Preprint - IAS

ON THE POISSON DISTRIBUTION OF LENGTHS OF LATTICE VECTORS 5Proposition 1. If 1 ≤ k ≤ n − 1, thenk∏∫ρ j (x) dx + ∑ ∑∫ ∫· · ·j=1R n (ν,µ) DR nR n j=1k∏ ( ∑m )ρ j d ij x i dx 1 . . . dx m=k∏i=1j=1V j + ∑ (ν,µ)∏mM ν,µ V νj ,where the sum over D is taken over all (ν, µ)-admissible matrices having entriesd ij ∈ {0, ±1} and exactly one non-zero entry in each column.Proof. The matrices D on this form satisfies by definition d j,νj = 1, 1 ≤ j ≤ m.If we furthermore let λ l be such that d λl ,µ l= ±1, 1 ≤ l ≤ k − m, we find that∫ ∫ k∏ ( ∑m )· · · ρ j d ij x i dx 1 . . . dx mR n R n∫ ∫= · · ·R nj=1m ∏R n j=1i=1k−m∏ρ νj (x j )l=1Since the volumes V j are increasing this integral equals∫ ∫∏m· · · ρ νj (x j ) dx 1 . . . dx m =R n R nwhich gives the desired result.j=1ρ µl (x λl ) dx 1 . . . dx m .m∏V νj ,We next give a bound on the contribution from most of the remaining terms in(6).Proposition 2. Let∫ ∫I(ν, µ, q, D) = · · ·R n(ν,µ) q=1DR n j=1k∏ ( ∑mρ ji=1j=1j=1d) ijq x i dx 1 . . . dx mand let M(D) be the largest value ( taken by any ) determinant of an m × m-minor ofD. Assume that n > max m(k − m) + 1 . Then1≤m≤k−1∑ ∞∑ ∑( e1q · · · em ) nI(ν, ∑ ∑µ, q, D) = I(ν, µ, q, D) + R(k),qwhere the remainder term satisfies(ν,µ)0 ≤ R(k) ≪ 2 −n .Dq=1M(D)=1The implied constant depends on k and the volumes V 1 , V 2 , . . . , V k but not on n.Proof. This is a trivial adaptation of [10, Sec. 9] to our situation. (We boundρ 1 , ρ 2 , . . . , ρ k−1 from above by ρ k .)□It is immediate to verify that matrices D with q = 1 and M(D) = 1 have all entriesd ij ∈ {0, ±1}. In particular all the matrices in Proposition 1 are on this form. In theremaining part of this section we will discuss the contribution to (6) coming from□