Representations of Integral QuadraticForms by Sums of Squares

Myung-Hwan Kim∗ and Byeong-Kweon Oh†

∗School of Mathematics, Seoul National UniversitySeoul 151-742, Korea. E-mail: mhkim@math.snu.ac.kr†Department of Applied Mathematics, Sejong University

Seoul 143-747, Korea. E-mail: bkoh@sejong.ac.kr

Abstract

Let g[n] be the minimum number of squares whose sum representsall positive definite integral quadratic forms of rank n which are rep-resented by sums of squares. In this article, we first discuss represen-tations of integers by unimodular lattices. We then estimate the orderof magnitude of the diameter of the 2-graph of unimodular lattices ofrank n. Combining these results we prove g[n] = O(3n/2 n log n). Wealso provide a lower bound for g[n]. Finally, we discuss s-integrablelattices as an application of our method.

1 Introduction

In 1770, Lagrange [L] proved the famous four square theorem, i.e., for everypositive integer n, there exists an integer solution of the equation

x2 + y2 + z2 + w2 = n.

∗The first author was partially supported by KOSEF Research Fund (00-0701-01-5-2).AMS Subject Classification (2000): 11E12, 11E06. Keywords: representation, quadraticforms, sums of squares

1

After Lagrange, this theorem was generalized in many directions. One in-teresting generalization is so called the new (or quadratic) Waring’s Problemdue to L. J. Mordell [M1], which is about sums of squares that represent allpositive integral quadratic forms of given rank. In particular, Mordell provedthat every binary positive integral quadratic form can be represented by asum of five squares. Later, C. Ko [K1] proved that every ternary (quaternaryor quinary) positive integral quadratic form can be represented by a sum ofsix (seven or eight, respectively) squares. So, they naturally expected thatevery positive n-ary integral quadratic form can be represented by a sum of(n + 3) squares. This, however, turned out to be false. The quadratic formdefined by the root lattice E6 cannot be represented by sums of squares (see[M2], [CS2] and [Pl]). After Mordell found this, they tried to determine theminimum number g[n] of squares whose sum represents all positive integralquadratic forms of rank n that are represented by sums of squares.

We adopt lattice theoretic language. A Z-lattice L is a finitely generatedfree Z-module in Rn equipped with a non-degenerate symmetric bilinear formB such that B(L,L) ⊆ Z. The corresponding quadratic map is denoted by Q,i.e., Q(e) = B(e, e) for every e ∈ L. The ideal of Z generated by B(ei, ej)’sis called the scale of L, denoted by s(L). For a Z-lattice

L = Ze1 + Ze2 + · · ·+ Zem

with basis {e1, e2, · · · , em}, we define its corresponding matrix

ML := (B(ei, ej)),

which is an m ×m symmetric integer matrix. We often identify a Z-latticeL with its corresponding matrix ML. If ML is diagonal, we often write

L = 〈Q(e1), Q(e2), · · · , Q(en)〉 or 〈Q(e1)〉 ⊥ 〈Q(e2)〉 ⊥ · · · ⊥ 〈Q(en)〉.

For a ∈ R×, we let

aL := Z(ae1) + Z(ae2) + · · ·+ Z(aen).

The lattice Zn equipped with the standard inner product is denoted by In.A Z-lattice L is said to be positive definite or simply positive if Q(e) > 0 forevery e ∈ L, e 6= 0. In this paper, we always assume the following unlessstated otherwise:

2

Every Z-lattice is positive definite.

A Z-lattice L is said to be even if Q(L) ⊆ 2Z, and odd, otherwise. Asusual,

dL := det(B(ei, ej))

is called the discriminant of L. L is said to be unimodular if dL = 1. For aZ-lattice L and a prime p,

Lp := ZpL = Zp ⊗Z L

is a Zp-lattice called the localization of L at p, where Zp is the p-adic integerring.

Let `, L be Z-lattices. We say that L represents ` if there is an injective Z-linear map from ` into L that preserves the bilinear forms, and write ` → L.Such a map is called a representation. A representation is called an isometryif it is surjective. We say that ` and L are isometric and write ` ∼= L if thereexists an isometry between them. For a Z-lattice L, we define the class andgenus of L by

cls(L) := {K : Z-lattice |K ' L over Z },gen(L) := {K : Z-lattice |Kp ' Lp over Zp for all p }.

It is well known that gen(L) contains a finite number of distinct classes.This number is called the class number of L, denoted by h(L). If `p → Lp

for all p, or equivalently ` → gen(L), then it is known that ` → K for someK ∈ gen(L).

Let

Sn := { ` : Z-lattices | ` → Ig for some g, rank(`) = n },

andg[n] := min{ g | ` → Ig for every ` ∈ Sn }.

Applying the results in [HKK], one can easily prove that g[n] exists for all n.The four square theorem of Lagrange, and the results of Mordell’s and Ko’smentioned above can be summarized as the following:

g[n] = n + 3 for 1 ≤ n ≤ 5.

3

In fact, this is an immediate consequence of local representation theory andthe fact that the class number of In is one for 1 ≤ n ≤ 8. The followingquestion arises quite naturally:

g[n] = n + 3 for all n ≥ 1 ?

Concerning this question, Ko conjectured in [K2] that g[6] = 9. However,the authors proved recently [KO1][KO2] that

g[n] ≥ b3n/2c+ 1 > n + 3 for all n ≥ 6, and g[6] = 10.

The first explicit upper bound for g[n] was given by Icaza [Ic]. She ob-tained her bound by computing the so called HKK-constant (see [HKK]).But her bound is too big containing a factor nn+12 4h(In+3), where h(In) de-notes the class number of the genus of In which can be shown to be of theform h(In) = nΘ(n2) for some arithmetic function Θ(n2) bounded betweenconstant multiples of n2 for all n >> 0.

In section 2 of this paper, we prove that every odd (even) unimodularZ-lattice of rank bigger than three represents all (even, respectively) integersgreater than or equal to (n/2)4. In section 3, we compute a lower bound andan upper bound of the diameter of the 2-graph of unimodular lattices of rankn. And then from these results, we provide a sharper upper bound, of sizeO(3n/2 n log n), for g[n] in section 4. Furthermore, we give a lower bound ofg[n] and φ(s) (for the definition of φ(s), see section 4).

For unexplained terminology, notation, and basic facts about Z-lattices,we refer the readers to O’Meara [O’M1], Conway and Sloane [CS1].

2 Representation of integers

by unimodular Z-lattices

Let L be a unimodular Z-lattice of rank n and µ(L) be the smallest positiveinteger that is represented by L. It was proved in [MOS] for even case andin [RS] for odd case that

µ(L) ≤ 2bn/24c+ 2

unless L is odd unimodular of rank 23. In the exceptional case, µ(L) ≤ 3(see [RS]). In contrast with these results, Conway and Thompson proved

4

(unpublished, see [MH]) that there exists a unimodular Z-lattice L of rank nwith µ(L) ≥ k(n), where k(n) denotes the closest integer to ( 5

3ωn)2/n. Here

ωn = πn/2 (Γ((n + 2)/2))−1

is the volume of the n-dimensional unit ball in Rn and therefore k(n) isasymptotically n

2πe. For a Z-lattice K, we define

c(K) := min{m | 〈k〉 → K for all (even, if K is even) integers k ≥ m }.Let

ce(n) := max{ c(K) | K : even unimodular of rank n },co(n) := max{ c(K) | K : odd unimodular of rank n },

andc(n) := max{ ce(n), co(n) }.

If n ≥ 4, then c(n) always exists (see [T]). For the exact values and upperbounds of ce(n) for lower rank even unimodular Z-lattices, see [CLR], [OS],[LR] and [Pe]. In this section, we give an upper bound for c(n) for n ≥ 4.We begin with the following trivial but useful lemma.

Lemma 2.1 Let K be any Z-lattice of any rank n with µ(K) = µ. For apositive integer t ≥ µ, let r(t, K) be the number of vectors x ∈ K such thatQ(x) = t. Then

r(t,K) ≤(2√

t/µ + 1)n

−(2√

t/µ− 1)n

.

Proof. We may assume that K ⊂ Rn. Let Vn(r) = ωnrn be the volume of then-dimensional ball of radius r in Rn. Let X := {x ∈ K | Q(x) = t}. Thenthe balls with radius

õ/2 centered at the points in X do not intersect each

other. Therefore

r(t, K) · Vn

(õ/4

)≤ Vn

(√t +

õ/4

)− Vn

(√t−

õ/4

).

The lemma follows immediately. ¤

Remark. We call r(t,K) the representation number of t by K. Clearlyr(1, K) ≤ 2n. For n ≥ 16, one can easily show that r(2, K) ≤ 2n(n− 1) byconsidering all root lattices. For n ≥ 5, the asymptotic behavior for r(t, K)is well known (see [H], [P] and [I]).

5

Theorem 2.1 For n ≥ 4,

c(n) ≤ (n/2)4 .

Proof. Let L be a unimodular Z-lattice of rank n. Since all unimodular Z-lattices of rank less than or equal to 24 are classified (see, for example, [CS1]),one can easily check the above assertion for those Z-lattices. Therefore, forcomputational convenience, we may assume that

n ≥ 20 and µ(L) ≥ 2. (1)

We follow Iwaniec [I, chapter 11], where he introduced the circle methodfor the representation number of an integer by a quadratic form of rank ≥ 4.For details, we refer the readers to his book. To prove the theorem, we mustcompute all the error terms quite accurately. Let k = n/2 and C ≥ 1 be anyreal number. As usual, e(z) = e2πiz. We define the adjoint quadratic form

Q∗ := (1/2)A−1[x]

of Q, where A = 2ML and A[x] := txAx.If we define

T (c, t; x) = |A|−1/2c−n

(i

z

)k ∑

m∈Zn

Tm(c, t; x)e

(−Q∗(m)

c2z

), (2)

where c is a positive integer less than or equal to C and

Tm(c, t; x) =∑

C<d≤c+Cgcd(c,d)=1, cdx<1

e

(t d

c

) ∑

h (mod c)

e

(−d

c(Q(h) + thm)

), (3)

then

r(t, L) = 2 Re

bCc∑c=1

∫ 1cC

0

T (c, t; x)e(−tz)dx. (4)

Here, z = x + iy ∈ C with y > 0, Re z = x, and d is the multiplicativeinverse of d (mod c). If 0 < x < 1

c(c+C), then Tm(c, t; x) is independent of x.

6

In that case, we simply write Tm(c, t). Now we divide (4) into the followingfive parts:

r(t, L) = |A|−1/2

∞∑c=1

c−nT0(c, t)

∫ ∞

−∞

(i

z

)k

e(−tz)dx · · · (r1)

−|A|−1/2

∞∑

c=bCc+1

c−nT0(c, t)

∫ ∞

−∞

(i

z

)k

e(−tz)dx · · · (r2)

+2 Re

bCc∑c=1

|A|−1/2c−n

∫ 1/cC

1/c(c+C)

T0(c, t; x)

(i

z

)k

e(−tz)dx · · · (r3)

+2 Re

bCc∑c=1

|A|−1/2c−n

∫ 1/cC

0

(i

z

)k

×∑

m∈Zn\{0}Tm(c, t; x)e

(−Q∗(m)

c2z

)e(−tz)dx · · · (r4)

−2 Re

bCc∑c=1

|A|−1/2c−nT0(c, t)

∫ ∞

1/c(c+C)

(i

z

)k

e(−tz)dx . · · · (r5)

First assume that L is odd unimodular. Let

Gc,d(Q,m) :=∑

h (mod c)

e

(−d

c(Q(h) + thm)

)(5)

be the Gauss sum factor in (3). By a suitable transformation, we may assumethat Q(h) ≡ h2

1 + h22 + · · ·+ h2

n (mod c). Then

Gc,d(Q,m) =n∏

i=1

∑

h (mod c)

e

(−d

c(h2 + mih)

),

where m = (m1,m2, · · · ,mn). Therefore by [S, Lemma 1], we obtain

|Gc,d(Q,m)| ≤ (gc)k, (6)

where g = gcd(2, c). From this and (3) follows

|Tm(c, t; x)| ≤ 2k−1ck+1 for all m, x and t. (7)

Now we take C =√

t and y = 1/t. Then

∫ ∞

−∞

(i

z

)k

e(−tz)dx = (2π)kΓ(k)−1tk−1,

7

and ∞∑c=1

c−nT0(c, t) =∏

p: prime

αp(t, L),

where αp(t, L) is the local density at p. Therefore

r1 = πkΓ(k)−1tk−1∏

p:prime

αp(t, L).

Now we compute the upper bounds of the absolute values for the other ri’sby using Lemma 2.1 and (7). One can easily show the following:

|r2| ≤ 2k−1πktk−1

Γ(k)

∞∑

c=b√tc+1

c1−k ≤ 22k−1πktk/2

Γ(k),

|r3| ≤b√tc∑c=1

e2π

k − 1

((c +

√t)k−1 −

√t

k−1)≤ 2k−1e2π

k − 1tk/2,

|r5| ≤ 2k−1e2π

k − 1tk/2.

Since L is unimodular,∣∣∣∣∣∣

∑

m∈Zn\{0}e

(−Q∗(m)

c2z

)∣∣∣∣∣∣≤

∑

m∈Zn\{0}e

( −πy

2c2(x2 + y2)Q−1[m]

)

=∞∑

m=1

r(m,L)e

( −πym

2c2(x2 + y2)

).

Let β =πy

2c2(x2 + y2). We may assume that r(m, L) ≤ (4m)k by the assump-

tion (1). Therefore

∞∑m=1

r(m,L)e−βm ≤ 4k

∫ ∞

0

xke−βxdx ≤ 4kΓ(k + 1)

βk+1. (8)

This implies

|r4| ≤ e2π23k+1Γ(k + 1)tk+1

πk+1

b√tc∑c=1

ck+3

∫ 1/c√

t

0

|z|k+2dx

≤ e2π2(7k+4)/2Γ(k + 1)

πk+1tk/2.

8

By Lemma 9.4 of [MH],

5

6≤

∏p

αp(t, L) ≤ 6

5.

Therefore

r(t, L) ≥(

5πk

6Γ(k)

)tk−1 −

(22k−1πk

Γ(k)+

2ke2π

k − 1+

e2π2(7k+4)/2Γ(k + 1)

πk+1

)tk/2.

Consequently, if t ≥ k4, then r(t, L) > 0, which proves the theorem for oddunimodular case.

Now assume that L is even. In this case, we may assume that Q(h) ≡2h1h2 + · · ·+2hn−1hn (mod c) by a suitable transformation. Then, one caneasily show that

|Tm(c, t; x)| ≤ 2k−1ck+1,

and25

16≤

∏p

αp(2t, L) ≤ 18

7,

just by comparing the 2-adic densities of even and odd cases. This completesthe proof. ¤

3 The 2-graph of unimodular Z-lattices

Let Un be the set of all unimodular (both even and odd) Z-lattices of rank n.Two Z-lattices K, L ∈ Un are said to be 2-neighbors if there exists a Z-latticeL′ isometric to L such that

[L′ : L′ ∩K] = [K : L′ ∩K] = 2.

In order to define the 2-graph G(Un : 2) of Un, we take for the vertices theisometric classes of unimodular lattices in Un. Two vertices are connectedby a simple edge when and only when they are 2-neighbors. This definitionis due to Kneser, which is slightly different from those of [BH] and [CS1]. Itis well known that G(Un : 2) is connected (see [Kn] or [O’M1; 106.4]). Thefollowing figure illustrates the 2-graph G(U15 : 2) of U15.

9

Figure 3.1.

E8 ⊥ I7 E7E7[11] ⊥ 〈1〉

I15 D12[1] ⊥ I3 A15[4]

­­

­­

­­

­­­ J

JJ

JJ

JJ

JJ ­­

­­

­­

­­­ J

JJ

JJ

JJ

JJ

For the gluing notations for Z-lattices and the 2-graph of even unimodularlattices of rank 24, see [CS1], [CS2]. By abuse of notations, we will regardindividual lattices in Un as representatives of their classes which are verticesof the graph.

For a vertex L ∈ Un, let ∆(L) denote the number of vertices adjacent toL. For two vertices L,K ∈ Un, define the distance between L and K by theminimum number of edges between L and K and write d(L,K). Let

d(L) := max{ d(L, K) |K ∈ Un } and d(n) := max{ d(L) |L ∈ Un }.We call d(n) the diameter of G(Un : 2). It is well known [Kn] that everyunimodular Z-lattice that is adjacent to In is of the form In−4k ⊥ D4k[1].Thus we obtain

∆(In) ≤ bn/4c.It is also well known [Kn] that

d(n) ≤ 2 for all n ≤ 16. (9)

Lemma 3.1 For L ∈ Un,

∆(L) ≤

2

((n4

)+

(n8

)+ · · ·+

(n

4bn/4c))

if L is odd,

2n + 2n/2 − 2 if L is even.

10

Proof. Let L = Zx1 + Zx2 + · · ·+ Zxn be any unimodular Z-lattice of rankn. First assume that L is odd. By the weak approximation theorem, wemay also assume that B(xi,xj) ≡ δij (mod 8). Any unimodular lattice Kadjacent to L is isometric to

Ly = Zy + {x ∈ L | B(x,y) ∈ Z}for some y ∈ (1/2)L− L and Q(y) ∈ Z (see [O’M1]). Note that

Ly = Ly′ if and only if y − y′ ∈ L and B(y,y′) ∈ Z.

Define

S := {a = (a1, a2, · · · , an) | ai ∈ {0, 1, 2, 3}, a 6∈ 2Zn,∑

a2i ≡ 0 (mod 4)},

and give an equivalence relation on S by

a ∼ b if and only if ai ≡ bi (mod 2) and∑

aibi ≡ 0 (mod 4).

If we definey(a) := (1/2)(a1x1 + a2x2 + · · ·+ anxn)

for a = (a1, a2, · · · , an) ∈ S, then Ly(a) is adjacent to L and every Ly adjacentto L can be obtained in this way. Furthermore, Ly(a) = Ly(b) if and only ifa ∼ b. Note that they might be isometric to each other. Thus we obtain

∆(L) ≤ |S/ ∼ |and hence the desired inequality follows.

Now assume that L is even. Then n ≡ 0 (mod 8). We may assume that

ML ≡(

0 11 0

)⊥

(0 11 0

)⊥ · · · ⊥

(0 11 0

)(mod 8).

We omit the rest, which is similar to the odd unimodular case. ¤

Lemma 3.2 For n ≥ 17,

d(In) ≤n∑

i=17

b2 log2 ic − n + 18.

11

Proof. Let L = Zx1 + Zx2 + · · ·+ Zxn ∈ Un for n ≥ 17. Since

4b2 log2 nc−1 ≥ (n/2)4 ≥ c(n),

L should represent 4b2 log2 nc−1 according to Theorem 2.1. So, there is somes ≤ b2 log2 nc − 1 such that L represents 4s primitively and hence we mayassume that

Q(x1) = 4s, B(x1,x2) ≡ 1 (mod 2) and B(x1,xi) ≡ 0 (mod 2)

for 3 ≤ i ≤ n. If s ≥ 1, then

K := Z(1/2)x1 + Z2x2 + Zx3 + · · ·+ Zxn ∈ Un

is adjacent to L and K represents 4s−1. By repeating this procedure, ifnecessary, we may conclude that there exists L′ ∈ Un−1 such that

d(L, 〈1〉 ⊥ L′) ≤ s ≤ b2 log2 nc − 1.

Therefored(In) ≤ b2 log2 nc − 1 + d(In−1).

The lemma follows from this and (9). ¤

We let for n ≥ 17,

τn :=n∑

i=17

b2 log2 ic − n + 18 = O(n log n). (10)

From Lemma 3.2 follows that

d(n) ≤ 2τn for n ≥ 17.

Corollary 3.1 Let ` be any Z-lattice of rank n, n ≥ 14, with s(`) ⊆ 4τn+3Z.Then ` → In+3.

Proof. Note that (1/2)τn+3` → gen(In+3) (see [O’M2]). So there exists L ∈gen(In+3) such that

(1/2)τn+3` → L.

Since 2τn+3L → In+3 by Lemma 3.2, we have ` → 2τn+3L → In+3. ¤

12

Theorem 3.2

1

4 log 2≤ lim

n→∞inf

d(In)

n log n≤ lim

n→∞sup

d(In)

n log n≤ 2

log 2.

Proof. The inequality in the middle is trivial. The last inequality is thedirect consequence of Lemma 3.2 and the Stirling’s formula. Now we provethe first inequality. Since ∆(L) < 2n+1 for all L ∈ Un by Lemma 3.1,

|Un/ ' | ≤d(In)∑i=1

2(n+1)i.

Let M(In) be the mass of the genus of In. Then

2M(In) ≤ |Un/ ' | ≤ 2(n+1)d(In) − 1

2n+1 − 1.

Since

M(In) ∼C

(n

2πe√

e

)n2/4 (8πe

n

)n/4

n1/24,

where C is a constant (see [MH]), the first inequality follows. ¤

Remark. Observe that from the proof of the above theorem, one can easilydeduce that

h(In) = nΘ(n2).

4 Upper and lower bounds for g[n]

Let ` = Zx1+Zx2+· · ·+Zxn ⊂ Ig be a Z-lattice, where xi = (ai1, ai2, · · · , aig)for 1 ≤ i ≤ n. For 1 ≤ j ≤ g, define

vj := t(a1j, a2j, · · · , anj),

where tA is the transpose of a matrix or a vector A. Since these vj’s alsocharacterize `, we define

Sn(v1,v2, · · ·vg) := ` ⊂ Ig. (11)

13

Note that for positive integers s1, s2, · · · , sg,

Sn(

s1 copies︷ ︸︸ ︷v1, · · ·v1,

s2 copies︷ ︸︸ ︷v2, · · ·v2, · · · ,

sg copies︷ ︸︸ ︷vg, · · ·vg) ⊂ Is1+s2+···+sg

is represented by the Z-lattice 〈s1〉 ⊥ 〈s2〉 ⊥ · · · ⊥ 〈sg〉 of rank g.We define

λ(n) := min{ g | ` → Ig ⊥ 〈2, · · · , 2〉 for all ` ∈ Sn }.For a sublattice ` of L ⊥ M of the form

` = Z(x1 + y1) + Z(x2 + y2) + · · ·+ Z(xn + yn)

with xi ∈ L and yi ∈ M , we define

`(L) := Zx1 + Zx2 + · · ·+ Zxn and `(M) := Zy1 + Zy2 + · · ·+ Zyn.

Lemma 4.1λ(n) ≤ 2n − 1.

Proof. Let ` = Sn(v1,v2, · · · ,vg) ⊂ Ig be any Z-sublattice of rank n. Assumethat g ≥ 2n. Then either (i) there exists i satisfying vi ≡ 0 (mod 2) or (ii)there exist i, j, i 6= j satisfying vi ≡ vj (mod 2). If we let

`0 = Sn ( · · · ,vi/2,vi/2,vi/2,vi/2, · · · )for the case (i), then

` ∼= `0 → Ig−1 ⊥ 〈4〉 → Ig−1 ⊥ 〈2, 2〉.Consider the case (ii). If vi = ±vj, then

` → Ig−1 ⊥ 〈2〉.If vi 6= ±vj, let

`1 = Sn ( · · · , (vi + vj)/2, (vi + vj)/2, · · · , (vi − vj)/2, (vi − vj)/2, · · ·) .

Then` ∼= `1 → Ig−2 ⊥ 〈2, 2〉.

The lemma follows. ¤

14

Theorem 4.1 For all n ≥ 14,

g[n] ≤ 2λ(n)τn+3 + n + 4.

Proof. Assume ` → Ig with g > λ(n). Then ∃λ1, g1 such that

` → Iλ1 ⊥√

2Ig1 ,

where λ1 ≤ λ(n). If g1 > λ(n), Then ∃λ2, g2 such that

` → Iλ1 ⊥√

2Iλ2 ⊥ 2Ig2 ,

where λ2 ≤ λ(n). Repeating this process, if applicable, we obtain

` → Iλ1 ⊥√

2Iλ2 ⊥ · · · ⊥ (√

2)t−1Iλt ⊥ (√

2)tIgt ,

where λj ≤ λ(n) for all 1 ≤ j ≤ t and t = 2τn+3. If we denote this repre-sentation by σ, then σ(`)((

√2)tIgt) is represented by In+3 by Corollary 3.1.

Observe that

Iλ1 ⊥ (√

2)2Iλ3 ⊥ · · · ⊥ (√

2)t−2Iλt−1 → Iλ1+λ3+···+λt−1 ,

and √2Iλ2 ⊥ (

√2)3Iλ4 ⊥ · · · ⊥ (

√2)t−1Iλt → Iλ2+λ4+···+λt+1.

Combining these, we obtain

` → Iλ1+···λt+n+4.

This proves the theorem. ¤

Note that Lemma 4.1 and the above theorem together with (10) implythat

g[n] = O(2n n log n). (12)

We now improve this upper bound. To do this, we have to improve the upperbound of λ(n) of Lemma 4.1. Let ` ∈ Sn. We call ` a terminal Z-lattice ifσ(`)(〈2, · · · , 2〉) = 0 for any representation σ : ` → Ig ⊥ 〈2, · · · , 2〉. Let Tn

be the set of all terminal Z-lattices of rank n, which is a subset of Sn.

15

Lemma 4.2 For all ` ∈ Sn, there exist an `′ ∈ Tn and a representationφ : ` → Ig ⊥ 〈2, · · · , 2〉 such that

φ(`)(Ig) = `′.

Proof. Let µi(`) be the i-th successive minimum of `. We will use an inductionon

m(`) :=n∏

i=1

µi(`).

If m(`) = 1, then clearly ` = In and hence ` itself is a terminal Z-lattice. Ifφ : ` → Ig ⊥ 〈2, · · · , 2〉 and φ(`)(〈2, · · · , 2〉) 6= 0, then m(φ(`)(Ig)) < m(`).Hence by induction hypothesis, there exists

φ′ : φ(`)(Ig) → Ih ⊥ 〈2, · · · , 2〉,such that φ′(φ(`)(Ig))(Ih) is a terminal Z-lattice. The lemma follows imme-diately. ¤

Remark. From above lemma, we have

λ(n) = min{ g | ` → Ig, for all ` ∈ Tn}.

Lemma 4.3λ(n)(λ(n)− 1) ≤ 2× 3n − 1.

Proof. For ` ∈ Tn, let ` = Sn(v1,v2, · · · ,vg) and vj 6= 0 for all 1 ≤ j ≤ g.Note that vi 6= vj for any i 6= j since ` /∈ Tn otherwise. We may assumethat g is the minimum rank in order for Ig to represent `. Suppose thatg(g − 1) ≥ 2 × 3n. Then, by the pigeon hole principle, either (i) there existtwo distinct indices i, j such that vi ≡ vj (mod 3), or (ii) there exist fourdistinct indices i, j, k, l such that

vi + vj ≡ vk + vl (mod 3).

For the case (i), note that

Sn(vi,vj) ' Sn

( 4 copies︷ ︸︸ ︷(vi − vj)/3, · · · , (vi − vj)/3, (vi + 2vj)/3, (−2vi − vj)/3

).

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This contradicts to the fact that ` ∈ Tn. Consider the case (ii). If vi + vj =vk + vl, then

Sn(vi,vj,vk,vl) → A3 → I3,

which is not possible. So, vi + vj 6= vk + vl. One can easily show that

Sn(vi,vj,vk,vl) ' Sn (w,w,w + vl,w + vk,w − vj,w − vi) ,

where w = (vi + vj − vk − vl)/3. This again contradicts to the fact that` ∈ Tn. ¤

Corollary 4.2 For n ≥ 14,

g[n] ≤ 3 · 3n/2 τn+3 + n + 4.

In particular,g[n] = O(3n/2 n log n).

Proof. This is an immediate consequence of (10), Theorem 4.1 and Lemma4.3. ¤

Remark. As we can see from Lemmas 4.1 and 4.3, it seems that this boundfor λ(n) can be improved further. But it is doubtful that our method maygive us a polynomial bound for λ(n) and hence for g[n].

We now provide a lower bound of g[n] in the following theorem.

Theorem 4.3g[n] ≥ 2n− dlog2 ne − 2.

Proof. Let m,n be positive integers satisfying n ≤ 2m. It suffices to show

g[n + m + 1] ≥ 2n + m. (13)

If we let N = {1, 2, · · · , n}, then we can always find m subsets T1, T2, · · · , Tm

of N such that for any {i, j} ⊂ N , there exists an index k for which

|Tk ∩ {i, j}| = 1.

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Let {e1, · · · , en, en+1, · · · , en+m, f1, · · · , fn} be an orthonormal basis of I2n+m.If we define

` :=n∑

i=1

Z(ei + fi) +m∑

k=1

(Z

(∑j∈Tk

ej + en+k

))+ Z(f1 + · · ·+ fn),

then one can easily show from the definition of Tk that

` → I2n+m but ` 6→ I2n+m−1.

From this follows (13) and the theorem. ¤

Remark. For small n’s, say 7 ≤ n ≤ 20, much sharper bounds are known(see [KO3]). For example, 11 ≤ g[7] ≤ 24, 13 ≤ g[8] ≤ 37, and

g[n] ≤{

n(n + 1)/2 + n + 3 for 9 ≤ n ≤ 11,5n3/2 for 12 ≤ n ≤ 20.

(14)

5 s-integrable Z-lattices

We apply our result to s-integrable Z-lattices introduced by Conway andSloane [CS3]. A Z-lattice ` is said to be s-integrable if

√s` can be represented

by a sum of squares, where s is a positive integer. Define

φ(s) := min{n | ∃ a Z-lattice ` of rank n such that√

s` /∈ Sn }.It is known [CS3] that

φ(1) = 6, φ(2) = 12, φ(3) = 14, · · ·and that

φ(s) ≥ 2

(ln ln s

ln ln ln s(1 + o(1))

).

We improve this bound in the following theorem.

Theorem 5.1 Let q(n) = (bn2/4c + 1) for a positive integer n. Then forany integer s satisfying

((q(n) + 1)2n − 1

)q(n)2n ≤ s ≤ (

(q(n + 1) + 1)2(n+1) − 1)q(n + 1)2(n+1),

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φ(s) ≥ n− 2. In particular, for large s,

φ(s) ≥ ln s

8 ln ln s.

Proof. Let L = Zx1 + Zx2 + · · · + Zxn be an odd unimodular Z-lattice ofrank n. Then by Theorem 2.1, L represents q2, where q := q(n). So, we canfind a divisor b of q such that b2 is primitively represented by L and henceby a suitable base change, if necessary, we may assume that

Q(x1) = b2, gcd(B(x1,x2), b) = 1, B(x1,xi) ≡ 0 (mod b)

for 3 ≤ i ≤ n. Furthermore we may assume that

bQ(x2) 6≡ B(x1,x2) (mod 2).

LetL := Z(1/b)x1 + Zbx2 + Zx3 + · · ·+ Zxn.

Then L ∼= 〈1〉 ⊥ L′, where L′ is an odd unimodular Z-lattice of rank n − 1.So

qL → 〈1〉 ⊥ L′.

Since L′ also represents q2 by Theorem 2.1, we may conclude that

qnL → In and similarly (q + 1)n` → In.

Thus for every integer s ≥ ((q + 1)2n − 1)q2n, L is s-integrable. Since everyZ-lattice ` of rank n − 3 is represented by gen(In), ` is s-integrable. Thetheorem follows. ¤

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