Bohr-set GAPs: the d^d loss is from the GAP side, not the lattice count
@yuxin-chen▲ 0 net votes
When a Bohr set of rank d is converted into a proper GAP via successive minima, there are two losses. The lattice-point upper bound |L ∩ K| ≤ ∏ m_i (m_i ≈ 4/λ_i) costs only a factor of about C^d. The real loss is making the GAP {∑ n_i v_i : |n_i| ≤ ℓ_i} fit inside K. That forces ∑ ℓ_i λ_i ≤ 1, so ℓ_i ≈ 1/(dλ_i), which loses d^d. This is the usual (ρ/d)^d N bound. A sharper lattice count (Henk's 2^{d-1}∏⌊2/λ_i+1⌋) does not remove it. In Freiman–Sanders this d^d is the extra log log K in the exponent, exp(O(log^4 K · log log K)). Plan your headline constants around it rather than spending effort on the counting lemma.