Definite unimodular lattices having an automorphism of given characteristic polynomial
Definite unimodular lattices having an automorphism of given characteristic polynomial
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具有给定特征多项式自同构的定单模格
DOI：
10.1007/bf02566364
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发表时间：
1984
影响因子：
0.9
通讯作者：
E. Bayer
中科院分区：
文献类型：
--
作者：
E. Bayer
格将是非零判别式的积分对称双线性形式。一个确定格的正交群是有限的。这意味着一个确定格的自同构的特征多项式是分圆多项式的乘积。相反，设f是分圆多项式的乘积。是否存在一个确定的幺模格，它的自同构具有特征多项式f？本文的第一部分是对这一问题的研究。我们将给出一个完整的解决方案的情况下，f是一个权力的分圆多项式。作为一个例子，我们讨论f= 4~ m的情况，即Ruth分圆多项式，其中m不是2的幂。本文给出了具有特征多项式ibm的自同构t的确定么模格（L，S）存在的必要条件。其中一个条件是m必须是混合的，即m不是pr或2 p”的形式，其中p是素数。实际上，如果m= pr或2 p，则det（1-t）det（1+ t）= 4~ m（1）q~ m（1 - 1）= p（cf. [13]第一章第八，第1和第3段）。因此，S '= S（t-t-1）的行列式是p。但这是不可能的，因为S'是反对称的，所以det（S '）一定是平方。另一方面，不难证明（L，S）必须是偶数，即S（x，x）对L中的所有x都能被2整除（见引理1.4）。一个偶数的秩，确定的格是可被8整除的（参见。例如[21]，Chapitre V，2.1），因此q~（m）= deg ibm必能被8整除。
A lattice will be an integral symmetric bilinear form of non-zero discriminant. The orthogonal group of a definite lattice is finite. This implies that the characteristic polynomial of an automorphism of a definite lattice is a product of cyclotomic polynomials. Conversely, let f be a product of cyclotomic polynomials. Does there exists a definite and unimodular lattice which has an automorphism with characteristic polynomial f? The first part of the present paper is devoted to the study of this problem. We shall give a complete solution in the case where f is a power of a cyclotomic polynomial. As an example, let us discuss the case f= 4~ m, the ruth cyclotomic polynomial, where m is not a power of 2. We shall give some necessary conditions for the existence of a definite unimodular lattice (L, S) having an automorphism t with characteristic polynomial ibm. One of these conditions is that m must be mixed, ie m is not of the form pr or 2p" where p is a prime. Indeed, if m= pr or 2p~ then det (1-t) det (1+ t)= 4~ m (1) q~ m (--1)= p (cf. eg [13] Chap. VIII, w 1 and 3). Therefore the determinant of S'= S (t-t-1) is p. But this is impossible because S'is skew-symmetric so det (S') must be a square. On the other hand it is not difficult to prove that (L, S) must be even, ie S (x, x) is divisible by 2 for all x in L (see Lemma 1.4). The rank of an even, definite lattice is divisible by 8 (cf. eg [21], Chapitre V, 2.1) therefore q~(m)= deg ibm must be divisible by 8.