Segments: which multisets j ≡ ±c (mod n) are stable under a unit #
For N(c) = #{1 ≤ j ≤ m | j ≡ ±c (mod n)} (with h = 2m) and n ≥ 3, N is stable under
multiplication by a unit a ≢ ±1 iff n ∣ h, n ∣ h + 1 or n ∣ h + 2; in that case it is
stable under every unit. The core is a lemma of Amiot: an interval of k elements of ℤ/n,
1 < k < n − 1, is not stable under multiplication by a unit a ≢ ±1. The proof reduces m to
m mod n (complete periods), reflects r ↦ n − 1 − r, and counts #{x ∈ I | x + d ∈ I}.
The odd case is the same content as NotaPuente.Families.prop_4_2_odd (through
SU(2m+1) = Sp(2m)), counted differently; no bridge between the two counts is formalized.
Main results #
Segmentos.amiot_interval: Amiot's lemma, for everynand every interval.Segmentos.seg_stable_of_dvd: ifn ∣ h,h + 1orh + 2,Nis stable under every unit.Segmentos.lem_seg: the equivalence forn ≥ 3anda ≢ ±1.Segmentos.familias,Segmentos.familias_m6: the columnm mod nof the table of families and its examples form = 6.
References #
- C. Marín, Conductors of orders generated by character values of Sp(2m), preprint (2026); programs and outputs: Zenodo, doi:10.5281/zenodo.23090540.
- C. Marín, Where the character values of Sp(2m) at a torsion element fail to generate the ring of integers, Zenodo (2026), doi:10.5281/zenodo.22813216 (a longer preprint of the same work).
Objects #
N^{(n)}(c) = #{1 ≤ j ≤ m : j ≡ ±c (mod n)}. It satisfies N(-c) = N(c), so it is the
multiset of the paper on the classes (ℤ/n)/±1.
Equations
- Segmentos.N n m c = {j ∈ Finset.Icc 1 m | ↑j = c ∨ ↑j = -c}.card
Instances For
N^{(n)} is stable under c ↦ a c (as a multiset: N(a c) = N(c) for every class).
Equations
- Segmentos.Stable n m a = ∀ (c : ZMod n), Segmentos.N n m (a * c) = Segmentos.N n m c
Instances For
Equations
Arithmetic helpers #
n consecutive integers m+1, …, m+n represent each class of ℤ/n once.
Equations
Amiot's lemma #
Amiot's lemma (core of lem:seg; [Ami09, Thm. 1]). Let 1 < k < n - 1 and let a be a
unit of ℤ/n with a ≢ ±1. Then no interval I = {c₀, c₀+1, …, c₀+k-1} of k elements of
ℤ/n satisfies a I ⊆ I (in particular a I ≠ I).