp(`1))p) ≥ 12,

(γ) gen(γs−1p (`1))→ γs−1

p (L1).

For K = ` or L, we define

K =

{γs−1p (K1) if γs−1

p (K1) is even,

γ2(γs−1p (K1)) otherwise.

Note that the second condition in the statement of the lemma follows fromconditions (α) and (β).

Let N be any even Z-lattice with rank less than or equal to 9 such that

Np→Lp. Since rq,0(γs−1p ((`1)q)) = rq,0((`1)q) ≥ 12, Nq→ ˜

q for any q 6= p.

Furthermore, from the fact that Np→Lp and (α) and (β), one may easily

show that Np→ ˜p. Therefore N→gen(˜), 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

r2,0(L2) < 12.

Then there is a Z-lattice L, not necessarily even, satisfying the followingproperties:

(i) r2,0(L2) < 12.

(ii) L2 satisfies one of the following rank conditions:

r2,0(L2)(type) r2,1(L2)(type) r2,2(L2) r2,0(L2)(type) r2,1(L2)(type) r2,2(L2)

1(odd) ≥ 12 1(odd) 0 ≥ 11

2(odd) ≥ 10 a(even) 0, 1(odd) ≥ 12− aa(even) 2(odd) ≥ 10− a (even) ≥ 12(even)

(even) ≥ 13(odd)

Here, we say the 2iZ2-modular component L2,i in a Jordan de-

composition of L2 is of odd type if s(L2,i) = n(L2,i), and of even typeotherwise.

(iii) Let

L2 = L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,t

be a Jordan decomposition and i be an integer such that

rank(L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,i−1) < 12 ≤ rank(L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,i).

Let N be any Z-lattice with rank less than or equal to 10 such that

N2→ L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,i−1 ⊥M,

10 BYEONG-KWEON OH

for any 2iZ2-modular lattice M with same type to L2,i, and

rank(M) = t− rank(L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,i−1),

where t = 13 if L2,i is of even type and rank(L2,0 ⊥ L2,1 ⊥ · · · ⊥L2,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 assumethat there are Z-lattices `1 and L1 such that

gen(`1)→ L1 and rp,0((`1)p) ≥[n

2

]≥ 13 for every odd prime p,

where rank(L1) = rank(L), rank(`1) = n or n− 1, and (`1)2 ≺ (L1)2. Now,by a similar reasoning to Lemma 2.4, there is an integer s such that

(α′) r2,0((γs−12 (L1))2) < 12 and r2,0((γsp(L1))2) ≥ r2,0((γsp(`1))2) ≥ 12.

(β′) (γs−12 (`1))2 and (γs−1

2 (L1))2 satisfy the condition (iv) of Lemma 2.4.

(γ′) gen(γs−12 (`1))→ γs−1

2 (L1).

We define L = γs−12 (L1) and ˜ = γs−1

2 (`1). Then, by using conditions(α′) and (β′), we may easily show that the rank condition (ii) in this lemmais satisfied. Let N be any Z-lattice satisfying the condition (iii). Since

rp,0(˜p) ≥ 13 for every odd prime p, Np→ ˜p. For p = 2, choose any 2iZ2-

modular sublattice M of the 2iZ2-modular component of ˜2 satisfying all

conditions given in the lemma such that L2,0 ⊥ L2,1 ⊥ · · · ⊥ L2,i−1 ⊥ M

is represented by ˜2. Note that this is always possible by Lemma 2.4 (iv).

Hence N2 is represented by ˜2 and N is represented by gen(˜). 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 or3.3. The following lemma is very useful.

Lemma 4.1. Let L be a Z-lattice. If ` is a sublattice of L and if x ∈L−Q` ∩ L, then

d(`+ Zx) ≤ d` ·Q(x).

Furthermore if the equality holds, then B(`, x) = 0.

Proof. See [1], p. 330. �

Assuming that Theorem 3.1 is false one has at least one prime p for whichk = rp,0(Lp) < 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 latticesof 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 Lp for anodd 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 Lp of dimensionlager 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 =

(N1, N2, . . . , Nt) of lattices represented by L, i.e., having N1, N2, . . . , Ni−1

allows to construct another lattice Ni represented by L and not yet amongN1, N2, . . . , Ni−1. Finally one finds (starting from Nt and using Lemma 4.1)

a sublattice N 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 thatevery (even) root lattice is an orthogonal direct sum of

An(n ≥ 1), Dn(n ≥ 4) and En(6 ≤ n ≤ 8).

For the definitions of theses lattices, see [5]. Note that

D49 An, An→Dn+1 and A8, D8→ E8,

for any positive integer n. We also define an indefinite root lattice E9. First,consider the following graph:

t t t t t t ttt

1 2 3

8

9

4 5 6 7

Graph of E9.

The Z-lattice E9 = Zx1 + Zx2 + · · · + Zx9 is defined by Q(xi) = 2 forevery i = 1, 2, . . . , 9, and

B(xi, xj) =

{−1 if i is directly connected to j in the graph E9,

0 otherwise.

Note that d(E9) = −6. We also denote by E9 the corresponding matrix(B(xi, xj)), if no confusion arises.

For any even Z-lattice L, RL denotes the sublattice of L generated byvectors of norm 2. In general, RL is not necessarily a primitive sublattice ofL. However, every root lattice that is represented by L is also representedby RL.En(k, s) = (eij) denotes the n× n elementary matrix such that

eij =

{1 if i = k and j = s,

0 otherwise.

Let a, b, k be any integers such that a + (k − 1)b > 0 and a > max(0, b).We denote by Mk(a, b) = Zx1 + Zx2 + · · · + Zxk the Z-lattice of rank ksatisfying

Q(xi) = a, B(xi, xj) = b for any i, j such that 1 ≤ i 6= j ≤ n.

12 BYEONG-KWEON OH

Note that det(Mk(a, b)) = (a + (k − 1)b)(a − b)k−1. For any odd prime psuch that p - kb and ordp(a− b) = 1, one may easily show that

Mk(a, b)p ' 〈a, p, p, . . . , p, pε〉,for some ε ∈ Z×p .

Finally, for any Z- (or Zp-) lattice L, we define

Lk =

k-times︷ ︸︸ ︷L ⊥ · · · ⊥ L .

Lemma 4.2. For an odd prime p, let L be a Z-lattice satisfying all condi-tions in Lemma 3.2. Then we have

1 ≤ rp,0(Lp) ≤ 4.

Proof. First assume that p 6= 3. Let rp,0(Lp) = 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 A2 ⊥ A2, D4 D4 ⊥ A2, D6, E6

6 A5, D5, A22 ⊥ A1, A3 ⊥ A2 D6 ⊥ A1, E6 ⊥ A1, E7, A5 ⊥ D5

7 E6, D6, D4 ⊥ A2 E8, E7 ⊥ A2, E6 ⊥ D6

8 E7, D7, A22 ⊥ A3

1 E8 ⊥ A1, E7 ⊥ D7

9 E8, A5 ⊥ A3 E8 ⊥ A3

k ≥ 10 E8, D9 E8 ⊥ D9

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 A5 and D5,

it should represent either D6 or E6 or A5 ⊥ D5. Assume that L represents

D6. Since A2 ⊥ A2 ⊥ A1 is represented by L but is not represented by D6,

L represents either D6 ⊥ A1 or E7. Finally assume that L represents E6.

In this case, since A3 ⊥ A2 is not represented by E6, L should representE6 ⊥ A1 or E7.

For each k, by a similar reasoning to the above, L should represent atleast one root lattice in the right hand side. However, every Z-lattice in theright hand side, whose rank is greater than k, are unimodular over Zq for any

prime q 6= 2, 3. This is a contradiction to the assumption that rp,0(Lp) = k.

Now assume that p = 3 and r3,0(L3) = 5. Since A4, D4→ L, we may

assume that D5→ L. Note that if A4 ⊥ D4→ L, then r3,0(L3) ≥ 8. Hence,

A4 ⊥ A1 is also represented by L. Therefore the only possibility is that

E6 is represented by L. This implies that r3,1(L3) ≥ 12 by Lemma 3.2 (ii).

Therefore A2 ⊥ A2 ⊥ A2 ⊥ A2→ L, which is a contradiction.

Assume that r3,0(L3) = 6. Let L3,0 be the unimodular component in a

Jordan decomposition of L3. Suppose that d(L3,0) is a nonsquare unit in

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 13

Z×3 . Then, D5 ⊥ A1 and A4 ⊥ A1 ⊥ A1 are represented by L. This implies

that E6 ⊥ A1 is also represented by L. By a similar reasoning to the above

argument, A52→L. This implies that r3,0(L3) ≥ 7, which is a contradiction.

For each remaining case, the sequence E in Table 3.2 gives a contradiction.

rank(L3,0)

(det(L3,0)

3

)E

6 1 D6, A6

7 −1 E7, A4 ⊥ A3

7 1 D7, A4 ⊥ A31

8 −1 E7, D7

8 1 E8, A4 ⊥ A3 ⊥ A1

9 −1 E8 ⊥ A1, D5 ⊥ A4

9 1 E8, D9

rank(L3,0) ≥ 10 ±1 E8, D9

Table 3.2.

Every computation is quite similar to the above case. �

Proof of Theorem 3.1. Now we show that the inequality rp,0(Lp) ≤ 4 is also

impossible for every odd prime p, and r2,0(L2) ≥ 12. Note that L is alwayseven when we consider an odd prime case.

Case (1) p ≥ 29. First assume that rp,0(Lp) = 1 and 〈1〉→Lp . Note that

for every positive even integer a such that(ap

)= 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 Zp-latticeZpx+Zpy is unimodular, which is a contradiction. For the remaining cases,every computation is quite similar to this case. For each case, we may take

E =

(〈2〉, 〈4〉) if 2 ∈ (Z×p )2,

(〈4〉, 〈12〉) if 3 ∈ (Z×p )2 and 2 6∈ (Z×p )2,

(〈4〉, 〈6〉) otherwise.

Assume that rp,0(Lp) = 1 and 〈∆p〉→Lp. If 2 6∈ (Z×p )2, then we take

E = (〈2〉, 〈12〉) when 3 6∈ (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 6∈ (Z×p )2, then 6, 12→ L. Choose vectors x, y ∈ L such thatQ(x) = 6, Q(y) = 12. Then the discriminant of the binary Z-lattice Zx+Zyis 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 tothe above, we may conclude that 8a2 ≥ 4p. Combining this and the aboveinequality, we have p ≤ 238. Furthermore since p ≡ 1 (mod 24), p is one ofthe following primes 73, 97, 193. One may easily show by direct computa-

tions that L contains a binary sublattice whose discriminant is not divisibleby p, by taking two suitable nonresidues for each remaining prime p. Thisis a contradiction.

Now assume that p ≡ 7 (mod 8). By a similar reasoning to the above, wemay assume that 3 ∈ (Z×p )2, that is, p ≡ 23 (mod 24). Let a be the smallestquadratic nonresidue positive integer modulo p. Note that

a < p25 + 12p

15 + 33

by [12]. Since 2a and 6a are represented by L, 3a2 ≥ p by a similar reasoningto the above argument. Therefore we have p ≤ 145. For each prime satis-fying this inequality, one may show by direct computations that there is anonresidue a such that 3a2 < p, except the following primes 47, 71, 311, 479.For each exceptional prime p, one may take

E =

{〈6〉, 〈12〉} if p = 41,

{〈14〉, 〈22〉} if p = 71,

{〈22〉, 〈34〉} if p = 311,

{〈26〉, 〈34〉} if p = 479.

Assume that rp,0(Lp) = 2. Since 2, 4 are represented by L, at least one ofthe following binary Z-lattices(

2 00 2

),

(2 11 2

),

(2 00 4

),

(2 11 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

may easily show that the binary lattice

(4 00 6

)is also represented by L.

This implies that the third successive minimum µ3(L) is less than or equal

to 6. Hence, L contains a ternary sublattice whose discriminant is less thanor equal to 8 · 6 = 48 by Lemma 4.1. This is a contradiction.

If rp,0(Lp) = 3, then both A1 ⊥ A1 and A2 are represented by L. Hence

A2 ⊥ A1→L or A3→L. For the former case, at least one of the followingternary lattices

A1 ⊥ A1 ⊥ A1, A3, A2 ⊥ 〈4〉is represented by L, and for the latter case, at least one of the followingternary lattices

A1 ⊥ A1 ⊥ A1, A1 ⊥ A2, A2 ⊥ 〈4〉

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 15

is represented by L. Therefore µ4(L) ≤ 4 for both cases, which implies that

L contains a quaternary sublattice whose discriminant is less than or equalto 24. This is a contradiction.

Finally assume that rp,0(Lp) = 4. Since A3, A2 ⊥ A1 and A1 ⊥ A1 ⊥ A1

are represented by L, A3 ⊥ A1 is represented by L. Furthermore, sincethere does not exist a root lattice of rank 5 whose discriminant is divisibleby p ≥ 29, the root sublattice R

Lof L is A3 ⊥ A1. Hence A4, A2 ⊥ A1 ⊥ A1

and A2 ⊥ A2 are not represented by L. Therefore we may assume that(2

p

)6=(

3

p

)=

(5

p

)= 1.

From this follows

〈2, 4〉 ⊥(

4 11 4

)→ L,

which implies that µ5(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 thesequence E.

rank(L23,0)

(det(L23,0)

23

)E

1 −1 〈10〉, 〈14〉, S623 = γ23(A4A1230[21 1

10]),M6(20,−3)

1 1 A1, 〈4〉

2 −1

(2 11 4

),

(4 11 4

)2 1 A2, A2

1

3 −1 A2, A21

3 1 A3, A1 ⊥ A2

4 −1 A4, A31

4 1 A1 ⊥ A3, A22

Table 3.3. p = 23.

In those tables, Snp is a Z-lattice of rank n such that det(Snp ) = pn−1 andrp,0((Snp )p) = 1. For the definition of these lattices, see [5]. In every casecontaining Snp , the pair of the first two integers, which are represented by

L, shows that rp,1(Lp) 6= 0. Hence by Lemma 3.2, Snp is represented by L.Furthermore, since the Z-lattice Mk(a, b) on the same line is also represented

by L but not by Snp , there is a vector x ∈ L − φ(Snp ) such that Q(x) = a.

Here φ is any representation from Snp to L. Therefore L has a sublattice Kof rank n+ 1 such that

φ(Snp ) ⊂ K and d(K) ≤ pn−1a.

Since n is even in every case, d(K) is divisible by 2pn. This gives a contra-diction.

16 BYEONG-KWEON OH

rank(L19,0)

(det(L19,0)

19

)E

1 −1 A1, 〈10〉, 〈12〉

1 1 〈4〉, 〈6〉

2 −1 A2, A1 ⊥ 〈4〉

2 1 A21,

(2 1

1 4

)3 −1 A3

1, A2 ⊥ 〈4〉

3 1 A3, A1 ⊥ A2

4 −1 A1 ⊥ A3, A21 ⊥ A2

4 1 A4, A22

Table 3.4. p = 19.

rank(L17,0)

(det(L17,0)

17

)E

1 −1 〈6〉, 〈10〉, S417 = γ17(A2A1102[11 1

6]),M4(14,−3)

1 1 A1, 〈4〉

2 −1 A2,

(2 1

1 4

)2 1 A2

1, A1 ⊥ 〈4〉

3 −1 A1 ⊥ A2, A1 ⊥(

2 1

1 4

), A2 ⊥ 〈4〉

3 1 A3, A31

4 −1 A4, A31

4 1 A1 ⊥ A3, A22

Table 3.5. p = 17.

rank(L13,0)

(det(L13,0)

13

)E

1 −1 A1, 〈6〉

1 1 〈4〉, 〈10〉, S413 = γ13(A352[1 1

4]),M5(12,−1)

2 −1

(2 11 4

), A1 ⊥ 〈4〉,

(2 11 8

),

2 1 1

1 10 4

1 4 10

2 1 A2, A2

1

3 −1 A1 ⊥ A2, A31

3 1 A3,

(2 11 4

), 〈4〉 ⊥

(4 22 4

)4 −1 A4, A1 ⊥ A3

4 1 D4, A1 ⊥ A2

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(L5,0) = 2 and(

det(L5,0)5

)= −1, as a

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 17

rank(L11,0)

(det(L11,0)

11

)E

1 −1 A1, 〈6〉,(

8 3

3 8

)1 1 〈4〉, 〈12〉, S6

11 = γ11(D544[1 14

]),M6(12, 1)

2 −1

(2 1

1 4

), A1 ⊥ 〈4〉

2 1 A2, A21

3 −1 A1 ⊥ A2, 〈4〉 ⊥(

2 11 4

)3 1 A3, A1 ⊥

(2 1

1 4

)4 −1 A1 ⊥ A3, A2

1 ⊥(

2 1

1 4

),

(2 1

1 4

)⊥(

4 1

1 4

)4 1 A4, A2 ⊥ A2

Table 3.7. p = 11.

rank(L7,0)

(det(L7,0)

7

)E

1 −1 〈6〉, 〈10〉, S67 = γ7(A6),M7(10, 3)

1 1 A1, 〈4〉,

4 2 22 8 1

2 1 8

2 −1 A2,

(4 2

2 4

)2 1 A2

1, A1 ⊥ 〈4〉,(

2 11 8

),

(6 11 6

)⊥ 〈12〉

3 −1 A1 ⊥ A2, A1 ⊥(

4 22 4

)3 1 A3, A3

1

4 −1 A4, A2 ⊥ 〈6〉, A2 ⊥(

4 11 4

)4 1 A1 ⊥ A3, A2

2

Table 3.8. p = 7.

representative one. Since A2 and A1 ⊥ 〈4〉 are represented by L in this case,

K(3) =

2 1 01 2 10 1 4

→ L.

Note that (4 22 4

)→ L and

(4 22 4

)9 K(3).

This implies that µ4(L) ≤ 4, and hence

K(4) =

2 1 0 11 2 1 10 1 4 21 1 2 4

→ L.

18 BYEONG-KWEON OH

rank(L5,0)

(det(L5,0)

5

)E

1 −1 A1, 〈12〉, S85 := γ5(E710[1 1

2]),M8(8, 3)

1 1 〈4〉, 〈6〉, S45 := γ5(A4),M6(6, 1)

2 −1 A2, A1 ⊥ 〈4〉,(

4 2

2 4

),

(2 1

1 8

)⊥(

6 1

1 6

)2 1 A2

1,

(2 11 6

),

(4 11 4

)3 −1 A2, A2

1

3 1 A3, A1 ⊥ A2

4 −1 A1 ⊥ A3, A21 ⊥ A2, A2 ⊥ 〈6, 6〉

4 1 D4, A22

Table 3.9. p = 5.

Note that d(K(4)) = 52. Since(2 11 8

)⊥(

6 11 6

)→ L,

(2 11 8

)⊥(

6 11 6

)9 K(4),

L contains a quinary sublattice whose determinant is less than or equal to52 ·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 · 53.

Case (3) p = 3. Note that

r3,1(L3) ≥ 12 or r3,1(L3) = 0 and r3,0(L3) + r3,2(L3) ≥ 12.

Hence if r3,1(L3) 6= 0, then we may assume that r3,1(L3) ≥ 12. For eachcase, the sequence E that gives a contradiction is given by Table 3.10. TheZ-lattices T9 and T8(314) appearing in Table 3.10 are defined as follows: TheZ-lattice

T8(314) = Zx1 + Zx2 + · · ·+ Zx8

is defined by{Q(x1) = 4, Q(x2) = 16, B(x1, x2) = 1, Q(xi) = 10 and

B(xi, xj) = 1, B(x2, xi) = −2B(x1, xi) = −4 for 3 ≤ i 6= j ≤ 8.

Note that d(T8(314)) = 314 and T8(314) is a sublattice of E8 such that

(T8(314))3 ' 〈1, 32, 32, 32, 32, 32, 32, 32〉.T9 = Zx1 +Zx2 + · · ·+Zx9 is the Z-lattice whose corresponding matrix is

2E9(4, 4)+3E9. Note that d(T9) = 2 ·5 ·39. Since (T9)3 has one dimensional

unimodular component, T9→ 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 A2, T9→ L and the unimodular component in a Jordan decomposition

of L3 is isometric to 〈∆3〉. The assumptions on L imply that it has theseproperties. Then the root sublattice R

Lis isometric to A2. Let

L = Zx1 + Zx2 + Zx3 + · · ·+ ZxN

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 19

rank(L3,0)

(det(L3,0)

3

)r3,1(L3) = 0? E

1 −1 No A2, T9

1 −1 Yes A1,

(8 11 8

)1 1 No γ3(E6),M7(4, 1)

1 1 Yes T8(314),M8(10, 1)

2 −1 No A2 ⊥ γ3(E6),M8(4, 1)

2 −1 Yes

2 0 1

0 4 2

1 2 6

,

(4 11 6

)2 1 No A2

2,M4(4, 1)

2 1 Yes A21,

(2 1

1 4

)3 −1 No A3

2,M7(4, 1)

3 −1 Yes A31, A1 ⊥

(2 1

1 4

)3 1 No A2

2, A3

3 1 Yes A3,

(2 11 4

)⊥ 〈4〉

4 −1 No A5, A32

4 −1 Yes A4, A31

4 1 No A42, D4

4 1 Yes D4, A21 ⊥

(2 1

1 4

)Table 3.10. p = 3.

and K = Zx1 + Zx2 ' A2. One may easily show that for every z ∈ L suchthat Q(z) = 6, z ∈ K or B(K, z) = 0. Assume that

M = Zy1 + Zy2 + · · ·+ Zy9

is a sublattice of L such that (B(yi, yj)) = 2E9(4, 4) + 3E9. Since vertices(1, 2, 3), (3, 8, 9) and (5, 6, 7) are connected in the graph E9,

yi ∈ K⊥ = {w ∈ L : B(w,K) = 0}

for every i 6= 4. Note that Q(y4) = 8 and y4 6∈ K. The corresponding matrixof K + Zy4 is of the form2 1 a

1 2 ba b 8

for some a, b such that −3 ≤ a, b ≤ 3,

and d(K + Zy4) is divisible by 9. Hence K + Zy4 ' A2 ⊥ 〈6〉. Thereforethere are vectors u1 ∈ K and u2 ∈ K⊥ such that Q(u1) = 2, Q(u2) = 6 andy4 = u1 + u2. Therefore the sublattice

Zy1 + Zy2 + Zy3 + Zu2 + Zy5 + · · ·+ Zy9

of L is isometric to√

3E9, which is indefinite. This is a contradiction.

20 BYEONG-KWEON OH

In the remaining case for p = 2, we will use notations Ma(b) and Na(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 Na(b). In this case, we only check that thereexists a genus containing Na(b).

Case (4) p = 2. For any even unimodular Z2-lattice ` of rank n,

n ≡ 0 (mod 2) and d` ≡ n+ 1 (mod 4).

If d` ≡ 1, 3 (mod 8) we say ` is of type 1, and of type 2 otherwise.

Let L be a Z-lattice satisfying all conditions in Lemma 3.3 and let

L2 ' L2,0 ⊥ L2,1 ⊥ L2,2 ⊥ · · · ⊥ L2,t

be a Jordan decomposition of L2 . Note that r2,0(L2) ≤ 2, or L2,0 is an evenZ2-lattice with rank less than or equal to 10 by Lemma 3.3.

Subcase (4-1) 8 ≤ r2,0(L2) ≤ 10. First, assume that r2,0(L2) = 10. Since

E8 and A6 ⊥ A2 are represented by L, E8 ⊥ A2→L. Therefore A10 is

represented by L, which is a contradiction. Assume that r2,0(L2) = 8 and

L2,0 is of type 1. Then E8→ L, and hence there is a Z-lattice M such that

L ' E8 ⊥M and s(M) ⊂ 2Z. Since

A22 ⊥

(2 11 4

)⊥(

2 11 4

)→ L and A2

2 ⊥(

2 11 4

)⊥(

2 11 4

)9 E8,

M should represent 2. Therefore A2 ⊥ A6→L, which is a contradiction.

Now assume that r2,0(L2) = 8 and the unimodular component is always of

type 2 for any Jordan decomposition of L2. Then either L2,1 = 0 or L2,1 isof even type. But this is impossible for E7 and A2 ⊥ A6 are represented by

L.

Subcase (4-2) 2 ≤ r2,0(L2) ≤ 6 and L2,0 is even. Note that, by Lemma

3.3 (ii), every possible structure of L2 up to the 4Z2-modular component

can be given by Table 3.11. In this table, for example, L2,2(o) is the 4Z2-

modular component of odd type and L2,2(e) is the 4Z2-modular component

of even type. In particular, L2,0(e, i) is the even unimodular component oftype i for i = 1, 2. Note that the 4Z2-modular component might be nonzeroin Number 10, 11 and 14 cases in Table 3.11. For those cases, the rank ofthe 2Z2-modular component is greater than or equal to 12 by Lemma 3.3(ii).

Subcase (4-2-1) r2,0(L2) = 6. In this case, we have Table 3.12. In thistable,

M2(7) =

(2 11 4

)and M4(16 · 5) =

2 1 1 11 4 0 01 0 4 01 0 0 4

.

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 21

Name L2,0 ⊥ L2,1 ⊥ L2,2 Name L2,0 ⊥ L2,1 ⊥ L2,2

1 L2,0(e, 1) ⊥ L2,2(o) 2 L2,0(e, 1) ⊥ L2,2(e)

3 L2,0(e, 1) ⊥ 〈2〉 ⊥ L2,2(o) 4 L2,0(e, 1) ⊥ 〈2〉 ⊥ L2,2(e)

5 L2,0(e, 1) ⊥ 〈6〉 ⊥ L2,2(e) 6 L2,0(e, 1) ⊥ 〈2, 2〉 ⊥ L2,2(o)

7 L2,0(e, 1) ⊥ 〈2, 2〉 ⊥ L2,2(e) 8 L2,0(e, 1) ⊥ 〈2, 6〉 ⊥ L2,2(e)

9 L2,0(e, 1) ⊥ 〈6, 6〉 ⊥ L2,2(e) 10 L2,0(e, 1) ⊥ L2,1(o)

11 L2,0(e, 1) ⊥ L2,1(e) 12 L2,0(e, 2) ⊥ L2,2(o)

13 L2,0(e, 2) ⊥ L2,2(e) 14 L2,0(e, 2) ⊥ L2,1(e)

Table 3.11. Possible structures of L2,0 ⊥ L2,1 ⊥ L2,2.

Name E Name E

1 E6, A21 ⊥ A4 2 E6(848)2[1 1

313

], A2 ⊥M2(7)2

3 E7, A1 ⊥ E6 4 E6, A6

5 E7, A22 ⊥M2(7) 6 E7 ⊥ A1, A2

1 ⊥ E6

7 E6, A6 8 E7 ⊥ A1, A1 ⊥ A22 ⊥M2(7)

9 E7, D7 10 E7, A22 ⊥ D4

11 E6, A22 ⊥ D4 12 D7, A7

13 A78[4 12

], A22 ⊥M4(16 · 5) 14 A7, D4 ⊥ D4

Table 3.12. r2,0(L2) = 6.

Since other cases can be done in a similar manner, we only provide aproof of Number 2 case. Note that d(E6(848)2[11

313 ]) = 16 (for the precise

definition of this lattice, see [5]) and(E6(848)2

[1

1

3

1

3

])2

'(

2 11 2

)⊥(

2 11 2

)⊥(

2 11 2

)⊥(

8 44 8

).

Therefore it is represented by L for this case. Let φ be a representation from

E6(848)2[113

13 ] to L. Since

A2 ⊥M2(7)2→L and A2 ⊥M2(7)29 E6(848)2

[1

1

3

1

3

],

there is a vector x ∈ L−φ(E6(848)2[113

13 ]) such that Q(x) = 4. This implies

that L has a sublattice of rank 9 whose discriminant is less than or equal to16 ·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) r2,0(L2) = 4. In this case we have Table 3.13.

22 BYEONG-KWEON OH

Name E Name E

1 A23,M4(16 · 5)2 2 N8(28),M4(16 · 13) ⊥ M4(16 · 5)

3 A5, A2 ⊥ A3 4 A1 ⊥ A4, A1 ⊥ A22

5 N7(25),M4(16 · 5)2 6 D6, A1 ⊥ A5

7 D5, A2 ⊥ A3 8 N7(25) ⊥ A1, A1 ⊥M4(16 · 5)2

9 D6,M2(7) ⊥M2(15) ⊥ 〈4, 4〉 10 D6, A3 ⊥ D4

11 D24 , A

91 12 D5, A2 ⊥M4(16 · 5)

13 N ′8(28),M4(16 · 5) ⊥M4(16 · 13) 14 D5, A2

1 ⊥ D4

Table 3.13. r2,0(L2) = 4.

In this table M2(15) =

(4 11 4

),

M4(16 · 5) =

2 1 1 01 2 0 11 0 6 10 1 1 6

and M4(16 · 13) =

2 1 0 11 6 1 30 1 6 31 3 3 6

.

N8(28) and N ′8(28) are Z-lattices of discriminant 28 such that

(N8(28))2 '(

2 11 2

)⊥(

2 11 2

)⊥(

8 44 8

)⊥(

8 44 8

),

(N ′8(28))2 '(

2 11 2

)⊥(

0 11 0

)⊥(

8 44 8

)⊥(

0 44 0

).

Note that each Z-lattice given above exists as a sublattice of E8. The Z-lattice N7(25) is a sublattice of E7 with discriminant 25 such that

(N7(25))2 '(

2 11 2

)⊥(

2 11 2

)⊥ 〈6〉 ⊥

(8 44 8

).

We only provide a proof of Number 1 case. In this case, note that

L2 '(

2 11 2

)⊥(

2 11 2

)⊥ L2,2(o) ⊥M,

where r2,2(L2) = rank(L2,2(o)) ≥ 8 and s(M) ⊂ 8Z2. Therefore RL

= A3 ⊥A3. Note that R

Lis a primitive sublattice of L. Let

L = Zx1 + Zx2 + · · ·+ Zx6 + · · ·+ ZxNand K = Zx1 + Zx2 + · · ·+ Zx6 ' A3 ⊥ A3. Assume that there is a vectorx ∈ L such that Q(x) = 4. If x 6∈ K then the rank of K ′ = Zx1 +Zx2 + · · ·+Zx6 + Zx is 7, and the discriminant dK ′ is less than or equal to 43. Since

the discriminant of any sublattice of L with rank 7 is divisible by 43, wehave dK ′ = 43. This also implies that B(xi, 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 Zw1 + Zw2 ⊂ L such that

(B(wi, wj)) =

(2 11 4

).

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 23

Since RL

= K, we have w1 ∈ K. Therefore w2 ∈ K by the above observa-tion. Now since

M4(16 · 5) ⊥M4(16 · 5) =

2 1 1 11 4 0 01 0 4 01 0 0 4

⊥

2 1 1 11 4 0 01 0 4 01 0 0 4

→ L,

M4(16 · 5) ⊥M4(16 · 5)→ K. This is a contradiction.

Subcase (4-2-3) r2,0(L2) = 2 and L2,0 is even. In this case, we haveTable 3.14.

Name E Name E

1 A3,M5(29) 2 N8(212),M9(6, 2)

3 A1 ⊥ A3,M4(16 · 5) 4 A1 ⊥ A2,M4(16 · 5)

5 N7(29),M3(25 · 5) ⊥M4(16 · 13) 6 A21 ⊥ A3, A2

1 ⊥M3(12)

7 A3, A21 ⊥ A2 8 N8(210), A1 ⊥M4(16 · 13) ⊥M3(25 · 5)

9 N6(26),M4(16 · 13) ⊥ 〈6, 6〉 10 D4, A516[11111 1

2]

11 D4, A51 12 N ′

8(212),M9(6, 2)

13 N8(26), A91 14 M5(26),M4(16 · 7)

Table 3.14. r2,0(L2) = 2 and L2,0 is even.

In Number 1 case,

M5(29) =

4 0 0 0 20 4 0 0 20 0 4 0 20 0 0 4 22 2 2 2 6

,

which is represented by L. The proof of this case is as follows: Let L =Zx1 + Zx2 + Zx3 + Zx4 + · · ·+ ZxN and K = Zx1 + Zx2 + Zx3 ' A3. Wecan 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 Zw1 +Zw2 ⊂ L be a binaryZ-lattice such that

(B(wi, wj)) =

(4 22 6

).

Assume that w1 6∈ K, i.e., w1 ∈ K⊥. Then w2 6∈ K. Hence K ′ = K + Zw2

is a quaternary sublattice of L. Since dK ′ is divisible by 16, K ′ ' A3 ⊥ 〈4〉.This implies that there are vectors z1 ∈ K and z2 ∈ K⊥ with Q(z1) = 2 andQ(z2) = 4 such that w2 = z1 + z2. Hence

K + Zw1 + Zz2 ' A3 ⊥(

4 22 4

)→ L,

which is a contradiction. Consequently w1 ∈ K. From this and the fact

that M5(29)→ L, A3 represents a quaternary Z-lattice 〈4, 4, 4, 4〉. This is acontradiction.

24 BYEONG-KWEON OH

In Number 2 and 12 cases, N8(212) and N ′8(212) are sublattices of E8 suchthat

(N8(212))2 '(

2 11 2

)⊥(

8 44 8

)⊥(

8 44 8

)⊥(

8 44 8

),

(N ′8(212))2 '(

0 11 0

)⊥(

8 44 8

)⊥(

8 44 8

)⊥(

0 44 0

).

Also note that

(M9(6, 2))2 ' 〈6〉 ⊥(

8 44 8

)⊥(

8 44 8

)⊥(

8 44 8

)⊥(

8 44 8

).

In Number 5 and 6 cases, N7(29) is a sublattice of E7 such that

(N7(29))2 '(

2 11 2

)⊥ 〈6〉 ⊥

(8 44 8

)⊥(

8 44 8

)and

M3(25 · 5) =

6 2 22 6 22 2 6

, M3(12) =

2 0 10 2 11 1 4

.

In Number 8 case, N8(210) is a sublattice of E8 such that

(N8(210))2 '(

2 11 2

)⊥ 〈2, 6〉 ⊥

(8 44 8

)⊥(

8 44 8

).

In Number 9 case, N6(26) is a sublattice of I6 such that

(N6(26))2 '(

2 11 2

)⊥ 〈6, 6〉 ⊥

(8 44 8

).

In Number 13 case, N8(26) is a sublattice of E8 such that

(N8(26))2 '(

0 11 0

)⊥(

0 22 0

)⊥(

0 22 0

)⊥(

0 22 0

).

In Number 14 case,

M5(26) =

2 0 0 0 10 2 0 0 10 0 4 0 20 0 0 4 21 1 2 2 4

, M4(16 · 7) =

4 2 1 22 4 0 01 0 4 02 0 0 4

.

Note that

(M5(26))2 '(

0 11 0

)⊥ 〈28, 28, 28〉 and (M4(16 · 7))2 '

(0 11 0

)⊥ 〈20, 20〉.

Hence they are represented by L in this case. One may easily show that

every vector x ∈ L of norm 4 is contained in M5(26) ⊥M5(26)⊥. This givesa contradiction.

Subcase (4-3) r2,0(L2) = 2 and L2,0 is odd. In this case, r2,1(L2) =

rank(L2,1) ≥ 10. We may easily show that for each case, the sequence E can

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 25

be given as follows:

E =

(〈1, 1〉, A1A1A16[1111

2 ]) if L2,0 ⊥ L2,1 ' 〈1, 1〉 ⊥ L2,1(o),

(〈1, 1〉, 〈2, 2, 2〉) if L2,0 ⊥ L2,1 ' 〈1, 1〉 ⊥ L2,1(e),(〈1〉 ⊥ N7(26), 〈1〉 ⊥M8(3, 1)

)if L2,0 ⊥ L2,1 ' 〈1, 3〉 ⊥ L2,1(e),

(A61[111111], A7

1) if L2,0 ⊥ L2,1 ' 〈3, 3〉 ⊥ L2,1(e).

Here N7(26) is a sublattice of I7 such that

(N7(26))2 ' 〈3〉 ⊥(

4 22 4

)⊥(

4 22 4

)⊥(

4 22 4

).

Subcase (4-4) r2,0(L2) = 1. In this case, we have the following Table3.15 on the possible local structures up to 4Z2-modular component and thesequence E.

L2,0 ⊥ L2,1 ⊥ L2,2 E L2,0 ⊥ L2,1 ⊥ L2,2 E

〈1〉 ⊥ L2,1(o) ⊥ L2,2 〈1〉,M5(3 · 24) 〈1〉 ⊥ L2,1(e) ⊥ L2,2 〈1〉,S9(2)

〈3〉 ⊥ L2,1(e) ⊥ L2,2 N7(26),M8(3, 1) 〈1〉 ⊥ L2,2(o) 〈1〉,S′9(2)

〈3〉 ⊥ L2,2(o) N7(212), 〈3〉 〈1〉 ⊥ L2,2(e) 〈1〉,S′9(2)

〈3〉 ⊥ L2,2(e) M3(24), 〈4〉4 〈5〉 ⊥ L2,2(e) N5(28), 〈4〉6

〈7〉 ⊥ L2,2(e) N7(212), 〈4〉8

Table 3.15. r2,0(L2) = 1.

In the first case,

M5(3 · 24) =

3 2 2 2 22 4 2 0 02 2 4 0 02 0 0 4 22 0 0 2 4

.

If 3 is replaced by 2 in the above matrix, then it is indefinite. S9(2) is the Z-lattice whose corresponding matrix is E9(3, 3) + 2E9. Note that d(S9(2)) =28 · 3 · 11 and

(S9(2))2 ' 〈1〉 ⊥(

4 22 4

)⊥(

4 22 4

)⊥(

4 22 4

)⊥(

4 22 4

).

S ′9(2) is the Z-lattice whose corresponding matrix is E9(3, 3) + 4E9 Notethat d(S9(2)) = 216 · 3 · 7 and

(S ′9(2))2 ' 〈1〉 ⊥(

8 44 8

)⊥(

8 44 8

)⊥(

8 44 8

)⊥(

0 44 0

).

N7(212) is a sublattice of I7 such that

(N7(212))2 ' 〈7〉 ⊥(

0 44 0

)⊥(

0 44 0

)⊥(

0 44 0

).

26 BYEONG-KWEON OH

N5(28) is a sublattice of I5 such that

(N5(28))2 ' 〈5〉 ⊥(

8 44 8

)⊥(

0 44 0

).

Finally

M3(24) =

3 1 11 3 −11 −1 3

,

which is a sublattice of I3.

5. n-regular Z-lattices

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 everyn-regular Z-lattice L and every prime p,

rp,0(Lp) ≥ n+ 4.

Proof. Let L be an n-regular Z-lattice. In Section 4 we proved that rp,0(Lp)is greater than or equal to 12, for every prime p. This implies that L is even9-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 D8k[1] isan indecomposable even unimodular Z-lattice of rank 8k, for every positiveinteger k. Since E8 = D8[1] is represented by L, there is a Z-lattice L′ suchthat L ' E8 ⊥ L′. Furthermore since D9 is also represented by L, we mayassume that D9→L′, which follows from the fact that every indecomposablesplitting of root systems is unique. From this, D16[1] is locally representedby L and hence is represented by L globally by regularity of L. Thereforewe have E8 ⊥ D16[1]→ L by a similar reasoning to the above. Now, by aninduction argument, we have

E8 ⊥ D16[1] ⊥ · · · ⊥ D8m[1]→ L.

Therefore rp,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 Zp-lattice of rank n+3 representsall (even, respectively) Zp-lattices of rank n, for every prime p. Hence thetheorem 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 genusof 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. Hencerp,0(L) ≥ n+ 3 by Theorem 5.1, which implies that the Zp-lattice Lp repre-sents all (even, if L2 is even) Zp-lattices of rank n, for every prime p. Thecorollary follows from this. �

POSITIVE DEFINITE n-REGULAR QUADRATIC FORMS 27

Theorem 5.4. For any integer n, define s(n) = 8[n

8

]. If s(n) ≥ 144, then

the minimal rank R(n) of n-regular forms satisfies

R(n) ≥ 2s(n)(1− αs(n)

)× wt(Ms(n)),

where wt(Ms(n)) =∑

cls(Mi)∈gen(Ms(n))1

O(Mi)is the mass of an even unimod-

ular lattice of rank s(n) and αs(n) = 2s(n)+1(√

2π)s(n)

Γ(s(n)2

).

Proof. Let L be an n-regular Z-lattice. Let {J1, J2, . . . , Ju} be the set ofall indecomposable even unimodular Z-lattices of rank s(n) up to isometry.From the fact that every indecomposable splitting of lattices is unique, wehave

J1 ⊥ J2 ⊥ · · · ⊥ Ju→ L.

Therefore, rank(L) ≥ s(n)u. Clearly u is greater than or equal to the numberof even unimodular Z-lattices of rank s(n) that have a trivial automorphismgroup. Hence by [2], u ≥ 2(1 − αs(n)) × wt(Ms(n)). The theorem followsfrom this. �

Remark 5.5. (1) Note that Theorem 5.4 can be improved by adding blocksof indecomposable unimodular lattices in smaller dimensions, e.g.,

[n8

]-times︷ ︸︸ ︷E8 ⊥ · · · ⊥ E8

has also occur in the splitting of L.(2) Note that αs(n) −→ 0 and for a constant C ∼ 0.705,

wt(Ms(n)) ∼C

(s(n)

2πe√e

)s(n)2/4( 8πe

s(n)

)s(n)/4

s(n)1/24,

as n goes to infinity (for this, see [16], p.50). Hence the minimal rank ofn-regular Z-lattices has an exponential lower bound for n as it increases.(3) Theorems 5.2 and 5.4 give a partial answer of the Problem 9 of Chapter6 in [14], for higher rank cases (see also [19]).(4) The even Z-lattice A2 ⊥ E8 is 10-regular, whereas it is not even 10-universal. Hence the minimum integer n satisfying Theorem 5.2 lies between11 and 27.(5) It was proved in [4] that the number of primitive n-regular forms of rankn + 3 is finite up to isometry. Theorem 5.2 implies that such an n-regularZ-lattice of rank n+ 3 does not exist, for every n ≥ 27.(6) There are exactly 22 indecomposable even unimodular Z-lattices of rank24. Since every 27-regular Z-lattice represents all even unimodular lattices ofrank 24 including E3

8 , D16[1] ⊥ E8, the minimal rank of 27-regular Z-latticeis greater than or equal to 22 · 24 + 24 + 16 = 568, i.e., R(27) ≥ 568.

Acknowledgement. The author deeply appreciates the referee’s commentsand suggestions.

28 BYEONG-KWEON OH

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Department of Applied Mathematics, Sejong University, Seoul, 143-747, Ko-rea

E-mail address: bkoh@sejong.ac.kr