2-universal O-lattices over real quadratic fields

manuscripta math. 119, 97–106 (2006)

© Springer-Verlag 2005

Hideyo Sasaki

2-universal O -lattices over real quadratic fields Received: 11 March 2004 / Published online: 2 December 2005 √ Abstract. Let F = Q( m) be a real quadratic field with m a square-free positive rational integer, and O be the ring of integers in F . An O-lattice L on a totally positive definite quadratic space V over F is called r-universal if L represents all totally positive definite O-lattices l with rank r over O. We prove that there exists no 2-universal O-lattice over F with rank less than 6, and there exists a 2-universal O-lattice over F with rank 6 if and only if m = 2, 5. Moreover there exists only one 2-universal O-lattice with rank 6, up to √ isometry, over F =Q( 2).

1. Introduction √ Let F = Q( m) be a real quadratic field with m a square-free positive rational integer, and O be the ring of integers in F . Let (V , Q) be a totally positive definite quadratic space over F and B a bilinear form associated with the quadratic form Q. Let L be an O-lattice on V , that is L be a finitely generated O-module in V and contains a basis of V over F . In the following we assume that all O-lattices L are integral, that is B(L, L) ⊂ O. Let l, L be O-lattices over quadratic spaces W , V , respectively. We say that L represents l over O if there is an isometry σ from W into V such that σ (l) ⊂ L and write l −→ L. An O-lattice L is called r-universal if L represents all totally positive definite O-lattices l with rank r over O. √Chan-Kim-Raghavan [1] proved that for real quadratic number fields F = Q( m), ternary (1-)universal classic quadratic forms over F exist if and only if m = 2, 3 and 5, and they also determined all universal ternary quadratic forms, up to equivalence, over each F above. In this paper we prove the following result for r = 2: √ Theorem 1. Let F = Q( m) be a real quadratic field with m > 0 a square-free positive rational integer. There exists no 2-universal O-lattice over F with rank less than 6, and there exists a 2-universal O-lattice over F with rank 6 if and only if m = 2, 5. Moreover√there exists only one 2-universal O-lattice of rank 6, up to isometry, over F =Q( 2). H. Sasaki: Otemae College, 2-2-2 Inano-cho, Itami, Hyogo 664-0861, Japan. e-mail: [email protected] Mathematics Subject Classification (2000): Primary 11E25; Secondary 11E12

DOI: 10.1007/s00229-005-0607-9

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In Section 2 we show that there exists no 2-universal lattice with rank 6 over √ F = Q( m) with a square-free integer m = 2, 5. In Section√3 and 4√we give the proof of the existence of 2-universal O-lattices over F = Q( 2), Q( 5), respectively. The class numbers of O-lattices in Theorem 7 and 18 are both greater than 1, so we cannot apply the Hasse principle on local- and global representations of O-lattices for them. We give terminologies and notations. See [2] for general terminologies and basic facts about quadratic lattices. We denote  the fundamental unit of F . For any two integers α, β ∈ O, we write α ∼ β when α = β 2n with n ∈ Z. α denote the conjugate of α in F . We denote the trace T r(α) := α +α and the norm N (α) := αα of α ∈ F , respectively. We denote by O+ the set of all totally positive integers, and 0 := O ∪ {0}. we put O+ + For any elements α, β, γ in F or its localization F℘ , we denote [α; β; γ ] a 2 × 2 symmetric matrix
 αβ [α; β; γ ] := . βγ We call a free O-lattice L with rank n additively indecomposable if for a corresponding symmetric matrix A ∈ Mn (O) to L, A = B +C with totally semi-positive symmetric matrices B, C ∈ Mn (O) implies either B or C = O. For an O-lattice L, we denote cls(L) and gen(L) the class and genus of L in a quadratic space V = F L, respectively. For L an O-lattice over F , h(L) denotes the class number of L, that is, the number of equivalence classes in gen(L). We denote L(a) for a scaling of L by a ∈ F × or F℘× . Put In :=⊥n Oei ∼ =⊥n 1 , with {e1 , · · · , en } an O-basis of IN with B(ei , ej ) = δij for all i, j = 1, · · · , n, where δij a Kronecker’s delta. Note that if L is r-universal, then In ⊥ L is also r-universal for any n ≥ 1. Thus if there exists no r-universal O-lattice with rank l then there exists no runiversal O-lattice with rank less √ than l, either. We put ur (m) := min{rankL|L is a r-universal O-lattice over Q( m)}. 2. Non-existence of 2-universal lattice with rank 6 In this section we show that there exists no 2-universal O-lattice with rank 6 if m = 2, 5. √ For m ≡ 2 or 3 mod 4 we put τm := k − m where k = km is the least positive √ rational integer such that τm ∈ O+ , and for m ≡ 1 mod 4 we put τm := (k− m)/2 where k = km is the least positive rational odd integer such that τm ∈ O+ . Note that τm is additively indecomposable and τm −→ 1 for all m = 5. Lemma 2. Suppose that either m ≥ 37 with m ≡ 1 mod 4 or m ≥ 7 with m ≡ 2, 3 mod 4. Let η ∈ O and a = 2 or 3. (i) If M = [a; η; τm ] is totally semi-positive definite then η = 0 or ±1. (ii) If M = [2; η; a] is totally semi-positive definite then η = 0, ±1 or ±2.

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Proof. We prove (i) for m ≥ 37 with m ≡ 1√ mod 4 and the rest cases can be obtained by the similar arguments. Put η = (s +t m)/2 with s, t ∈Z, s ≡ t mod 2. Note that k ≥ 7 for given m’s. √ 2 By the definition of τm , we√have τm −1 ∈ O+ and (k−2)2 − m = k 2 −4k+4− √ √ m < 0. We have |M| = a · k−2 m −( s+t2 m )2 = 41 (2ak −s 2 −t 2 m−2(a +st) m). If |t| ≥ 2 then 2ak − s 2 − t 2 m < 6k − 4m < −4k 2 + 22k − 16 < 0 for 0 . Suppose |t| = 1. If a = 2 then a + st = 0 k ≥ 7 and we have |M| ∈ O+ and 2ak − s 2 − t 2 m − (k − 2) ≤ 4k − 1 − m − k + 2 = 3k + 1 − m < 0 . If a = 3 and |s| = 3 −k 2 + 7k − 3 < 0 for k ≥ 7. We also have |M| ∈ O+ 2 2 2 then 2ak − s − t m = 6k − 9 − m = −k + 10k − 13 < 0 for k ≥ 9, and 6k − 9 − m ≤ 6 · 7 − 9 − 37 < 0 for k = 7. If a = 3 and |s| = 3 then a + st = 0 and 2ak − s 2 − t 2 m − (k − 2) ≤ 6k − 1 − m − k + 2 = 5k + 1 − m < −k 2 + 9k − 3 < 0 0 for k ≥ 9, and 5k + 1 − m ≤ 5 · 7 + 1 − 37 < 0 for k = 7. Then we have |M| ∈ O+ again. Thus we obtain t = 0 and η ∈Z with |η| ≤ 1.   Lemma 3. Let m > 5 and S = {1, 2, 3, 4}. Suppose for x ∈ S, x = y + z with y, z ∈ O+ . Then y, z ∈ S. Proof. It is obtained from [1, Proposition 2.1].

 

Proposition 4. Let m ≥ 37 with m ≡ 1 mod 4 or m ≥ 7 with m ≡ 2, 3 mod 4. Then u2 (m) > 6. Proof. Suppose that there exists a 2-universal O-lattice L with rank 6. L represents a binary unimodular O-lattice I2 = Oe1 ⊥ Oe2 . By [2, 82:15] I2 splits L, say L = I2 ⊥ L1 for a submodule L1 in L. If T r(Q(y)) > 4 for any vector x ∈ L1 then L cannot represent [2; 1; 2] . By Lemma 3 L1 contains the vector y1 with Q(y1 ) = 1, 2. If Q(y1 ) = 1 then Oy1 splits L1 and L = I3 ⊥ L2 for a submodule L2 in L. If T r(Q(y)) > 4 for any vector y ∈ L2 then L cannot represent [2; 1; 3] . Thus L2 contains a vector y2 with Q(y2 ) = 1 or 2. In the case of Q(y2 ) = 1 we have L = I4 ⊥ L3 for some sublattice L3 . If T r(Q(y)) > 6 for any vector x ∈ L3 then L cannot represent [2; 1; 4] , thus L3 contains a vector y3 with Q(y3 ) = 1, 2 or 3. Suppose I1 −→ L2 and Q(y2 ) = 2. Since an O-lattice I3 ⊥ Oy2 cannot represent 3, 3 , L2 also contains a vector y3 = ±y2 with Q(y3 ) = 2 or 3. Since L is totally poisitive definite we obtain B(y2 , y3 ) = 0, ±1 by Lemma 2 (if Q(y2 ) = 3 and B(y1 , y2 ) = ±2 then 1 is represented by O[y1 , y2 ] which contradicts the assumption). Suppose I1 −→ L1 and Q(y1 ) = 2. We may assume [2; 1; 2] is represented by L1 and there is a vector y2 with Q(y2 ) = 2 and B(y1 , y2 ) = 1. Since I2 ⊥ (Oy1 + Oy2 ) cannot represent 2, 3 , L1 also contains a vector y3 with Q(y3 ) = 2, 3 which is not represented by Oy1 + Oy2 and B(y3 , yi ) = 0, 1 (i = 1, 2). L also represents an O-lattice τm , τm . L contains two vectors y4 , y5 with Q(yi ) = τm for i = 4, 5 and B(y4 , y5 ) = 0. Since τm is additively indecomposable and τm −→ I1 , B(yi , e) = 0 for i = 4, 5 where e ∈ L with Q(e) = 1.

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Consider the free O-lattice L0 = I2 ⊥ (Oy1 + · · · + Oy5 ). By Lemma 2 i = 1, 2, 3, j = 4, 5. B(yi , yj ) = 0, ±1 for √ Put dL0 = (u + v m)/2 with u, v ∈ Z with u ≡ v mod 2. By direct computations we can verify v = 0 for each yi ’s and we obtain dim(F L) ≥ 7, which contradicts the assertion. Thus u2 (m) > 6.   We next show √ that u2 (m) > 6 for m = 3, 6, 13, 17, 21, 29, 33. The ideal class number of Q( m) for those m’s are all one. We can easily check the following by direct calculations: Lemma 5. These O-lattices are additively indecomposable in each F . √ √ √ √ ∼ [3 + 3; 3; 3 − 3] in F = Q( 3). (1) N3 := √ √ √ 17 7+ 17 (2) N17 :∼ = [2; 1+√ 2 ; 2 ] in F = Q(√17). √ 21 9+ 21 (3) N21 :∼ = [2; 1+√ 2 ; 2√ ] in F = Q( √21). 1+ 29 11+ 29 ∼ (4) N29 := [2; ; ] in F = Q( 29). 2

2

Remark. We have N (dNm ) = 3, 2, 7, 5 for m = 3, 17, 21, 29, respectively. Thus if there exists an O-lattice K such that Nm −→ K then Nm ∼ = K for each m by comparing the volume ideals of two lattices. Suppose that there exists a 2-universal O-lattice L with rank 6 for each m above. Let m = 17, 29. Then L √ represent I2 , an additively indecomposable binary unimodular O-lattice J = [3; m; (1 + m)/3] , and a binary O-lattice Nm in the above Lemma. Thus by remark in above we have Nm −→ J and L = I2 ⊥ J ⊥ Nm . However, this L cannot represent [2; 1; 2] . This contradicts the assumption. Thus u2 (m) > 6 for m = 17, 29. For m = 3, 6, 21, 33, noting that N () = 1, we have L ∼ = 1, 1, ,  ⊥ J where J be an additively indecomposable binary unimodular O-lattice J ∼ = √ √ √ 1+ 21 [2; 1; 12 + 2 33] for m = 33, [2; 2 ; 4] for m = 21, [5; 2 6; 5] for √ m = 6, [2; 3; 2] for m = 3, respectively. These L cannot √ √ represent √ Nm in Lemma 5 for m = 3, 21, [2; 1; 3 +√ 6] for m = 6, 6 − 33, 6 − 33 for m= √ 33, respectively (note that 6 − 33 is additively indecomposable integer in Q( 33)). Thus u2 (m) > 6 for m = 3, 6, 21, 33. Suppose m = 13. L represents binary O-lattices I2 , [2; 1; 2] , τm , τm and √  τm , τm . Note that τm (= 5−2 13 ) = τm in F × /(F × )2 . Then for L = I2 ⊥ L ,  applying the similar argument in Proposition 4, L contains the vectors y1 , . . . , y5 with Q(y1 ) = 1 or 2, and if Q(y1 ) = 1 then B(y1 , yi ) = 0 for i = 2, . . . , 5; Q(yi ) = τm for i = 2, 3 with B(y2 , y3 ) = 0; Q(yj ) = τm for j = 4, 5 with B(y4 , y5 ) = 0. If Q(y1 ) = 2 then we have B(vi , vj ) = 0, ±1. We also obtain B(vi , vj ) = 0, ±1 for i = 2, 3, j = 4, 5. Then for each free O-module U = I2 ⊥ (Ov1 + · · · + Ov5 ) we obtain that dU = 0 by calculating dU directly. It yields that U is non-degenerate and we have u2 (13) > 6. Thus we conclude Theorem 6. u2 (m) > 6 for m = 2, 5.

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√ 3. The 2-universal lattice over Q( 2) In this section√we prove the existence of the totally positive 2-universal √ O-lattice rank 6. We put the fundamental unit  = 1 + over F = Q( 2) with 2. √ √ 2; 1; 2 − 2]

be the totally positive definite binary unimod [2 + Let K  ∼ = ular O-lattice. We can easily see that K  is additively indecomposable. First we show that there exists no 2-universal O-lattice with rank = 5. Suppose L be an 2-universal O-lattice with rankL = 5. Since L represents two unimodular O-lattices I2 and K  we have L = I2 ⊥ K  ⊥ a

for some a ∈ O+ . We can easily see that L cannot represent A ∼ = [2; 1; 2] unless a ∼ 1. Thus √ we have L = I3 ⊥ K  . But this L cannot represent an O-lattice √  A ∼ = [2 − 2; 1; 4 + 2 2] . Thus there exists no 2-universal O-lattice with rank = 5. Next we give the candidates for 2-universal O-lattices with rank 6. Let L be a 2-universal O-lattice with rank = 6. Then we have L = I2 ⊥ K  ⊥ L0 for some binary O-submodule L0 of L.  Suppose that L0 represents 1 . Then we have L = √ I3 ⊥ K ⊥ a for some  a ∈ O+ . This L does not represent A unless a ∼ 2 + 2. Thus we have √ L = I3 ⊥ K  ⊥ 2 + 2 . Next suppose that L0 does not represent 1 . Since (3O) (= dAO) is a prime ideal in O we obtain L0 ∼ = A and L ∼ = I2 ⊥ K  ⊥ A . But this L cannot represent  A and we conclude that this L is not 2-universal. Thus we will prove the following assertion. Theorem 7. There exists no 2-universal O-lattice with rank less than 6. There exists only one 2-universal O-lattice L with rank = 6, up to isometry, that is: √ L∼ = I3 ⊥ K  ⊥ 2 + 2 . We have already proved the first part of Theorem. Note that the class√number  of L is greater than 1 since an O-lattice L = I1 ⊥ K  ⊥ K  ⊥ 2 + 2 is in   L. We put W := I2 ⊥ K  . gen(L) and L ∼ = Lemma 8. Let L be a unimodular O-lattice with rank = 4, and q be any nondyadic prime spot in F . Then (L)q is a 2-univesal Oq -lattice, that is, (L)q represents all binary Oq -lattices. Proof. For any non-dyadic prime spot q in F , Lq ∼ = H ⊥ H , where H = [0, 1, 0] , and Lq represents all binary Oq -lattices.   Lemma 9. The class numbers of O-lattices I4 , W and K  ⊥ K  are all one. Proof. In [7, Theorem 19], the facts h(I4 ) = 1 and h(W ) = 1 are shown. It also says that h(K  ⊥ K  ) = 2, that is, there exists an indecomposable O-lattice N4,2 in the genus of (K  ⊥ K  ) ([7, Proposition 11]). But B(y1 , y2 ) = 0 when m = 2 and n = 4 in the proof of [7, Proposition 11]. Thus Takada’s proof is invalid for N4,2 . Hence N4,2 = K  ⊥ K  and we have h(K  ⊥ K  ) = 1.  

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Lemma 10. For the O-lattice L in the theorem, L(2+

√ 2)

−→ L.

  √ Proof of Theorem 7. Let ℘ = ( 2O) be the dyadic prime spot in F in the following. Let M be a binary O-lattice. By Lemma 10, we may assume that s(M℘ ) = O℘ . Then we consider three cases in the following. Case (i): M℘ ∼ = α, β (α, β ∈ O℘ ). Applying [5, Third Main Theorem] and Lemma 8 and 9 we have M −→ W ⊂ L. Case (ii): M℘ ∼ = [2; 1; 2β] (β ∈ O℘ ). In this case we have M −→ I4 . Let {x1 , x2 } be an O-basis of M such that xi =⊥4i=1 αij ej with αij ∈ O, i = 1, 2, √ j = 1, . . . , 4. We can easily see that (α1k , α2k ) ≡ (0, 0) mod 2 for some k = 1, . . . , 4 and assume k = 4 by changing the basis of I4 if necessary. Noting that K  represents 2 , take a vectory ∈ K  with Q(y)√ = 2. Then for the vectors    3 x1 , x2 ∈ I3 ⊥ K  with xi = j =1 αij ej + (αi4 / 2)y for i = 1, 2, we have   ∼ M = O[x1 , x2 ] = O[x1 , x2 ] −→ I3 ⊥ K  ⊂ L. √ √ Case (iii): M℘ ∼ = [ 2η; 1; 2β] (η ∈ O℘× , β ∈ O℘ ). We have M −→ K  ⊥ K  in this case. Suppose {x1 , x2 } be an O-basis of M such that xi = (ai1 v1 + ai2 v2 ) + (bi1 u1 + bi2 u2 ) with aij , bij ∈ O for i, j = 1, 2, where vi , ui be√the O-basis of K  ⊥ K  with Q(v1 ) = Q(v2 ) = Q(u1 ) = Q(u2 ) = 2 + 2, B(v √ B(u1 , u2 )√=  and B(v √ i , uj ) = 0 for i, j = 1, 2 (Note that K = [2 + √ 1 , v2 ) = ∼ 2; 1; 2− 2] = [2+ 2; ; 2+ 2]). Put Mv := O[a11 v1 +a12 v2 , a21 v1 +a22 v2 ] and Mu := O[b11 u1 + b12 u2 , b21 u1 + b22 u2 ]. Since s(M℘ ) = O℘ we may assume B(a11 v1 + a12 v2 , a21 v1 + a22 v2 ) ≡ 1 and √ B(b11 u1 + b12 u2 , b21 u1 +√ b22 u2 ) ≡ 0 mod 2. By the second equation we have b11 b22 + b12 b21 ≡ 0√ mod 2. √ √ If b11 ≡ 0 mod 2 then b12√or b21 ≡ √ √ 0 mod √ 2. If b21 ≡√0 mod √2 then ∼ 2; 2√+ 2] ∼ Mu ⊂ O[ 2u√ = 2 + 2, 2 + 2 . If 1 , u2 ] = [4 + 2 2; 2 + b12 √ ≡ 0 mod 2 and b ≡ b ≡ 1 mod 2 then we √ have Mu ⊂ O[u1 − 21 22 √ √ can apply u2 , 2u2 ] ∼ = [2; − 2; 4 + 2 2] −→ I2 . If b22 ≡ 0 mod 2 then we √ the same argument to√the previous cases. Suppose b11 ≡ √ b22 ≡ 1 mod 2. Then b12 ≡ b21 ≡ 1 mod 2 and it implies Mu ⊂ O[u1 − u2 , 2u2 ] −→ I2 √ If√Mu −→ I2 then M −→ W ⊂ L. Now we assume Mu√−→ 2 + 2 ⊥ 2 + 2 . Then M −→ K  ⊥ Oz1 ⊥ Oz2 with Q(zi ) = 2 + 2 for i = 1, 2 and we put Proof. We can easily verify it.

xi = (ai1 v1 + ai2 v2 ) + ci1 z1 + ci2 z2 (i = 1, 2) with cij ∈ O, i, j = 1, 2.√Changing the basis of M if necessary we may assume both Q(ai1 v1 + ai2 v2 ) ∈ 2O \ 2O for i=1, 2. Put Mzi := O[x1 − c1i , x2 − c2i ] (i = 1, 2), Z := O[c11 z1 + c12 z2 , c21 z1 + c22 z2 ]. √ O[c1i zi , c2i zi ] −→ 4 + √ If c1i ≡ c2i ≡ 0 mod 2 for either i = 1, 2 then √ 2 2 −→ I and it implies M −→ I ⊥ K ⊥ 2 + 2 ⊂ L. If c1i ≡ c2i ≡ 2 2 √ 1 mod 2 for either i = 1, 2 then either (Mz1 )℘ or (Mz2 )℘ is an even binary

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O℘ -lattice and applying the argument in the case (ii) we have Mzi −→ I3 ⊥ 2

√ and√M −→ L. If √ ci1 ≡ ci2√mod 2√for both i = 1, 2 then Z −→ O[z1 − z2 , 2z2 ] ∼ = [4 + 2 2,√2 + 2 2, 2 + 2] −→ I2 and√it implies M −→ W ⊂ L. c11 ≡ c22 ≡ 1(0) mod 2 and c12 ≡ c21 ≡ 0(1) mod 2 does not occur since M is an even O-lattice and it contradicts the assertion. Thus we proved the theorem.   √ 4. A 2-universal lattice over Q( 5) In this section we prove the existence of a 2-universal O-lattice with rank 6 over √ F = Q( 5). √ We put a fundamental unit  = 1+2 5 in F . We denote ℘ a dyadic prime spot (2O), and O℘× the unit group in O℘ . We also denote a binary O-lattice G :∼ = [2; ; 2] . Note that G is an additively indecomposable and G℘ ∼  H℘ = [0; 1; 0]

= since dG℘ = dH℘ . Proposition 11. There exists no 2-universal O-lattice with rank 5 over F . Proof. Suppose there exists such lattice L. We apply the similar argument in section 3. Since L represents I2 and G there exists ternary O-submodule L0 of L such that L = I2 ⊥ L0 and G −→ L0 . Since  L also represents [2; 1; 2] we have either  2  0  L = I3 ⊥ G or I2 ⊥ L0 , where L0 ∼ =   2 1  . But [2 + ; 1; 3 + ] −→ L 012 in the first case and [2; 1; 2 + ] −→ L in the second. Thus there exists no 2-universal O-lattice with rank 5.   We take an O-lattice L := I4 ⊥ G as a candidate of 2-universal O-lattice with rank 6. The class number of this lattice is greater than one, since an O-lattice    20 0  2 1 0   ∼ I2 ⊥   L = 0 1 2 1  . 0 0 1 2+ is in gen(L) and L ∼ =L. 

Lemma 12. Let M be a binary O-lattice. Then M is represented by I4 unless M℘ ∼ = G℘ over O℘ . Proof. By [3] we have h(I4 ) = 1. Since Mq −→ (I4 )q at all non-dyadic prime spots q and M℘ −→ (I4 )℘ unless M℘ ∼   = G℘ , we obtain the lemma. Proposition 13. The class numbers of O-lattices I3 ⊥ 2 +  , G ⊥ G, G ⊥ G(2) are all one.

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Proof. First we see the class number of the O-lattice K1 := I3 ⊥ 2+ is one. We note that h(I5 ) = 2 and gen(I5 ) consists of two classes cls(I5 ) and cls( 1 ⊥ E4 ), where E4 is the unique even quaternary unimodular O-lattice in F ([3]). If K0 ∈ gen(K1 ), then K0 ⊥ 2 +  must be represented by I5 . But then K0 must be orthogonal to a vector of length 2 +  in I5 and hence K0 ∼ = K1 . Thus we have h(K1 ) = 1. Now we show h(G ⊥ G), h(G ⊥ G(2) ) = 1. For any positive O-lattice L, the mass of L is defined to be 1 m(L) = |O(Li )| Li

where Li runs through a complete set of representatives of cls(L) in gen(L), and |O(Li )| is the order of the orthogonal group of Li . If we can show that m(L) is 1 equal to |O(L , then L must have class number 1. Siegel [6] proved that m(L) i )| can be expressed as an infinite product of local densities. Applying Satz 1 and Hilfssatz 8 in [4] for the calculating local densities of the both lattices, we have m(G ⊥ G) = 1/800 and m(G ⊥ G(2) ) = 1/400. On the other hand we obtain |O(G ⊥ G)| = 800 and |O(G ⊥ G(2) )| = 400. Thus we have h(G ⊥ G) = h(G ⊥ G(2) ) = 1.   Lemma 14. Let M be a totally positive definite binary O-lattice with M℘ ∼ = G℘ over O℘ . Suppose that M is represented by I3 ⊥ 2 +  . Then M is represented by I3 ⊥ 2, 2 . 

Proof. Let L = Ox1 ⊥ · · · ⊥ Ox4 be an O-lattice with Q(xi ) = 1 (i = 1, 2, 3) 4 4 and Q(x4 ) = 2 + . By the assumption we have M ∼ =O i=1 ai xi , j =1 bj xj for some ai and bj in O(i, j = 1, 2, 3, 4) and moreover we may assume that


 a1 a2 a3 a4 1100 1100 ≡ or mod 2 011 011 b1 b2 b3 b4 by changing the bases of M and I3 if necessary. Since 2 +  −→ I2 over O we may assume  that M  −→ I5 = Oe1 ⊥ 5 5   ∼ · · · ⊥ Oe5 , and M = O i=1 ai ei , j =1 bj ej where ai = ai , bi = bi for 



i = 1, 2, 3, 4 and a5 = a4 and b5 = b4 .

     Then for either k = 4 or 5 the scale of Mk∗ = O a3 e3 + ak ek , b3 e3 + bk ek be in 2O. It yields Mk∗ −→ 2, 2 and M −→ I3 ⊥ 2, 2 .  

Lemma 15. Let α ∈ O+ . Then there exist aα , bα ∈Z for α such that α ∼ aα + bα  and aα > bα ≥ 0. Proof. Put α = a + b (a, b ∈Z and a > 0). If a > b > 0 then putting a = aα , b = bα we obtain the assertion. Suppose 0 < a ≤ b. Then α ∼ (a + b) −2 = (a + b)(2 − ) = (2a − b) + (b − a). If a = b then put aα := 2a − b = a and bα := b−a = 0 and these yield the assertion. Suppose a < b. Then a > 2a−b > 0

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and b − a > 0. Repeating multiplicating  −2 to a + b we have α −2n = a + b      with a > b ≥ 0 for some n ∈Z, n > 0. aα = a and bα = b satisfy the assertion. 2 If b < 0 then by (a + b)e = (a + b)(1 + ) = (a + b) + (a + 2b) we   have a + b > 0 and a + 2b > b. Repeating multiplicating  2 , α 2n = a + b   with b ≥ 0 for some n ∈Z, n > 0. Thus we can apply the previous arguments and obtain the assertion.   We call aα + bα  ∈ O+ in above lemma the reduced integer of α. Lemma 16. Let α ∈ O+ the reduced integer. Then α = 1 + γ for some γ ∈ O0+ Proof. Put α = aα + bα  with aα > bα ≥ 0. If α = 1 then the assertion is clear. Suppose α = 1. Then aα ≥ 2 thus aα + bα  = 1 + {(aα − 1) + bα } and   γ := (aα − 1) + bα  ∈ O+ since aα − 1 ≥ bα ≥ 0. Lemma 17. Let 1 , 2 , 3 be in O℘× . Then at least one of i + j (1 ≤ i < j ≤ 3), 1 + 2 + 3 is in 2O℘ . Proof. Since α ≡ 1, , 1 +  mod 2 for α ∈ O℘× , the assertion is easily implied.   Theorem 18. L = I4 ⊥ G is a 2-universal O-lattice with rank 6 over F . Proof. Let M be a totally positive definite binary O-lattice. If M℘ is not isometric to G℘ , then M −→ I4 ⊂ L by Lemma 12. So we may assume that M℘ ∼ = G℘ in the following. We may also assume that Mq is Oq -maximal at all non-archimedian prime spots q. If dM(2+) ∈ (O(2+) )2 , then Mq −→ (G ⊥ G(2) )q over Oq at all prime spots q. By Lemma 13 we have M −→ G ⊥ G(2) ⊂ L. Thus we assume dM(2+) ∈ (O(2+) )2 in the following. Let S be the set of all non-dyadic prime spots q such that s(Mq ) = q. We take an element πq ∈ O, πq > 0 with (πq O) = q for each q ∈ S. We may assume that all πq ∈ S are in O+ by changing πq to πq  if necessary. By Lemma 15 we put α the reduced integer of q∈S πq and α := 1 if S = ∅. By Proposition 13, M −→ (I3 ⊥ 2 + e )(α) . Then M −→ (I3 ⊥ 2, 2 )(α) by Lemma 14. For an O-basis x1 , x2 of M we put xi = z + σi1 y1 + σi2 y2 (i = 1, 2), (α) where z ∈ I3 , Q(y1 ) = Q(y2 ) = 2α, B(y1 , y2 ) = 0 and σij ∈ O (i, j = 1, 2). If all σij ∈ 2O then M℘ ∼ = G℘ which contradicts the assumption. Thus we may assume σ11 ∈ O \ 2O by changing the basis of M and replacing y1 and y2 each other if necessary. By Lemma 16, α = 1 + γ for some γ ∈ O0+ . Thus we may put xi = σi1 y + ni where Q(y) = 2 and ni with B(y, ni ) = 0 (i = 1, 2) the O-basis of 
2    σ11 σ11 σ21 ∗ ∼ M = B(xi , xj ) − 2 . 2 σ11 σ21 σ21 2 (i = 1, 2), B(n , n ) = B(x , x ) − 2σ σ . with Q(ni ) = Q(xi ) − 2σi1 1 2 1 2 11 21 ∗ ∗ ∼  G℘ then M −→ I4 by Lemma 12 and M −→ L. Suppose M℘∗ ∼ If M℘ = = ∗ (2) G℘ . Then by Proposition 13 M is represented by either G ⊥ G or G ⊥ G .

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Suppose M ∗ −→ G ⊥ G(2) . For a binary O-lattice N which is represented by 2 ⊥ G(2) , we have s(N) ⊆ 2O and N −→ I4 . Thus we obtain M −→ L. Suppose that M ∗ −→ G1 ⊥ G2 with G1 ∼ = G2 ∼ = G. Put ni = gi1 + gi2 with gij ∈ Gj for i, j = 1, 2. If Q(g1j ) ∈ 4O for j = 1 or 2 then s(O[g1j , g2j ]) ⊆ 2O for such j since G℘ ∼ = H℘ in O℘ . Moreover we have s(O[σ11 y + g1j , σ21 y + g2j ]) ⊆ 2O, hence we obtain O[σ11 y + g1j , σ21 y + g2j ] −→ I4 and M −→ L. Suppose that Q(g1j ) ≡ 0 mod 4 for both j = 1, 2. By Lemma 17 one of Q(σ11 y + g1j ) (j = 1, 2) and Q(g11 + g12 ) is in 4O (if Q(σ11 y + g11 + g12 ) ∈ 4O then it yields that M℘ ∼  G℘ and contradicts the assumption). Then one of O= lattices O[σ11 y + g1j , σ21 y + g2j ] (j = 1, 2) and O[g11 + g12 , g21 + g22 ] is not equivalent to G℘ over O℘ and represented by I4 over O. Hence we obtain M −→ L. Thus we conclude the proof of Theorem.   References [1] Chan, W-K., Kim, M-H., Raghavan, S.: Ternary universal integral quadratic forms over real quadratic fields: Japan J. Math. 22, 263–273 (1996) [2] O’Meara, O. T.: Introduction to Quadratic Forms: Berlin-Heidelberg-New York: Springer-Verlag, 1973 [3] Mimura, Y.: On the class number of a unit lattice over a ring of real quadratic integers: Nagoya Math. J. 92, 89–106 (1983) [4] Pfeuffer, H.: Einklassige Geschlechter totalpositiver quadratischer Formen in totalreellen algebraischen Zahlkörpern: J. Number Theory 3, 371–411 (1971) [5] Riehm, C.: On the representation of quadratic forms over local fields: Am. J. Math. 86, 25–62 (1964) [6] Siegel, C.L.: Über die analytische Theorie der quadratischen Formen III: Ann. Math. 38, 212–291 (1937) [7] Takada, √ I.: On the classification of definite unimodular lattices over the ring of integers in Q( 2): Math. Japonica 30, 423–433 (1985)