aa r X i v : . [ m a t h . C O ] J un Nonexistence for extremal Type II Z k -Codes Tsuyoshi Miezaki ∗ Abstract
In this paper, we show that an extremal Type II Z k -code of length n dose not exist for all sufficiently large n when k = 2 , , Let Z k (= { , , , . . . , k − } ) be the ring of integers modulo 2 k , where k isa positive integer. We sometimes regard the elements of Z k as those of Z . A Z k -code C of length n (or a code C of length n over Z k ) is a Z k -submoduleof Z n k . A code C is self-dual if C = C ⊥ where the dual code C ⊥ of C isdefined as C ⊥ = { x ∈ Z n k | x · y = 0 for all y ∈ C } under the standard innerproduct x · y . The Euclidean weight of a codeword x = ( x , x , . . . , x n ) is P ni =1 min { x i , (2 k − x i ) } . The minimum Euclidean weight d E ( C ) of C is thesmallest Euclidean weight among all nonzero codewords of C .A binary doubly even self-dual code is often called Type II. For Z -codes,Type II codes were first defined in [4] as self-dual codes containing a ( ± ± Z k -codes was defined in [3] as a self-dualcode with the property that all Euclidean weights are divisible by 4 k . It isknown that a Type II Z k -code of length n exists if and only if n is divisibleby eight.In [9], we show the following theorem: ∗ Division of Mathematics, Graduate School of Information Sciences, TohokuUniversity, 6-3-09 Aramaki-Aza-Aoba, Aoba-ku, Sendai 980-8579, Japan. email:[email protected] Key Words:
Type II code, Euclidean weight, extremal code, theta series 2000
Math-ematics Subject Classification . Primary 94B05; Secondary 11F03. heorem 1.1 (cf. [9]) . Let C be a Type II Z k -code of length n . If k ≤ then the minimum Euclidean weight d E ( C ) of C is bounded by (1) d E ( C ) ≤ k j n k + 4 k. Remark . The upper bound (1) is known for the cases k = 1 [13] and k = 2 [4]. For k ≥
3, the bound (1) is known under the assumption that ⌊ n/ ⌋ ≤ k − Z k -code meeting the bound (1) withequality is extremal for k ≤ Theorem 1.2.
For k ≤ , an extremal Type II Z k -code of length n dose notexist for all sufficiently large n .Remark . For the case k = 1, the above result in Theorem 1.2 was shownin [13]. An n -dimensional (Euclidean) lattice Λ is a subset of R n with the propertythat there exists a basis { e , e , . . . , e n } of R n such that Λ = Z e ⊕ Z e ⊕· · · ⊕ Z e n , i.e., Λ consists of all integral linear combinations of the vectors e , e , . . . , e n . The dual lattice Λ ∗ of Λ is the lattice { x ∈ R n | h x, y i ∈ Z for all y ∈ Λ } , where h x, y i is the standard inner product. A lattice withΛ = Λ ∗ is called unimodular . The norm of x is h x, x i . A unimodular latticewith even norms is said to be even , otherwise odd . An n -dimensional evenunimodular lattice exists if and only if n ≡ m , the shell Λ m of norm m is defined as { x ∈ Λ | h x, x i = m } .The theta series of Λ isΘ Λ ( z ) = Θ Λ ( q ) = X x ∈ Λ q h x,x i = ∞ X m =0 | Λ m | q m , q = e πiz , Im( z ) > . E -lattice. Then,Θ Λ ( q ) = E ( q ) = 1 + 240 ∞ X m =1 σ ( m ) q m = 1 + 240 q + 2160 q + 6720 q + 17520 q + · · · , where σ ( m ) is a divisor function σ ( m ) = P 6, namely, d E ( C ) =4 k ( µ + 1). We remark that a codeword of Euclidean weight 4 km gives avector of norm 2 m in A k ( C ). Then we choose the a , a , . . . , a µ so thatΘ A k ( C ) ( q ) = θ + X r ≥ µ +1) β ∗ r q r . Here, we set b s as E − j θ = P ∞ s =0 b s (∆ /E ) s . That is, θ = P ∞ s =0 b s E j − s ∆ s .Then µ X s =0 a s E j − s ∆ s = Θ A k ( C ) ( q ) = ∞ X s =0 b s E j − s ∆ s + X r ≥ µ +1) β ∗ r q r . Comparing the coefficients of q i (0 ≤ i ≤ µ ), we get a s = b s (0 ≤ s ≤ µ ).Hence we have − X r ≥ ( µ +1) b r E j − r ∆ r = X r ≥ µ +1) β ∗ r q r . (2)In (2), comparing the coefficients of q µ +1) and q µ +2) , we have (cid:26) β ∗ µ +1) = − b µ +1) ,β ∗ µ +2) = − b µ +2) + b µ +1) (24 µ − ν + 744) . (3)All the series are in q = t , and B¨urman’s formula [15, page 128] shows that b s = 1 s ! d s − dt s − (cid:18)(cid:18) ddt ( E − j θ ) (cid:19) ( tE / ∆) s (cid:19) { t =0 } . (4)In [9], we show that β ∗ µ +1) > Theorem 2.2 (cf. [14, page 18, Theorem 1.64]) . Let η ( z ) = t / Q ∞ m =1 (1 − t m ) be the Dedekind eta function, where t = e πiz , the same for several placesand Im ( z ) > . If f ( z ) = Q δ | N η ( δz ) r δ with k = (1 / P δ | N r δ ∈ Z , with theadditional properties that X δ | N δr δ ≡ and X δ | N Nδ r δ ≡ , then f ( z ) satisfies f (cid:18) az + bcz + d (cid:19) = χ ( d )( cz + d ) k f ( z ) for every (cid:18) a bc d (cid:19) ∈ Γ ( N ) . Here the character χ is defined by χ ( d ) := (cid:16) ( − k sd (cid:17) , where (cid:16) ·· (cid:17) is the usual Jacobi symbol and s := Q δ | N δ r δ . Theorem 2.3 (cf. [14, page 18, Theorem 1.65]) . Let c , d and N be positiveintegers with d | N and gcd( c, d ) = 1 . If f ( z ) = Q δ | N η ( δz ) r δ satisfying theconditions of Theorem 2.2 for N , then the order of vanishing of f ( z ) at thecusp c/d is N X δ | N gcd( d, δ ) r δ gcd( d, Nd ) dδ . In this section, we give a proof of Theorem 1.2. Our proof is an analogue ofthat of [12]. Before we give the proof of Theorem 1.2, we give two lemmas.First, we quote the following lemma from [12]. In [11], Ibukiyama remarkedthat in [12, Lemma 1] 2 π (p. 70, l. − 1) should be (2 π ) / .5 emma 3.1 ([12, Lemma 1], [11, Theorem 12]) . Suppose that G ( q ) , H ( q ) are analytic inside the circle | q | = 1 and satisfy :(i) H ( q ) = ∞ X s =0 H s q s with H > , H > , and H s ≥ f or s ≥ , (ii) if F ( y ) = e πy H ( e − πy ) , then F ′ ( y ) = 0 has a solution y = y in the rangey > , with F ( y ) = c > , F ′′ ( y ) /F ( y ) = c > , G ( e − πy ) = 0 . Then β r , the coefficient of q r in G ( q ) H ( q ) r , satisfies β r ∼ (2 π ) / ( rc ) / G ( e − πy ) c r , as r → ∞ . Second, we show the following lemma: Lemma 3.2. We set t = q = e πiz and f ( k, t ) = P x ∈ Z t kx . Let Z ( k, t ) :=[ f ( k, t ) , E ( t )] / f ( k, t ) E ( t ) ′ − ( f ( k, t ) ) ′ E ( t ) , where [ , ] is the Rankin-Cohen bracket and f ( t ) ′ = t ( df /dt ) . Then, for ≤ k ≤ and a positive realnumber y , Z ( k, e ( − πy ) ) = 0 .Proof. Let f (resp. g ) be a modular form of weight k (resp. ℓ ) for a groupΓ. Then, [ f, g ] := kf g ′ − ℓf ′ g is a modular form of weight k + ℓ + 2 for Γ [6,page 53].We remark that f (1 , t ) is a modular form of weight 1 / (4) [14, page12]. Therefore, f (1 , t ) is a modular form of weight 2 for Γ (4). Moreover, f ( k, t ) is a modular form of weight 2 for Γ (4 k ) [14, page 28, Proposition2.22]. • The case of k = 1:We remark that Z (1 , t ) ∈ Γ (4) and define the functions: ∆ ∞ ( t ) = η (4 z ) /η (2 z ) , ∆ ( t ) = η ( z ) /η (2 z ) ,J ( t ) = ∆ ( t ) / ∆ ∞ ( t ) , Note that J ( t ) is an isomorphism from a fundamental domain of Γ (4)to the Riemann sphere C ∪ {∞} and a generator of the function field6f H ∗ / Γ (4), where H be the upper half plane and H ∗ / Γ (4) is a com-pactification of H / Γ (4) [5, page 407], [2, page 16]. Then, we have thefollowing equality: Z (1 , t )∆ ∞ ( t ) = 224 X + 11264 X + 188416 X + 1048576 X, (6)where X := J ( t ). It is easy to check that there are no positive realroots of the right-hand side (6). Here, we remark that J ( e (2 πiz ) ) takesa real on the imaginary axis. Using Theorem 2.2 and 2.3, we have∆ ∞ ( e (2 πi ) = 0 and ∆ ( e (2 πi ) = 0, namely J ( e (2 πi ) = 0. Therefore,the values of the J ( t ) on the imaginary axis are positive real numbersand we have Z (1 , t ) = 0 on the imaginary axis.The other cases can be proved similarly. We only mention the functionswhich could be used for the proofs of the cases k = 2, 3 and 4. • The case of k = 2: ∆ ∞ ( t ) = η (8 z ) /η (4 z ) , ∆ ( t ) = η ( z ) /η (2 z ) ,J ( t ) = ∆ ( t ) / ∆ ∞ ( t ) , where J ( t ) is Hauptmodul for type “8 − ” [7, page 331]. Z (2 , t ) / ∆ ∞ ( t ) =240 X + 12928 X + 283136 X + 3358720 X + 23883776 X + 105086976 X + 281018368 X + 419430400 X + 268435456 X where X := J ( t ). • The case of k = 3: ∆ ∞ ( t ) = η (2 z ) η − (4 z ) η − (6 z ) η (12 z ) , ∆ ( t ) = η ( z ) η − (2 z ) η − (3 z ) η (6 z ) ,J ( t ) = (∆ ( t ) / ∆ ∞ ( t )) / , J ( t ) is Hauptmodul for type “12 − ” [7, page 331]. Z (3 , t ) =240 X + 18000 X + 616032 X + 12860832 X + 184227840 X + 1927623168 X + 15293558784 X + 94189206528 X + 456914313216 X + 1760257683456 X + 5401844490240 X + 13181394788352 X + 25400510447616 X + 38149727846400 X + 43699899727872 X + 36857648775168 X + 21565588635648 X + 7815347306496 X + 1320903770112 X where X := J ( t ). • The case of k = 4: ∆ ∞ ( t ) = η (16 z ) /η (8 z ) , ∆ ( t ) = η ( z ) /η (2 z ) ,J ( t ) = ∆ ( t ) / ∆ ∞ ( t ) , where J ( t ) is Hauptmodul for type “16 − ” [7, page 331]. Z (3 , t ) =240 X + 13440 X + 339840 X + 5259776 X + 56422912 X + 448143360 X + 2741043200 X + 13230211072 X + 51153629184 X + 159735971840 X + 403939164160 X + 825259589632 X + 1351740293120 X + 1750333390848 X + 1751407132672 X + 1305938493440 X + 682899800064 X + 223338299392 X + 34359738368 X where X := J ( t ). Proof of Theorem 1.2. Using the equation (4) and the fact that θ = θ j where θ is the theta series of the lattice (2 k Z ) / √ k , we have b s = − js ! d s − dt s − (cid:0) E s − j − θ j − ( θ E ′ − θ ′ E )( t/ ∆) s (cid:1) { t =0 } , where f ′ is the derivation of f with respect to t = q .8e show that β ∗ µ +2) < n . When we set h ( t ) = Q ∞ r =1 (1 − t r ) − , we have b µ +1) = − j ( µ + 1)! d µ dt µ (cid:0) E − ν θ j − ( θ E ′ − θ ′ E )( h ( q )) µ +1 (cid:1) { t =0 } ,b µ +2) = − j ( µ + 2)! d µ +1 dt µ +1 (cid:0) E − ν θ j − ( θ E ′ − θ ′ E )( h ( q )) µ +2 (cid:1) { t =0 } . We show that | b µ +2) /b µ +1) | is bounded, which implies that β ∗ µ +2) < n → ∞ since the equations (3) and the inequality (5) hold.We now apply Lemma 3.1 with G ( t ) = G ( t ) = E − ν θ j − ( θ E ′ − θ ′ E ) h ( t )and H ( t ) = h ( t ). Then, as is shown in [12], and using Lemma 3.2, thehypotheses (i) and (ii) in Lemma 3.1 are satisfied. So, b µ +1) ∼ − (2 π ) / jc − / µ − / G ( e − πy ) c µ , as r → ∞ . where c and c are constants. Similarly with G ( q ) = G ( q ) = E − ν θ j − ( θ E ′ − θ ′ E ) h ( q ) and H ( q ) = h ( q ). b µ +2) ∼ − (2 π ) / jc − / µ − / G ( e − πy ) c µ +11 , as r → ∞ . Hence | b µ +2) /b µ +1) | is bounded (In fact, it approaches a limit of about1 . × as µ → ∞ ). Remark . Using the equations (3), the coefficient β ∗ µ +2) first goes negativewhen n is about 1 . × . Remark . For k = 5 and 6, we could not show G ( e − πy ) = 0 in thehypothesis (ii) in Lemma 3.1. The method of Lemma 3.2 does not workbecause there are no Hauptmoduls for the groups Γ (20) and Γ (24) sincethe groups are not genus zero. Acknowledgment. The author would like to thank Masaaki Harada, KenichiroTanabe and Junichi Shigezumi for useful discussions. The author would alsolike to thank the referee for informing us the reference [11] and its helpfulcomments. 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